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Finish Strong!
It is early in May, but nearing the end of the school year.
It is important to be able to finish strong and be as prepared as possible to score well on Final Exams that are coming up.
NOW is the time to make sure you completely understand all the theorems, postulates and rules for the last few sections in the Geometry year. Things like Circles, Surface Area, Volume are the key
concepts for this part of the course. As always, the Pythagorean Theorem and understanding all the tricks with special triangles will help to make you successful during the Final Exam.
Even though I have many students who I have been regularly helping during this year, I understand how important it is to cram in extra help during this time - - - so don't hesitate to contact me for
help. Whether you are a current student of mine, or someone who is looking for last minute help, or even to rescue the semester, contact me here and we will make it happen.
I love Geometry and want all of you to have success with it too!!!
- Kelvin
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Implementation of path finder algorithm in C
11-22-2009 #1
Registered User
Join Date
Nov 2009
Implementation of path finder algorithm in C
Hey all,
I'm looking for a C implementation of the Lee path algorithm. It has to satisfy the following:
* The algorithm is to work inside a two dimensional space
* From each point, one can only traverse to the next point in four directions; up, down, left, right
* Points can only be either blocked or nonblocked, only nonblocked points can be traversed
Basically on a 2D rectangular grid, there can be any number for blocks (one block per x-y position). The path from one block to another must be found. The end result is that the algorithm must
find the shortest path from one block to another (source and goal).
It gets a little more complicated in that a given source A can connect to multiple goals B C D and so on. That is to say within a given path (whether already found or not) there are "sub-paths".
Sort of like a highway that has multiple lanes. Several cars starting from the same point (source) can reach different destinations (goal) by choosing a lane which corresponds to a desired exit.
For example, from source A we can use lane A to connect to goal B; lane B to goal C; lane C to goal B.
I understand the main idea behind it. We have to expand out from the source to the goal, and then trace back to the start, the path that has the smallest cost function. It is also a requirement
that we should not waste time with expanding from the source in all directions, instead we should use some sort of predictor to somehow narrow our expansion.
C is my first programming and I very recently started to learn it. Now I understand that I wont be given code in its entirety. But I am just so confused on where to begin. There are so many
requirements. I would appreciate any design hints you are willing to share.
Thank you.
I would take this time to read up on Lee's algorithm on Wikipedia (if it's there), and Google it as well.
Your main idea sounds too vague to put into pseudo code, let alone a program, just yet.
We can assist you with the code portion (most generally), but you need to know the algorithm, and give it a starting code portion, at least. Be ready to ask specific questions and provide
specific answers.
Lee's is basically a flood fill.
here = start = 1
while here is not the exit
for each open move
open move = here + 1
It looks something like this:
Here, the green path is the shortest to the exit (green-R). Assuming that both paths are being processed at the same time (it makes the two C marks at the same time, then the two D marks, etc),
you can see that it would hit green-R (the exit in the lower right) long before the other path would have completed. Were you to let it keep running, red-V to green-Q would be red W, and instead
of green-R, red-X would be the exit.
So, what you end up doing is backtracking by simply moving from the exit to whatever neighbor has the lowest value. R to Q, Q to P and so on. Whatever value the exit has (in this case R, is the
length of the shortest path.
Do this for every sub block. Then store their values in a grid corresponding to the sub-block's location:
Move from the S in the upper left to the K in the lower right using the shortest sum possible. At least that sounds to me like what you're trying to do.
Hope is the first step on the road to disappointment.
11-22-2009 #2
Registered User
Join Date
Sep 2006
11-22-2009 #3
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Posts from April 2011 on viXra log
Octonions in String Theory
April 29, 2011
John Baez and his student John Huerta have an article in Scientific American this month about octonions in string theory and M-theory. Peter Woit has given it a bit of a cynical review describing it
as hype. The defence from John in the comments is worth reading. Here is a bit of what he says
“So, don’t try to make it sound like an obscure yawn-inducing technicality about “some supersymmetry algebra working out a certain way in a certain dimension”. It’s a shocking and bizarre fact, which
hits you in the face as soon as you start trying to learn about superstrings. It’s a fact that I’d been curious about for years. So when John Huerta finally made it really clear, it seemed worth
explaining — in detail in some math papers, and in a popularized way in Scientific American.”
The article entitled “The Strangest Numbers in String Theory” is about an early observation from the study of superstring theories that the four division algebras are related to four classical
formulations of superstring theory in 3, 4, 6 and 10 dimensions. The four division algebras are the reals, complex numbers, quaternions and octonions with dimensions 1,2,4 and 8 and the dimensions of
the superstring theories are 2 dimensions higher in each case.
Many of our readers will be familiar with the internet writings of John Baez where he has described these things so I wont attempt to cover any details. His student John Huerta has just finished off
his thesis in which he clarifies these observations using higher dimensional lie algebras. The results extend to one dimension higher when strings are replaced by membranes. In the quantum theory
only the highest dimensional versions related to the octonions hold up consistently giving us the 10 dimensional superstring theories and M-theory in 11 dimensions. Of course this is not complete
since we still don’t know what the full formulation for M-theory is. Even these higher dimensional observations are not new, see for example Mike Duffs brane-scan from 1987 where the relationships
were already plotted out. This new work clarifies these results using the concept of 3-lie-algebras from n-category theory.
The Scientific American version does not go into great detail but is a very well written introduction to the ideas. If you don’t have access to it don’t worry, John says he will be allowed to post an
online version after a month or so. You can also explore what has been posted already starting here which is more advanced than the article but still very pedagogical.
Personally I find these algebraic ideas for M-theory very enticing. It is a major goal to formulate a complete non-perturbative version of string theory that encompasses all its forms and I think the
purely algebraic approach is the best line of attack. It is especially intriguing that the octonions have such a direct relationship to the dimensions in which these theories work, but ultimately the
algebraic structures we need to understand it fully are probably much more complex.
The work of Mike Duff and his collaborators which brings in the algebra of hyperdeterminants and qubits to understand a slightly different role of the octonions in string theory is one of the areas
to follow. This work brings in the duality algebras found in string theory black holes. I know that several of our regular commenters are very familiar with this already so I need not give more
details. Indeed it is the fact that the same algebraic structures keep appearing in different contexts that is so intriguing, yet so confusing, as if we are missing some principle of unification that
relates these things.
LHC Luminosity Estimates for 2011 (and poll)
April 28, 2011
Last night the Large Hadron Collider clocked up another record luminosity using 624 bunches per beam. In ATLAS the peak luminosity was measured at 737/μb/s while CMS saw about 10% less at 660/μb/s.
In theory they should be seeing about the same amount. Paul Collier (head of the beams department at CERN) has said that they suspect the ATLAS figure is too high. A more accurate measurement will be
made after the next technical stop in two weeks time when a Van der Meer scan will be run through. Meanwhile I will quote the lower CMS numbers and will be updating the figure in the viXra Log banner
An important date coming up in the LHC calendar is the 6th June when the “Physics at LHC 2011″ conference starts. The experiment teams will be keen to show results using as much data as possible
collected before that date. That will be challenging given the rate at which data will then be accumulating and the need to get about 3000 physicists in each collaboration to sign off on any results
before they are presented.
A few weeks ago I said the LHC should be able to deliver 200/pb before the conference. In fact they have already surpassed that figure for 2011 and can add last years 40/pb as well to make nearly 240
/pb. I will also be tracking this number above. Two weeks ago I upped the estimate to 500/pb for the conference but even that is looking a little unambitious now. According to the plan there are
about 30 days left for physics before the 6th June and they can now easily collect about 30/pb each day. My new estimate is at least 1000/pb which is 1/fb in time for the conference assuming they
don’t lose too many days to machine development or faults. Recall that 1/fb is the official target for the whole of 2011! It may sound like their original expectations were a bit low but really this
amazing result is due to the exceptional early performance of the collider when compared with previous models.
The 6th June also marks the point when commissioning of the LHC for this year should officially stop. If all goes well they will be aiming to reach this years maximum luminosity by this date. Which
will mean filling with 1404 bunches per beam to provide something like 70/pb per day. From 6th June there are 124 days marked on the plan for running proton physics. Do the sums and you will get my
new estimate for this years total of 10/fb, easily enough to find the Higgs boson whatever mass it has, or rule it out.
Update: If you dont like my estimate of 10/fb here is your chance to say how much luminosity you think the LHC will deliver this year
The future of science news reporting in the MSM
April 26, 2011
The Main Stream Media have increasing trouble reporting science news because journalists are rarely sufficiently expert in specialized fields and just make mistakes. Newsy.com (who is not really
MSM), has found the solution. Just quote lots of snippets from science blogs and with next to no effort they have an incredibly well balanced report!
LHC takes hadron collider luminosity record
April 22, 2011
Until yesterday the highest luminosity measured in a hadron collider was 402.4/μb/s achieved at the Tevatron last year. Last night the Large Hadron Collider smashed this record with a luminosity of
481/μb/s using a 480 bunch filling scheme. This takes them nearly half way to this years target of 1/nb/s . In fact they are likely to reach that level in the next few weeks and can go well beyond if
they stick to the 50ns bunch spacing.
The superior energy of the LHC over the Tevatron means that it now has every advantage. Strong hints of new physics from about 70/pb of collision data are already showing up, yet CERN could collect
200 times that amount before the end of 2011. This could be a good year to be doing particle physics! (The data collected in 2011 now amounts to 90/pb in each of ATLAS and CMS to add to the 40/pb
from 2010)
CERNs new record follows a visit on Wednesday by Jose Barosso, president of the European Commission who said “CERN is a true European centre of excellence: a place where our collective talent is
pooled and produces cutting-edge research, European in its foundations but global in its outlook.”
Update 23-Apr-2011: The latest run collected a record 15/pb taking this year’s total to over 100/pb. Until at least Monday they plan to stay with 480 bunches. At this level they should collect 20/pb
per day. With about 150 days of proton physics left this year that would already provide 3/fb. However it will probably not be long before they continue to push up the luminosity. Another factor of
four should be possible if they can go all the way to 1404 bunches and use nominal intensities.
Update 27-Apr-2011: Now they are filling with 624 bunches. This could take luminosities up to 750/μb/s
New New Particle Rumour
April 22, 2011
There is a new rumour about a new particle doing the rounds of the physics blogs. So far it is reported at Not Even Wrong, Resonaances and The Reference Frame.
I do not claim to be the best source of details about new phenomenology,but sometimes my intuition is good so here are a few points.
• This is not a hoax. It was revealed in an internal memo that was apparently more pubic than it should have been.
• The memo described a bump in diphoton channel at 115 GeV
• This is where SUSY prefers the Higgs, but said Higgs does not give such a high rate of diphoton decays.
• As seen recently bumps can be made to disappear by adjusting the analysis, and this analysis is not completed and approved so it could go away.
• However, the signal seems to be consistent with the old LEP observation and with faint hints in Tevatron results, so this suggests something real.
• If it is real then CMS must see it too. Therefore Tommaso Dorigo must remain silent.
• Although the old LEP result was always considered a hint of a Higgs and this seems likely to be the same thing, we should be open-minded about what it really is.
• On the one hand I want it to be SUSY and this is possible. We expect to hear about variants of SUSY that fit the bill at the next arXiv hep-ph update.
• On the other hand, a high photon coupling suggests a neutral composite particle made of charged constituents, like a meson. I don’t know if this is plausible but someone will be suggesting such
possibilities soon.
• Whatever it is, if it is real it is almost certainly Beyond Standard Model and therefore very exciting.
• The data collected by ATLAS is already quite a bit more than the amount used in this analysis and it will increase by a factor in the coming weeks. There will be more news soon.
Update: Well I was wrong about Tommaso remaining silent. He is sceptical because of comparison with Tevatron and also because of look-elsewhere effect, but the Tevatron can just about
accommodate this and there is no look-elsewhere effect if you were specifically looking for a state at 115 GeV :)
Update 23-Apr-2011: Tommaso is now willing to offer $1000 against your $500 that this is not a genuine observation. If you are tempted to take him on you should consider that he may have access to
data from CMS that would decisively resolve the question :)
Update 25-Apr-2011: Channel 4 news UK picked up this story and included a well balanced interview with John Butterworth
LHC back to physics runs with 50ns bunch spacing
April 15, 2011
The Large Hadron Collider today returned to physics runs. Over the last few days they had been filling the collider for scrubbing runs in an effort to remove the e-cloud problem. In fact they only
did a few hours of scrubbing runs before surprisingly returning to physics with 50ns bunch spacing.
On the face of it this is great news because it suggests that they have concluded that the problems are not too severe and physics with 50ns spacing will be the norm for this year. However I have
heard no official word so I can’t confirm that they wont change back to 75ns spacing.
With the 50ns spacing they will be able to get 1400 bunches into the collider. This is half the design limit of the machine which may only be achievable when they move to the higher energy of 14 TeV
in 2014. During the scrubbing runs they were already able to circulate 1020 bunches, so perhaps it will not be long before they reach similar numbers with colliding beams. It is very impressive that
they have reached this point so quickly.
For today they have been running with 228 bunches providing a luminosity of 250/ub/s just below the records of a few weeks ago. They need three runs and 20 hours of stable beams before they can go to
the next step at about 300 bunches. Just one more runs of about 5 hours should do it. Today’s runs have already added about 10/pb to the total integrated luminosity delivered to each of ATLAS and
Update 16-Apr-2011: The next step is set as 336 bunches and still with 50ns spacing. Expected peak luminosity is about 360/ub/s which will be a new record, fingers crossed. Total luminosity delivered
for 2011 has passed the 40/pb delivered in 2010.
Update 17-Apr-2011: After a few false starts a good long run with 336 bunches is now in progress. It started with a peak luminosity of 380/ub/s in ATLAS and 360/ub/s for CMS. After 9 hours the
integrated luminosity has passed 10/pb, the highest yet for a single run. This may be a good moment to speculate about how much data they can collect before the “Physics at LHC” conference to be held
in Italy from 6-11 June. The answer depends on many things but I don’t think 0.5/fb is overly optimistic at this stage assuming they proceed with continuous physics runs and increasing luminosity
according to the plan which has six weeks of running before the conference. To give some meaning to that amount of data here is the expected Higgs limits for ATLAS.
0.5/fb would exclude the Higgs between 140GeV and 180GeV, unless it is there. Of course even if they do collect 0.5/fb before the conference they would have to do the analysis very quickly to be able
to report it in time and I don’t expect that to be possible. Other results are more likely to be shown and the Higgs limit should just be taken as an indication of how much progress has been made
when these landmarks are reached.
Spring Cleaning and a Scary Moment for the LHC
April 9, 2011
Spring is very much in the air here in Europe with warm temperatures and sunny skies over most areas. As many of us spring clean our homes the Large Hadron Collider is also undergoing a thorough
scrubbing of its pipes to make it ready for it’s first full scale physics runs. For the collider this does not involve teams of cleaning ladies with dusters and scrubbing brushes. The pipes are
scoured clean by sending high intensity beams through the pipes for a period of several days.
Before the scrubbing could begin the LHC had to recover from a scheduled technical stop at the start of the month. This process can sometimes seem painfully slow to us outsiders watching the progress
but on Wednesday we were given a stark reminder of why they have to be so cautious. During the build up to the scrubbing runs as they attempted to pump in a record 600 bunches to produce the high
intensity needed for the scrubbing runs a signal was generated that triggered an immediate quench of the magnets in one sector, sending high currents through the high tension lines. As far as I know
it was the first time such an unplanned quench had happened since the disastrous events of 2008 when the splices proved unable to take the high load leading to an explosive release of helium that
shut down the collider for over a year. This time round they had the added complication of high intensity beams in the rings which had to be dumped safely. Failure to do so could have sent trains of
proton bunches into sensitive areas of the collider causing damage that would have meant another long shutdown. The problem was caused by a faulty temperature sensor where some cables had been
plugged in the wrong way. It was lucky that there were two identical systems functioning and the problem only existed with one of them. This made it possible to detect the fault and fortunately the
protection systems all worked as designed and no harm was done. It must have been a heart stopping moment for the operators who reported that without the double protection system they could have seen
“beyond repair damage”
The collider was stopped for more than a day while they checked all similar connections to ensure that no further faults were still present. The scrubbing runs have now restarted.
These runs were planned at the end of last year after they first reached high intensity with the proton beam runs. They noticed excess background events in the ATLAS detector which were found to be
caused by a build-up of an electron cloud in the pipes where the beams approached the detectors. The electrons are released from residual gas molecules in the walls of the pipes. These electrons then
strike back on the surface releasing a cascade of further electrons. The circulating proton beams collide with the electrons sending particles towards the detectors that have to be filtered out of
the analysis. Too much background of this sort can hide the signals they are looking for.
One step taken to resolve the problem is the installation of solenoids that produce electric fields around the affected areas steering the electrons away from the pipe to avoid the cascades. But in a
addition they scheduled these scrubbing runs in which high intensity proton beams are circulated at 450 GeV to tare the residual gas molecules out from the walls of the pipes. They can then be pumped
out using the vacuum systems.
The process will last about a week and could see as many as 1200 bunches being circulated. The protons will not be ramped up to the full energy of 3.5 TeV at which collisions have been made but the
runs will also allow the operators to see how well the injection systems work to produce the beam intensities that could be used in physics runs later this year.
The extent to which these scrubbing runs are successful will determine how much collision data can be collected this year. If they manage to remove most of the electron clouds they will progress to
running with the 50ns bunch spacing that will make it possible to inject up to 1400 bunches for the physics runs. Otherwise they will stick with the 75ns spacing currently in use for physics but that
would limit the number of bunches to about 900 and will mean correspondingly lower luminosities for the runs this year.
update 11-apr-2011: For the scrubbing runs they have injected 1020 bunches per beam. This is near to the maximum intensities that they can get to with 50ns spacing. Although they are not ramping up
to 3.5 TeV or colliding particles at this time it is good to see that this step can already be done successfully.
New Particle?
April 6, 2011
Other people do the phenomenology stuff much better than me so I don’t try to compete. See e.g. here, here and here, and now it’s also here, here, here, here, here, here and finally here . However,
sometimes I like to see my plots without those fitted lines that lead the eye
I can always add my own alternative bumps
Hidden Variables and the 24 Rays of Peres
April 4, 2011
A lot of people like to worry about the measurement problem and wave function collapse in quantum mechanics. How can a physical outcome depend in such a fundamental way on how we observe it? Many of
us have been happy to accept that quantum mechanics works as described and that no real paradoxes arise from its interpretation, but ever since Einstein challenged Bohr’s Copenhagen interpretation, a
minority of physicists from Bohm to ‘t Hooft have tried to find a hidden variable explanation that avoids the philosphical problems. Even if you don’t worry about such things, the maths and physics
behind such ideas can be quite interesting. Last week I came across the paper arXiv:1103.6058 by Waegell and Aravind that relates a proof of the Kochen-Specker theorem to the 24-cell.
The Kochen-Specker theorem is a no-go result for hidden variable theories and the 24-cell is a unique mathematical structure that comes up in the context of systems of qubits as I discussed just
Any hidden variable theory must avoid certain no-go theorems of this type. The most well known is Bell’s inequality that follows from the assumption of locality and has been shown to be violated in
experiments. This is consistent with quantum mechanics but rules are local hidden variable theories. It is quite a strong refutal of Einstein’s philosphical stance against quantum mechanics, because
he had a strong belief in locality that followed from his work on relativity where he established the principle that no signal can be sent faster than light. It turns out that quantum mechanics does
not violate this principle in the practical sense, yet it is formulated in such a way that wave-function collapse describes an apparent non-local effect. To some physicists including Einstein this
seemed philosophically unsatisfactory. Too bad! Sometimes the universe does not respect our philosophical preferences and this is one of them. The experimental verification of the violation of Bell’s
inequality shows that we have to accept non-locality in some form.
But non-locality is not the only philosophical objection that physicists have. The bigger problem is that the rules of wave-function collapse depend on what is measured. This seems to give observers
a special role in the laws of physics, but there is no good way to define what an observer is from first principle. In a hidden variable theory we would postulate the existence of state variables
with definite values that cannot be easily seen but which determine the outcome of quantum mechanical measurements. So could there be a hidden variable theory of quantum mechanics which is non-local
but where no variable depends on the context of the measurement? The Kochen-Specker theorem proved by Simon B. Kochen and Ernst Specker in 1967 rules out such a theory and it does so in a very
interesting way.
The original proof was quite complex but in 1991 quantum information expert Asher Peres gave a simpler proof. Although he did not mention it at the time, his proof relies on the symmetry of the root
system of the exceptional Lie algebra F4. This comprises 48 vectors in 4D space which can be interpreted as the vertices of a 24-cell and its dual. You don’t need to know anything about root systems
or Lie-groups or the 24-cell to understand the proof so don’t be put off.
Each root vector is paired with its negative to define a line through the origin in 4d space. These 24 lines are the 24 rays of Peres. I listed these points in my previous post on the 24-cell but
here again are the 24 rays with numbers as in the new paper so that I can refer to them.
│ 1 │ (2,0,0,0) │ 2 │ (0,2,0,0) │ 3 │ (0,0,2,0) │ 4 │ (0,0,0,2) │
│ 5 │ (1,1,1,1) │ 6 │ (1,1,-1,-1) │ 7 │ (1,-1,1,-1) │ 8 │ (1,-1,-1,1) │
│ 9 │ (-1,1,1,1) │ 10 │ (1,-1,1,1) │ 11 │ (1,1,-1,1) │ 12 │ (1,1,1,-1) │
│ 13 │ (1,1,0,0) │ 14 │ (1,-1,0,0) │ 15 │ (0,0,1,1) │ 16 │ (0,0,1,-1) │
│ 17 │ (0,1,0,1) │ 18 │ (0,1,0,-1) │ 19 │ (1,0,1,0) │ 20 │ (1,0,-1,0) │
│ 21 │ (1,0,0,-1) │ 22 │ (1,0,0,1) │ 23 │ (0,1,-1,0) │ 24 │ (0,1,1,0) │
Suppose these (when normalised) are 24 quantum states |ψ[i]> in a 4 dimensionl Hilbert Space e.g. it might be a system of two qubits. For each state we can define a projection operator
P[i] = |ψ[i]><ψ[i]|
These are Hermitian operators with three eigenvlaues of 0 and one of 1. They can be considered as observables and we could set up an experimental system where we prepare states and measure these
observables to check that they comply with the rules of quantum mechanics. One thing that we observe is that there are sets of 4 operators which commute because the 4 rays they are based on are
mutually orthogonal. An example would be the four operators P[1], P[2], P[3], P[4].
We know from the theory of quantum mechanics that if we measure these observables in any order we will end up with a state which is a common eigenvector i.e. one of the first four rays. The values of
the observables will always be given by 1,0,0,0 in some order. This can be checked experimentally. There are actually 36 sets of 4 different rays that are mutually orthogonal, but we just need nine
of them as follows:
{P[2], P[4], P[19], P[20]}
{P[10], P[11], P[21], P[24]}
{P[7], P[8], P[13], P[15]}
{P[2], P[3], P[21], P[22]}
{P[6], P[8], P[17], P[19]}
{P[11], P[12], P[14], P[15]}
{P[6], P[7], P[22], P[24]}
{P[3], P[4], P[13], P[14]}
{P[10], P[12], P[17], P[20]}
At this point you need to check two things, firstly that each of these sets of 4 observables are mutually commuting because the rays are othogonal, secondly that there are 18 observables each of
which appears in exactly two sets.
imagine now that there is some hidden variable theory that explains this system and which reproduces all the predictions of quantum mechanics. At any given moment the system would be in a definite
state and values for each of the 18 operators would be determinate, even if it is hard for us to see what they are directly. The values must be 0 or 1 but they must also comply with the rules that
they are equal to 1 for exactly one observable in each of the nine sets. the other three values in each set will be 0. So there must be nine values set to one overall. But this is impossible! each
observable appears twice so which ever observables have the value of 1 there will always be an even number of ones in total, and nine is not even. This proves the Kochen-Specker theorem. (This
version of the proof using only 18 of the 24 vectors is a later refinement due to Kernaghan, Cabello, Estebaranz, and Garcia-Alcaine, see paper linked for references)
The conclusion is that it you want to believe in hidden variable theories you had better find a way of implementing it that does not comply with the assumptions of this theorem. It is not enough to
look for non-local hidden variable theories. You have to avoid the conditions that the variables have definite values or that they are independent of the context of the measurement. For my money that
takes you away from the philosophical objections that hidden variable theories are supposed to answer, but if you want to dispute that you can do it here :).
Mike Duff Finds New Way to Test Strings
April 1, 2011
String theorist Mike Duff famed for his pioneering work on M-theory has announced a novel and practical way to test the theory of strings. The Abdus-Salam Professor of Theoretical Physics at Imperial
College delivered a talk on “Black Holes and Qubits” in Durban earlier this month. His work with Borsten, Dahanayake, Ebrahim, Marrani and Rubens caused a stir last year because of the
widely misinterpreted claim that it provides a test of the mathematics of string theory. “Two different branches of theoretical physics, string theory and quantum information theory (QIT), share
many of the same features, allowing knowledge on one side to provide new insights on the other. In particular the matching of the classification of black holes and the classification of four-qubit
entanglement provides a falsifiable prediction of string theory in the field of QIT.” he said.
During the workshop, Duff teamed up with new collaborators from various fields of quantum physics to try out a completely new way of testing strings that came to light during the interdisciplinary
discussions, and this time it was very much for real. Andrzej Dragan, Jason Doukas, Ivette Fuentes, Mike Duff and Nick Menicucci demonstrated the method that involves placing long strings under
tension. The objective was to observe the temperature that results from uniform acceleration known as the Unruh effect. Some of their critics have already described it as “highly risky” and “jumping
to wild conclusions.”, but Duff has responded by challenging them to test it for themselves. ViXra Log has exclusive video of how the amazing experiment turned out.
Update: I probably did not fool many people with this post, but in case anyone is wondering, that really is Mike Duff and other participants of the Relativistic Quantum Information Workshop doing the
bungy jumping. The jump-off point is 106 meters above the Moses Mabhida Stadium in Durban. If anyone has experienced any equally unusual activities organised at conferences and workshops please do
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Another random problem
Post reply
Another random problem
a point is randomly selected with a rectangle whose vertices are (0,0), (2,0), (2,3) and (0,3). What is the probability that the x-coordinate of the point is less than the y-coordinate?
Re: Another random problem
I think it is 1/3
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Another random problem
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Another random problem
Hi bobbym
how did you get that?
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Another random problem
x ∈ [0,2], y ∈ [0,3] both are uniform distributions. There are 3 y's for every 2 x's.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Another random problem
Hi bobbym
i think i switched x and y coordinates.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Another random problem
Or the answer is the area enclosed by the rectangle on top and the line y = x.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Another random problem
Hi bobbym
yes that's how i did it,i just flipped the rectangle around the y=x line.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Another random problem
Hi anonimnystefy;
It is late and time for you to rest. I will see you tomorrow.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Another random problem
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Another random problem
Is it not 2:20 AM?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Another random problem
Yes it is.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Another random problem
Yawning, sleepy eyelids, a general buildup of toxins and a lowering of body temperature all signalling the need for sleep...
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Another random problem
And yet math is awaiting.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Another random problem
Let the problems of the day be sufficient for the day.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Re: Another random problem
Don't understand.
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Another random problem
Math will wait.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Full Member
Re: Another random problem
actually, bobbym
The point (x,y) satisfies x<y if and only if it belongs to the shaded triangle bounded by the lines x=y,y=2 , and , x=0 the area of which is 2. The rectangle has area 6, so the probability in
question is 1/3
I see you have graph paper.
You must be plotting something
Re: Another random problem
Hi cooljackiec
Bobbym's answer is correct. Take a closer look and try plotting a few values fir (x,y) to see which area the points will belong.
And, welcome to the forum!
The limit operator is just an excuse for doing something you know you can't.
“It's the subject that nobody knows anything about that we can all talk about!” ― Richard Feynman
“Taking a new step, uttering a new word, is what people fear most.” ― Fyodor Dostoyevsky, Crime and Punishment
Re: Another random problem
Hi cooljackiec;
The area above the triangle that the line y = x makes is twice as large as the area below the triangle. y is a 2 to 1 favorite. There is only one probability that is twice its complement. That is 2 /
3. The probability the P(y > x ) = 2 / 3 so P(x < y ) = 2 / 3.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Post reply
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Sequence within a non-empty, bounded above set
January 30th 2013, 11:35 AM #1
Junior Member
Mar 2011
Sequence within a non-empty, bounded above set
Let S be a non-empty subset of R (real numbers) that is bounded above. Show that there exists a sequence (xn, n is a natural number), contained in S (that is, xn is an element of S for all n in
the set of natural numbers) and which is convergent with limit equal to sup S.
Any help would be greatly appreciated
Re: Sequence within a non-empty, bounded above set
Suppose that $\sigma=\sup(S)$. Two cases:
1) $\sigma\in S$, what is a constant sequence that works?
2) $\sigma otin S$ and $\exists s_1\in S$ such that $s<\sigma$. You explain why!
If $n>1$ then $\exists s_n\in S$ such that $\sigma-\tfrac{1}{n}\le s_n<\sigma$. You explain why!
January 30th 2013, 11:57 AM #2
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Squares and Square Roots Recall from Section 1-3 that to square a number means to multiply the number by itself.To square 8,we write: 8 2 8 8 64
Review of Algebra
SOLUTION Using the Binomial Theorem with ,,, we have Radicals The most commonly ... A 10 * REVIEWOF ALGEBRA EXAMPLE 17 (a) (b) (c) Alternative solution: (d) (e) * x y * 3 * y 2 x z * 4 * x 3 y 3 *
Main Menu for Tutorials: Math Placement
Introduction Patience Dedication Concentration Supreme Concern These algebra tutorials were written for those students who comtemplate coming to The Univer ...
Sample Test Questions A Guide for Students and Parents
Algebra Placement Test The Algebra Placement Test is composed of items from three ... Exponents and Radicals 7. Rational Expressions 8. Linear Equations in Two Variables ...
Simplifying Radical Expressions
... DATE_____PERIOD_____ 11-1 11-1 Glencoe/McGraw-Hill 83 Glencoe Algebra 1 Chapter 11 ... finding the principal square root of an expression containing variables, be Radicals ...
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SOLUTION Using the Binomial Theorem with ,,, we have Radicals The most commonly ... A 10 * REVIEWOF ALGEBRA EXAMPLE 17 (a) (b) (c) Alternative solution: (d) (e) * x y * 3 * y 2 x z * 4 * x 3 y 3 *
ModuMath Algebra Learning Objectives
Algebra 24 - Roots Radicals II 1) Multiply and divide two radicals. 2) Add or subtract two radicals. 3) Rationalize the denominator.
Algebra 2 Lesson on Solving Quadratic and Radical Equations
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Basic Rules of Algebra, Algebraic Fractions, Laws of Exponents ...
Basic Rules of Algebra, Algebraic Fractions, Laws of Exponents, and Roots/Radicals by Charles Ormsby If you find yourself having difficulty with basic algebraic ...
Simplifying Radical Expressions Date Period
Create your own worksheets like this one with Infinite Algebra 1. Free trial ... Simplifying Radicals Author: Mike Created Date: 12/23/2011 9:46:56 AM
Integrated Algebra 1
Integrated Algebra 1 is a new text for high school algebra that continues the ... OPERATIONS WITH RADICALS 469 12-1 Radicals and the Rational Numbers 470 12-2 Radicals and ...
Integrated Algebra 1
Integrated Algebra 1 is a new text for high school algebra that continues the ... OPERATIONS WITH RADICALS 469 12-1 Radicals and the Rational Numbers 470 12-2 Radicals and ...
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Algebra Units and Lessons
... 5788508, 6064856, 6758674 and Des. 385431 and European Patent No. 0656139 Algebra ... Simplifying Square and Cube Roots Simplifying Sums and Differences of Radicals ...
ALGEBRA REVIEW LEARNING SKILLS CENTER The Review Series in Algebra is taught at the ... k b n = k(b n) k b n (kb) n (b) n m = b ( n m) b n m () b nm Exponents and Radicals: Practice ...
Algebra Cheat Sheet Basic Properties Facts
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Mental Math Grade 5 - Timberdoodle Co - Homeschool Math
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Product Code: 332-805
Learn to Make Math Pictures in Your Mind with Singapore Math - Mental Math
Math researchers agree that students need the ability to make math pictures in their minds. Mental Math is a workbook devoted to helping your children master mental calculation, working out math
problems in their minds. Important computation, strategies, quick tips, and thinking shortcuts are provided. Mental Math strategies will help your children solve math problems quickly and
accurately while developing a foundation for future math encounters.
Mental Math features one page for each week of the year. Each strategy is introduced with an example and guided steps that help students break down problems and compute answers without the aid of
a calculator or written computation. This collection was written in Singapore and adapted from the world-renowned Singapore math curriculum. There are sixty-four reproducible pages and an answer
Suitable For: Grade 5/Level 4
Pages: 64 pages
Printed in the USA
Copyright: 2011
Topics covered:
Addition: Breaking Up Numbers
Addition: Rounding Numbers
Subtraction: Breaking Up Numbers
Subtraction: Rounding Numbers
Multiplying 2-Digit Numbers by 11
Multiplying 3-Digit Numbers by 11
Multiplication: Breaking Up Numbers (Part 1-3)
Multiplication: Rounding Numbers Ending with 9
Multiplication: Identical First Digits, Sum of Last Digits Is 10
Multiplication: Identical Last Digits, Sum of First Digits Is 10
Multiplication: Identical First Digits for 2-Digit Numbers
Multiplication: Identical First Digits, Sum of Last Digits is 5
Multiplication: Multiplying 2-Digit Numbers by Hundreds
Division: Breaking Up Numbers
Division: Finding Remainders When Dividing by 3
Division: Finding Remainders When Dividing by 4
Adding Fractions with 1 as the Numerator
Adding Fractions with the Same Numerator
Subtracting Fractions with 1 as the Numerator
Decimals: Multiplying by 10
Decimals: Multiplying by 100
Decimals: Multiplying 2-Digit Numbers by Decimals Ending with 0.9
Decimals: Multiplying 2-Digit Numbers by 1.1
Decimals: Breaking Up Numbers to Multiply
Decimals: Breaking Up Numbers Ending in 0 to Multiply
Decimals: Dividing by 10
Decimals: Dividing by 100
Decimals: Breaking Up Numbers to Divide
Squaring Numbers Ending with 0
Squaring Even Numbers
Squaring Odd Numbers
Squaring Numbers Ending with 1
Squaring Numbers Ending with 2
Squaring Numbers Ending with 3
Squaring Numbers Beginning with 5
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Maze Generation: Sidewinder algorithm
Posted by Jamis on February 03, 2011 @ 07:20 AM
Last of the (eleven!) maze algorithms on my list is the Sidewinder algorithm. With a name like that, it’s got to be cool, right?
Well, possibly. It’s got its problems, but it’s quick and easy to implement, and allows arbitrarily tall mazes. It’s closely related to the Binary Tree algorithm, but manages to get away with only
one side being spanned by a passage, instead of two.
In a nutshell, it goes like this:
1. Work through the grid row-wise, starting with the cell at 0,0. Initialize the “run” set to be empty.
2. Add the current cell to the “run” set.
3. For the current cell, randomly decide whether to carve east or not.
4. If a passage was carved, make the new cell the current cell and repeat steps 2-4.
5. If a passage was not carved, choose any one of the cells in the run set and carve a passage north. Then empty the run set, set the next cell in the row to be the current cell, and repeat steps
6. Continue until all rows have been processed.
So, while it’s related to the Binary Tree algorithm, it’s a bit more complicated. However, words don’t do it justice; it really is a lot more straightforward than it sounds.
Let’s walk through an example.
An example
I’ll use a 4×4 grid, and I’ll move fairly quickly, so pay attention. (You in the back! Eyes to the front!)
I’ll start with the first (top) row, although the algorithm works just fine if you start with the last one and work up. Starting at the top lets us get the one special case out of the way: the entire
first row must be a single passage (because you can’t carve north from the northernmost row):
There, we’ve got that done with. Now, let’s get to the interesting bit. Let’s say we manage to carve 3/4 of the way across the row before we decide (randomly and arbitrarily) not to carve eastward:
The cells in red are the current “run set”. Since we’ve ended the current run of horizontal cells, we now choose one of these cells at random, and carve a passage north. Thus:
Then we clear out the run set:
Make the next cell the current cell:
And try to decide whether to carve eastward or not. However, at this point, we can’t carve east. It would take us outside the bounds of the maze, which is not allowed. So we have to end the run
there, and choose one of the cells in the set to carve north from. The decision is out of our hands: with only a single cell in the set, we have to choose it. So we carve north:
Then we start the next row. Notice how each row is independent of the row preceding it, much like Eller’s algorithm.
For the next row, we decide immediately that we want to stop carving east on the very first cell:
So we stop, choose a cell from the run set (easy) and carve north:
Continuing, we connect the next two cells:
Randomly choose one of them to carve north from:
And continue:
And since we’re at the end of the row, we’re again forced to stop and carve north:
For the last row, let’s say we connect the first two cells:
Carve north and clear the set:
Then connect the last two cells:
And finish it off by adding a final northward connection:
And that’s our maze, courtesty of the Sidewinder.
Feel free to play with the following demos, to get a feel for how the Sidewinder works on larger grids. The single corridor spanning the top row is hard to miss, but see if you can spot any other
properties of the algorithm.
All this talk of “run sets” might make you think you actually need to use a set. Well, you can, sure, if you want to. No one’s going to stop you, right? But if you notice that each set consists of
consecutive cells, you can take a short-cut and just keep track of the cell that started the run. The “set” is then implied as the sequence of cells between the start of the run, and the current
Aside from that, the algorithm has very few surprises. As you’d expect you need to iterate over each row:
1 height.times do |y|
2 run_start = 0
3 # ...
4 end
At the start of each row I initialize the beginning of the current run to the first cell.
Then, within that loop, I iterate over each cell in the current row:
1 width.times do |x|
2 # ...
3 end
Within that loop is where the magic really happens. There are basically two conditions:
1 if y > 0 && (x+1 == width || rand(2) == 0)
2 # end current run and carve north
3 elsif x+1 < width
4 # carve east
5 end
Remember how the first row is always a single corridor? That “y > 0” guard prevents us from ever ending the current run if we’re on the first row. Similarly, the “elsif” clause prevents us from
carving east if we’re at the last cell in a row.
The second condition in the first clause makes sure that we always end the current run at the end of the current row (x+1 == width), but also allows us to randomly end the current run (rand(2) == 0).
Ending a run requires us to randomly choose a cell from the current run set, and carve north from it:
1 cell = run_start + rand(x - run_start + 1)
2 grid[y][cell] |= N
3 grid[y-1][cell] |= S
4 run_start = x+1
We also reset the run set to begin at the next cell.
The last bit of code, from the elsif clause, is simple: carving east is merely setting a bit:
1 grid[y][x] |= E
2 grid[y][x+1] |= W
And that’s all there is to it. Like I said, it’s easier to implement than describe.
With a name like “Sidewinder” you can’t help but wish the algorithm was cooler than it is. I’m actually not sure where the name comes from; a Google search reveals a routing algorithm with the same
name, but I wasn’t able to find any solid information to compare with the one I describe here. (If anyone has more information, I’m definitely curious.)
The Sidewinder’s greatest “flaw”, if it can be called that, is that it is all but trivial to start at the bottom row of the maze, and work your way to the top. Any maze generated by the Sidewinder
algorithm will never have any north-facing dead-ends, which means you can never get stuck when moving from south to north. (This is similar to the Binary Tree algorithm, which will never have any
dead-ends in the direction of its bias.) And for quickly generating arbitrarily tall mazes, Eller’s algorithm is probably better (with fewer drawbacks), but is harder to implement.
Would I actually use the Sidewinder algorithm for anything? I’m not sure. I hesitate to say “never”, but I certainly can’t think of anything I’d use it for that another algorithm wouldn’t do better!
But it was fun to implement, so it has entertainment value, if nothing else.
Give it a try yourself and share how it goes! For reference, my complete implementation follows:
03 Feb 2011
1. Martijn said...
Funny, it looks really cool and the implementation is easy but it’s not really useful. Your posts are a cool read, though :)
04 Feb 2011
2. Macuyiko said...
Interesting post, as always. This page^1 suggests the name comes from the fact that ‘Like binary tree, a sidewinder Maze can be solved deterministically without error from bottom to top, because at
each row, there will always be exactly one passage leading up. A solution to a sidewinder Maze will never double back on itself or visit a row more than once, although it will “wind from side to
side”.’ So as you traverse the rows, you’ll ‘wave’ from side to side.
You’re right however that there also exists an integer programming-based routing pattern approach called Sidewinder. The relevant paper is available online^2, together with slides^3. There is no
direct link between the two algorithms, at least not as far as I can tell after a quick read through the otherwise interesting paper, although it can be noted that the routing problem as described in
the paper also tends to require looking for maze-like patterns :).
I’m looking forward to more posts in this style, be it about mazes or other topics. Your code samples in particular are a nice contribution to anyone wanting to explore maze generation.
3. Jamis said...
@Macuyiko, the “Think Labyrinth” site inspired all of these maze articles I’ve written, and yet I managed to overlook that bit about a solution to a Sidewinder maze winding “from side to side”!
Thank-you for pointing that out; that makes a lot of sense.
Thanks also for researching the other, routing-based algorithm. I had thought there was a chance of a connection there, because as you said, routing problems relate to mazes (both being associated
with spanning trees).
And thank-you, especially, for the kind words!
06 Feb 2011
4. Tim said...
I stumbled upon your posts awhile ago and just decided to start implementing them in flash and adding in some ui to generate mazes from 3×3 cells to 100×100 cells. Very informative, I had no idea how
mazes could be generated prior.
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The case for base-twelve, Part 2
(If you haven't already, read Part 1, where I explain why we're going through all this trouble.)
An important thing to understand before we move on: what we tend to call a "number" is actually comprised of several different aspects. It has a
, which is the count of things it represents; it has a
, which is made up of the numerals 0-9 which can be arranged in single- and multi-digit fashion; and it has a
. In day-to-day life, when we don't bother with anything other than the standard base-ten system, the three aspects seem essentially tied to each other. However, once we start considering other
systems we find that the three aspects are actually separate. In base-ten the symbol "10" coincides with this many things: (o o o o o o o o o o), but in another system it wouldn't (and it may or may
not be called "ten").
In light of those distinctions, I'm going to employ a certain formatting scheme in this post that will hopefully make it easier to compare base-ten with other systems. If I write out the name of a
number with letters then I'm referring to the
. So by "ten" I mean a group of this many things: (o o o o o o o o o o). On the other hand, if I were to write "10" then I'm referring to the
formed by the digits themselves, and not a specific value or count of something. It gets a little less clear when we get to the
, since I'm already using the spelled-out word to indicate the value. But generally, if I put the word for the number in quotation marks I'm probably talking about it specifically as a name rather
than as a representation of the value. A little confusing, but hopefully it'll make sense in context.
Now, if we were planning to switch to any system between base-one and base-nine we'd be all set for symbols: we'd have a few extra hanging around that we wouldn't need anymore. But since we're
talking about switching to a system higher than base-ten, we're gonna need a couple more single-digit numerals to fill in the new gap before we reach "10" (which, remember, is twelve things in
base-twelve). For now, we can just swipe a couple symbols from the Greek alphabet*. Doing so could give us a series of number symbols that looks like this:
1, 2, 3, 4, 5, 6, 7, 8, 9, γ, ψ, 10
Two values that were formerly represented by double-digit symbols, ten and eleven, are now represented by single-digit symbols, and the symbol "10" has been reassigned to represent a dozen of
Obviously, reciting that as "one, two, three, four, five, six, seven, eight, nine, ten, eleven, ten" isn't going to work. We need some new names for ten (γ, or this many things: o o o o o o o o o o)
and eleven (ψ, or γ plus one) and to reduce confusion we should probably also rename twelve, which is now of course represented by the symbol "10." We could make them sound similar to their old names
to ease the transition: γ (the old ten) can be "ben" and ψ (the old eleven) can be "elv." And 10 (the old twelve) can be "doz." (I'm pronouncing the "oz" in "doz" like the "aws" in "jaws.")
So, "1, 2, 3, 4, 5, 6, 7, 8, 9, γ, ψ, 10" would be read as "one, two, three, four, five, six, seven, eight, nine, ben, elv, doz."
Weird, definitely, but hopefully you can see what I'm getting at.
Our two new single-digit numerals would work just like the others, and so, for instance, the sequence would continue: 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1γ, 1ψ, 20. This is where things start to
get really messy, though, because now none of those symbols (or indeed any symbols from here on out) correspond with their values in base-ten. If we really wanted to separate base-ten and base-twelve
we could potentially go further than γ and ψ and come up with a new symbol for
numeral for one through nine (zero could stay 0, I suppose) so that you wouldn't have to worry about any symbols having different values between the two systems (because the two systems simply
wouldn't share any). Maybe that could be the better way to do it, but for this post I'm going to stick to just adding γ and ψ. It means you don't have to memorize a whole new list of numerals, but it
also means having to realize that, say, "14" in base-twelve does not represent the same value as it does in base-ten. What
consistent, though, is that it still represents the notion of "one set + four;" we've just changed the definition of a "set."
Since we already assigned the
([oooooooooo], [oooooooooo]o, [oooooooooo]oo) new
(ben, elv, doz), you might argue that we could therefore let the
that used to represent them ("11" and "12") keep the old names ("eleven" and "twelve"). But that's just asking for miscommunication, I think, so the values now represented by "11" and "12" need new
names. Deciding what they should be could be interesting/complicated. The names "eleven" and "twelve" are actually kind of weird to start with, because they don't fit the pattern of the rest of the
nearby numbers. Based on the others, you'd expect them to be "oneteen" and "twoteen" or something like that ("thirteen" and "fifteen" are almost what you'd expect, which would be "threeteen" and
"fiveteen"). So maybe we could just call them that? But then there's the fact that
the names in English for the values between ten and twenty (in base-ten) don't match the pattern that's subsequently followed. Really, you'd expect them to be "ten, ten-one, ten-two, ten-three,
ten-four..." Or, if "ten" was named like the rest of the two-digits: "tenty, tenty-one, tenty-two..." Or perhaps something like "unty" would be even more in-line with the pattern, so: "unty,
unty-one, unty-two..."
[S:Y'know... maybe we (okay, I) should resist the temptation to fix all the irregularities in English number names; we've already got plenty of things to wrap our heads around without deciding that
that's a problem for us to solve. (And just like we already decided not to give every numeral a new symbol, we're also not going to give every numeral a new name; so "20" will stay "twenty" even
though it's a different value, "30" will stay "thirty," and so on). So we'll just go with the first idea and call them "oneteen" and "twoteen." Hardly elegant, of course, but it should serve us well
enough.:S] [S: :S]
Fuck it. I went and wrote that fairly sensible paragraph, but I've changed my mind. We're still not gonna redo the numerals or give them new names, but we are totally going to rename the resulting
, because that'll make talking about them a lot less ambiguous.
In our base-twelve system, two-digit numbers after doz (10) are named like this: [first numeral]duz-[second numeral]. Therefore "11" is oneduz-one, "25" is twoduz-five, and so on. The number names in
English for three-digits and above also have a slew of problems, but I've gotta draw the line somewhere, so we'll just adopt them into our system more-or-less how they are. For three- and four-digit
numbers, replace "hundred" with "grondred" and "thousand" with "zousand." Yes, it sounds dumb; I don't care, it makes (some sort of) sense. Sadly we can't just have million, billion, trillion, etc.
all just go by "zillion," so they'll be mizion, bizion, and so on. Again, dumb; again, I don't care. And I think that should cover everything. Hooray! The old "teens" get a little clunkier, but the
rest of the two-digits (plus the three- and above) all have the same number of syllables as before. Saying the names obviously won't be nearly as natural, but the extra bit of thought you have to put
into it is probably a good thing.
Shut up, it's not like that at all...
So, to summarize. The sequence of symbols in our base-twelve system looks like this: "1, 2, 3, 4, 5, 6, 7, 8, 9, γ, ψ, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1γ, 1ψ, 20, 21, 22..." and is read like
this: "one, two, three, four, five, six, seven, eight, nine, ben, elv, doz, oneduz-one, oneduz-two, oneduz-three, oneduz-four, oneduz-five, oneduz-six, oneduz-seven, oneduz-eight, oneduz-nine,
oneduz-ben, oneduz-elv, twoduz, twoduz-one, twoduz-two..." So that means "5,964" is "five-zousand, nine-grondred sixduz-four." Likewise, "33,1γ2,ψ0ψ,810" is "threeduz-three bizion, one-grondred
benduz-two mizion, elv-grondred elv zousand, eight-grondred doz."
Let the beauty of Paul Rudd assuage all your doubts...
My hope is that my naming system for the two- and three-digit numbers helps highlight the nature of being in base-twelve. The fact that "27" is called "twoduz-seven" reflects the idea that it
signifies "two(dozen) + seven" (making it 31 in base-ten), just how in base-ten the same "27" symbol indicates "two(ten) + seven." Similarly, the name "grondred" is meant to reflect the fact that,
instead of the "ten tens" of a "hundred" in base-ten, "100" in base-twelve is a "dozen dozens," also known as a "gross" (144 in base-ten).
Now, with the business of names and symbols out of the way, it's time to address the issue that's been sitting menacingly in the corner: Changing our number system automatically changes math. In a
base-twelve system, 10 ÷ 4 = 3, and 6 × 3 = 16. Half of something is 60%. It's
The realization that our system of math isn't absolute may lead you to wonder whether it means anything at all. All these equations and formulas that we use to describe our world through physics and
chemistry... is it all pointless? Are we just taking abstract concepts that have no bearing on reality and pretending they combine in meaningful ways?
Is everything I know a lie?!
No. Now get a grip. The fact of the matter is that regardless of which number you base your system on, this many (o o) of something plus this many (o o) of something will always give you this many (o
o o o). And this many things (o o o o o o o o o o o o) divided into this many (o o o) groups will always look like this (oooo oooo oooo). The values stay the same; all that's changing are the names
and symbols you attribute to the values. Your number system and its inherent math are just a language to describe relationship of quantities. And just like the sun's gonna rise no matter what you've
named it, the values are going to combine the same way no matter what you've decided to call a full 'set.'
Admitting that the values are constant but that the symbols for them are arbitrary may very well break a few hearts, though. The much-loved ratio of a circle's circumference to its diameter,
π, is no longer 3.
1415926535... Those symbols themselves have no objective significance, only the underlying the ratio of values that they express does. In base-ten that ratio is roughly 22/7, but in our nomenclature
for base-twelve it becomes 1γ/7, which comes out as 3.184809493ψ... To be clear: those two different-looking strings of numbers are actually describing the same value; they're just in different
So, this was a long post, and no one will blame you if it hurt your head a little bit (hopefully all the pain was from the concepts themselves rather than me choosing suboptimal ways to relate them).
Now that we sort of see what a base-twelve system would look like, you might be asking yourself,
Would it really be worth it to switch?
Or maybe there's no question for you, and you've decided,
No, it would not be worth it to switch
. And if that's where you're at, I've gotta say I think I agree with you, despite all the effort I've put into describing it. I fear there's probably no plausible way to switch our number system at
this point; we've come too far with base-ten. It's a shame, but it's the truth. That doesn't mean I think it's worthless to understand what it would look like, though. As many people can surely
attest, learning a foreign language can be very helpful in understanding the nuances of your own.
I have a few more miscellaneous thoughts on this subject, but I'll save those for
another post
* I should note that my choice of letters from the Greek alphabet was largely arbitrary; I basically just chose two that weren't too similar to the existing numerals and which I thought looked nice
with the font I'm using. There have of course been may other proposals for the symbols which should represent the values ten and eleven in base-twelve. A and B are common ones, as are T and E. I went
off on my own and selected Greek letters because I wanted symbols that had very little meaning for most people outside of how they're being used here. Obviously all the Greek letters are used heavily
in math and science so the solution is hardly ideal, but I think it's decently illustrative of the points I'm trying to make. Also, the names I've listed—for those two numerals, "10," and the
double-digits—are entirely my own. I think they're sensible, but I'm sure there are better-known (and maybe better overall) alternatives out there. And I could have just called "100" in base-twelve a
"gross" and "1,000" a "great gross" since those are existing names for those quantities, but I wanted their names to emphasize how they're related to the values with the same symbols in base-ten,
hence "grondred" and "zousand."
1 comment:
1. Yes, please change the world to Base Twelve. I am thinking in two bases now, and that is too difficult. It is easier and simpler to think in one base and that one Base is the Base the Veca Time
Matrix Generally uses and that is Base Twelve. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, X, E, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 1X, 1E, 20, ... 30, 40, 50, 60, 70, 80, 90, X0, E0, 100,.... and so one
to infinity. Our planet is infinitesimal in comparison to the 15 Dimensional Veca Time Matrix and is more likely to be destroyed if we don't change to Base Twelve because Base Ten is out of synch
with the rest of the 15 Dimensional Veca Time Matrix.
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To achieve interstellar travel, the Kline Directive instructs us to be bold, to explore what others have not, to seek what others will not, to change what others dare not. To extend the boundaries of
our knowledge, to advocate new methods, techniques and research, to sponsor change not status quo, on 5 fronts, Legal Standing, Safety Awareness, Economic Viability, Theoretical-Empirical
Relationships, and Technological Feasibility.
In this post I discuss three concepts, that if implemented should speed up the rate of innovation and discovery so that we can achieve interstellar travel within a time frame of decades, not
Okay, what I’m going to say will upset some physicists, but I need to say it because we need to resolve some issues in physics to distinguish between mathematical construction and conjecture. Once we
are on the road to mathematical construction, there is hope that this will eventually lead to technological feasibility. This post is taken from my published paper “Gravitational Acceleration Without
Mass And Noninertia Fields” in the peer reviewed AIP journal, Physics Essays, and from my book An Introduction to Gravity Modification.
The Universe is much more consistent than most of us (even physicists) suspect. Therefore, we can use this consistency to weed out mathematical conjecture from our collection of physical
hypotheses. There are two set of transformations that are observable. The first, in a gravitational field at a point where acceleration is a compared to a location at 0 an infinite distance from the
gravitational source, there exists Non-Linear transformations Γ(a) which states that time dilation t[a]/t[0], length contraction x[0]/x[a], and mass increase m[a]/m[0], behave in a consistent manner
such that:
The second consistency is Lorentz-Fitzgerald transformations Γ(v) which states that at a velocity v compared to rest at 0, time dilation t[v]/t[0], length contraction x[0]/x[v], and mass increase m
[v]/m[0], behave in a consistent manner such that:
Now here is the surprise. The Universe is so consistent that if we use the Non-Linear transformation, equation (1) to calculate the free fall velocity (from infinity) to a certain height above the
planet’s or star’s surface, and it’s corresponding time dilation, we find that it is exactly what the Lorentz-Fitzgerald transformation, equation (2) requires. That there is this previously
undiscovered second level of consistency!
You won’t find this discovery in any physics text book. Not yet anyway. I published this in my 2011 AIP peer reviewed Physics Essays paper, “Gravitational Acceleration Without Mass And Noninertia
Now let us think about this for a moment. What this says is that the Universe is so consistent that the linear velocity-time dilation relationship must be observable where ever velocity and time
dilation is present, even in non-linear spacetime relationships where acceleration is present and altering the velocity and therefore the time dilation present.
Or to put it differently, where ever Γ(a) is present the space, time, velocity and acceleration relationship must allow for Γ(v) to be present in a correct and consistent manner. When I discovered
this I said, wow! Why? Because we now have a means of differentiating hypothetical-theoretical gravitational fields, and therefore mathematical conjectures, from natural-theoretical gravitational
fields, which are correct mathematical constructions.
That is, we can test the various quantum gravity & string hypotheses and any of the tensor metrics! Einstein’s tensor metrics should be correct, but from a propulsion perspective there is something
more interesting, Alcubierre tensor metrics. Alcubierre was the first, using General Relativity, to propose the theoretical possibility of warp speed (note, not how to engineer it). Alcubierre’s work
is very sophisticated. However, the concept is elegantly simple. That one can wrap a space craft in gravitational-type deformed spacetime to get it to ‘fall’ in the direction of travel.
The concept suggest that both equations (1) and (2) are no longer valid as the relative velocity between the outer edges of the spacetime wrap and an external observer is either at c, the velocity of
light or greater – one needs to do the math to get the correct answer. Even at an acceleration of 1g, and assuming that this craft has eventually reached c, equation (1) and (2) are no longer
consistent. Therefore, my inference is that Alcubierre metrics allows for zero time dilation within the wrap but not velocities greater than the velocity of light. Therefore, it is also doubtful that
Dr. Richard Obousy hypothesis that it is possible to achieve velocities of 1E30c with a quantum string version of Alcubierre warp drive is correct.
Previous post in the Kline Directive series.
Next post in the Kline Directive series.
Benjamin T Solomon is the author & principal investigator of the 12-year study into the theoretical & technological feasibility of gravitation modification, titled An Introduction to Gravity
Modification, to achieve interstellar travel in our lifetimes. For more information visit iSETI LLC, Interstellar Space Exploration Technology Initiative.
Solomon is inviting all serious participants to his LinkedIn Group Interstellar Travel & Gravity Modification.
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Math Teacher Mambo
Probably in math classes across the country no matter what level (algebra through calculus), right now, if you asked a student, "how many feet squared is 100 inches squares?", the majority of them
would JUST divide by 12 to get their answer. Right? Right?
I've tried various things throughout the years, and things "stick" for the unit, but later, say the following year or years, I ask the same question, the student reverts back to JUST dividing by 12.
Must be hardwired into their heads.
Anyway, this came up again yesterday in geometry class with the following problem:
You want to paint the exterior of a cylindrical container with a 4 inch radius and 15 inch height. Paint costs 86 cents per square foot, how much would it cost.
I had an answer bank on the sheet, and LO AND BEHOLD, their answer was not on there. Hmmmmmm. Then I prompted: be careful with your units. OH! Okay, convert convert. OH! the answer is STILL not on
there. Hmmmmmm. The dreaded JUST dividing by 12 dilemma. Anyway, I held up a piece of white paper and basically did what you see here below.
It SEEMED to make sense to the students. I liked the visual and the methodical dividing the side by 12 AND the algebraic equation by 12 right afterward, so they see what happens. It SEEMED to stick,
but I'm not going to fall for that again. I'll quiz them again next year or two to see. My optimistic self thinks, "YES! I've solved the problem of world peace." Don't burst my bubble. Anyway, one
more example to add to the arsenal.
8 comments:
1. A funny (and relevant) thing happened a month ago. I was working with a girl (who is home-schooled) who is learning math out of a 7th grade Singapore text, and we got to the section on converting
square units (square meters to square cm, etc), and I thought--we should do some similar problems with inches and feet. So I went to my department's library of middle school math texts (3 text
book series), and looked for a section that included converting square units. Guess what? That section wasn't there! So from my short survey, most US middle school texts don't teach this at all
(whereas they've been converting linear units since second grade). I'm thinking this could be part of the problem.
2. Lsquared:
That's just crazy! Hopefully, the teachers are supplementing. Unfortunately, it doesn't seem to stick much.
3. I prefer to start with a square foot. "How many inches is this side? "12in". How many inches on this side?" 12in" So how many square inches in a square foot? "144sqin."
Then switch to metric and 1g H2O = 1ml, 1ml = 1 cm3; 1000cm3 = 1dm3; 1000L = 1m3;
__ cm3 = 1m3;
__ cm3 = 1km3
How many Liters is that 383ci. hotrod engine?
For what it's worth, I don't think anyone teaches square units or cubic units in middle school. Too busy pretending they're teaching algebra -- raising standards, don't cha know.
4. nice posting . . keep blogging!!
5. I hang out all day long with problem-solving geeks, most of whom are really visual thinkers. Also, I write problem-solving support for middle-school and elementary-school students. Again, we tend
to go visual or hands-on with those groups. If you had posed this scenario to me 5 years ago, I would have been thinking algebraically like you did in that nicely-illustrated document. My newly
found visual problem-solver got stuck on the following question: does it matter how those 100 square inches are arranged? Can you always fit 25/36 of a square foot into 100 square inches? What if
it's a 1-inch by 144-inch square foot? Is that still considered a *square* foot? Is it easier to see in a squarer square foot? And what are the shapes in the original context anyway?
Funny what different questions and answers come up when you approach with a different problem-solving strategy!
6. Max,
Those are great questions. They would be nice extensions. ... I'm guessing you know the answers and are just posing the fun thinking questions. I'll have to enhance the next go-round with my
geometry kids.
7. carolann12:45 AM
Why did you have to put that sneaky " what about volume" question at the end?
8. Well, I wanted them to do more than just blindly plug into formulas. Also, I wanted to show a realistic situation of how this can be used. For example, if company is mass producing this objects,
they need the cost information for budgets.
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Natural Sciences & Mathematics
Degree Programs
Why major in math?
First of all, you like it and/or you're good at it. Do you really need another reason? If you need a little more convincing, read on.
Professional graduate schools (business, law, medicine) think it's a great major because they realize that studying mathematics develops analytical skills and the ability to work in a problem solving
environment; these are skills and experience which rank high on their list of assets.
Jobs in the private sector abound; careers where a mathematics major is particularly well suited include:
• Actuarial Science
• Computer analyst or programmer
• Cryptographer
• Economist
• Elementary or secondary school teaching, college professor
• Engineering analyst
• Information scientist
• Marketing research analyst
• Mathematician
• Meteorologist
• Numerical analyst
• Operations research analyst
• Statistician
• Systems analyst
Whether you're interested in developing models and interpreting their results, or are interested in developing efficient algorithms to expedite known processes, mathematics and computer science are
the tools of choice.
Models are needed to investigate air flow across the surface of aircraft wings, chemical and biological processes, astronomical trajectories and urban development. These models need to be designed,
created, the data from them collected and analyzed, conclusions drawn and predictions made from them.
Math and Top Jobs
In the recent book The Jobs Rated Almanac author Les Krantz ranks 250 jobs according to six criteria: income, stress, physical demands, potential growth, job security and work environment. He
obtained his data from the government, trade groups and telephone surveys. The to ten jobs according to Krantz are:
1. Web site manager
2. Actuary
3. Computer systems analyst
4. Software engineer
5. Mathematician
6. Computer programmer
7. Accountant
8. Industrial engineer
9. Hospital administrator
10. Web developer
Note that the jobs rated higher than Mathematician also involve significant mathematical reasoning and knowledge.
Links to jobs for math majors
Here are various links to nonacademic job opportunities for mathematics majors.
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How many ounces is 120ml?
You asked:
How many ounces is 120ml?
4.05768272412 US fluid ounces
the volume 4.05768272412 US fluid ounces
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Need help proving the inverse of a bijective function is bijective
December 4th 2012, 03:50 PM #1
Need help proving the inverse of a bijective function is bijective
I've a math problem that I can't solve which is:
How do I prove that if a function is bijective then it's inverse is also bijective?
I know I've to prove that if $f^{-1}(a) = f^{-1}(b)$ then $a = b$ So then the inverse is injective.
Then to prove a function $f^{-1} : B \to A$ is surjective I've to show that each element of $A$ is the image of at least one element of $B$
If I can show these two to be valid then the inverse is bijective.
Can anyone kindly tell me how do I show this?
How do I show that a inverse of a bijective function is injective and surjective?
Re: Need help proving the inverse of a bijective function is bijective
Re: Need help proving the inverse of a bijective function is bijective
I've a math problem that I can't solve which is:
How do I prove that if a function is bijective then it's inverse is also bijective?
I know I've to prove that if $f^{-1}(a) = f^{-1}(b)$ then $a = b$ So then the inverse is injective.
Then to prove a function $f^{-1} : B \to A$ is surjective I've to show that each element of $A$ is the image of at least one element of $B$
If I can show these two to be valid then the inverse is bijective.
Can anyone kindly tell me how do I show this?
How do I show that a inverse of a bijective function is injective and surjective?
suppose a is in A. to show f^-1 is surjective, we need to exhibit a b in B with f^-1(b) = a. b = f(a) works.
Re: Need help proving the inverse of a bijective function is bijective
Re: Need help proving the inverse of a bijective function is bijective
it's not obvious that the inverse of a bijective morphism is a bijective morphism. in fact, in the category of topological spaces (where "morphism" means homeomorphism) this is false. it just
happens to be true in SOME categories (Set is one of them).
the definition only tells us a bijective function has an inverse function. it doesn't explicitly say this inverse is also bijective (although it turns out that it is). it's pretty obvious that in
the case that the domain of a function is FINITE, f^-1 is a "mirror image" of f (in fact, we only need to check if f is injective OR surjective). with infinite sets, it's not so clear.
the problem is with "surjectiveness". it only asserts the existence of a pre-image for a co-domain element, it doesn't say how we should go about producing one. to even say we CAN always produce
one is tantamount to the axiom of choice (something one "just has to take on faith"). this is in stark contrast to the situation with injective functions, where there is a clear-cut test: if f(x)
= f(y), see if x = y is true.
insofar as "proving definitions go", i am sure you are well-aware that concepts which are logically equivalent (iff's) often come in quite different disguises. some texts define a bijection as a
function for which there exists a two-sided inverse. some texts define a bijection as an injective surjection. the equivalence of the two isn't "something you can prove", in fact there are those
who doubt it's even TRUE.
Re: Need help proving the inverse of a bijective function is bijective
Thanks Deveno and Plato for answer.
Sorry if I'm missing something obvious Deveno. If $f^{-1}$ is surjective why is it enough to only show injective or surjective in order to prove that the inverse function is bijective for finite
To prove bijection don't we need to prove that the function is both injective and surjective? Isn't that the definition of bijection?
Can you kindly shed light on this?
Re: Need help proving the inverse of a bijective function is bijective
my bad. i meant from a set to itself (don't know what i was thinking. i'd blame the whiskey, but i don't actually HAVE any).
Re: Need help proving the inverse of a bijective function is bijective
Thanks Plato and Deveno for help.
December 4th 2012, 04:05 PM #2
December 4th 2012, 04:29 PM #3
MHF Contributor
Mar 2011
December 4th 2012, 04:46 PM #4
December 4th 2012, 05:26 PM #5
MHF Contributor
Mar 2011
December 4th 2012, 06:15 PM #6
December 4th 2012, 10:32 PM #7
MHF Contributor
Mar 2011
December 5th 2012, 03:39 AM #8
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Random Matrix Theory and Life over Finite Fields
Isaac Newton Institute for Mathematical Sciences
Random Matrix Theory and Life over Finite Fields
1st July 2004
Author: NIcholas M. Katz (Princeton Univ.)
We will give an introductory survey of some of the relations between random matrix theory and various diophantine questions over finite fields.
• Isaac Newton Institute for Mathematical Sciences
We will give an introductory survey of some of the relations between random matrix theory and various diophantine questions over finite fields.
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Algorithm of the Week: Binary Search Tree
Constructing a linked list is a fairly simple task. Linked lists are a linear structure and the items are located one after another, each pointing to its predecessor and its successor. Almost every
operation is easy to code in few lines and doesn’t require advanced skills. Operations like insert, delete, etc. over linked lists are performed in a linear time. Of course on small data sets this
works fine, but as the data grows these operations, especially the search operation becomes too slow.
Indeed searching in a linked list has a linear complexity and in the worst case we must go through the entire list in order to find the desired element. The worst case is when the item doesn’t belong
to the list and we must check every single item of the list even the last one without success. This approach seems much like the sequential search over arrays. Of course this is bad when we talk
about large data sets.
In terms of arrays, we could perform binary search and go directly in the middle of the array, then jump back or forward. That is because we can access array items directly using their index. However
as we saw the linked lists unlike arrays can’t benefit of a direct access and we must go item by item.
Because of this natural problem of linked lists searching is slow and obviously we can’t make it better. The only way to improve searching over dynamic data structures is to use different data
The tree is a data structure where each item, except of keeping some data, keeps a reference (pointer) to its children and its parent.
Of course if the item doesn’t have children, they are NIL, then this is considered a leaf in the tree terminology. In the other hand if the item doesn’t have parent item it is considered the root.
If there is no item in the tree the tree is considered empty.
In these terms only the root has no parent, and each item can have as many children as possible. Here are some trees in form of a diagrams.
If we’re looking at the root of the tree we can assume there are two sub-trees – one left and one right. However if we isolate only one of these sub-trees we can again think of it as a tree and
assume that it has one left and one right sub-trees and go recursively with this definition.
A binary tree is a tree where each item can have at most two children.
Binary trees are especially important because they can contain ordered data in a specific manner. Building a binary tree isn’t difficult at all and it’s very similar to building a linked list.
However a binary tree isn’t more successful in searching than any other tree or data structure. If the items aren’t placed in a specific order we must go through the entire tree in order to find the
searched item. This isn’t a great optimization, so we must put an order in it to improve the searching process.
Binary Search Tree
The binary search tree is a specific kind of binary tree, where the each item keeps greater elements on the right, while the smaller items are on the left.
Constructing a binary search tree is easy, because we can go for inserting each item only by comparing it with the root and decide where to go (left or right) based on its value.
The following code in PHP describes the basic principles of a binary search tree.
class Node
public $parent = null;
public $left = null;
public $right = null;
public $data = null;
public function __construct($data)
$this->data = $data;
public function __toString()
return $this->data;
class BinaryTree
protected $_root = null;
protected function _insert(&$new, &$node)
if ($node == null) {
$node = $new;
if ($new->data <= $node->data) {
if ($node->left == null) {
$node->left = $new;
$new->parent = $node;
} else {
$this->_insert($new, $node->left);
} else {
if ($node->right == null) {
$node->right = $new;
$new->parent = $node;
} else {
$this->_insert($new, $node->right);
protected function _search(&$target, &$node)
if ($target == $node) {
return 1;
} else if ($target->data > $node->data && isset($node->right)) {
return $this->_search($target, $node->right);
} else if ($target->data <= $node->data && isset($node->left)) {
return $this->_search($target, $node->left);
return 0;
public function insert($node)
$this->_insert($node, $this->_root);
public function search($item)
return $this->_search($item, $this->_root);
$a = new Node(3);
$b = new Node(2);
$c = new Node(4);
$d = new Node(7);
$e = new Node(6);
$t = new BinaryTree();
echo $t->search($e);
Search Complexity
Searching in binary search trees is supposed to be faster than searching into linked list. However the searching process in a BST can be very fast, but also can be as slow as on linked list. That is
because depending on the input of items they can be placed only on the one side of the root.
That makes the worst-case searching as slow as on linked list which is linear O(n). However if the tree is somehow balanced we can search very quickly with O(log(n)) time.
Further Optimization
We now see how ineffective binary search trees can be, so the only thing we must care is how to keep them balanced, so the search will be faster. The answer is to maintain (during insertion) a
balanced binary search tree, which is another very handy data structure.
A balanced binary search tree, or only balanced tree, is a data structure where the height of left and the right sub-trees can vary by one level at most.
Binary search trees are easy to build and maintain. The great thing is that if the data is well balanced they can be very useful for searching. The only problem is that these structures can be
ineffective depending on the insertion order. However if we are somehow sure that the items aren’t ordered on the input, we may expect some optimized searching compared to a linked list. Compared to
balanced binary search trees, BST require much less time to build and maintain (insert, delete).
Trees are very useful when working with graphs. Actually one of the very common tasks is walking through the entire tree, which can be done in several ways. First we can go to the left sub-tree, then
the root and then the right sub-tree. Or right-root-left. Or root-left-right.
However we can go in depth first often called depth-first-search or a breadth-first-search.
These two methods are designed to walk through the items in a specific order, which is very handy for some specific tasks – at least each tree is also a graph.
Published at DZone with permission of Stoimen Popov, author and DZone MVB. (source)
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Tikz Introduction
September 27, 2011
By Ralph
The pgf drawing package for LaTeX provides facilities for drawing simple of complicated pictures within a LaTeX document. There are many options available within the package and in this post we
consider some of the basics to get up and running.
Fast Tube by Casper
As with all LaTeX documents we need to select a document class and include some preamble material prior to the body of our document. A blank template for a document with a single tikz picture is
shown here:
The tikz picture has a coordinate system similar to that which you would expect where moving from left to right on the page corresponds to increasing the x value and bottom to top increases the y
value. A line can be drawn between two points wit the \draw command:
To draw a line between multiple points these can be chained together in a single draw command:
\draw (0,0) -- (1,0) -- (1, 4);
The line style can be altered by adding various options in square brackets directly after the draw command. So to change to a dashed red line we would write the following code:
\draw[red,dashed] (0,0) -- (2,0);
A circle of a given radius can be draw using the \draw command and we specify the radius of the circle in round brackets:
\draw (0,0) circle (2.5cm);
This will draw a circle with radius of 2.5 cm. The circle could be changed into an ellipse and we would then need to specify the radius in two directions, an example of this:
\draw (0,0) ellipse (2cm and 3.5cm);
Other useful resources are provided on the Supplementary Material page.
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formula from curves for two inputs
June 19th 2010, 02:10 AM #1
Jun 2010
formula from curves for two inputs
dear all,
I need your help in this matter,
the discription of the subject is:
we have two input vectors ( included in one vector):
T ( temperature ) and H ( humidity ), there is a corresponding one output
P ( power demand ), and all these data are normalized
I have thousands of data for these values done for many years.
for sample example: ( l assume that we have the data for 5 hours )
T_H=[0.3333 0.3333 0.3542 0.3750 0.3750 ; 0.6970 0.8384 0.6869 0.7980 0.7677]
and the corresponding P is:
P=[0.6561 0.6618 0.6568 0.6694 0.6622]
I need to draw a curves relating all data,
and THE MOST IMPORTANT THING WHICH IS THE AIM OF MY PROBLEM IS :
finding ( inventing ) a formula depending on the curves drawn that relates
the power as an output with the temperature & humidity as inputs,
P = f (T,H)
for example: the formula model may be:
P = a0 + a1 * T + a2*H*sin(T) + a3*T^3*H + a4*T*H
then finding a0,a1,a2 and a3 and this is easy and I
know how to find them.
BUT, my problem is how can I find this above example model function P ?
is it by dividing the curve to parts and then finding a function
for each part ? but how is this done by MATLAB ?
Any idea ?
Can anybody help me? I am appreciating any help.
and remember, any service will be valuable for me
Best regards
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Time Filter Technical Explanation
A movie camera is a sampling system in time
Motion picture cameras acquire images sequentially in time, with each image representing a sample of the real world in time.
In both digital and film based motion picture cameras, the time varying signal is measured at a fixed frame rate, usually 24 frames per second (fps). The sampling rate is 24 cycles per second (or 24
hertz), the same as the frame rate in frames per second.
This type of system can be considered a time-sampling system, and the performance of such a sampling system can be analyzed and predicted with the well-known Nyquist-Shannon sampling theorem.
To understand the time-domain sampling of a motion picture camera, consider a simple light source, such as a light bulb, which is being photographed with a motion picture camera. If the intensity of
the light bulb is modulated sinusoidally, the intensity recorded by the film or digital sensor should correspondingly represent samples of the time-varying brightness of the light bulb, and upon
playback the light intensity varying over time should match the sine wave of the original light bulb. The real world continuously varying intensity of the light bulb was recorded as a finite string
of discrete values, one value for every frame of the movie.
Aliasing can occur in sampling systems
The Nyquist frequency is defined as half the sampling frequency. For example, in a 24 frame per second (or 24 cycles per second, or 24 hertz) motion picture camera, the Nyquist frequency would be 12
A well-understood property of sampling systems is aliasing, which the Nyquist-Shannon theorem predicts when real-world signals with frequencies above the Nyquist frequency are sampled. Any real world
signal frequency above the Nyquist rate will be aliased, or shifted into another—false—frequency that can be represented by the sampling system.
Since motion picture cameras are sampled systems, aliasing can and does occur when the real-world frequencies exceed the Nyquist rate. Motion picture cameras measure in three dimensions: two spatial
dimensions (the two-dimensional image produced for every frame) and also time. For our purposes, we are considering only the time sampling, i.e. temporal sampling. Aliasing in time dimension sampling
is known as temporal aliasing.
In the sinusoidally varying light bulb example, if the frequency of the sine wave were 10 hertz, and the light was sampled with a normal 24 frame per second camera system, the 10 hertz signal would
be accurately recorded and reproduced because it is less than the Nyquist frequency of 12 hertz.
However, if the light bulb were varied sinusoidally at 14 hertz, the recorded and reproduced frequency from a 24 frame per second camera would incorrectly be 10 hertz. This is because 14 hertz is 2
hertz above the Nyquist frequency, so the resulting frequency is 2 hertz below the Nyquist frequency. This is an example of signal aliasing when a frequency higher than the Nyquist frequency is
Examples of aliasing in motion picture cameras
Temporal aliasing in motion picture cameras is exhibited in many ways. The most common and popularly understood manifestation of temporal aliasing is known as the “wagon wheel” effect. This is a
rapidly moving wagon wheel captured by a motion picture camera appears to stop, reverse direction, or move slowly. The higher frequencies of the motion are aliased, or falsely shifted, to appear as
different frequencies.
Temporal aliasing in motion picture cameras is exhibited in nearly all situations where fast movement is being captured.
Prefiltering as a method for eliminating aliasing
In sampling systems, the method employed to eliminate aliasing is to band limit the real-world signal before the sampling takes place to ensure that no frequencies above the Nyquist frequency are
allowed to enter the sampling system. This is known as prefiltering.
Prefiltering is a procedure performed on the unsampled signal before sampling. It is usually a low-pass frequency filter.
The ideal low-pass frequency filter for prefiltering would be unity (signal unaffected) below the Nyquist frequency, and 0 (no signal allowed) above the Nyquist frequency.
Normal motion picture cameras have some frequency prefiltering, but it doesn't work that well
Motion picture cameras do have some inherent prefilter, as the amount of time the shutter is open causes some motion blurring on a single frame (sample).
Exposure time for a frame is typically indicated as a shutter angle. A 360 degree shutter angle indicates the frame is exposed for the entire time of the sample, while a 180 degree shutter angle
indicates the frame is exposed for half of the time between samples. For example, in a 24 frame per second motion picture system, a 180 degree shutter would expose each frame for 1/48 of a second,
while a 360 degree shutter would expose each frame for 1/24 of a second.
If the amount of light allowed to pass to the sensor (film or digital sensor) during the time of the frame is plotted as a function of time, the resulting plot describes how the incoming image
intensity changes over time. This change in intensity over time is called the exposure window function, or simply the window function.
It can be seen that the exposure window functions for motion picture shutters have a sharp transition between 0 (no light) and 1 (full exposure). Existing motion picture cameras do not implement any
other values other than 0 and 1, as the shutter is either open or closed.
Filters can be represented by their response to a given frequency, and one such representation is called the modulation transfer function, or MTF. The modulation transfer function when expressed
linearly is normalized between 0 and 1, where 1 is full response to a given frequency and 0 is no response.
There is a direct mathematical relationship between the exposure window function and the prefiltering. If an exposure window function is known, the resulting modulation transfer function of the
prefilter can be calculated.
The following figure shows the MTFs of the effective prefiltering of a 180 degree and a 360 degree shutter angle compared with an ideal prefilter for a 24 frame per second system (Nyquist frequency
is therefore 12 hertz).
The Tessive Time Filter
If the exposure window function were shaped differently, and had transitional values other than 0 and 1 (fully closed and fully open), the resulting modulation transfer function of the prefilter
could be shaped in new ways.
The patent pending Tessive Time Filter has a continuously varied exposure windows which creates a prefilter with a MTF that better reduces aliasing frequencies. The following figure shows the
exposure window and the resulting MTF, compared to a normal 180 degree shutter MTF. The resulting MTF has substantially less response above the Nyquist frequency, but a slightly reduced response
below Nyquist.
The Tessive Time Compensator
The slight reduction in response in the region below Nyquist frequency can be adjusted with a postfilter. A postfilter is a digital finite impulse response (FIR) convolutional filter. The following
figure shows the combined MTF of a prefilter and a postfilter employed to provide good response below Nyquist frequency while reducing response above Nyquist, compared with the MTF of a typical
180-degree shutter.
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[Numpy-discussion] strange divergence in performance
Ernest Adrogué eadrogue@gmx....
Wed Jan 20 15:56:40 CST 2010
I have a function where an array of integers (1-d) is compared
element-wise to an integer using the greater-than operator.
I noticed that when the integer is 0 it takes about 75% more time
than when it's 1 or 2. Is there an explanation?
Here is a stripped-down version which does (sort of)show what I say:
def filter_array(array, f1, f2, flag=False):
if flag:
k = 1
k = 0
m1 = reduce(np.add, [(array['f1'] == i).astype(int) for i in f1]) > 0
m2 = reduce(np.add, [(array['f2'] == i).astype(int) for i in f2]) > 0
mask = reduce(np.add, (i.astype(int) for i in (m1, m2))) > k
return array[mask]
Now let's create an array with two fields:
a = np.array(zip( np.random.random_integers(0,10,size=5000), np.random.random_integers(0,10,size=5000)), dtype=[('f1',int),('f2',int)])
Now call the function with flag=True and flag=False, and see what happens:
In [29]: %timeit filter_array(a, (6,), (0,), flag=False)
1000 loops, best of 3: 536 us per loop
In [30]: %timeit filter_array(a, (6,), (0,), flag=True)
1000 loops, best of 3: 245 us per loop
In this example the difference seems to be 1:2. In my program
is 1:4. I am at a loss about what causes this.
More information about the NumPy-Discussion mailing list
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Legislative Analyst's Office
Analysis of the 2001-02 Budget Bill
Algebra Incentive Program
The 2001-02 Governor's Budget includes $30 million from the General Fund (Proposition 98) for incentive payments to school districts to increase the number of students enrolled in algebra classes.
School districts would be permitted to use these payments to attract and retain qualified algebra teachers including the use of salary differentials, training, and reduction of class "loads" for
these teachers. The funds also could be used to improve pre-algebra skills of students falling behind in mathematics. The administration intends to seek legislation to establish this program of
algebra-related incentives but had not provided the Legislature with draft bill language at the time of this Analysis.
Shortage of Math Teachers
Student proficiency in algebra has played a central role in the state's recent education reform efforts. For instance, the State Board of Education approved a proposal to include the state's algebra
content standards in the upcoming High School Exit Exam (HSEE) that students must pass in order to graduate from high school, beginning with the graduating class of 2004. In part due to this upcoming
high-stakes requirement, the Legislature has taken steps to ensure students receive instruction in algebra. For example, Chapter 1024, Statutes of 2000 (SB 1354, Poochigian), requires students in
grades 7 through 12 to complete an algebra course as a condition of receiving a high school diploma, commencing with the 2003-04 school year. This requirement gives students a reasonable opportunity
to learn this subject before they will be tested on the HSEE. Given the requirement placed on schools to offer algebra courses, Chapter 1024 creates a potentially reimbursable mandate for schools to
claim some additional costs associated with providing algebra instruction.
The budget proposal seeks to address a problem that districts face in helping students meet the new algebra requirement a shortage of qualified math teachers in the state to teach algebra. Many
schools are experiencing difficulty in recruiting and retaining math teachers for various reasons, including the availability of more lucrative job opportunities elsewhere. Based on current
enrollments in algebra, the Office of the Secretary for Education estimates that over the next three years, schools will need to add about 1,300 algebra teachers to instruct about 160,000 students
who would not have otherwise taken algebra.
Governor's Budget Proposal
The budget provides $30 million in Item 6110-121-0001 to establish the Algebra Incentive Program to assist school districts in attracting and retaining math teachers for algebra. Of this total, $17
million would provide districts with $50 for each student in grades 7 through 12 enrolled in algebra and who takes the standards-based algebra test. (For the budget year, the test-taking requirement
would apply to the test administered in spring 2000. The standards-based algebra test is included in the content standards-aligned portion of the state's Standardized Testing and Reporting STAR
exam.) The remaining $13 million would pay districts $100 for each additional student, beyond the current-year level, in grades 7 through 12 taking algebra and the standards-based algebra test in
subsequent years.
Districts would be permitted to use their algebra-incentive funds for locally determined initiatives to attract and retain algebra teachers including salary differentials, training, smaller class
sizes, or other forms of reducing teaching "loads." Districts also could use funds for pre-algebra training for students having difficulty learning math. As of this writing, the administration has
provided little detail on exactly how funds would be distributed to schools for increasing algebra enrollments.
Concerns With Governor's Proposal
The Governor has identified an important problem facing many public schools. However, we see two problems with his proposed solution, as discussed below.
Misguided Targeting. About $17 million of the $30 million proposed for the algebra incentive program would be awarded at a rate of $50 per student for students already taking algebra. Thus, these are
"windfall" payments that will not induce any behavioral changes. In addition, schools facing little challenge in getting students enrolled in algebra would receive a disproportionate share of the
estimated $17 million. Instead, we believe that this funding should be targeted to schools with the highest levels of algebra "under-enrollment" and the most severe shortages of qualified math
Problem Narrowly Defined. We agree that many school districts face challenges in attracting and retaining math teachers for algebra. The administration's approach, however, focuses on the algebra
problem to the exclusion of addressing other important shortage areas. For example, a study by SRI International found that serious shortages exist for fully qualified teachers for special education,
science, and services for English language learners in many public schools across the state.
• Special Education. During the 1998-99 fiscal year, the Commission on Teacher Credentialing (CTC) issued about 6,000 emergency permits for individuals hired to be education specialists and
resource specialists in special education. (Emergency permits authorize individuals who lack a teaching credential to provide classroom instruction. Permits are issued to individuals on an annual
basis for a maximum of five years.)
• Science. Science has long been a subject area lacking sufficient numbers of qualified teachers. The SRI study pointed out that this subject area accounts for the largest number of single-subject
emergency permit holders. For example, the CTC issued about 2,400 emergency permits in science in 1998-99. According to the State Department of Education, school districts expect to hire over
16,000 new teachers for 15 different subject areas in 2000-01. About 12 percent of these teachers will fill new or vacant positions in science.
• English Language Learner (ELL) Pupils. In 1999-00, about 30 percent of the 124,000 teachers in California public schools providing: (1) primary language assistance, (2) English language
development (ELD), or (3) specifically designed academic instruction in English to ELL pupils did not have a CTC certificate to provide such instruction. (This total does not include teachers who
instruct ELL pupils in mainstream classroom settings.)
Given the serious shortages of fully qualified teachers in subject areas other than algebra, we believe that the Governor's initiative is too narrowly drawn.
Legislative Analyst's Recommendation
We recommend that the Legislature redirect $17 million of the $30 million for the Algebra Incentive Program to augment the existing Teaching as a Priority (TAP) block grant because (1) the proposed
allocation of the $17 million would largely benefit schools lacking teacher recruitment/retention problems and (2) other schools need resources to recruit and retain teachers not only in algebra but
in other key areas of shortage. We further recommend that the Legislature broaden the permissible uses of the TAP block grant to include teacher salary differentials for subject areas with critical
shortages of teachers. (Reduce Item 6110-121-0001 by $17 million and augment Item 6110-134-0001 by same amount.)
We believe the Governor's proposed algebra incentives program has important elements that can help address the teacher shortages in this one subject area. For example, the proposal gives considerable
flexibility, including authority for salary differentials. We believe the proposal would be improved, however, by extending this type of flexibility to help districts attract/retain teachers in all
subject areas with critical needs, not just algebra. Below, we present an alternative approach that allows the Legislature to retain the best elements of the budget proposal, yet also address the
supply of teachers for other hard-to-staff subjects.
Approve $13 Million for Algebra Incentive Program. We believe that the Governor's proposal to provide $100 for each additional student enrolling in math would help address teacher shortages in
algebra in the schools most in need. Given the challenges facing districts in helping students meet the state's new algebra requirement, we recommend that the Legislature approve $13 million for the
Algebra Incentive Program. In adopting legislation to establish this program, we also recommend that the Legislature approve language specifying that funds provided to districts under the Algebra
Incentive Program should be deemed to offset any possible state-mandated costs related to the state's algebra course requirement.
Redirect $17 Million to Teaching as a Priority. In addition to funds for the Algebra Incentive Program, the Governor's budget also includes $118.7 million in Proposition 98 funds under Item
6110-134-0001 for the TAP block grant. This block grant program, established by Chapter 70, Statutes of 2000 (SB 1666, Alarcon), awards funds to school districts for the purpose of attracting and
retaining credentialed teachers in low-performing schools. Funding is allocated to school districts based on the number of pupils enrolled in schools ranked in the bottom half of the Academic
Performance Index. Schools ranked in the bottom 30 percent receive one and one-half times the per-pupil funding of schools ranked between the 40^th and 50^th percentiles. Schools have flexibility to
use the funds for different recruitment and retention incentives in order to reduce the number of teachers on emergency permits. Such incentives can include signing bonuses, improved work conditions,
teacher compensation, and housing subsidies.
In view of our concerns that the Governor's initiative is too narrowly focused and that about $17 million of the proposed funds would be allocated to schools with the least need of recruiting and
retaining fully qualified math teachers, we recommend that the Legislature redirect the $17 million to the TAP block grant. The block grant not only targets funding to low-performing schools, but it
also gives school districts broad discretion to recruit and retain teachers in any subject area. Under our approach, school districts could use the funds to recruit and retain algebra teachers, but
they also would be given the flexibility to address shortages in other subject areas, based on local needs.
The Governor's proposal specifies that school districts would be able to use algebra incentive payments for salary differentials. We believe that such an approach for attracting and retaining
teachers should be clarified as one of the permissible uses of TAP funds as well. Therefore, we further recommend the enactment of legislation to (1) clarify that TAP funds can be used for salary
differentials for subject areas with critical teacher shortages and (2) specify that funding provided to districts under the TAP program be deemed available to offset any possible state-mandated
costs related to the state's algebra course requirement.
Return to Education Table of Contents, 2001-02 Budget Analysis
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7. Differentiating Powers of a Function
by M. Bourne
Function of a Function
If y is a function of u, and u is a function of x, then we say
"y is a function of the function u".
Example 1
Consider the function
y = (5x + 7)^12.
If we let u = 5x + 7 (the inner-most expression), then we could write our original function as
y = u^12
We have written y as a function of u, and in turn, u is a function of x.
This is a vital concept in differentiation, since many of the functions we meet from now on will be functions of functions, and we need to recognise them in order to differentiate them properly.
Chain Rule
To find the derivative of a function of a function, we need to use the Chain Rule:
`(dy)/(dx) = (dy)/(du) (du)/(dx)`
This means we need to
1. Recognise `u` (always choose the inner-most expression, usually the part inside brackets, or under the square root sign).
2. Then we need to re-express `y` in terms of `u`.
3. Then we differentiate `y` (with respect to `u`), then we re-express everything in terms of `x`.
4. The next step is to find `(du)/dx`.
5. Then we multiply `dy/(du)` and `(du)/dx`.
Example 2
Find `dy/dx` if `y = (x^2+ 3)^5`.
Play with the graph of this example on the Differentiation Java applet page and explore what it means.
Example 3
Find `dy/dx` if `y=sqrt(4x^2-x)`.
You can play with this example on the Differentiation Java applet page.
The Derivative of a Power of a Function (Power Rule)
An extension of the chain rule is the Power Rule for differentiating. We are finding the derivative of u^n (a power of a function):
`d/dxu^n=n u^(n-1)(du)/dx`
Example 4
In the case of `y=(2x^3-1)^4` we have a power of a function.
Play with this example on the Differentiation Java applet page.
Find the derivative of `y=(x^2(3x+1))/(x^4+2)`
Play with this challenge example on the Differentiation Java applet page.
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Hamming Distance
godavemon davefowler at gmail.com
Fri Jun 20 01:27:58 CEST 2008
I need to calculate the Hamming Distance of two integers. The hamming
distance is the number of bits in two integers that don't match. I
thought there'd be a function in math or scipy but i haven't been able
to find one. This is my function but it seems like there should be a
faster way. I do this computation many times and speed up is
def hamdist( a, b , bits = 32):
def _hamdist( x, bits):
if bits:
return (x & 1) + _hamdist(x >> 1, bits-1)
return x & 1
return _hamdist( a ^ b, bits)
Another alternative would be to convert the XOR to a binary string and
count the # of 1's.
Which would be fastest? Are there better alternatives?
More information about the Python-list mailing list
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Electronic Circuits and Diagram-Electronics Projects and Design
Types of Chopper Circuits
In chopper circuits, unidirectional power semiconductors are used. If these semiconductor devices are arranged appropriately, a chopper can work in any of the four quadrants. we can classify chopper
circuits according to their working in any of these four quadrants as type A, type B, type C, type D and type E. Let us now take a look of these classifications and the characteristics of various
Type A Chopper or First–Quadrant Chopper
This type of chopper is shown in the figure. It is known as first-quadrant chopper or type A chopper. When the chopper is on, v[0 = ]V[S ]as a result and the current flows in the direction of the
load. But when the chopper is off v[0 ]is zero but I[0 ] continues to flow in the same direction through the freewheeling diode FD, thus average value of voltage and current say V[0 ]and I[0 ]
will be always positive as shown in the graph.
In type A chopper the power flow will be always from source to the load. As the average voltage V[0 ]is less than the dc input voltage V[s ]
Type B Chopper or Second-Quadrant Chopper
In type B or second quadrant chopper the load must always contain a dc source E . When the chopper is on, v[0 ] is zero but the load voltage E drives the current through the inductor L and the
chopper, L stores the energy during the time T[on ] of the chopper . When the chopper is off , v[0 ]=( E+ L . di/dt ) will be more than the source voltage V[s ]. Because of this the diode D2 will
be forward biased and begins conducting and hence the power starts flowing to the source. No matter the chopper is on or off the current I[0] will be flowing out of the load and is treated
negative . Since V[O ] is positive and the current I[0 ] is negative , the direction of power flow will be from load to source. The load voltage V[0 ] = (E+L .di/dt ) will be more than the
voltage V[s] so the type B chopper is also known as a step up chopper .
Type -C chopper or Two-quadrant type-A Chopper
Type C chopper is obtained by connecting type –A and type –B choppers in parallel. We will always get a positive output voltage V[0 ] as the freewheeling diode FD is present across the load. When
the chopper is on the freewheeling diode starts conducting and the output voltage v[0] will be equal to V[s ]. The direction of the load current i[0 ]will be reversed. The current i[0 ] will be
flowing towards the source and it will be positive regardless the chopper is on or the FD conducts. The load current will be negative if the chopper is or the diode D2 conducts. We can say the
chopper and FD operate together as type-A chopper in first quadrant. In the second quadrant, the chopper and D2 will operate together as type –B chopper.
The average voltage will be always positive but the average load current might be positive or negative. The power flow may be life the first quadrant operation ie from source to load or from load to
source like the second quadrant operation. The two choppers should not be turned on simultaneously as the combined action my cause a short circuit in supply lines. For regenerative braking and
motoring these type of chopper configuration is used.
Type D Chopper or Two-Quadrant Type –B Chopper
The circuit diagram of the type D chopper is shown in the above figure. When the two choppers are on the output voltage v[0 ]will be equal to V[s ] . When v[0 ]= – V[s ] the two choppers will be off
but both the diodes D1 and D2 will start conducting. V[0] the average output voltage will be positive when the choppers turn-on the time T[on ]will be more than the turn off time T[off ] its shown
in the wave form below. As the diodes and choppers conduct current only in one direction the direction of load current will be always positive.
The power flows from source to load as the average values of both v[0 ]and i[0 ]is positive. From the wave form it is seen that the average value of V[0 ] is positive thus the forth quadrant
operation of type D chopper is obtained.
From the wave forms the Average value of output voltage is given by
V0= (V[s] T[on]-V[s]T[off])/T = V[s].(T[on]-T[off])/T
Type –E chopper or the Fourth-Quadrant Chopper
Type E or the fourth quadrant chopper consists of four semiconductor switches and four diodes arranged in antiparallel. The 4 choppers are numbered according to which quadrant they belong. Their
operation will be in each quadrant and the corresponding chopper only be active in its quadrant.
During the first quadrant operation the chopper CH4 will be on . Chopper CH3 will be off and CH1 will be operated. AS the CH1 and CH4 is on the load voltage v[0] will be equal to the source voltage
V[s ] and the load current i[0 ] will begin to flow . v[0] and i[0 ]will be positive as the first quadrant operation is taking place. As soon as the chopper CH1 is turned off, the positive current
freewheels through CH4 and the diode D2 . The type E chopper acts as a step- down chopper in the first quadrant.
In this case the chopper CH2 will be operational and the other three are kept off. As CH2 is on negative current will starts flowing through the inductor L . CH2 ,E and D4. Energy is stored in the
inductor L as the chopper CH2 is on. When CH2 is off the current will be fed back to the source through the diodes D1 and D4. Here (E+L.di/dt) will be more than the source voltage V[s ] . In second
quadrant the chopper will act as a step-up chopper as the power is fed back from load to source
In third quadrant operation CH1 will be kept off , CH2 will be on and CH3 is operated. For this quadrant working the polarity of the load should be reversed. As the chopper CH3 is on, the load gets
connected to the source V[s] and v[0] and i[0] will be negative and the third quadrant operation will takes place. This chopper acts as a step-down chopper
CH4 will be operated and CH1, CH2 and CH3 will be off. When the chopper CH4 is turned on positive current starts to flow through CH4, D2 ,E and the inductor L will store energy. As the CH4 is turned
off the current is feedback to the source through the diodes D2 and D3 , the operation will be in fourth quadrant as the load voltage is negative but the load current is positive. The chopper acts as
a step up chopper as the power is fed back from load to source.
7 Responses to “Types of Chopper Circuits”
• EVERYTHING IS HERE FANTASTIC SITE THANKS
• Design the voltage commutated chopper circuit for operating the 42 volts battery powered electric car with starting current of 35 A
• Thanks.
• I need your permission to use the document for choppers, a use on my website. What we thank you in advance.
□ OK
• how are you?please,my problem is to know how trace the fault of tv or power supply.and to have a tv circuit
• thanks
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1. There Is No Function DecimalToRoman 2. No Overloading... | Chegg.com
1. there is no function decimalToRoman
2. No overloading of insertion and extraction operator.
3. No derived class extRomanType
4. No overloaded arithmetic operators for derived class
20. In Programming Exercise 1 Chapter 1, we defined a class romanType to implement Roman numerals in a program. In that exercise, we also implemented a function, romanToDecimal, to convert a Roman
numeral into its equivalent decimal number.
Modify the definition of the class romanType so that the data members are declared as protected. Use the class string to manipulate the strings. Furthermore, overload ther stream insertion and stream
extraction operators for easy input and output. The stream insertion operator outputs the Roman numeral in the Roman format.
Also, inclue a member function, decimalToRoman, that converts the decimal number( the decimal nmber must be a positive integer) to an equivalent Roman numeral format. Write the definition of the
member function decimalToRoman.
For simplicity, we assume that only the letter I can appear in front of another letter and that it appears only in front of the letters V and X. For example, 4 is represented as IV, 9 is represented
as IX, 39 is represented as XXXIX, etc.
b. Derive a class extRomanType from the class romanType to do the following. In the class extRomanType, overload the arithmetic operators +, -, *, and / so that arithmetic operations can be performed
on Roman numerals. Also, overload the pre- and postincrement and decrement operators as member functions of the class extRomanType.
To add(subtract, multiply, or divide) Roman numerals, add(subtract, multiply, or divide respectively) their decimal representations and then convert the result to Roman numeral format. For
subtraction, if the first number is smaller than the second number, output a message saying that, "Because the first number is smaller than the second, the numbers can not be subtracted". Similarly,
for division, the numerator must be larger than the denominator. Use similar conventions for the increment and decrement operators.
c. Write the definitions of the functions to overload the operators described in part b
d. Write a program to test your class extRomanType.
void romanType::convertNum( int& total )
const int SIZE = std::strlen(num);
int count[SIZE] = { 0 } ;
for( int i = 0 ; i < SIZE ; ++i )
switch( ::toupper( num[i] ) )
case 'M':
count[i] = 1000;
case 'D':
count[i] = 500;
case 'C':
count[i] = 100;
case 'L':
count[i] = 50;
case 'X':
count[i] = 10;
case 'V':
count[i] = 5;
case 'I':
count[i] = 1;
total = 0;
for (int i = 1 ; i < SIZE ; ++i )
if( count[i] > count[i-1] )
total -= count[i-1] ;
total += count[i-1] ;
total += count[ SIZE - 1 ] ;
Computer Science
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Romer 4th Down Study Underestimates Itself
study of when NFL offenses should go for it on 4th down rather than kick is possibly the most original, clever, and conclusive research of its kind. Romer's study concludes that offenses should go
for the 1st down far more often than they typically do.
Beyond all the economist jibber-jabber in the paper, it is devilishly simple and compelling. Based on the concept of expected points (the average points scored from a first down at each yard line),
Romer compares the average expected point value resulting from three options: punts, field goal attempts, and going for the 1st down. Ultimately, given the field position and to go distance, Romer
recommends when to choose each option.
But I think his conclusions are wrong. Or perhaps I should say they're even more right than previously thought.
To explain the research itself and why I think it misses the mark, I'll illustrate an example. Say a team faces a 4th down and 4 from its opponent's 40 yard line in the first quarter.
A team would typically punt in that situation. On average, punting from the 40 results in an opponent having field position at its own 12 yard line. The opponent would have about -0.2 expected points
in that situation. (+0.2 points for our team) That's the conventional decision.
Attempting a field goal wouldn't be wise. On average, a 57-yard FG attempt is only successful 36% of the time. You'd think the expected points of a made FG is obviously 3, but it isn't. It's actually
2.3 because the ensuing kickoff hands the ball to the opponent at, on average, their own 27. That situation has an expected point value of 0.7. (TDs are likewise worth only 6.3 points.) The other 64%
of the time, the unsuccessful FG would give the ball to the opponent at their own 47, with an expected value of 2.1 points. The resulting total expected value of an attempted FG is:
0.36 * 2.3 + 0.64 * (-2.1) = -0.5 exp pts
So far, punting is the better option from the 40. But what about going for the 1st down? Attempting the 1st down conversion is risky, but the payoff may outweigh the risk. With 4 yards to go, the
probability of conversion is 0.52. So 48% of the time, the offense would be unsuccessful and turn the ball over at (about) the 40. That would be 1.5 expected points for the opponents, or -1.5 points
to our team. But 52% of the time we'd get the first down yielding 2.5 expected points. The total expected value of going for the 1st down is:
0.52 * 2.5 + 0.48 * (-1.5) = 0.6 exp pts.
All things considered, going for it is worth +0.6 pts, a FG attempt is worth -0.5 pts, and punting is worth +0.2 pts. We can conclude that going for the 1st down on 4th and 4 from the 40 is usually
the right decision.
The only problem is that going for it really isn't worth +0.6 expected points. It's actually worth more. I think Romer underestimates the value of the "go for it" strategy.
Remember that the entire expected point curve is based on an empirical observation of actual NFL drives. As a rule, those drives were characterized by the "kick on anything longer than 4th and
inches" strategy.
If a team actually employed Romer's recommendations, their expected point curve would steepen, resulting in ever more favorable values for 1st downs.
In my example above, a team shouldn't go for it on the 40 only to revert to conventional tactics at the 30 or 20 or 10 yard line. A "Romer offense's" expected point value at the 40 would not really
be 2.5, but something even higher.
In other words, it's not a static process. The Romer strategy is a dynamic, recursive process that redefines the expected point curve itself. Brilliant as it is, the Romer paper underestimates its
own implications. The paper's recommendations are a departure from the conservative style of football currently played in the NFL. And if I'm right, it would mean the optimum strategy should be even
more radical.
14 Responses to “Romer 4th Down Study Underestimates Itself”
Anonymous says:
Very clever.
Steve H says:
You're right, but you're also opening a can of worms. Romer is not trying to define the proper strategy, but to show that coaches are afraid to deviate from the conventional wisdom. So if
nothing else changes, going for it on 4th down tends to be a better strategy, also known as a partial equilibrium analysis. If coaches started going for it on 4th down, as per the
recommendations, defenses might also change and that's your can of worms. For example, defenses might blitz even more on 3rd down in order to make the 4th down attempt tougher. It's next
to impossible to extrapolate the partial equilibrium to a general equilibrium model.
BTW, where did the numbers on the diagram come from?
Brian Burke says:
True. Offenses might run more often too knowing 4th down was available. It's not too hard to get 10 yds in 4 tries running the ball. Defenses would have to respond to that as well. The
resulting equilibrium wouldn't resemble what we see today. It could be the Anti-CFL. Lots of running, very little punting, maybe less short passing and more bombs. Who knows?
Re the graph: I traced an approximate actual expected point curve from 1st and 3rd quarters of all regular season games 2000-2008. The "Romer offense" curve is just an illustration.
Anonymous says:
You're assuming that the difference is one-sided. If the opposing team is also playing aggressively on 4th down, then the difference in field position will also be giving more points when
the 4th down play fails, since the difference in expected points with an aggressive strategy would, I expect, be more pronounced starting on the 40 vs. the 12, since you can't really play
that much more aggressively on 4th deep in your own end.
Brian Burke says:
The "aggressiveness" of the strategies is in the decision to kick or not. The defense doesn't get a vote.
But you're right in that the defense could sell out on a blitz or something. But they would be making a big mistake if they did that.
We can safely say that the offense's chance of converting on 3rd down is definitely *no worse* than they had on 3rd down and the same to go distance. All they need to do is choose the
same type of plays they would have on 3rd down. If the defense does something different than they would normally would have on 3rd, that only helps the offense.
Assuming the defense's strategy is rational on 3rd down, if they alter their strategy equilibrium on 4th down they would be choosing a sub-optimum decision. The result would be an even
higher probability of conversion for the offense.
In game theory, the 4th down decision is considered a "2-player zero-sum game with perfect information." It's perfect information because the defense knows whether or not the offense is
going to kick or go for it.
Anonymous says:
unless of course it's a fake.
Steve H says:
Actually, Romer already beat you to your point. I was reading the 2005 version (I had only read the older version) and I don't remember reading this before:
Partial Equilibrium versus General Equilibrium.
The analysis looks at decisions on individual fourth downs, taking all other decisions as given. But these decisions could affect other choices. If both teams follow the recommendations
of the dynamic-programming analysis, their offenses will do better on average. This suggests that the value of having the ball anywhere on the field will be greater than the
partial-equilibrium estimates imply. In this case, the benefits of going for it relative to kicking will be even larger than the preceding analysis suggests, particularly when teams are
far from their opponents’ goal line. If only one team is more aggressive on fourth downs, on the other hand, it will on average score more points; this implies that any situation
(including ones where its opponent has the ball) will be more valuable to it. There is no evident reason for this to have an important effect on optimal choices. It appears likely,
however, that the increase in value will be larger in situations where the team is more likely to face a fourth down soon, such as when it has the ball in near its own goal line. If this
is correct, it implies that the analysis tends to understate the value of going for it near one’s goal line.
Anonymous says:
I'd point out that the team to team variability on this analysis would be huge, and negate any "rule of thumb" to go for it.
By definition of an average success rate of 52% for 4th and 4, half the teams are below that.(if i may assume a mean is a median, but you get my point)
Case in point, on thanksgiving day, the lions should not go for it against the titans (the lions should probably punt on third down. haha.)
On the other hand, i bet the titans could always go for 4th down against the lions.
Brian Burke says:
Steve-Thanks. I should have known!
Bob-I don't think it would negate the rule of thumb, but it would definitely need to be tailored to a team's strengths and weaknesses.
Regarding weak vs strong teams, the opposite may be true. It may be the weaker teams that need this strategy the most. The strategy would increase the variance of points scored and
allowed, which is exactly what a weaker opponent wants. It might be more like baseball, with larger game-to-game variance, which is what allows the Royals to somewhat frequently beat the
Red Sox.
Anonymous says:
"Regarding weak vs strong teams, the opposite may be true. It may be the weaker teams that need this strategy the most. "
according to the rule of thumb, yes weaker teams need it the most - but that could be wrong.
We can figure out the %success requ'd to make it a 'good gamble'.
x * 2.5 - (1-x) * (1.5) = -0.5 exp pts.
or x = 0.25%
so if a team has a worse than 25% chance then they should kick. We can take this number, to see what the max distance for 4th down can be, in order to go4it and kick to give the same
expected points.
Brian Burke says:
Romer does that in his paper. There's a nice graph at the back that shows recommended maximum to go distances on 4th downs.
I agree with this analysis, except when there is a mismatch in team strength. At some point in a game, a weaker team that is trailing is going to need some luck to win. So even when the
percentages say 'punt,' a weaker team should roll the dice and go for it if they want a realistic chance of winning.
Anonymous says:
"Assuming the defense's strategy is rational on 3rd down, if they alter their strategy equilibrium on 4th down they would be choosing a sub-optimum decision. The result would be an even
higher probability of conversion for the offense."
False. This is because the offense shouldn't run the same types of plays on 3rd & 4 as it would on 4th & 4 - for the very reasons set out here. The expected cost of failure of, say, a run
play to get four yards isn't spectacularly high. If you get 3 1/2, you can go for it on 4th. Or punt. On 4th down, though, the cost to the offense of a gain of less than four is much
higher, so they should pick a play with a higher expected gain, meaning that the optimal defensive strategy is likely different.
Brian Burke says:
Ok. I see your point, but I don't know if that's going to make any difference. I've never heard of a coach making 3rd down play calls so that he can just get closer for a 4th down
Assuming the goal is the same--to get the first down--then nothing should change. If your strategy mix is exactly the same as it would be for 3rd down, you are guaranteed at least the
same rate of success.
Z-Dog says:
@Brian Burke
It's not that coaches call a play on 3rd down just to get them closer on 4th down. It's that they'll call plays that are higher-risk on 3rd down, knowing that they still have a 4th down
to get the conversion. This is particularly true in 3rd down passing situations.
Think of it this way: the probability of converting 3rd & 10 followed by 4th and 10 is less than the probability of converting 3rd & 10 followed by 4th & 2. So on 3rd & 10, you might call
something (say, a draw play) that has a low chance (but not chance) of getting 10 yards, but has a very high chance of getting you 6-8 yards, rather than a play that has an ok chance of
getting you 10+ yards (a pass play) but also a very good chance of getting you zero or negative yards.
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and numbers
A look at the law and numbers, Metcalf, Reed, Dunbar, and others
In order to understand the value and power of networks, such as a community, a number of individuals have come up with equations and analysis. It’s worth a bit of time to understand what they are
talking about and how it may be useful to a CM.
Metcalf’s Law: Holds that the number of connections in a group grows very quickly as the number of members in the group increases. A group of two can have one connection, 3 members has 3, but 4
members has 6, 5 has 10, and so on.
• For a CM, this shows how quickly opportunities for relationships increase as the group size gets bigger.
Reed’s Law: Reed adds the idea of a sub-group, or connections to a set of multiple people. Thus, a member may have a connection with a sub-groups made up of 2, 3, 4, 5, or more members. When this
notion is taken up with Metcalf’s the number of possible connections grows exponentially. A group of two can have one connection, 3 members has 4, 4 members has 11, 5 has 26, and so on.
• For a CM this means a member of a large group with subgroup ability has the benefit of many more possible connections than a member without subgroup ability.
• A community that allows subgroups increases their possible interconnectedness exponentially.
• It also means that a small increase in group size may quickly overwhelm anyone trying to keep a handle on various connections.
Dunbar’s Number: Is the number of individuals a person can maintain a relationship with. So, although Reed’s Law shows how quickly a person’s possible connections grow in a group, there may be a
limit as to the number of connections a person can actually maintain.
Around 150 people is what Dunbar estimates a person may be able to maintain at any one time. This would include both online and offline. While others have come up with higher numbers, for a CM it
means there is a limit.
• Since there is a limit, it is best to put effort into making those connections of the highest value.
• At only 8 members, Reed’s law put it at 242 possible connections. That may help explain the number of subgroups that don’t gain traction. Even with very few members, it is possible to have more
connections than people can work with.
Perfect group size: Others have tackled this problem from a different angles and while there is no definitive number, or law, one could go with the number 6 as a rule of thumb. Part of this depends
on the task and individuals.
• When looking at creating subgroups, especially ones oriented toward tasks, it may be that these sort of smaller numbers are better.
How these laws and numbers work within your communities context will probably differ. This is even more true when you look at how active your members are, or not. My big take-away was just how
quickly the opportunities for connections scale up, and can quickly overwhelm what an individual can use.
Much of this was taken from my exploration in and around Reed’s Law in Wikipedia.
This entry was posted in Collaboration / Community. Bookmark the permalink.
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Gravitational Flexion
Gravitational Flexion (pronounced "flecks-shun" for the uninitiated, and yes, it is a real word) is a new technique for measuring 2nd order gravitational lensing signals in background galaxies and
radio lobes.
Unlike shear, flexion directly probes variations of the potential field. Moreover, the information contained in flexion is orthogonal to what is found in the shear. Thus, we get the information "for
Some Recent Results
Shear map of A1689. Strong+Weak Reconstruction of A1689 (Using the Bradac et al. reconstruction method)
Parametric density reconstruction using flexion. Contour map of the same reconstruction.
Flexion Papers
• The Galaxy Octopole Moment as a Probe of Weak Lensing Shear Fields, DM Goldberg & P Natarajan, 2002, Astrophys. J., 564, 65., astro-ph/0107187
This is a paper I wrote with Priya Natarajan, while working for her at Yale. In it, we lay out the basic formalism for using 2nd order perturbation theory to study gravitational lensing. The math
is a bit messy, I'm afraid.
• Galaxy-Galaxy Flexion: Weak Lensing to Second Order, DM Goldberg & DJ Bacon, 2005, Astrophys. J. 619, 741, astro-ph/0406376
This paper was written with David Bacon of the Royal Observatory, Edinburgh. In it, we use the shapelets formalism developed by Alexandre Refregier to flesh out the shapelets technique. We also
show its viability by measuring the galaxy-galaxy flexion signal from the Deep Lens Survey.
• Weak Gravitational Flexion, DJ Bacon, DM Goldberg, BTP Rowe, and AN Taylor, 2006, MNRAS, 365, 414 astro-ph/0504478
This is a theoretical paper which explores many potential future avenues of flexion analysis: measurements in clusters, galaxy-galaxy flexion, cosmic flexion, and so on. Plus, Andy Taylor
introduces a really nice complex space formalism that not only works for flexion, but for shear as well. You'll love it!
• A New Measure for Weak Gravitational FlexionY Okura, K Umetsu & T Futamase, astro-ph/0607288
This paper does a good job showing that flexion is expressible as octopole moments (HOLICs). We have an implementation of the basic technique available in the flexion package above.
• Measuring Flexion, D Goldberg & A Leonard, astro-ph/0607602
We introduce a KSB-type inversion technique for the HOLICs approach, and do an observational comparison between shapelets and HOLICs.
• Weak Gravitational shear and flexion with polar shapelets, R Massey, B Rowe, A Refregier, DJ Bacon & J Berge. astro-ph/0609795
Richard Massey and the gang derive flexion and shear relations with a new set of polar basis functions.
• Gravitational Shear, Flexion, and Strong Lensing in Abell 1689, A Leonard, DM Goldberg, J Haaga & R Massey astro-ph/0702242
Adrienne led this effort to combine non-parametric reconstructions using shear (shapelets), with multiple images (strong lensing), as well as parametric reconstructions using flexion in the rich
cluster A1689.
• Analytic models of plausible gravitational lens potentials, EA Baltz, P Marshall, & M Oguri, astro-ph/0705.0682
This paper addresses analytic estimates of flexion in elliptical potentials.
• Weak lensing goes bananas: What flexion really measures, P Schneider & Xinzhong Er, astro-ph/0709.1003
Schneider and Er address the tricky question of reduced flexion as the true observable.
• A Method for Weak Lensing Flexion Analysis by the HOLICs Moment Approach, Y. Okura, K. Umetsu, & T Futamase, arxiv/0710.2262
Okura et al. further refine the flexion analysis using HOLICs, including development of formalism to deal with anisotropic shear. They also produce a remarkable analysis of A1689. Using Flexion
only, they identify to central density peaks which clearly correspond to peaks in galaxy counts. Remember, folks, this is on a scale of only 1 arcminute!
Fig. 9 from Okura et al. (2007)
Flexion Code
I know what you're thinking: "Looking at the flexion papers, there's no way that I can code all of that!" Well, you're in luck, I have some idl code (which, presumably, can be translated into C or
fortran or whatever with a judicious use of search and replace), which calculates flexion:
• flexion.tar.gz (7/26/06) -- a tarball containing the basic flexion codes. This assumes you have the "shapelets" code as well. This also has a little code called "test_flexion.pro" which will show
you how flexion is called, as well as psf_test.pro, which does a similar thing for the PSF inversion. In addition, we have a separate code "flexion_moments.pro", which calculates the flexion
using octopole moments using an adaptation from Okura et al. (2006), and our adaptation (Goldberg & Leonard, 2006). There are also a bunch of subsidiary codes. Please let me know if you have any
• flexion_minimal.tar.gz (2/28/06) -- a tarball containing the minimum shapelets codes necessary to run the sample codes above. I swear, this really works with idl v5.6.
• Richard Massey's Shapelets Page -- Go here to get the latest shapelets idl code. A warning, the code is not always backwards compatible, so you may need to do a bit of tweaking to get it to work
with the examples above. Please let me know if you are having any trouble.
Flexion People
Drexel folks
Dave Goldberg -- which is to say, me. I'll be happy to answer any and all of your shapelets questions.
Sanghamitra Deb -- Sanghamitra is a graduate student working on combining strong and weak lensing techniques in novel ways.
Jason Haaga -- Jason is an undergraduate co-op student who worked with us on the reconstruction of A1689.
Cambridge folks
Adrienne Leonard -- Adrienne did her master's thesis with me at Drexel, and is now at Cambridge doing her Ph.D. with Lindsay King. She led the effort to measure shear and flexion in A1689.
Edinburgh folks
David Bacon -- David is a PPARC Fellow at the Royal Observatory. He and I worked out the basic flexion formalism and made the first real measurements.
Barnaby Rowe -- Barney is a graduate student working with David on measuring flexion in the GEMS Survey.
Andy Taylor -- As mentioned above, Andy came up with a very nice way of expressing shear, first and second flexion using complex notation.
Caltech folks
Richard Massey -- Richard is the author of some very fine shapelets codes. He is also working on measuring flexion.
Related Work
John Irwin & Marina Shmakova have done related work measuring the Sextupole moments. What we term "Flexion" is referred to in their work as displacements (1st flexion) and cardioids (2nd flexion).
They have a novel technique of correlating these terms with the shear as a means of detecting the signal, and have done so in the Hubble Ultra-Deep Field with co-author Jay Anderson.
Yuki Okura, Keiichi Umetsu, and Toshifumi Futamase have developed a very intuitive approach to measuring flexion -- direct measurement of the octopole moments. They have done the full inversion of
these moments (which we did not do in Goldberg & Natarajan). They refer to their method as Higher Order Lensing Image's Characteristics or HOLICs, and it may hold the key to fast estimate of flexion
in large fields.
Indeed, they have applied there technique to A1689 (see above), and have done a remarkable job on direct detection of substructure within the cluster using flexion alone.
Edward Baltz, Phil Marshall, and Msaamune Oguri explored analytic models of flexion for elliptical potentials.
Peter Schneider and his student, Xinzhong Er have recently addressed a very thorny issue -- the reduced flexion. As with any lensing system, the reconstructed flexion field actually corresponds to a
single solution among an entire degenerate class of solutions. This is the well-known "mass-sheet degeneracy." Schneider and Er relate the observed 3rd moments of images to the reduced flexion and
show how these can be used to reconstruct a density field.
This page is written, maintained, and edited by Dave Goldberg. Last edited 9/18/07.
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Multiplicative Number Theory, Second Edition, revised by H.L
Results 1 - 10 of 26
- IEEE Transactions on Information Theory , 1992
"... A new technique, based on the pseudo-random properties of certain graphs, known as expanders, is used to obtain new simple explicit constructions of asymptotically good codes. In one of the
constructions, the expanders are used to enhance Justesen codes by replicating, shuffling and then regrouping ..."
Cited by 117 (24 self)
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A new technique, based on the pseudo-random properties of certain graphs, known as expanders, is used to obtain new simple explicit constructions of asymptotically good codes. In one of the
constructions, the expanders are used to enhance Justesen codes by replicating, shuffling and then regrouping the code coordinates. For any fixed (small) rate, and for sufficiently large alphabet,
the codes thus obtained lie above the Zyablov bound. Using these codes as outer codes in a concatenated scheme, a second asymptotic good construction is obtained which applies to small alphabets
(say, GF (2)) as well. Although these concatenated codes lie below Zyablov bound, they are still superior to previously-known explicit constructions in the zero-rate neighborhood.
, 2006
"... For every ɛ > 0 and every integer m > 0, we construct explicitly graphs with O(m/ɛ) vertices and maximum degree O(1/ɛ²), such that after removing any (1 − ɛ) portion of their vertices or edges,
the remaining graph still contains a path of length m. This settles a problem of Rosenberg, which was moti ..."
Cited by 107 (14 self)
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For every ɛ > 0 and every integer m > 0, we construct explicitly graphs with O(m/ɛ) vertices and maximum degree O(1/ɛ²), such that after removing any (1 − ɛ) portion of their vertices or edges, the
remaining graph still contains a path of length m. This settles a problem of Rosenberg, which was motivated by the study of fault tolerant linear arrays.
, 2003
"... The problem of quickly determining whether a given large integer is prime or composite has been of interest for centuries, if not longer. The past 30 years has seen a great deal of progress,
leading up to the recent deterministic, polynomial-time algorithm of Agrawal, Kayal, and Saxena [2]. This new ..."
Cited by 18 (0 self)
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The problem of quickly determining whether a given large integer is prime or composite has been of interest for centuries, if not longer. The past 30 years has seen a great deal of progress, leading
up to the recent deterministic, polynomial-time algorithm of Agrawal, Kayal, and Saxena [2]. This new “AKS test ” for the primality of n involves verifying the
, 1993
"... For odd square-free n > 1 the cyclotomic polynomial n (x) satises the identity of Gauss 4 n (x) = A 2 n ( 1) (n 1)=2 nB 2 n : A similar identity of Aurifeuille, Le Lasseur and Lucas is n (( 1)
(n 1)=2 x) = C 2 n nxD 2 n or, in the case that n is even and square-free, n=2 ( x 2 ) ..."
Cited by 14 (5 self)
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For odd square-free n > 1 the cyclotomic polynomial n (x) satises the identity of Gauss 4 n (x) = A 2 n ( 1) (n 1)=2 nB 2 n : A similar identity of Aurifeuille, Le Lasseur and Lucas is n (( 1) (n 1)=
2 x) = C 2 n nxD 2 n or, in the case that n is even and square-free, n=2 ( x 2 ) = C 2 n nxD 2 n ; Here A n (x); : : : ; D n (x) are polynomials with integer coecients. We show how these coef- cients
can be computed by simple algorithms which require O(n 2 ) arithmetic operations and work over the integers. We also give explicit formulae and generating functions for A n (x); : : : ; D n (x), and
illustrate the application to integer factorization with some numerical examples.
- AMER. MATH. SOC , 1985
"... ..."
- J. Number Theory
"... Abstract. This paper studies the non-holomorphic Eisenstein series E(z, s) for the modular surface PSL(2, Z)\H, and shows that integration with respect to certain non-negative measures µ(z)
gives meromorphic functions Fµ(s) that have all their zeros on the line ℜ(s) = 1 2. For the constant term a0( ..."
Cited by 8 (2 self)
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Abstract. This paper studies the non-holomorphic Eisenstein series E(z, s) for the modular surface PSL(2, Z)\H, and shows that integration with respect to certain non-negative measures µ(z) gives
meromorphic functions Fµ(s) that have all their zeros on the line ℜ(s) = 1 2. For the constant term a0(y, s) of the Eisenstein series the Riemann hypothesis holds for all values y ≥ 1, with at most
two exceptional real zeros, which occur exactly for those y> 4πe −γ = 7.0555+. The Riemann hypothesis holds for all truncation integrals with truncation parameter T ≥ 1. At the value T = 1 this
proves the Riemann hypothesis for a zeta function Z2,Q(s) recently introduced by Lin Weng, associated to rank 2 semistable lattices over Q. 1.
"... Abstract. Let pn denote the nth prime. Goldston, Pintz, and Yıldırım recently proved that (pn+1 − pn) lim inf =0. n→ ∞ log pn We give an alternative proof of this result. We also prove some
corresponding results for numbers with two prime factors. Let qn denote the nth number that is a product of ex ..."
Cited by 8 (2 self)
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Abstract. Let pn denote the nth prime. Goldston, Pintz, and Yıldırım recently proved that (pn+1 − pn) lim inf =0. n→ ∞ log pn We give an alternative proof of this result. We also prove some
corresponding results for numbers with two prime factors. Let qn denote the nth number that is a product of exactly two distinct primes. We prove that lim inf n→ ∞ (qn+1 − qn) ≤ 26. If an appropriate
generalization of the Elliott-Halberstam Conjecture is true, then the above bound can be improved to 6. 1.
- IEEE Trans. Inform. Theory , 1997
"... We study codes over GF (q) that can correct t channel errors assuming the error values are known. This is a counterpart to the well-known problem of erasure correction, where error values are
found assuming the locations are known. The correction capabilities of these so called t-location correcting ..."
Cited by 7 (1 self)
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We study codes over GF (q) that can correct t channel errors assuming the error values are known. This is a counterpart to the well-known problem of erasure correction, where error values are found
assuming the locations are known. The correction capabilities of these so called t-location correcting codes (t-LCCs) are characterized by a new metric, the decomposability distance, which plays a
role analogous to that of the Hamming metric in conventional error-correcting codes (ECCs). Based on the new metric, we present bounds on the parameters of t- LCCs that are counterparts to the
classical Singleton, sphere packing and GilbertVarshamov bounds for ECCs. In particular, we show examples of perfect LCCs, and we study optimal (MDS-like) LCCs that attain the Singleton-type bound on
the redundancy. We show that these optimal codes are generally much shorter than their erasure (or conventional ECC) analogues: The length n of any t-LCC that attains the Singleton-type bound for t ?
1 is bounde...
- Annales Inst. Fourier
"... This paper studies a two-variable zeta function ZK(w,s) attached to an algebraic number field K, introduced by van der Geer and Schoof [9], which is based on an analogue of the Riemann-Roch
theorem for number fields using Arakelov divisors. We term it the Arakelov zeta function. When w = 1 this func ..."
Cited by 6 (0 self)
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This paper studies a two-variable zeta function ZK(w,s) attached to an algebraic number field K, introduced by van der Geer and Schoof [9], which is based on an analogue of the Riemann-Roch theorem
for number fields using Arakelov divisors. We term it the Arakelov zeta function. When w = 1 this function becomes the completed Dedekind zeta ˆ ζK(s) function of the field K. The function is an
meromorphic function of two complex variables with polar divisor s(w − s), and it satisfies the functional equation ZK(w,s) = ZK(w,w − s). We consider the special case K = Q, where for w = 1 this
function is ˆ s ζ(s) = π 2Γ ( s
- J. Combinatorial Theory, Ser. A
"... Let C denote the complex field. A vector v in the tensor product ⊗ m i=1 Cki is called a pure product vector if it is a vector of the form v1 ⊗ v2 · · · ⊗ vm, with vi ∈ C ki. A set F of pure
product vectors is called an unextendible product basis if F consists of orthogonal nonzero vectors, and t ..."
Cited by 6 (0 self)
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Let C denote the complex field. A vector v in the tensor product ⊗ m i=1 Cki is called a pure product vector if it is a vector of the form v1 ⊗ v2 · · · ⊗ vm, with vi ∈ C ki. A set F of pure product
vectors is called an unextendible product basis if F consists of orthogonal nonzero vectors, and there is no nonzero pure product vector in ⊗ m i=1 Cki which is orthogonal to all members of F. The
construction of such sets of small cardinality is motivated by a problem in quantum information theory. Here it is shown that the minimum possible cardinality of such a set F is precisely 1 + � m i=1
(ki − 1) for every sequence of integers k1, k2,..., km ≥ 2 unless either (i) m = 2 and 2 ∈ {k1, k2} or (ii)1+ � m i=1 (ki −1) is odd and at least one ki is even. In each of these two cases, the
minimum cardinality of the corresponding F is strictly bigger than 1 + � m i=1 (ki − 1). 1
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Can Climate Projections Models Be Applied to Baseball Projections ?
In Nate Silver’s new book, The Signal and the Noise, he had a chapter on climate projection. The chapter showed projections had a sweet spot at sometime in the future where they were the most
accurate. Three uncertainty factors were at work affecting the projection and the sweet spot.
• The first is is initial variability. With climate, a location may experience a very cold winter in the first year of the model. The extra cold and hot winters will hopefully even out at some
point. This error starts out high and eventually goes to zero.
• The second error is long term unknowns. With climate, maybe a new scientific invention is created which removes CO2 from the air or a couple of volcanoes go off at once cooling the earth. This
value starts at 0 and grows steadily over time.
• The third is an underlying unpredictability. This value is the most steady of the three. Say the climatologist want create a 40 year prediction. From the beginning they will have a certain level
of uncertainty. The further they want to predict into the future, the higher level of underlying uncertainty exists.
The sum of the 3 factors is the total uncertainty in the projection. The total error amount will be the lowest at the point in the future where the projection is aimed to be the the most correct.
Here is a sample graph:
The total error is lowest 40 years into the future.
This same idea can be applied to baseball projections. Right now, projections only look at how a player will do next year and have the least amount of error set for a single season’s worth of data.
The one year projection is good, but not in all instances.
Would it be nice to have a 2, 3, 4, or more year projection when looking at free agents? By the time a player has reached free agency, there would have enough information on the player to see what
is expected from them over a set number of years. This projection would have high initial level on underlying variability, but by limiting the unknowns curve it has lowest level of error for the
years in question. The error graph would look like this with the lowest amount of error at time 5 or 6:
Now, how about a pitcher’s projection once it is known they are throwing 2 MPH slower?Something like a 30 game (6 start) projection could be created which would take into account all the known
factors of the pitcher right now. It would be really accurate for a short time, but would have more and more error over the time compared to the long term projection. Here is an example:
Ideally, in my opinion, I think would be nice to have a 30 game, 160 game, 1 year, 2 year, 3 year, 4 year and 5 year projections for all players all the time. The 30 and 160 game projections will be
a little tough to pull off, but multiple year projections should be available now. Stay tuned.
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URGENT HELP WITH some problems. :x
September 2nd 2008, 12:58 PM #1
Junior Member
Aug 2008
Okay it starts with how to use algebraic tests to check for symmetry with respect to both axes and the origin.
13. y = -4x+1
14. y= 5x-6
15. y=5-x^squared
16.y=x^squared - 10
17. y=x cubed + 3
18. y= -6 -x cubed
19. y = square root of x +5
20. y=|x| +9
I know theres this formula for this its (x-h) squared + (y-k) squared = r squared but
How to find the center of a radius of a circle when they give u just this
21. x squared + y squared = 9
22. x squared + y squared = 4
23. (x+2)squared + y squared = 16
24.x squared + (y-8)squared = 81
25. (x - 1/2) squared + (y + 1)squared = 36
26. (x+4) squared + (y - 3/2)squared = 100
Whoever answers them all first gets thanks IF they helped me understand how to do them correctly
much love
If a function is even (i.e. symmetric about the y-axis), then: $f(x) = f(-x)$, i.e. if you plug in -x to your function, you'll get the same function if you plugged in x.
If a function is odd (i.e. symmetric about the origin), then: $f(-x) = -f(x)$, i.e. if you plug in -x to your function, then you'll get the negative of f(x).
Show us your work and we'll point out any errors.
As for the circle, given $(x-a)^2 + (y-b)^2 = r^2$, the centre is given by $(a,b)$ (note the minus signs in front of them in the equation). For example, $(x - 5)^2 + (y + 2)^2 = 9$ has centre $
(5, {\color{red}-}2)$
If a function is even (i.e. symmetric about the y-axis), then: $f(x) = f(-x)$, i.e. if you plug in -x to your function, you'll get the same function if you plugged in x.
If a function is odd (i.e. symmetric about the origin), then: $f(-x) = -f(x)$, i.e. if you plug in -x to your function, then you'll get the negative of f(x).
Show us your work and we'll point out any errors.
As for the circle, given $(x-a)^2 + (y-b)^2 = r^2$, the centre is given by $(a,b)$ (note the minus signs in front of them in the equation). For example, $(x - 5)^2 + (y + 2)^2 = 9$ has centre $
(5, {\color{red}-}2)$
Im confused by your first explanation but here if i have say y=-4x + 1
is this function negative or odd? i don't understand.. do i have to plug in (1) to x and see because that is what i understand from your explanations..
or would the function y=5x - 6 be positive because its slope is 5x and its starting point or b is -6?
I am confused in that aspect and if i had something like
y=x cubed + 3 to me this would be positive or am i not correct? if it positive then for me to find out if its symmetric about the origin how would i know this? and if its odd its symmetric about
the origin is that what your saying because i get that if its odd its symmetric by the origin and if its even it is symmetric about the y axis but what i also don't understand on something like y
=-4x +1 is how i found out what number is odd do i plug in 1 to find out what just y equals and determine if its symmetric about the y axis or origin? So confuseed.
Or do i determine if the slopes odd / or the starting point?
For the 2nd part I understand that something like x squared + y squared = 9
is (x-1)squared + (y - 1) squared = 9 or 3 squared so is that how that is graphed at (-1,-1) and goes 3 to the sides to form the circle on a graph?
and for something like 23. (x+2)squared + y squared = 16 how would i take out (x+2) squared to make it work with the formula square root (x+2) and Square root 16 to get 4 and then subtract 2 and
have x + y squared = 2 Sorry if i confused you there. but im completely confused ha.
September 2nd 2008, 01:11 PM #2
September 2nd 2008, 03:33 PM #3
Junior Member
Aug 2008
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DOCUMENTA MATHEMATICA, Vol. 2 (1997), 31-46
DOCUMENTA MATHEMATICA
, Vol. 2 (1997), 31-46
Amnon Besser
On the Finiteness of $\Sha$ for Motives Associated to Modular Forms
Let $f$ be a modular form of even weight on $\Gamma_0(N)$ with associated motive $\mathcal{M}_f$. Let $K$ be a quadratic imaginary field satisfying certain standard conditions. We improve a result of
\Nekovar{} and prove that if a rational prime $p$ is outside a finite set of primes depending only on the form $f$, and if the image of the Heegner cycle associated with $K$ in the $p$-adic
intermediate Jacobian of $\mathcal{M}_f$ is not divisible by $p$, then the $p$-part of the Tate-\shafarevic{} group of $\mathcal{M}_f$ over $K$ is trivial. An important ingredient of this work is an
analysis of the behavior of ``Kolyvagin test classes'' at primes dividing the level $N$. In addition, certain complications, due to the possibility of $f$ having a Galois conjugate self-twist, have
to be dealt with.
1991 Mathematics Subject Classification: 11G18, 11F66, 11R34, 14C15.
Full text: dvi.gz 33 k, dvi 83 k, ps.gz 97 k.
Home Page of DOCUMENTA MATHEMATICA
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Does primitive (resp. to comultiplication) homology classes comes from Hurewicz map?
up vote 2 down vote favorite
I heard this statement for loop spaces, but can't find proof for that or counter-example for following question. consider ordinary rational homologies, there are primitive elements with respect to
coalgebra structure, it is almost obvious that homology classes released by spheres (with rational coefficient) are such elements, is it all?
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1 Answer
active oldest votes
Yes. This is classical, maybe originally in Milnor and Moore's paper on Hopf algebras. For a recent exposition see for example "More concise algebraic topology" by Kate Ponto and
up vote 10 myself. If $X$ is a connected $H$-space (say with finitely generated rational homology groups), then the rationalized Hurewicz homomorphism is a monomorphism with image the primitive
down vote elements of $H_*(X;\mathbf Q)$. The essential point is that the rationalization of $X$ is equivalent to a product of Eilenberg-MacLane spaces.
Thank you. What for spaces without h-group structure? Is a Hurewicz map to primitive elements surjective? – Bad English Mar 25 '13 at 19:32
Don't have time to think about a counterexample, but it is clear from how Sullivan rational homotopy theory works that they are plentiful. That theory gives a rational DGA A(X) for a
nilpotent (or simply connected in the original) space X such that the cohomology of A(X) is the rational cohomology of X and the indecomposables of A(X) are dual to the rationalized
homotopy groups of X. The cohomology of A(X) can have indecomposables that are decomposable in A(X). No reason the Hurewicz homomorphism should hit their duals. – Peter May Mar 27 '13
at 2:18
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Not the answer you're looking for? Browse other questions tagged at.algebraic-topology or ask your own question.
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Posts by
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I'm not sure exactly what you are asking. In psychology, positive control (or at least an attempt at it) involves incentives and rewards. Negative aspects involve punishment or lack/removal of
reward. If this is what you are seeking, you should be able to think of examples.
You got them all correct.
Multiply first equation by 2. Assuming that y was accidentally eliminated in the second equation. 4x - 10y = 6 -4x + 10y = -6 Adding the two together: 0 = 0. One equation is just the negative of the
other. In other words, if you multiply one equation by -1, you get the other.
Statistics (?)
What is your question?
Science (?)
What type of wave?
4/5 + 4/5 = ?
3/5 * 10 = ?
1/2 * 1/3 = ?
3 * 4 = ?
statistics (?)
What is your question? If you want to know Kelsey's score: Z = (score-mean)/SD -1.85 = (score-72)/4
(100*10)/(7*60) = ?
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
(1) Let x = standard shipments and y = express. y = 108 - x 3.5x + 7.5y = 506 Substitute 108-x for y in second equation and solve for x. Insert that value into the first equation and solve for y.
Check by inserting both values into the second equation. (2) Use similar process ...
I assume you only need to determine the sex chromosomes. The daughter has XX sex chromosomes. Since she has hemophilia, a recessive gene on the X chromosome, she must have inherited the hemophilia
from both parents. However, since the son also receives an X chromosome from mom...
See your previous post.
62,400/(100*#workdays in a year) = ?
You might want to round it to the nearest person. It rounds to 17, but if you want all the records to be covered, you might round it up to 18. This would give the FTEs a little slack.
8600/(100*5) = ?
The girls all receive normal X chromosome from dad, meaning they cannot be more than carriers. If males get recessive gene, they will have hemophilia since there is no dominant gene on the Y
chromosome to supress the hemophilia gene. How do you get six variations? XCXC, XcY XC...
Online, * is used to indicate multiplication to avoid confusion with x as an unknown. Assuming the comma has the same meaning as a period in the USA: 2.3 * -5 = -11.5
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 13/52 * 12/51 * 11/50 * 10/49 * (13*3)/48 = ?
Statistics 2
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
1600 + 17600*.081 + (29700-17600)*.09 = ?
algebra 1
With answer from B, Mean(for median) = (B + ?)/2 = 227
algebra 1
A. mean = ∑x/n = (225+245+222+230+x)/5 = 221 Solve for x. B. Arrange 5 score in order of value. Middle score is the median. C. With even number of scores, mean of the two middle scores = median
Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
Math.stat help :(
Ho: mean1 = mean2 Ha: mean1 ≠ mean2 Since these are just samples rather than the whole population, µ is an inappropriate designation. µ is parameter rather than a statistic. Z = (mean1 - mean2)/
standard error (SE) of difference between means SEdiff = √(...
It is the Z test (google it). Where is your confusion centered?
Z = (mean1 - mean2)/standard error (SE) of difference between means SEdiff = √(SEmean1^2 + SEmean2^2) SEm = SD/√n (n = # of trips) If only one SD is provided, you can use just that to determine
SEdiff. Find table in the back of your statistics text labeled somethin...
Health (Ms. Sue)
1. They could view genetics as a boon or threat, which would influence their dealings greatly. What would do if you thought something was a boon or a threat? 2. I would need to know first what the
news reports actually say.
Child development
Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question. I would say B.
Z = (mean1 - mean2)/standard error (SE) of difference between means SEdiff = √(SEmean1^2 + SEmean2^2) SEm = SD/√n If only one SD is provided, you can use just that to determine SEdiff. Find table in
the back of your statistics text labeled something like "area...
Since this is not my area of expertise, I searched Google under the key words " bacterial vs. mammalian transcription" to get these possible sources: https://www.google.com/search?client=safari&rls=
en&q=bacterial+vs.+mammalian+transcription&ie=UTF-8&oe=UTF-8 In the f...
Clouds of what? http://scienceline.ucsb.edu/getkey.php?key=884
The sun does not have an atmosphere or crust.
Online, * is used to indicate multiplication to avoid confusion with x as an unknown. However, the sign is not needed, if a parenthesis is present. 2y(-28)=48 -56y = 48 y = 48/-56 Reduce.
A parameter is data about the whole population, while a statistic deals with a part of the population, a sample. What does that tell you?
It's possible, but what type of genes are going to be inserted? Who will determine who will get what genes? As with medicine, will there be unintended side effects? That should give you a start.
This is asking for your response to an unknown artifact. Our psychic abilities are limited or nonexistent.
statistics psy/315
Z = (mean1 - mean2)/standard error (SE) of difference between means SEdiff = √(SEmean1^2 + SEmean2^2) SEm = SD/√n If only one SD is provided, you can use just that to determine SEdiff. Find table in
the back of your statistics text labeled something like "area...
math (?)
No information above. Original price of chair missing.
math (?)
Data missing. A parameter is a measure on the total population, while a statistic is a measure on a portion of the population.
A normal distribution approximates a bell-shaped curve. Outliers to one side would skew the distribution. If the distribution was flat, having about the same frequencies throughout the distribution,
this would be non-normal. Also a bimodal distribution would be non-normal, sug...
% = mean ± Z(SEm) Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (%/2) from the mean and its Z score.
Substitute values into above equation and calculate. If you d...
math (?)
What is your question?
pre-algebra (?)
What model?
Since this is not my area of expertise, I searched Google under the key words "Lamarck vs. Darwin differences" to get these possible sources: https://www.google.com/search?client=safari&rls=en&q=
lamarck+vs+darwin+differences&ie=UTF-8&oe=UTF-8 In the future, you can f...
statistics (?)
What following? Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.
This may help you decide.
I am assuming the Eric is ON the train. 24/3 = 8 ft./sec. 8 + 80 = ?
I would assume so. Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in
another post. Thank you.
algebra 2
Except for multiplying/dividing by a negative number, treat just like an equation. First get rid of the parentheses by multiplying. 2x - 10 > 6x -18 Subtract 2x from both sides and add 18 to both
sides. You should be able to work it from there.
medical needs help
Indicate your specific subject in the "School Subject" box, so those with expertise in the area will respond to the question.
Cannot sketch on these posts. Z = (score-mean)/SD Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions/probabilities related
to the Z scores.
What type of mutations are you referring to? Dominant? Recessive? Other?
1. B 2. B
statistics (?)
Where is your data?
math (?)
Powers not shown.
Z = (mean1 - mean2)/standard error (SE) of difference between means SEdiff = √(SEmean1^2 + SEmean2^2) SEm = SD/√n If only one SD is provided, you can use just that to determine SEdiff. Find table in
the back of your statistics text labeled something like "area...
College Algebra (?)
What line? Graphics cannot be posted.
Null hypotheses propose no effect.
College Algebra
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
See your later post.
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
I answered this before. Carriers have the recessive gene in their genotype, b...
% = mean ± ? SEm SEm = SD/√n Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability ([1-%]/2) to find the Z score.
Insert that into ? position. Example: 95% = 74.8 &...
% = mean ± ? SEm Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability ([1-%]/2) and its Z score. Enter Z score
for ?. Solve for intervals.
A null hypothesis expects no difference.
Algebra 1
72 + 4(12-6) = ?
Since this is not my area of expertise, I searched Google under the key words "atom structure picture" to get these possible sources: https://www.google.com/search?client=safari&rls=en&q=
atom+structure+picture&ie=UTF-8&oe=UTF-8 In the future, you can find the informa...
math(check answers please)
1. B 2. A 3. 3f + 2f = 5f
1, 2. If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events. 3. 8 lots total. With each selection, there
is one less large lot and one less total. 6/8 * 5/7 * 4/6 = ? 4. 6/15 * 5/14 = ?
I searched Google under the key words "motivation" to get these possible sources: https://www.google.com/search?client=safari&rls=en&q=motivation&ie=UTF-8&oe=UTF-8 In the future, you can find the
information you desire more quickly, if you use appropriate key words t...
The smallest is most significant.
physical (?)
What statements?
(58.7 * 10^16)/(9.3 * 10^4) = ? seconds To convert to hours, multiply by 360. For days, multiply again by 24. For years, multiply again by 365. for centuries, multiply again by 100.
Not sure what this statement means. The light from the sun gets stronger at sunrise and weaker at sunset, but the radiation from the sun is about the same under both conditions.
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
See your later post.
Ho: Smpking does not effect the period of gestation. Generalize from the above to other null hypotheses.
math (?)
What are your choices?
To carry the disease means that it is in the genotype but not the phenotype. That would happen with a recessive gene. It will occur.
Math (?)
We cannot view graphs on these posts.
Please, where is the specific question?
5 + 4.75 + 2.25 = ? Convert to feet and inches.
Subtract 3x from both sides, then divide both sides by 6.
1.483.56 has 6 significant figures.
A square has all sides of equal length. You do not have a square. For a rectangle, 2[(4x+3)+3(2x - 5)] = ?
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
See your later post.
$14,260/24 = $594.17 19.63/594.17 = ?
statistics (?)
What is your question?
Online, * is used to indicate multiplication to avoid confusion with x as an unknown. How did you get zero? In all = 9 * 3 = ? Twice = 2 times 9 * 2 = 18
Right. See later post for second problem. Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already
been answered in another post. Thank you.
All correct except 3. Each basket in basketball counts as two points. Also, online * is used to indicate multiplication to avoid confusion with x as an unknown.
Conclude H1 at the 5% level of significance, since it is a one-tailed test and it is well beyond the 90% confidence limits.
If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.
a, b. No inferences being made. c. There is no data for other nights, so this is an inference.
Please only post your questions once. Repeating posts will not get a quicker response. In addition, it wastes our time looking over reposts that have already been answered in another post. Thank you.
See your later post.
We do not do your homework for you. Although it might take more effort to do the work on your own, you will profit more from your effort. We will be happy to evaluate your work though.
To get help, you need to post specific questions.
Proportion = # in category/total number
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The Mary P. Dolciani Award Winner Announced
Inaugural Mary P. Dolciani Award Presented at MAA MathFest 2012
William G. McCallum (University of Arizona) received the first Mary P. Dolciani Award during the MAA Prize Session on Friday, August 3, 2012, at the 2012 MAA MathFest in Madison, Wisconsin. Full
citation and biographical information is available below. Award brochure (pdf).
Established in 2012, the Dolciani Award recognizes a research mathematician in pure or applied mathematics who is making a distinguished contribution to the mathematical education of K-16 students.
This award is administered by the Mathematical Association of America on behalf of the Mary P. Dolciani Halloran Foundation.
Mary P. Dolciani Halloran (1923-1985) was a gifted mathematician, educator, and author. She received her Ph.D. in mathematics from Cornell University in 1947, and devoted her life to developing
excellence in mathematics education through her own teaching, mentoring, and writing. A leading author in the field of mathematical textbooks at the college and secondary school levels, she published
under her professional name Dr. Mary P. Dolciani. At a time when it was uncommon for women to enter the field of mathematics, Dr. Dolciani advanced and excelled.
MAA President Paul Zorn presents inaugural Mary P. Dolciani Award to William G. McCallum during the MAA MathFest Prize Session.
Professor William G. McCallum exemplifies the Mary P. Dolciani Award as a research mathematician who is making distinguished contributions to mathematical education across all levels from primary to
secondary to college. McCallum’s Ph.D. from Harvard was followed by a year as a fellow at the Mathematical Sciences Research Institute at Berkeley. He then joined the faculty at the University of
Arizona, where he currently chairs the Department of Mathematics. For his contributions to mathematical education at the primary and secondary level, the Committee cites his work as one of three lead
writers of the Common Core State Standards in mathematics. More than 40 states have agreed to these Standards and mathematicians read them with approval. McCallum writes a blog ’Tools for the Common
Core Standardsâ? giving interesting references and reminders of his many presentations on the Standards.
Through his writing and speaking, McCallum has been a champion in getting university faculty to help in solving problems of mathematics education. To further productive collaboration between research
mathematicians and mathematics educators, McCallum founded in 2006 the Institute for Mathematics and Education at the University of Arizona and remained its director until 2009; he continues as chair
of the Institute’s Advisory Board. This Institute supports local, national, and international projects in mathematics education that focus on both the mathematics and the students. At the college
level, he has been a forceful advocate for the calculus reform effort and is the lead author of the Harvard Calculus Consortium’s multivariable calculus book. McCallum is a voice to and for all
stakeholders in mathematics education: students, teachers, college and university faculty.
Biographical Information
William G. McCallum is a University Distinguished Professor of Mathematics and Head of the Department of Mathematics at the University of Arizona. Born in Sydney, Australia in 1956, he received his
Ph.D. in 1984 under the supervision of Barry Mazur. After spending two years at the University of California, Berkeley, and one at the Mathematical Sciences Research Institute in Berkeley, he joined
the faculty at the University of Arizona in 1987. In 1993-94 he spent a year at the Institut des Hautes Ã?tudes Scientifiques, and in 1995-96 he spent a year at the Institute for Advanced Study on a
Centennial Fellowship from the American Mathematical Society. In 2005 he received the Director's Award for Distinguished Teaching Scholars from the National Science Foundation. In 2006 he founded the
Institute for Mathematics and Education at the University of Arizona. He was Director of the Institute until 2009 and now chairs its advisory board. In 2009?2010 he was one of the lead writers for
the Common Core State Standards in Mathematics. His professional interests include arithmetical algebraic geometry and mathematics education. He has received grants and written articles, essays, and
books in both areas.
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Regression Experimental Designs
Next: Observational Studies Up: Regression : Second Pass Previous: Regression Experimental Designs: A
Consider a response which is related to an independent variable. For the last example, suppose we measure the LDL level of a quail given a specific dose level of the drug. Now we vary the dose level
for different quail. This would be an example of a regression experimental design. Instead of introducing a lot of notation, here's a simple definition of such a design. Read through it and then read
the examples that follow.
Controlled Regression Design . We want to investigate a response over several different levels of an independent variable. Randomly select n experimental units and randomly assign a preassigned
number to each level of the independent variable. Keep all other variables which could influence the response at a predetermined fixed level. At the end of the experiment time period measure the
Suds Example. Here is another simple example (From, Draper and Smith (1966), Applied Regression Analysis, New York: Wiley): For a manufacturer of dishwasher detergent, the height of soap suds in the
dishpan is important, even though it is a psychological factor. The suds height should depend on the amount of detergent used. So 7 pans of water were prepared. To each (by random assignment) an
amount of dishwasher detergent was added. Then the dishpan was agitated for a set amount of time and the height of the suds was measured. Some of the variables controlled here were: temperature of
water, time of agitation, type of dishpan, and measurement of the height conducted in the same way. The data are:
Grams of Product (X): 4 4.5 5.0 5.5 6.0 6.5 7.0
Height of Suds mm(Y): 33 42 45 51 53 61 62
The plot of interest is a scatter plot of Height versus Grams:
- * *
Height -
- *
- *
- *
- *
- *
------+---------+---------+---------+---------+---------+ Grams
4.20 4.80 5.40 6.00 6.60 7.20
There is an increasing relationship between height of suds and grams of detergent. It looks fairly linear except it seems to taper off for the high suds levels. Using the regression module, we fit
the linear model:
Height of Suds = a + b*(Grams of detergent) + error
We used the Wilcoxon option. The prediction equation is
Predict Height = -3.33 + 9.67*(Grams of detergent)
The estimate of slope is 9.67, that is we estimate the height of suds to increase 9.67 mm for each additional gram of detergent. We could also use the equation to predict the height of the suds level
for values of grams of detergent. For instance, for 6 gm of detergent we predict the suds level to be
Predicted height = -3.33 + 9.67*6 = 54.69
Inference. The only inference we will consider is a confidence interval for the slope parameter. The estimation of slope is just that, an estimate. We need to estimate how much it missed the true
slope by. We will also use this confidence interval to test the hypotheses: Our decision rule is simple, we reject H[0] in favor of H[A] if 0 is not in the confidence interval for b.
We will use a Central Limit Theorem confidence interval for b. Besides the estimation class code prints out the standard errors of the estimates. These are in the table which follows the regression
equation. The first numerical column gives the estimate and the second column gives the estimated standard deviation of the estimate (i.e., the standard error). Our confidence interval is then of the
Suds Example, continued. From the class code, the estimated slope was 9.67 with Stdev = 1.21. Hence the confidence interval is: Hence we estimate the height of the suds to increase 7 to 12 mm in
height for every gram of additional detergent. The confidence interval does not include 0 so we reject H[0] in favor of H[A] and we conclude that there is a positive linear relationship between the
height of suds and the amount of detergent.
Concrete Example . (From Vardeman (1994), Statistics for Engineering Problem Solving, Boston: PWS.) A study was performed to investigate the relationship between the strength (psi) of concrete and
water/cement ratio. Three settings of water to cement were chosen (.45, .50, .55). For each setting 3 batches of concrete were made. Each batch was measured for strength 14 days later. All other
variables were kept constant (mix time, quantity of batch, same mixer used (which was cleaned after every use), etc.). Here's the data:
Water/cement 0.45 0.45 0.45 0.50 0.50 0.50 0.55 0.55 0.55
Strength 2954 2913 2923 2743 2779 2739 2652 2607 2583
Here's a scatter plot:
Strength - *
- 2
- *
- 2
- *
- *
- *
--------+---------+---------+---------+---------+-------- water/cement
0.460 0.480 0.500 0.520 0.540
The plot indicates a decreasing relationship between strength of concrete and water to cement ratio,; i.e., the more water one uses, the weaker the cement. Clicking on regression module, and using
the Wilcoxon estimate, we obtain the prediction equation
• Strength = 4345 - 3160*(w/c).
What does the estimate of the slope mean?
Keeping the range of x in mind (.1), it is best to phrase this as for each additional tenth of water to cement, we estimate the strength of the concrete to drop by 316 psi. From the class code, we
form a confidence interval for slope by:
Since 0 is not in the confidence interval we reject H[0]. One way of concluding would be: for each additional tenth of water to cement, we estimate the strength of the concrete to drop from 262 to
370 psi.
There is a lot more to experimental designs than we have covered in this chapter. The effects of more than one variable at a time changing on the response can be analyzed. These variables are set at
certain values (the design of the experiment) and other variables are controlled. If they cannot be controlled then they are recorded. These will be used as covariates to adjust the analysis. These
items are beyond this course. In fact there are several courses you can take at Western on experimental design.
There are many situations, though, where we can not design an experiment, (set the levels of the independent variables). These are basically observational studies which we discuss in the next
Exercise 12.3.1
1. (From Bhattacharyya and Johnson (1977), Statistical Concepts and Methods, New York: Wiley). A study was performed to investigate the relationship between speed and stopping distance for an
automobile. 10 cars were selected (same year, model, etc.). Each was driven at preassigned speed and when the driver attained that speed the he applied the brakes. The distance to a complete stop
was then measured. The data are:
Speed (X) : 20 20 30 30 30 40 40 50 50 60
Distance (Y): 16.3 26.7 39.2 63.5 51.3 98.4 65.7 104.1 155.6 217.2
(a) Assuming this was a designed experiment what other variables besides car were controlled?
(b) Scatter plot this data (Y versus X). Comment on the plot. Does it look linear?
(c) Regardless of your discussion in the last part, use the regression module to fit the model. Predict the stopping distance for an initial speed of 35. Predict the stopping distance for an
initial speed of 55.
(d) Use your predictions in the last part to plot your fit on the scatter plot. Comment? Interpret the estimate of slope.
(e) Obtain a confidence interval for the slope parameter. What does it mean in terms of the problem? Use it to test H[0] . Conclude in terms of the problem.
(f) Determine the fit and the residual for the response 98.4 at x = 40.
(g) Next obtain the residual plot. Does the observation (40, 98.4) seem to be an outlier? Is the scatter random? See the next problem for the answer.
2. Here is the residual plot for the last problem:
25+ *
- *
Ehat -
- * * *
0+ * *
- *
- *
- *
+---------+---------+---------+---------+---------+------ Yhat
It is not a random scatter. Sometimes a simple transformation will help. Consider the square root of the stopping distances. These are given by:
Speed (X) : 20 20 30 30 30 40 40 50 50 60
SqrtDistance 4.03 5.16 6.26 7.96 7.16 9.91 8.10 10.20 12.47 14.73
Repeat the last problem using these responses. Notice interpretation changes. As you will see, the residual plot improves considerably but there are still problems with it.
3. (From Vardeman (1994), Statistics for Engineering Problem Solving, Boston: PWS.) A study was performed to investigate the relationship between the carburetor jetting size and the time of a Camaro
for a quarter-mile run. The data are:
Jet Size 76 68 70 72 74 76
Time 15.08 14.60 14.50 14.53 14.79 15.02
(a) Assuming this was a designed experiment what other variables besides car model were controlled?
(b) Scatter plot this data. Comment on the plot. Does it look linear?
(c) Regardless of your discussion in the last part, use the regression module to fit the model. Predict the time for a jet size of 76. Predict the time for a jet size of 68.
(d) Use your predictions in the last part to plot your fit on the scatter plot. Comment? Interpret the estimate of slope.
(e) Obtain a confidence interval for the slope parameter. What does it mean in terms of the problem? Use it to test H[0]. Conclude in terms of the problem.
Next: Observational Studies Up: Regression : Second Pass Previous: Regression Experimental Designs: A
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Transients in a transmission line
May 8th 2012, 02:14 AM #1
May 2012
Transients in a transmission line
Hi there,
I attached a pdf with the problem and some basic background theory. It is not necessary to understand everything since the problem itself is of pure mathematical nature.
The problem I face is how I am suppose to write equation (7) and (8) as a sum?
Last edited by Mavven; May 8th 2012 at 02:17 AM.
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[FOM] Harvey on invariant maximality
Harvey Friedman friedman at math.ohio-state.edu
Wed Mar 28 23:37:46 EDT 2012
> On Mar 27, 2012, at 7:43 PM, Timothy Y. Chow wrote:
> Harvey Friedman wrote:
>> This has nothing directly to do with raising social status of f.o.m.
>> That is a by product.
> Similarly, he wrote:
>> It is my view that "naturalness" and "inevitability" are NOT
>> sociological. In particular, these notions are timeless and
>> independent
>> of the human condition. The only extent that they may depend on the
>> human condition is the overall brain capacity of humans, given by
>> numerical quantities.
> On the other hand, he also wrote:
>> I am NOT doing Concrete Mathematical Incompleteness for the purpose
>> of
>> showing that large cardinals exist. I am attacking Conventional
>> Wisdom
>> concerning the profound and intrinsic irrelevance of so called
>> Abstract
>> Nonsense of which higher set theory is generally included.
>> Conventional Wisdom supports the total disregard of the
>> Incompleteness
>> Phenomena as a silly distraction from real mathematics.
>> First this Conventional Wisdom must be profoundly destroyed. One is
>> then beginning to be armed with new tools needed for dealing with
>> further issues about which nothing convincing is being currently
>> said.
> Frankly, I think it is disingenuous to claim that all your talk of
> naturalness has nothing to do with sociology. Why the obsession with
> mathematicians who have won prestigious awards? Why the use of the
> term
> "victory"? Victory in what kind of battle, if not a sociological one?
> Mathematicians do not usually use the term "victory" to refer to their
> technical achievements.
Naturalness, fundamentalness, and inevitability are not sociological,
but fundamental.
I use interaction with core mathematicians to provide evidence that I
am not making errors in judgment concerning naturalness,
fundamentalness, and inevitability.
Core mathematicians, particularly luminaries, accepting naturalness,
etc., under the present circumstances, strongly suggests that the
statements are in fact natural, etc.
However, under the present circumstances, if core mathematicians do
not accept statements as natural, then that is rather poor evidence
that they are not natural.
Recall that I asked the FOM subscriber list what form of "Victory"
this constitutes, and indicated that I have not settled on the terms
in which this "Victory" should be cast.
> Suppose you devise a theory of "naturalness" and show that according
> to
> the notion of naturalness explicated by the theory, a certain
> statement is
> both natural and independent of ZFC. Then you might go around
> trumpeting
> the fact that you have solved the longstanding problem of exhibiting a
> statement that is both natural and independent of ZFC. This is a free
> country, after all; we can all say what we want. However, unless the
> statement in question is *accepted by the mathematical community* as
> natural---either because it directly affirms it as such, or because it
> accepts your theory of "naturalness" and accepts that the statement in
> question is natural in your sense---such a "victory" will be a
> hollow one.
> In particular, the Conventional Wisdom will remain the dominant
> point of
> view, and sociologically, all you will have accomplished is to
> convince
> *yourself* even more strongly that the Conventional Wisdom is wrong.
You are talking about sociology, and I am talking about naturalness,
fundamentalness, and inevitability. So if I or anyone else "devises" a
really good theory, then that is a great achievement of fundamental
> Call that a "victory" if you want, but I would reserve that term for
> a sea
> change in the way mathematicians in general think about f.o.m.
> Using the
> term "victory" for what is admittedly a very impressive technical
> achievement, but that does not convince anyone who is not already
> convinced, is a tactic that in my opinion will ultimately be
> detrimental
> to the social status of f.o.m. And even if you declare that the
> social
> status of f.o.m. is only of secondary interest, it is still important
> enough that it should not be ignored.
Well, I asked for advice about how to state "victory" here, and you
have given it. Thank you.
> I would go even further and say that the tactic of arguing that the
> word
> "natural" is not sociological, *even in the context of the search for
> "natural" independent statements*, is also detrimental to the social
> status of f.o.m.
Just the opposite is true. The notion of "natural", which is
fundamental, and crucial to the development of mathematics, in the
past, now, and in the future, begs to have some nontrivial analysis. I
haven't seen any interesting analysis of this, even in limited
contexts. In the present environment, we cannot look to the
mathematics community to do this sort of work.
I have no idea how you view what you call the "social status of
f.o.m." It appears that very very few mathematicians are aware that
there is an area of mathematically rich research called f.o.m. They
are somewhat aware that there is what they call a "branch of
mathematics" called mathematical logic. Very very few mathematicians
have any idea what is going on in mathematical logic.
When you say "detrimental to the social status of f.o.m.", I think you
are talking about the social status of f.o.m. in the mathematics
community, in which case we are starting with essentially a total lack
of awareness. It doesn't make any sense to me that this can become
worse by talking about truths like " "natural" is fundamental and not
> Of course, there's nothing wrong with trying to develop
> a theory of mathematical naturalness that captures many of the
> intuitions
> we have about it. However, when people ask for a natural statement
> independent of ZFC, most of them are probably looking for something
> that
> has already occurred in the literature of core mathematics, or
> connects
> strongly to it.
Invariant Maximality connects "strongly" to existing mathematics, in
some very clear senses of "strongly connects". This kind of "strong
connection" can probably be subject to illuminating theory - or at
least this is promising.
> In particular, they are using the word "natural" in a
> sociological sense.
There are of course some sociological meanings of "natural" which
Invariant Maximality falls under. E.g., the sociological notion of
"combining two fundamental notions, both in common use, in simple ways".
> If you respond to them that such-and-such a proposed
> statement is "natural" in a non-sociological sense, and respond to
> their
> protests that that's not what they meant by telling them that their
> notion
> of "natural" is wrong, it will strike them as a semantic trick.
> They will
> not be persuaded.
I don't know whether you are talking about core mathematicians,
logicians, or observers of f.o.m. On the contrary, I expect many
people to be rather convinced by Invariant Maximality and the
fascinating prospect that there might well be a theory behind
fundamental notions of naturalness in mathematics and other facets of
human thought.
EVERY INVARIANT SUBSET OF Q[0,1]^2k HAS AN INVARIANT' MAXIMAL SQUARE
for which basic notions of invariance is this statement true?
is convincing enough to meet many sociological and non sociological
notions of natural, fundamental, and inevitable.
The examples given, where this is independent, and Pi01, are enough to
launch the full investigation. The full investigation of course is at
a very early stage.
Harvey Friedman
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Motion with Uniform Acceleration
October 4th 2009, 02:27 AM #1
Sep 2009
North West England
Motion with Uniform Acceleration
Though id post this one here becuase its partly mechanics so physcis based etc. This thing is im working with SuVat as in S=displacement u=uniform acceleration V=velocity a=acceleration and t=
time. Ive manged to answer most of the questions I have on this work but have become stuck on the final one and was wondernig if anyone can help. The question is A stone fall past a window, 3m
high, in 0.5s. Taking acceleration due to gravity, g (a) = 9.8ms-2, find the height from which the stone fell. It would be great help if you could help me out. The formulas I need to use are v=
u+at s=ut+1/2at squared. v squared = u squared + 2as and s=1/2(u+v)t. sorry if it not clear.
Though id post this one here becuase its partly mechanics so physcis based etc. This thing is im working with SuVat as in S=displacement u=uniform acceleration V=velocity a=acceleration and t=
time. Ive manged to answer most of the questions I have on this work but have become stuck on the final one and was wondernig if anyone can help. The question is A stone fall past a window, 3m
high, in 0.5s. Taking acceleration due to gravity, g (a) = 9.8ms-2, find the height from which the stone fell. It would be great help if you could help me out. The formulas I need to use are v=
u+at s=ut+1/2at squared. v squared = u squared + 2as and s=1/2(u+v)t. sorry if it not clear.
You have to combine 2 equations:
$\left|\begin{array}{rcl}s+3&=&\frac12 \cdot 9.81 \cdot (t+0.5)^2 \\ s&=& \frac12 \cdot 9.81 \cdot t^2 \end{array} \right.$
Expand the first equation and subtract columnwise the second equation from the first one. You'll get an equation in t. Solve for t.
I've got $t \approx 0.3616\ s$ which means that the stone starts to fall 0.64 m above the window.
Though id post this one here becuase its partly mechanics so physcis based etc. This thing is im working with SuVat as in S=displacement u=uniform acceleration V=velocity a=acceleration and t=
time. Ive manged to answer most of the questions I have on this work but have become stuck on the final one and was wondernig if anyone can help. The question is A stone fall past a window, 3m
high, in 0.5s. Taking acceleration due to gravity, g (a) = 9.8ms-2, find the height from which the stone fell. It would be great help if you could help me out. The formulas I need to use are v=
u+at s=ut+1/2at squared. v squared = u squared + 2as and s=1/2(u+v)t. sorry if it not clear.
I'm assuming the stone was released from rest.
the $d$ = distance from where the stone was dropped to the top of the window
$t$ = elapsed time from when the stone was released until it gets to the top of the window
$d = \frac{1}{2}gt^2$
$d+3 = \frac{1}{2}g(t+0.5)^2<br />$
substitute $\frac{1}{2}gt^2$ for $d$ in the second equation ...
$\frac{1}{2}gt^2 + 3 = \frac{1}{2}g(t+ 0.5)^2$
$\frac{1}{2}gt^2 + 3 = \frac{1}{2}g(t^2 + t + 0.25)$
$\frac{1}{2}gt^2 + 3 = \frac{1}{2}gt^2 + \frac{1}{2}gt + \frac{1}{8}g$
$3 = \frac{1}{2}gt + \frac{1}{8}g$
$24 = 4gt + g$
$t = \frac{24-g}{4g}$
$d = \frac{1}{2}g\left(\frac{24-g}{4g}\right)^2$
Oh thankyouu soo much. One things for sure I deff havent picked this up well in class that working out loooks sooo confusing but Im sure ill get the correct answer if I follow youre working out.
October 4th 2009, 04:44 AM #2
October 4th 2009, 05:01 AM #3
October 4th 2009, 05:08 AM #4
Sep 2009
North West England
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Homework Help
Posted by Cole Brown on Monday, March 4, 2013 at 1:06am.
1. For the following scores,
X Y
Sketch a scatter plot and estimate the value of the
Pearson correlation.
b. Compute the Pearson correlation
2. For the following set of scores,
X Y
a. Compute the Pearson correlation.
b. Add 2 points to each X value and compute the correlation for the modified scores. How does adding a constant to every score affect the value
of the correlation?
c. Multiply each of the original X values by 2 and compute the correlation for the modified scores. How does multiplying each score by a constant
affect the value of the correlation?
3. To simplify the weight variable, the women are classified into five categories that measure actual weight relative to height, from 1 thinnest
to 5 heaviest. Income figures are annual income (in thousands), rounded to the nearest $1,000.
a. Calculate the Pearson correlation for these data.
b. Is the correlation statistically significant? Use a two-tailed test with .05.
Weight (X) Income (Y)
4. Assume a two-tailed test with .05.
(Note: The table does not list all the possible df values. Use the sample size corresponding to the appropriate
df value that is listed in the table.)
a. A correlation of r 0.30.
b. A correlation of r 0.25.
c. A correlation of r 0.20.
5. A professor obtains SAT scores and freshman grade point averages (GPAs) for a group of n 15 college students. The SAT scores have a mean of M 580
with SS 22,400, and the GPAs have a mean of 3.10 with SS 1.26, and SP 84.
a. Find the regression equation for predicting GPA from SAT scores.
b. What percentage of the variance in GPAs is accounted for by the regression equation? (Compute the correlation, r, then find r2.)
c. Does the regression equation account for a significant portion of the variance in GPA? use a=.05 to evaluate the F-ratio.
6. a. One set of 20 pairs of scores, X and Y values, produces a correlation of r 0.70. If SSY 150, find the standard error of estimate for the regression line.
b. A second set of 20 pairs of X and Y values produces of correlation of r 0.30. If SSY 150, find the standard error of estimate for the regression line.
7. A researcher obtained the following multiple-regression equation using two predictor variables:
Yˆ 0.5X1 4.5X2 9.6. Given that SSY 210, the
SP value for X1 and Y is 40, and the SP value for X2 and Y is 9, find R2, the percentage of variance accounted for by the equation.
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D1 - Neural Networks for Data Structures Principles and Applications
Sunday, AM
Paolo Frasconi & Alessandro Sperduti
The purpose of the tutorial is to examine the state of the art in the use of connectionist networks for processing data structures and to present a unified view of formalisms and tools for dealing
with rich data representations, covering connectionist architectures for data structures, learning algorithms, and applications. In particular, we will show that it is possible to represent and
classify structured information very naturally. Moreover, it is possible to formalize several supervised models for classification of structures which stem very naturally from well known models, such
as back propagation through time networks, real-time recurrent networks, simple recurrent networks, recurrent cascade correlation networks, and neural trees.
Because many concepts and formal tools are inherited from the theoretical framework of recurrent networks for sequence processing, the tutorial will begin with a review of basic concepts underpinning
recurrent neural networks, for those attendees which are not familiar with such a class of models. Algorithms for training recursive neural networks are presented as generalizations of gradient
computation algorithms for recurrent nets, and complexity as well computational issues are discussed. Finally, examples of applications in chemistry, structural pattern recognition, and theorem
proving are presented.
Prerequisite knowledge:
Although we will briefly review the essential concepts for data structure and neural networks, we will assume that the attendants will be familiar with data structures; we also assume basic knowledge
of linear algebra and calculus for the treatment of some neural network paradigms.
Paolo Frasconi received the M.Sc. degree in Electronic Engineering in 1990 and the Ph.D. degree in Computer Science in 1994, both from the University of Florence, Italy. He is currently Associate
Professor with Dipartimento di Ingegneria Elettrica ed Elettronica at the University of Cagliari, Italy. He was Assistant Professor with the Dipartimento di Sistemi e Informatica at the University of
Florence, Italy. In 1992 he was a Visiting Scholar in the Department of Brain and Cognitive Science at the Massachusetts Institute of Technology, Cambridge. In 1994 he was a Visiting Scientist at
Centro Studi e Laboratori Telecomunicazioni (CSELT), Turin. His current research interests include neural networks, Markovian models, and graphical models, with particular emphasis on problems
involving learning about sequential and structured information. Paolo Frasconi is the author of around 50 refereed papers mainly in the areas of graphical models for learning, neural networks,
pattern recognition, artificial intelligence.
Alessandro Sperduti received his university education from the University of Pisa, Italy ("Laurea" and Doctoral degrees in 1988 and 1993, respectively, all in Computer Science). In 1993 he spent a
period at the International Computer Science Institute, Berkeley, supported by a postdoctoral fellowship. In 1994 he moved back to the Computer Science Department, University of Pisa, where he was
Assistant Professor, and where he presently is Associate Professor. His research interests include pattern recognition, image processing, neural networks, hybrid systems. In the field of hybrid
systems his work has focused on the integration of symbolic and connectionist systems. He contributed to the organization of several workshops on this subject and he served also in the program
committee of conferences on neural networks. Alessandro Sperduti is the author of around 50 refereed papers mainly in the areas of neural networks, fuzzy systems, pattern recognition, and image
Webmaster: Sven Olofsson, sveno@dsv.su.se Last modified: Mar 16, 1999
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Physics engines for dummies
Hello and welcome back to my blog!
This time i’m going to talk about the basic components that make up a physics engine and how to put them together; this tutorial is aimed at programmers who have a basic grasp of maths and geometry
but would like to step into the world of simulation.
It is my hope that if, at the beginning of this article, you are able to code the physics behind the 1972 game Pong, by the end of the article you will be equally happy writing your own constraints
to use in your own physics solver!
I’ve always found the title of those books of from the ‘…for dummies’ series reassuring; after all, if a dummy can learn this stuff you should stand a good chance, right?
Hence the title of this article.
Plan of action
Ok, so i’m going to cover a few of the things you might want in a physics engine:
• Rigid bodies
• Collisions
• Resting contact
• Constraints (Joints)
Bear with me, this is going to start out really basically, but hopefully later on it will become clear why.
Starting out with particles, which we assume will have a position P and a velocity V. Each frame we advance the position of P by adding the velocity V on to it. This is called integration.
You can use whatever units you like for you simulation; the usual choice is one unit/meter which is what i’m using here – the screen is two metres by two metres (in our world), so our velocities are
specified in metres/second. To make this work, we must know how many seconds there are per frame, then we can do P += V*dt in our integration to advance the particles properly, where dt is the number
of seconds per frame.
We can simulate multiple particles by storing an array of positions and velocities, and looping over integrating each one. To prevent our particles going off screen we would like to collide them
against the screen’s edges.
To bounce the particles off the edges we simply detect collisions against the edges and reverse the velocity in the offending axis, thus:
for all i
if (P[i].x < -1 && V[i].x < 0)
V[i].x = -V[i].x;
if (P[i].y > 1 && V[i].y > 0)
V[i].y = -V[i].y;
if (P[i].x > 1 && V[i].x > 0)
V[i].x = -V[i].x;
if (P[i].y < -1 && V[i].y < 0)
V[i].y = -V[i].y;
The first condition of each case checks for intersection of the particle in space and the second checks that the particle is actually heading towards the edge. The second condition is important
otherwise we would do the collision response again next frame erronusly should the particles move more than one pixel per frame. This second condition carries though all of impulse based rigid body
simulation and is what separates inequality contraints (e.g contacts) and equality constraints (most joints).
This is the kind of example code you will no doubt have written a long time ago when you first started getting interested in simulation. I’m including it here because I think there is a natural
progression from this easy example to more complex physics code.
Whats going on in this simple simulation?
Well, without knowing it, we have assumed a physical matieral type for our particles including the coefficient of restituition and followed Newton’s principle of conservation of momentum. We have
also chosen that the world have infinite mass such that it doesn’t feel a collision response when the particles impact on it. By choosing to reflect particles on impact we are also preseving their
momentum (even though we have ignored their mass in our calculations), indicating that we have chosen a coefficient of restitution of 1, i.e. perfectly elastic – like a super-ball. Additionally, we
have chosen an impulse/velocity model for the collision response, rather than a force/acceleration model – we did this simply by changing the velocity instantaneously, rather than over a period of
What we have here is actually a highly optimised special case physics simulation. Optimised how, you ask? Well, let me explain:
If we were to code the above “properly”, not taking any short cuts, we would have to assume the following. Our environment is defined by its boundaries which are four axis aligned planes (actually,
they’re lines because we are in 2d). Each one has a normal (which points inwards) and a distance to the origin. They look like this:
Planes[4] =
(1, 0, 1),
(0, -1, 1),
(-1, 0, 1),
(0, 1, 1)
So, now we have to detect our collisions as we did before, but we can’t take any shortcuts this time. In order to detect if our particles have penetrated the planes we must perform a dot product and
add the plane’s distance to origin:
for all particles i
for all planes j
distance = P[i].x*Planes[j].a + P[i].y*Planes[j].b + Planes[j].c;
if (distance < 0)
// collision responce
What this code is doing is finding, by projection, how much of the vector from the plane to the particle is in the direction of the plane normal. Because our plane’s normals are unit length, this is
a measure of the closest distance from the particle to the plane. Obviously, if this distance is less than 0, our particle has penetrated the plane and we need to take action to perform the collision
Now, if we take a closer look at that distance check above, including each plane’s coefficents:
plane0dist = P[i].x*1 + P[i].y*0 + 1;
plane1dist = P[i].x*0 + P[i].y*-1 + 1;
plane2dist = P[i].x*-1 + P[i].y*0 + 1;
plane3dist = P[i].x*0 + P[i].y*1 + 1;
If we simplify that down a bit:
plane0dist = P[i].x + 1;
plane1dist = -P[i].y + 1;
plane2dist = -P[i].x + 1;
plane3dist = P[i].y + 1;
Re-arranging slightly we get these plane tests conditions:
if (P[i].x < -1)
if (-P[i].y < -1) = if (P[i].y > 1)
if (-P[i].x < -1) = if (P[i].x > 1)
if (P[i].y < -1)
Which are exactly the same as those which we used in our first simple example. The second condition for each plane must also be satisfied so we only do the collision response once per collision. This
can be done by performing a dot product between the particle’s velocity and the normal for each plane;
if (P[i].x < -1 && V[i]•N[i] < 0)
We call this the normal velocity since its the particle’s velocity in the direction of the normal. If this results in a value less than zero then the particle is moving towards the plane and we allow
the collision. Once the collision is resolved, the normal velocity will be >= 0 depending on the coefficient of restitution. I could perform the same analysis on the second condition as i did the
first to prove that this general solution is the same as the first specific version, but i will leave that as an excercise for the reader.
This shows our two approaches are physically the same.
Ok, so what do we do now for the collision response?
We need a solution which achieves the same result as the original program but is more general. We can do this using the reflection vector from the law of reflection. The reflection vector is
calculated thus:
R = V – 2*N*(V•N)
Where V is the velocity vector and N the surface normal. We need to do a similar comparison with this operator as we did with the collision checks:
plane0vel x = V.x - 2* 1*(V.x* 1 + V.y* 0) = V.x -2*V.x = -V.x
plane0vel y = V.y - 2* 0*(V.x* 1 + V.y* 0) = V.y - 0 = V.y
plane1vel x = V.x - 2* 0*(V.x* 0 + V.y*-1) = V.x - 0 = V.x
plane1vel y = V.y - 2*-1*(V.x* 0 + V.y*-1) = V.y - 2*V.y = -V.y
plane2vel x = V.x - 2*-1*(V.x*-1 + V.y* 0) = V.x - 2*V.x = -V.x
plane2vel y = V.y - 2* 0*(V.x*-1 + V.y* 0) = V.y
plane3vel x = V.x - 2* 0*(V.x* 0 + V.y* 1) = V.x
plane3vel y = V.y - 2* 1*(V.x* 0 + V.y* 1) = V.y - 2*V.y = -V.y
Which, you should be able to see, are the exact same resulting velocities after collision with each plane in the first simple example.
You will have noticed that mass has been ignored in our calculations so far, this is because it just drops right out of the equations when you are dealing with collision against a body of infinite
mass (i.e. our simple world).
So what does this new version look like so far in pseudo-code?
for all particles i
for all planes j
N = {Planes.a, Planes.b};
distance = P[i]•N + Planes[j].c;
if (distance < 0 && V[i]•N < 0)
// collision response, reflect particle
V[i] -= 2*N*N•V[i];
Great, so we’ve turned a nice simple example into one that’s far more complicated than it needs to be. Why?
Firstly, to demonstrate that the roots of all our inital assumptions were grounded in physics and secondly, to demonstrate the advantages of the more complex implementation. Because we now have a way
to deal with arbitrary 2d planes instead of axis aligned planes, it means we can rotate our world and still have the simulation work correctly:
Move the mouse over the demo to rotate the planes. Note that in this demo, i have done some correction to the points as the planes are rotated to make sure the points still lie within the planes.
Ok, so this is all very well but its not a very real world simulation. There is no gravity and the coefficient of restistution is far too high – in the real world, most everyday things have a near
plastic coefficient of restitution, i.e near 0.
So, adding gravity is pretty simple, we just subtract gravity from our particle’s velocities before we do any collision, remembering to account for the fact that gravity is an acceleration:
V[i] += G*dt
In this case G is the vector {0, -9.8} and dt is the number of seconds per frame as before.
We would also like to use a different coefficient of restitution as well because our particles might not be made of super-rubber. Recall the reflection vector equation that we used earlier:
R = V – 2*N*(V•N)
We can actually re-write this equation to include the coefficient of restitution:
R = V – (1+e)*N*(V•N)
Where e is the coefficient of restitution which varies from 0 (totally plastic) to 1 (totally elastic). You can see that these two equations are equivelent; if we set e to 1, we get the original
equation. The particles in this example have coefficients of restitution varying from 0 to 1.
The only thing we can now do to this particle simulation to make it more realistic (without going to a full rigid body simulation with rotation) is to handle masses in our collisions. To accomplish
this, we will have to use circles instead of particles, since its quite unlikely that two particles will ever collide what with them having no actual size!
As per Figure 1, two circles A and B intersect if the distance between them is less than the sum of their radii. The collision normal is simply the vector d between the two, but normalised.
In order to deal with the masses of our particles we need to do some thinking regarding the reflection equation that we already have. What we really need is some kind of mass ratio to apply a bias to
the equation so that lighter particles get pushed out of the way by heavier particles, but we need it to conserve momentum as well.
ratio[a] = M[b] / (M[a] + M[b])
ratio[b] = M[a] / (M[a] + M[b])
The above two ratios will do the trick. Here is an example:
M[a] = 1.0
M[b] = 0.5
ratio[a] = 0.5 / (1.0 + 0.5) = 1/3
ratio[b] = 1.0 / (1.0 + 0.5) = 2/3
Obviously 1/3 + 2/3 = 1, so you can see that this conserves momentum correctly, and since body b is half the mass of body a, it should experience twice the reaction upon collision, which it does.
So we can plug these two ratios into our reflection vector equation to obtain two new equations for the reflected velocity of each body after collision:
V[a] = V[a] – (1+e)*N*((V[b]-V[a]) • N)*(M[b] / (M[a]+M[b]))
V[b] = V[b] – (1+e)*-N*((V[b]-V[a]) • -N)*(M[a] / (M[a]+M[b]))
Our collision code gives us the normal pointing from body b towards body a, so we must reverse the normal to calculate body b‘s reflected velocity. You will have noticed that there are a lot of
duplicated calculations in the two equations above. We would like to simplify this down a bit. What we can do is use the relative velocity of the two objects in question:
V[r] = V[a ]- V[b]
Now that we have only one velocity to deal with the equation gets simpiler.
I = (1+e)*N*(V[r] • N)
V[a] = -I*(M[b] / (M[a]+M[b]))
V[b] = +I*(M[a] / (M[a]+M[b]))
What we are doing here is to treat one object as being stationary relative to the other moving object. But there is still more we can do here to simplify these equations:
I = (1+e)*N*(V[r] • N)*(M[a]*M[b])/([ ]M[a]+M[b])
V[a] – = I / M[a]
V[b] + = I / M[b]
By combining the ratios we can save some more calculation and reduce the final step to a simple divide though by the mass of the objects in question to produce the final velocities for those objects.
You can see this works because
(M[a]*M[b])/([ ]M[a]+M[b]) / M[a ]= M[b]/([ ]M[a]+M[b])
(M[a]*M[b])/([ ]M[a]+M[b]) / M[b ]=[ ]M[a] / (M[a]+M[b])
Finally we can show that (from fractions)
(M[a]+M[b])/([ ]M[a]*M[b]) = 1/M[a ]+ 1/M[b]
Which means we can re-write the final linear impulse equation
I = (1+e)*N*(V[r] • N) / (1/M[a ]+ 1/M[b])
This is handy because we can just store 1/M[a ]and 1/M[b] inside the definition of our circle rigid bodies so we don’t have to re-calculate. It also means that we now have a way to represent
infinitly massive objects, such as our world (by storing the inverse mass as 0).
V[a] – = I * 1/M[a]
V[b] + = I * 1/M[b]
In these demos you can manipulate the circles with the mouse.
Ok, so you probably already noticed we are starting to see problems due to penetration caused by the collision detection system detecting the collision after its already too late. This problem only
gets more severe as we add more rigid-bodies, and is compounded by the fact that the system is currently doing nothing to correct the penetration:
To combat this problem, we need to be able to detect collisions before they actually happen and deal with penetration. Luckily i’ve covered this technique already in a previous article which i
suggest you read.
Once this technique is applied the problem just goes away; and there are only a few lines of code required. The only caveat is that we have to assume that the coefficient of restitution is never
required to be non-0; in the majority of real-world cases this isn’t too much of a limitation, especially given the benefits.
Software engineering
In physics engines, the mathsy part is only half the story; the second half is how you organise your physics engine in terms of classes hierarchy and how you actually code it.
I’m going to present one possible solution here which has worked for me in the past; it may be over-simplified for your requirements but it should give you an idea. The language here is actionscript,
but the principles apply to any OO language.
public class RigidBody
protected var m_pos:Vector2;
protected var m_vel:Vector2;
protected var m_invMass:Number;
public function RigidBody( pos:Vector2, vel:Vector2, invMass:Number )
public function GenerateContact( rb:RigidBody ):Contact
throw new Error("Not implemented on base class");
public function Integrate( dt:Number ):void
m_pos=m_pos.Add( m_vel.MulScalar( dt ) );
So, i’ve defined a RigidBody base class containing the three parameters that we need so far in our example. There is an integrate function to move foward in time and (what should be a virtual)
function called GenerateContact() which will generate a collision normal and distance between any two RigidBodys.
public class Circle extends RigidBody
private var m_radius:Number;
public function Circle( pos:Vector2, radius:Number, invMass:Number )
m_radius = radius;
super(pos, new Vector2(), invMass);
public override function GenerateContact( rb:RigidBody ) :Contact
if ( rb is Circle )
else if (rb is ...)
throw new Error("unahandled case!");
And we have a Circle which derives from RigidBody and extends its functionality by having a radius parameter. Also, we have implemented the GenerateContact() function to return the required
information when needed.
By separating functionality out into base class and concrete implementations we will be able to deal with multiple RigidBodys in the same lists, which greatly simplifies the code.
public class Plane extends RigidBody
private var m_n:Vector2;
private var m_d:Number
public function Plane(n:Vector2, d:Number )
super(n.MulScalar(-d), new Vector2(), 0);
public override function GenerateContact( rb:RigidBody ):Contact
if ( rb is Particle )
else if ( rb is Circle )
throw new Error("unhandled case!");
Above is another concrete implementation, this time representing an infinite plane (our screen edges).
public class Contact
public var m_normal:Vector2;
public var m_dist:Number;
public function Contact( n:Vector2, dist:Number )
m_normal = n;
m_dist = dist;
Above is the definition for the Contact class, which represents the information the system will need to resolve one collision.
private function Update( e:GameLoopEvent ):void
const dt:Number = Math.min(e.m_elapsed, 1.0/15.0);
// apply gravity
for each ( var p:RigidBody in m_rigidBodies )
if ( p.m_InvMass>0 )
p.m_vel=p.m_vel.Add( Constants.kGravity.MulScalar( dt ) );
// collide
for ( var i:int=0; i<0||rbj.m_InvMass>0 )
const c:Contact=rbi.GenerateContact( rbj );
//resolve collision
// integrate
for each ( p in m_rigidBodies )
p.Integrate( dt );
Above is what our update loop currently looks like. The order here is fairly important;
1. Firstly, external forces are applied – i.e. gravity.
2. Then, all the collision detection is done to generate a contact which needs to be resolved for each pair of interacting rigid bodies.
3. Finally, each RigidBody is integrated forward in time.
The time-step clamping at the beginning is so that as we debug the code, we don’t end up with such a massive time-step the next frame that everything explodes!
Ok, so we have a pile of spheres up and running and our physics engine is starting to take shape quite nicely, what about constraints?
Before i talk about this, i feel i should introduce some concepts needed to understand the problem.
Loosely, this is something which enforces a condition between two rigid bodies. So, when there is a collision we create a constraint which enforces the rule that the colliding rigid body may not move
through the object it collides with but also that it do this by only pushing. We already derrived the code for this above (not taking into account rotation yet):
I = (1+e)*N*(V[r] • N) / (1/M[a ]+ 1/M[b])
All the constraints i’m going to work with are impulse velocity level constraints. Forces and accellerations will not come into it.
Inequality constraint
This is a constraint which only acts in one direction. So a collision constraint is an inequality constraint because it is only allowed to push and never pull. If it were allowed to pull, it would
stick the rigid body to the object it collided with. The way we enforce this inequality is simply the check i mentioned at the beginning:
if (V•N < 0)
// handle constaint
Little Big Planet contained a lot of constraints of this type; winch, piston, string are a few examples – they only allowed motion in one direction within the limits of the constraint. So a piston,
for example was a length constraint with lower and upper bounds on the allowed length.
Equality constraint
This constraint type is allowed to pull and push equally. An example is a point to point constraint, where two rigid bodies are connected by a single point – one point on each body is forced to lie
in the same place. With a hinge constraint the whole hinge axis is constrained to lie in the same place on both rigid bodies. A rod constraint would be a a length constraint where the length is never
allowed to change.
Designing a constraint
The way to approach how to design a particular constraint is to think about what restricting effect it has on the two rigid-bodies its connected between.
For example, a distance constraint (a rod in LBP) simply prevents the two end points of the rod getting any closer than (or any further away from) the specified distance. Putting the end points
exactly at the centre of mass of each rigid body, as we are about to describe keeps simple circular rigid bodies at a constant distance from each other, while still allowing both of them to continue
to move around in 2d space.
In Figure 2, bodies A and B must not get any closer than distance d, but are still allowed to move around independently otherwise – this resolves as each body being able to orbit the other as they
fly through space.
Impulse/velocity level
As mentioned previously, the simulator i’m describing is an impulse/velocity level simulator, rather than a force/acceleration level one. This means that all the constraints must only be solved by
applying impulses to change velocities of bodies.
What this means when designing a constraint is that once you’ve worked out how it behaves on a position/distance level (i.e. on paper, with a diagram), you need to then think about how that
translates into velocities.
So, to recap: our distance constraint stops the end points getting any closer or further away from each other. In velocity terms this is describing a one dimensional constraint; the relative velocity
of the bodies is constrained to be zero in one particular axis. That axis is defined by vector between the end points.
In Figure 3, A and B have starting velocities which violate the length constraint that we’ve placed between them (from Figure 2). The first step in solving a constraint is to figure out where that
velocity actually is. In this case you can see its the projection of A[v] and B[v] on to the length constraint axis.
Once we have the velocities we want to remove, we convert into a single, relative velocity (as we did before with the contact constraint) and we already derived the maths required to solve one
dimensional constraints (from the contact constraint):
I = (1+e)*N*(Vr • N) / (1/Ma + 1/Mb)
Lets re-write that so it becomes less of a mouth-full:
I = RelativeVelocityMagnitudeToRemove / (1/Ma + 1/Mb)
In addition to the velocity part, there will also be some positional fix-up to do, since with more than one constraint acting in the system simultaneously it will often fail to be resolved with one
one solver iteration. In this simple example we just add an artificial correction to the velocity (current length – desired length) / time-step.
Above is the result – you can manipulate the circles with the mouse as before, although one is fixed in place to demonstrate the constraint chain.
Software engineering
So, lets cover the additions to the code.
public class Constraint
protected var m_bodyA:RigidBody;
protected var m_bodyB:RigidBody;
public function Constraint( bodyA:RigidBody, bodyB:RigidBody )
m_bodyA = bodyA;
m_bodyB = bodyB;
Assert( m_bodyA.m_InvMass>0||m_bodyB.m_InvMass>0, "Constraint between two infinite mass bodies not allowed" );
public function ApplyImpulse( I:Vector2 ):void
m_bodyA.m_vel = m_bodyA.m_vel.Add( I.MulScalar(m_bodyA.m_InvMass) );
m_bodyB.m_vel = m_bodyB.m_vel.Sub( I.MulScalar(m_bodyB.m_InvMass) );
public function Solve( dt:Number ):void
throw new Error("base class doesn't implement!");
Above you can see we’ve added another base class; this one defines the basic part of a constraint but doesn’t actually define any concrete details – as before, we leave this to the derived classes.
It does however provide a framework for us – each constraint must implement the Solve() function which does the meat of the work. Additionally, there is an ApplyImpulse() function defined to save us
some work in the concrete implementations.
public class DistanceConstraint extends Constraint
private var m_distance:Number;
public function DistanceConstraint( bodyA:RigidBody, bodyB:RigidBody, distance:Number )
super(bodyA, bodyB);
m_distance = distance;
public override function Solve( dt:Number ):void
// get some information that we need
const axis:Vector2 = m_bodyB.m_Pos.Sub(m_bodyA.m_Pos);
const currentDistance:Number = axis.m_Len;
const unitAxis:Vector2 = axis.MulScalar(1/currentDistance);
// calculate relative velocity in the axis, we want to remove this
const relVel:Number = m_bodyB.m_vel.Sub(m_bodyA.m_vel).Dot(unitAxis);
const relDist:Number = currentDistance-m_distance;
// calculate impulse to solve
const remove:Number = relVel+relDist/dt;
const impulse:Number = remove / (m_bodyA.m_InvMass + m_bodyB.m_InvMass);
// generate impulse vector
const I:Vector2 = unitAxis.MulScalar(impulse);
// apply
Above is the implementation for the actual distance constraint, it does the work as described above.
private var m_joints:Vector.<Constraint>;
In the solver we define a list of constraints – they are different to contacts in that they’re more permanent, so they get their own list.
const dt:Number = Math.min(e.m_elapsed, 1.0/15.0);
// apply gravity
for each ( var p:RigidBody in m_rigidBodies )
// collide
for ( var i:int=0; i<m_rigidBodies.length-1; i++ )
const rbi:RigidBody=m_rigidBodies[i];
for ( var j:int=i+1; j<m_rigidBodies.length; j++ )
const rbj:RigidBody=m_rigidBodies[j];
if ( rbi.m_InvMass>0||rbj.m_InvMass>0 )
// solve constraints
for each(var jt:Constraint in m_joints)
// integrate
for each ( p in m_rigidBodies )
p.Integrate( dt );
The solver loop now looks like the above – we have gained a little loop for solving the constraints – i’ve chosen to do that after solving for contacts which makes the joints ‘stronger’ than the
contacts (in that because they’re resolved last, they will be more correct than the contacts which are done first), you can do it the other way around if you like.
This article has uncovered the physics lurking behind the most simple of 2d games (pong), shown that the principles therein are grounded in physics and expanded on the technique used in that game
step by step all the way up to the point where designing a constraint in a physics engine can be seen as a simple extension of colliding two objects.
I hope this article has whet your appetite for physics simulation while also being easy to understand. I have run out of time with this article, but depending on its popularity i can write a lot more
on the subject. Register your interest by posting in the comments!
Source code
As ever, if you enjoyed reading this article and would like to see more, please consider buying the source code which accompanies it; this will allow an indie developer like me to pay the rent and
buy more food to eat!
You will receive all the framework code that i’ve laid out in this article, including the code behind the demos on this page which should serve as a good base to expand upon. They are written in
actionscript 3.0 but the techniques are applicable to all OO languages.
After purchasing, you will be redirected to a page where you can download the source immediately.
SALE! USD 9.99 – was USD 14.99!
Subscribers can access the source here
Until next time, have fun!
Cheers, Paul.
183 Responses to Physics engines for dummies
1. Another superb article, Paul!
Love it!
(Seriously, people – These aren’t the ramblings of some deranged hobo, Paul is one clever muffin indeed!)
Keep posting this stuff Paul. You will have enough for a book soon enough!
□ You sure? I read:
“this will allow an indie developer like me to pay the rent and buy more food to eat! ”
“will code for food”.
☆ Any indie developer able to live off his own business has to be clever. Working for food so he can follow his passion is a common start, even after experience on big companies.
2. thanks for the article, it’s a really great introduction!
I’ve also read your previous article and really like little big planet, but I have the feeling that the physics in lbp are a bit “fluffy”. I can’t really describe it, but I have the feeling that
something is a really tiny bit off. May it be because of the speculative contacts and the missing energy?
□ Hi questor,
Regarding the fluffy feeling physics in LBP – this might be down to the character controller (the thing which drives sackboy’s motion); this came in for a bit of criticism after the game was
launched. In the PSP version we changed that a bit to make sackboy feel more solid… Could that be what you’re experiencing?
Cheers, Paul.
3. Awesome!
I’ve got a question about the position correction; you’re doing “error = relVel + relDist/dt” which seems like a Baumgarte factor of 1… surely this is way too much and would result in lots of
noticeable added velocity?
□ Hi raigan,
Yes, this is probably true – it worked ok in the sample on the page though, which was only running 1 iteration. It may need some tweaking when deployed in production code, of course
In LBP PSP we were using split impulses for this, but i wanted to keep the tutorial as simple as possible…
Cheers, Paul.
☆ Ah, good — I was worried that I was missing something and/or doing it wrong!
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5. I happen to be a physicist and have done some programming in my day… excellent write from both fields of study. good job.
6. Cheers man, you just made my multi media class so much simpler.
7. Thanks for this article. It’s very lucid and non-threatening.
I used to write code like this. Yours are good principles to follow. But at the time, I wanted to make a rigid body, and could never figure out how to do it. Real rigid bodies consist of about 10
^23 little masses coupled to each other with springs. Since I never quite grasped the whole Hamiltonian thing, I just stuck with the masses and springs I understood. Unfortunately, that is the
long, slow way to create a quasi-rigid body.
Can the methods you use here be extended to make something 2-D or 3-D and rigid?
□ Hi Ralph,
Yes, these methods extend into 3d very easily
The masses and springs model is very like what Jacobsen talked about in his early article about rigid bodies composed of length constrants:
Cheers, Paul.
☆ That looks like a good article by Jacobsen. I used to build all kinds of little computable objects from masses and springs. You can get interesting effects by varying individual masses
and spring constants, and of course gravity is easy to add.
I used to worry about approximating the square root. Eventually I decided it wasn’t worth it to over-optimize. Especially with today’s processors there is almost no speed penalty for
using floating point and square roots. Writing code for fixed point is not a great way to enjoy life.
Spring models are outstandingly stable, and objects can in principle be made very realistic (with some work), but it’s a lot of computation.
The simplest spring has zero equilibrium length, but it’s easy to give them non-zero length. I never did very much with collisions. With lots of little masses and springs, fully general
collision tests would take far too much time. I will study Jacobsen’s method for making that practical.
Thanks again for your article.
○ Hi Ralph,
I was very excited when i first read that Jacobsen article and indeed the way he pushes the constraint solver onto the geometry that make up the interacting rigid bodies is an amazing
Cheers, Paul.
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9. Wow! Amazin tuts.. Thanks
10. Nice article! I might pass it on to the next person asking about how to do simulations.
One suggestion: In “The collision normal is simply the vector d between the two, but normalised” it might be worth explaining what “normalised” means and that it’s different from “normal vector”.
□ Thanks for the feedback, Kevin
11. I dont like PHYSICS…!!!!
12. Hey, lovely write-up!
I’ve got 2 weeks off, and writing my own one of these is just what I need to keep me entertained!
One type I noticed: “You can use whatever units you like for you simulation; the usual choice is one unit/meter ” – I think you meant unit/second or unit/frame.
□ Thanks!
The units thing – i was referring to units of space, rather than speed
13. Outstanding.
Thanks you for putting this up!
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16. Consider adding flattr.com to your sources of revenue, I’d have clicked that button right away!
□ Hi Mikeal,
Thanks for pointing that site out! I will definitely take a look at that
Cheers, Paul.
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18. Your physics simulations have some issues with oscillations and constraint enforcements, and because of that you made them really “soft”. There is a solution to solve these problems without
making everything feel packed in wool, which is verlet integration. I’ve written two articles about that (including examples and code as well). I also discuss a hybrid approach to verlet
integration with impulse preservation and stable at-rest acceleration.
□ Hi Florian,
The reason things look ‘soft’ as you say, is that the solver is only running 1 iteration
Cheers, Paul.
□ Hi there Florian, your sim looks very good. How many iterations are you using for the “bridge builder” style steel girder sim?
19. You lost me at the “Planes[4] =” part.
What do all the 1, 0, and -1 values mean? Why are there 12 numbers there?
“Each one has a normal (which points inwards)” — What does this mean?
□ Hi Wolter,
The planes array is describing the planes which make up the edges of the screen, the values {x,y, d} in each plane are {direction x, direction y, distance to plane}.
The first two components describe a unit length direction vector (also called a normal, sometimes), the third represents the distance from the plane to the origin
Hope that makes sense,
Cheers, Paul.
☆ Thanks for this clarification. There is one point that I’m still a little unclear on. Maybe you can confirm my understanding.
At first, I thought that the origin point of each of these normals was the middle of the screen. If that were the case, I think Planes[0] would represent the right plane, Planes[1] would
represent the bottom plane, and so on (clockwise).
However, after re-reading “the normals point inwards”, it seems to me that the middle of the screen is actually the endpoint of each normal. If this is the case, then Planes[0] would
represent the left plane, Planes[1] would represent the top plane, and so on (still clockwise).
Is my second interpretation correct? It makes sense, but it’s a bit different from other vector representations I’ve seen, so I want to make sure I’m understanding correctly.
○ Hi Jake,
The planes are
Planes[] =
The coordinate system in flash has down the screen = positive, which is why 0,-1 points up (and is hence the bottom plane).
Try not to think about the end-points of the normals – they’re just a unit length direction vector, so they would all exist exactly in the middle of the screen (0,0), except that we
add the distance parameter which ‘pushes’ them out along their direction vector by the distance to origin. e.g:
point on plane = direction vector * -distance
For the left plane (1, 0, 1):
point on plane = (1,0) * -distance = -1,0
For the right plane (-1, 0, 1):
point on plane = (-1,0) * -distance = 1,0
We use negative distance because the planes are all pointing inwards, so to approach them we must move against their direction vectors
Cheers, Paul.
20. Paul,
I’ve been enjoying your tutorials so far.
I’ve noticed in some of the demos that the circles go outside of the bounding boxes (or overlap, etc). When the change in x,y is greater than 1 pixel, this can happen. What can be done to fix
this ?
□ Hi Derrick,
Glad you’ve been enjoying the tutorials
Did you happen to read my article on continuous collision detection?
This covers a simple technique you can use to solve these issues
Cheers, Paul.
21. Hello, Paul. Thanks for tutorials. Sadly that you didn’t mention friction. Or maybe there will be separate tutorial?
□ Hi Volgogradetzzz,
I didn’t get around to friction yet – but don’t worry it will get covered eventually
Cheers, Paul.
i have done some correction to the points as the planes are rotated to make sure the points still lie within the planes`
Can you give us information about corrections you’ve done to make sure the points still lie within the planes?
I’m trying to code this examples in Pygame(Python), and having difficult times to hold points between planes.
□ if ( dist<0 )
if ( relNv<0 )
p.m_vel = p.m_vel.Sub( n.MulScalar( 2*relNv ) );
p.m_correct = n.MulScalar( -dist );
p.m_correct is added on in the integration phase, then zeroed
☆ Can you please more explanatory? What is relNv, p.m_vel, Sub(), m_correct, etc.
Actually it would be better if you just tell me the way you correct the positions of points
○ Sorry, I find the penetration distance for each point, multiply by the contact normal, store that value and then add it on to the position in the integration phase…
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24. This is a very informative and well written article.
Thank you for spending the time to educate us!
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26. This is a great article, it really made the math easy enough to implement. However, I although I have it implemented up to the coefficient of restitution, I don’t really understand how the math
relates to the macroscopic representation (i.e. the simulation). I’ve never taken linear algebra so although I can manipulate the math I don’t comprehend how it relates. If you are looking for
more material for articles in the future I would be extremely grateful if you would cover that topic.
Here is my (rudimentary) implementation of this in Javascript / HTML 5:
Click to make bunches of balls that float around. Press r to make them random colors. Press g to turn on gravity and watch them bounce.
□ Hi Brandon,
When you ask how the maths relates to the simulation, which part are you talking about? Is it just a software engineering question, or more of a mathsy one?
You demo look good, though!
Cheers, Paul.
27. Hi Paul. Thanks for the article – really inspiring.
One thing I would like to ask you to clarify is why add distance to the origin in “In order to detect if our particles have penetrated the planes we must perform a dot product and add the plane’s
distance to origin …” instead of subtracting from it?
Given that the normals are inverted as in originating at the origin and pointing toward the planes; the dot product, assuming normals are of unit length, gives component of particle vector onto
one of the normals. Why not subtract it from plane’s distance to the origin? In this case if a particle is closer to plane A than to plane B, distance to plane A is 1 – comp_An_ p and distance to
plane B is 1 + comp_Bn_ p because the component is negative now. Here An, An are normals to planes A and B respectively, p is a particle’s position vector.
Since your demos work and no-one complained about this part, I’m probably just rushing through the article and it is misunderstanding on my side (hope I didn’t make a fool of myself here
Thanks a bunch!
□ Hi Ivan,
The reason the distance is added on to the dot of the particle and the plane is that the planes all have their normals pointing towards the origin (so the world appears to be an inside out
box, with all normals pointing inwards):
Consider the left plane, at distance d=1 from the origin with the normal 1,0. It actually sits at -1 in the x axis, since its on the left.
Particle P at -0.5, 0.5 is 0.5 units distance from this plane.
P . N = -0.5
P . N + 1 = -0.5 + 1 = 0.5
Hope that helps,
Cheers, Paul.
28. This is a great article, it really made the math easy enough to implement. (with some work). Keep posting this kind of material . And i can’t wait to buy your book in the future.
29. Great article Paul, really takes alot of the confusion out of the whole collision & response ideas that I’ve been stuck on.
One question I have though, (and I don’t know if this would be compatible with this style of engine) would be to have bodies with bounding shapes made up of squares & circles, forming a custom
collision shape… how would this be done? or is it something completely different?
□ Hi Stephen,
Glad you enjoyed the article
If you want a custom collision shape it should be quite easy to add one – for example you could add another class which extends RigidBody called CompoundObject which would actually contain a
list of other shapes – it might be worth separating the direct relationship of Shape->RigidBody so that you can have a list of shapes corresponding to only one RigidBody (which is what you
Then you can just query the individual shapes within the CompoundObject to generate your contact information as before
Cheers, Paul.
30. Great article Paul. I have a few friends who are physicists and I think they would really enjoy this. I will have to share this with them.
31. Hi!
First of all, thanks for the great article. I’m new to game development and this article helped a lot.
Now for my newbie question, you said:
“GenerateContact() will generate a collision normal and distance between any two RigidBodys”
Then you have this code:
// collide
for ( var i:int=0; i0 )
const c:Contact=rbi.GenerateContact( rbj );
//resolve collision
In your 1st example collision of a particle in a plane, what is the collision normal and the distance?
Is the collision normal equal to the vector perpendicular to the plane? (which is always pointing to the origin (0, 0))?
Is the distance equal to 1? Is it because you declared the planes to be:
(1, 0, 1),
(0, -1, 1),
(-1, 0, 1),
(0, 1, 1)
Again, thanks a lot!
□ Hi Walter,
Yes, you have it exactly right
Cheers, Paul.
32. Hi Paul,
Another question regarding the “GenerateContact” method, say you have a class Plane and a class Circle, in its GenerateContact method, will it have the same logic? Please see code below:
public class Plane extends RigidBody {
public override function GenerateContact( rb:RigidBody ):Contact {
if ( rb is Plane )
else if ( rb is Circle )
//is the logic here the same as in the class Circle?
else if ( rb is Particle )
throw new Error("unhandled case!");
public class Circle extends RigidBody {
public override function GenerateContact( rb:RigidBody ):Contact {
if ( rb is Plane )
//is the logic here the same as in the class Plane?
else if ( rb is Circle )
else if ( rb is Particle )
throw new Error("unhandled case!");
Thanks again!
□ Hi walter,
The logic is the same, but the objects ra and rb are swapped around… This is what I do in the 2nd case:
if (rb is Plane)
var contact:Contact = rb->GenerateContact(this);
// flip normal, and A and B references around
So I avoid having to write the same code twice by simply swapping the inputs around
Cheers, Paul.
33. Hi Paul,
Thank you for this great article!
I have a question about your equations for the reflected velocity of two objects after collision:
Va = Va – (1+e)*N*(Va • N)*(Mb / (Ma+Mb))
Vb = Vb – (1+e)*-N*(Vb • -N)*(Ma / (Ma+Mb))
If Va=[1,0], Vb=[-1,0], N=[1,0], Ma=1, Mb=1, e=1.
I imagine the two objects’ velocity should be [-1,0] and [1,0] after collision, but your equation always returns 0. Am I doing something wrong?
Could you also explain a bit more about the equations below. Why there is a minus sign in Va and opposite in Vb?
I = (1+e)*N*(Vr • N)
Va = -I*(Mb / (Ma+Mb))
Vb = +I*(Ma / (Ma+Mb))
Excuse my poor English…
□ Hi Liy,
I think you found a mistake!
I think it should be:
Va = Va – (1+e)*N*((Vb-Va) • N)*(Mb / (Ma+Mb))
Vb = Vb – (1+e)*-N*((Vb-Va) • -N)*(Ma / (Ma+Mb))
But I would use the equations just below in the article; they are less complex to follow…
There is a – + here
Va = -I*(Mb / (Ma+Mb))
Vb = +I*(Ma / (Ma+Mb))
because we’re applying an impulse equal and oppositely to each object
Cheers, Paul.
☆ I’ll use your final impulse equations.
But I’m trying to understand your two equations:
Va = Va – (1+e)*N*((Vb-Va) • N)*(Mb / (Ma+Mb))
Vb = Vb – (1+e)*-N*((Vb-Va) • -N)*(Ma / (Ma+Mb))
Still quite confusing…
Can you show me how you plug the mass ratio into original R = V – (1+e)*N*(V•N) and formed up the two equations? Any hints will be much appreciated!
○ The mass ratios are Mb / (Ma+Mb) and Ma / (Ma+Mb), and simply scale the response for each body based on the relative masses. That’s why the reflection vector equation is multiplied by
the individual ratio for each. I guess they don’t ‘plug in’ exactly because there is no mass-ratio term in the reflection vector equation
Cheers, Paul.
34. Hi Paul,
Great articles. I have a couple of questions.
1. In modeling physics using constraints, is the reflection of a ball off of a wall or floor considered a ‘constraint’? I wasn’t sure if I simply used the reflection math at the collision or if I
should turn it into a constraint and solve it with the other constraints.
2. In the formula for reflection, there is “(1+e)”. I understand that e=coefficient of restitution of one of the objects. Does that mean “1″ is the COR of the other? So, if I wanted a floor that
deadened the reflection some, would I do “…(0.5+e)…”.
□ Hi Tim,
1) Reflection of a ball off a wall would be considered a contact, which is like a transient constraint which only exists for one frame
2) (1+e) is from the original reflection vector equation (well, it was 2 originally)… You need to combine the two CORs into one – how you combine the COR of two objects during a collision is
a bit hand-wavey and vague… I *think* in previous engines, I’ve just picked the minimum of the two
Cheers, Paul.
☆ Thanks. That helps.
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36. Hi Paul,
I’ve been working with the impulse based physics engine, but I’m a little stuck. What I noticed with the “contact” is that things seem to come to a stop instead of bounce (when not rotating).
I wanted to make a ball bounce, so I created a new type of contact called “bounce”. The challenge I had was that if I simply applied the reflection equation at the point where the motion
bounding-box collides, it doesn’t look like the ball hits the ground.
So, in my “bounce” contact, I modified the velocity based on your original “contact” to get the ball to the ground this frame and then I take the remaining reflection velocity and apply it the
next frame.
95% of the time, it works perfectly. However, every once in a while my “delayed velocity” adds a little extra jump to the ball. I started looking at the original “contact” again and noticed that
the ball does bounce, but only if it’s rotating.
I was hoping you could enlighten me a bit on the best way to get a ball to bounce (whether it’s rotating or not) while still having it look like it impacts the ground.
I hope this all makes sense.
□ Hi Tim,
Which source code are you using?
Having a non 0 coefficient of restitution is hard if you want to use speculative contacts. I’d recommend against trying to use them together because they’re not really suited…
At a push I would suggest you try and do what you have already attempted – though I’ve never done this myself!
Sorry I can’t be more help,
Cheers, Paul.
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solving for delta, formal limit definition
January 28th 2009, 09:34 AM #1
Jan 2009
solving for delta, formal limit definition
We started on this in class today and my professor gave no more than a quick glance at how to solve and possible questions that would arise when trying to complete the homework.
When graphed and explained the proof made perfect sense. It is when I try to complete the homework that I run into problems. The instructions for the problem read:
Numerically and graphically determine a delta corresponding to (a) epsilon = 0.1 and (b) epsilon = 0.05. Graph the function in the epsilon - delta window [xrange is (a-delta, a+delta) and yrange
is (L-epsilon, L+epsilon)] to verify that your choice works.
1) lim of (x^2 + 1) as x approaches 0 (the value of L is given in these problems, and is of course 1 in this case)
The only way in which the professor showed us how to solve the problems is such:
the proof states that |F(x) - L| < epsilon
so my problem is such:
|[x^2 + 1] - 1| < 0.1
We break down the absolute value down into:
-0.1 < (x^2 + 1) - 1 < 0.1
which ends up looking like this:
-0.1 < X^2 < 0.1
Basically from there he jumped to solving X^2 = 0.1 to get x = (0.1)^1/2, which is the correct answer. Now for all problems in which one of the sides of the inequality is outside the range of the
function (ex. -1.1 as a value for cosx, or -0.1 as a value for x^2), this gives me the correct answer...however it is obvious to me that I'm missing a piece of the puzzle.
The second problem on the homework is this:
lim of (x^2 + 1) as x approaches 2, L=5 epsilon=0.1
i start with |(x^2 + 1) - 5|<0.1 which breaks down into
3.9 < X^2 < 4.1, now where do I go from here to find a correct delta?
I've tried reading the book (which is useless), my companion ("The Calculus Lifesaver"), and the internet and none of the resources seem to explain to me how to solve the problem...at least in a
fashion that I can grasp.
We started on this in class today and my professor gave no more than a quick glance at how to solve and possible questions that would arise when trying to complete the homework.
When graphed and explained the proof made perfect sense. It is when I try to complete the homework that I run into problems. The instructions for the problem read:
1) lim of (x^2 + 1) as x approaches 0 (the value of L is given in these problems, and is of course 1 in this case)
The only way in which the professor showed us how to solve the problems is such:
the proof states that |F(x) - L| < epsilon
so my problem is such:
|[x^2 + 1] - 1| < 0.1
We break down the absolute value down into:
-0.1 < (x^2 + 1) - 1 < 0.1
which ends up looking like this:
-0.1 < X^2 < 0.1
Basically from there he jumped to solving X^2 = 0.1 to get x = (0.1)^1/2, which is the correct answer. Now for all problems in which one of the sides of the inequality is outside the range of the
function (ex. -1.1 as a value for cosx, or -0.1 as a value for x^2), this gives me the correct answer...however it is obvious to me that I'm missing a piece of the puzzle.
The second problem on the homework is this:
lim of (x^2 + 1) as x approaches 2, L=5 epsilon=0.1
i start with |(x^2 + 1) - 5|<0.1 which breaks down into
3.9 < X^2 < 4.1, now where do I go from here to find a correct delta?
I've tried reading the book (which is useless), my companion ("The Calculus Lifesaver"), and the internet and none of the resources seem to explain to me how to solve the problem...at least in a
fashion that I can grasp.
Take the square root of the inequality
$1.9748 < x < 2.0248$
$-.0251< x - 2 < 0.0248$ and for $| x - 2 | < \delta$
so $\delta = \min\{.0251,.0248\}$
Why does the 2 come back into the equation? Because originally it gives you the points around the value x is approaching (2 in this case) and you need to find the distance from the value a and
not the actual x values? So that I end up with a range of correct deltas (0.0248 to 0.0251) in this case?
I'll post another problem and maybe I'm doing it correctly.
limit of (x+3)^1/2 as x approaches 1, epsilon=0.1
|(x+3)^1/2 - 2| < 0.1
-0.1 < (x+3)^1/2 - 2 < 0.1
1.9 < (x+3)^1/2 < 2.1
(1.9)^2 < (x+3) < (2.1)^2
(1.9)^2 - 3 < x < (2.1)^2 - 3
0.61 < X < 1.41
0.61 < x-1 < 1.41
|0.61 - 1| = 0.39 and |1.41-1| = 0.41
The book only gives 0.39 as the answer, but if I understand correctly both are correct and the book just chooses the smaller delta of the two?
Why does the 2 come back into the equation? Because originally it gives you the points around the value x is approaching (2 in this case) and you need to find the distance from the value a and
not the actual x values? So that I end up with a range of correct deltas (0.0248 to 0.0251) in this case?
I'll post another problem and maybe I'm doing it correctly.
limit of (x+3)^1/2 as x approaches 1, epsilon=0.1
|(x+3)^1/2 - 2| < 0.1
-0.1 < (x+3)^1/2 - 2 < 0.1
1.9 < (x+3)^1/2 < 2.1
(1.9)^2 < (x+3) < (2.1)^2
(1.9)^2 - 3 < x < (2.1)^2 - 3
0.61 < X < 1.41
0.61 < x-1 < 1.41
|0.61 - 1| = 0.39 and |1.41-1| = 0.41
The book only gives 0.39 as the answer, but if I understand correctly both are correct and the book just chooses the smaller delta of the two?
Actually, both don't work. Only the smaller of the two. Let's consider both cases.
Case 1: $\delta = 0.39$
Here the interval is $[.61, 1.39]$. Substituting the endpoints into the function gives
$\sqrt{.61 + 3} = 1.900,\;\;\;\sqrt{1.39 + 3} = 2.0952,$
Here, both are within the given L, (L = 0.1)
Case 2: $\delta = 0.41$
Here the interval is $[.59, 1.41]$. Substituting the endpoints into the function gives
$\sqrt{.59+ 3} = 1.8947,\;\;\;\sqrt{1.41 + 3} = 2.1,$
Here, both are not within the given L (L=0.1). The first one is outside.
See the difference?
I do, makes much more sense now. Thank you!
one final question. In the event that I have an inequality such as
0.9 < cosx < 1.1
-0.1 < x2 < 0.1
Is there any special way to show why I disregard part of the inequality (the square root of a negative number, or a value outside the range of cosine) or can it generally just be assumed the
person will understand why? This seems like an instructor specific question but I prefer to know any and all steps so that I can show all of my work when possible...and because I'll be the one
explaining how to work the homework to the class before the professor shows up in the morning
one final question. In the event that I have an inequality such as
0.9 < cosx < 1.1
-0.1 < x2 < 0.1
Is there any special way to show why I disregard part of the inequality (the square root of a negative number, or a value outside the range of cosine) or can it generally just be assumed the
person will understand why? This seems like an instructor specific question but I prefer to know any and all steps so that I can show all of my work when possible...and because I'll be the one
explaining how to work the homework to the class before the professor shows up in the morning
You might have some problems with those examples because there are no x values such that
$\cos x = 1.1$ or $x^2 = - 0.1$
I'd stick with problems where the range of the function is defined.
January 28th 2009, 09:59 AM #2
January 28th 2009, 10:26 AM #3
Jan 2009
January 28th 2009, 10:39 AM #4
January 28th 2009, 10:45 AM #5
Jan 2009
January 28th 2009, 11:28 AM #6
Jan 2009
January 28th 2009, 12:28 PM #7
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Topic: Need help on this interesting fact
Replies: 3 Last Post: Mar 14, 2005 10:50 PM
Messages: [ Previous | Next ]
Re: Need help on this interesting fact
Posted: Oct 21, 2003 4:54 PM
On Tue, 21 Oct 2003 18:42:32 GMT, MB <mel@invalid.invalid> wrote, in part:
> I saw this on a t-shirt:
> (pi^4 + pi^5)^(1/6) = e
> This piqued my curiosity. I confirmed the truism on my calculator, but I
> can't seem to prove it.
> Can someone supply the proof or give me some hints?
Apparently it's only approximately true: see
(near the top; use your Web browser to search in the page for the string
'pi^' (no quotes)).
Michael Hamm Since mid-September of 2003,
BA scl Math, PBK, NYU I've been erasing too much UBE.
msh210@math.wustl.edu Of a reply, then, if you have been cheated,
http://math.wustl.edu/~msh210/ Likely your mail's by mistake been deleted.
submissions: post to k12.ed.math or e-mail to k12math@k12groups.org
private e-mail to the k12.ed.math moderator: kem-moderator@k12groups.org
newsgroup website: http://www.thinkspot.net/k12math/
newsgroup charter: http://www.thinkspot.net/k12math/charter.html
Date Subject Author
10/21/03 Need help on this interesting fact MB
10/21/03 Re: Need help on this interesting fact Michael Hamm
10/21/03 Re: Need help on this interesting fact Kevin Karplus
3/14/05 Re: Need help on this interesting fact Fred
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Category: algorithms Component type: function
template <class ForwardIterator, class T>
void iota(ForwardIterator first, ForwardIterator last, T value);
Iota assigns sequentially increasing values to a range. That is, it assigns value to *first, value + 1 to *(first + 1) and so on. In general, each iterator i in the range [first, last) is assigned
value + (i - first). [1]
Defined in the standard header numeric, and in the nonstandard backward-compatibility header algo.h. This function is an SGI extension; it is not part of the C++ standard.
Requirements on types
• ForwardIterator is a model of Forward Iterator.
• ForwardIterator is mutable.
• T is Assignable.
• If x is an object of type T, then x++ is defined.
• T is convertible to ForwardIterator's value type.
• [first, last) is a valid range.
Linear. Exactly last - first assignments.
int main()
vector<int> V(10);
iota(V.begin(), V.end(), 7);
copy(V.begin(), V.end(), ostream_iterator<int>(cout, " "));
cout << endl;
[1] The name iota is taken from the programming language APL.
See also
fill, generate, partial_sum
STL Main Page
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Column Addition: Addition with Regrouping (Traditional Method)
It’s an exciting time when your child begins to explore some of the more complicated 2^nd grade math skills. Addition with Regrouping is a concept that can be a bit tricky for children to pick up. As
a parent, there are several things you can do at home to help reinforce this new math skill.
What is Addition with Regrouping?
Addition with Regrouping is adding a 2-digit number (or larger) to other 2-digit (or larger) numbers, such that at least one of the individual columns adds up to 10 or more. You might have called it
Addition with Carrying when you were in school.
In a previous blog, I explained how children begin Column Addition with simple 2-digit numbers such as the equation below. In this example, the child has two options.
He could use the Traditional Method to solve it, or he could use the Partial Sums Method. Today, we are going to discuss the last step in learning Column Addition – when regrouping (or carrying) is
• It is important that children understand the concept of Place Value. Be sure your child understands that the column to the far right is called the Ones column, and the next column over is called
the Tens column. Take a piece of paper and draw a line down the middle. Write “Tens” in the left column, and “Ones” in the right column. Write the Addition with Regrouping problems on this
paper so your child sees that some numbers fall into the Tens column, while others fall into the Ones column.
• Special blocks called “Base Ten Blocks” are useful when teaching this skill. Most parents don’t have these blocks at home, so you can make your own by taking drinking straws and a marker and
marking 10 lines on each straw at equal intervals. You will want to do this on about 12 straws. Cut ONE of the straws into the 10 little bits. The full-length straws will represent the “Longs,”
or the groups of 10, and the little bits will represent the Ones.
• Ask your child to “build” an Addition question such as 27 + 45 using the straw “Longs” and the straw “Bits”.
1. Using a blank place value chart (see first bullet point), your child would place 2 straw “Longs” in the Tens column, and 7 straw “Bits” in the Ones column (to represent 27).
2. Directly under this, he would place 4 straw “Longs” in the Tens column, and 5 straw “Bits” in the Ones column (to represent 45).
3. Explain to your child that he always starts in the Ones column when using the Traditional Method. He adds 7 + 5 “Bits” and gets an answer of 12. Tell him that the number 12 is larger than
10, so we have to send some of those Bits over to the Tens column. Remind him that only groups of 10 are allowed in the Tens column (No Bits allowed!).
4. Your child should then count out 10 Bits or Ones, and trade them in for a “Long,” which he then places in the Tens column. He should have 2 “Bits” leftover in the Ones column.
5. He is now ready to add the “Longs” in the Tens column. 2 Longs + 4 Longs + the 1 Long we brought over from the Ones column = 7 Longs, or 70. The 7 Longs with the 2 Bits (Ones) makes an
answer of 72.
• Once your child understands the concept of Addition with Regrouping, he will be ready to do the same method on paper without using the straws.
1. Write the equation in column form (one number on top of the other).
Remind him to begin in the Ones column. He would add 7+5, and get the answer of 12.
2. Remind him that we can’t put 12 in the Ones column, that we have to send the bundle of 10 (the Long) over to the Tens column. That leaves 2 in the Ones Column. He should write 2 under the
3. Show him how we put a small “1” on top of the Tens column to represent the Long that we brought over from the Ones column.
4. Now add up all the digits in the Tens column and write the answer below the line.
What to Watch For:
• If your child has learned both the Partial Sums Method and the Traditional Method of Column Addition, watch that he doesn’t confuse the two methods. The Partial Sums Method has the child
beginning in the Tens Column, whereas the Traditional Method begins in the Ones column.
• Also, the Partial Sums Method has the child thinking about numbers in the Tens column as real values of that number as opposed to a digit (ex: the number 2 in the Tens column is worth 20 and the
child thinks of this number as 20). This can sometimes be confusing for children as they move to the Traditional Method.
• When your child first sees an Addition problem such as 79 + 54, he may be confused because the final answer spills over into the Hundreds column. You will need to talk about the Hundreds column
so he knows where to “carry the 1.”
• Have questions or ideas about this story?
• Need help or advice about your child’s learning?
• Have ideas for future Parent Homework Help stories?
Go to “Leave a Reply” at the bottom of this page. I’d love to help.
Leave a Comment
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November 18th 2008, 02:41 PM
To cancel an electronic alarm, a 5-digit code number must be entered into the code box. Assuming that digits may be repeated, how many codes are possible?
November 18th 2008, 04:33 PM
There are ten digits: 0, 1, 2,...8, 9
Think of this question from perspective of the person who sets the code.
When he picks the first of the 5 digits, how many choices does he have? Exactly 10 choices!
When he picks the second of his 5 digits, he can again choose from 10 options.
For digits 3, 4 and 5 he also has 10 choices.
So he has 10*10*10*10*10 choices or 10^5.
To contract: if repetition is not allowed, there are
10*9*8*7*6 choices.
For the first digit he can choose any number 0-9.
For the second digit he can choose any number 0-9 EXCEPT the number he chose for the first digit, so he has 9 options.
And so on.
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Function extension in a Sobolev space
up vote 1 down vote favorite
Let $\Omega$ be a domain of $R^n$ and let $H^2(\Omega)$ be the usual Sobolev space.
Let $\emptyset\ne \omega_1\subset\omega_2$ be open subsets of $\Omega$, and let $\theta \in H^2(\omega_1)$.
I am wondering about the existence of a function $\tilde{\theta} \in H^2(\Omega)$ such that :
1) $\tilde{\theta}=\theta$ on $\omega_1$,
2) $\tilde{\theta} $ is constant on $\Omega-\omega_2$.
sobolev-spaces fa.functional-analysis
add comment
1 Answer
active oldest votes
The answer is yes, with the caveat indicated in the comment below. Consider an arbitrary extension $\hat{\theta}\in H^2(\Omega)$. Now choose a compactly supported smooth function $\
eta$ such that
up vote 1 down $${\rm supp}\;\eta\subset \Omega\setminus \omega_2,\;\;\eta\equiv 1 \;\;\mbox{on $\omega_1$}. $$
The function $\tilde{\theta}=\eta\cdot \hat{\theta}$ has the properties you asked for.
Provided that the function $\theta \in H^2(\omega_1)$ has a Sobolev-extension to begin with. – Tapio Rajala Jul 13 '12 at 14:07
If there is no extension, then the answer is obviously negative. However extensions do exist under mild regularity assumption. See e.g. Adams' book on Sobolev spaces. – Liviu
Nicolaescu Jul 13 '12 at 14:36
add comment
Not the answer you're looking for? Browse other questions tagged sobolev-spaces fa.functional-analysis or ask your own question.
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Instant Analysis: Closing Address by Justice Scalia at #FedSoc2012
Posted in Uncategorized
Instant Analysis: Closing Address by Justice Scalia at #FedSoc2012
I am over my tweet limit?! My running comments of Justice Scala’s address, will begin at 4:15 till 5:00 here.
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
“Court must follow statute, but Congress can’t enact general law to
dictate rules of interpretation that courts must apply. “that’s our
job” ”
I believe that one of the first Acts of Congress, the Judiciary Act of 1789, enacted “general law to dictate rules of interpretation.”
I know for a fact that Congress can, in any given statute, spell out how the text should be interpreted (or not). I fail to see why Congress lacks the authority to declare how the words that it
duly enacts into law should be construed. For instance, why can it not lay down a principle that, when there is doubt, a particular point of view should prevail?
|
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|
Probability: A Philosophical Introduction
Synopses & Reviews
Publisher Comments:
Probability: A Philosophical Introduction introduces and explains the principal concepts and applications of probability. It is intended for philosophers and others who want to understand probability
as we all apply it in our working and everyday lives. The book is not a course in mathematical probability, of which it uses only the simplest results, and avoids all needless technicality.
The role of probability in modern theories of knowledge, inference, induction, causation, laws of nature, action and decision-making makes an understanding of it especially important to philosophers
and students of philosophy, to whom this book will be invaluable both as a textbook and a work of reference.
In this book D. H. Mellor discusses the three basic kinds of probability physical, epistemic, and subjective and introduces and assesses the main theories and interpretations of them. The topics and
concepts covered include:
* chance
* frequency
* possibility
* propensity
* credence
* confirmation
* Bayesianism.
Probability: A Philosophical Introduction is essential reading for all philosophy students and others who encounter or need to apply ideas of probability.
"Probability: A Philosophical Introduction "presents the basic concepts of probability to philosophy students who are new to this area of the subject.
Probability plays an important part in many areas of philosophy, such as causation in metaphysics, induction in philosophy of science, and belief and justification in epistemology. This book:
* assumes no mathematical background and keeps the technicalities to a minimum
* explains the most important applications of probability theory to philosophy
* outlines the different kinds of probability and assesses the strengths and weaknesses of each
* introduces and discusses key terms such as propensity, frequency, evidence and credence
D.H. Mellor presents and compares theories of probability in a simple and impartial way, making "Probability: A Philosophical" "Introduction" essential reading for any philosophy student confronted
with probability for the first time.
"Probability: A Philosophical Introduction "presents the basic concepts of probability to philosophy students who are new to this area of the subject. Probability plays an important part in many
areas of philosophy, such as causation in metaphysics, induction in philosophy of science, and belief and justification in epistemology. This book: * assumes no mathematical background and keeps the
technicalities to a minimum * explains the most important applications of probability theory to philosophy * outlines the different kinds of probability and assesses the strengths and weaknesses of
each * introduces and discusses key terms such as propensity, frequency, evidence and credence D.H. Mellor presents and compares theories of probability in a simple and impartial way, making
"Probability: A Philosophical" "Introduction" essential reading for any philosophy student confronted with probability for the first time.
What Our Readers Are Saying
Be the first to add a comment for a chance to win!
|
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|
Provided by:
rrdcreate - Set up a new Round Robin Database
rrdtool create filename [--start|-b start time] [--step|-s step]
[--no-overwrite] [DS:ds-name:DST:dst arguments] [RRA:CF:cf arguments]
The create function of RRDtool lets you set up new Round Robin Database
(RRD) files. The file is created at its final, full size and filled
with *UNKNOWN* data.
The name of the RRD you want to create. RRD files should end with the
extension .rrd. However, RRDtool will accept any filename.
--start|-b start time (default: now - 10s)
Specifies the time in seconds since 1970-01-01 UTC when the first value
should be added to the RRD. RRDtool will not accept any data timed
before or at the time specified.
See also AT-STYLE TIME SPECIFICATION section in the rrdfetch
documentation for other ways to specify time.
--step|-s step (default: 300 seconds)
Specifies the base interval in seconds with which data will be fed into
the RRD.
Do not clobber an existing file of the same name.
DS:ds-name:DST:dst arguments
A single RRD can accept input from several data sources (DS), for
example incoming and outgoing traffic on a specific communication line.
With the DS configuration option you must define some basic properties
of each data source you want to store in the RRD.
ds-name is the name you will use to reference this particular data
source from an RRD. A ds-name must be 1 to 19 characters long in the
characters [a-zA-Z0-9_].
DST defines the Data Source Type. The remaining arguments of a data
source entry depend on the data source type. For GAUGE, COUNTER,
DERIVE, and ABSOLUTE the format for a data source entry is:
DS:ds-name:GAUGE | COUNTER | DERIVE | ABSOLUTE:heartbeat:min:max
For COMPUTE data sources, the format is:
In order to decide which data source type to use, review the
definitions that follow. Also consult the section on "HOW TO MEASURE"
for further insight.
is for things like temperatures or number of people in a room or
the value of a RedHat share.
is for continuous incrementing counters like the ifInOctets counter
in a router. The COUNTER data source assumes that the counter never
decreases, except when a counter overflows. The update function
takes the overflow into account. The counter is stored as a per-
second rate. When the counter overflows, RRDtool checks if the
overflow happened at the 32bit or 64bit border and acts accordingly
by adding an appropriate value to the result.
will store the derivative of the line going from the last to the
current value of the data source. This can be useful for gauges,
for example, to measure the rate of people entering or leaving a
room. Internally, derive works exactly like COUNTER but without
overflow checks. So if your counter does not reset at 32 or 64 bit
you might want to use DERIVE and combine it with a MIN value of 0.
NOTE on COUNTER vs DERIVE
by Don Baarda <don.baarda@baesystems.com>
If you cannot tolerate ever mistaking the occasional counter reset
for a legitimate counter wrap, and would prefer "Unknowns" for all
legitimate counter wraps and resets, always use DERIVE with min=0.
Otherwise, using COUNTER with a suitable max will return correct
values for all legitimate counter wraps, mark some counter resets
as "Unknown", but can mistake some counter resets for a legitimate
counter wrap.
For a 5 minute step and 32-bit counter, the probability of
mistaking a counter reset for a legitimate wrap is arguably about
0.8% per 1Mbps of maximum bandwidth. Note that this equates to 80%
for 100Mbps interfaces, so for high bandwidth interfaces and a
32bit counter, DERIVE with min=0 is probably preferable. If you are
using a 64bit counter, just about any max setting will eliminate
the possibility of mistaking a reset for a counter wrap.
is for counters which get reset upon reading. This is used for fast
counters which tend to overflow. So instead of reading them
normally you reset them after every read to make sure you have a
maximum time available before the next overflow. Another usage is
for things you count like number of messages since the last update.
is for storing the result of a formula applied to other data
sources in the RRD. This data source is not supplied a value on
update, but rather its Primary Data Points (PDPs) are computed from
the PDPs of the data sources according to the rpn-expression that
defines the formula. Consolidation functions are then applied
normally to the PDPs of the COMPUTE data source (that is the rpn-
expression is only applied to generate PDPs). In database software,
such data sets are referred to as "virtual" or "computed" columns.
heartbeat defines the maximum number of seconds that may pass between
two updates of this data source before the value of the data source is
assumed to be *UNKNOWN*.
min and max define the expected range values for data supplied by a
data source. If min and/or max any value outside the defined range will
be regarded as *UNKNOWN*. If you do not know or care about min and max,
set them to U for unknown. Note that min and max always refer to the
processed values of the DS. For a traffic-COUNTER type DS this would be
the maximum and minimum data-rate expected from the device.
If information on minimal/maximal expected values is available, always
set the min and/or max properties. This will help RRDtool in doing a
simple sanity check on the data supplied when running update.
rpn-expression defines the formula used to compute the PDPs of a
COMPUTE data source from other data sources in the same <RRD>. It is
similar to defining a CDEF argument for the graph command. Please refer
to that manual page for a list and description of RPN operations
supported. For COMPUTE data sources, the following RPN operations are
not supported: COUNT, PREV, TIME, and LTIME. In addition, in defining
the RPN expression, the COMPUTE data source may only refer to the names
of data source listed previously in the create command. This is similar
to the restriction that CDEFs must refer only to DEFs and CDEFs
previously defined in the same graph command.
RRA:CF:cf arguments
The purpose of an RRD is to store data in the round robin archives
(RRA). An archive consists of a number of data values or statistics for
each of the defined data-sources (DS) and is defined with an RRA line.
When data is entered into an RRD, it is first fit into time slots of
the length defined with the -s option, thus becoming a primary data
The data is also processed with the consolidation function (CF) of the
archive. There are several consolidation functions that consolidate
primary data points via an aggregate function: AVERAGE, MIN, MAX, LAST.
the average of the data points is stored.
MIN the smallest of the data points is stored.
MAX the largest of the data points is stored.
the last data points is used.
Note that data aggregation inevitably leads to loss of precision and
information. The trick is to pick the aggregate function such that the
interesting properties of your data is kept across the aggregation
The format of RRA line for these consolidation functions is:
RRA:AVERAGE | MIN | MAX | LAST:xff:steps:rows
xff The xfiles factor defines what part of a consolidation interval may
be made up from *UNKNOWN* data while the consolidated value is still
regarded as known. It is given as the ratio of allowed *UNKNOWN* PDPs
to the number of PDPs in the interval. Thus, it ranges from 0 to 1
steps defines how many of these primary data points are used to build a
consolidated data point which then goes into the archive.
rows defines how many generations of data values are kept in an RRA.
Obviously, this has to be greater than zero.
Aberrant Behavior Detection with Holt-Winters Forecasting
In addition to the aggregate functions, there are a set of specialized
functions that enable RRDtool to provide data smoothing (via the Holt-
Winters forecasting algorithm), confidence bands, and the flagging
aberrant behavior in the data source time series:
o RRA:HWPREDICT:rows:alpha:beta:seasonal period[:rra-num]
o RRA:MHWPREDICT:rows:alpha:beta:seasonal period[:rra-num]
o RRA:SEASONAL:seasonal period:gamma:rra-
o RRA:DEVSEASONAL:seasonal period:gamma:rra-
o RRA:DEVPREDICT:rows:rra-num
o RRA:FAILURES:rows:threshold:window length:rra-num
These RRAs differ from the true consolidation functions in several
ways. First, each of the RRAs is updated once for every primary data
point. Second, these RRAs are interdependent. To generate real-time
confidence bounds, a matched set of SEASONAL, DEVSEASONAL, DEVPREDICT,
and either HWPREDICT or MHWPREDICT must exist. Generating smoothed
values of the primary data points requires a SEASONAL RRA and either an
HWPREDICT or MHWPREDICT RRA. Aberrant behavior detection requires
FAILURES, DEVSEASONAL, SEASONAL, and either HWPREDICT or MHWPREDICT.
The predicted, or smoothed, values are stored in the HWPREDICT or
MHWPREDICT RRA. HWPREDICT and MHWPREDICT are actually two variations on
the Holt-Winters method. They are interchangeable. Both attempt to
decompose data into three components: a baseline, a trend, and a
seasonal coefficient. HWPREDICT adds its seasonal coefficient to the
baseline to form a prediction, whereas MHWPREDICT multiplies its
seasonal coefficient by the baseline to form a prediction. The
difference is noticeable when the baseline changes significantly in the
course of a season; HWPREDICT will predict the seasonality to stay
constant as the baseline changes, but MHWPREDICT will predict the
seasonality to grow or shrink in proportion to the baseline. The proper
choice of method depends on the thing being modeled. For simplicity,
the rest of this discussion will refer to HWPREDICT, but MHWPREDICT may
be substituted in its place.
The predicted deviations are stored in DEVPREDICT (think a standard
deviation which can be scaled to yield a confidence band). The FAILURES
RRA stores binary indicators. A 1 marks the indexed observation as
failure; that is, the number of confidence bounds violations in the
preceding window of observations met or exceeded a specified threshold.
An example of using these RRAs to graph confidence bounds and failures
appears in rrdgraph.
The SEASONAL and DEVSEASONAL RRAs store the seasonal coefficients for
the Holt-Winters forecasting algorithm and the seasonal deviations,
respectively. There is one entry per observation time point in the
seasonal cycle. For example, if primary data points are generated every
five minutes and the seasonal cycle is 1 day, both SEASONAL and
DEVSEASONAL will have 288 rows.
In order to simplify the creation for the novice user, in addition to
supporting explicit creation of the HWPREDICT, SEASONAL, DEVPREDICT,
DEVSEASONAL, and FAILURES RRAs, the RRDtool create command supports
implicit creation of the other four when HWPREDICT is specified alone
and the final argument rra-num is omitted.
rows specifies the length of the RRA prior to wrap around. Remember
that there is a one-to-one correspondence between primary data points
and entries in these RRAs. For the HWPREDICT CF, rows should be larger
than the seasonal period. If the DEVPREDICT RRA is implicitly created,
the default number of rows is the same as the HWPREDICT rows argument.
If the FAILURES RRA is implicitly created, rows will be set to the
seasonal period argument of the HWPREDICT RRA. Of course, the RRDtool
resize command is available if these defaults are not sufficient and
the creator wishes to avoid explicit creations of the other specialized
function RRAs.
seasonal period specifies the number of primary data points in a
seasonal cycle. If SEASONAL and DEVSEASONAL are implicitly created,
this argument for those RRAs is set automatically to the value
specified by HWPREDICT. If they are explicitly created, the creator
should verify that all three seasonal period arguments agree.
alpha is the adaption parameter of the intercept (or baseline)
coefficient in the Holt-Winters forecasting algorithm. See rrdtool for
a description of this algorithm. alpha must lie between 0 and 1. A
value closer to 1 means that more recent observations carry greater
weight in predicting the baseline component of the forecast. A value
closer to 0 means that past history carries greater weight in
predicting the baseline component.
beta is the adaption parameter of the slope (or linear trend)
coefficient in the Holt-Winters forecasting algorithm. beta must lie
between 0 and 1 and plays the same role as alpha with respect to the
predicted linear trend.
gamma is the adaption parameter of the seasonal coefficients in the
Holt-Winters forecasting algorithm (HWPREDICT) or the adaption
parameter in the exponential smoothing update of the seasonal
deviations. It must lie between 0 and 1. If the SEASONAL and
DEVSEASONAL RRAs are created implicitly, they will both have the same
value for gamma: the value specified for the HWPREDICT alpha argument.
Note that because there is one seasonal coefficient (or deviation) for
each time point during the seasonal cycle, the adaptation rate is much
slower than the baseline. Each seasonal coefficient is only updated (or
adapts) when the observed value occurs at the offset in the seasonal
cycle corresponding to that coefficient.
If SEASONAL and DEVSEASONAL RRAs are created explicitly, gamma need not
be the same for both. Note that gamma can also be changed via the
RRDtool tune command.
smoothing-window specifies the fraction of a season that should be
averaged around each point. By default, the value of smoothing-window
is 0.05, which means each value in SEASONAL and DEVSEASONAL will be
occasionally replaced by averaging it with its (seasonal period*0.05)
nearest neighbors. Setting smoothing-window to zero will disable the
running-average smoother altogether.
rra-num provides the links between related RRAs. If HWPREDICT is
specified alone and the other RRAs are created implicitly, then there
is no need to worry about this argument. If RRAs are created
explicitly, then carefully pay attention to this argument. For each RRA
which includes this argument, there is a dependency between that RRA
and another RRA. The rra-num argument is the 1-based index in the order
of RRA creation (that is, the order they appear in the create command).
The dependent RRA for each RRA requiring the rra-num argument is listed
o HWPREDICT rra-num is the index of the SEASONAL RRA.
o SEASONAL rra-num is the index of the HWPREDICT RRA.
o DEVPREDICT rra-num is the index of the DEVSEASONAL RRA.
o DEVSEASONAL rra-num is the index of the HWPREDICT RRA.
o FAILURES rra-num is the index of the DEVSEASONAL RRA.
threshold is the minimum number of violations (observed values outside
the confidence bounds) within a window that constitutes a failure. If
the FAILURES RRA is implicitly created, the default value is 7.
window length is the number of time points in the window. Specify an
integer greater than or equal to the threshold and less than or equal
to 28. The time interval this window represents depends on the
interval between primary data points. If the FAILURES RRA is implicitly
created, the default value is 9.
The HEARTBEAT and the STEP
Here is an explanation by Don Baarda on the inner workings of RRDtool.
It may help you to sort out why all this *UNKNOWN* data is popping up
in your databases:
RRDtool gets fed samples/updates at arbitrary times. From these it
builds Primary Data Points (PDPs) on every "step" interval. The PDPs
are then accumulated into the RRAs.
The "heartbeat" defines the maximum acceptable interval between
samples/updates. If the interval between samples is less than
"heartbeat", then an average rate is calculated and applied for that
interval. If the interval between samples is longer than "heartbeat",
then that entire interval is considered "unknown". Note that there are
other things that can make a sample interval "unknown", such as the
rate exceeding limits, or a sample that was explicitly marked as
The known rates during a PDP's "step" interval are used to calculate an
average rate for that PDP. If the total "unknown" time accounts for
more than half the "step", the entire PDP is marked as "unknown". This
means that a mixture of known and "unknown" sample times in a single
PDP "step" may or may not add up to enough "known" time to warrant a
known PDP.
The "heartbeat" can be short (unusual) or long (typical) relative to
the "step" interval between PDPs. A short "heartbeat" means you require
multiple samples per PDP, and if you don't get them mark the PDP
unknown. A long heartbeat can span multiple "steps", which means it is
acceptable to have multiple PDPs calculated from a single sample. An
extreme example of this might be a "step" of 5 minutes and a
"heartbeat" of one day, in which case a single sample every day will
result in all the PDPs for that entire day period being set to the same
average rate. -- Don Baarda <don.baarda@baesystems.com>
u|02|----* sample1, restart "hb"-timer
u|03| /
u|04| /
u|05| /
u|06|/ "hbt" expired
|08|----* sample2, restart "hb"
|09| /
|10| /
u|11|----* sample3, restart "hb"
u|12| /
u|13| /
step1_u|14| /
u|15|/ "swt" expired
|17|----* sample4, restart "hb", create "pdp" for step1 =
|18| / = unknown due to 10 "u" labled secs > 0.5 * step
|19| /
|20| /
|21|----* sample5, restart "hb"
|22| /
|23| /
|24|----* sample6, restart "hb"
|25| /
|26| /
|27|----* sample7, restart "hb"
step2__|28| /
|22| /
|23|----* sample8, restart "hb", create "pdp" for step1, create "cdp"
|24| /
|25| /
graphics by vladimir.lavrov@desy.de.
Here are a few hints on how to measure:
Usually you have some type of meter you can read to get the
temperature. The temperature is not really connected with a time.
The only connection is that the temperature reading happened at a
certain time. You can use the GAUGE data source type for this.
RRDtool will then record your reading together with the time.
Mail Messages
Assume you have a method to count the number of messages
transported by your mail server in a certain amount of time, giving
you data like '5 messages in the last 65 seconds'. If you look at
the count of 5 like an ABSOLUTE data type you can simply update the
RRD with the number 5 and the end time of your monitoring period.
RRDtool will then record the number of messages per second. If at
some later stage you want to know the number of messages
transported in a day, you can get the average messages per second
from RRDtool for the day in question and multiply this number with
the number of seconds in a day. Because all math is run with
Doubles, the precision should be acceptable.
It's always a Rate
RRDtool stores rates in amount/second for COUNTER, DERIVE and
ABSOLUTE data. When you plot the data, you will get on the y axis
amount/second which you might be tempted to convert to an absolute
amount by multiplying by the delta-time between the points. RRDtool
plots continuous data, and as such is not appropriate for plotting
absolute amounts as for example "total bytes" sent and received in
a router. What you probably want is plot rates that you can scale
to bytes/hour, for example, or plot absolute amounts with another
tool that draws bar-plots, where the delta-time is clear on the
plot for each point (such that when you read the graph you see for
example GB on the y axis, days on the x axis and one bar for each
rrdtool create temperature.rrd --step 300 \
DS:temp:GAUGE:600:-273:5000 \
RRA:AVERAGE:0.5:1:1200 \
RRA:MIN:0.5:12:2400 \
RRA:MAX:0.5:12:2400 \
This sets up an RRD called temperature.rrd which accepts one
temperature value every 300 seconds. If no new data is supplied for
more than 600 seconds, the temperature becomes *UNKNOWN*. The minimum
acceptable value is -273 and the maximum is 5'000.
A few archive areas are also defined. The first stores the temperatures
supplied for 100 hours (1'200 * 300 seconds = 100 hours). The second
RRA stores the minimum temperature recorded over every hour (12 * 300
seconds = 1 hour), for 100 days (2'400 hours). The third and the fourth
RRA's do the same for the maximum and average temperature,
EXAMPLE 2
rrdtool create monitor.rrd --step 300 \
DS:ifOutOctets:COUNTER:1800:0:4294967295 \
RRA:AVERAGE:0.5:1:2016 \
This example is a monitor of a router interface. The first RRA tracks
the traffic flow in octets; the second RRA generates the specialized
functions RRAs for aberrant behavior detection. Note that the rra-num
argument of HWPREDICT is missing, so the other RRAs will implicitly be
created with default parameter values. In this example, the forecasting
algorithm baseline adapts quickly; in fact the most recent one hour of
observations (each at 5 minute intervals) accounts for 75% of the
baseline prediction. The linear trend forecast adapts much more slowly.
Observations made during the last day (at 288 observations per day)
account for only 65% of the predicted linear trend. Note: these
computations rely on an exponential smoothing formula described in the
LISA 2000 paper.
The seasonal cycle is one day (288 data points at 300 second
intervals), and the seasonal adaption parameter will be set to 0.1. The
RRD file will store 5 days (1'440 data points) of forecasts and
deviation predictions before wrap around. The file will store 1 day (a
seasonal cycle) of 0-1 indicators in the FAILURES RRA.
The same RRD file and RRAs are created with the following command,
which explicitly creates all specialized function RRAs.
rrdtool create monitor.rrd --step 300 \
DS:ifOutOctets:COUNTER:1800:0:4294967295 \
RRA:AVERAGE:0.5:1:2016 \
RRA:HWPREDICT:1440:0.1:0.0035:288:3 \
RRA:SEASONAL:288:0.1:2 \
RRA:DEVPREDICT:1440:5 \
RRA:DEVSEASONAL:288:0.1:2 \
Of course, explicit creation need not replicate implicit create, a
number of arguments could be changed.
EXAMPLE 3
rrdtool create proxy.rrd --step 300 \
DS:Total:DERIVE:1800:0:U \
DS:Duration:DERIVE:1800:0:U \
DS:AvgReqDur:COMPUTE:Duration,Requests,0,EQ,1,Requests,IF,/ \
This example is monitoring the average request duration during each 300
sec interval for requests processed by a web proxy during the interval.
In this case, the proxy exposes two counters, the number of requests
processed since boot and the total cumulative duration of all processed
requests. Clearly these counters both have some rollover point, but
using the DERIVE data source also handles the reset that occurs when
the web proxy is stopped and restarted.
In the RRD, the first data source stores the requests per second rate
during the interval. The second data source stores the total duration
of all requests processed during the interval divided by 300. The
COMPUTE data source divides each PDP of the AccumDuration by the
corresponding PDP of TotalRequests and stores the average request
duration. The remainder of the RPN expression handles the divide by
zero case.
Tobias Oetiker <tobi@oetiker.ch>
|
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|
Distributive property and negative monomial
1. 222294
Distributive property and negative monomial
Give an example of using the distributive property for a negative monomial times a trinomial with different signs on the terms[for example:-3x(2xy+3y-2x)] and show each step of the distribution. Why
do you think many students make sign errors on this type of problem? What would be your advice to a student who has trouble with signs?
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integrator, instrument for performing the mathematical operation of integration, important for the solution of differential and integral equations and the generation of many mathematical functions.
The earliest integrator was a mechanical instrument called the planimeter. The Encyclopædia Britannica, Inc. (top) shows a simple mechanical integrator of the disk-and-wheel variety, which has
essential parts mounted on mutually perpendicular shafts, with a means of positioning the wheel in frictional contact with the disk, or turntable. In use, an angular displacement of the disk causes
the wheel to turn correspondingly. The radius of the integrating wheel introduces a scale factor, and its positioning on the disk represents the integrand. Thus the rotations of the disk and the
wheel are related through multiplicative factors and the number of turns made by the integrating wheel (for any number of turns of the disk) will be expressed as a definite integral of the function
represented by the variable position of the wheel on the disk.
Electronic integrators or electrical integrating circuits have largely displaced mechanical integrators. The (bottom) shows an electrical circuit that acts as an integrator. For time-varying input,
if the resistance R shown in the schematic diagram is very large compared with the capacitive reactance X[C] of the capacitor C, the current will be almost in phase with the input voltage E[IN], but
the output voltage E[OUT] will lag the phase of the input voltage E[IN] by almost 90°. Thus the output voltage E[OUT] is the time integral of the input voltage E[IN], as well as the product of the
current and the capacitive reactance, X[C].
Viewed as analogues, many common devices can be considered as integrators—examples being the odometer and the watt-hour meter. See also analog computer; differential analyzer.
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Matches for:
In the late summer of 1893, following the Congress of Mathematicians held in Chicago, Felix Klein gave two weeks of lectures on the current state of mathematics. Rather than offering a universal
perspective, Klein presented his personal view of the most important topics of the time. It is remarkable how most of the topics continue to be important today. Originally published in 1893 and
reissued by the AMS in 1911, we are pleased to bring this work into print once more with this new edition.
Klein begins by highlighting the works of Clebsch and of Lie. In particular, he discusses Clebsch's work on Abelian functions and compares his approach to the theory with Riemann's more geometrical
point of view. Klein devotes two lectures to Sophus Lie, focussing on his contributions to geometry, including sphere geometry and contact geometry.
Klein's ability to connect different mathematical disciplines clearly comes through in his lectures on mathematical developments. For instance, he discusses recent progress in non-Euclidean geometry
by emphasizing the connections to projective geometry and the role of transformation groups. In his descriptions of analytic function theory and of recent work in hyperelliptic and Abelian functions,
Klein is guided by Riemann's geometric point of view. He discusses Galois theory and solutions of algebraic equations of degree five or higher by reducing them to normal forms that might be solved by
non-algebraic means. Thus, as discovered by Hermite and Kronecker, the quintic can be solved "by elliptic functions". This also leads to Klein's well-known work connecting the quintic to the group of
the icosahedron.
Klein expounds on the roles of intuition and logical thinking in mathematics. He reflects on the influence of physics and the physical world on mathematics and, conversely, on the influence of
mathematics on physics and the other natural sciences. The discussion is strikingly similar to today's discussions about "physical mathematics".
There are a few other topics covered in the lectures which are somewhat removed from Klein's own work. For example, he discusses Hilbert's proof of the transcendence of certain types of numbers
(including \(\pi\) and \(e\)), which Klein finds much simpler than the methods used by Lindemann to show the transcendence of \(\pi\). Also, Klein uses the example of quadratic forms (and forms of
higher degree) to explain the need for a theory of ideals as developed by Kummer.
Klein's look at mathematics at the end of the 19th Century remains compelling today, both as history and as mathematics. It is delightful and fascinating to observe from a one-hundred year
retrospect, the musings of one of the masters of an earlier era.
Graduate students, research mathematicians, and mathematical historians.
"It is a noble example that Professor Klein has set all ages of mathematicians that, beginning his activity at a time when the contributions of the immediate past were so rich and so unrelated, he
was able to uncover the essential bonds that connect them and to discern the fields to whose development the new methods were best adapted ... His instinct for that which is vital in mathematics is
sure, and the light with which his treatment illumines the problems here considered may well serve as a guide for the youth who is approaching the study of the problems of a later day."
-- William F. Osgood, President of the AMS, 1905-1906
• Lecture I.: Clebsch
• Lecture II.: Sophus Lie
• Lecture III.: Sophus Lie
• Lecture IV.: On the real shape of algebraic curves and surfaces
• Lecture V.: Theory of functions and geometry
• Lecture VI.: On the mathematical character of space-intuition and the relation of pure mathematics to the applied sciences
• Lecture VII.: The transcendency of the numbers \(e\) and \(\pi\)
• Lecture VIII.: Ideal numbers
• Lecture IX.: The solution of higher algebraic equations
• Lecture X.: On some recent advances in hyperelliptic and Abelian functions
• Lecture XI.: The most recent researches in non-Euclidean geometry
• Lecture XII.: The study of mathematics at Göttingen
• The development of mathematics at the German Universities
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|
e, in mathematics, irrational number occurring widely in mathematics and science, approximately equal to the value 2.71828; it is the base of natural, or Naperian, logarithms. The number e is defined
as the limit of the expression (1+1/ n )^ n as n becomes infinitely large, or;e13;none;1;e13;;;block;;;;no;1;4224n;90919n;;;;;eq13;comptd;;center;stack;;;;;CE5In 1873 the French mathematician C.
Hermite proved that e was transcendental, i.e., not a root of any algebraic equation; this proof constituted a great contribution to the growth of mathematics. The number e is also known as Euler's
number, for Leonhard Euler, who discovered the famous formula e ^ i π = - 1, where i = - 1art/square-root-of-negative-1.gif the square root of negative 1, thus expressing the relationship between the
numbers e, i, and π. The exponential function e^ x , often written exp( x ), occurs in various applications ranging from statistics to nuclear physics.
See study by E. Maor (1994).
The Columbia Electronic Encyclopedia, 6th ed. Copyright © 2012, Columbia University Press. All rights reserved.
See more Encyclopedia articles on: Mathematics
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Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and
Author Message
Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and [#permalink] 15 Feb 2007, 17:09
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Joined: 07 Jan 2007
Question Stats:
Posts: 4
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(01:32) correct
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given: 0 20% (00:00)
based on 5 sessions
Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each
other, in how many different arrangements
Last edited by
on 15 Feb 2007, 20:11, edited 1 time in total.
6x5x4x3x2x1 = total number of arrangements = 720
total arrangements of J and M sitting together = 4x3x2x1x5x2 = 240
Joined: 29 Nov 2006
total arrangement of J and M not sitting together = 480
Posts: 85
Followers: 1 480
Kudos [?]: 2 [0],
given: 0
Senior Manager 1
Joined: 12 Mar 2006 This post received
Posts: 372
total ways of arranging 6 ppl = 6! = 6*5*4*3*2 = 720
Schools: Kellogg
School of ways in which m&j sit together = 5!*2 = 240
ans = 720 - 240 = 480
Followers: 2
Kudos [?]: 27 [1] ,
given: 3
This post received
terp26 KUDOS
VP 480 as well
Joined: 22 Oct 2006 total number of arrangements 6*5*4*3*2*1 = 720
Posts: 1443 want to count criteria we dont want
Schools: Chicago If Marcia sits in first chair, and Jan sits in 2nd chair there are 4! = 24 arrangements for the other people
Booth '11
Jan could also sit in first chair and Marcia in 2nd, so there are also 24 different arrangments
Followers: 7
24+24 = 48
Kudos [?]: 141 [1]
, given: 12 so we can have 1st and 2nd, 2nd and 3rd, 3rd and 4th, 4th and 5th, or 5th and 6th
therefore there are 5 ways each have 48 combinations so 48*5=240
720-240 = 480
Re: Advanced Constraint Combinatorics [#permalink] 04 Aug 2011, 20:12
This post received
Expert's post
GMAT100 wrote:
The following Question:
Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theatre. If Marcia and Jan will not sit next to each
other, in how many different arrangements
Answer will be posted tomorrow
krishp84 wrote:
why cannot we apply symmetry here ?
We cannot apply symmetry because the number of cases where M and J are sitting together is not equal to the number of cases where M and J are not sitting together.
Consider 3 people: A, B and C
They can be arranged in 3! = 6 ways
Veritas Prep GMAT
Instructor BAC
Joined: 16 Oct 2010 BCA
Posts: 4178 CAB
Location: Pune, CBA
In how many of these are A and B sitting together? 2
Followers: 895
In how many are they not sitting together? 4
Kudos [?]: 3795 [2]
, given: 148 When we arrange 6 people here in 6! (= 720) ways, in how many of those will M and J sit together?
Consider M and J to be one and arrange 5 people in 5! ways. Also, M and J can exchange places so multiply by 2.
You get 5!*2 = 240 ways
Out of 720, M and J will be next to each other in only 240 ways.
krishp84 wrote:
I recall a question related to this - "6 people standing in a line and one person(A) cannot stand behind another person(B). What is the total number of ways of forming the line? "
Cannot remember the complete question, something like this...But remember that it was pure symmetry : 6!/2=360 ways
That question would be something like this: 6 people go to a movie and sit next to each other in 6 adjacent seats in the front row of the theatre. If Marcia will not sit to the
right of Jan, how many different arrangements are possible?
'to the right of Jan' means anywhere on the right, not necessarily on the adjacent seat. Here we see symmetry because there are only 2 ways in which Marcia can sit. She can sit
either to the left of Jan (any seat on the left) or to the right of Jan (any seat on the right). There is nothing else possible. The number of cases in which she will sit to the
left will be same as the number of cases in which she will sit to the right. That is why the answer here will be 6!/2.
But the original question talks about sitting right next to each other on adjacent seats. The probability of sitting 'next to each other' is less than the probability of sitting
'not next to each other'. Hence we cannot apply the symmetry principle there.
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krishp84 Re: Advanced Constraint Combinatorics [#permalink] 05 Aug 2011, 16:38
Manager @Karishma - Can you please confirm the solutions and approach ?
Status: On... I am posting 3 more variants of the same question.
Joined: 16 Jan 2011 Let me know whether the symmetrical approach can be applied here ?
Posts: 190 _________________
Followers: 3 Labor cost for typing this post >= Labor cost for pushing the Kudos Button
Variant-2.1 of this question -
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the
Status: On... left of E in how many different arrangements ?
Joined: 16 Jan 2011 _________________
Posts: 190 Labor cost for typing this post >= Labor cost for pushing the Kudos Button
Followers: 3
Variant-3.1 of this question -
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different
Status: On... arrangements ?
Joined: 16 Jan 2011 _________________
Posts: 190 Labor cost for typing this post >= Labor cost for pushing the Kudos Button
Followers: 3
Variant-4.1 of this question -
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the
Status: On... left of E in how many different arrangements ?
Joined: 16 Jan 2011 _________________
Posts: 190 Labor cost for typing this post >= Labor cost for pushing the Kudos Button
Followers: 3
Re: Advanced Constraint Combinatorics [#permalink] 16 Aug 2011, 21:28
Expert's post
Variant-1 of this question -
7 people(A,B,C,D,E,F.H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different
arrangements ?
VeritasPrepKarishma 7! - 2((7-2+1)!) = 7!-2(6!)=7(6!)-2(6!) = 5(6!) ways
Veritas Prep GMAT Yes, that's fine. You make them sit together and subtract that out of 7!.
Variant-2 of this question -
Joined: 16 Oct 2010
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A,F and E will not sit next to each other in how many
Posts: 4178 different arrangements ?
Location: Pune, 7!-(3!)(7-3+1)! = 7!- 6(5!) ways
A little ambiguity here.
Followers: 895 I would say - A, F and E will not all sit together in how many ways?
The solution is fine.
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Re: Advanced Constraint Combinatorics [#permalink] 16 Aug 2011, 21:32
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Expert's post
Variant-3 of this question -
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different
arrangements ?
8P7-2[(8-2+1)P(7-2+1)] = 8P7-2(7P6) ways = 8! - 2(7!) = 6(7!) ways
The answer is correct. An alternative approach could be this:
Veritas Prep GMAT
Instructor I will just think of the vacant seat as a person V.
So 8 people, 8 seats in 8! ways.
Joined: 16 Oct 2010 2 people should not be together so 8! - 2*7!
Posts: 4178 Another interesting variant could be this:
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one
Location: Pune, person between A and F?
Try it.
Followers: 895
Variant-4 of this question -
Kudos [?]: 3795 [1]
, given: 148 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A ,F and E will not sit next to each other in how many
different arrangements ?
Correct solution. Alternative approach parallel to the one above:
The vacant spot is a person V.
Required arrangements: 8! - 3!*6!
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Re: Advanced Constraint Combinatorics [#permalink] 16 Aug 2011, 21:42
Expert's post
Variant-1.1 of this question -
7 people(A,B,C,D,E,F.H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different
arrangements ?
7!/2 ways
Variant-2.1 of this question -
Veritas Prep GMAT
Instructor 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the
left of E in how many different arrangements ?
Joined: 16 Oct 2010
Variant-3.1 of this question -
Posts: 4178
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different
Location: Pune, arrangements ?
Variant-4.1 of this question -
Followers: 895
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the
left of E in how many different arrangements ?
You can apply symmetry in all these questions. Tell me how you will do it. I will confirm the answer.
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Re: Advanced Constraint Combinatorics [#permalink] 18 Aug 2011, 17:30
VeritasPrepKarishma wrote:
Variant-3 of this question -
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different
arrangements ?
8P7-2[(8-2+1)P(7-2+1)] = 8P7-2(7P6) ways = 8! - 2(7!) = 6(7!) ways
The answer is correct. An alternative approach could be this:
I will just think of the vacant seat as a person V.
So 8 people, 8 seats in 8! ways.
krishp84 2 people should not be together so 8! - 2*7!
Manager WoW - Loved this Approach --> Kudos +1
Status: On... VeritasPrepKarishma wrote:
Joined: 16 Jan 2011 Another interesting variant could be this:
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one
Posts: 190 person between A and F?
Followers: 3 Try it.
This is equivalent to saying 8 people sit in 8 adjacent seats with an imaginary girl Ms. X with them
Subtract the number of arrangements when Ms. X sits between A and F
All the above are equivalent - No person sits between A and F
So answer will be 8!-(3!)(6!) waysNow let me apply my traditional approach and confirm this -
8P7-(3!)((8-3+1)P(7-3+1)) = 8P7-(3!)(6P5) = 8!-(3!)(6!) ways
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Re: Advanced Constraint Combinatorics [#permalink] 18 Aug 2011, 17:43
This post received
krishp84 KUDOS
Manager krishp84 wrote:
Status: On... Variant-2.1 of this question -
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 7 adjacent seats in the front row of the theater. A will not sit to the left of F and F will not sit to the
Joined: 16 Jan 2011 left of E in how many different arrangements ?
Posts: 190 Let me post my solution -
This will be equivalent as below -
Followers: 3
A can come to left/right of F
Kudos [?]: 31 [1] , F can come to left/right of E
given: 62 (I diagrammed, but it is time-taking to create a picture and post here)
So number of different arrangements = (1/2)(7!/2) = 7!/4 ways
Labor cost for typing this post >= Labor cost for pushing the Kudos Button
Re: Advanced Constraint Combinatorics [#permalink] 18 Aug 2011, 17:56
This post received
VeritasPrepKarishma wrote:
Variant-3.1 of this question -
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F in how many different
arrangements ?
This will be 8!/2 ways
Manager or
(8P7)/2 = 8!/2 ways
Status: On...
VeritasPrepKarishma wrote:
Joined: 16 Jan 2011
Variant-4.1 of this question -
Posts: 190 7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A will not sit to the left of F and F will not sit to the
left of E in how many different arrangements ?
Followers: 3
This will be (1/2)(8!/2) = 8!/4 ways
Kudos [?]: 31 [1] , or
given: 62 (1/2)(1/2)(8P7) = 8!/4 ways
Logic - combination of variant 2.1 and 3.1
VeritasPrepKarishma wrote:
You can apply symmetry in all these questions. Tell me how you will do it. I will confirm the answer.
I understand this is a HUGE post - But could not find another way to link all these related concepts.Welcome to any new variant. I LOVE these puzzles.
Labor cost for typing this post >= Labor cost for pushing the Kudos Button
Re: Advanced Constraint Combinatorics [#permalink] 18 Aug 2011, 21:39
This post received
Expert's post
VeritasPrepKarishma wrote:
Variant-3 of this question -
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. A and F will not sit next to each other in how many different
arrangements ?
I will just think of the vacant seat as a person V.
So 8 people, 8 seats in 8! ways.
2 people should not be together so 8! - 2*7!
Another interesting variant could be this:
7 people(A,B,C,D,E,F,H) go to a movie and sit next to each other in 8 adjacent seats in the front row of the theatre. In how many different arrangements will there be at least one
person between A and F?
krishp84 wrote:
This is equivalent to saying 8 people sit in 8 adjacent seats with an imaginary girl Ms. X with them
VeritasPrepKarishma Subtract the number of arrangements when Ms. X sits between A and F
Veritas Prep GMAT FXA
Instructor XAF
Joined: 16 Oct 2010 AFX
Posts: 4178 All the above are equivalent - No person sits between A and F
So answer will be 8!-(3!)(6!) ways
Location: Pune,
India This variant wants you to put at least one person between A and F. This means that all those cases where A and F are together are not acceptable and also those cases where A and F
have V (or your Ms. X!) between them are not acceptable.
Followers: 895
Number of cases where A and F are together = 2*7!
Kudos [?]: 3795 [1]
, given: 148 Number of cases where A and F have V between them = 2*6!
[AVF and FVA]
These are the cases that are not acceptable.
Answer should be 8! - 2*7! - 2*6! = 8! - 16*6!
Think about it another way: Compare the two questions.
One where you don't want them to be together,
the other where you don't want them to be together and you don't want the vacant spot between them.
Obviously, in the second case, the number of cases you do not want are higher.
In the first question, you subtracted a total of 2*7! arrangements i.e. 14*6! arrangements.
In the second question, you need to subtract some more. You cannot subtract 3!*6! = 6*6! arrangements only.
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Re: Advanced Constraint Combinatorics [#permalink] 18 Aug 2011, 21:47
Expert's post
krishp84 wrote:
VeritasPrepKarishma I understand this is a HUGE post - But could not find another way to link all these related concepts.[/color]
Welcome to any new variant. I LOVE these puzzles.
Veritas Prep GMAT
Instructor Your solutions are correct.
Joined: 16 Oct 2010 In a few weeks, I am going to start Permutation Combination on my blog. I already intend to make a post on this question and all its variants.
Posts: 4178 If you want to try out further complications, try putting in 2 or 3 vacant spots. Now there will be 2-3 identical people named 'V' so adjustments will be needed.
Location: Pune, _________________
Followers: 895 Veritas Prep | GMAT Instructor
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Re: Advanced Constraint Combinatorics [#permalink] 19 Aug 2011, 15:08
VeritasPrepKarishma wrote:
This variant wants you to put at least one person between A and F. This means that all those cases where A and F are together are not acceptable and also those cases where A and F
have V (or your Ms. X!) between them are not acceptable.
krishp84 Number of cases where A and F are together = 2*7!
Number of cases where A and F have V between them = 2*6!
Manager [AVF and FVA]
These are the cases that are not acceptable.
Status: On... Answer should be 8! - 2*7! - 2*6! = 8! - 16*6!
Joined: 16 Jan 2011 Think about it another way: Compare the two questions.
One where you don't want them to be together,
Posts: 190 the other where you don't want them to be together and you don't want the vacant spot between them.
Obviously, in the second case, the number of cases you do not want are higher.
Followers: 3 In the first question, you subtracted a total of 2*7! arrangements i.e. 14*6! arrangements.
In the second question, you need to subtract some more. You cannot subtract 3!*6! = 6*6! arrangements only.
Yes - I made that mistake of assuming X/V as a part of the group with A,F instead of thinking them separately. Stupid me !!!
Labor cost for typing this post >= Labor cost for pushing the Kudos Button
Re: Advanced Constraint Combinatorics [#permalink] 19 Aug 2011, 15:12
VeritasPrepKarishma wrote:
If you want to try out further complications, try putting in 2 or 3 vacant spots. Now there will be 2-3 identical people named 'V' so adjustments will be needed.
Status: On...
Yes - I was thinking on these lines....I will post some variants on this once I am free.
Joined: 16 Jan 2011
Also thought of combining Probability with this because it is so related especially - the symmetry part.
Posts: 190
Followers: 3
Labor cost for typing this post >= Labor cost for pushing the Kudos Button
Re: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and [#permalink] 15 Dec 2013, 03:47
This post received
Expert's post
GMAT100 wrote:
Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and sit next to each other in 6 adjacent seats in the front row of the theater. If Marcia and Jan will not sit next to each
other, in how many different arrangements
The best way to deal with the questions like this is to find the total # of arrangements and then subtract # of arrangements for which opposite of restriction occur:
Total # of arrangements of 6 people (let's say A, B, C, D, E, F) is
# of arrangement for which 2 particular persons (let's say A and B)
are adjacent
Math Expert
can be calculated as follows: consider these two persons as one unit like {AB}. We would have total 5 units: {AB}{C}{D}{E}{F} - # of arrangement of them 5!, # of arrangements of A
Joined: 02 Sep 2009 and B within their unit is 2!, hence total # of arrangement when A and B are adjacent is
Posts: 17321 5!*2!
Followers: 2875 .
Kudos [?]: 18404 [1 # of arrangement when A and B are not adjacent is
] , given: 2350
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25 extra-hard Quant Tests
gmatclubot Re: Greg, Marcia, Peter, Jan, Bobby and Cindy go to a movie and [#permalink] 15 Dec 2013, 03:47
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Another Superfluous Run Estimator
It is hard to muster a lot of enthusiasm for looking at run estimators other then Linear Weights or Base Runs, unless somebody manages to improve the score rate estimator for BsR someday, or somebody
comes up with an alternative model of the scoring process that works at extremes as well as everyday teams.
The method I am looking at here is not one of them[LW or BsR]. It is a faulty run estimator, with those flaws often being obvious. However, if we set James’ Runs Created as the minimum standard a run
estimator must meet in order to be taken seriously, this one does that. I believe that it may in fact be superior to Runs Created, but of course inferior to LW and BsR. It probably also falls below
Eric Van’s Contextual Runs as well (Contextual Runs is sort of a poor man’s BsR. It takes out homers, but the score rate estimator does not hold up nearly as well at theoretical extremes as BsR’s
does. This is probably partially because it is based on an advancement to out ratio instead of advancement as a percentage of advancement plus outs as is done by David Smyth).
Anyway, here I will take another look at what I call Appraised Runs (all of the good names are already taken after all), which I developed a few years ago as an adaptation of Mike Gimbel’s Run
Production Average. RPA started with a set of linear weights (which Gimbel never fully explained where they came from, except that they represent “run-driving” value) which clearly did not measure
the same thing as traditional LW as they severely underrated on base events and overrated homers. He then adjusted these by a “set up” rating which adjusted for the value of the event in setting up
further scoring opportunities for the team. Homers actually reduced the set up rating because they take runners off base.
The set up rating was then divided by the league average, and that was used to scale 50% of the run-driving value. That plus 50% of the run-driving total was the estimated number of runs scored. This
step still does not make sense to me because the team’s runs scored should not depend on the league rate, nor should the number of runners they have on base after an event.
Gimbel also included a number of categories that aren’t widely available, such as balks and wild pitches (for offenses), and so it was difficult to test his work and apply it with limited data. So I
came up with my own method that started with his basic values and then applied a set up rating that was not dependent on the league average.
My starting point was Gimbel’s run-driving values, scaled to equal total runs scored for the dataset I developed the formula on:
RD = .289S + .408D + .697T + 1.433HR + .164W
Then the set up figure, which I abbreviated UP, was found through regression and some trial and error, trying to get realistic intrinsic linear weight values:
UP = (5.7S + 8.6(D+T) + 1.44HR + 5W)/(AB+W) - .821
Then AR = UP*RD*.5 + RD*.5
Immediately you can see some of the flaws in this equation. A team that gets a very small number of runners on base will get a negative UP and negative runs. A team that hits 500 HRs will have a RD
of 716.5, and UP of .619, and an AR of 580, when we know they will score 500 runs. It may be better at the extremes then RC, but it is not correct at the extremes and is not nearly as good as BsR.
Please note, again, that I am not claiming this method should remain in use in sabermetrics. I am claiming that it is probably as good as or better then Runs Created. If you read Gimbel’s book in
1993 and did not yet know about BsR, it may well have been the best dynamic run estimator at that time.
I also came up with a version that included SB and CS:
RD = .288S + .407D + .694T + 1.428HR + .164W + .099SB - .164CS
UP = (5.7S + 8.6(D+T) + 1.44HR + 5W + 1.5SB - 3CS)/(AB+W) - .818
AR = UP*RD*.5 + RD*.5
Anyway, if you do an accuracy test, using the SB version, on all teams in the 1980s excepting 1981(the same sample I used in the post about Mann’s RPA), the RMSE of AR is 23.09, compared with 23.64
for ERP, 22.85 for BsR, and 25.15 for RC. So it is just as accurate as any of the other methods with normal teams. If you use just the RD portion, you get a RMSE of 31.06. Obviously it is flawed, not
only from under-weighting the value of on base events but also not considering outs at all. This construction necessitates the use of the UP factor or some other adjustment.
There is more than a similarity of names between the Gimbel and Mann RPAs. Both start with coefficients that do not capture the importance of avoiding outs and do not properly credit getting on base,
then apply adjustments in order to make the formula useful. Gimbel’s adjustments seem to be much more thought out and reasonable then Mann’s. I might be very wrong, but I get the impression that
Gimbel developed his system knowing that he would need to apply adjustments and actually wanting that to be the way his estimator worked, whereas it seems that Mann kept adding stuff in order to give
his estimator some semblance of accuracy.
Of course, I like to differentiate everything so that we can see the intrinsic linear weights. In order to do this, we need to differentiate UP and RD. So the derivative of UP with respect to an
event will be dUP, and for RD we’ll call it dRD(which will simply be the coefficient of the event in the RD equation). We can write UP as U/P, where U is numerator and P is the denominator. Since we
are dealing with a constant subtraction of .818, we can ignore this, since the derivative of a constant is zero. We will call the coefficient of any event in the UP numerator u and the coefficient of
any event in the UP denominator p. Then dUP = (P*u - U*p)/P^2. Knowing that value, the formula for the AR intrinsic LW is simple:
dAR = .5*(UP*dRD + RD*dUP) + .5*dRD
For example, using the non-SB version of AR, the 1979 Red Sox had an RD of 809.529, an UP of 1.17269, with U = 11663.06 and P = 5850. So the dUP for a walk was:
dUP = (5850*5 - 11663.06*1)/5850^2 = 5.139*10^(-4). We can plug into the dAR formula now:
dAR(wrt W) = .5*(1.17629*.164 + 809.529*5.139*10^(-4)) + .5*.164 = .3862
So the intrinsic LW value of a walk for the 1979 BoSox, according to AR, is .3862 runs. We can likewise calculate the intrinsic LWs for the other offensive events.
I have calculated the LW for each of the three methods based on the 1980s sample, and gotten these results(displayed as S, D, T, HR, W, SB, CS, O):
RC: .58, .89, 1.21, 1.52, .35, .16, -.39, -.122
AR: .51, .80, 1.09, 1.41, .35, .08, -.44, -.105
BsR: .47, .77, 1.08, 1.45, .33, .23, -.32, -.094
The AR weights are a better match for the BsR weights then are the RC weights. Although it seems that I have not properly factored in the value of the stolen base in the AR SB version. Maybe they and
CS should be given more weight in the UP factor.
If you apply the linear versions in order to do an accuracy test, the RMSE for RC is 43.65 (this is not really a reflection on RC, it is that I used the technical version without using all of the
categories, so you should probably take the intrinsic LW for RC with a grain of salt too), 31.27 for AR, and 23.02 for BsR (interestingly, the BsR linear weight error is only .17 higher then the
error for the multiplicative BsR equation).
I tried the derivative formulas on an extreme player, one with 500 AB, 180 H, 40 D, 60 HR, 150 W, and no SB or CS. That is a batting line of 360/508/800, which of course is ridiculously awesome. The
point was to see how the intrinsic LW from AR would match up with those from RC or BsR. I used the basic versions of AR and BsR but the technical version of RC, just so that RC would include walks in
the B factor. This may cause other problems, since it would be unrealistic to have no DP, SF, IW, etc. for a player like this, but I think the walks not being included at all in advancement problem
is bigger. The estimated runs created for the player by each method are 192.4 for RC, 202.3 for AR, and 191.2 for BsR. Here are the resulting intrinsic weights from the three equations:
RC: .84, 1.35, 1.86, 2.36, .46, -.34
AR: .76, 1.25, 1.64, 1.81, .51, -.29
BsR: .66, 1.01, 1.36, 1.52, .49, -.21
As you can see, the AR weights are a better match for the BsR weights then the RC weights, but the AR and RC estimates are closer to each other then they are to BsR. AR performs better largely
because, while it does not treat the HR correctly by counting it as an automatic run as BsR does, it does recognize that homers reduce the number of runners on base for the following batters(this is
what the UP factor in it’s original form was supposed to capture--the replacement for it that I have used here is empirical and not theoretical as Gimbel’s was), and therefore does not allow the
value of the HR to keep compounding as RC does.
Although I have not tested it, I think that is possible that AR shares with BsR one flaw that occurs at particularly high OBAs that you could call the “triple jump” flaw--the value of a triple will
jump the value of a HR at some point. This is obviously incorrect, but the distortion that comes from having the triple valued slightly higher then the home run in BsR is of a much smaller magnitude
then the RC flaw of allowing a HR to continue to grow in value, up to a maximum of four runs a pop. It is unfortunate that we have this imperfection in BsR, and hopefully future innovation will allow
us to eliminate it.
I am sure that one could improve the accuracy of this formula by fiddling around with the coefficients, particularly for UP. Perhaps one could develop an UP with a completely different structure that
would be useable as well, or change the percentage of RD that is scaled by UP, etc. I’m not sure what the point would be though. This type of run estimator is best accepted for what it is: a clever
but ultimately not very useful approximation of the scoring process that displays ingenuity on the part of its inventor and may have once been state of the art, but is no longer. As far as I’m
concerned, this is exactly the same category that Runs Created falls into. It, however, shows no signs of being consigned to the dustbin any time soon.
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Patent application title: ZERO-KNOWLEDGE PROOF SYSTEM, ZERO-KNOWLEDGE PROOF DEVICE, ZERO-KNOWLEDGE VERIFICATION DEVICE, ZERO-KNOWLEDGE PROOF METHOD AND PROGRAM THEREFOR
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Provided is a zero-knowledge proof system that allows a discrete-logarithm zero-knowledge proof. The zero-knowledge proof device includes a temporary memory unit that stores pseudorandom numbers and
previously determined hash values, a first processing unit that calculates multiple pseudorandom numbers and performs multiple iterations of processing to calculate hash values based on the
calculated pseudorandom numbers and the information stored in the temporary memory unit, a second processing unit that determines some of the multiple pseudorandom numbers based on the hash values,
and a third processing unit that re-calculates some of the pseudorandom numbers and sends the hash values obtained to a zero-knowledge verification device. The zero-knowledge verification device
includes a temporary memory region, a data receiving module that sequentially receives new input data, and a processing module that overwrites hash values including variables and input data, as
variables into the temporary memory region each time the input data are received.
A zero-knowledge proof system constituted with a zero-knowledge proof device and a zero-knowledge verification device, which performs discrete-logarithm zero-knowledge proof regarding whether or not
"the zero-knowledge proof device knows x that satisfies H=G
" as a verification under a state where the zero-knowledge verification device does not know the x, wherein the zero-knowledge proof device comprises: a temporary memory unit which stores
pseudorandom numbers and hash values acquired in the past; a first processing unit which calculates a plurality of pseudorandom numbers from an arbitrary random number sequence and a pseudorandom
function, and performs a plurality of iterations of processing to calculate hash values based on the calculated pseudorandom numbers and information stored in the temporary memory unit and to
overwrite the calculated pseudorandom numbers and the hash values to the temporary memory unit; a second processing unit which determines a part of the plurality of pseudorandom numbers based on the
hash values; and a third processing unit which transmits the hash values acquired by re-calculating the part of the pseudorandom numbers to the zero-knowledge verification device, and the
zero-knowledge verification device comprises: a data receiving module which sequentially receives new input data from the zero-knowledge proof device; a processing module which overwrites the hash
values of data containing a variable and input data stored in a temporary memory unit provided in advance as a new variable onto the temporary memory unit every time the data receiving module
receives the input data; and a judging unit which judges whether to authenticate or to reject the zero-knowledge proof device based on the variable, and returns a result of the judgment to the
zero-knowledge proof device.
The zero-knowledge proof system as claimed in claim 10, wherein: the first processing unit of the zero-knowledge proof device has a repeat processing function which repeats N-times (N is a natural
number of 2 or larger) of processing to read elements G, H of a group, to define initial values of data V showing the pseudorandom numbers, to calculate first and second data pseudorandom function
values, and to update the hash values of the data containing V and the first and second data pseudorandom function values Y0 and R0 as new V; the second processing unit of the zero-knowledge proof
device has a repeat processing function which repeats N-times of processing to give some kind of initial values to data Y that shows a part of the plurality of pseudorandom numbers, and to take a
value acquired by adding a value based on a hash value U of data containing the V and the j to the Y as new Y in j-th processing (
ltoreq.j≦N); and the third processing unit of the zero-knowledge proof device has a hash-value output processing function which: repeats N-times of processing to take a residue of the Y-th power to
the G as A, to calculate first and second hash values T0 and T1 based on the Y0 and the R0 in the j-th processing (
ltoreq.j≦N), to calculate 1-bit data c of from data containing the V, the A, and the j, and to transmit the Y0, the R0, the T1 to the zero-knowledge verification device when the c is 0 while
transmitting the Y1, the R1, the T0 when c is
1. 12.
The zero-knowledge proof system as claimed in claim 11, wherein the processing module of the zero-knowledge verification device comprises: a first processing unit which repeats N-times of processing
to read elements G, H of a group, to give some kind of initial values to data V that shows the variables, to receive data c, data Y, data T, and data R from the zero-knowledge proof device as inputs,
to calculate first and second hash values containing the Y and the R, and to overwrite the values on the temporary memory unit as data V anew; and a second processing unit which repeats N-times of
procedure to perform initial setting of data W to 0 and data C to 0, respectively, to take the hash value of data containing the V and the j as U in the j-th processing containing (
ltoreq.j≦N), to take a result acquired by adding the W to a product of the Wj and the U as new W, and to take a result acquired by adding the C to a product of the Cj and the U as new C; and the
judging unit of the zero-knowledge verification device comprises a third processing unit which repeats a procedure to take a result acquired by multiplying a residue of W-th power to G by a residue
of -C-th power to H as A and to output data indicating rejection to stop the zero-knowledge proof device when the hash values containing the V, the A, and the j do not match with the Cj in the j-th
processing (
ltoreq.j≦N), while outputting data indicating to authenticate the zero-knowledge proof device when the data indicating rejection is not outputted after repeating such procedure for N-times.
The zero-knowledge proof system as claimed in claim 10, wherein: the zero-knowledge proof device comprises a storage device which collectively stores sets of the pseudorandom numbers and the hash
values outputted to the zero-knowledge verification device and outputs the sets to the zero-knowledge verification device; and the zero-knowledge verification device comprises a storage device which
collectively stores the sets of the pseudorandom numbers and hash values received at the data receiving module from the zero-knowledge proof device.
A zero-knowledge proof device which works in cooperation with a zero-knowledge verification device to let the zero-knowledge verification device that does not know x verify whether or not "the
zero-knowledge proof device itself knows the x that satisfies H=G
", the zero-knowledge proof device comprising: a temporary memory unit which stores pseudorandom numbers and hash values acquired in the past; a first processing unit which calculates a plurality of
pseudorandom numbers from an arbitrary random number sequence and a pseudorandom function, and executes a plurality of iterations of processing to calculate hash values based on the calculated
pseudorandom numbers and information stored in the temporary memory unit and to overwrite the calculated pseudorandom numbers and the hash values on the temporary memory unit; a second processing
unit which determines a part of the plurality of is pseudorandom numbers based on the hash values; a third processing unit which transmits the hash values acquired by re-calculating a part of the
pseudorandom numbers to the zero-knowledge verification device; and a data receiving unit which receives data indicating whether to authenticate or to reject from the zero-knowledge verification
device after transmitting the hash values to the zero-knowledge verification device.
A zero-knowledge verification device which, under a state where the verification device itself does not know x, verifies whether or not "a zero-knowledge proof device knows the x that satisfies H=G
" according to a request from the zero-knowledge proof device, the zero-knowledge verification device comprising: a data receiving module which sequentially receives hash values generated by the
zero-knowledge proof device based on a plurality of pseudorandom numbers calculated from an arbitrary random number sequence and a pseudorandom function as new input data; a processing module which
overwrites hash values of data containing a variable and the input data stored in a temporary memory unit provided in advance as a new variable onto the temporary memory unit every time the data
receiving module receives the input data; and a judging unit which judges whether to authenticate or to reject the zero-knowledge proof device based on the variable, and returns a result of the
judgment to the zero-knowledge proof device.
A method used in a zero-knowledge proof system constituted with a zero-knowledge proof device and a zero-knowledge verification device for performing discrete-logarithm zero-knowledge proof regarding
whether or not "the zero-knowledge proof device knows x that satisfies H=G
" as a verification under a state where the zero-knowledge verification device does not know the x, wherein on the zero-knowledge proof device side: a first processing unit calculates a plurality of
pseudorandom numbers from an arbitrary random number sequence and a pseudorandom function, and performs a plurality of iterations of processing to calculate hash values based on the calculated
pseudorandom numbers and pseudorandom numbers stored in a temporary memory unit provided in advance and hash values acquired in the past and to overwrite the calculated pseudorandom numbers and the
hash values onto the temporary memory unit; a second processing unit determines a part of the plurality of pseudorandom numbers to be outputted to the zero-knowledge verification device based on the
hash values; and a third processing unit re-calculates the part of the pseudorandom numbers and transmits the calculated pseudorandom numbers to the zero-knowledge verification device, and on the
zero-knowledge verification device side: a data receiving module sequentially receives new input data from the zero-knowledge proof device; a verification module overwrites the hash values of data
containing a variable and input data stored in the temporary memory unit as the new variable onto the temporary memory unit provided in advance every time the data receiving module receives the input
data; and a judging unit judges whether to authenticate or to reject the zero-knowledge proof device based on the variable, and returns a result of the judgment to the zero-knowledge proof device.
A non-transitory computer readable recording medium storing a zero-knowledge proof program used for a zero-proof knowledge device which works in cooperation with a zero-knowledge verification device
to let the zero-knowledge verification device that does not know x verify whether or not "the zero-knowledge proof device itself knows the x that satisfies H=G
", the zero-knowledge program causing a computer to execute: a procedure to store pseudorandom numbers and hash values acquired in the past to a temporary memory unit provided in advance; a procedure
to calculate a plurality of pseudorandom numbers from an arbitrary random number sequence and a pseudorandom function and to execute a plurality of iterations of processing to calculate hash values
based on the calculated pseudorandom numbers and information stored in the temporary memory unit and to overwrite the calculated pseudorandom numbers and the hash values on the temporary memory unit;
a procedure to determine a part of the plurality of pseudorandom numbers based on the hash values; a procedure to transmit the hash values acquired by re-calculating a part of the pseudorandom
numbers to the zero-knowledge verification device; and a procedure to receive data indicating whether to authenticate or to reject from the zero-knowledge verification device after transmitting the
hash values to the zero-knowledge verification device.
A non-transitory computer readable recording medium storing a zero-knowledge proof program used for a zero-knowledge verification device which, under a state where the verification device itself does
not know x, verifies whether or not "a zero-knowledge proof device knows the x that satisfies H=G
" according to a request from the zero-knowledge proof device, the zero-knowledge proof program causing a computer to execute: a procedure to sequentially receive hash values generated by the
zero-knowledge proof device based on a plurality of pseudorandom numbers calculated from an arbitrary random number sequence and a pseudorandom function as new input data; a procedure to overwrite
hash values of data containing a variable and the input data stored in a temporary memory unit provided in advance as the new variable onto the temporary memory unit every time the data receiving
module receives the input data; and a procedure to judge whether to authenticate or to reject the zero-knowledge proof device based on the variable, and returns a result of the judgment to the
zero-knowledge proof device.
A zero-knowledge proof system constituted with a zero-knowledge proof device and a zero-knowledge verification device, which performs discrete-logarithm zero-knowledge proof regarding whether or not
"the zero-knowledge proof device knows x that satisfies H=G
" as a verification under a state where the zero-knowledge verification device does not know the x, wherein the zero-knowledge proof device comprises: temporary memory means for storing pseudorandom
numbers and hash values acquired in the past; first processing means for calculating a plurality of pseudorandom numbers from an arbitrary random number sequence and a pseudorandom function, and
performing a plurality of iterations of processing to calculate hash values based on the calculated pseudorandom numbers and information stored in the temporary memory means and to overwrite the
calculated pseudorandom numbers and the hash values to the temporary memory means; second processing means for determining a part of the plurality of pseudorandom numbers based on the hash values;
and third processing means for transmitting the hash values acquired by re-calculating the part of the pseudorandom numbers to the zero-knowledge verification device, and the zero-knowledge
verification device comprises: data receiving means for sequentially receiving new input data from the zero-knowledge proof device; processing means for overwriting the hash values of data containing
a variable and input data stored in temporary memory means provided in advance as a new variable onto the temporary memory means every time the data receiving means receives the input data; and
judging means for judging whether to authenticate or to reject the zero-knowledge proof device based on the variable, and returning a result of the judgment to the zero-knowledge proof device.
A zero-knowledge proof device which works in cooperation with a zero-knowledge verification device to let the zero-knowledge verification device that does not know x verify whether or not "the
zero-knowledge proof device itself knows the x that satisfies H=G
", the zero-knowledge proof device comprising: temporary memory means for storing pseudorandom numbers and hash values acquired in the past; first processing means for calculating a plurality of
pseudorandom numbers from an arbitrary random number sequence and a pseudorandom function, and executing a plurality of iterations of processing to calculate hash values based on the calculated
pseudorandom numbers and information stored in the temporary memory means and to overwrite the calculated pseudorandom numbers and the hash values on the temporary memory means; second processing
means for determining a part of the plurality of pseudorandom numbers based on the hash values; third processing means for transmitting the hash values acquired by re-calculating a part of the
pseudorandom numbers to the zero-knowledge verification device; and data receiving means for receiving data indicating whether to authenticate or to reject from the zero-knowledge verification device
after transmitting the hash values to the zero-knowledge verification device.
A zero-knowledge verification device which, under a state where the verification device itself does not know x, verifies whether or not "a zero-knowledge proof device knows the x that satisfies H=G
" according to a request from the zero-knowledge proof device, the zero-knowledge verification device comprising: data receiving means for sequentially receiving hash values generated by the
zero-knowledge proof device based on a plurality of pseudorandom numbers calculated from an arbitrary random number sequence and a pseudorandom function as new input data; processing means for
overwriting hash values of data containing a variable and the input data stored in temporary memory means provided in advance as a new variable onto the temporary memory means every time the data
receiving means receives the input data; and judging means for judging whether to authenticate or to reject the zero-knowledge proof device based on the variable, and returning a result of the
judgment to the zero-knowledge proof device.
TECHNICAL FIELD [0001]
The present invention relates to a discrete-logarithm zero-knowledge proof system. More specifically, the present invention relates to a zero-knowledge proof system, a zero-knowledge proof device, a
zero-knowledge verification device, a zero-knowledge proof method and a program therefore, which can decrease a device storage capacity required for zero-knowledge proofs.
BACKGROUND ART [0002]
First, the discrete-logarithm zero-knowledge proof will be described. The discrete-logarithm zero-knowledge proof herein is a technique used when a prover as an entity (element) proves to a verifier
as an entity a fact that "the prover knows x that satisfies H=G
" without revealing x.
Note here that n is a natural number, G is an element for Z/nZ, and x is an integer. Further, in the lines other than numerical expressions, "G to the power of x" is expressed as "G
". This technique is utilized greatly in techniques related to cryptography, and applied in various scenes such as public-key cryptography, digital signatures, group signatures, and electronic
Various methods are known as the discrete-logarithm zero-knowledge proofs. Among those, there is a method which excludes message m from the method depicted in Non-Patent Document 1. In order to prove
the knowledge of the element x of Z/nZ which satisfies H=G
with this method, the prover first selects r from Z/nZ randomly, and calculates u by a following expression.
[Expression 1]
Subsequently, the prover calculates hash value c=H(u) and calculates ρ=r+cx, and sends a set of (c, ρ) to the verifier. Upon that, the verifier calculates u by a following expression. When it is
found to satisfy c=H(u), the verifier returns "accept" to the prover indicating that the verifier judges the fact the prover is to proof is adequate. If not, the verifier returns "reject" to the
prover indicating that the verifier judges the proof is inadequate.
c [Expression 2]
However, with the method depicted in Non-Patent Document 1, even a prover who does not know x can commit perjury that "the prover knows x that satisfies H=G
" with a probability of about 1/q.sup.(1/3), provided that the order of G is q. Thus, normally, it is necessary to repeat the above-described processing for a plurality of times in order to reduce
the possibility of perjury. Therefore, the processing becomes complicated and time-consuming.
Patent Document 1 discloses a zero-knowledge proof method capable of improving that aspect, which selects N-pieces of random numbers Y1, - - - YN, calculates data Ti from each Yi, calculates hash
value Θ (T1, - - - Tn, - - - ), and further performs a calculation using Y1, - - - YN thereafter. With this method, it is possible to reduce the possibility of perjury committed by the prover who
does not know x to pretend that "the prover knows x that satisfies H=G
" to about 1/q.
Further, in addition, there are following documents as technical documents related to the zero-knowledge proofs. Patent Document 2 discloses a technique regarding reception of electronic money, which
is regarding a technique for reducing the storage capacity of ID transmitted for a challenge. Patent Document 3 discloses a technique which sequentially generates challenges for a single hash value.
Patent Document 4 discloses a technique with which a capacitance for the final receiver to verify the trueness of data is not spoiled even if a middle person erases a signature part of a data stream
on which a digital signature has been given.
Patent Document 1: Domestic Republication of PCT International Application 2006-077701
Patent Document 2: Japanese Unexamined Patent Publication 2000-067141
Patent Document 3: Japanese Unexamined Patent Publication 2003-218858
Patent Document 4: Japanese Patent Application Publication 2007-503134 Non-Patent Document 1: Claus-Peter Schnorr. Efficient Signature Generation by Smart Cards. J. Cryptology 4(3): 161-174 (1991)
DISCLOSURE OF THE INVENTION Problems to be Solved by the Invention [0013]
However, the zero-knowledge proof method depicted in Patent Document 1 can be employed, when it is assumed that all the sets of Yi and Θ (T1, - - - Tn, - - - ) are stored in a main storage device. Yi
is data of 1300 to 2500 bits, and N is a value of about 1000 to 2000. Thus, storage capacitance of about 130 to 500 megabits is required in the main storage device for achieving this method.
Therefore, it is difficult to use the method depicted in Patent Document 1 with devices such as mobile phone terminals and PDAs (Personal Digital Assistants) which only have 1.0 a small-capacity main
memory device. Patent Documents 2 to 4 and Non-Patent Document 1 disclose no structure that can overcome such issue.
An object of the present invention is to provide a zero-knowledge proof system, a zero knowledge proof device, a zero-knowledge proof verification device, a zero-knowledge proof method and a program
therefore, which can perform discrete-logarithm zero-knowledge proof even with a device that has only a small-capacity main memory device.
Means for Solving the Problems [0015]
In order to achieve the foregoing object, the zero-knowledge proof system according to the present invention is characterized as a zero-knowledge proof system constituted with a zero-knowledge proof
device and a zero-knowledge verification device, which performs discrete-logarithm zero-knowledge proof regarding whether or not "the zero-knowledge proof device knows x that satisfies H=G
" as a verification under a state where the zero-knowledge verification device does not know the x, wherein, the zero-knowledge proof device includes: a temporary memory unit which stores
pseudorandom numbers and hash values acquired in the past; a first processing unit which calculates a plurality of pseudorandom numbers from an arbitrary random number sequence and a pseudorandom
function, and performs a plurality of iterations of processing to calculate hash values based on the calculated pseudorandom numbers and information stored in the temporary memory unit and to
overwrite the calculated pseudorandom numbers and the hash values to the temporary memory unit; a second processing unit which determines a part of the plurality of pseudorandom numbers based on the
hash values; and a third processing unit which transmits the hash values acquired by re-calculating the part of the pseudorandom numbers to the zero-knowledge verification device, and the
zero-knowledge verification device includes: a data receiving module which sequentially receives new input data from the zero-knowledge proof device; a processing module which overwrites the hash
values of data containing a variable and input data stored in a temporary memory unit provided in advance as a new variable onto the temporary memory unit every time the data receiving module
receives the input data; and a judging unit which judges whether to authenticate or to reject the zero-knowledge proof device based on the variable, and returns a result of the judgment to the
zero-knowledge proof device.
In order to achieve the foregoing object, the zero-knowledge proof device according to the present invention is characterized as a zero-proof knowledge device which works in cooperation with a
zero-knowledge verification device to let the zero-knowledge verification device that does not know x verify whether or not "the zero-knowledge proof device itself knows the x that satisfies H=G
", and the zero-knowledge proof device includes: a temporary memory unit which stores pseudorandom numbers and hash values acquired in the past; a first processing unit which calculates pseudorandom
numbers from an arbitrary random number sequence and a pseudorandom function, and executes a plurality of iterations of processing to calculate hash values based on the calculated pseudorandom
numbers and information stored in the temporary memory unit and to overwrite the calculated pseudorandom numbers and the hash values on the temporary memory unit; a second processing unit which
determines a part of the plurality of pseudorandom numbers based on the hash values; a third processing unit which transmits the hash values acquired by re-calculating a part of the pseudorandom
numbers to the zero-knowledge verification device; and a data receiving unit which receives data indicating whether to authenticate or to reject from the zero-knowledge verification device after
transmitting the hash values to the zero-knowledge verification device.
In order to achieve the foregoing object, the zero-knowledge verification device according to the present invention is characterized as a zero-knowledge verification device which, under a state where
the verification device itself does not know x, verifies whether or not "a zero-knowledge proof device knows the x that satisfies H=G
" according to a request from the zero-knowledge proof device, and the zero-knowledge verification device includes: a data receiving module which sequentially receives new input data from the
zero-knowledge proof device; a processing module which overwrites hash values of data containing a variable and the input data stored in a temporary memory unit provided in advance as a new variable
onto the temporary memory unit every time the data receiving module receives the input data; and a judging unit which judges whether to authenticate or to reject the zero-knowledge proof device based
on the variable, and returns a result of the judgment to the zero-knowledge proof device.
In order to achieve the foregoing object, the zero-knowledge proof method according to the present invention is characterized as a method used in a zero-knowledge proof system constituted with a
zero-knowledge proof device and a zero-knowledge verification device for performing discrete-logarithm zero-knowledge proof regarding whether or not "the zero-knowledge proof device knows x that
satisfies H=G
" as a verification under a state where the zero-knowledge verification device does not know the x, wherein, on the zero-knowledge proof device side: a first processing unit calculates a plurality of
pseudorandom numbers from an arbitrary random number sequence and a pseudorandom function, and performs a plurality of iterations of processing to calculate hash values based on the calculated
pseudorandom numbers and pseudorandom numbers stored in a temporary memory unit provided in advance and hash values acquired in the past and to overwrite the calculated pseudorandom numbers and the
hash values onto the temporary memory unit; a second processing unit determines a part of the plurality of pseudorandom numbers outputted to the zero-knowledge verification device based on the hash
values; and a third processing unit re-calculates the part of the pseudorandom numbers and transmits the calculated pseudorandom numbers to the zero-knowledge verification device, and on the
zero-knowledge verification device side: a data receiving module sequentially receives new input data from the zero-knowledge proof device; a verification module overwrites the hash values of data
containing a variable and input data stored in the temporary memory unit as the new variable onto the temporary memory unit provided in advance every time the data receiving module receives the input
data; and a judging unit judges whether to authenticate or to reject the zero-knowledge proof device based on the variable, and returns a result of the judgment to the zero-knowledge proof device.
In order to achieve the foregoing object, the zero-knowledge proof program according to the present invention is characterized as a zero-knowledge proof program used for a zero-proof knowledge device
for performing discrete-logarithm zero-knowledge proof which works in cooperation with a zero-knowledge verification device to let the zero-knowledge verification device that does not know x verify
whether or not "the zero-knowledge proof device itself knows the x that satisfies H=G
", and the zero-knowledge program causes a computer to execute: a procedure to store pseudorandom numbers and hash values acquired in the past to a temporary memory unit provided in advance; a
procedure to calculate a plurality of pseudorandom numbers from an arbitrary random number sequence and a pseudorandom function and to execute a plurality of iterations of processing to calculate
hash values based on the calculated pseudorandom numbers and information stored in the temporary memory unit and to overwrite the calculated pseudorandom numbers and the hash values on the temporary
memory unit; a procedure to determine a part of the plurality of pseudorandom numbers based on the hash values; a procedure to transmit the hash values acquired by re-calculating a part of the
pseudorandom numbers to the zero-knowledge verification device; and a procedure to receive data indicating whether to authenticate or to reject from the zero-knowledge verification device after
transmitting the hash values to the zero-knowledge verification device.
In order to achieve the foregoing object, another zero-knowledge proof program according to the present invention is characterized as a zero-knowledge proof program used for a zero-knowledge
verification device for performing discrete-logarithm zero-knowledge proof, which, under a state where the verification device itself does not know x, verifies whether or not "a zero-knowledge proof
device knows the x that satisfies H=G
" according to a request from the zero-knowledge proof device, and the zero-knowledge proof program causes a computer to execute: a procedure to sequentially receive new input data from the
zero-knowledge proof device; a procedure to overwrite hash values of data containing a variable and the input data stored in a temporary memory unit provided in advance as the new variable onto the
temporary memory unit every time the data receiving module receives the input data; and a procedure to judge whether to authenticate or to reject the zero-knowledge proof device based on the
variable, and returns a result of the judgment to the zero-knowledge proof device.
Effect of the Invention [0021]
As described above, the present invention is structured to perform the calculation processing for calculating the hash values while re-utilizing the memory that stores the pseudorandom number and the
hash values. Therefore, it is unnecessary to store all the sets of the plurality of random numbers and the corresponding hash values. This makes it possible to perform the discrete-logarithm
zero-knowledge proof even with a device that has only a small-capacity main memory device.
BRIEF DESCRIPTION OF THE DRAWINGS [0022] FIG. 1
is an explanatory illustration showing the structure of a zero-knowledge proof system according to a first embodiment of the present invention;
[0023] FIG. 2
is a flowchart showing operations of a proof module shown in
FIG. 1
[0024] FIG. 3
is a flowchart continued from
FIG. 2
[0025] FIG. 4
is a flowchart continued from
FIG. 2
FIG. 3
[0026] FIG. 5
is a flowchart showing operations of a verification module shown in
FIG. 1
[0027] FIG. 6
is a flowchart continued from
FIG. 5
FIG. 7 is a flowchart continued from
FIG. 5
FIG. 6
[0029] FIG. 8
is an explanatory illustration showing the structure of a zero-knowledge proof system according to a second embodiment of the present invention;
[0030] FIG. 9
is a flowchart showing operations of a proof module shown in
FIG. 8
; and
[0031] FIG. 10
is a flowchart showing operations of a verification module shown in
FIG. 8
BEST MODES FOR CARRYING OUT THE INVENTION First Embodiment [0032]
Hereinafter, structures of embodiments of the present invention will be described by referring to the accompanying drawings.
The basic contents of the embodiment will be described first, and more specific contents thereof will be described thereafter.
A zero-knowledge proof system 1 according to the embodiment is constituted with a zero-knowledge proof device (prover device 10) and a zero-knowledge verification device (verifier device 20). The
zero-knowledge proof device (prover device 10) includes: a temporary memory unit (RAM 12) which stores pseudorandom numbers and hash values acquired in the past; a first processing unit 14a which
calculates a plurality of pseudorandom numbers from an arbitrary random number sequence and a pseudorandom function, and performs a plurality of iterations of processing to calculate hash values
based on the calculated pseudorandom numbers and information stored in the temporary memory unit and to overwrite the calculated pseudorandom numbers and the hash values to the temporary memory unit;
a second processing unit 14b which determines a part of the plurality of pseudorandom numbers based on the hash values; and a third processing unit 14c which transmits the hash values acquired by
re-calculating the part of the pseudorandom numbers to the zero-knowledge verification device. The zero-knowledge verification device (verifier device 20) includes: a data receiving module
(communication interface 23) which sequentially receives new input data from the zero-knowledge proof device; processing modules (first and second processing units 24a, 24b) which overwrite the hash
values of data containing a variable and input data stored in a temporary memory unit (RAM 22) provided in advance as a new variable onto the temporary memory unit every time the data receiving
module receives input data; and a judging unit (third processing unit 24c) which judges whether to authenticate or to reject the zero-knowledge proof device based on the variable, and returns the
judgment result to the zero-knowledge proof device.
Note here that the first processing unit 14a of the zero-knowledge proof device (prover device 10) has a repeat processing function which repeats N-times (N is a natural number of 2 or larger) of
processing to read elements G, H of a group, to define initial values of data V showing the pseudorandom numbers, to calculate first and second data pseudorandom function values, and to update the
hash values of the data containing V and the first and second data pseudorandom function values Y0 and R0 as new V.
The following second processing unit 14b has a repeat processing function which repeats N-times of processing to give some kind of initial values to data Y that shows a part of the plurality of
pseudorandom numbers, and to take a value acquired by adding a value based on a hash value U of data containing V and j to Y as new Y.
Further, the following third processing unit 14c has a hash-value output processing function which: repeats N-times of processing to take a residue of Y-th power to G as A, to calculate first and
second hash values T0 and T1 based on Y0 and R0 in the j-th processing (1≦j≦N), to calculate 1-bit data c from data containing V, A, and j, and to transmit Y0, R0, T1 to the zero-knowledge
verification device when c is 0 while transmitting Y1, R1, T0 when c is 1. Further, the zero-knowledge proof device includes a data receiving unit (communication interface 13) which receives data
indicating whether to reject or to authenticate, which is returned from the zero-knowledge verification device after transmitting the hash values to the zero-knowledge verification device.
In the meantime, the processing module (verification module 24) of the zero-knowledge verification device (verifier device 20) includes a first processing unit 24a which repeats N-times of processing
to read elements G, H of a group, to give some kind of initial values to data V that shows variables; receive data c, data Y, data T, and data R from the zero-knowledge proof device as inputs, to
calculate first and second hash values containing Y and R, and to overwrite the values on the temporary memory unit as data V anew.
Further, the processing module includes a second processing unit 24b which repeats N-times of procedure to perform initial setting of data W to 0 and data C to 0, respectively, to take the hash
values of data containing V and j as U in the j-th processing (1≦j≦N), to take a result acquired by adding W to a product of Wj and U as new W, and to take a result acquired by adding C to a product
of Cj and U as new C.
The processing module further includes a third processing unit 24c which repeats a procedure to take a result acquired by multiplying a residue of W-th power to G by a residue of -C-th power to H as
A and to output data indicating rejection to stop the zero-knowledge proof device when the hash values of data containing V, A, and j do not match with Cj in the j-th processing (1≦j≦N), while
outputting data indicating to authenticate the zero-knowledge proof device when the data indicating rejection is not outputted after repeating such procedure for N-times.
With such structure, it is unnecessary to store all the sets of the plurality of random numbers and each of corresponding hash values. Therefore, the embodiment makes it possible to perform
discrete-logarithm zero-knowledge proof even with the device having only a small-capacity main memory device.
This will be described in more details hereinafter.
Each of α, N, π, ζ, λ, κ, and n is defined as a security parameter. In practical use, α, N, π, ζ, λ, κ, and n may be defined as 160, 1304, 60, 60, 1244, 1024, and 1024, respectively. However, for the
better security, it is also possible to define α, N, π, , λ, κ, and n as 192, 2496, 112, 112, 2384, 2048, and 2048, respectively.
It is defined that n is a nonnegative integer whose bit number is α, G is an element of Z/nZ, x is a nonnegative integer of λ-bit, and H=G
. Here, it is supposed that the zero-knowledge proof system according to the present invention is to prove a fact that "the zero-knowledge proof device knows x that satisfies H=G
"Fλ+ζ" is taken as a pseudorandom function which generates an output of "λ+ζ" bits, and the output of "Fλ+ζ" when data X and a key K are inputted is written as "Fλ+ζ(K, X)". Similarly, "Fλ+π" is
taken as a pseudorandom function which generates an output of "λ+π" bits, and the output of "Fλ+π" when data X and a key K are inputted is written as "Fλ+π(K, X)". Note that "λ+ζ" and "λ+π" are
subscript letters in actual expressions.
As long as the output satisfies the above-described condition, the output may be used as any kinds of "Fλ+ζ" and "Fλ+π". For example, functions for corresponding the hash values to (K, X) can be
defined as "Fλ+ζ" and "Fλ+π". Θλ, Θκ, and Θ1 are defined as hash functions whose outputs are λ-bit, κ-bit, and 1-bit, respectively. Note that each of the letters λ, κ, and l of Θλ, Θκ, and Θ1 are
subscript letters in actual expressions.
[0047] FIG. 1
is an explanatory illustration showing the structure of the zero-knowledge proof system according to the first embodiment of the present invention. The zero-knowledge proof system 1 is constituted
with the prover device 10 that is a computer device operated by the prover and the verifier device 20 that is a computer device operated by the verifier, and the prover device 10 and the verifier
device 20 are connected mutually.
The prover device 10 includes: a CPU (Central Processing Unit) 11 as a main body for executing computer programs; a RAM (Random Access Memory) 12 to which computer programs executed by the CPU 11 are
loaded and stored; and a communication interface 13 which exchanges data with other computers. A proof module 14 that is a computer program executed by the CPU 11 is stored in the RAM 12 and
executed. An input device 16 is used for inputting initial data and the like required for operations of the proof module 14.
Similarly, the verifier device 20 includes a CPU 21, a RAM 22, and a communication interface 23. Further, a verification module 24 as a computer program executed by the CPU 21 is stored in the RAM 22
and executed. An input device 26 is used for inputting initial data and the like required for operations of the verification module 24. In
FIG. 1
, the proof module 14 and the verification module 24 are illustrated to exist on the CPUs 11 and 21 to be executed thereby, respectively, for convenience' sake.
There is a loop 15 existing in the algorithm of the proof module 14, and the proof module 14 outputs (c, Y, R, T) to the verifier device 20 via the communication interface 13 every time the loop 15
is executed. There is a loop 25 existing in the algorithm of the verification module 24, and the verification module 24 receives (c, Y, R, T) outputted from the verification module 14 every time the
loop 25 is executed. The loops 15 and 25 can be achieved by for text, while text, do-while text, or the like in C++ language, for example. However, it is also possible to achieve the loops 15 and 25
with other program languages by using texts according to each language.
Each of the parameters G, H, n is inputted to the prover device 10 by the prover and to the verifier device 20 by the verifier by using the input devices 16 and 26, respectively. Further, x is
inputted to the prover device 10 by the prover by using the input device 16. The prover or the proof module 14 secures storage regions such as STORE[G] 12g, STORE[H] 12h, STORE[n]12n, and STORE[x]12x
on the RAM 12, and writes G, H, n, x to each of those regions, respectively. Similarly, the verifier or the verification device 24 secures storage regions such as STORE[G]22g, STORE[H]22h, and STORE
[n]22n on the RAM 22, and writes G, H, n to each of those regions, respectively.
The proof module 14 secures storage regions such as STORE[V]12v, STORE[RandX]12rx, and STORE[RandR] 12rr, STORE[Y]12y, and STORE[A] 12a on the RAM 12. Those storage regions are regions required only
during executions of the proof module, so that those regions may be dynamically secured when executing the proof module 14. Further, while explanations are provided hereinafter on assumption that the
regions are different regions from each other on the RAM 12, a same region may be used for STORE[Y]12y and STORE[A] 12a since those regions are not used simultaneously.
Similarly, the verification module 24 secures storage regions such as STORE[V]22v, STORE[W1]22w1, - - - , STORE[WN]22wn, STORE[C1]22c1, - - - , STORE[CN]22cn, STORE[W]22w, and STORE[C]22c on the RAM
22. Those storage regions are regions required only during executions of the verification module 24, so that those regions may be dynamically secured when executing the verification module 24.
(Operations of Proof Module)
[0054] FIG. 2
FIG. 4
are flowcharts showing operations of the proof module 14 shown in
FIG. 1
. In
FIG. 2
FIG. 4
, a section of the proof module 14, which performs operations (steps S101 to 114) illustrated in
FIG. 2
, is referred to as a "first processing unit" in Claims, a section of the proof module 14, which performs operations (steps S115 to 123) illustrated in
FIG. 3
, is referred to as a "second processing unit" in Claims, and a section of the proof module 14, which performs operations (steps S124 to 132) illustrated in
FIG. 4
, is referred to as a "third processing unit" in Claims.
When processing is started, the proof module 14 first calculates V=Θκ(G, H) by a following expression, and stores V in STORE[V]12v (step S101).
=Θ.sub.κ(G,H) [Expression 3]
Subsequently, the proof module 14 randomly selects RandX and RandR which are both bit sequences of α-bit, and writes selected RandX, RandR to STORE[RandX]12rx, STORE[RandR]12rr, respectively (Steps
S102 to 103).
Then, the proof module 14 sets initial value of variable j to 0 (step S104), and executes steps S107 to 114 to be described later for each case of j=1 to N by incrementing j by 1. When it reaches
j≧N+1, the proof module 14 advances to step S115 to be described later (steps S105 to 106).
The proof module 14 reads RandX from STORE[RandX]12rx, and calculates Y0 by a following expression (step S107).
=F.sub.λ+ξ(R.sub.andX,j) [Expression 4]
Subsequently, the proof module 14 reads RandR from STORE[RandR]12rr, and calculates R0 by a following expression (step S108).
=F.sub.λ+π(R.sub.andR,0∥j) [Expression 5]
Further, the proof module 14 calculates T0 by a following expression (step S109).
) [Expression 6]
Then, the proof module 14 reads V from STORE[V]12v, calculates V by a following expression, and overwrites acquired V on STORE[V]12v (step S110).
) [Expression 7]
Subsequently, the proof module 14 reads x from STORE[x]12x, and calculates Y1 by a following expression (step S111).
[Expression 8]
Then, the proof module 14 reads RandR from STORE[RandR]12rr, and calculates R1 by a following expression (step S112).
=F.sub.λ+π(R.sub.andR,1∥j) [Expression 9]
The proof module 14 calculates T1 by a following expression from acquired Y1 and R1 (step S113).
) [Expression 10]
Further, the proof module 14 reads V from STORE[V] 12v, calculates V by a following expression, and overwrites acquired V on STORE[V]12v (step S114).
) [Expression 11]
Thereafter, the processing of the proof module 14 returns to step S105, and the processing of steps S107 to 114 is repeated by incrementing j by 1 until it reaches j≧N+1.
When it is judged in step S106 as j≧N+1, the proof module 14 secures the storage region STORE[Y] 12y on the RAM 12, sets the initial value of Y as Y=0, and writes Y to STORE[Y]12y (step S115).
Subsequently, the proof module 14 sets the initial value of the variable j to 0 (step S116), and executes steps S119 to 121 to be described later for each case of j=1 to N by incrementing j by 1.
When it reaches j≧N+1, the proof module 14 advances to step S122 to be described later (steps S117 to 118).
The proof module 14 reads RandX from STORE[RandX]12rx, and calculates Y0 by a following expression (step S119).
=F.sub.λ+ξ(R.sub.andX,j) [Expression 12]
Subsequently, the proof module 14 calculates U by a following expression (step S120).
=Θ.sub.κ(V,j) [Expression 13]
Then, the proof module 14 reads Y from STORE[Y] 12y, calculates Y by a following expression, and overwrites acquired Y on STORE[Y]12y (step S121).
[Expression 14]
When it is judged in step S118 as j≧N+1, the proof module 14 reads each of G, Y, n from STORE[G] 12g, STORE[Y] 12y, STORE[n]12n to calculate A by a following expression (step S122), and writes
acquired A to STORE[A] (step S123). In a case where STORE[Y]12y and STORE[A]12a use different storage regions, STORE[Y] 12y may be released since STORE[Y]12y is not used in the following steps.
mod n [Expression 15]
Subsequently, the proof module 14 sets the initial value of the variable j to 0 (step S124), and executes steps S127 to 132 to be described later for each case of j=1 to N by incrementing j by 1.
When it reaches j≧N+1, the proof module 14 ends the processing (steps S125 to 126). The steps 127 to 132 are the loop 15 described above.
The proof module 14 reads RandX from STORE[RandX]12rx, and calculates Y0 by a following expression (step S127).
=F.sub.λ+ξ(R.sub.andX,j) [Expression 16]
Thereafter, the proof module 14 reads x from STORE[x]12x, and calculates Y1 by a following expression (step S128).
[Expression 17]
Then, the proof module 14 reads RandR from STORE[RandR] 12rr, and calculates R0 and R1 by a following expression (step S129).
=F.sub.λ+π(R.sub.andR,1∥j) [Expression 18]
Further, the proof module 14 calculates T0 and T1 by a following expression (step S130).
) [Expression 19]
Then, the proof module 14 reads each of V and A from STORE[V]12v and STORE[A]12a, and calculates c by a following expression (step S131).
(V,A,j) [Expression 20]
At last, the proof module 14 calculates each of Y, T, R by a following expression, and outputs (c, Y, R, T) to the verifier device 20 via the communication interface 13 (step S132).
-c) [Expression 21]
(Operations of Verification Module)
[0079] FIG. 5
to FIG. 7 are flowcharts showing operations of the verification module 24 shown in
FIG. 1
. In
FIG. 5
to FIG. 7, a section of the verification module 24, which performs operations (steps S201 to 211) illustrated in
FIG. 5
, is referred to as a "first processing unit" in Claims, and a section of the verification module 24, which performs operations (steps S212 to 218) illustrated in
FIG. 6
, is referred to as a "second processing unit" in Claims. The "first processing unit" and the "second processing unit" are collectively referred to as a "processing module". A section of the
verification module 24, which performs operations (steps S219 to 226) illustrated in FIG. 7, is referred to as a "judging unit" or a "third processing unit" in Claims.
When processing is started, the verification module 24 first calculates V by a following expression from G and H given in advance, and writes acquired. V to STORE[V]22v (step S201).
=Θ.sub.κ(G,H) [Expression 22]
Then, the verification module 24 sets the initial value of the variable j to 0 (step S202), and executes steps S205 to 211 to be described later for each case of j=1 to N by incrementing j by 1. When
it reaches j≧N+1, the verification module 24 advances to step S212 to be described later (steps S203 to 204). The steps 205 to 211 are the loop 25 described above.
The verification module 24 receives each numerical value of c, Y, T, R transmitted from the proof module 14 as inputs (step S205). Subsequently, the verification module 24 judges whether c=1 or c=0,
and calculates T0 and T1 by using Expression 23 when c=1 and by using Expression 24 when c=0 (steps S206 to 208).
=T [Expression 23]
=Θ.sub.λ(Y,R) [Expression 24]
Subsequently, the verification module 24 reads the value of V stored in STORE[V]22v, calculates V by a following expression, and overwrites acquired V on STORE[V]22v (step S209).
) [Expression 25]
Further, the verification module 24 reads the value of V stored in STORE[V]22v again, calculates V by a following expression, and further overwrites acquired V on STORE[V]22v (step S210).
) [Expression 26]
Then, the verification module 24 defines Cj and Wj as in following expressions, and writes defined Cj and Wj to STORE[Cj]22cj and STORE[Wj]22wj, respectively (step S211).
=Y [Expression 27]
Thereafter, the processing of the verification module 24 returns to step S203, and the processing of steps S205 to 211 is repeated until it reaches j≧N+1 by incrementing j by 1.
When it is judged in step S204 as j≧N+1, the verification module 24 defines as W=0, C=0, and writes W, C to STORE[W]22w, STORE[C]22c, respectively (step S212). Then, the verification module 24 sets
the initial value of the variable j to 0 (step S213), and executes steps S216 to 218 to be described later for each case of j=1 to N by incrementing j by 1. When it reaches j≧N+1, the verification
module 24 advances to step S212 to be described later (steps S214 to 215).
The verification module 24 calculates U by a following expression (step S216).
=Θ.sub.κ+ξ(V,j) [Expression 28]
Then, the verification module 24 reads W, Wj from STORE[W]22w, STORE[Wj]22wj, respectively, calculates W by a following expression by using the value of U calculated in step S216, and overwrites
acquired W on STORE[W] (step S217).
U [Expression 29]
Similarly, the verification module 24 reads C, Cj from STORE[C]22c, STORE[Cj]22cj, respectively, calculates C by a following expression, and overwrites acquired C on STORE[C] 22c (step S218).
U [Expression 30]
Thereafter, the processing of the verification module 24 returns to step S214, and the processing of steps S216 to 218 is repeated by incrementing j by 1 until it reaches j≧N+1.
When it is judged in step S214 as j≧N+1, the verification module 24 reads G, H, n from STORE[G]22g, STORE[H]22h, STORE[n]22n, respectively, and calculates A by a following expression (step S219).
C mod n [Expression 31]
Then, the verification module 24 sets the initial value of the variable j to 0 (step S220), and executes steps S223 to 224 to be described later for each case of j=1 to N by incrementing j by 1. When
it reaches j≧N+1, the verification module 24 advances to step S226 to be described later (steps S221 to 222).
The verification module 24 reads V from STORE[V]22v, and judges whether or not the condition regarding Cj shown by a following expression applies (steps S223 to 224). When judged that it applies, the
verification module 24 outputs "reject" to the prover device 10 (details thereof will be described later), and ends the processing (step S225).
(V,A,j) [Expression 32]
When the condition shown in Expression 32 does not apply, the verification module 24 executes nothing special but returns to the processing of step S221, and repeats the processing of steps S223 to
224 until it reaches j≧N+1 by incrementing j by 1.
If "reject" has not been outputted till then even when it is judged in step S222 as j≧N+1, the verification module 24 outputs "accept" to the prover device 10 (details thereof will be described
later) and ends the processing (step S226).
"Reject" outputted from the verification module 24 to the prover device 10 means that the verifier device 20 has judged that the fact to be proved by the prover device 10 is inadequate, while
"accept" means that it is adequate. Upon receiving "accept" or "reject" output from the verifier device 20, the prover device 10 may inform so to the user via a display device, for example. Further,
it is also possible to inform so to another program that uses the zero-knowledge proof as a sub-routine within the prover device 10 so as to continue the processing by the program upon receiving an
"accept" output and to stop the processing there upon receiving a "reject" output.
(Overall Operations of First Embodiment)
Next, the overall operations of the embodiment will be described. The operations according to the embodiment are as follows. On the zero-knowledge proof device side of the zero-knowledge proof system
1 constituted with the zero-knowledge proof device (prover device 10) and the zero-knowledge verifier device (verifier device 20), the first processing unit 14a calculates pseudorandom numbers from
an arbitrary random number sequence and a pseudorandom function (
FIG. 2
: steps S107 to 108), and the first processing unit executes a plurality of times of processing which calculates the hash values based on the calculated pseudorandom numbers, the pseudorandom numbers
stored in the temporary memory unit (RAM 12) provided in advance, and the hash values acquired in the past, and overwrites the calculated pseudorandom numbers and hash values on the temporary memory
unit (
FIG. 2
: steps S109 to 114), the second processing unit 14b determines a part of the plurality of pseudorandom numbers outputted to the zero-knowledge proof device based on the hash values (
FIG. 3
: steps S115 to 122), and the third processing unit transmits the hash values acquired by re-calculating a part of the pseudorandom numbers to the zero-knowledge verification device (
FIG. 4
: steps S125 to 132). On the zero-knowledge verification device side, the data receiving module (communication interface 23) receives new input data sequentially from the zero-knowledge proof device
FIG. 5
: step S205), the verification module 24 overwrites the hash value of input data containing the variable stored in the temporary memory unit (RAM 22) provided in advance and the input data as a new
variable on the temporary memory unit every time the data receiving module receives the input data (
FIG. 5
: steps S209 to 210), judges whether to authenticate or to reject the zero-knowledge proof device based on the variable, and returns the result of judgment to the zero-knowledge proof device (FIG. 7:
steps S219 to 226).
Note here that the first processing unit 14a of the zero-knowledge proof device repeats N-times (N is a natural number of 2 or larger) of processing to read elements G, H of a group (
FIG. 2
: steps S101), to read integer x, to define initial values of data V showing the pseudorandom numbers, to calculate first and second pseudorandom function values, and to update the hash values of the
data containing V and the first and second data pseudorandom function values Y0 and R0 as new V (
FIG. 2
: steps S107 to 114).
Subsequently, the following second processing unit 14b of the zero-knowledge proof device repeats N-times of processing to give some kind of initial value to data Y that shows a part of the plurality
of pseudorandom numbers, and to take a value acquired by adding a value based on a hash value U of data containing V and j to Y as new Yin the j-th processing (1≦j≦N) (
FIG. 3
: steps S119 to 121).
Further, the third processing unit 14c of the zero-knowledge proof device repeats N-times of processing to take a residue of Y-th power to G as A (
FIG. 4
: steps S127 to 132), to calculate first and second hash values T0 and T1 based on Y0 and R0 in the j-th processing (1≦j≦N) (
FIG. 4
: steps S127 to 130), to calculate 1-bit data c from data containing V, A, and j (
FIG. 4
: step S131), and to transmit Y0, R0, T1 to the zero-knowledge verification device 20 when c is 0, while transmitting Y1, R1, T0 when c is 1 (
FIG. 4
: step S132).
The first processing unit 24a provided to the processing module 24 of the zero-knowledge verification device (verifier device 20) repeats N-times of processing to read elements G, H of a group, to
give some kind of initial values to data V that shows variables (
FIG. 5
: step S201), receives data c, data Y, data T, and data R from the zero-knowledge proof device 10 as inputs in the j-th processing (1≦j≦N) (
FIG. 5
: step S205), to calculate first and second hash values containing Y and R, and to overwrite the values on the temporary memory unit as data V anew (
FIG. 5
: steps S207 to 211).
Subsequently, the second processing unit 24h repeats N-times of procedure which performs initial setting of data W to 0 and data C to 0, respectively (
FIG. 6
: step S212), takes the hash value of data containing V and j as U in the j-th processing (1≦j≦N) (
FIG. 6
: step S216), takes a result acquired by adding W to a product of Wj and U as new W (
FIG. 6
: step S217); and takes a result acquired by adding C to a product of Cj and U as new C (
FIG. 6
: step S218).
Subsequently, the third processing unit 24c repeats N-times of procedure to take a result acquired by multiplying a residue of W-th power to G by a minus-power residue of the data C calculated by the
second processing unit to H as A (FIG. 7: step S219), and output data (reject) indicating rejection to stop the zero-knowledge proof device when the hash value of data containing V, A, and j does not
match with q in the j-th processing (1≦j≦N) (FIG. 7: steps S224 to 225), while outputting data (accept) indicating to authenticate the zero-knowledge proof device when "reject" is not outputted after
repeating that procedure for N-times (FIG. 7: step S226).
Each of the above-described operation steps may be put into programs that can be executed by a computer, and those programs may be executed by the prover device 10 and the verifier device 20 which
are computers directly executing each of the above-described steps. With those operations, the embodiment can provide following effects.
This embodiment makes it possible to reduce the required storage capacity compared to the technique depicted in Patent Document 1. The reasons are as follows. The first reason is that some of data
such as Yi in the embodiment are generated by the pseudorandom function using a same key. As a result, it becomes possible to calculate those data from the key as necessary only through storing the
key of the pseudorandom function. Therefore, it is unnecessary to store those data, thereby making it possible to reduce the required storage capacity.
The second reason is that the method for calculating the hash value is changed. With the technique depicted in Patent Document 1, it is necessary to store all of T1 to TN simultaneously and calculate
hash values V=Θ(T1, - - - , TN) at once. Thus, it is necessary to provide the storage capacity proportional to N. In the meantime, the embodiment is designed to calculate the hash value by
initializing V to an empty sequence first, and calculating V=Θ(V, Ti) for i=1 to N. Thus, Ti can be erased after V=Θ(V, Ti) is calculated. Therefore, it is not necessary to store all of T1, - - - ,
TN simultaneously, and the value with which calculation is completed can be erased in order.
The third reason is that the data output is done in a subdivided manner. With the technique depicted in Patent Document 1, the data is outputted after completing all the calculations of the data to
be outputted. Thus, the amount of data to be outputted becomes proportional to N. In the meantime, with the embodiment, the data with which the calculation is completed can be erased from the storage
region after the output is done at that point.
Therefore, the required storage capacity can be reduced to be small with the embodiment. Thereby, the embodiment can be utilized even with the device whose storage capacity is small.
Second Embodiment [0110]
Structures of a network, hardware, and a rough structure of software according to a second embodiment of the present invention are the same as those of the first embodiment described by referring to
FIG. 1 to FIG. 7. The different points of the second embodiment with respect to the first embodiment described above are that the zero-knowledge proof device (prover device 10) includes a first
storage device (storage 317) which collectively stores and outputs sets of pseudorandom numbers and hash values outputted to the zero-knowledge verification device (verifier device 20), and that the
zero-proof verification device (verifier device 20) includes a second storage device (storage 327) which collectively stores the sets of the pseudorandom numbers and hash values received at the date
receiving module from the zero-knowledge proof device (prover device 10).
The storage device has a larger storage capacity per unit price compared to a volatile storage module, so that it is easy to achieve a mass storage capacity. It is possible with this embodiment to
also achieve the same effects as those of the first embodiment by having such structure.
This will be described in more details hereinafter.
[0113] FIG. 8
is an explanatory illustration showing the structure of a zero-knowledge proof system 301 according to the second embodiment of the present invention. The zero-knowledge proof system 301 is
constituted with a prover device 310 that is a computer device operated by the prover and a verifier device 320 that is a computer device operated by the verifier, and the prover device 310 and the
verifier device 320 are connected mutually. The same names and reference numerals are applied to same elements as those of the first embodiment.
The prover device 310 includes a CPU 11, a RAM 12, a communication interface 13, and an input device 16 as in the case of the prover device 10 according to the first embodiment. In addition, the
prover device 310 includes the storage 317 that is a mass-capacity non-volatile storage module. The storage 317 specifically is a hard disk or a flash memory, for example. A proof module 314 as a
computer program executed by the CPU 11 is stored in the RAM 12 and executed.
Similarly, the verifier device 320 includes a CPU 21, a RAM 22, a communication interface 23, and an input device 26. In addition, the verifier device 320 includes the storage 327 that is a
mass-capacity non-volatile storage module such as a hard disk or a flash memory. A verification module 324 as a computer program executed by the CPU 21 is stored in the RAM 22 and executed.
FIG. 9
is a flowchart showing the operations of the proof module 314 shown in
FIG. 8
FIG. 9
only shows the different points with respect to the operations of the proof module 14 shown in
FIG. 2
FIG. 4
. The operations of steps S101 to 131 are the same as the operations of the proof module 14 shown in
FIG. 2
FIG. 4
. However, (c, Y, R, T) calculated in step S132 are not transmitted to the verifier device 20 but saved in the storage 317 (step S132b). Further, when it reaches j≧N+1 in step S126, all (c, Y, R,T)
saved in the storage 317 are transmitted to the verifier device 320 via the communication interface 13 (step S133b), and the processing is ended thereafter.
[0116] FIG. 10
is a flowchart showing the operations of the verification module 324 shown in
FIG. 8
. After step S201, the verification module 324 receives all (c, Y, R, T) from the proof module 314, and saves those in the storage 327 (step S201b). The operations thereafter are the same as the
operations of the verification module 24 shown in
FIG. 5
to FIG. 7, i.e., steps S202 to 226, except that step S205 is changed to "read data from the storage 327" (step S205b).
The storages 317 and 327 as the non-volatile storage modules have a larger storage capacity per unit price compared to the RAMs 12 and 22 as the volatile storage modules, so that it is easy to
achieve a mass storage capacity. Therefore, it is also possible with the second embodiment to achieve the same effects as those of the first embodiment.
While the present invention has been described heretofore by referring to the specific embodiments illustrated in the drawings, the present invention is not limited only to the embodiments shown in
the drawings. Any known structures can be employed, as long as the effects of the present invention can be achieved therewith.
While cases of building the embodiments of the present invention as hardware are described in the above, the functions of the zero-knowledge proof system, the zero-knowledge proof device, and the
zero-knowledge verification device may also be built as programs achieved on software. In that case, the programs are recorded on a recording medium and can be treated as commercial dealings.
This Application claims the Priority right based on Japanese Patent Application No. 2008-316022 filed on Dec. 11, 2008 and the disclosure thereof is hereby incorporated by reference in its entirety.
INDUSTRIAL APPLICABILITY [0121]
The present invention can be utilized broadly in scenes where the discrete-logarithm zero-knowledge proofs are utilized. More specifically, the present invention can be utilized for public-key
cryptography, digital signatures, group signatures, electronic voting, and the like. Particularly, the present invention is suited for utilizing such technique in devices with a small storage
capacity such as mobile phone terminals and PDAs (Personal Digital Assistants).
REFERENCE NUMERALS [0122]
1, 301 Zero-knowledge proof system
10, 310 Prover device
11, 21 CPU
12, 22 RAM
12g STORE[G]
12h STORE[H]
12n STORE[n]
12x STORE[x]
12v STORE[V]
12rx STORE [RandX]
12rr STORE[RandR]
12y STORE[Y]
12a STORE[A]
13, 23 Communication interface
14, 314 Proof module
14a, 24a, 324a First processing unit
14b, 24b Second processing unit
14c, 24c, 314c Third processing unit
15, 25 Loop
16, 26 Input device
20, 320 Verifier device
22g STORE[G]
22h STORE[H]
22n STORE[n]
22v STORE[V]
22w1 STORE[W1]
22wn STORE[WN]
22c1 STORE[C1]
22cn STORE[CN]
22w STORE[W]
22c STORE[C]
24, 324 Verification module
317, 327 Storage
Patent applications by Isamu Teranishi, Tokyo JP
Patent applications in class Authentication by digital signature representation or digital watermark
Patent applications in all subclasses Authentication by digital signature representation or digital watermark
User Contributions:
Comment about this patent or add new information about this topic:
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Size the cylinder right for proper servo operation
No matter how much attention is given to motion controller selection and programming, hydraulic system performance may be limited by incorrect sizing or physical location of components. A common
design oversight is the use of undersized cylinders. Designers sometimes intend to increase piston rod speed by specifying a cylinder with a smaller bore — based on the assumption that for a given
amount of oil flow, a smaller cylinder will produce quicker accelerations and higher velocities.
However, this only works for very light loads. For actuators moving moderate to heavy masses, acceleration, velocity, and deceleration are limited by the av ailable force — not by oil flow. Because
cylinder bore determines the force it can produce, if the bore is too small, the cylinder may be incapable of attaining the desired speeds or cycle times required by the application.
A real world example
We were recently asked to diagnose problems with a system operating incorrectly. After obtaining specific application data, we performed a hydraulic system simulation. Our mathematical model
suggested-that more force was necessary. The simulation was changed to evaluate cylinders with larger bores. The dramatic results, graphed in Figures 1 and 2, suggested replacing the existing 2-in.
bore cylinders with 3¼-in. bore cylinders and replacing the hydraulic valves with appropriately larger ones.
Figure 1 shows how an undersized cylinder may not provide enough force to quickly decelerate a large load when extending. The cylinder cavitates (pressure goes to zero) on the cap end, and pressure
on the rod end exceeds system pressure. The system is out of control because oil on the rod side is going back into the oil supply, thereby actually decreasing deceleration!
When the cylinder diameter is increased to 3¼-in., Figure 2, the piston rod accelerates the load at the same rate, but the pressure doesn't change as much because of the greater piston surface area.
The cylinder pressures throughout the cycle are in the middle of the pressure range, with plenty of pressure drop across the piston to maintain positive control.
Increasing the cylinder bore increases the natural frequency (stiffness)of the system, allowing the motion controller to manage faster accelerations and decelerations, and yielding higher system
performance when properly tuned. However, larger cylinders require larger valves and higher flow. Very large cylinders increase cost. And because large valves tend to be slower, at some point
increasing valve size will no longer increase system response.
Force makes the system go
Force is directly related to cylinder size. Often, novice designers apply the familiar, but oversimplified, formula:
v = Q ÷ A,
where v is the piston velocity,
Q is the flow, and
A is the piston area.
But this equation is accurate only if the mass is zero. It should only be used to calculate flow: Q = v x A. The following "rule of thumb" formula takes into account the force needed to produce
acceleration and the pressure drop required by servovalves for control. (Typically, servovalves are rated at 70 bar, approximately equal to 1000 psi.)
A = L[P] ÷ (P[S] — 1000 psi),
where L[P] is load pressure, and
P[S] is the system pressure.
This formula assumes that peak load occurs at peak speed. The peak load includes the force necessary to accelerate or decelerate the load, friction, and the load's weight if the system is vertical.
Minimum system pressure should be used. This formula, which should be applied for both extend and retract directions, neglects the opposing force required to push oil out the opposite end of the
cylinder. Therefore, the estimated size should be considered a minimum.
For high-performance motion, a second check should be made to ensure that the natural frequency of the system is higher than the frequency of motion generated by the motion profile. For example, if
the frequency of acceleration is 5 Hz; the actuator's natural frequency should be three to four times higher:
A[avg] = (f × 4)^2× π × L[S] × W[L] ÷ (g × ß)
where A[avg] is the average area of the piston,
f is the cycle frequency,
L[S] is the stroke length,
W[L] is the weight of the load,
g is the acceleration due to gravity (32 ft/sec^2), and
is the bulk modulus (incompressibility constant) of oil (~200,000 psi).
This formula also tends to underestimate the correct cylinder bore because it makes some optimistic assumptions. The most significant is that the valve is mounted right on the cylinder. If this isn't
the case, the cylinder stroke should be increased by about the length of the hose connecting the valve to the cylinder.
The area of the hose is smaller than that of the cylinder, but hose is more compliant than cylinder tubing. Hose or extra pipe between the valve and the cylinder complicate these calculations and
reduce performance.
For the most critical applications, the VCCM equation, developed by Jack Johnson, takes into account additional variables and produces an excellent estimation of performance.
Peter Nachtwey is president of Delta Computer Systems Inc., Battle Ground, Wash. For additional technical information, e-mail him at support@deltamotion.com or visit www.deltamotion.com.
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Archimedean Algorithm for Approximating Square Roots
October 5th 2010, 10:43 PM #1
Mar 2010
Archimedean Algorithm for Approximating Square Roots
An algorithm for approximating the sqrt (square root) of A is the following: To approximate sqrt A, begin with the initial approximation (obtained by "eyeballing it") of x0. Then, let the next
approximation be x1 = ((x0)^2 + A)/(2*x0), then x2 = ((x1)^2 + A)/(2*x1) and generally, use the recursive definition xn+1 = ((xn)^2 + A)/(2*xn).
Assuming that the sequence of successive approximations converges to a limit L, show that L = sqrt A.
I think that I need to take limits of both sides of xn+1 = ((xn)^2 + A)/(2*xn). I do not know how I would go about doing this.
Hint -
What can you say about
(sqrt(A) - x_n) and (sqrt(A) - x_n+1)
It should be a bounded and monotonic sequence and hence will converge
Yes, that is exactly what you need to do. You are told that $x_n$ converges to $\scriptstyle L$ as $n\to\infty$, and that means that $x_{n+1}$ also converges to $\scriptstyle L$. So let $n\to\
infty$ on both sides of that equation, and it will give you an equation for $\scriptstyle L.$
I am sorry but I am still not getting anywhere. I have a weak understanding of limits. Would you be so kind as to show me how I would get the equation for L?
I think what Opalg is suggesting is the in limit n->infinity; x_n = x_n+1 = L (though you have not proved it but that is not needed as this is given in the question)
Now look at x_n+1 = (x_n^2 + A) / 2*x_n
In the limit n-> infinity this eq will become
L = (L^2 + A) / 2L
solve for L
Thank you so much!
October 6th 2010, 12:10 AM #2
Super Member
Apr 2009
October 6th 2010, 10:40 AM #3
October 6th 2010, 06:15 PM #4
Mar 2010
October 6th 2010, 08:15 PM #5
Super Member
Apr 2009
October 6th 2010, 10:05 PM #6
Mar 2010
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ALEX Lesson Plans
Subject: Mathematics (3 - 5)
Title: What's the Question Again?
Description: This lesson focuses on problem solving strategies with word problems that are 2 or more steps. Students are forced through the strategies to focus on the content of a problem before
attempting to answer the "question" which tells what they are to solve. Visualization and communication are reinforced throughout the lesson. The strategies are generic. Included are problems for
3rd, 4th, and 5th grade computation. However, these same strategies can be used for all grade levels. This is a College- and Career-Ready Standards showcase lesson plan.
Subject: Mathematics (4)
Title: What's Your Question? Multiplication Word Problems
Description: Students will explore multiplication and word problems. They will work cooperatively to examine a story scenarioand list possible questions to develop a completed word problem and solve
it using a variety of strategies. The students will also create a word problem using an online story creator.
Subject: Mathematics (4), or Technology Education (3 - 5)
Title: Little Red Riding Hood's Journey to Grandma's
Description: The students will listen to the story of Little Red Riding Hood and work subtraction word problems connected to the story. The problems are related to distance traveled. They will work
cooperatively with partners and share subtraction strategies with their peers. At the end of the lesson they will work a subtraction story problem and show strategies for solving.
Subject: Mathematics (4), or Technology Education (3 - 5)
Title: Nursery Rhyme Word Problems
Description: In this lesson students will create basic addition, subtraction, multiplication, and division word problems using common nursery rhymes. Students will distinguish the key elements of
word problems and incorporate those elements into a nursery rhyme.
Subject: Mathematics (4)
Title: All squared up!
Description: In this AMSTI 5 E lesson plan students will be engaged in an investigation to discover the patterns in squared numbers then be able to apply the concepts needed to evaluate the
expression. This lesson plan was created by exemplary Alabama Math Teachers through the AMSTI project.
Subject: Mathematics (4 - 5)
Title: Exploring Prime Numbers
Description: Exploring Prime Numbers is a hands-on, minds-on math lesson that engages students in the discovery of prime numbers through the construction of multiplication arrays. Students then
extend on their background knowledge of multiples and their new understanding of prime and composite numbers to complete an Eratosthenes’ Sieve. This lesson plan was created by exemplary Alabama Math
Teachers through the AMSTI project.
Subject: Mathematics (4), or Technology Education (3 - 5)
Title: Math at Dinner!
Description: Each group of students will use word processing software to create a restaurant menu complete with food titles and prices. Groups will then exchange menus and use the menus to order a
meal. Each student will total their meal, add 10% sales tax, 15% gratuity and figure their change from $20.00. This lesson plan was created as a result of the Girls Engaged in Math and Science
University, GEMS-U Project.
Subject: English Language Arts (5), or English Language Arts (5), or Mathematics (3 - 5), or Technology Education (3 - 5)
Title: Fun with Problem Solving
Description: The object of this lesson is to teach students that problem solving can be interesting and fun. It is important to pay attention to what is asked and information that is in the
problem.This lesson plan was created as a result of the Girls Engaged in Math and Science, GEMS Project funded by the Malone Family Foundation.
Subject: Character Education (K - 12), or Mathematics (3 - 4)
Title: Estimation Station
Description: During this lesson students will learn the difference between estimation and guessing. Knowing how to estimate is an essential skill that will help students determine approximate totals
as well as check the reasonableness of their answer to a problem. After listening to the book Betcha by Stuart J. Murphy, the students will be given opportunities to practice the strategies that are
presented. This lesson plan was created as a result of the Girls Engaged in Math and Science, GEMS Project funded by the Malone Family Foundation.
Subject: Mathematics (3 - 4)
Title: Number Clues
Description: This lesson allows students to observe and participate in problem solving activites involving mystery numbers, clues, and characteristics of numbers.This lesson plan was created as a
result of the Girls Engaged in Math and Science, GEMS Project funded by the Malone Family Foundation.
Subject: Mathematics (4 - 7)
Title: Comparing Fuel Economy (adapted from CMP "Comparing and Scaling" Investigation 4.1)
Description: In this lesson students will explore rates. They will use the concept of rates to compute and compare fuel economy.This lesson plan was created as a result of the Girls Engaged in Math
and Science, GEMS Project funded by the Malone Family Foundation.
Thinkfinity Lesson Plans
Subject: Mathematics,Science
Title: Classroom Paper Add Bookmark
Description: In this lesson, one of a multi-part unit from Illuminations, students investigate data in connection with recyclable materials and develop plans to help the environment. Students gather
and graph information about their classroom paper use over a period of several days and then interpret graphical information and use the data as a basis for future planning.
Thinkfinity Partner: Illuminations
Grade Span: K,PreK,1,2,3,4,5
Subject: Mathematics
Title: Mathematics and Football: Unit Overview Add Bookmark
Description: In this five-lesson unit, from Illuminations, students participate in activities in which they analyze information represented graphically. Students are asked to discuss, describe, read,
and write about the graphs and the information they contain. These activities focus on connections between mathematics and football by using the Super Bowl. Students are asked to look at the Super
Bowl not just as the big game but as an opportunity to apply mathematics to some interesting problems. This unit includes an individual activity for four different levels plus one for parents to
complete with their child at home.
Thinkfinity Partner: Illuminations
Grade Span: K,PreK,1,2,3,4,5,6,7,8
Subject: Mathematics
Title: Picture This Add Bookmark
Description: In this lesson, from Illuminations, students compose number sentences based on mathematics stories and then conduct a survey to determine how various people interpret pictures
mathematically. They analyze and summarize the data they collect, and draw conclusions from their findings.
Thinkfinity Partner: Illuminations
Grade Span: 3,4,5
Subject: Mathematics
Title: Number and Operations Web Links Add Bookmark
Description: This collection of Web links, reviewed and presented by Illuminations, offers teachers and students information about and practice in concepts related to arithmetic. Users can read the
Illuminations Editorial Board's review of each Web site, or choose to link directly to the sites.
Thinkfinity Partner: Illuminations
Grade Span: K,1,2,3,4,5,6,7,8,9,10,11,12
Subject: Mathematics
Title: Communicating about Mathematics Using Games: Playing Fraction Tracks Add Bookmark
Description: Mathematical games can foster mathematical communication as students explain and justify their moves to one another. In addition, games can motivate students and engage them in thinking
about and applying concepts and skills. This e-example from Illuminations contains an interactive version of a game that can be used in the grades 3-5 classroom to support students' learning about
fractions. e-Math Investigations are selected e-examples from the electronic version of the Principles and Standards of School Mathematics (PSSM). The e-examples are part of the electronic version of
the PSSM document. Given their interactive nature and focused discussion tied to the PSSM document, the e-examples are natural companions to the i-Math investigations.
Thinkfinity Partner: Illuminations
Grade Span: 3,4,5
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Need help with Elimination (Linear Equation)
September 27th 2008, 11:12 AM
Need help with Elimination (Linear Equation)
So, i'm having a very difficult time understanding elimination and would appreciate anyones help on that matter (most of my class feels the same, so i'm not alone in this, for once :P)
The equation that i'm stumped on is.
2x + 4y = 7
4x - 3y = 3
I understanding the graphing and subsitution methods pretty well but not this method.
A step-by-step walkthrough would help alot with this.
Any help?
September 27th 2008, 11:16 AM
So, i'm having a very difficult time understanding elimination and would appreciate anyones help on that matter (most of my class feels the same, so i'm not alone in this, for once :P)
The equation that i'm stumped on is.
2x - 4y = 7
4x - 3y = 3
I understanding the graphing and subsitution methods pretty well but not this method.
A step-by-step walkthrough would help alot with this.
Any help?
the point is, you want to eliminate one of the variables so that you can solve for the other. to do this, you must have the coefficients of the variable you want to eliminate to be equal to each
other (except maybe by a factor of a minus sign). now, lets say we want to eliminate x. we want the coefficients of x in both equations to be equal, in which case we can subtract one equation
from the other and get rid of the x. (if we make one coefficient the negative of the other, then we simply add both equations).
so lets do that. eliminate the x. what would you do to get the coefficients of x the same in both equations?
September 27th 2008, 11:33 AM
Opps, I made a mistake writing the equation out.
It's 2x + 4y = 7, not 2x - 4y = 7.
Does that change anything?
September 27th 2008, 11:39 AM
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Mathematical approximations and how wrong they are
36 Responses to “Mathematical approximations and how wrong they are”
1. Pessimist! I see it as “how right they are” ;-)
□ Greg Miller says:
Honestly, in most engineering uses, they’re all good enough. That’s something that’s forgotten a lot, but a problem in engineering is people striving for perfection, when good enough will
get the job done sooner and cheaper. There are some things where it’s necessary, but in most cases it isn’t.
☆ Paul Renault says:
I kinda recall that knowing π to a fairly low number of decimal places is plenty accurate for just about anything.
For example, if you memorize π to 31 places (it’s really not that difficult), and you used that number (3.1415926535897932384626433832795) to calculate the circumference of the Milky Way
galaxy, you’d be off by about 1/20th of the diameter of a proton.
Just using “How I wish I could enumerate pi” (3.141592) when calculating Earth’s circumference from its diameter, you’d be off by a little more than 7 1/2 meters.
If you used the Katharevousa Greek mnemonic for π, giving you 22 decimal places, for that same calculation, you’d be back down to sub-atomic particle ‘error’ territory:
Ἀεὶ ὁ Θεὸς ὀ Μέγας γεωμετρεῖ,τὸ κύκλου μῆκος ἵνα ὁρίσῃ διαμέτρῳ,παρήγαγεν ἀριθμὸν ἀπέραντον,καὶ ὅν, φεῦ, οὐδέποτε ὅλον θνητοὶ θὰ εὕρωσι
○ Jorpho says:
”How I need a drink, alcoholic of course, after the heavy chapters involving quantum mechanics” will always be my favorite.
2. davide405 says:
Further evidence that e and pi are part of God’s address.
□ Guest says:
i like pi/e, pi*e, pi+e, and even pi-e. My favorite is pi^e.
☆ PaulDavisTheFirst says:
e^(i * pi) = -1
the universe – it works, bitches!
○ Steve Lord says:
In its alternate form, e^(i*pi) + 1 = 0, that equation includes five fundamental constants:
The additive identity, 0. [0 + any number = that number]
The multiplicative identity, 1. [ 1 * any number = that number]
One of the square roots of -1. [i^2 = -1]
The base of the natural logarithm.
The ratio of the circumference of a circle to its diameter.
■ Camp Freddie says:
I agree with most of the beautiful equation, but Pi is a terrible idea for a fundamental constant. Tau is much more sensible (the ratio of circumference to radius).It makes
trigonometry and calculus much more intuitive (tau = 360°, 1/2(tau)r^2 = area, etc.) Also, e^(i×tau)+0=1
Google “Pi is wrong” and have your mind blown. Lots of simple equations based on Pi need an extra factor of 2 in order to balance, since Pi treats diameter as the fundamental
property of a circle.
3. semiotix says:
As a woodworker, I’m constantly having to switch between metric and English length measurements. 5-over-eth-root-π is so much simpler to remember than 1m≈3′ 3 3/8″. Thanks, XKCD!
□ James B says:
I build a good bit of furniture, and like my combo metric/inch tape measure and combination square. So much easier to find half of 13-9/16″ when it is expressed as 344 mm. And I’m somewhat
partial to multiplying inches by 25.4 to get to mm.
☆ timquinn says:
Half of 13 9/16 is 6 1/2 +9/32. The fractional system is built for division by two. You don’t have to add it together and reduce the fractions to find it on the tape measure. If you
don’t need 1/32′s say in between 1/4 and 5/16. On the other hand if you were to scribble 344 mms on a piece of paper and then came back later to find you couldn’t tell if it was 394 or
349 or 399 or . . . with a fraction you know when there is a mistake and you have a good idea of what the number should be. If it says 5/10 or 6/16 you know there is problem and what it
probably was. Unless, of course, you did not grow up using fractions and just can’t see the advantages at all because you have been propagandized to accept top down solutions to very low
level problems.
○ James B says:
Just saying what works for me. And maybe you caught where I like measuring instruments marked both ways. (Preferably both facing the same direction.) For carpentry work I tend
more towards using fractional inch measurements: I intuitively know what 3-1/2″ inches looks like, and most of the work is just measuring and cutting to length. Plus, like you sort
of imply, it is less prone to errors. But for laying out mortise and tenons, or when I mark metal for machine work, I still like metric.
4. jerwin says:
You’d think that any calculator powerful enough to compute those values would have enough space for “physical constants”.
How precisely is “mean earth radius” known?
I’m at awe at the “proton-electron mass ratio” given that “phi” is (1+sqrt(5))/2
□ SamSam says:
How precisely is “mean earth radius” known?
Unfortunately there seems to be no snarky site called “let me Wikipedia that for you” like there is for Google, so I guess I’ll just be your lazyweb for you instead: http://en.wikipedia.org/
☆ Ambiguity says:
Perhaps what the Internet needs is a good meta-snark site — a site search site that will find snarky sites (like lmgtfy) for any occasion.
Then we could make ironic spin-offs of that site, and the eschaton would finally come in the form of a meta-singularity!
○ Guest says:
there is, or will be, an xkcd for that.
■ MarcVader says:
I like turtles.
☆ jerwin says:
I don’t understand.
I’ve calculated e*6*8^5 as 6370973.03545089 m
Your “source” gives the mean radius of the earth as 6,371.009 km, which deviates substantially from Randall’s calculation. Given that Randall never lies, he must be using a more accurate
reference for his numbers than some crummy wiki.
○ Ambiguity says:
The difference between 6370,973 km and 6371.009 km is about 100m. Saying that is falls “within actual variation” may be accurate. The earth is an oblate sphereoid, not a sphere, so
there could be some variation in the way it is defined and measured.
■ jerwin says:
Ah, you are correct. The various methods of calculating the radii give values ranging from 6367.445 m (Meridional Earth radius) to 6,372.797 km (Great Circle radius)– so the
actual value is only known to one in a thousand. Not very impressive.
The written form of Randall’s equation is more complicated than the actual value it represents.
○ timquinn says:
If you read the wiki, not just skim it for numbers, you would see that; #1 radii exist only on spheres, which the earth is not one. #2 Any idea of the mean radius of the earth
depends hugely on your needs and is slippery in the best circumstance. So your statement that you do not understand is the most relevant part of your post.
■ timquinn says:
In other words precision is useless with out accuracy.
5. Ambiguity says:
I love the hover-over text on this one: “If you ever find yourself raising log(anything)^e or taking the pi-th root of anything, set down the marker and back away from the whiteboard; something
has gone horribly wrong.”
6. nixiebunny says:
Fortunately, you can replace all occurrences of pi with 355/113 and achieve pretty good results. Because who (especially an XKCD reader) would be bothered to remember the numeric value of pi?
8. Mark Dow says:
A better approximation to the fine structure constant is (i^i)^pi, or i^(i*pi). It rolls off the tongue too, “eye to the eye to the pi”.
9. juan says:
xln(x)-x = ln(x!)
10. glittalogik says:
I’m kind of surprised he didn’t include the Golden Ratio/Fibonacci Sequence for approximate mile-to-km conversion. Each number in miles is close (within .2) to the next number in km.
□ invictus says:
Except for those pesky first members of the set. 1mi. is 1.609km.
11. Halloween_Jack says:
Gotta be honest here: this is one of those xkcds that just goes straight over my head, mostly, and I’ve always done well on the math portions of standardized tests. Well enough, anyway.
12. Jellodyne says:
Jenny’s Constant for the win . I got it, I got it!
□ timquinn says:
How many 7 digit numbers do you know by heart? Right? and this one is a member of a twin prime pair!
13. The approximation “kilometres in a mile = φ” is accurate to about one part in 185 and is actually useful.
□ Sparrow says:
I had an old car mileage computer that indicated φmiles rather than kilometres as a metric indication. Now it makes sense.
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Eugene L. Allgower and Kurt Georg
Classics in Applied Mathematics 45
Numerical continuation methods have provided important contributions toward the numerical solution of nonlinear systems of equations for many years. The methods may be used not only to compute
solutions, which might otherwise be hard to obtain, but also to gain insight into qualitative properties of the solutions. Introduction to Numerical Continuation Methods, originally published in
1979, was the first book to provide easy access to the numerical aspects of predictor corrector continuation and piecewise linear continuation methods. Not only do these seemingly distinct methods
share many common features and general principles, they can be numerically implemented in similar ways. Introduction to Numerical Continuation Methods also features the piecewise linear approximation
of implicitly defined surfaces, the algorithms of which are frequently used in computer graphics, mesh generation, and the evaluation of surface integrals.
To help potential users of numerical continuation methods create programs adapted to their particular needs, this book presents pseudo-codes and Fortran codes as illustrations. Since it first
appeared, many specialized packages for treating such varied problems as bifurcation, polynomial systems, eigenvalues, economic equilibria, optimization, and the approximation of manifolds have been
written. The original extensive bibliography has been updated in the SIAM Classics edition to include more recent references and several URLs so users can look for codes to suit their needs or write
their own based on the models included in the book.
This book continues to be useful for researchers and graduate students in mathematics, sciences, engineering, economics, and business looking for an introduction to computational methods for solving
a large variety of nonlinear systems of equations. A background in elementary analysis and linear algebra are adequate prerequisites for reading this book; some knowledge from a first course in
numerical analysis may also be helpful.
Table of Pseudo Codes; Preface to the Classics Edition; Foreword; Chapter 1: Introduction; Chapter 2: The Basic Principles of Continuation Methods; Chapter 3: Newton's Method as Corrector; Chapter 4:
Solving the Linear Systems; Chapter 5: Convergence of Euler-Newton-Like Methods; Chapter 6: Steplength Adaptations for the Predictor; Chapter 7: Predictor-Corrector Methods Using Updating; Chapter 8:
Detection of Bifurcation Points Along a Curve; Chapter 9: Calculating Special Points of the Solution Curve; Chapter 10: Large Scale Problems; Chapter 11: Numerically Implementable Existence Proofs;
Chapter 12: PL Continuation Methods; Chapter 13: PL Homotopy Algorithms; Chapter 14: General PL Algorithms on PL Manifolds; Chapter 15: Approximating Implicitly DeŻned Manifolds; Chapter 16: Update
Methods and their Numerical Stability; Appendix 1: A Simple PC Continuation Method; Appendix 2: A PL Homotopy Method; Appendix 3: A Simple Euler Newton Update Method; Appendix 4: A Continuation
Algorithm for Handling Bifurcation; Appendix 5: A PL Surface Generator; Appendix 6: SCOUT | Simplicial Continuation Utilities; Bibliography; Index and Notation.
2003 / xxvi + 388 pages / Softcover / ISBN-13: 978-0-898715-44-6 / ISBN-10: 0-89871-544-X /
List Price $74.50 / SIAM Member Price $52.15 / Order Code CL45
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Graphing Functions
12.5: Graphing Functions
Created by: CK-12
Roller Coaster Speed
Jana loves roller coasters. She can’t wait to ride some of the roller coasters at the amusement park for the class trip. Jana is so curious about roller coasters that she starts to do some research
about them. For example, Jana wonders whether or not the speed of the roller coaster is connected to the height of the roller coaster or the length of the roller coaster. She thinks that the speed of
the roller coaster is a function of its height.
After doing some research, here is what Jana discovers.
The Timber Terror Roller Coaster
Height $= 85 \ ft$
Speed $= 55 \ mph$
Kingda Ka Roller Coaster
Height $= 456 \ feet$
Speed $= 128 \ mph$
Top Thrill Dragster Roller Coaster
Height $= 420 \ ft.$
Speed $= 120 \ mph$
Jana wants to show how this data appears in a chart. She wants to be able to prove that the speed of the roller coaster is a function of its height.
This lesson is all about graphing functions. Pay close attention and at the end of this lesson you will be able to help Jana organize and graph her function.
What You Will Learn
By the end of this lesson you will be able to demonstrate the following skills:
• Graph linear functions in the coordinate plane.
• Distinguish between linear and non-linear functions.
• Use function graphs to relate perimeter, area and volume to linear dimensions of objects.
• Model and solve real-world problems involving patterns of change, using multiple representations of functions.
Teaching Time
I. Graph Linear Functions in the Coordinate Plane
In the last lesson you learned about functions. We actually didn’t really call them “functions,” but we called them input/output tables. Let’s look at what it means for the data in an input/output
table to be a function.
What is a function?
A function is a set of data that has a specific relationship. One variable in the data set is related to or depends on a different variable in the same data set. Each input matches with only one
Let’s look at a table to look at this.
Do you see something different in this table?
In this table we use the letters $x$$y$$x$$y$
Here the $x$$y$$y$$x$$x$$y$This means that this table forms a function.
Here is another table.
Do you see something different here?
The $x$
What does it mean when real life data forms a function?
It means that one variable depends on or is a function of the other variable in the data.
Let’s look at an example.
Felix has a job cutting grass in the summer time. He earns $10.00 per lawn that he cuts.
This is an example of a function.
The amount of money that Felix makes is related to the number of lawns that he cuts. If Felix cuts 10 lawns, then he will make $100.00. The amount of money is a function of the number of lawns.
We can look at some data about Felix and then show how this forms a function.
Felix cut the following lawns on four different days.
Day 1 = 1 lawn = $10.00
Day 2 = 2 lawns = $20.00
Day 3 = 3 lawns = $30.00
Day 4 = 4 lawns = $40.00
How can we organize this data in a table?
Well, the number of lawns would be the $x$$y$$x$$y$$x$
Here is our table.
$x$ $y$
1 $10
2 $20
3 $30
4 $40
We can say that the amount of money that Felix earns is a function of the number of lawns that he mows.
We can also graph functions on the coordinate grid. We do this by using the values in each column to form our ordered pairs.
Notice that we have an $x$$y$$x$$y$
Let’s write this data as ordered pairs.
(1, 10)
(2, 20)
(3, 30)
(4, 40)
Now we can graph our data.
We create a graph by plotting the $x$$x$$y$$y$
Wow! This graph forms a line!
Yes it does. This graph forms what we call a linear function. Anytime that a graph forms a line like this it is called a linear graph-and a linear graph is a graph of a linear function.
In the next section you will learn more about linear and non-linear functions and their graphs.
Practice graphing the following functions.
$x$ $y$
$x$ $y$
Take a few minutes to check your work with a partner. Do your graphs show linear graphs?
II. Distinguish Between Linear and Non-Linear Functions
In the last section you learned to identify a linear function. Let’s identify a linear function now.
What is a linear function?
A linear function has a graph that is straight line.
Let’s look at this table.
Notice that each $x$$y$$y$
Let’s be sure that it does. Here is the graph of this function.
That’s a great question.
What is a non-linear function?
A non-linear function is a function where the data does not increase or decrease in a systematic or sequential way. In short, a non-linear function does not form a straight line when it is graphed.
Let’s look at a non-linear function in a table.
Do you notice anything different about this function?
The data does not move in a sequential way. This graph will not form a straight line.
Let’s graph this function to be sure. Here is the graph of a non-linear function.
We could connect these points, but it does not change the fact that this is a non-linear function.
Practice identifying whether each represents a linear or a non-linear function.
Take a few minutes to check your work with a peer.
III. Use Function Graphs to Relate Perimeter, Area and Volume to Linear Dimensions of Objects
In an earlier lesson, you learned about how to calculate the perimeter, area and volume of different figures. Well now that you have learned about functions, we can apply functions to our work with
perimeter, area and volume.
Perimeter, area and volume are a function of their dimensions.
What does this mean?
It means that when you know the dimensions of the figures, you can calculate the perimeter area and volume. We can also create a function table to show possible dimensions for certain areas,
perimeters and volumes of figures.
Let’s start by looking at area.
Let’s say that I have the area of a rectangle. You know that we can calculate the area of a rectangle by multiplying the length times the width.
$A = lw$
The length and width are the two variables. These can change, and depending on how they change, our area can also change.
What are the possible dimensions for a rectangle with an area of 36 square units?
How can we tackle this problem? Well, we can start by creating a table. We know that the length and width are going to be the variables, just like $x$$y$
Now we have to choose values for the table. We need to choose values that we can multiply together to give us an area of 36 sq. units. Everything goes back to the area of the figure given the
We can use the formula for our rule. Watch how this works. If I put two in for the width, then the length has to be 18, because two times 18 is 36.
Next, we need to find three other values for dimensions that will give us 36 square units for an area.
$w$ $l$
What would a graph of this function look like?
We can graph this function and use $w$$l$$x$$y$
Here is a graph of the function showing that a fixed area of 36 has several possible lengths and widths. Notice that it is a linear function.
What about perimeter?
We can create a function to match the dimensions of a rectangle with regards to perimeter too. Let’s say that we have a fixed perimeter of 12 units. We use the formula for perimeter of a rectangle to
create our table of values.
$P = 2L + 2W$
Next, we select a set of values for the width and then figure out the length using the formula. Remember that no matter which values we choose, that we have to end up with a perimeter of 12 units.
Now let’s graph the function and see if this is a linear function.
This is also a linear graph. A fixed perimeter of a rectangle forms a linear function.
What about volume?
Volume is the amount of space inside a three-dimensional figure.
We find the volume of a rectangular prism by multiplying the length, the width and the height.
$V = lwh$
If we have a rectangular prism, we can keep the width and the height the same and only change the length. Watch what happens to the volume of the rectangular prism if we keep doubling the length of
the prism.
Width = 4
Height = 2
How did we get these numbers?
We got them by using the formula for volume and by keeping the width and the height the same and we kept changing the length. In fact, we kept doubling the length.
What happened to the volume each time the length was doubled?
Each time the length was doubled, the volume also doubled.
Volume is a function of the relationship between the length, width and height of a prism.
Is this a linear function?
Notice that the increment that each value changes by is consistent and sequential. If we were to graph this function it would create a linear function.
Practice applying this information.
1. Create a table for a rectangle with a fixed area of twenty-four square units. Then graph the results.
Check your table and graph with a friend. Is this a linear function?
Take a few notes on linear and non-linear functions and how to tell the difference before moving on to the next section.
Real Life Example Completed
Roller Coaster Speed
Here is the original problem once again. Reread the problem and then work on creating a table and function graph of Jana’s data.
Jana loves roller coasters. She can’t wait to ride some of the roller coasters at the amusement park for the class trip. Jana is so curious about roller coasters that she starts to do some research
about them. For example, Jana wonders whether or not the speed of the roller coaster is connected to the height of the roller coaster or the length of the roller coaster. She thinks that the speed of
the roller coaster is a function of its height.
After doing some research, here is what Jana discovers.
The Timber Terror Roller Coaster
Height $= 85 \ ft$
Speed $= 55 \ mph$
Kingda Ka Roller Coaster
Height $= 456 \ feet$
Speed $= 128 \ mph$
Top Thrill Dragster Roller Coaster
Height $= 420 \ ft.$
Speed $= 120 \ mph$
To create a table of Jana’s data we must use the height as one variable and the speed as the other. Here is a table of our data.
$H$ $S$
You can see that as the height increases so does the speed. Using this information, Jana can conclude that the speed of a roller coaster is a function of its height.
Let’s create a graph of the function.
Notice that this graph is a non-linear graph. Even though the speed increases with the height of the roller coaster, the interval that it increases is not even. Therefore, the graph of this function
is non-linear.
Here are the vocabulary words that are found in this lesson.
one variable is dependent on another. One variable matches exactly one other value.
Linear Function
the graph of a linear function forms a straight line.
Non-Linear Function
the graph of a non-linear function does not form a straight line.
the distance around the outside edge of a figure.
the measure of the surface of a two-dimensional figure
the measure of the space contained inside a three-dimensional figure.
Technology Integration
James Sousa, Plotting Points on the Coordinate Plane
James Sousa, Graphing Equations by Plotting Points, Part 1
James Sousa, Graphing Equations by Plotting Points, Part 2
You can learn more about roller coasters at these websites.
Time to Practice
Directions: Look at each table and determine whether the function is linear or non-linear.
$x$ $y$
$x$ $y$
Directions: Now use each table in 1 – 10 and graph each function. You should have 10 graphs for this section. Number these graphs 11 – 20. If the graph is a linear graph, then please connect the
points with a line.
Files can only be attached to the latest version of None
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Frequency and Period of a
Lesson 2: Properties of Waves
Frequency and Period of a Wave
The nature of a wave was discussed in Lesson 1 of this unit. In that lesson, it was mentioned that a wave is created in a slinky by the periodic and repeating vibration of the first coil of the
slinky. This vibration creates a disturbance which moves through the slinky and transports energy from the first coil to the last coil. A single back-and-forth vibration of the first coil of a slinky
introduces a pulse into the medium. But the act of continually vibrating the first coil with a back-and-forth motion in periodic fashion introduces a wave into the slinky.
Suppose that a hand holding the first coil of a slinky is moved back-and-forth two complete cycles in one second. The rate of the hand's motion would be 2 cycles/second. The first coil, being
attached to the hand, in turn would vibrate at a rate of 2 cycles/second. The second coil, being attached to the first coil, would vibrate at a rate of 2 cycles/second. In fact, every coil of the
slinky would vibrate at this rate of 2 cycles/second. This rate of 2 cycles/second is referred to as the frequency of the wave. The frequency of a wave refers to how often the particles of the medium
vibrate when a wave passes through the medium. Frequency is a part of our common, everyday language. For example, it is not uncommon to hear a question like "How frequently do you mow the lawn during
the summer months?" Of course the question is an inquiry about how often the lawn is mowed and the answer is usually given in the form of "1 time per week." In mathematical terms, the frequency is
the number of complete vibrational cycles of a medium per a given amount of time. Given this definition, it is reasonable that the quantity frequency would have units of cycles/second, waves/second,
vibrations/second, or something/second. Another unit for frequency is the Hertz (abbreviated Hz) where 1 Hz is equivalent to 1 cycle/second. If a coil of slinky makes 2 vibrational cycles in one
second, then the frequency is 2 Hz. If a coil of slinky makes 3 vibrational cycles in one second, then the frequency is 3 Hz. And if a coil makes 8 vibrational cycles in 4 seconds, then the frequency
is 2 Hz (8 cycles/4 s = 2 cycles/s).
The quantity frequency is often confused with the quantity period. Period refers to the time which it takes to do something. When an event occurs repeatedly, then we say that the event is periodic
and refer to the time for the event to repeat itself as the period. The period of a wave is the time for a particle on a medium to make one complete vibrational cycle. Period, being a time, is
measured in units of time such as seconds, hours, days or years. The period of orbit for the Earth around the Sun is approximately 365 days; it takes 365 days for the Earth to complete a cycle. The
period of a typical class at Glenbrook South (non-block and non-AP) is 55 minutes; every 55 minutes a class cycle begins (50 minutes for class and 5 minutes for passing time means that a class begins
every 55 minutes). The period for the minute hand on a clock is 3600 seconds (60 minutes); it takes the minute hand 3600 seconds to complete one cycle around the clock. When a physics teacher is
regular with his stools, the period of the stools is 24 hours. That doesn't mean he spends 24 hours on the stool, it merely means that it takes 24 hours before he must return to the stools to repeat
the daily cycle (of course, this assumes that a trip to the stools is a periodic event for that teacher).
Frequency and period are distinctly different, yet related, quantities. Frequency refers to how often something happens; period refers to the time it takes something to happen. Frequency is a rate
quantity; period is a time quantity. Frequency is the cycles/second; period is the seconds/cycle. As an example of the distinction and the relatedness of frequency and period, consider a woodpecker
that drums upon a tree at a periodic rate. If the woodpecker drums upon a tree 2 times in one second, then the frequency is 2 Hz; each drum must endure for one-half a second, so the period is 0.5 s.
If the woodpecker drums upon a tree 4 times in one second, then the frequency is 4 Hz; each drum must endure for one-fourth a second, so the period is 0.25 s. If the woodpecker drums upon a tree 5
times in one second, then the frequency is 5 Hz; each drum must endure for one-fifth a second, so the period is 0.2 s. Do you observe the relationship? Mathematically, the period is the reciprocal of
the frequency and vice versa. In equation form, this is expressed as follows.
Since the symbol f is used for frequency and the symbol T is used for period, these equations are also expressed as:
The quantity frequency is also confused with the quantity speed. The speed of an object refers to how fast an object is moving and is usually expressed as the distance traveled per time of travel.
For a wave, the speed is the distance traveled by a given point on the wave (such as a crest) in a given period of time. So while wave frequency refers to the number of cycles occurring per second,
wave speed refers to the meters traveled per second. A wave can vibrate back and forth very frequently, yet have a small speed; and a wave can vibrate back and forth with a low frequency, yet have a
high speed. Frequency and speed are distinctly different quantities. Wave speed will be discussed in more detail later in this lesson.
Check Your Understanding
Throughout this unit, internalize the meaning of terms such as period, frequency, and wavelength. Utilize the meaning of these terms to answer conceptual questions; avoid a formula fixation.
1. A wave has an amplitude of 2 cm and a frequency of 12 Hz, and the distance from a crest to the nearest trough is measured to be 5 cm. Determine the period of such a wave.
2. A fly flaps its wings back and forth 150 times each second. The period of a wing flap is
a. 150 sec b. 2.5 sec c. 0.040 sec d. 0.0067 sec
3. A tennis coach paces back and forth along the sideline 10 times in 2 minutes. The frequency of her pacing is ________.
a. 5.0 Hz b. 0.20 Hz c. 0.12 Hz d. 0.083 Hz
4. The frequency of rotation of a second hand on a clock is _______.
a. 1/60 Hz b. 1/12 Hz c. 1/2 Hz d. 1 Hz e. 60 Hz
5. A kid on a playground swing makes a complete to-and-fro swing each 2 seconds. The frequency of swing is _________.
a. 0.5 Hz b. 1 Hz c. 2 Hz
6. In problem #5, the period of swing is __________.
a. 0.5 second b. 1 second c. 2 second
7. A period of 5.0 seconds corresponds to a frequency of ________ Hertz.
a. 0.2 b. 0.5 c. 0.02 d. 0.05 e. 0.002
8. A pendulum makes 40 vibrations in 20 seconds. Calculate its period?
9. A child in a swing makes one complete back and forth motion in 4.0 seconds. This statement provides information about the child's
a. speed b. frequency c. period
10. The period of a 440 Hertz sound wave is ___________.
11. As the frequency of a wave increases, the period of the wave ___________.
a. decreases b. increases c. remains the same
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Algebra of Functions
of functions.
Basic function operations
Function, domain, & range
domain of a function is the set of all “first
coordinates” of the ordered pairs of a relation.
range of a function is the set of all “second
coordinates” of the ordered pairs of a relation.
unique (they do not repeat).
line test.
of a relation in more than one point, the relation is not
a function.
Example 1
Example 2
in calories, can be calculated by finding the ratio of
two functions of heat input, D and N, where
││D(i) = i – 5700 and N(i) = i . ││
││ ││
││the engine in terms of heat input (i), in calories. ││
Composition of functions
application of the functions in a specific order.
is f o g defined by
“f of g of x.”
domain of g such that g(x) is in the domain of f.
g must be in the domain of function f.”
A composite function
Example 3
f (x) = x − 3
g (x) = 2x^2 −1
Example 4
(Since a radicand can’t be negative in the set of real
numbers, the radicand must be greater than or equal to
zero. This is what limits the domain.)
Example 5
each day is a function of the number of hours (t) the
assembly line is in operation that day and is given by
n = P(t) = 75t – 2t^2.
the number of helmets produced and is given by
C(n) = 7n +1000.
helmets in terms of the number of hours the assembly line is
functioning on a given day.
when the assembly line was functioning 12 hours.
(solution on next slide)
C(n) = 7n + 1000
n = P (t ) = 75t − 2t^2
Solution to Example 5:
the helmets in terms of the number of hours the
assembly line is functioning on a given day.
when the assembly line was functioning 12 hours.
^2 , find f (g (x)).
Answers to review:
Domain of g (f (x)) is {x : x ∈
The domains of the two functions are different because
the denominator of b(x) cannot be zero.
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Electrostatic potential from the Poisson equation
Prof. Jens Nöckel, University of Oregon
This is the HTML version of a Mathematica 8 notebook. You can copy and paste the following into a notebook as literal plain text. An executable notebook is linked here: PoissonDielectricSolver2D.cdf.
This web page is intended for educational use and copyright Jens U. Nöckel (2011). No claims are made as to the accuracy and reliability of the results. Any commercial use requires the permission of
the author. There are no restrictions on personal or educational use, but also no warranties.
The notebook implements a relaxation procedure to determine the potential in the presence of arbitrarily shaped conductors, linear dielectrics, and charges. No iterative refinement or other numerical
optimizations are performed. The goal is to keep the code as simple as possible while still allowing it to cover a wide range of problems in planar electrostatics with a simple user interface.
The main Mathematica-specific innovation in this code is the fact that you can create all the objects defining the problem by drawing them graphically. This drawing process is completely decoupled
from the actual calculation, so it could in principle be done in any external program. However, I will use Mathematica's graphics commands to set the stage in the example below.
defaultGridSize = 120;
step = Compile[{{phi, _Real, 2 }, {orig, _Real, 2 }, {mask, _Real,
2 }, {dchiX, _Real, 2 }, {dchiY, _Real, 2 }},
Module[{f = (RotateRight[phi] + RotateLeft[phi] +
RotateRight[phi, {0, 1}] + RotateLeft[phi, {0, 1}])*mask +
orig, ex, ey},
ex = (RotateRight[phi, {0, 1}] - RotateLeft[phi, {0, 1}]);
ey = (RotateRight[phi] - RotateLeft[phi]);
f - (dchiX*ex + dchiY*ey)*mask
iterate = Compile[{{gridArray, _Real, 2 }, {originalArray, _Real,
2 }, {maskArray, _Real, 2 }, {dchiX, _Real, 2 }, {dchiY, _Real,
2 }, {tol, _Real}},
FixedPoint[step[#, gridArray, maskArray, dchiX, dchiY] &,
originalArray, 100000,
SameTest -> (Max@Abs@Flatten[#1 - #2] < tol &)]];
digitize[gr_, n_] :=
Image[Show[gr, Background -> Black,
BaseStyle -> {Antialiasing -> False}], ImageSize -> n],
createLandscape[conductors_, chargePlus_, chargeMinus_,
suceptibility_, nGrid_] :=
Module[{gridConductors, gridRho, gridChi, maskList},
gridConductors = digitize[conductors, nGrid];
maskList = N[1. - Unitize[gridConductors]];
gridRho = (digitize[chargePlus, nGrid] -
digitize[chargeMinus, nGrid])*maskList;
gridChi = digitize[suceptibility, nGrid]*maskList;
{gridConductors, gridRho, gridChi, maskList}
poissonSolver[conductors_, chargePlus_, chargeMinus_, suceptibility_,
nGrid_: defaultGridSize, tolerance_: 10^(-6)] :=
Block[{averagePotential, gridConductors, gridRho, gridChi, gridEps,
gridList, maskList, mask4List, dChiYList, dChiXList,
{gridConductors, gridRho, gridChi, maskList} =
createLandscape[conductors, chargePlus, chargeMinus,
suceptibility, nGrid];
averagePotential = Mean[Select[Flatten@gridConductors, Positive]];
initialGrid = averagePotential*maskList + gridConductors;
gridEps = 1. + gridChi;
gridList = gridConductors + gridRho/(4.*gridEps);
mask4List = maskList/4.;
dChiYList = (RotateLeft[gridChi] - RotateRight[gridChi])/(
2. gridEps);
dChiXList = (
RotateLeft[gridChi, {0, 1}] - RotateRight[gridChi, {0, 1}])/(
2. gridEps);
iterate[gridList, initialGrid, mask4List, dChiXList, dChiYList,
The function poissonSolver takes four two-dimensional graphics objects (all must correspond to a spatial area of equal dimensions), and outputs a two-dimensional array of horizontal dimension nGrid.
If the graphics are not square, the vertical dimension is adjusted to maintain the same aspect ratio as the input graphics.
All the ingredients are supplied separately as 2D Graphics objects:
• conductors (black = no conductor, all other GrayLevel values give the value of the fixed potential). The potential on each conductor should be constant. Because I use GrayLevel to encode the
potential, all potentials are positive (as can always be arranged with a suitable additive constant).
• chargeMinus (black = zero charge, white = large negative charge)
• chargePlus (black = zero charge, white = large positive charge)
• susceptibility (susceptibility; black = 0)
In the function digitize above, we create all the grids needed for the simulation. A mask is applied to zero out all charges and susceptibilities that coincide with a conducting region. The input to
digitize for this example is given below:
conductors = Graphics[{
{GrayLevel[.5], Rectangle[1.1 {-1, -1}, 1.1 {1, 1}]},
{Black, Disk[{0, 0}, 1]}, {GrayLevel[.1],
Disk[{-.25, 0}, .15]}, {GrayLevel[1], Disk[{.1, -.25}, .1]}
}, PlotRangePadding -> 0, ImagePadding -> None];
chargePlus =
Graphics[{{GrayLevel[.1], Disk[{-.5, .3}, .05]}},
PlotRange -> (PlotRange /. FullOptions[conductors])];
chargeMinus =
Graphics[{{GrayLevel[.1], Disk[{0., .4}, .05]}},
PlotRange -> (PlotRange /. FullOptions[conductors])];
susceptibility =
Graphics[{GrayLevel[.4], Rectangle[{0.3, -1}, {1, 1}]},
PlotRange -> (PlotRange /. FullOptions[conductors])];
Column[{Style[#1, FontColor -> Blue, FontFamily -> "Helvetica"],
Show[#2, Background -> Black]]}] &, {{"conductor potentials (click to flip)",
"dielectric susceptibility", "positive charges",
"negative charges"}, {conductors, susceptibility, chargePlus,
chargeMinus}}], Background -> Orange]
This shows the superposition of all four images, to illustrate (from left to right) the relative position of conductors (a round boundary in gray surrounding one bright and one almost black disk),
the dielectric (a bright rectangle) and charges: there is one image for positive and one for negative charges, so that I can keep the convention that black corresponds to zero charge (black means
zero in all images, except for the conductors where black means simply the absence of conductors). In the last two images, the charges are small disks. In the conductor configuration, the gray
boundary represents a hollow conductor that provides a Dirichlet boundary condition which we can consider the "ground" level with respect to which the potentials on the other two conductors are
Finally, we run the simulation and plot it. The optional Timing command yields 3.2s on a Macbook Pro with a 2.4 Ghz Intel Core I7.
Timing[potential =
poissonSolver[conductors, chargePlus, chargeMinus,
ListPlot3D[potential, PlotRange -> All,
PlotStyle -> {Orange, Specularity[White, 10]}]
The charges create the two spikes, and the conductors create flat plateaus. A kink in the potential can be seen at the dielectric interface.
noeckel@uoregon.edu Last modified: Mon Oct 8 20:36:53 PDT 2012
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1. Integration: The General Power Formula
by M. Bourne
In this section, we apply the following formula to trigonometric, logarithmic and exponential functions:
`intu^ndu=(u^(n+1))/(n+1)+C\ \ \ (n!=-1)`
(We met this substitution formula in an earlier chapter: General Power Formula for Integration.)
Example 1: Integrate: `intsin^(1/3)x\ cos\ x\ dx`
Example 2: Integrate: `int(sin^(-1)4x)/sqrt(1-16x^2)dx`
Example 3: Integrate: `int((3+ln\ 2x)^3)/xdx`
Example 4: Integrate: `int2sqrt(1-e^(-x))e^(-x)dx`
Example 5: Find the equation of the curve for which `(dy)/(dx)=((ln\ x)^2)/x` if the curve passes through `(1, 2)`.
Integrate each of the following functions:
1. `int((cos^(-1)2x)^4)/sqrt(1-4x^2)dx`
2. `int_1^e((1-2\ ln\ x))/xdx`
3. `int(e^x+e^(-x))^(1/4)(e^x-e^(-x))dx`
4. `int_(pi//3)^(pi//2)(sin\ theta\ d theta)/(sqrt(1+cos\ theta)`
5. Find the equation of the curve for which `(dy)/(dx)=(1+tan\ 2x)^2sec^2 2x` if the curve passes through `(2, 1)`.
6. A space vehicle is launched vertically from the ground such that its velocity v (in km/s) is given by
where t is the time in seconds. Find the altitude of the vehicle after 10.0 s.
The graph of `v=[ln^2(t^3+1)](t^2)/(t^3+1)` is as follows:
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hypothesis/critical region help
October 8th 2008, 05:30 PM #1
Oct 2008
hypothesis/critical region help
So I have this word problem about herbs, scores on the test form a normal-shaped distribution with a mean of u=70 SD= 15 n=25 people take a herbal supplement everyday for 30 days, at the end each
person takers a memory test and the mean score for the sample is calculated to be M=78.
Identify the critical region for the test assuming that the researcher is using =.05
I don't understand what the critical region is, how do I figure that out? Or how you know to use .05?? In this case it says too use .05, but what if it doesn't say?
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Here's the question you clicked on:
In 2004 there were 72 million voters registered as Democrats, 55 million American voters registered as Republicans, and 42 million registered Independents. What is the probability that 3 randomly
selected, registered voters would be Independents?
• 8 months ago
• 8 months ago
Best Response
You've already chosen the best response.
The total number of voters is (72 + 55 + 42) million. The probability of a randomly selected voter being an independent is \[P(Independent)=\frac{42}{72+55+42}=you\ can\ calculate\] When you have
posted the result of the above calculation we can move on to the final step in the solution.
Best Response
You've already chosen the best response.
i got .24852071
Best Response
You've already chosen the best response.
0.15 0.077 0.015 0.036 but idk how when these are the ones i have to choose from its frustrating
Best Response
You've already chosen the best response.
Correct! When selecting each of the voters the probability that they are an Independent is 0.24852071. The result of each selection is independent of the other results. Therefore the probability
of all three being Independents is given by \[P(all\ 3\ Independents)=(0.248520710)^{3}=you\ can\ calculate\]
Best Response
You've already chosen the best response.
so would i do 169 by what i got
Best Response
You've already chosen the best response.
You just need to calculate 0.24852071 * 0.24852071 * 0.24852071 = ?
Best Response
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so it would be .015 right
Best Response
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Good work! Your result is correct.
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thank you
Best Response
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You're welcome :)
Your question is ready. Sign up for free to start getting answers.
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Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.
This is the testimonial you wrote.
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Putnam proof
August 29th 2006, 12:42 PM #1
MHF Contributor
Oct 2005
Putnam proof
I am studying for the Putnam competition in December and I want just a hint at how to do this problem, not a full solution please. I'm trying to develop a good methodology to approaching proofs.
A1. Determine, with proof, the number of ordered triples $(A_1,A_2,A_3)$ of sets which have the property that
(i) $A_1 \cup A_2 \cup A_3 = \{1,2,3,4,5,6,7,8,9,10\}$, and
(ii) $A_1 \cap A_2 \cap A_3 = \emptyset$. Express the answer in the form of $2^a3^b5^c7^d$, where a,b,c, and d are non-negative integers.
What's the best way to approach this?
Last edited by Jameson; August 30th 2006 at 11:52 AM.
I am studying for the Putnam competition in December and I want just a hint at how to do this problem, not a full solution please. I'm trying to develop a good methodology to approaching proofs.
A1. Determine, with proof, the number of ordered triples $(A_1,A_2,A_3)$ of sets which have the property that
(i) $A_1 \cup A_2 \cup A_3 = {1,2,3,4,5,6,7,8,9,10}$, and
(ii) $A_1 \cap A_2 \cap A_3 e \emptyset$. Express the answer in the form of $2^a3^b5^c7^d$, where a,b,c, and d are non-negative integers.
What's the best way to approach this?
The way I like to is to procede the opposite.
Meaning do the problem:
$A_1\cup A_2\cup A_3 = \{\}$
Then from the full total subtract this number.
I do not understand why you chose to place this Algebra section.
I do not understand why you chose to place this Algebra section.
Ooops! I didn't really think about where to put it. I'll move it.
Thanks for your help guys. I'll think about it and post back what I have.
The way I like to is to procede the opposite.
Meaning do the problem:
$A_1\cup A_2\cup A_3 = \{\}$
Then from the full total subtract this number.
Ok. I'll try to start this way. My problem is that I have trouble going from specific cases and setting up a pattern/generalizing behavior that's needed for a proof.
So if $A_1\cup A_2\cup A_3 = \{\}$, then obviously the sets can not have any items in common. So I need to find the number of ways to split up the numbers 1-10 into 3 groups.
I see this as having $A_1={_{10} C _ i}$, $A_2={_{10} C _j}$, and $A_3={_{10} C _k}$ , where i+j+k=10. On the right track or should I stop and consider another method? If I'm going in the right
direction, any suggestions after the step I gave?
Thanks guys.
PH - Are you going to take the Putnam exam?
Sorry, I meant to write.
$A_1\cap A_2 \cap A_3$
(Still number theory is not appropriate. I would place it in the High School Probabily question. Yes I know this is very complicated but as far as level is concerned it deserves to be there. I
guess I am the only one of the forum who is careful where everything goes).
I see this as having $A_1={_{10} C _ i}$, $A_2={_{10} C _j}$, and $A_3={_{10} C _k}$ , where i+j+k=10. On the right track or should I stop and consider another method? If I'm going in the right
direction, any suggestions after the step I gave?
Not true.
The first set you can select.
${{10}\choose n}$ for $1\leq n\leq 8$
The second set you can select.
${{10-n}\choose m}$ for $1\leq m \leq 9-n$
And the third set you can only take the remaining which is one.
$\frac{10!}{n!(10-n!)}\cdot \frac{(10-n)!}{(10-n-m)!m!}$
Which simplifies to,
Now you add each one.
I am studying for the Putnam competition in December and I want just a hint at how to do this problem, not a full solution please. I'm trying to develop a good methodology to approaching proofs.
A1. Determine, with proof, the number of ordered triples $(A_1,A_2,A_3)$ of sets which have the property that
(i) $A_1 \cup A_2 \cup A_3 = \{1,2,3,4,5,6,7,8,9,10\}$, and
(ii) $A_1 \cap A_2 \cap A_3 e \emptyset$. Express the answer in the form of $2^a3^b5^c7^d$, where a,b,c, and d are non-negative integers.
What's the best way to approach this?
I solved this problem and got the answer $7^{10} - 6^{10},$ which does not have the required form. I found the reason for this is that the problem is not stated correctly: there is an equal sign
in the second condition.
Your methodology so far is off track. Look carefully at what conditions (i) and the correct (ii) mean. This is not a problem of how to place 10 objects in 3 containers. Up to 30 objects could be
Consider the sets $A_1,\ A_2,\ A_3$ to be containers into which
counters labled $1,\ 2,\ 3,\ \dots \,\ 10$ are to be placed.
Of course after reading JakeD's response I realise that I was solving the
problem with:
$A_i \cap A_j = \emptyset,\ \ ie j$
Ok, I finally got it.
Drawing a Venn Diagram with three circles there are seven distinct regions. From condition 1 all of these regions may be used, but because of condition 2 the center region is excluded.
So there are 6 regions which we can place 10 elements in. Thus the answer is $10^6$ ways, but in the form they requested $2^{10}3^{10}$
August 29th 2006, 01:45 PM #2
Global Moderator
Nov 2005
New York City
August 29th 2006, 02:38 PM #3
MHF Contributor
Oct 2005
August 29th 2006, 03:05 PM #4
MHF Contributor
Oct 2005
August 29th 2006, 04:12 PM #5
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Nov 2005
New York City
August 30th 2006, 08:22 AM #6
Global Moderator
Nov 2005
New York City
August 30th 2006, 11:30 AM #7
August 30th 2006, 11:47 AM #8
Grand Panjandrum
Nov 2005
August 31st 2006, 01:46 PM #9
MHF Contributor
Oct 2005
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Find the distance between the pair of points. Give an exact answer and a three-decimal-place approximation - WyzAnt Answers
Find the distance between the pair of points. Give an exact answer and a three-decimal-place approximation.
(2.8,-1.8) and (-8.3,9.6)
Find the exact distance
Find the decimal approximation (type the integer or decimal rounded to three decimal places as needed)
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1 Answer
Start with the distance formula:
Plugging your x and y values in you get:
Simplify inside before you square:
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elementary algebra :: Algebraic expressions
elementary algebra
Article Free Pass
Any of the quantities mentioned so far may be combined in expressions according to the usual arithmetic operations of addition, subtraction, and multiplication. Thus, ax + by and axx + bx + c are
common algebraic expressions. However, exponential notation is commonly used to avoid repeating the same term in a product, so that one writes x^2 for xx and y^3 for yyy. (By convention x^0 = 1.)
Expressions built up in this way from the real and complex numbers, the algebraic quantities a, b, c, …, x, y, z, and the three above operations are called polynomials—a word introduced in the late
16th century by the French mathematician François Viète from the Greek polys (“many”) and the Latin nominem (“name” or “term”). One way of characterizing a polynomial is by the number of different
unknown, or variable, quantities in it. Another way of characterizing a polynomial is by its degree. The degree of a polynomial in one unknown is the largest power of the unknown appearing in it. The
expressions ax + b, ax^2 + bx + c, and ax^3 + bx^2 + cx + d are general polynomials in one unknown (x) of degrees 1, 2, and 3, respectively. When only one unknown is involved, it does not matter
which letter is used for it. One could equally well write the above polynomials as ay + b, az^2 + bz + c, and at^3 + bt^2 + ct + d.
Because some insight into complicated functions can be obtained by approximating them with simpler functions, polynomials of the first degree were investigated early on. In particular, ax + by = c,
which represents a straight line, and ax + by + cz = e, which represents a plane in three-dimensional space, were among the first algebraic equations studied.
Polynomials can be combined according to the three arithmetic operations of addition, subtraction, and multiplication, and the result is again a polynomial. To simplify expressions obtained by
combining polynomials in this way, one uses the distributive law, as well as the commutative and associative laws for addition and multiplication (see the table). Until very recently a major drawback
of algebra was the extreme tedium of routine manipulation of polynomials, but now a number of symbolic algebra programs make this work as easy as typing the expressions into a computer.
By extending the operations on polynomials to include division, or ratios of polynomials, one obtains the rational functions. Examples of such rational functions are 2/3x and (a + bx^2)/(c + dx^2 + e
x^5). Working with rational functions allows one to introduce the expression 1/x and its powers, 1/x^2, 1/x^3, … (often written x^−1, x^−2, x^−3, …). When the degree of the numerator of a rational
function is at least as large as that of its denominator, it is possible to divide the numerator by the denominator much as one divides one integer by another. In this way one can write any rational
function as the sum of a polynomial and a rational function in which the degree of the numerator is less than that of the denominator. For example, (x^8 − x^5 + 3x^3 + 2)/(x^3 − 1) = x^5 + 3 + 5/(x^3
− 1). Since this process reduces the degrees of the terms involved, it is especially useful for calculating the values of rational functions and for dealing with them when they arise in calculus.
For theoretical work and applications one often needs to find numbers that, when substituted for the unknown, make a certain polynomial equal to zero. Such a number is called a “root” of the
polynomial. For example, the polynomial −16t^2 + 88t + 48represents the height above Earth at t seconds of a projectile thrown straight up at 88 feet per second from the top of a tower 48 feet high.
(The 16 in the formula comes from one-half the acceleration of gravity, 32 feet per second per second.) By setting the equation equal to zero and factoring it as (4t − 24)(−4t − 2) = 0, the
equation’s one positive root is found to be 6, meaning that the object will hit the ground about 6 seconds after it is thrown. (This problem also illustrates the important algebraic concept of the
zero factor property: if ab = 0, then either a = 0 or b = 0.)
The theorem that every polynomial has as many complex roots as its degree is known as the fundamental theorem of algebra and was first proved in 1799 by the German mathematician Carl Friedrich Gauss.
Simple formulas exist for finding the roots of the general polynomials of degrees one and two (see the table), and much less simple formulas exist for polynomials of degrees three and four. The
French mathematician Évariste Galois discovered, shortly before his death in 1832, that no such formula exists for a general polynomial of degree greater than four. Many ways exist, however, of
approximating the roots of these polynomials.
Do you know anything more about this topic that you’d like to share?
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Reply to comment
Submitted by Anonymous on December 22, 2010.
Thank you for this amazingly informative article. I hardly like to mention that, in equation (4) above, where you have given the equations of the foci of an ellipse, the second set of co-ordinates
should refer to F2, and not F3 as listed. I am sure that this is just a typing error and does not detract from the excellence of this treatise.
Sorry for making such a minute correction to this spendid article.
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Mathematical Reasoning for Elementary Teachers
ISBN: 9780321286963 | 0321286960
Edition: 4th
Format: Hardcover
Publisher: Addison Wesley
Pub. Date: 1/1/2006
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Functional differentiation
From HaskellWiki
1 Introduction
Functional differentiation means computing or approximating the derivative of a function. There are several ways to do this:
• Approximate the derivative f'(x) by $\frac{f(x+h)-f(x)}{h}$ where h is close to zero. (or at best the square root of the machine precision $\varepsilon$.
• Compute the derivative of f symbolically. This approach is particularly interesting for Haskell.
2 Functional analysis
If you want to explain the terms Higher order function and Currying to mathematicians, this is certainly a good example. The mathematician writes
$D f (x) = \lim_{h\to 0} \frac{f(x+h)-f(x)}{h}$
and the Haskell programmer writes
derive :: (Fractional a) => a -> (a -> a) -> (a -> a)
derive h f x = (f (x+h) - f x) / h .
approximates the mathematician's
In functional analysis D is called a (linear) function operator, because it maps functions to functions.
In Haskell
is called a higher order function for the same reason.
D is in curried form. If it would be uncurried, you would write D(f,x).
3 Blog Posts
There have been several blog posts on this recently. I think we should gather the information together and make a nice wiki article on it here. For now, here are links to articles on the topic.
4 Code
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In The Picture Below, A 1 Kg Bar3.5 M Long Is Set ... | Chegg.com
In the picture below, a 1 kg bar3.5 m long is set up to rotate about itscenter (the dark circle). Each arrow represents a force that isbeing applied to the bar. F[A] = 48 N, F[B] = 29 N,F[C] = 27 N,
& F[D]= 25 N. F[B] is applied halfwaybetween the center and the end and F[C] is applied rightat the end. Using the center as the pivot point, answer thefollowing questions.
a.) Determine the net torque on the bar.
b.) Determine the moment of inertia of thebar.
c.) Determine the angular acceleration of thebar.
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You are currently browsing the tag archive for the ‘stable commutator length’ tag.
I have just uploaded a paper to the arXiv, entitled “Scl, sails and surgery”. The paper discusses a connection between stable commutator length in free groups and the geometry of sails. This is an
interesting example of what sometimes happens in geometry, where a complicated topological problem in low dimensions can be translated into a “simple” geometric problem in high dimensions. Other
examples include the Veronese embedding in Algebraic geometry (i.e. the embedding of one projective space into another taking a point with homogeneous co-ordinates $x_i$ to the point whose
homogeneous co-ordinates are the monomials of some fixed degree in the $x_i$), which lets one exhibit any projective variety as an intersection of a Veronese variety (whose geometry is understood
very well) with a linear subspace.
In my paper, the fundamental problem is to compute stable commutator length in free groups, and more generally in free products of Abelian groups. Let’s focus on the case of a group $G = A*B$ where
$A,B$ are free abelian of finite rank. A $K(G,1)$ is just a wedge $X:=K_A \vee K_B$ of tori of dimension equal to the ranks of $A,B$. Let $\Gamma: \coprod_i S^1 \to X$ be a free homotopy class of $1$
-manifold in $X$, which is homologically trivial. Formally, we can think of $\Gamma$ as a chain $\sum_i g_i$ in $B_1^H(G)$, the vector space of group $1$-boundaries, modulo homogenization; i.e.
quotiented by the subspace spanned by chains of the form $g^n - ng$ and $g-hgh^{-1}$. One wants to find the simplest surface $S$ mapping to $X$ that rationally bounds $\Gamma$. I.e. we want to find a
map $f:S \to X$ such that $\partial f:\partial S \to X$ factors through $\Gamma$, and so that the boundary $\partial S$ wraps homologically $n(S)$ times around each loop of $\Gamma$, in such a way as
to infimize $-\chi(S)/2n(S)$. This infimum, over all maps of all surfaces $S$ of all possible genus, is the stable commutator length of the chain $\sum_i g_i$. Computing this quantity for all such
finite chains is tantamount to understanding the bounded cohomology of a free group in dimension $2$.
Given such a surface $S$, one can cut it up into simpler pieces, along the preimage of the basepoint $K_A \cap K_B$. Since $S$ is a surface with boundary, these simpler pieces are surfaces with
corners. In general, understanding how a surface can be assembled from an abstract collection of surfaces with corners is a hopeless task. When one tries to glue the pieces back together, one runs
into trouble at the corners — how does one decide when a collection of surfaces “closes up” around a corner? The wrong decision leads to branch points; moreover, a decision made at one corner will
propogate along an edge and lead to constraints on the choices one can make at other corners. This problem arises again and again in low-dimensional topology, and has several different (and not
always equivalent) formulations and guises, including -
• Given an abstract branched surface and a weight on that surface, when is there an unbranched surface carried by the abstract branched surface and realizing the weight?
• Given a triangulation of a $3$-manifold and a collection of normal surface types in each simplex satisfying the gluing constraints but *not* necessarily satisfying the quadrilateral condition
(i.e. there might be more than one quadrilateral type per simplex), when is there an immersed unbranched normal surface in the manifold realizing the weight?
• Given an immersed curve in the plane, when is there an immersion from the disk to the plane whose boundary is the given curve?
• Given a polyhedral surface (arising e.g. in computer graphics), how can one choose smooth approximations of the polygonal faces that mesh smoothly at the vertices?
I think of all these problems as examples of what I like to call the holonomy problem, since all of them can be reduced, in one way or another, to studying representations of fundamental groups of
punctured surfaces into finite groups. The fortunate “accident” in this case is that every corner arises by intersecting a cut with a boundary edge of $S$. Consequently, one never wants to glue more
than two pieces up at any corner, and the holonomy problem does not arise. Hence in principle, to understand the surface $S$ one just needs to understand the pieces of $S$ that can arise by cutting,
and the ways in which they can be reassembled.
This is still not a complete solution of the problem, since infinitely many kinds of pieces can arise by cutting complicated surfaces $S$. The $1$-manifold $\Gamma$ decomposes into a collection of
arcs in the tori $K_A$ and $K_B$ which we denote $\tau_A,\tau_B$ respectively, and the surface $S \cap K_A$ (hereafter abbreviated to $S_A$) has edges that alternate between elements of $\tau_A$, and
edges mapping to $K_A \cap K_B$. Since $K_A$ is a torus, handles of $S_A$ mapping to $K_A$ can be compressed, reducing the complexity of $S_A$, and thereby $S$, so one need only consider planar
surfaces $S_A$.
Let $C_2(A)$ denote the real vector space with basis the set of ordered pairs $(t,t')$ of elements of $\tau_A$ (not necessarily distinct), and $C_1(A)$ the real vector space with basis the elements
of $\tau_A$. A surface $S_A$ determines a non-negative integral vector $v(S_A) \in C_2(A)$, by counting the number of times a given pair of edges $(t,t')$ appear in succession on one of the
(oriented) boundary components of $S_A$. The vector $v(S_A)$ satisfies two linear constraints. First, there is a map $\partial: C_2(A) \to C_1(A)$ defined on a basis vector by $\partial(t,t') = t -
t'$. The vector $v(S_A)$ satisfies $\partial v(S_A) = 0$. Second, each element $t \in \tau_A$ is a based loop in $K_A$, and therefore corresponds to an element in the free abelian group $A$. Define
$h:C_2(A) \to A \otimes \mathbb{R}$ on a basis vector by $h(t,t') = t+t'$ (warning: the notation obscures the fact that $\partial$ and $h$ map to quite different vector spaces). Then $h v(S_A)=0$;
moreover, a non-negative rational vector $v \in C_2(A)$ satisfying $\partial v = h v = 0$ has a multiple of the form $v(S_A)$ for some $S_A$ as above. Denote the subspace of $C_2(A)$ consisting of
non-negative vectors in the kernel of $\partial$ and $h$ by $V_A$. This is a rational polyhedral cone — i.e. a cone with finitely many extremal rays, each spanned by a rational vector.
Although every integral $v \in V_A$ is equal to $v(S_A)$ for some $S_A$, many different $S_A$ correspond to a given $v$. Moreover, if we are allowed to consider formal weighted sums of surfaces, then
even more possibilities. In order to compute stable commutator length, we must determine, for a given vector $v \in V_A$, an expression $v = \sum t_i v(S_i)$ where the $t_i$ are positive real
numbers, which minimizes $\sum -t_i \chi_o(S_i)$. Here $\chi_o(\cdot)$ denotes orbifold Euler characteristic of a surface with corners; each corner contributes $-1/4$ to $\chi_o$. The reason one
counts complexity using this modified definition is that the result is additive: $\chi(S) = \chi_o(S_A) + \chi_o(S_B)$. The contribution to $\chi_o$ from corners is a linear function on $V_A$.
Moreover, a component $S_i$ with $\chi(S_i) \le 0$ can be covered by a surface of high genus and compressed (increasing $\chi$); so such a term can always be replaced by a formal sum $1/n S_i'$ for
which $\chi(S_i') = \chi(S_i)$. Thus the only nonlinear contribution to $\chi_o$ comes from the surfaces $S_i$ whose underlying topological surface is a disk.
Call a vector $v \in V_A$ a disk vector if $v = v(S_A)$ where $S_A$ is topologically a disk (with corners). It turns out that the set of disk vectors $\mathcal{D}_A$ has the following simple form: it
is equal to the union of the integer lattice points contained in certain of the open faces of $V_A$ (those satisfying a combinatorial criterion). Define the sail of $V_A$ to be equal to the boundary
of the convex hull of the polyhedron $\mathcal{D}_A + V_A$ (where $+$ here denotes Minkowski sum). The Klein function $\kappa$ is the unique continuous function on $V_A$, linear on rays, that is
equal to $1$ exactly on the sail. Then $\chi_o(v):= \max \sum t_i\chi_o(S_i)$ over expressions $v = \sum t_i v(S_i)$ satisfies $\chi_o(v) = \kappa(v) - |v|/2$ where $|\cdot|$ denotes $L^1$ norm. To
calculate stable commutator length, one minimizes $-\chi_o(v) - \chi_o(v')$ over $(v,v')$ contained in a certain rational polyhedron in $V_A \times V_B$.
Sails are considered elsewhere by several authors; usually, people take $\mathcal{D}_A$ to be the set of all integer vectors except the vertex of the cone, and the sail is therefore the boundary of
the convex hull of this (simpler) set. Klein introduced sails as a higher-dimensional generalization of continued fractions: if $V$ is a polyhedral cone in two dimensions (i.e. a sector in the plane,
normalized so that one edge is the horizontal axis, say), the vertices of the sail are the continued fraction approximations of the boundary slope. Arnold has revived the study of such objects in
recent years. They arise in many different interesting contexts, such as numerical analysis (especially diophantine approximation) and algebraic number theory. For example, let $A \in \text{SL}(n,\
mathbb{Z})$ be a matrix with irreducible characteristic equation, and all eigenvalues real and positive. There is a basis for $\mathbb{R}^n$ consisting of eigenvalues, spanning a convex cone $V$. The
cone — and therefore its sail — is invariant under $A$; moreover, there is a $\mathbb{Z}^{n-1}$ subgroup of $\text{SL}(n,\mathbb{Z})$ consisting of matrices with the same set of eigenvectors; this
observation follows from Dirichlet’s theorem on the units in a number field, and is due to Tsuchihashi. This abelian group acts freely on the sail with quotient a (topological) torus of dimension
$n-1$, together with a “canonical” cell decomposition. This connection between number theory and combinatorics is quite mysterious; for example, Arnold asks: which cell decompositions can arise? This
is unknown even in the case $n=3$.
The most interesting aspect of this correspondence, between stable commutator length and sails, is that it allows one to introduce parameters. An element in a free group $F_2$ can be expressed as a
word in letters $a,b,a^{-1},b^{-1}$, e.g. $aab^{-1}b^{-1}a^{-1}a^{-1}a^{-1}bbbbab^{-1}b^{-1}$, which is usually abbreviated with exponential notation, e.g. $a^2b^{-2}a^{-3}b^4ab^{-2}$. Having
introduced this notation, one can think of the exponents as parameters, and study stable commutator length in families of words, e.g. $a^{2+p}b^{-2+q}a^{-3-p}b^{4-q}ab^{-2}$. Under the correspondence
above, the parameters only affect the coefficients of the linear map $h$, and therefore one obtains families of polyhedral cones $V_A(p,q,\cdots)$ whose extremal rays depend linearly on the exponent
parameters. This lets one prove many facts about the stable commutator length spectrum in a free group, including:
Theorem: The image of a nonabelian free group of rank at least $4$ under scl in $\mathbb{R}/\mathbb{Z}$ is precisely $\mathbb{Q}/\mathbb{Z}$.
Theorem: For each $n$, the image of the free group $F_n$ under scl contains a well-ordered sequence of values with ordinal type $\omega^{\lfloor n/4 \rfloor}$. The image of $F_\infty$ contains a
well-ordered sequence of values with ordinal type $\omega^\omega$.
One can also say things about the precise dependence of scl on parameters in particular families. More conjecturally, one would like to use this correspondence to say something about the statistical
distribution of scl in free groups. Experimentally, this distribution appears to obey power laws, in the sense that a given (reduced) fraction $p/q$ appears in certain infinite families of elements
with frequency proportional to $q^{-\delta}$ for some power $\delta$ (which unfortunately depends in a rather opaque way on the family). Such power laws are reminiscent of Arnold tongues in dynamics,
one of the best-known examples of phase locking of coupled nonlinear oscillators. Heuristically one expects such power laws to appear in the geometry of “random” sails — this is explained by the fact
that the (affine) geometry of a sail depends only on its $\text{SL}(n,\mathbb{Z})$ orbit, and the existence of invariant measures on a natural moduli space; see e.g. Kontsevich and Suhov. The
simplest example concerns the ($1$-dimensional) cone spanned by a random integral vector in $\mathbb{Z}^2$. The $\text{SL}(2,\mathbb{Z})$ orbit of such a vector depends only on the gcd of the two
co-ordinates. As is easy to see, the probability distribution of the gcd of a random pair of integers $p,q$ obeys a power law: $\text{gcd}(p,q) = n$ with probability $\zeta(2)^{-1}/n^2$. The rigorous
justification of the power laws observed in the scl spectrum of free groups remains the focus of current research by myself and my students.
Mapping class groups (also called modular groups) are of central importance in many fields of geometry. If $S$ is an oriented surface (i.e. a $2$-manifold), the group $\text{Homeo}^+(S)$ of
orientation-preserving self-homeomorphisms of $S$ is a topological group with the compact-open topology. The mapping class group of $S$, denoted $\text{MCG}(S)$ (or $\text{Mod}(S)$ by some people) is
the group of path-components of $\text{Homeo}^+(S)$, i.e. $\pi_0(\text{Homeo}^+(S))$, or equivalently $\text{Homeo}^+(S)/\text{Homeo}_0(S)$ where $\text{Homeo}_0(S)$ is the subgroup of homeomorphisms
isotopic to the identity.
When $S$ is a surface of finite type (i.e. a closed surface minus finitely many points), the group $\text{MCG}(S)$ is finitely presented, and one knows a great deal about the algebra and geometry of
this group. Less well-studied are groups of the form $\text{MCG}(S)$ when $S$ is of infinite type. However, such groups do arise naturally in dynamics.
Example: Let $G$ be a group of (orientation-preserving) homeomorphisms of the plane, and suppose that $G$ has a bounded orbit (i.e. there is some point $p$ for which the orbit $Gp$ is contained in a
compact subset of the plane). The closure of such an orbit $Gp$ is compact and $G$-invariant. Let $K$ be the union of the closure of $Gp$ with the set of bounded open complementary regions. Then $K$
is compact, $G$-invariant, and has connected complement. Define an equivalence relation $\sim$ on the plane whose equivalence classes are the points in the complement of $K$, and the connected
components of $K$. The quotient of the plane by this equivalence relation is again homeomorphic to the plane (by a theorem of R. L. Moore), and the image of $K$ is a totally disconnected set $k$. The
original group $G$ admits a natural homomorphism to the mapping class group of $\mathbb{R}^2 - k$. After passing to a $G$-invariant closed subset of $k$ if necessary, we may assume that $k$ is
minimal (i.e. every orbit is dense). Since $k$ is compact, it is either a finite discrete set, or it is a Cantor set.
The mapping class group of $\mathbb{R}^2 - \text{finite set}$ contains a subgroup of finite index fixing the end of $\mathbb{R}^2$; this subgroup is the quotient of a braid group by its center. There
are many tools that show that certain groups $G$ cannot have a big image in such a mapping class group.
Much less studied is the case that $k$ is a Cantor set. In the remainder of this post, we will abbreviate $\text{MCG}(\mathbb{R}^2 - \text{Cantor set})$ by $\Gamma$. Notice that any homeomorphism of
$\mathbb{R}^2 - \text{Cantor set}$ extends in a unique way to a homeomorphism of $S^2$, fixing the point at infinity, and permuting the points of the Cantor set (this can be seen by thinking of the
“missing points” intrinsically as the space of ends of the surface). Let $\Gamma'$ denote the mapping class group of $S^2 - \text{Cantor set}$. Then there is a natural surjection $\Gamma \to \Gamma'$
whose kernel is $\pi_1(S^2 - \text{Cantor set})$ (this is just the familiar Birman exact sequence).
The following is proved in the first section of my paper “Circular groups, planar groups and the Euler class”. This is the first step to showing that any group $G$ of orientation-preserving
diffeomorphisms of the plane with a bounded orbit is circularly orderable:
Proposition: There is an injective homomorphism $\Gamma \to \text{Homeo}^+(S^1)$.
Sketch of Proof: Choose a complete hyperbolic structure on $S^2 - \text{Cantor set}$. The Birman exact sequence exhibits $\Gamma$ as a group of (equivalence classes) of homeomorphisms of the
universal cover of this hyperbolic surface which commute with the deck group. Each such homeomorphism extends in a unique way to a homeomorphism of the circle at infinity. This extension does not
depend on the choice of a representative in an equivalence class, and one can check that the extension of a nontrivial mapping class is nontrivial at infinity. qed.
This property of the mapping class group $\Gamma$ does not distinguish it from mapping class groups of surfaces of finite type (with punctures); in fact, the argument is barely sensitive to the
topology of the surface at all. By contrast, the next theorem demonstrates a significant difference between mapping class groups of surfaces of finite type, and $\Gamma$. Recall that for a surface
$S$ of finite type, the group $\text{MCG}(S)$ acts simplicially on the complex of curves $\mathcal{C}(S)$, a simplicial complex whose simplices are the sets of isotopy classes of essential simple
closed curves in $S$ that can be realized mutually disjointly. A fundamental theorem of Masur-Minsky says that $\mathcal{C}(S)$ (with its natural simplicial path metric) is $\delta$-hyperbolic
(though it is not locally finite). Bestvina-Fujiwara show that any reasonably big subgroup of $\text{MCG}(S)$ contains lots of elements that act on $\mathcal{C}(S)$ weakly properly, and therefore
such groups admit many nontrivial quasimorphisms. This has many important consequences, and shows that for many interesting classes of groups, every homomorphism to a mapping class group (of finite
type) factors through a finite group. In view of the potential applications to dynamics as above, one would like to be able to construct quasimorphisms on mapping class groups of infinite type.
Unfortunately, this does not seem so easy.
Proposition: The group $\Gamma'$ is uniformly perfect.
Proof: Remember that $\Gamma'$ denotes the mapping class group of $S^2 - \text{Cantor set}$. We denote the Cantor set in the sequel by $C$.
A closed disk $D$ is a dividing disk if its boundary is disjoint from $C$, and separates $C$ into two components (both necessarily Cantor sets). An element $g \in \Gamma$ is said to be local if it
has a representative whose support is contained in a dividing disk. Note that the closure of the complement of a dividing disk is also a dividing disk. Given any dividing disk $D$, there is a
homeomorphism of the sphere $\varphi$ permuting $C$, that takes $D$ off itself, and so that the family of disks $\varphi^n(D)$ are pairwise disjoint, and converge to a limiting point $x \in C$.
Define $h$ to be the infinite product $h = \prod_i \varphi^i g \varphi^{-i}$. Notice that $h$ is a well-defined homeomorphism of the plane permuting $C$. Moreover, there is an identity $[h^{-1},\
varphi] = g$, thereby exhibiting $g$ as a commutator. The theorem will therefore be proved if we can exhibit any element of $\Gamma'$ as a bounded product of local elements.
Now, let $g$ be an arbitrary homeomorphism of the sphere permuting $C$. Pick an arbitrary $p \in C$. If $g(p)=p$ then let $h$ be a local homeomorphism taking $p$ to a disjoint point $q$, and define
$g' = hg$. So without loss of generality, we can find $g' = hg$ where $h$ is local (possibly trivial), and $g'(p) = q e p$. Let ${}E$ be a sufficiently small dividing disk containing $p$ so that $g'
(E)$ is disjoint from ${}E$, and their union does not contain every point of $C$. Join ${}E$ to $g'(E)$ by a path in the complement of $C$, and let $D$ be a regular neighborhood, which by
construction is a dividing disk. Let $f$ be a local homeomorphism, supported in $D$, that interchanges ${}E$ and $g'(E)$, and so that $f g'$ is the identity on $D$. Then $fg'$ is itself local,
because the complement of the interior of a dividing disk is also a dividing disk, and we have expressed $g$ as a product of at most three local homeomorphisms. This shows that the commutator length
of $g$ is at most $3$, and since $g$ was arbitrary, we are done. qed.
The same argument just barely fails to work with $\Gamma$ in place of $\Gamma'$. One can also define dividing disks and local homeomorphisms in $\Gamma$, with the following important difference. One
can show by the same argument that local homeomorphisms in $\Gamma$ are commutators, and that for an arbitrary element $g \in \Gamma$ there are local elements $h,f$ so that $fhg$ is the identity on a
dividing disk; i.e. this composition is anti-local. However, the complement of the interior of a dividing disk in the plane is not a dividing disk; the difference can be measured by keeping track of
the point at infinity. This is a restatement of the Birman exact sequence; at the level of quasimorphisms, one has the following exact sequence: $Q(\Gamma') \to Q(\Gamma) \to Q(\pi_1(S^2 - C))^{\
The so-called “point-pushing” subgroup $\pi_1(S^2 - C)$ can be understood geometrically by tracking the image of a proper ray from $C$ to infinity. We are therefore motivated to consider the
following object:
Definition: The ray graph $R$ is the graph whose vertex set is the set of isotopy classes of proper rays $r$, with interior in the complement of $C$, from a point in $C$ to infinity, and whose edges
are the pairs of such rays that can be realized disjointly.
One can verify that the graph $R$ is connected, and that the group $\Gamma$ acts simplicially on $R$ by automorphisms, and transitively on vertices.
Lemma: Let $g \in \Gamma$ and suppose there is a vertex $v \in R$ such that $v,g(v)$ share an edge. Then $g$ is a product of at most two local homeomorphisms.
Sketch of proof: After adjusting $g$ by an isotopy, assume that $r$ and $g(r)$ are actually disjoint. Let $E,g(E)$ be sufficiently small disjoint disks about the endpoint of $r$ and $g(r)$, and $\
alpha$ an arc from ${}E$ to $g(E)$ disjoint from $r$ and $g(r)$, so that the union $r \cup E \cup \alpha \cup g(E) \cup g(r)$ does not separate the part of $C$ outside $E \cup g(E)$. Then this union
can be engulfed in a punctured disk $D'$ containing infinity, whose complement contains some of $C$. There is a local $h$ supported in a neighborhood of $E \cup \alpha \cup g(E)$ such that $hg$ is
supported (after isotopy) in the complement of $D'$ (i.e. it is also local). qed.
It follows that if $g \in\Gamma$ has a bounded orbit in $R$, then the commutator lengths of the powers of $g$ are bounded, and therefore $\text{scl}(g)$ vanishes. If this is true for every $g \in \
Gamma$, then Bavard duality implies that $\Gamma$ admits no nontrivial homogeneous quasimorphisms. This motivates the following questions:
Question: Is the diameter of $R$ infinite? (Exercise: show $\text{diam}(R)\ge 3$)
Question: Does any element of $\Gamma$ act on $R$ with positive translation length?
Question: Can one use this action to construct nontrivial quasimorphisms on $\Gamma$?
A basic reference for the background to this post is my monograph.
Let $G$ be a group, and let $[G,G]$ denote the commutator subgroup. Every element of $[G,G]$ can be expressed as a product of commutators; the commutator length of an element $g$ is the minimum
number of commutators necessary, and is denoted $\text{cl}(g)$. The stable commutator length is the growth rate of the commutator lengths of powers of an element; i.e. $\text{scl}(g) = \lim_{n \to \
infty} \text{cl}(g^n)/n$. Recall that a group $G$ is said to satisfy a law if there is a nontrivial word $w$ in a free group $F$ for which every homomorphism from $F$ to $G$ sends $w$ to $\text{id}$.
The purpose of this post is to give a very short proof of the following proposition (modulo some background that I wanted to talk about anyway):
Proposition: Suppose $G$ obeys a law. Then the stable commutator length vanishes identically on $[G,G]$.
The proof depends on a duality between stable commutator length and a certain class of functions, called homogeneous quasimorphisms.
Definition: A function $\phi:G \to \mathbb{R}$ is a quasimorphism if there is some least number $D(\phi)\ge 0$ (called the defect) so that for any pair of elements $g,h \in G$ there is an inequality
$|\phi(x) + \phi(y) - \phi(xy)| \le D(\phi)$. A quasimorphism is homogeneous if it satisfies $\phi(g^n) = n\phi(g)$ for all integers $n$.
Note that a homogeneous quasimorphism with defect zero is a homomorphism (to $\mathbb{R}$). The defect satisfies the following formula:
Lemma: Let $f$ be a homogeneous quasimorphism. Then $D(\phi) = \sup_{g,h} \phi([g,h])$.
A fundamental theorem, due to Bavard, is the following:
Theorem: (Bavard duality) There is an equality $\text{scl}(g) = \sup_\phi \frac {\phi(g)} {2D(\phi)}$ where the supremum is taken over all homogeneous quasimorphisms with nonzero defect.
In particular, $\text{scl}$ vanishes identically on $[G,G]$ if and only if every homogeneous quasimorphism on $G$ is a homomorphism.
One final ingredient is another geometric definition of $\text{scl}$ in terms of Euler characteristic. Let $X$ be a space with $\pi_1(X) = G$, and let $\gamma:S^1 \to X$ be a free homotopy class
representing a given conjugacy class $g$. If $S$ is a compact, oriented surface without sphere or disk components, a map $f:S \to X$ is admissible if the map on $\partial S$ factors through $\partial
f:\partial S \to S^1 \to X$, where the second map is $\gamma$. For an admissible map, define $n(S)$ by the equality $[\partial S] \to n(S) [S^1]$ in $H_1(S^1;\mathbb{Z})$ (i.e. $n(S)$ is the degree
with which $\partial S$ wraps around $\gamma$). With this notation, one has the following:
Lemma: There is an equality $\text{scl}(g) = \inf_S \frac {-\chi^-(S)} {2n(S)}$.
Note: the function $-\chi^-$ is the sum of $-\chi$ over non-disk and non-sphere components of $S$. By hypothesis, there are none, so we could just write $-\chi$. However, it is worth writing $-\chi^
-$ and observing that for more general (orientable) surfaces, this function is equal to the function $\rho$ defined in a previous post.
We now give the proof of the Proposition.
Proof. Suppose to the contrary that stable commutator length does not vanish on $[G,G]$. By Bavard duality, there is a homogeneous quasimorphism $\phi$ with nonzero defect. Rescale $\phi$ to have
defect $1$. Then for any $\epsilon$ there are elements $g,h$ with $\phi([g,h]) \ge 1-\epsilon$, and consequently $\text{scl}([g,h]) \ge 1/2 - \epsilon/2$ by Bavard duality. On the other hand, if $X$
is a space with $\pi_1(X)=G$, and $\gamma:S^1 \to X$ is a loop representing the conjugacy class of $[g,h]$, there is a map $f:S \to X$ from a once-punctured torus $S$ to $X$ whose boundary represents
$\gamma$. The fundamental group of $S$ is free on two generators $x,y$ which map to the class of $g,h$ respectively. If $w$ is a word in $x,y$ mapping to the identity in $G$, there is an essential
loop $\alpha$ in $S$ that maps inessentially to $X$. There is a finite cover $\widetilde{S}$ of $S$, of degree $d$ depending on the word length of $w$, for which $\alpha$ lifts to an embedded loop.
This can be compressed to give a surface $S'$ with $-\chi^-(S') \le -\chi^-(\widetilde{S})-2$. However, Euler characteristic is multiplicative under coverings, so $-\chi^-(\widetilde{S}) = -\chi^-(S)
\cdot d$. On the other hand, $n(S') = n(\widetilde{S})=d$ so $\text{scl}([g,h]) \le 1/2 - 1/d$. If $G$ obeys a law, then $d$ is fixed, but $\epsilon$ can be made arbitrarily small. So $G$ does not
obey a law. qed.
As an experiment, I plan to spend the next five weeks documenting my current research on this blog. This research comprises several related projects, but most are concerned in one way or another with
the general program of studying the geometry of a space by probing it with surfaces. Since I am nominally a topologist, these surfaces are real $2$-manifolds, and I am usually interested in working
in the homotopy category (or some rational “quotient” of it). I am especially concerned with surfaces with boundary, and even (occasionally) with corners.
Since it is good to have a “big question” lurking somewhere in the background (for the purposes of motivation and advertising, if nothing else), I should admit from the start that I am interested in
Gromov’s well-known question about surface subgroups, which asks:
Question (Gromov): Does every one-ended word-hyperbolic group contain a closed hyperbolic surface subgroup?
I don’t have strong feelings about whether the answer to this question is “yes” or “no”, but I do think the question can be sharpened usefully in many ways, and it is my intention to do so. Gromov’s
question is certainly inspired by questions such as Waldhausen’s conjecture and the virtual fibration conjecture in $3$-manifold topology, but it is hard to imagine that a proof of one of these
conjectures would shed much light on Gromov’s question in general. At least one essential tool in $3$-manifold topology — namely Dehn’s lemma — has no meaningful analogue in geometric group theory,
and I think it is important to try to imagine different methods of constructing surface groups from “first principles”.
Another long-term project that informs much of my current research is the problem of understanding stable commutator length in free groups. The interested reader can learn something about this from
my monograph (which can be downloaded from this page). I hope to explain why this is a fundamental and interesting problem, with rich structure and many potential applications.
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Probability For Dummies
Cheat Sheet
Probability For Dummies
Successfully working your way through probability problems means understanding some basic rules of probability along with discrete and continuous probability distributions. Use some helpful study
tips so you're well-prepared to take a probability exam.
Principles of Probability
The mathematics field of probability has its own rules, definitions, and laws, which you can use to find the probability of outcomes, events, or combinations of outcomes and events. To determine
probability, you need to add or subtract, multiply or divide the probabilities of the original outcomes and events. You use some combinations so often that they have their own rules and formulas. The
better you understand the ideas behind the formulas, the more likely it is that you'll remember them and be able to use them successfully.
Probability rules
Probability definitions
Probability laws
Counting rules
Discrete Probability Distributions
In probability, a discrete distribution has either a finite or a countably infinite number of possible values. That means you can enumerate or make a listing of all possible values, such as 1, 2, 3,
4, 5, 6 or 1, 2, 3, . . .
There are several kinds of discrete probability distributions, including discrete uniform, binomial, Poisson, geometric, negative binomial, and hypergeometric.
Continuous Probability Distributions
When you work with continuous probability distributions, the functions can take many forms. These include continuous uniform, exponential, normal, standard normal (Z), binomial approximation, Poisson
approximation, and distributions for the sample mean and sample proportion.
When you work with the normal distribution, you need to keep in mind that it's a continuous distribution, not a discrete one. A continuous distribution's probability function takes the form of a
continuous curve, and its random variable takes on an uncountably infinite number of possible values. This means the set of possible values is written as an interval, such as negative infinity to
positive infinity, zero to infinity, or an interval like [0, 10], which represents all real numbers from 0 to 10, including 0 and 10.
Probability Study Tips
If you're going to take a probability exam, you can better your chances of acing the test by studying the following topics. They have a high probability of being on the exam.
• The relationship between mutually exclusive and independent events
• Identifying when a probability is a conditional probability in a word problem
• Probability concepts that go against your intuition
• Marginal, conditional, and joint probabilities for a two-way table
• The Central Limit Theorem:
• When to use a permutation and when to use a combination
• Finding E(X) from scratch and interpreting it
• Sampling with replacement versus without replacement
• The Law of Total Probability and Bayes' Theorem
• When the Poisson and exponential are needed in the same problem
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FIG. 1.
(a) Schematic of experimental setup: Soap solution at constant pressure head flows through a nozzle between two parallel nylon wires that are a width W apart and over a length 1 m to form a soap
film. An annular ring of external radius R = 1.1 × 10−2 m hangs from a flexible string that is pivoted outside the soap film. (b) For a given flow rate, the pendulum exhibits oscillatory motion above
a critical length L c . Multiple images of oscillations over several periods record the pendulum amplitude A. (c) A longer pendulum exhibits more complex oscillations, such as one with a node as
marked with the blue circle. (d) Velocity field superposed on raw images shows when it is vertical and (e) when it is at its extremal position (enhanced online). [URL: http://dx.doi.org/10.1063/
1.4800057.1] [URL: http://dx.doi.org/10.1063/1.4800057.2]doi: 10.1063/1.4800057.1.
doi: 10.1063/1.4800057.2.
FIG. 2.
Dimensionless critical length L c /L 0 vs. dimensionless ring radius R/L 0, (where L 0 is the smallest length of a fiber without an attached ring that spontaneously oscillates in a flow) at constant
mass M R and flow rate, obtained experimentally (solid squares). The solid line corresponds to the best fit obtained from the first four points, with L c /L 0 = 1 + 1.2R/L 0.
FIG. 3.
(a) Dimensionless frequency ω/ω0 vs. dimensionless length L/R (bottom horizontal axis) when M R > M A (M R = 1.48 × 10−4 kg). Experimental data are shown as solid circles and a fit based on Eq. (2)
is shown as a solid line; ω0 = 9.44 s−1 is the natural frequency for a gravity pendulum of length L = 10R. Top horizontal axis shows the regime when M R < M A , with the ring radius held constant at
R = 1.1 × 10−2 m while increasing mass M R by adding plasticine, with M ρ = πR 2ρh being the mass of a disc of liquid of radius R. Experiment (solid squares) and theory fit (dashed line) from Eq. (4)
. (b) Dimensionless frequency ω/ω r vs. rescaled velocity based on Eq. (4) in the added mass regime M A > M R & M s . Here ω r = 19.08 s−1 is the offset frequency corresponding to the mode where the
ring oscillates around its pivot.
FIG. 4.
Comparison between experimentally obtained angular dynamics shown with open circles (red) and the simple theory based on a driven oscillator model (solid black curve) given by Eq. (5) with the
following parameters: M R = 0.12 × 10−3 kg, h = 10μm, V = 1 m/s, L = 5.5 × 10−2 m, γ = 80 s−1, R = 1.1 × 10−2 m. The numerical curve is intentionally offset by Δθ = 0.1 for clarity.
The following multimedia file is available, if you log in: 1.4800057.original.v1.mov
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Independent transformations [Archive] - OpenGL Discussion and Help Forums
02-24-2004, 08:04 PM
Hi All,
I am a beginner to OpenGL programming. I am stuck with a problem. I will
explain it with a simple example.
I wanted to draw a tilted V using two lines and some rotations. I wanted
to do in the following way.
0) Initial position.
1) Draw line along Y-axis.
2) Rotate it anticlickwise 45.0
3) Come to initial position. Such that transformation done in the step (2) does not
apply next.
4) Now draw one more line along Y-axis.
5) Rotate it clockwise 45.0. Now we should have a shape 'V'
6) Now rotate thsi whole model(, I mean the V shape), 90.0
That is I wanted to apply independent transformations to each object and one more transformation for the entire scene.
I tried to use pusmatrix and popmatrix. But it didn't help. May be I am not
using it in proper way. Can you please let me know how to acheive this.
A pseudo code will be really help ful.
Expecting your help.
Thanks in advance
Deek**** M
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[OT] stable algorithm with complexity O(n)
Lie Ryan lie.1296 at gmail.com
Sun Dec 14 22:42:33 CET 2008
On Sat, 13 Dec 2008 19:17:41 +0000, Duncan Booth wrote:
> "Diez B. Roggisch" <deets at nospam.web.de> wrote:
>> David HlÃ¡Ä ik schrieb:
>>> Hi guys,
>>> i am really sorry for making offtopic, hope you will not kill me, but
>>> this is for me life important problem which needs to be solved within
>>> next 12 hours..
>>> I have to create stable algorithm for sorting n numbers from interval
>>> [1,n^2] with time complexity O(n) .
>>> Can someone please give me a hint. Would be very very thankful!
>> Unless I grossly miss out on something in computer science 101, the
>> lower bound for sorting is O(n * log_2 n). Which makes your task
>> impossible, unless there is something to be assumed about the
>> distribution of numbers in your sequence.
>> Who has given you that assignment - a professor? Or some friend who's
>> playing tricks on you?
> I think you must have fallen asleep during CS101. The lower bound for
> sorting where you make a two way branch at each step is O(n * log_2 n),
> but if you can choose between k possible orderings in a single
> comparison you can get O(n * log_k n).
> To beat n * log_2 n just use a bucket sort: for numbers with a known
> maximum you can sort them digit by digit for O(n log_k n), and if you
> don't restrict yourself to decimal then k can be as large as you want,
> so for the problem in question I think you can set k=n and (log_n n)==1
> so you get O(n)
I'm sure someday, there will be a student who comes to class late and
sees this on the board: "Design a comparison sorting algorithm that has
better than O(n * log n) lower bound complexity." The unsuspecting
student copied it, thinking it's a homework. He crunched to the problem,
going to various meditations and yoga classes before finding a way that
works just before deadline, handing out the work a bit late. Six weeks
later, his professor called and said: "You know what you just did? You've
just found a problem that was supposed to be an example of unsolvable
It has happened before, why not again?
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Perfect Numbers
When playing taxman, there are numbers like 6 where the taxman will tie your first move. Another number where this happens is 28 (try it out). When a number equals the sum of its proper factors, it
is called a "perfect number".
Perfect numbers are surprisingly few and far between, and mathematical knowledge about perfect numbers is also surprisingly incomplete.
The first few perfect numbers are: 6, 28, 496, 8128, and 33550336; and they quickly get big enough to stretch the limits of computers. Only a few dozen perfect numbers are known. And while
mathematicians since the Greeks have suspected that there are an infinite number of perfect numbers, that conjecture has never been proven. Futhermore, all known perfect numbers are even, but it has
never been established that an odd perfect number is impossible.
Despite all the mathematical intrigue, any middle schooler can do the computations to understand the pattern underlying the known perfect numbers. All you have to do is try writing the first few
perfect numbers out in binary.
Following Euclid
It is easy to build intuition about all the known perfect numbers by working out a specific example.
The perfect number 496 is binary 1111000. In this form it is clear that 496 is the product of a prime made of all binary ones "1111" (a.k.a. 31 in ten-fingered society) and a power of two with the
same number of digits "1000" (a.k.a. 16).
So let's add up the complete set of prime factors of 496.
Since 1111 is a prime (let's abbreviate it p), the complete list of factors of (1111*1000) is pretty short and easy to understand. First, we list the factors of 1000, and then p times each of those
(still working in binary:)
1, 10, 100, 1000
p*1, p*10, p*100, p*1000
The first row sums to 1111.
The second row sums to p*1111.
And the total is (p+1)*(1111)=(1+1111)*1111=10000*1111=11110000.
With one extra zero, that sum of the factors is double our original number. But since we have included the number itself in the sum of factors, the number is half of its sum of factors. So it is
All 46 perfect numbers that have ever been found follow exactly this same pattern. Each one is the product of an all-ones prime (called a Mersenne prime) and a power of two with the same number of
It is not known if there are an infinite number of Mersenne primes. And it is not known if there are any perfect numbers that do not follow this pattern. Mathematical fame awaits anybody who can find
an odd perfect number.
Posted by David at December 11, 2008 10:57 PM
496 is binary 111110000.
1111000 is decimal 120.
Posted by: Matt Canning at January 3, 2010 07:18 PM
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Reply to comment
Submitted by Anonymous on December 22, 2010.
Thank you for this amazingly informative article. I hardly like to mention that, in equation (4) above, where you have given the equations of the foci of an ellipse, the second set of co-ordinates
should refer to F2, and not F3 as listed. I am sure that this is just a typing error and does not detract from the excellence of this treatise.
Sorry for making such a minute correction to this spendid article.
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• Choice of what to plot:
[-quantity=columnName | -equation=rpnExpression | -columnMatch=indepColumnName,matchingExpression | -waterfall=parameter=<parameter>,independentColumn=<xColumn>,colorColumn=<colorColumn>[,scroll=
□ quantity -- Specifies the name of the column to make a contour or color map of.
□ equation -- Specifies a rpn expression to make a contour or color map of. The expression may refer to the values in the columns by the appropriate column name, and may also refer to the
variable values by name.
□ columnMatch -- Specifies plotting of all columns matching matchingExpression as a function of the column indepColumnName. Each matching column is displayed as a horizontal color bar.
□ waterfall -- Specifies plotting of colorColumn in all pages as a function of the independentColumn. The parameter in each page is displayed as horizontal or vertical (provided by the scroll)
bar, the default is horizontal. The independentColumn should be the same in each page.
In the case of the first two choices, the file must contain tabular data with at least one numeric column, which will be organized into a 2d array with R rows and C columns. By default, the
values are assumed to come in row-major order (i.e., the file should contain a series of R sequences each containing the C values of a single row). The parameters of the 2d grid over which the
plot is to be made are communicated to the program in one of two ways:
1. The string parameters Variable1Name and Variable2Name contain the names of the x and y axis variables, which I'll represent as x and y respectively. The program expects to find six more
parameters, with names xMinimum, xInterval, and xDimension, and similarly for y. These parameters must be numeric, and contain the minimum value, the interval between grid points, and the
number of points, respectively, for the dimension in question. The data must be arranged so that y varies fastest as the row in the file increases. Put another way, variable 1 is the row
index and variable 2 is the column index.
2. The numeric parameters NumberOfRows and NumberOfColumns contain the values of R and C, respectively.
• rpn control:
[-rpnDefinitionsFiles=filename[,filename...]] [-rpnExpressions=setupExpression[,setupExpression...]]
□ rpnDefinitionsFiles -- Specifies the names of files containing rpn expressions to be executed before any other processing takes place.
□ rpnExpressions -- Specifies rpn expressions to be executed before any other processing takes place, immediately after any definitions files.
• Shade and contour control:
{-shade=number[,min,max,gray] | -contours=number[,min,max]} [-labelContours=interval[,offset]]
□ shade -- Specifies that a color (or grey-scale) map should be produced, with the indicated number of shades mapped onto the range from min to max. If min and max are not given, they are taken
to be equal to the minimum and maximum data values.
□ contours -- Specifies that contour lines should be drawn, with the indicated number of lines for the range from min to max. If min and max are not given, they are taken to be equal to the
minimum and maximum data values.
□ labelContours -- Specifies that every intervaloffset
• Image processing:
[-interpolate=nx,ny[,floor | ceiling | antiripple]] [-filter=xcutoff,ycutoff]
□ interpolate -- Specifies that the 2d map should be interpolated to have nx times more rows (or x grid points) and ny times more columns (or y grid points). Since FFTs are used to do the
interpolation, the original number of grid points must be a power of 2, as must the factor. Giving a factor of 1 disables interpolation for the dimension in question. floor, ceiling, and
antiripple specify image processing of the interpolated map. floor and ceiling respectively force values below (above) the minimum (maximum) value of the data to be set equal to that value.
antiripple causes the map to be altered so that non-zero values in the new map between zero values on the original map are set to zero; this suppresses ripples that sometimes occur in regions
where the data was originally all zero.
□ filter -- Applies low-pass filters to the data with the specified normalized cutoff frequencies. The integer cutoff values give the number of frequencies starting at the Nyquist frequency
that are to be eliminated.
• Plot labeling:
[-xLabel=string|@<parameter-name>] [-yLabel=string|@<parameter-name>] [-title=string|@<parameter-name>|file[,edit=<string>]] [-topline=string|@<parameter-name>|file[,edit=<string>]] [-topTitle]
[-noLabels] [-noScales] [-dateStamp]
□ xLabel, yLabel, title, topline -- These specify strings to be placed in the various label locations on the plot. If @<parameter-name> is provided, the value of given parameter will be
printed; If -topline=file[,edit=<string>] or -title=file[,edit=<string>] option is provided, then the input file name or edited file name (if edit command is also provided) will be printed to
the topline or title.
□ topTitle -- Requests that the title label be placed at the top of the plot, rather than at the bottom.
□ noLabels -- Requests that no labels be placed on the plot.
□ noScales -- Requests omission of the numeric scales.
□ noBorder -- Requests omission of the border around the data. Implies -no_scales.
□ dateStamp -- Requests that the date and time be placed on the pot.
• Plot tick labeling: (only valid for -columnMatch plot)
[-xrange=mimum=value|@parameterName,maximum=value|@parameterName] [-yrange=mimum=value|@parameterName,maximum=value|@parameterName] [-yaxis=scaleValue=<value>|scaleParameter=<name>[,offsetValue=
□ xrange -- specifies the minimum and maximum value of x axis, the value can be provided or obtained from parameters. If -xrange is provided, the indepentColumn will be ignored.
□ yrange -- specifies the minimum and maximum value of y axis, the value can be provided or obtained from parameters. If -yrange is provided, the y tick labels will be numberically labeled with
provided range.
□ yaxis -- specifies the scale and offset value of y axis, the value can be provided or obtained from parameters. Only one of the -yrange and -yaxis can be provided. If -yaxis is provided, the
y tick labels will be numberically labeled with provided scale and offset.
For example, origin1, delta1, max_ext1, origin2, delta2 and max_ext2 are the parameters in sddscontour.input1 file, origin1, delta1 and max_ext1 represent the minimum, delta and maximum values of
x coordinate, origin2, delta2, and max_ext2 represents the minimum, delta and maximum values of y coordinate. The Index column represents the index of x coordinate, i.e. value of x=Index * delta1
+ origin1; The Ex_n column represents the Ex field at nth y value, where y=(n-1)*delta2 + origin2. If no -xrange and -yrange provided as in following command, the actual value of x and y will not
be shown in the plot. (click the show_plot button will show you the corresponding plot.)
sddscontour sddscontour.input1 -columnMatch=Index,Ex* -ystring=sparse=10 -ylabel=y -shade show_plot
We can use -ystring to remove the string part of y label as following:
sddscontour sddscontour.input1 -columnMatch=Index,Ex* -ystring=sparse=10,edit=%/Ex_// -ylabel=y -shade show_plot
The above y tick label still shows the index of y coordinate, not y values. Following command allows us to see the y values:
sddscontour sddscontour.input1 -columnMatch=Index,Ex* -yrange=minimum=@origin2,maximum=@max_ext2 -ylabel=y -shade show_plot
Now, for the x tick labels, the above plot shows the Index value. Following command will show the values of x coordinate:
sddscontour sddscontour.input1 -column=Index,Ex* -yrange=min=@origin2,max=@max_ext2 -xrange=min=@origin1,max=@max_ext1 -xlabel=x -ylabel=y -shade show_plot
The independent column - Index in the above command is useless. Therefore, -xrange provides a way for plotting a set of columns with contour without indepent column. If use sddsprocess to create
x column through x=Index * delta1 + origin1, the above plot can be created using following command, note that the titles in two plots are different because the independent column names are
different since the title is automatically generated from input column names if it is not provided.
sddscontour sddscontour.input1 -column=x,Ex* -yrange=min=@origin2,max=@max_ext2 -ylabel=y -shade show_plot
Here shows the examples of providing xrange and yrange from parameters, however, they can be provided by fixed values from commandline also.
• Data scaling:
[-deltas[={fractional | normalize}]] [-logscale[=floor]]
□ deltas -- For use with -columnMatch and -waterfall option. Specifies plotting only differential values (relative to the mean of each column). If the fractional qualifier is given, then the
differential values normalized to the individual means are plotted. If the normalize qualifier is given, then all differential values are normalized to the range [-1, 1] before plotting.
□ logscale -- Specifies plotting the base-10 logarithm of the values. If a floor value is given, it is added to each value prior to taking the logarithm; this can help prevent taking the log of
zero, for example.
• Miscellaneous plot control:
[-scales=xl,xh,yl,yh] [-device=name[,deviceArguments]] [-swapxy] [-equalAspect[=-1,1]] [-noBorder]
□ scales -- Specifies the extent of the plot region.
□ device -- Specifies the device name and optional device-specific arguments. png devices take rootname and template identifiers. rootname=string specifies a rootname for automatic filename
generation; the resulting filenames are of the form rootname.DDD, where DDD is a three-digit integer. template=string provides a more general facility; one uses it to specify an sprintf-style
format string to use in creating filenames. For example, the behavior obtained using rootname=name may be obtained using template=name.%03ld.
□ swapxy -- Requests that the horizontal and vertical coordinates be interchanged.
□ equalAspect -- Requests plotting with an aspect ratio of 1. If the '1' qualifier is given, then the aspect ratio is achieved by changing the size of the plot region within the window; this is
the default. If the '-1' qualifier is given, then the aspect ratio is achieved by changing the size of the plot region in user's coordinates.
□ noBorder -- Specifies that no border will be placed around the graph.
• Miscellaneous:
[-output=filename] [-verbosity[=level]]
□ output -- Requests SDDS output of a new file containing the data with any modifications resulting in the processing requested.
□ verbosity -- Sets the verbosity level of informational printouts. Higher integer values of the level parameter result in more output.
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What are the higher homotopy groups of Spec Z ?
up vote 25 down vote favorite
The homotopy groups of the étale topos of a scheme were defined by Artin and Mazur. Are these known for Spec Z? Certainly π[1] is trivial because Spec Z has no unramified étale covers, but what is
known about the higher homotopy groups?
homotopy-theory ag.algebraic-geometry topos-theory nt.number-theory
add comment
2 Answers
active oldest votes
Since nobody else is answering this question I will give a few vague thoughts. Caveat: this is a completely speculative answer for various reasons, not least of which because I don't know
Artin-Mazur's definition of pi_i.
Since X = Spec(Z) is "simply connected", one can pretend that the Hurewicz theorem applies.
up vote 5 I believe that H^i(X,Z/nZ) is trivial for all i, as a consequence of class field theory and the fact that Z has neither many units nor n-th roots of unity. (I'm not completely sure about i
down vote = 3 and n = 2 here.) One can then squint and imagine that the higher homotopy groups of X are trivial. This seems a little dodgy. Another direction one could go is to note that the groups H
^i(X,G_m) vanish unless i = 3, and H^3(X,G_m) = Q/Z. From this (and other) facts it has been argued that X is analogous to the 3-sphere.
For what it is worth, both computations suggest that pi_2(X) is trivial. If one wanted to turn this comment into mathematics, one should try to define an algebraic Hurewicz map.
Indeed H^i(X) is supposed to point out that X looks like a 3-sphere and it's not like homotopy groups of S^3 are easy. – Ilya Nikokoshev Nov 3 '09 at 9:03
My memory is vague, but does H^3(X,G_m) = Q/Z imply that there are nontrivial elements in H^3(X,Z/2) arising from the kernel of the squaring map G_m -> G_m, since Z/2 and the 2nd roots of
unity represent the same etale sheaf on this site? – Tyler Lawson Nov 3 '09 at 12:23
\mu_2 is not etale over Spec(Z). – moonface Nov 3 '09 at 13:52
The word "represent" may not have been the best to use. No, it's not etale, but it's a sheaf on the etale site as the kernel of the map G_m -> G_m. (The cokernel, if I remember correctly,
is isomorphic the direct image of the additive group on the etale site of Z/2.) – Tyler Lawson Nov 3 '09 at 15:02
Oh, I see what you mean. Thanks for clarifying! – moonface Nov 3 '09 at 17:20
add comment
If etale pi_1 classifies obstructions to trivializing finite flat unramified Z-algebras, it would be nice if the whole etale homotopy type classified obstructions to trivializing simplicial
commutative Z-algebras that were finite, flat, and unramified in a homotopy sense. I think all of these notions make sense: "finite" means that the homotopy groups vanish in high degrees
and are finitely generated, "flat" means that these homotopy groups have no torsion, and "unramified" means that the cotangent complex is zero. Is that right?
Presumably algebraic topologists have thought about the sphere spectrum version of this question. Are there any connective E-infinity ring spectra that are finite, flat, and unramified over
up vote 1 the sphere?
down vote
After Tyler's comments, I see that this is a bad analogy. The dictionary between etale locally constant sheaves of sets and finite flat unramified algebras (which in one direction takes an
algebra and associates the sheaf of sections of its spectrum over Spec Z) just doesn't extend to a dictionary between homotopy-style locally constant sheaves and homotopy-style finite flat
unramified algebras.
With regards to your second question: This appears in John Rognes' paper on Galois theory of structured ring spectra, at least in part. There are none if you ask that the generators all
live in pi_0. If you allow new generators in positive degrees, there are square-zero extensions and their ilk, and more following those that are harder to classify. I am not sure about
your first question. – Tyler Lawson Nov 3 '09 at 21:52
Are you really saying that there are square-zero extensions of the sphere spectrum that are unramified? This isn't possible with plain rings. – David Treumann Nov 3 '09 at 22:55
I guess I was thinking by analogy with your "homotopy vanishing in high degrees" examples above. No, those can't ever have trivial cotangent complex, and thickenings of a commutative
1 Z-algebra to something with positive homotopy groups usually can't either; e.g. if R -> S is an isomorphism on pi_0 then the first nonvanishing relative homotopy group coincides with the
first nonzero homotopy group of the cotangent complex. – Tyler Lawson Nov 4 '09 at 0:35
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Not the answer you're looking for? Browse other questions tagged homotopy-theory ag.algebraic-geometry topos-theory nt.number-theory or ask your own question.
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"Twin strings" problem - Codeforces
I'm trying to solve the following problem:
Two strings can be shuffled by interleaving their letters to form a new string (original order of letters is preserved). We will consider a shuffle of two identical strings (twins).
For instance, the string AAABAB can be obtained by shuffling two strings AAB.
For a given string we should check if it can be obtained by shuffling two twins.
At first, we should check the parity of each letter (XOR of all letters == 0).
Next, i can't think up anything except brute-force solution with complexity O(2^N) :(
Is there a way to solve the problem efficiently? Thanks!
9 months ago, # |
» ← Rev. 3 -8
Note that the first letter of the given string also has to be the first letter of each twin. We can remove the first two copies of that letter, and then recurse. This should give an O(n)
aquamongoose algorithm if implemented efficiently.
Edit: it appears this problem is NP-hard. My solution is wrong.
• »
9 months ago, # ^ |
» 0
I've posted similar solution to the russian branch. But if I understand you correctly, your solution fails on the test "ABCCAB". Answer for this test is "no"
□ 9 months ago, # ^ |
← Rev. 3 0
No, maybe I didn't explain my solution well enough. Here is pseudocode of my idea:
def is_twin_shuffle(string g of length n):
s = ""
» t = ""
s_index = 0
» for i from 0 to n-1:
if s_index = length of s or g[i] != s[s_index]:
» s += g[i]
if s_index == -1:
s_index = 0
aquamongoose else:
t += g[i]
s_index += 1
return (s == t)
edit: fixed a bug in pseudocode
edit: I'm wrong, sorry.
9 months ago, # |
• »
9 months ago, # ^ |
» 0
Thank you for the link. Haven't found it, very interesting
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Cambridge Journal of Mathematics
ISSN Print 2168-0930 ISSN Online 2168-0949
4 issues per year
David Jerison (Department of Mathematics, Massachusetts Institute of Technology)
Mark Kisin (Department of Mathematics, Harvard University)
William Minicozzi (Department of Mathematics, Massachusetts Institute of Technology)
Wilfried Schmid, editor-in-chief (Department of Mathematics, Harvard University)
Horng-Tzer Yau (Department of Mathematics, Harvard University)
Publisher’s Representative
Shing-Tung Yau (Department of Mathematics, Harvard University)
|
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Maximum Likelihood Estimator with binomial distribution question
March 6th 2011, 03:40 PM #1
Mar 2011
Maximum Likelihood Estimator with binomial distribution question
Q. There are N fish in a lake. Six fishermen fished for 3 hours and caught 8 fish between them. They fished for another 3 hours and caught 3 more fish between them.
Assume the following model: Each fish is equally hard to catch; the probability of catching a fish is p. Let Y_1 be the amount of fish caught in the first 3 hours, Y_1 ~ B(N,p)
Let Y_2 be the amount of fish caught in the second 3 hours, Y_2 ~ B( N-Y_1, p).
Find the MLE of N and p.
Hi everyone, please could you take a look at what I've done so far:
Let Ξ = L(N)/L(N-1), ...where L(N) = (N c Y).p^Y.(1-p)^(N-Y)
= N(Y-p)/(N-1), ≥1
Then N ≤ Y/p
So for Y_1 = 8, N ≤ 8/p
Doing the same with Y_2, we find that:
N ≤ 11/p + 8
My question is how do I get an actual value for MLE of N (and hence p) for, as it stands, I only have a range of values?
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help beginning programming
05-16-2009 #1
Registered User
Join Date
May 2009
I have a homework assignment that I have no idea where to go or what to do next. Could you please help? Thanks!
The assignment reads:
your program is to read from a file students.dat which may contain a list of student names that could contain as many as eleven characters. In addition to the list of names, your program should
read four sets of numbers which represent two exams worth 20%, a final exam worth 30%, and homework worth 30%. The student's name and grades is on the first line, with no title lines. Find the
weighted average grade for all students. The average grade is to berounded to the nearest integer, not to truncated to an integer. Then, sort the list of students by grade, and assign letter
grades to each student. The sorting can be a selection or a bubble sort, and is to be done in function sub-program. Say also if the student passes or fails. Put the sorted list into a file called
outstu.dat. Put a two line title above the output.
The grade distribution is
85-100 A
70-84 B
55-69 C
40-54 D
0-39 F
This is what i have so far:
/* Homework 12 */
#include FILENAME "students.dat"
#include <stdio.h>
#include <math.h>
#include <string.h>
struct record
char name[12];
int ex1, ex2, final, hw, ave, grade passf1[5];
int main(void)
struct record s[50];
int nos, i=0;
FILE *students;
while (fscanf(students, "%s %i %i %i %i," s[i].name, s[i].ex1, s[i].ex2, s[i].final, s[i].hw)==5)
if(s[i].ave>=85) s[i].grade='A';
if(s[i].ave>=70) s[i].grade='B';
if(s[i].ave>=55) s[i].grade='C';
if(s[i].ave>=40) s[i].grade='D';
if(s[i].ave>=0) s[i].grade='F';
strcpy(s[i].passf1, "pass");
else strcpy(s[i].pass1, "fail");
sort(s, nos);
for(i=0, i<nos, ++i);
fprintf(outf, "%-11s %5i %5i %5i %5i %5i %c %4sh," s[i].name, s[i].ex1, s[i].ex2, s[i].final,
void sort(record s[], int n);
int k,l m;
record hold;
for(k=0, k<=2, ++k);
for(i=k+1, j<=n);
if(s[j] < x[m], ave);
/* Homework 12 */ /*Does this mean this is your twelfth homework? */
#include FILENAME "students.dat" /*You know by now, surely, how to open a file; and this isn't it*/
#include <stdio.h>
#include <math.h>
#include <string.h>
struct record
char name[12];
int ex1, ex2, final, hw, ave, grade passf1[5]; /*What type do you want passf1 to be?*/
int main(void)
struct record s[50];
int nos, i=0;
FILE *students; /*Remember opening a file? Maybe you should do that now.*/
while (fscanf(students, "%s %i %i %i %i," s[i].name, s[i].ex1, s[i].ex2, s[i].final, s[i].hw)==5)
++i; /*Now you're not working on the data you just read in, but the one after that*/
if(s[i].ave>=85) s[i].grade='A';
if(s[i].ave>=70) s[i].grade='B';
if(s[i].ave>=55) s[i].grade='C';
if(s[i].ave>=40) s[i].grade='D';
if(s[i].ave>=0) s[i].grade='F'; /*Hooray! Everybody gets an F! */
strcpy(s[i].passf1, "pass");
else strcpy(s[i].pass1, "fail");
sort(s, nos);
for(i=0, i<nos, ++i);
fprintf(outf, "%-11s %5i %5i %5i %5i %5i %c %4sh," s[i].name, s[i].ex1, s[i].ex2, s[i].final,
s[i].hw) /*You have to finish this off before you can start doing another function*/
void sort(record s[], int n);
int k,l m;
record hold;
for(k=0, k<=2, ++k); /*Only going to sort two records? */
for(i=k+1, j<=n);
if(s[j] < x[m], ave);
And you definitely need to indent.
This code is so far from resembling something that might be called a C program it is not funny. Really, really, really bad. Horrific.
It is a very serious mistake to think you can just start conceptualizing and typing and hope that when you are finished, that is the time to compile and test. You should be compiling and testing
the code you are writing every 2 or 3 lines. That means breaking the task down into "sub tasks".
You have a struct, and a datafile to read. Before you proceed to the more complicated needs of the program, you should write something that will read the file into the the struct array, and then
print out all the data in the array, without performing any calculations or sorting. At least then you will have some foundation to build on, which right now you do not.
If you cannot do that, drop the class now rather than failing later. You are a long way from completing this assignment.
I'm sorry if this seems mean. I just want to make it very clear to you that, for whatever reason, you are screwing this up big time. AND YOU DEFINITELY NEED TO INDENT.
Last edited by MK27; 05-16-2009 at 06:02 PM.
C programming resources:
GNU C Function and Macro Index -- glibc reference manual
The C Book -- nice online learner guide
Current ISO draft standard
CCAN -- new CPAN like open source library repository
3 (different) GNU debugger tutorials: #1 -- #2 -- #3
cpwiki -- our wiki on sourceforge
okay here is my retry
/* Homework 12 */
#define FILENAME "students.dat"
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <string.h>
#define FILENAME2 "outstu.dat"
struct record
{char name[12]; int ex1, ex2, final, hw, ave, grade passfail[5];} record;
int main(void)
struct record s[50];
int nos=0, i=0;
struct record h[50]
FILE *grades
FILE *students;
/* Open file. */
void sort(record);
students=fopen(FILENAME, "r");
grades=fopen(FILENAME2, "w");
/*Read file. */
for(i=0; i<50; i++)
while (fscanf(students, "%s %i %i %i %i," s[i].name, s[i].ex1, s[i].ex2, s[i].final, s[i].hw)==5)
/*Determine Average for each Student */
if(s[i].ave>=85) s[i].grade='A';
if(s[i].ave>=70) s[i].grade='B';
if(s[i].ave>=55) s[i].grade='C';
if(s[i].ave>=40) s[i].grade='D';
if(s[i].ave>=0) s[i].grade='F';
/*Determine if the student passed or failed*/
strcpy(s[i].passfail, "pass");
else strcpy(s[i].passfail, "fail");
sort(s, nos);
for(i=0, i<nos, ++i);
fprintf(grades, "%11s %5i %5i %5i %c %4s \n", s[i].name, s[i].ex1, s[i].ex2, s[i].final, s[i].hw, s[i].grade, s[i].passfail);
/* Sorting Grades */
void sort(record s[], int nos);
int k, j, m, nos, s[50];
struct record hold;
for(k=0; k<=nos-2; k++);
for(j=k+1; j<=nos-1; j++)
if(s[j] < s[m], ave);
And when you compile that, what happens?
C programming resources:
GNU C Function and Macro Index -- glibc reference manual
The C Book -- nice online learner guide
Current ISO draft standard
CCAN -- new CPAN like open source library repository
3 (different) GNU debugger tutorials: #1 -- #2 -- #3
cpwiki -- our wiki on sourceforge
i just got a lot of errors and warnings...
hcp-129-8-135-3% gcc hw12.c -o hw12.out
hw12.c:11: warning: no semicolon at end of struct or union
hw12.c:11: error: syntax error before "passfail"
hw12.c:11: warning: data definition has no type or storage class
hw12.c: In function `main':
hw12.c:15: error: storage size of 's' isn't known
hw12.c:18: error: syntax error before "FILE"
hw12.c:22: warning: parameter names (without types) in function declaration
hw12.c:23: error: `students' undeclared (first use in this function)
hw12.c:23: error: (Each undeclared identifier is reported only once
hw12.c:23: error: for each function it appears in.)
hw12.c:24: error: `grades' undeclared (first use in this function)
hw12.c:29: error: syntax error before "s"
hw12.c:35: error: syntax error before "s"
hw12.c:53: error: syntax error before ')' token
hw12.c:58: error: syntax error before "s"
hw12.c:61: error: storage size of 'hold' isn't known
hw12.c:68: error: `ave' undeclared (first use in this function)
i just got a lot of errors and warnings...
hcp-129-8-135-3% gcc hw12.c -o hw12.out
hw12.c:11: warning: no semicolon at end of struct or union
hw12.c:11: error: syntax error before "passfail"
hw12.c:11: warning: data definition has no type or storage class
hw12.c: In function `main':
hw12.c:15: error: storage size of 's' isn't known
hw12.c:18: error: syntax error before "FILE"
hw12.c:22: warning: parameter names (without types) in function declaration
hw12.c:23: error: `students' undeclared (first use in this function)
hw12.c:23: error: (Each undeclared identifier is reported only once
hw12.c:23: error: for each function it appears in.)
hw12.c:24: error: `grades' undeclared (first use in this function)
hw12.c:29: error: syntax error before "s"
hw12.c:35: error: syntax error before "s"
hw12.c:53: error: syntax error before ')' token
hw12.c:58: error: syntax error before "s"
hw12.c:61: error: storage size of 'hold' isn't known
hw12.c:68: error: `ave' undeclared (first use in this function)
And do you intend the fix the errors? Start at the top: no semicolon at end of struct is pretty straightforward, as is syntax error before passfail (I even mentioned it in my comments the first
time). Fixing these will help with some of the later ones.
Imagine that. I would say you should read this post. Carefully.
C programming resources:
GNU C Function and Macro Index -- glibc reference manual
The C Book -- nice online learner guide
Current ISO draft standard
CCAN -- new CPAN like open source library repository
3 (different) GNU debugger tutorials: #1 -- #2 -- #3
cpwiki -- our wiki on sourceforge
alright i have gotten all the errors to leave except for two and five warnings. I am unsure how to fix these.
hw12.c:20: warning: parameter names (without types) in function declaration
hw12.c:27: warning: passing arg 1 of `fscanf' from incompatible pointer type
hw12.c:27: warning: passing arg 2 of `fscanf' from incompatible pointer type
hw12.c:47: warning: passing arg 1 of `strcpy' from incompatible pointer type
hw12.c:48: warning: passing arg 1 of `strcpy' from incompatible pointer type
hw12.c:68: error: incompatible types in assignment
hw12.c:70: error: incompatible types in assignment
here is my current code.
/* Homework 12 */
#define FILENAME "students.dat"
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#include <string.h>
#define FILENAME2 "outstu.dat"
struct record {char name[12]; int ex1, ex2, final, hw, ave, grade, passfail[5];} record;
int main(void)
struct record s[50];
int nos=0, i=0;
/* Open file. */
void sort(record);
students=fopen(FILENAME, "r");
grades=fopen(FILENAME2, "w");
/*Read file. */
for(i=0; i<50; i++)
while (fscanf("students, %s %i %i %i %i", &s[i].name, &s[i].ex1, &s[i].ex2, &s[i].final, &s[i].hw)==5)
/*Determine Average for each Student */
if(s[i].ave>=85) s[i].grade='A';
if(s[i].ave>=70) s[i].grade='B';
if(s[i].ave>=55) s[i].grade='C';
if(s[i].ave>=40) s[i].grade='D';
if(s[i].ave>=0) s[i].grade='F';
/*Determine if the student passed or failed*/
strcpy(s[i].passfail, "pass");
else strcpy(s[i].passfail, "fail");
sort(s, nos);
for(i=0; i<nos; ++i);
fprintf(grades, "%11s %5i %5i %5i %c %4s \n", s[i].name, s[i].ex1, s[i].ex2, s[i].final, s[i].hw, s[i].grade, s[i].pas$
/* Sorting Grades */
void sort(int s[], int nos);
int k, j, m, ave, nos, s[50];
struct record hold;
for(k=0; k<=nos-2; k++);
for(j=k+1; j<=nos-1; j++)
if(s[j] < s[m])
Just delete those two lines, which will at least allow you to compile and witness what a monstrosity this is. It won't matter at all, since the flow of execution leading up to lines 68 & 70 is
An hour and a half ago I gave you some advice that you have not followed at all. Perhaps you believe this mess is somehow salvageable -- I don't blame you for sticking it out until you are
convinced yourself. It is really just a question of how much longer that will take.
C programming resources:
GNU C Function and Macro Index -- glibc reference manual
The C Book -- nice online learner guide
Current ISO draft standard
CCAN -- new CPAN like open source library repository
3 (different) GNU debugger tutorials: #1 -- #2 -- #3
cpwiki -- our wiki on sourceforge
honestly, i have to finish this class out and i will this semester with the week left. if you want to help me, please do, but who sits on a computer programming message board at 6pm on a saturday
to trash talk a student who is trying to get her assignment done? don't help me if you just want to belittle me. i know i am not good at this...that is precisely why i am on this site trying to
learn from those who know far better than i do.
You've gotten about as far as you can with just blind typing. From here you need to (1) read and (2) think. The reading part comes in with the errors -- if you read both the error message, and
the line 12 (or 20, etc.) that it refers to, then you will know what the error is.
Also, you need to read the program you have written carefully -- because what you want it to do, and what it is going to do are two wildly different things. It would appear to me that you have
the basic outline correct, but the implementation needs work. To be honest, I agree with MK here: the best thing you can do with this code (not the idea, but the code) is to burn it. Start again,
and take more care with the implementation.
honestly, i have to finish this class out and i will this semester with the week left. if you want to help me, please do, but who sits on a computer programming message board at 6pm on a saturday
to trash talk a student who is trying to get her assignment done? don't help me if you just want to belittle me. i know i am not good at this...that is precisely why i am on this site trying to
learn from those who know far better than i do.
I have felt this way before**. I know it is hard to take criticism like that*. But I am not going to lie to you, either. It is also hard to admit to yourself when you have wasted your time and
have to scrap something and start again. I've been there too. Programming is fun when it goes well -- when it doesn't, it's just frustration and staring at a stupid monitor. And no one will feel
sorry for you in your failed attempts to be a programmer (there are much worse tragedies in the world). It is very difficult at first.
If you want to "to learn from those who know far better", then do what I told you before: write a simple program to read the data into a struct array, and print the data in the array out to make
sure you have done it correctly. Then proceed from there. This is not expressionist painting; there really is only two catagories here: do it right, or do it wrong. There are an infinite number
of ways to do it wrong. There are probably a very limited number of ways to do it right.
*it may be harder to take it from your prof, so consider yourself lucky.
**I think that was tabstop's fault
Last edited by MK27; 05-16-2009 at 07:21 PM.
C programming resources:
GNU C Function and Macro Index -- glibc reference manual
The C Book -- nice online learner guide
Current ISO draft standard
CCAN -- new CPAN like open source library repository
3 (different) GNU debugger tutorials: #1 -- #2 -- #3
cpwiki -- our wiki on sourceforge
Well to the original poster. I think your trying to think to big. You should be breaking your problems down to where any one thing you have to do is a small function.
edit: I notice you've also posted your homework 14 in another thread. Also it should be mentioned that your professor/teacher/mentor told you specificaly to use a function sub-program for your
sort. This mean I think you were sleeping in class to much.
AND BTW:
Giving up a saturday night is nothing if you want to learn programming. Sun, Mon, Tues, Wed, Thurs, Friday nights should all be set aside until you reach a decent level of programming profiecency
or just give up now.
Did you think you were special and could learn this material in your drooling-zombie-mode when you signed-up for the class?
I would say it does get easier, But at the same time it gets easier it also requires even more time. Things only get bigger and require more time as they get more complex. You got into
programming for what reason exactly?
And on a honest note. I took a C programming class and before week two 25 % of the students dropped it. By the end of the class less than 33% were still there and even fewer actually passed it at
all. Thats truely staggering. I guess it was to much for them to give up a saturday night. The professor said it was normal and to reason that they pack the class so full to start. At the
beginning we had to get extra chairs to suit all the students. Anyone else here have stuff like this in there programming classes?(if you actually had to take them)
Last edited by strickyc; 05-16-2009 at 08:11 PM.
Read this A development process
Then start again with an empty file which you build up SLOWLY with lots of compiling and testing as you go.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
I support http://www.ukip.org/ as the first necessary step to a free Europe.
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Integral test problem
April 22nd 2007, 04:18 PM #1
Oct 2006
Integral test problem
This problem was confusing me so any help would be greatly appreciated!
For the following series, use the Integral Test to determine whether it converges or diverges:
sigma starting at j=2 and going to infinity of 5 divided by j*ln(j)
April 22nd 2007, 04:44 PM #2
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