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Electronics Tools
Mobile friendly electronics tools and reference materials.
## Electronics ToolBox
34 Tools
Welcome to our electronics toolbox! In this toolbox, we have many tools to help hobbyist and professionals with their electronics calculations. Solve typical circuit configurations with our parallel and series equivalence calculators, voltage divider calculator, and delta-wye calculator or find the Q points and resonant frequencies with our RLC calculator, and finally, take advantage of our free, powerful, and easy-to-use scientific calculator.
In addition to these tools, we have unit converters for a wide variety of electronics and non-electronics units, from energy and power unit conversion to volume and speed unit converters. Are there a few converters that you would like to use frequently? Login and highlight them to save them to your profile for easy access.
Finally, we’re continuously adding new, ready-to-print reference materials as we work through different topics while creating tutorials. These references let you get quick access to important information in those topics in a clean and clear way.
We’re constantly trying to come up with the best calculators to help people of all skill levels and make them both desktop and mobile friendly. If anything is broken or you have any ideas for tools that we need to make our collection complete, reach out to us and let us know!
## Featured Electronics Calculators
Parallel Equivalence
Calculating the parallel equivalence of a resistor, especially when there are more than two of them, is certainly more complicated than finding the series equivalence. However, you can use the formula above to do it yourself and then check to make sure you did it correctly with this convenient calculator!
Voltage Divider
A simple, dirty way to get a lower voltage is to use a voltage divider - putting two (or more!) resistors in series and tapping the voltage between them. If you know the value of the resistors and voltage input, you can calculate the output voltage. If the resistors are the same value, then you’ll get half the input voltage. Otherwise, the math is going to be slightly more complicated but you won’t have any issues - just use the equation above and then use this fancy tool to double-check your work! Or just use the tool, we’re cool with that, too.
Delta-Wye Calculator
Wye, oh wye, would we need a calculator like this? Well, if you do much with three phase power, it could be very helpful to transform a wye circuit into a delta (or pi) circuit or vice versa. As you can see from the equations above, these calculations aren’t necessarily hard but they are tedious. By using this calculator, it should make things much faster for you!
Get the latest tools and tutorials, fresh from the toaster.
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# decimal point
Also found in: Dictionary, Thesaurus, Wikipedia.
## decimal point
a full stop or a raised full stop placed between the integral and fractional parts of a number in the decimal system
## decimal point
[′des·məl ‚pȯint]
(mathematics)
A dot written either on or slightly above the line; used to mark the point at which place values change from positive to negative powers of 10 in the decimal number system.
## decimal point
(character)
"." ASCII character 46. Common names are: point; dot; ITU-T, USA: period; ITU-T: decimal point. Rare: radix point; UK: full stop; INTERCAL: spot.
References in periodicals archive ?
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The traders, from an American bank, were trying to make a last-minute deal when they put a decimal point in the wrong place.
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8 inch) character-height seven-segment displays with decimal point, offering viewing distances up to 10 meters (33 feet) and a 50-degree viewing angle.
The embarrassing blunder was caused by a misplaced decimal point in a bank loan application.
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The morphine was drawn up by junior doctor Hilary Evans who put a decimal point in the wrong place.
In cases where product purities end with digits other than nine, the number(s) appear to the right of the decimal point (e.
Refractive index for typical CVD films, such as nitride, oxynitride, PSG, BPSG, and TEOS are measured with four decimal point precision.
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However, if you want to explicitly specify the increment, you can write: range (3,10,2) Here, the third argument considers the range from 3-10 while incrementing numbers by 2. Numeric Ranges This kind of for loop is a simplification of the previous kind. IS2240 Python – Lab 7 For Loops Intended Learning Outcomes Upon completion of this tutorial, students will be able to: 1. For in loops. Python for loop examples For Loops. This is one of the tricky and most popular examples. In the body, you need to add Python logic. Python For Loop With List. Let’s understand the usage of for loop with examples on different sequences including the list, dictionary, string, and set. The second method to iterate through the list in python is using the while loop. To make it much clear, I am going to explain the concept of Python For Loops in detail with some examples, they are. Ways to increment Iterator from inside the For loop in Python. If the loop-control statement is true, Python interpreter will start the executions of the loop body statement(s). 22, Apr 20. 10, Dec 20. Let us take a look at the Python for loop example for better understanding. Note: While Loop in python works same as while loop in C/C++. while (loop-control statement): #loop body statement(s) How to perform decrement in while loop in Python. Maybe not as easy as Python, but certainly much better than learning C. Neal Hughes. To increment or decrement a variable in python we can simply reassign it. Il existe for et while les opérateurs de boucle en Python, dans cette leçon , nous couvrons for. Statement n. If you observe the above Python for loop syntax, Object may be anything you want to iterate. Perform traditional for loops in Python 2. It's a counting or enumerating loop. # python for9.py john raj lisa for loop condition failed! Eg, for i in (2, 3, 5, 7, 11): print(i) i = 10 * i print(i) output. Specifying the increment in for-loops in Python. But, the next example will clarify bit more on what is the advantage of “else” inside for-loop. Now let’s talk about loops in Python. It's a counting or enumerating loop. It’s like the print() function in the sense that it’s provided by default. Now, let us understand about Python increment operator using an example.. In case the start index is not given, the index is considered as 0, and it will increment the value by 1 till the stop index. For loops are used for sequential traversal. In this tutorial, let’s look at for loop of decrementing index in Python. In Python, there is not C like syntax for(i=0; i This for loop is useful to create a definite-loop. In Python we can have an optional ‘else’ block associated with the loop. Example of a for loop. For loops. After the value incremented it will again check the condition. Python program to Increment Suffix Number in String. Code language: Python (python) In this syntax, the index is called a loop counter. Starting with a start value and counting up to an end value, like for i = 1 to 100 Python doesn't use this either. For example factorial of 4 is 24 (1 x 2 x 3 x 4). The body of the for loop, like the body of the Python while loop, is indented from the rest of the code in the program.. Go for this in-depth job-oriented Python Training in Hyderabad now!. It works like this: for x in list : do this.. do this.. Specifying the increment in for-loops in Python. Python does not provide multiple ways to do the same thing . A for loop in python is used to iterate over elements of a sequence. 3 hours ago Last Lesson Recap In our previous lab: 1. Vectorized for loops Python For Loop Examples. So, the “++” and “–” symbols do not exist in Python.. Python increment operator. Any such set could be iterated using the Python For Loop. Python For Loop Syntax. As depicted by the flowchart, the loop will continue to execute until the last item in the sequence is reached. First we’ll look at two slightly more familiar looping methods and then we’ll look at the idiomatic way to loop in Python. When do I use for loops? 3 hours ago How to prompt for user input and read command-line arguments? For instance String or Python Lists so on. Last Updated: June 1, 2020. It is mostly used when a code has to be repeated ‘n’ number of times. a += 1. to decrement a value, use− a -= 1 Example >>> a = 0 >>> >>> #Increment >>> a +=1 >>> >>> #Decrement >>> a -= 1 >>> >>> #value of a >>> a 0. Usage in Python. Perhaps this seems like a lot of unnecessary monkey business, but the benefit is substantial. What if you want to decrement the index.This can be done by using “range” function. In the following code I replaced your infinite while loop with a small for loop. After that, we need to use an Arithmetic Operator/Counter to increment or decrement it’s value. It might sound like, we might not really need a “else” inside “for” if it only gets executed at the end of for loop iteration. Perform Python-specific for loops 1. This is less like the for keyword in other programming languages, and works more like an iterator method as found in other object-orientated programming languages.. With the for loop we can execute a set of statements, once for each item in a list, tuple, set etc. An example of this kind of loop is the for-loop of the programming language C: for (i=0; i <= n; i++) This kind of for loop is not implemented in Python! To start, here is the structure of a while loop in Python: while condition is true: perform an action In the next section, you’ll see how to apply this structure in practice. Don't let the Lockdown slow you Down - Enroll Now and Get 3 Course at 25,000/-Only. Python | Increment value in dictionary. Now, you are ready to get started learning for loops in Python. Like other programming languages, for loops in Python are a little different in the sense that they work more like an iterator and less like a for keyword. The name of the loop counter doesn’t have to be index, you can use whatever you want.. 25, Sep 20. Python; AWS; … Python For Loops. For loop with else block. The for statement in Python is a bit different from what you usually use in other programming languages.. Rather than iterating over a numeric progression, Python’s for statement iterates over the items of any iterable (list, tuple, dictionary, set, or string).The items are iterated in the order that they appear in the iterable. Python For Loop On List. 22, Apr 20. 3 hours ago Creating an empty Pandas DataFrame, then filling it? Python does not have unary increment/decrement operator( ++/--). There is “for in” loop which is similar to for each loop in other languages. for x in sequence: statements Here the sequence may be a string or list or tuple or set or dictionary or range. In Python, there is no C style for loop, i.e., for (i=0; i
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# Linear Integrated Circuit Questions and Answers – Closed-Loop Frequency Response and Circuit Stability
This set of Linear Integrated Circuit Multiple Choice Questions & Answers (MCQs) focuses on “Closed-Loop Frequency Response and Circuit Stability”.
1. Open loop configuration is not preferred in op-amps because
a) First break frequency is too large
b) First break frequency is very small
c) Second break frequency is too large
d) All of the mentioned
Explanation: The bandwidth of the op-amp is simply the first break frequency. As the value of bandwidth is very small, the open loop configuration is of little use or not preferred in op-amp.
2. Calculate the value of open loop frequency response curve at any point beyond break frequency in 741C op-amp?
a) 1000000Hz
b) 1000Hz
c) 10000000Hz
d) 100000Hz
Explanation: For a 741C op-amp the gain=200000 and the break frequency = 5Hz. Therefore, the unity gain bandwidth or the product of the coordinates (gain and frequency) of any point beyond the break frequency = 200000×5Hz =1MHz =1000000Hz.
3.
Frequency response of µA709C for various closed loop gain is shown. Choose the curve that has wider bandwidth and high gain?
a) Curve 4
b) Curve 3
c) Curve 2
d) Curve 1
Explanation: To get a high gain and relatively wider bandwidth, the compensating components for curve 1 should be used. Curve 1 has the gain equal to 60dB and bandwidth nearly equivalent to 10MHz.
4. When does a system said to be stable?
a) Output reaches a minimum value at finite time
b) Output reaches a maximum value at any time
c) Output reaches a fixed value at finite time
d) Output reaches a fixed value at any time
Explanation: A circuit or a group of circuits connected together as a system is said to be stable, if its output reaches a fixed value in a finite time.
5. Why unstable systems are considered to be impractical?
a) None of the mentioned
b) Output decreases with time
c) Output reaches fixed value
d) Output increases with time
Explanation: A circuit is said to be unstable if its output increases with time instead of achieving fixed value. In fact the output keeps on increasing until the system breaks down. Therefore, unstable systems are impractical and need to be made stable.
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6. How is the criterion for the system determined?
a) Graphical method
b) Theoretical method
c) Analytical method
d) All of the mentioned
Explanation: The criterion for stability of system is tested practically using one of the above mentioned methods.
7. A standard block diagram of closed loop system composes of
a) Two blocks
b) Single block
c) Three blocks
d) None of the mentioned
Explanation: In closed loop system (control system) the standard form of representing a system is composed of two blocks.
(i) Forward block and
(ii) Feedback block.
8. “Transfer function” in control system refers to
a) Feedback block
b) Content of each block
c) Forward block
d) Input and output blocks
Explanation: The transfer function in control system refers the content of each block and it depends on the complexity of a system.
9. Which method is considered to be a graphical method in testing the system?
a) Bode plot
b) Routh-Hurwitz criteria
c) Circuit testing
d) None of the mentioned
Explanation: A graphical method used in determining the stability / testing the system is the bode plots. It is composed of magnitude versus frequency and phase angle versus frequency plots.
10. Measure taken to increase the bandwidth of an op-amp.
a) Increase the frequency for the configuration
b) Reduce the gain of the configuration
c) Closed loop configuration is used
d) Open loop configuration is used
Explanation: Closed loop configuration is preferred to increase the bandwidth of an op-amp.
11. Name the block connected in the feedback path
a) Feedback block
b) Forward block
c) Output block
d) Summation of feedback, forward and output block
Explanation: The block in the feedback path is referred to as the feedback blocks, which connects the block between the output signal and the feedback signal.
12. When the magnitude of (AoL )×( ß ) = 0dB, state the condition at which system become stable?
a) Phase angle is > -180o
b) Phase angle is < -180o
c) Phase angle is > +180o
d) Phase angle is < +180o
Explanation: Closed loop voltage gain AF = AoL/ (1+AoL×ß)
At lower frequency the contribution of AoL is zero. So, AoL×ß >0 and obviously AF < AoL and system is stable.
13. Mention the phase condition that leads to sustained oscillation
a) ∠-Aß =0
b) All of the mentioned
c) ∠-Aß = multiple of 2π
d) ∠Aß = odd multiple of π
Explanation: The circuit leads to oscillation if the characteristic equation (1+AoL×ß )=0
=> Loop gain, AoL×ß =1
Since AoL*ß is a complex quantity , the magnitude is |AoL×ß |=1 and the phase condition are ∠-Aß=0 (or multiple of 2π) or ∠Aß= π (or odd multiple of π).
14. Determine the state of the system
a) Stable system
b) Unstable system
c) Casual system
d) Bounded system
Explanation: The given circuit is used in inverting mode and produce a phase shift of 180o. For two corner frequency, the maximum phase shift associated with gain AoL is -180o which makes total phase shift = 0 and for some value of ß, the magnitude of Aß becomes unity. In this case the amplifier begins to oscillate as both magnitude and phase conditions are satisfied. Thus, the circuit enter into verge of instability.
15. Find out the system stability when a system has three RC poles pairs
a) Attain stability at low frequency
b) Attain stability at high frequency
c) Attain instability at high frequency
d) Attain instability at low frequency
Explanation: For a system with three pole pairs, AoL have three corner frequencies and contribute to maximum of 270o phase shift, at this condition AoL×ß is negative and instability occurs at high frequencies.
Sanfoundry Global Education & Learning Series – Linear Integrated Circuits.
To practice all areas of Linear Integrated Circuits, here is complete set of 1000+ Multiple Choice Questions and Answers.
If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]
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# testing array of logicals
P: n/a Hi list, Is there a more elagant way of doing this? # logflags is an array of logicals test=True for x in logflags: test = test and x print test -- Thanks, Jul 12 '06 #1
17 Replies
P: n/a John Henry wrote: Is there a more elagant way of doing this? # logflags is an array of logicals test=True for x in logflags: test = test and x print test Py2.5: test = all( logflags ) Py2.4 (although somewhat ugly): try: test = itertools.ifilterfalse( logflags ).next() except StopIteration: test = True otherwise: your above code will do just fine. Note that you can shortcut, though, if any of the flags evaluates to False: test = True for x in logflags: if not x: test = False break Stefan Jul 12 '06 #2
P: n/a John Henry wrote: Is there a more elagant way of doing this? # logflags is an array of logicals test=True for x in logflags: test = test and x print test your code checks all members, even if the first one's false. that's not very elegant. here's a better way to do it: def all(S): for x in S: if not x: return False return True print all(logfiles) if you upgrade to 2.5, you can get rid of the function definition; "all" is a built-in in 2.5. also see: http://www.artima.com/weblogs/viewpost.jsp?thread=98196 Jul 12 '06 #3
P: n/a John Henry wrote: Hi list, Is there a more elagant way of doing this? # logflags is an array of logicals test=True for x in logflags: test = test and x print test There's reduce, but it's not as explicit, and see F's post RE efficiency: >>x = [True, True, True]y = [True, False, True]print reduce(lambda a, b: a and b, x) True >>print reduce(lambda a, b: a and b, y) False >>> Jul 12 '06 #4
P: n/a John Henry wrote: Hi list, Is there a more elagant way of doing this? # logflags is an array of logicals test=True for x in logflags: test = test and x print test -- Thanks, The builtin "reduce" does that kind of thing for any function you wish to apply across the list. So then its only a matter of giving it a function that "and"s two arguments: Either: reduce(lambda a,b: a and b, logFlags) or def and2(a,b): return a and b reduce(and2, logFlags) Gary Herron Jul 12 '06 #5
P: n/a John Henry
P: n/a "John Henry"
P: n/a On 12 Jul 2006 11:14:43 -0700, John Henry
P: n/a Simon Brunning wrote: On 12 Jul 2006 11:14:43 -0700, John Henry
P: n/a On 13 Jul 2006 05:45:21 -0700, John Henry Simon Brunning wrote: min(logflags) !!! Be aware that not only is this an outrageous misuse of min(), it's also almost certainly much less efficient than /F's suggestion, 'cos it always iterates through the entire list. -- Cheers, Simon B, si***@brunningonline.net, http://www.brunningonline.net/simon/blog/ Jul 13 '06 #10
P: n/a Simon Brunning a écrit : On 13 Jul 2006 05:45:21 -0700, John Henry >Simon Brunning wrote: > min(logflags) !!! Be aware that not only is this an outrageous misuse of min(), +1 QOTW Ho, my, I've already proposed another one today :( Jul 13 '06 #11
P: n/a John Henry wrote: Hi list, Is there a more elagant way of doing this? # logflags is an array of logicals test=True for x in logflags: test = test and x print test -- Thanks, So many ways.... *drool* How about: False not in logflags (Anybody gonna run all these through timeit? ;P ) Jul 14 '06 #12
P: n/a > False not in logflags Or, if your values aren't already bools False not in (bool(n) for n in logflags) Peace, ~Simon Jul 14 '06 #13
P: n/a Simon Forman wrote: False not in logflags Or, if your values aren't already bools False not in (bool(n) for n in logflags) Peace, ~Simon Very intriguing use of "not in"... Thanks, Jul 14 '06 #14
P: n/a John Henry wrote: Simon Forman wrote: > False not in logflags > Or, if your values aren't already bools False not in (bool(n) for n in logflags) Very intriguing use of "not in"... Is there a reason why you didn't write True in (bool(n) for n in logflags) Aug 5 '06 #15
P: n/a Janto Dreijer wrote: John Henry wrote: Simon Forman wrote: False not in logflags > Or, if your values aren't already bools > False not in (bool(n) for n in logflags) Very intriguing use of "not in"... Is there a reason why you didn't write True in (bool(n) for n in logflags)
P: n/a Janto Dreijer wrote: Janto Dreijer wrote: John Henry wrote: Simon Forman wrote: > False not in logflags > Or, if your values aren't already bools False not in (bool(n) for n in logflags) > Very intriguing use of "not in"... Is there a reason why you didn't write True in (bool(n) for n in logflags)
P: n/a "Janto Dreijer" | Or, if your values aren't already bools | > | False not in (bool(n) for n in logflags) | | Very intriguing use of "not in"... | > | Is there a reason why you didn't write | True in (bool(n) for n in logflags) | | From once every six days or so if you are no good, to once in a lifetime if you are brilliant, and never only if you are a genius... First time it bit me I was an apprentice writing in Cobol. - Hendrik Aug 7 '06 #18
### This discussion thread is closed
Replies have been disabled for this discussion.
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# choclo.prism.magnetic_field#
choclo.prism.magnetic_field(easting, northing, upward, prism, magnetization)[source]#
Magnetic field due to a rectangular prism
Returns the three components of the magnetic field due to a single rectangular prism on a single computation point.
Note
Use this function when all the three component of the magnetic fields are needed. Running this function is faster than computing each component separately. Use one of magnetic_e, magnetic_n, magnetic_u if you need only one of them.
Parameters
• easting (float) – Easting coordinate of the observation point. Must be in meters.
• northing (float) – Northing coordinate of the observation point. Must be in meters.
• upward (float) – Upward coordinate of the observation point. Must be in meters.
• prism (1d-array) – One dimensional array containing the coordinates of the prism in the following order: west, east, south, north, bottom, top in a Cartesian coordinate system. All coordinates should be in meters.
• magnetization (1d-array) – Magnetization vector of the prism. It should have three components in the following order: magnetization_easting, magnetization_northing, magnetization_upward. Should be in $$A m^{-1}$$.
Returns
b_e, b_n, b_u (floats) – The three components of the magnetic field generated by the prism on the observation point in $$\text{T}$$. The components are returned in the following order: b_e, b_n, b_u.
Notes
Consider an observation point $$\mathbf{p}$$ and a prism $$R$$ with a magnetization vector $$\mathbf{M}$$. The magnetic field $$\mathbf{B}$$ it generates on the observation point $$\mathbf{p}$$ is defined as:
$\mathbf{B}(\mathbf{p}) = - \frac{\mu_0}{4\pi} \nabla_\mathbf{p} \left[ \int\limits_R \mathbf{M} \cdot \nabla_\mathbf{q} \left( \frac{1}{\lVert \mathbf{p} - \mathbf{q} \rVert} \right) dv \right]$
Since the magnetization vector is constant inside the boundaries of the prism, we can write the easting component of $$\mathbf{B}$$ as:
$B_x(\mathbf{p}) = - \frac{\mu_0}{4\pi} \left[ M_x \int\limits_R \frac{\partial}{\partial x_p} \left[ \frac{\partial}{\partial x_q} \left( \frac{1}{\lVert \mathbf{p} - \mathbf{q} \rVert} \right) \right] dv + M_y \int\limits_R \frac{\partial}{\partial x_p} \left[ \frac{\partial}{\partial y_q} \left( \frac{1}{\lVert \mathbf{p} - \mathbf{q} \rVert} \right) \right] dv + M_z \int\limits_R \frac{\partial}{\partial x_p} \left[ \frac{\partial}{\partial z_q} \left( \frac{1}{\lVert \mathbf{p} - \mathbf{q} \rVert} \right) \right] dv \right]$
where $$M_x$$, $$M_y$$ and $$M_z$$ are the components of the magnetization vector. The other components can be expressed in an analogous way.
It can be proved that
$\frac{\partial}{\partial x_q} \left( \frac{1}{\lVert \mathbf{p} - \mathbf{q} \rVert} \right) = - \frac{\partial}{\partial x_p} \left( \frac{1}{\lVert \mathbf{p} - \mathbf{q} \rVert} \right)$
and that it also holds for the two other directions. Therefore, we can rewrite $$B_x$$ as:
$B_x(\mathbf{p}) = + \frac{\mu_0}{4\pi} \left[ M_x \frac{\partial^2}{\partial x_p^2} \int\limits_R \frac{1}{\lVert \mathbf{p} - \mathbf{q} \rVert} dv + M_y \frac{\partial^2}{\partial x_p \partial y_p} \int\limits_R \frac{1}{\lVert \mathbf{p} - \mathbf{q} \rVert} dv + M_z \frac{\partial^2}{\partial x_p \partial z_p} \int\limits_R \frac{1}{\lVert \mathbf{p} - \mathbf{q} \rVert} dv \right]$
Solutions to each one of the integrals in the previous equation and their second derivatives are given by [Nagy2000].
Following [Oliveira2015] we can define a symmetrical 3x3 matrix $$\mathbf{U}$$ whose elements are the second derivatives of the previous integrals, such as:
$u_{ij} = \frac{\partial}{\partial i} \frac{\partial}{\partial j} \int\limits_R \frac{1}{\lVert \mathbf{p} - \mathbf{q} \rVert} dv$
with $$i, j \in \{x, y z\}$$.
We can then express the magnetic field $$\mathbf{B}(\mathbf{p})$$ generated by the prism in a compact form:
$\mathbf{B}(\mathbf{p}) = \frac{\mu_0}{4\pi} \mathbf{U} \cdot \mathbf{M}$
References
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The Noble Art of De-Coupling
Carl G. Lodström, EE. SM6MOM & KQ6AX
Reprinted with the permission of "VHF Communications" magazine
See also "The Log Probe Logarithmic Detector"
______________________________
1.
Introduction
______________________________
Much of the electronics built today will require de-coupling of various points for various frequencies. What first comes to mind may be the V+ of IC's. If the IC is an aggressive one, providing fast transitions with a lot of drive capability, the problem is no longer a simple one.
Nevertheless it is treated as if it were! "Ah, let's toss in a 10000 'puff' there! Eh, wait a moment, better make it a 1000 puff and a 'point-one'!" That is: a 1 nF and a 0.1µF capacitor.
For one thing, 1 nF or 0.1µF will make no difference (to RFI) above 500 MHz and almost none at 200 MHz! The capacitors are already well above their self-resonance frequencies.
The readers of VHF-Com do not have to be reminded about what will happen as clock frequencies are reaching the GHz mark! Even if the signals are not perfectly square, it is reasonable to assume that they are not sinusoidal either. Rise times may be on the order of 100 ~ 200 ps. It is thus proper to consider frequencies up to at least 2 GHz. (As things are going, do you too have the feeling that someone will laugh at this in a few years?!)
______________________________
2.
Elements
______________________________
My measurements have shown a 1206 capacitor to have ~ 1.8 nH and a 0805 seems to have ~ 1.5 nH. In some other applications it seemed like the 0805 had ~ 1 nH. Right or wrong, we can probably agree that it will have at least 1 nH and I will use it in the modeling. The Q of the inductance is probably not stellar, let me guess at 30. The Q of the capacitance may well be 50.
The source impedance of the transient generator is probably low. Let us use 10 Ω for this and 100 Ω for the other end, the "load", going to the power supply. We will treat the circuit like a filter, considering S11, S21, and S22 over frequencies to 3 GHz.
_______________________________
3.
Models
_______________________________
For the modeling I have used the Eagleware =Superstar= program, version 5.2
The first model we may want to take a look at is the one with a single de-coupling capacitor on the V+ pin of the IC. It is imagined here as "IN>"
The model, showing a piece of 40 mils (1 mm) wide trace to the capacitor, another trace to the 100 Ω load. Their lengths are mostly inconsequential. The capacitor is pictured with its inductance and it is grounded through a via hole of 24 mils dia. with 1.5 mils metal. The substrate is 32 mils Rogers 4003, with er=3.38 nominally, but various substrates would not make any great difference here.
_______________________________
4.
Simulations
_______________________________
On the left diagram below are S11 and S22 with 0 ~ 1 dB vertically. On the right is S21 on a 0 ~ 100 dB scale, showing the efficiency of the de-coupling. Four marker frequencies are set under each graph and their "dB" results can be read along the bottom.
One set of graphs is dashed and one set is solid. They represent a capacitance of 100 pF and 10 nF respective. Sure, there is a nice "suck-out" moving from ~500 MHz to ~30 MHz, but the rest is nothing to write home about. About 20 dB attenuation appears to be the rule, and it is not much of a de-coupling on an intense source of RF.
It appears likely that several stages, compounded, of de-coupling may do better. Let us look at a two stage solution and let =SuperStar= try to optimize it for -60 dB from 100 ~ 1000 MHz and -40 dB from 50 ~ 2000!
Well, now we are getting somewhere! The attenuation is below 30 dB up to 1420 MHz! Two "suck-outs" are visible, corresponding to the self-resonance of the capacitors. The program decided that they better be 587 and 3418 pF respective, with 802 mils from the left and a 645 mils long lines in-between. The third line, at the right end, to V+, is best when short.
I expanded the circuit to three capacitors, with a 0.5 mm line before each, ran =SuperStar= on it again for the following configuration:
Indeed, this net shows a very good performance with 50dB from 150 ~ 400 MHz and 40 dB from 80 ~ 2212 MHz. Notice that the max frequency is changed from 2 to 3 GHz!
On the S11 numbers can we see how pin 8 on the IC finds a near total reflex (<0.1 dB Return Loss) on almost all frequencies although the phase rotates via inductive to near open.
One possible practical realization of this net with a Small Outline 8-pin IC and 0805 components.
_______________________________
5.
Conclusions
_______________________________
Use of self-resonant capacitors for de-coupling has been known for most of the radio era. I have a book (that I cannot find just now) from the 1930's. In it is a graph of the "ideal length of de-coupling capacitors vs. frequency". From the values, clearly some 5 ~ 10 nH per cm was anticipated. I am not so sure it was considered in old IF strips and other circuits, but some designers surely knew about it.
As frequencies move up in the VHF and UHF range, so do the number of capacitor values one do no longer have to consider using! The 1 nF capacitor in a 1206 surface mount package will become self resonant at 130 MHz. In a leaded package it will resonate at a much lower frequency. 1 nF can possibly be used at 146 MHz if it is a disk that is soldered direct between ground and the point to be de-coupled.
As the frequencies go up the problem becomes more difficult. At 1296 MHz the 10 pF capacitor nears resonance!
This article may not have much of a solution, if there is one, but should at least have brought attention to the phenomena, and got the reader thinking.
It is possible that the problem can be divided up. Since fairly good results was reached down to ~100 MHz, shielded compartments and separate de-coupling of the "lower" frequencies may well be possible. Since it will take up a lot of real estate, it has to be planned for early in the design.
A novel and interesting treatise of the inductor, and it's self-resonance(s) by Randy Rhea (Founder of Eagleware) is recommended reading.
His findings and conclusions are profound. The articles can be down-loaded on PDF format. Go to:
http://www.amwireless.com/AMWhome.html and select 1997, the Nov-Dec issue from the Archive! The title of the article is: "A Multimode High-Frequency Inductor Model"
For the second article, select the November 2000 issue! The title of this article is: "Filters and an Oscillator Using a New Solenoid Model".
Carl G. Lodström has been a Consultant for many years in the San Diego area.
About RF Cafe Copyright: 1996 - 2024Webmaster: Kirt Blattenberger, BSEE - KB3UON RF Cafe began life in 1996 as "RF Tools" in an AOL screen name web space totaling 2 MB. Its primary purpose was to provide me with ready access to commonly needed formulas and reference material while performing my work as an RF system and circuit design engineer. The Internet was still largely an unknown entity at the time and not much was available in the form of WYSIWYG ... All trademarks, copyrights, patents, and other rights of ownership to images and text used on the RF Cafe website are hereby acknowledged. My Hobby Website: AirplanesAndRockets.com
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# Test Statistics for Hypothesis Testing in Statistics with Functions in R Programming
In my previous article, we talked about the terminology in the statistical world of hypothesis testing.
Moving on, once we have our hypothesis about the population parameters (based on the sample) in place, we need some sophisticated ways to practically support one claim (either H0 or H1) by using statistical inference, Based on the type of statistics, i.e., population mean, population variance, population correlations, testing for normality or testing for attributes, we have separate test statistics for each hypothesis testing.
Lets go over some of the commonly used test statistics in hypothesis testing and their R-programming codes:
1. Testing for the significance of the population mean (based on the sample or empirical observation)
Case (i): When population variance is known (generally not the case)
The test statistic
Here, we use the sample mean, the population mean as the estimate and the population variance (as given) to find the value of the test statistic.
R code: For a two tailed test for population mean
z.test (observations of sample X, alternative=”two.sided” ,mu=mu0, sigma.x=sigma)
Case (ii): When population variance is unknown (usual scenario)
This calls for a t-test, since, we use the estimate of the population variance, the sample variance, given by:
Here, we use the sample mean, the population mean as the estimate and the sample variance (as given) to find the value of the test statistic.
R code: For a two tailed test for population mean
t.test (observations of sample X, alternative=”two.sided” ,mu=mu0)
Result: As we know, when the p-value for the t-test for testing if the population average of the Sepal Length is equal to 6 cm is equal to 0.02186. Since, p-value is less than the general significance level (=0.05), we reject H0 and hence, we conclude that, the population average of sepal length for the iris data in R is not equal to 6cm.
2. Testing for the significance of the population variance
here,
There is no inbuilt function to perform this hypothesis testing in R. Therefore, it requires for us to code from scratch.
3. Testing for the independence of Attributes
This testing calls for the use of a chi square distribution. This is used for attributes, not variables, where, we have categories and some classifications (in the form of genders and other characteristics). This testing is used in cases like:
(i) When we would like to test the independence of two attributes, for example, we are given the attentive index of the students in the class based on the duration of the class (in groups of, Short and Lengthy) and the quality of interaction. Now in order to perform a test of independence between the attributes duration of the class and the quality of interaction in the class we use chi square test.
R code: For a two single tailed independence of attributes
chisq.test (matrix of the observed frequencies)
(ii) When we wish to perform the test for goodness of fit. There are times when we would like to know the kind of distribution (a pre determined pattern, with certain specific characteristics) the sample data tends to follow, we then make use of chi square goodness of fit test.
R code: For a two single tailed goodness of fit test
chisq.test (observations of sample X)
Suppose our question is:
Question: Use chi square test to test if the admissions on each day during the week are the same.
Here, we follow the code and obtain
Result: Since the p-value for testing the goodness of fit is equal to 2.2e^-16 <0.05(alpha or level of significance), we reject H0. Hence, we conclude, that the admissions in the hospital during the dates of the week is not equal.
Further topic of Discussion
We would further cover the testing for normality with various tests and forms of plots, like, PP plots and QQ plots.
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+0
what is 3/2 of 50, 000
0
149
2
What is 3/2 of 50, 000
Oct 9, 2022
#1
0
What is 3/2 of 50, 000
1.5 x 50,000 = 75,000
Oct 9, 2022
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Integral of a radial symmetric function.
I've found an article that essentially states that an integral of the form
$$I[f] = \int_{\mathbb{R}^N} f(x_1,\ldots,x_N) dx_1 \ldots dx_N = \int_{\mathbb{R}^N} f(r) dx_1 \ldots dx_N$$
where $$r = \sqrt{x_1^2 + \ldots x_N^2}$$, can always be split as product of two integrals, where the first integral is the surface of the $$n$$ dimensional sphere, say $$A_n$$ and the second integral is the radial integral
$$R[f] = \int_0^1 f(r) r^{N-1}dr$$
The article I've found only states the result, but it doesn't actually prove it, but assuming this results holds it proves that
$$A_N = \frac{2 \pi^{N/2}}{\Gamma\left( \frac{N}{2}\right)}$$
Is there a way to prove the statement which doesn't rely on generalized polar coordinates, which are a bit hard to remember and manipulate, if there's no other way is there an easy way to derive to generalized polar coordinates and again derive the result?
I would assume the proof is a calculus fact.
Since $$f$$ is radially symmetric, $$f(\vec{x})d^N\vec{x}=r^{N-1}f(r)drd\Omega$$, with $$d\Omega$$ an $$(N-1)$$-dimensional infinitesimal element over the angle coordinate(s). We don't need to work out how $$d\Omega$$ looks at all, so don't get out your $$\sin\theta$$s here. The claimed factorisation is immediate.
If you prefer a different proof that an $$(N-1)$$-sphere of radius $$R$$ has measure $$\frac{2\pi^{N/2}}{\Gamma(N/2)}R^{N-1}$$, let's prove instead that an $$N$$-ball of radius $$R$$ has volume $$\int_0^R\frac{2\pi^{N/2}}{\Gamma(N/2)}r^{N-1}dr=\frac{\pi^{N/2}}{\Gamma(N/2+1)}R^N$$. (To clarify, an $$n$$-ball is $$n$$-dimensional, but an $$n$$-sphere is the $$n$$-dimensional surface of an $$(n+1)$$-ball.) Equivalently, the unit $$N$$-ball has measure $$V_N:=\frac{\pi^{N/2}}{\Gamma(N/2+1)}$$. We'll proceed by induction.
The result is correct for $$N=0$$; a "$$0$$-sphere" is the single point in $$0$$-dimensional space, and the definition of measure is counting that point. (If that argument seems to strange, take $$N=1$$ as your base step, for which measure is a line segment's width, or failing that $$N=2$$, for which you just want a circle's area.) By slicing an $$N$$-ball into hypercylinders of infinitesimal thickness with $$(N-1)$$-ball cross sections of radius $$\sqrt{1-x^2}$$,$$V_N=\int_{-1}^1V_{N-1}(1-x^2)^{(N-1)/2}dx=2V_{N-1}\int_0^1(1-x^2)^{(N-1)/2}dx\\=V_{N-1}\int_0^1y^{-1/2}(1-y)^{(N-1)/2}dx=V_{N-1}\operatorname{B}\left(\frac12,\,\frac{N+1}{2}\right).$$(The intuition of this is easiest with slicing a sphere into cylinders, since you can visualise that.) So the inductive step is$$V_{N-1}=\frac{\pi^{(N-1)/2}}{\Gamma((N+1)/2)}\implies V_N=\frac{\pi^{(N-1)/2}}{\Gamma((N+1)/2)}\frac{\sqrt{\pi}\Gamma((N+1)/2)}{\Gamma(N/2)+1},$$which is what we needed.
• @user8469759 Well, that's just the polar form of $d^N\vec{x}$. It needs to be proportional to $d(r^N)$, because an $N$-ball of radius $R$ wouldn't otherwise have measure $\propto R^N$. – J.G. Oct 19 '19 at 18:24
• @user8469759 Yes, like when you write $f(x)dx=f(x)g^\prime(u)du$ if $x=g(u)$. – J.G. Oct 19 '19 at 18:30
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# Simple Machines Worksheet
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Cannot be combined to make compound machines. b. are tools that make work easier. c. reduce the amount of energy needed to do work. d. are made up of multiple moving parts. force motion.Define machine. do machines change the work done what do they change what are the types of simple machines define lever.
### 9. Simple Machines Interactive Notebook Science Doodle Doodles
About this quiz worksheet. use these resources to assess your understanding of compound machines. answer quiz questions on things like identifying simple machines Simple kitchen machines page of developed by as part of www.tryengineering.org simple kitchen machines provided by www.
### 10. Simple Machines Review Activity Activities Flip Book
There are six of them lever wheel axle pulley inclined plane screw teachers resources teaching worksheets, activities, and technology ideas for k correlates to standards of learning first grade second grade third grade fourth grade ideas lessons sol resources for first, second, third, and fourth grade.
### 12. Simple Machine Worksheet Grade 5 Google Search Magic School Bus Bill
When a caveman had to move a rock that was too heavy to lift, he might have used a big stick to make it easier. a long, long time ago, someone invented the wheel, and that made.Displaying top worksheets found for simple machines. some of the worksheets for this concept are name searching for simple machines, simple machines, name simple machines, the simple machines, compound words work a l, name simple machines, rube machines workbook, second grade work simple machines unit.
### 13. Simple Machines Study Guide
Copy attached below individual quiz with a word bank.for special needs. individual quiz with no word bank. reusable quiz with a word bank.for special needs.It is your entirely own time to decree reviewing habit. in the course of guides you could enjoy now is simple machines worksheet answers below.
### 14. Simple Machines Task Cards Community Activities Matter Science
Simple machines are basically classified into two groups, levers and inclined planes. simple machines like the pulley, screw, wheel and axle and wedge come under these two categories. the seesaw in the playground is a simple machine.Click simple machines worksheet.
### 15. Teaching Simple Machines Ideas Unit 6
Mechanical advantage. Enrich your study of simple and compound machines with this physics printable. students will refer to the diagram in this worksheet to answer questions about bicycles. then they will create their own diagram of a compound machine.
### 17. Types Simple Machines Activities Science Notes
Download file work and simple machines worksheet answers worksheet packet simple machines simple machines increase our ability to do work. the basic types of simple machines include the lever, inclined plane, pulley, screw, wedge, wheel page. Student worksheet name(s) section date simple machines activity two the inclined plane and pulley lesson one the inclined plane getting started.
### 18. Types Simple Machines Chart Machine
If you are using this worksheet, your.Worksheet simple machines a simple machine is some examples of machines that can do work but use electricity are the six kinds of simple machines are include name and a sketch for each your teacher has six examples of simple machines classify them here category machine names inclined.
### 19. Vocabulary Terms Simple Machines Lever Wedge Screw Wheel Axle Inclined Interactive Science Notebook Notebooks
Reading comprehension worksheets for middle school a collection of fiction and nonfiction passages written for middle school students in students in,, and grades. these printable article and stories have comprehension questions to check students reading comprehension skills.
### 20. Wheel Axle Worksheet Helps Students Identify Simple Machines Apologia General Science
A lot of our own articles or blog posts include layouts along with examining materials easily accessible in formatting with regard to convenience as well as swift get a hold of terrific for young students, school staff, as well as mom and dad who are usually for the go.
### 22. Simple Machine Worksheet Kids Machines Choice Homework Freebie Interactive Science Notebook Matter
Simple machine activities and worksheets simple machines game simple machines quiz simple machines word scramble game simple machines order game simple machines word search game simple machines word search worksheet simple machines word scramble worksheet force formula spring constant formula simple machines, simple machines and mechanical advantage from simple machines and mechanical advantage worksheet answer key, sourceyumpu.
### 23. Simple Machine Represents Object Shown Resolution Version Machines Eets Kids Activities
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### 24. Bananas Simple Machines Activities Machine Projects
A structure consisting of a framework and various fixed and moving parts, for doing some kind of work.Simple machines worksheet. match the simple machine with its correct definition by writing the corresponding number in the answer column. simple machines.
answer. definitions. lever. something that reduces the friction of moving something. inclined plane. something that can hold things together or lift an object.Compound machines a machine that is made of two or more simple machines is called a compound machine.
### 25. Lever Identification Types Simple Machines Activities Science Worksheets
what type of simple machine is found on a water bottle cap a. lever b. pulley c. wheel and axle d. screw. how is a wedge like an inclined plane how is it different. on which type of simple machine would you find a fulcrum explain what a fulcrum is.
### 26. Bill Motion Worksheet Lovely Science Guy Simple Machines Answers Super Teacher Worksheets Nursing Student Tips
That will remaining explained, all of us supply you with a assortment of simple nevertheless enlightening content plus design templates designed well suited for any.Selection file type icon file name description size revision time user radioactive decay practice.
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### 28. Everyday Simple Machines Activities Machine Projects
Created date a list of the simple machine discussed in the video and examples of specific devices or machines based on these. s i mp l e ma c h i n e e x a mp l e. a ramp or inclined plane can make lifting something or climbing a certain height easier use less force.
but there is a trade off. what is machines sort cut and paste examples and a quick way to assess their learning or do some review this is it versions of this worksheet for students to cut and paste examples and definitions of the different simple machines lever, pulley, wheel and axle, inclined plane, sc.
### 29. Find Simple Machines Worksheet Science Classroom Worksheets
Worksheet to conduct a simple machines scavenger hunt in which students find examples of simple machines used in the classroom and at home. in other lessons of this unit, students study each simple machine in more detail and see how each could be used as a tool to build a pyramid or a modern building.
This is a short eight page coloring booklet about simple machines introducing students to examples of simple machines wheel, screw, pulley, slope inclined plane version included, wedge and lever. it can be used as a color and take home activity for students learning about machines.
### 30. Free Simple Machines Poster Unit
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### 31. Gift Curiosity Unit Ideas Simple Machines Machine Projects
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### 32. Grade 4 Wheels Levers Simple Machines
Simple machines activity packet. here is a useful activity packet that includes different versions of vocabulary wall posters, mini activity book, homework activity sheet, word search, and different versions of activities to use with an interactive science journal when teaching simple.
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### 33. Info Leading Site Net Simple Machines Machine Ts Worksheets Kids
This unit teaches students about the six simple machines lever, inclined plane, wedge, wheel and axle, screw, and pulley.Id language school subject natural science grade age main content machines other contents simple machines add to my workbooks download file embed in my website or blog add to google worksheets www.
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### 34. Lever Simple Machine Combines Stem Writing Integrated Language Arts Science Machines Grade Projects
. this simple machine can be used to lift a weight. it has a fulcrum, or pivot point, which can be located in the center, near the end or at the end. examples of this simple machine are used to hold things together. it is made up of an inclined plane wrapped around a cylinder.
a heavy object could be rolled up this simple machine,Simple machines click the simple machines id language school subject natural science grade age. more machines interactive worksheets. simple machines by simple machines by machines by simple and complex machines by simple machines displaying top worksheets found for this concept.
### 35. Simple Machine Quiz Machines Activities Unit
Mechanical advantage.Unit use simple machines safely lesson focus read through the instruction sentences and give pupils time to complete the worksheet. pupils are to Machines a device that makes work easier. a machine can change the size, the direction, or the distance over which a force acts.
### 36. Lever Worksheets Simple Machines Activities Science
Grade science worksheets free printable fifth grade. Beside that, we also come with more related ideas like community worksheets middle schools, periodic table grade science worksheet and science simple machines worksheets. our goal is that these simple machines worksheet middle school images collection can be useful for you, give you more ideas and most important present you an awesome day.
Physical science graders, be the leaders i know you can be physical science is broken into main units astronomy, chemistry and physics. the best advice i can give to you is to budget your time properly, wait until the last minute the night before to get your work done.
### 38. Note Video Needed Completion Access Prior Science Guy Bill Simple Machines
A simple machine is a mechanical device used to apply increased force. there are six simple machines lever, inclined plane, wedge, pulley, wheel and axle, and screw. simple machines are the building blocks for creating more complex machines. for instance, a bicycle is created using wheels, levers,Simple machine explain that a simple machine is a device that makes work easier.
wedge a simple machine is a kind of inclined plane where the pointed edges are used to do work. wheel and axle a simple machine that is a kind of inclined plane that moves objects distances. the axle is a rod that goes through the wheel.Simple machines are simple tools used to make work easier.
### 40. Pulleys Worksheet Simple Machines Science Worksheets
Packets, to be clear, are simply a bunch of worksheets stapled together. These science simple machines worksheets are great for any classroom. engage your students with these science simple machines worksheets. members receive unlimited access to, cross-curricular educational resources, including interactive activities,, and custom worksheet generators.
### 41. Resources Worksheets Simple Machines Reading Comprehension Science Fair
Start and stop an object in motion skill science experiment to try. force it or, perhaps, force it
### 43. Science Worksheet Simple Machines Mailbox Machine Projects Worksheets
This guide provides a lesson planning worksheet see page, which can assist you in setting up your instruction around a topic.the following sections of this implementation guide are offered to assist your planning process. simple machines. bill the science guy.
. Simple machine a machine with few or no moving parts. simple machines make work easier. examples screw, wheel and axle, wedge, pulley, inclined plane, lever compound machine two or more simple machines working together to make work easier. examples wheelbarrow, can opener, bicycle the is that an object must be moved a longer.
### 44. Simple Compound Machines Unit Activities Machine Projects
Worksheets are mixtures work answer key elements mixtures and compounds mixtures and solutions review for test key classification of matter relative dating. This compound events - practice. worksheet is suitable for - grade. probability pros consider the roll of a die and the spin of a spinner labeled x-z to find probabilities for the independent compound events.
### 45. Worksheet Simple Machines Riddles Upper Elem Writing Forms
Outsourcing to a expert telecommunications and contact support assistance ensures that your customer service is second to none. some of the worksheets displayed are bill the science guy static electricity bill electricity work bill genes video work answers ambassadors guide safety t science with bill the bill the science guy electrical current grade science.
Simple machines basic machines worksheet lever, screw, wheel and wedge students name the simple machines shown in the picture.Simple machines in everyday use in these worksheets, students identify which simple machine types wheel, screw, lever or wedge are used in various common objects.
Worksheet packet simple machines identify the class of each lever shown below. label the effort force, resistance force, and fulcrum.Simple machines,,, and efficiency worksheet ideal mechanical advantage. a simple machine would be considered ideal if it had no friction.
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## How Many Tbsp In 3/4 Cup?
Jul 25, 2023
### What is 3 4 cup in tablespoons?
There are approximately 12 tablespoons in 3/4 cup.
### How can I measure 3 4 of a cup?
How to make 3 4 cup with measuring cups? – The best way to make 3/4 cup with measuring cups is to first fill a 1/2 cup measuring cup. Then, add another 1/4 cup until the measuring cup is full. This should get you close to an exact measurement of 3/4 cup.
## How many tablespoons is 3 4 cups of butter?
How many tablespoons is 3 4 cup of butter? – When measuring butter for baking purposes, it is essential to use accurate measurements to achieve the desired texture and taste of the final product. In this regard, knowing the equivalent number of tablespoons in a specific amount of butter is crucial.
If you need 3/4 cup of butter for your recipe, it is essential to note that this is equal to 12 tablespoons. Therefore, if you do not have a measuring cup, you can measure 12 tablespoons of butter using a spoon. However, it is crucial to ensure that the spoon is leveled off to avoid using more or less butter than required.
It is worth noting that using too little or too much butter can hugely impact the texture and taste of your baked goods. Using too little butter can result in dry and hard cookies, while too much butter may result in greasy finished products. How many tablespoons is 3 4 cup of butter?
### What is 3 4 cup plus 2 tablespoons?
3/4 cup = 12 tablespoons OR 6 fluid ounces.7/8 cup = 3/4 cup + 2 tablespoons.
### Is 3 4 cup the same as 12 tablespoons?
Different cup sizes – The next confusion you may get is the cup measurements around the world. The United States uses the United States customary system (USCS) with its own measurement convention called ‘US cups,’ where ¾ customary or ¾ standard cup is equal:
¾ US Cup = 177.44 mL¾ US Cup = 12 US tablespoons (tbsp)¾ US Cup = 11.83 metric tablespoons (United Kingdom, international)
An American tablespoon equals 0.0625 American cups. Legal US cup is used in cooking units, serving sizes, and nutrition labeling in the USA.
¾ US Legal Cup = 180 mL¾ US Legal Cup = 12.17 US tablespoons (tbsp)¾ US Legal Cup = 12 metric tablespoons (UK, international)
A US cup differs slightly from other English-speaking countries such as Australia, New Zealand, Canada, and South Africa. This is because these countries use metric measurements:
¾ Metric Cup = 187.5 mL.
The old British recipes use the British imperial system, where an Imperial (UK) cup is a non-metric liquid measurement unit of volume. Modern recipes in the UK use UK metric cup.
¾ Imperial Cup = 213 mL ¾ UK Metric Cup = 187.5 mL
1 cup Milliliters ¾ cup Milliliters
US Customary Cup 236.59 ml US Customary Cup 177.44 ml
US Legal Cup 240 ml US Legal Cup 180 ml
UK Metric Cup 250 ml UK Metric Cup 187.5 ml
UK Imperial Cup 284.13 ml UK Imperial Cup 213 ml
Australian Cup 250 ml Australian Cup 187.5 ml
South African Cup 250 ml South African Cup 187.5 ml
#### Is 12 tablespoons 3 4 of a cup?
This means that there are 12 US Customary tablespoons in the 3/4 US Customary Cup. Similarly, there are 12 metric tablespoons per metric cup. This is true for both wet and dry ingredients.
## What is 1 4 cup in tablespoons?
The answer to how many tablespoons are in ¼ cup is 4. There are 16 tablespoons in 1 cup, so there are 4 tablespoons in ¼ cup.
#### How much is a 3 4 cup in mL?
Convert 3/4 Cups to mL 3/4 US customary cups to mL = 177.44 mL.
#### How many teaspoons are in a 3 4 of a cup?
How can I convert to the metric system? – Whether you want to convert those cups of sugar to dry ounces or measure out orange juice in fluid ounces but only have measuring spoons, there is a simple way of figuring out how many teaspoons you need. As a general rule, 8 fl oz is equal to one cup, which makes it easy to convert to teaspoons. Whether you’re looking to measure a cup of sugar, cup of butter, or cup of milk, there are some specific measurements that you can learn for educational purposes so that you no longer need to search for a conversion chart every time you cook. You can avoid consequential damages of any kind when you’re able to follow these useful tips as an alternative way to properly convert your measurements from teaspoons to cups.
#### How much is two 3 4 cups?
Answer and Explanation: For cup measurements, the amounts are added the same way as fractions. So doubling a 3/4 cup would yield 3 4 × 2 = 3 2 = 1.5 c u p s.
## What is half of 3 ⁄ 4 cup?
How Much is Half of 3/4 cup – Using the information above, it’s easy to cut nice, even numbers in half, such as a full cup or half of a cup. But what is half of 3/4 cup? Technically, since 1/4 cup equals 4 Tablespoons, then we know 3/4 cups equals 12 Tablespoons.
This means that half of 3/4 cup is 6 Tablespoons. This is the same as 1/4 cup PLUS 2 Tablespoons. If you don’t want to dirty even more measuring cups and spoons, I will often use 1/3 cup as half of 3/4 cup. That’s because 3/4 cup is the same as 0.75 cup. If you divide 0.75 in half, you’ll get 0.375, which is very close to the 0.333 which is 1/3 cup.
Personally, I don’t think a few hundredths of a cup is going to make a big difference in most recipes, so I generally halve 3/4 cup to 1/3 cup.
#### What is 3 4 in spoons?
How to Measure ¾ Teaspoon: Tips & Helpful Hacks
• Use a and to measure your ingredient, or use a 1 / 4 teaspoon 3 times.
• Scoop up your ingredient with a teaspoon but don’t fill it up all the way.
• Buy a if you anticipate needing it for future recipes.
• Weigh your ingredients with a to get more accurate measurements.
1. 1 Scoop the ingredient with both a and a, Grab the 1 / 2 teaspoon and 1 / 4 teaspoon from your regular set of measuring spoons; together, they’ll create a 3 / 4 teaspoon in total. Simply scoop up and level off your ingredients in each spoon and add them to the rest of your mixture.
2. 2 Use a 1 / 4 teaspoon 3 times to get 3 / 4 total. Grab your trusty 1 / 4 teaspoon and scoop up the first portion of your ingredient. Simply repeat this process 2 more times, and you’ll have a 3 / 4 tsp in total added to your recipe! Advertisement
3. 3 Fill up a regular teaspoon so it’s three-quarters of the way full. A 3 / 4 teaspoon is 75% of a single teaspoon—with this in mind, fill up a traditional teaspoon about three-quarters of the way with the required ingredient. You can also fill up the teaspoon completely and then scoop a little bit off the top.
4. 4 Buy a, Smaller measuring spoon sets only come with 4-6 spoons (like 1 / 8 tsp, 1 / 4 tsp, 1 / 2 tsp, 1 tsp, 1 / 2 tbsp, and 1 tbsp), while larger come with a 3 / 4 tsp as well. Feel free to buy a single 3 / 4 teaspoon if you already have other measuring spoons at home, or upgrade your kitchen with a full set.
• are another great option—with this tool, you use a movable tab and printed lines to make a customized measuring spoon.
Approximate a 3 / 4 tsp by adding the ingredient in pinches. The average “pinch” of a dry ingredient ranges between 1 / 16 tsp and 1 / 8 tsp. If your pinches are pretty small, add up to 12 pinches of the required ingredient. If your pinches are on the larger side, use closer to 6 pinches in your recipe.
1. Use a if you’d like to be extra accurate. Kitchen scales measure down to the ounce (or gram) of an ingredient, making them the best option for exact measurements. We’ve put together a cheat sheet of common dry ingredient weights (in both ounces and grams), so you know how much you need to measure out for your recipe:
• 3 / 4 tsp of baking soda: 0.13 ounce (3.7 grams)
• 3 / 4 tsp of salt: 0.16 ounce (4.5 grams)
• 3 / 4 tsp of sugar: 0.11 ounce (3.14 grams)
• 3 / 4 tsp of baking powder: 0.13 ounce (3.6 grams)
Ask a Question Advertisement Co-authored by: Private Chef & Food Educator This article was co-authored by and by wikiHow staff writer,, Ollie George Cigliano is a Private Chef, Food Educator, and Owner of Ollie George Cooks, based in Long Beach, California.
1. With over 20 years of experience, she specializes in utilizing fresh, fun ingredients and mixing traditional and innovative cooking techniques.
2. Ollie George holds a BA in Comparative Literature from The University of California, Berkeley, and a Nutrition and Healthy Living Certificate from eCornell University.
• Co-authors: 2
• Updated: December 13, 2022
• Views: 19,510
Categories:
Thanks to all authors for creating a page that has been read 19,510 times. : How to Measure ¾ Teaspoon: Tips & Helpful Hacks
## What is a 3 4 cup?
Volume Equivalents (liquid)*
12 tablespoons 3/4 cup 6 fluid ounces
16 tablespoons 1 cup 8 fluid ounces
2 cups 1 pint 16 fluid ounces
2 pints 1 quart 32 fluid ounces
### Is 1 4 cup equal to 2 tablespoons?
A 1/4 cup is equivalent to 4 tablespoons or 2 fluid ounces. It is also equivalent to ½ of a half-cup (or 8 tablespoons).
## Does 3 tablespoons equal 1 cup?
3 tbsp makes 0.1875 cup in a measuring cup.
## Is 1 8 cup equal to 3 tablespoons?
Conclusion how many tablespoons are in a 1/8 cup – In the United States, one-eighth of a cup of flour is equivalent to 1.6 tablespoons. Therefore, there are 1.6 US tablespoons in 1/8 cups of flour. In Australia, Canada and New Zealand, one-eighth of a cup of flour is equivalent to 2.4 tablespoons.
Therefore, there are 2.4 tablespoons in 1/8 cups of flour. By using measuring spoons and understanding how to convert between cups and tablespoons, you can measure those complex measurements with ease! There are plenty of online resources available to help you understand the differences between the measurements so that you can accurately follow a recipe for success.
Picked For You:
#### Is 12 tablespoons half a cup?
Basic Tablespoon Conversions – The short answer for how many tablespoons are in a cup is 16 tablespoons, So when you need 4 tablespoons, you can use ¼ cup. For 8 tablespoons, use ½ cup. Or for 12 tablespoons use ¾ cup. The chart below shares all of these, plus more, as well as the fluid ounces of each cup equivalency.
Tablespoons (TBSP) Dry Measurement (Cups) Liquid Measurement (fl. oz.) 16 tablespoons 1 cup 8 fluid ounces 12 tablespoons ¾ cup 6 fluid ounces 10 tablespoons + 2 teaspoons ⅔ cup 4 ⅔ fluid ounces 8 tablespoons ½ cup 4 fluid ounces 5 tablespoons + 1 teaspoon ⅓ cup 2 ⅓ fluid ounces 4 tablespoons ¼ cup 2 fluid ounces 2 tablespoons + 2 teaspoons ⅙ cup 1 ⅓ fluid ounce 2 tablespoons ⅛ cup 1 fluid ounce 1 tablespoon (or 3 teaspoons) 1/16 cup ½ fluid ounce
### What is 3 4 in spoons?
How to Measure ¾ Teaspoon: Tips & Helpful Hacks
• Use a and to measure your ingredient, or use a 1 / 4 teaspoon 3 times.
• Scoop up your ingredient with a teaspoon but don’t fill it up all the way.
• Buy a if you anticipate needing it for future recipes.
• Weigh your ingredients with a to get more accurate measurements.
1. 1 Scoop the ingredient with both a and a, Grab the 1 / 2 teaspoon and 1 / 4 teaspoon from your regular set of measuring spoons; together, they’ll create a 3 / 4 teaspoon in total. Simply scoop up and level off your ingredients in each spoon and add them to the rest of your mixture.
2. 2 Use a 1 / 4 teaspoon 3 times to get 3 / 4 total. Grab your trusty 1 / 4 teaspoon and scoop up the first portion of your ingredient. Simply repeat this process 2 more times, and you’ll have a 3 / 4 tsp in total added to your recipe! Advertisement
3. 3 Fill up a regular teaspoon so it’s three-quarters of the way full. A 3 / 4 teaspoon is 75% of a single teaspoon—with this in mind, fill up a traditional teaspoon about three-quarters of the way with the required ingredient. You can also fill up the teaspoon completely and then scoop a little bit off the top.
4. 4 Buy a, Smaller measuring spoon sets only come with 4-6 spoons (like 1 / 8 tsp, 1 / 4 tsp, 1 / 2 tsp, 1 tsp, 1 / 2 tbsp, and 1 tbsp), while larger come with a 3 / 4 tsp as well. Feel free to buy a single 3 / 4 teaspoon if you already have other measuring spoons at home, or upgrade your kitchen with a full set.
• are another great option—with this tool, you use a movable tab and printed lines to make a customized measuring spoon.
Approximate a 3 / 4 tsp by adding the ingredient in pinches. The average “pinch” of a dry ingredient ranges between 1 / 16 tsp and 1 / 8 tsp. If your pinches are pretty small, add up to 12 pinches of the required ingredient. If your pinches are on the larger side, use closer to 6 pinches in your recipe.
1. Use a if you’d like to be extra accurate. Kitchen scales measure down to the ounce (or gram) of an ingredient, making them the best option for exact measurements. We’ve put together a cheat sheet of common dry ingredient weights (in both ounces and grams), so you know how much you need to measure out for your recipe:
• 3 / 4 tsp of baking soda: 0.13 ounce (3.7 grams)
• 3 / 4 tsp of salt: 0.16 ounce (4.5 grams)
• 3 / 4 tsp of sugar: 0.11 ounce (3.14 grams)
• 3 / 4 tsp of baking powder: 0.13 ounce (3.6 grams)
Ask a Question Advertisement Co-authored by: Private Chef & Food Educator This article was co-authored by and by wikiHow staff writer,, Ollie George Cigliano is a Private Chef, Food Educator, and Owner of Ollie George Cooks, based in Long Beach, California.
With over 20 years of experience, she specializes in utilizing fresh, fun ingredients and mixing traditional and innovative cooking techniques. Ollie George holds a BA in Comparative Literature from The University of California, Berkeley, and a Nutrition and Healthy Living Certificate from eCornell University.
• Co-authors: 2
• Updated: December 13, 2022
• Views: 19,510
Categories:
Thanks to all authors for creating a page that has been read 19,510 times. : How to Measure ¾ Teaspoon: Tips & Helpful Hacks
## What is a 3 4 cup?
Volume Equivalents (liquid)*
12 tablespoons 3/4 cup 6 fluid ounces
16 tablespoons 1 cup 8 fluid ounces
2 cups 1 pint 16 fluid ounces
2 pints 1 quart 32 fluid ounces
### What is the equivalent of 3 4 cup in ml?
3/4 US customary cups to mL = 177.44 mL.
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# Population Growth
Population Growth. Logistic and Exponential Growth Problems. 2 Types:. Exponential Growth Population increases rapidly (resources unlimited) r > 0 Is the sky the limit? Logistic Growth Happens after exponential growth Population will reach a carry capacity (K) or crash.
Télécharger la présentation
## Population Growth
E N D
### Presentation Transcript
1. Population Growth Logistic and Exponential Growth Problems
2. 2 Types: • Exponential Growth • Population increases rapidly (resources unlimited) • r > 0 • Is the sky the limit? • Logistic Growth • Happens after exponential growth • Population will reach a carry capacity (K) or crash
3. Exponential Growth • N = population size (changes through time) • r = growth rate (factor by which population increases) • r = _________ • Formula: dN/dt = rN
4. Exponential Growth Example
5. Exponential Growth • What if you wanted to figure out the population of the rabbits after 30 years? • Darwin said exponential growth can’t last forever . . . leads to . . .
6. Logistic Growth • K = carrying capacity (max amount in a population that the ecosystem can support) • Why? Limited resources/limiting factors slow population growth • dN/dt = change in the population size over time • FORMULA: dN/dt = rN(K-N/K)
7. Examples: • Take a quick look at how a rabbit population would grow exponentially…
8. Exponential GrowthHow many years does it take to reach K? Recall: dN/dt = rN(K-N/K)
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Surface Area
# Surface Area - Surface Area In this section we are going to...
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Surface Area In this section we are going to look once again at solids of revolution. We first looked at them back in Calculus I when we found the volume of the solid of revolution . In this section we want to find the surface area of this region. So, for the purposes of the derivation of the formula, let’s look at rotating the continuous function in the interval about the x -axis. Below is a sketch of a function and the solid of revolution we get by rotating the function about the x -axis. We can derive a formula for the surface area much as we derived the formula for arc length . We’ll start by dividing the integral into n equal subintervals of width . On each subinterval we will approximate the function with a straight line that agrees with the function at the endpoints of the each interval. Here is a sketch of that for our representative function using . Now, rotate the approximations about the x -axis and we get the following solid.
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The approximation on each interval gives a distinct portion of the solid and to make this clear each portion is colored differently. Each of these portions are called frustums and we know how to find the surface area of frustums. The surface area of a frustum is given by, where, and l is the length of the slant of the frustum.
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Easy and Practical Tips for Roulette Secrets
Let us get to the point. You want tips for roulette secrets, so we will give you tips for roulette secrets. Let us enumerate.
Easy and Practical Tips for Roulette Secrets # 1: There is no 100 percent guarantee in winning roulette.
Easy and Practical Tips for Roulette Secrets # 2: People, websites or books that tell you there is a 100 percent way to win at roulette is lying or trying to get money from you.
Easy and Practical Tips for Roulette Secrets # 3: Even if you are a mathematician, there is no complicated mathematical guide that will let you win in roulette.
Easy and Practical Tips for Roulette Secrets # 4: Roulette is a game of chance, plain and simple. You might think that the current spin is related to each other, but it is not.
Easy and Practical Tips for Roulette Secrets # 5: When a number is sleeping, it doesn’t mean that it will be the one to come up on the next spin.
Easy and Practical Tips for Roulette Secrets # 6: When you manage your money, you give yourself an advantage but you do not decrease the house’s advantage. Managing your money so you can play for a long time is the best way to boost your odds. Why? The more spins you bet on, the more chances you get. The more chances you get to win, the bigger the odds of your making money.
Easy and Practical Tips for Roulette Secrets # 7: Money management works when you increase your bets at times when you see that the odds are in your favor and decrease it when it is not. It is a pretty logical way to win and we just had to state it.
Easy and Practical Tips for Roulette Secrets # 8: Let reality set it and know this: the house always has the edge to win so control yourself and be smart in handling your bankroll.
Easy and Practical Tips for Roulette Secrets # 9: Play European roulette because the house has less edge compared to American roulette.
Easy and Practical Tips for Roulette Secrets # 10: Place the bets when you are inching closer to payouts. This means you put money on whose odds are similar to their payouts. This includes even money which comprises of even, odd, low numbers, high numbers, red and black. These payouts are 1 is to 1 and there are better chances of winning these odds.
Easy and Practical Tips for Roulette Secrets # 11: Take advantage of the bet en prison which happens when the ball ends up on zero. You do not lose your bet nor win, but you lessen the house advantage to 1.35 percent. This is a good opportunity to add more bets to your wager on the table as the house
Easy and Practical Tips for Roulette Secrets
Easy and Practical Tips for Roulette Secrets # 4: Roulette is a game of chance, plain and simple. You might think that the current spin is r...
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# MIS41120 Statistical Learning
Broadly speaking, we think of Statistical and Machine Learning as computational methods that use (learn from) experience to improve performance or prediction accuracy. They arose in different research communities but have significant overlap. Statistical Learning focusses more on linear models, for which there is stronger theoretical foundation, and (to an extent) on inference; Machine Learning focusses more on nonlinear methods, founded more on experimental evidence, and is often more associated with prediction.
This Statistical Learning course discusses these, and also investigates the foundations of these methods: how well they work, error estimates, tradeoffs involved, etc: the principles underpinning algorithmic learning - the methods used in Knowledge Discovery and Data Mining.
Statistical learning refers to supervised and unsupervised learning, especially regression, classification, clustering, and especially with structured numerical data. These are the most common techniques used for modelling, with the goals of inference and prediction in business (and elsewhere); hence, their statistical theory is well-developed.
This module aims to develop both theory and practice to expert level.
Show/hide contentOpenClose All
Curricular information is subject to change
Learning Outcomes:
On completion of the module students should be able to:
● Distinguish between supervised and unsupervised learning and define regression, classification and clustering problems formally;
● Describe bias, variance and the bias-variance trade-off;
● Describe common loss functions and performance measures;
● Define the problem of overfitting and how to overcome it;
● Distinguish among common models, from linear regression to artificial neural networks to generalised linear models, and execute them with the help of a software library;
● Describe the main ideas of statistical learning theory, including the theory of the VC dimension.
Indicative Module Content:
Topics of the course are drawn from:
● Motivation: goals of prediction and inference/understanding
● Supervised and unsupervised learning: Regression, Classification, Clustering
● Measuring performance: accuracy and interpretability
● Bias, variance and the bias-variance tradeoff
● Generalisation and stability
● Model selection
● Loss functions
● The problem of Overfitting: Regularisation
● Sparse models including the lasso, elastic net and support vector machine
● Generalised linear models
● Artificial neural networks
● Deep nets
● Model capacity, shattering and VC dimension
Student Effort Type Hours
Specified Learning Activities
40
Autonomous Student Learning
100
Lectures
36
Total
176
Requirements, Exclusions and Recommendations
Not applicable to this module.
Module Requisites and Incompatibles
Not applicable to this module.
Assessment Strategy
Description Timing Open Book Exam Component Scale Must Pass Component % of Final Grade
Assignment: Project work on data analysis Throughout the Trimester n/a Graded No
25
Examination: Main Examination 2 hour End of Trimester Exam No Standard conversion grade scale 40% No
75
Carry forward of passed components
Yes
Resit In Terminal Exam
Autumn Yes - 2 Hour
Feedback Strategy/Strategies
• Feedback individually to students, post-assessment
• Group/class feedback, post-assessment
How will my Feedback be Delivered?
Feedback on strengths and weaknesses of assignment submission
Name Role
Assoc Professor Peter Keenan Lecturer / Co-Lecturer
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View the step-by-step solution to:
# A book publisher has the following demand function for the firm's novels (Qx):
A book publisher has the following demand function for the firm’s novels (Qx):
Qx = 12,000 – 5,000Px + 5I + 500Pc
where Px is the price charged for the firm’s novels, I is income per capita, and Pc is the price of books from competing publishers.
Assume that the initial values of Px, I, and Pc are \$5, \$10,000, and \$6, respectively
(A) Calculate the price elasticity and determine what effect a marginal (small) price increase from the initial price would have on total revenues
A book publisher has the following demand function for the firm’s novels (Qx):
Qx = 12,000 – 5,000Px + 5I + 500Pc
In this function we derivate Qx with respect to Px and rest other terms are...
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# KVPY (Kishore Vaigyanik Protsahan Yojana) Stream-SA (Class 11) Math: Questions 65 - 69 of 265
Get 1 year subscription: Access detailed explanations (illustrated with images and videos) to 265 questions. Access all new questions we will add tracking exam-pattern and syllabus changes. View Sample Explanation or View Features.
Rs. 200.00 or
## Question number: 65
» Basic Mathematics » Geometry » Advanced Therorems and Properties
Appeared in Year: 2012
MCQ▾
### Question
In triangle ABC, let AD, BE and CF be the internal angle bisectors with D, E and F on the sides BC, CA and AB respectively. Suppose AD, BE and CF concur at I and B, D, I, F are concyclic, then ∠IFD has measure.
### Choices
Choice (4) Response
a.
any value ≤ 90
b.
15
c.
45
d.
30
## Question number: 66
» Basic Mathematics » Arithmetic » Rates
Appeared in Year: 2010
MCQ▾
### Question
A closed conical vessel is filled with water fully and is placed with its vertex down. The water is flow out at a constant speed. After 21 minutes, it was found that the height of the water column is half of the original height. How much more time in minutes does it require to empty the vessel?
### Choices
Choice (4) Response
a.
21
b.
7
c.
14
d.
3
## Question number: 67
» Basic Mathematics » Geometry » Triangle and Properties
Appeared in Year: 2014
MCQ▾
### Question
In a quadrilateral ABCD, which is not a trapezium, it is known that ∠DAB = ∠ABC = 600. Moreover, ∠CAB = ∠CBD. Then
### Choices
Choice (4) Response
a.
b.
AB = BC + CD
c.
d.
## Question number: 68
Appeared in Year: 2010
MCQ▾
### Question
22. If x, y are real numbers such that, then the value of is
### Choices
Choice (4) Response
a.
3
b.
2
c.
0
d.
1
## Question number: 69
» Basic Mathematics » Arithmetic » Ratio and Proportion
MCQ▾
### Question
In a concert, the charge of pass per person is Rs. 500. On the first day, only 40 % of the passes were sold. The owner decided to reduce the price by 30 % and there was an increase of 60 % in the number of passes sold on the next day. The percentage increase in the revenue on the second day was
### Choices
Choice (4) Response
a.
13
b.
12
c.
10
d.
11
f Page
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# Mole/Hour to Megamole/Second Converter
1 Mole/Hour = 2.7777777777778e-10 Megamole/Second
## One Mole/Hour is Equal to How Many Megamole/Second?
The answer is one Mole/Hour is equal to 2.7777777777778e-10 Megamole/Second and that means we can also write it as 1 Mole/Hour = 2.7777777777778e-10 Megamole/Second. Feel free to use our online unit conversion calculator to convert the unit from Mole/Hour to Megamole/Second. Just simply enter value 1 in Mole/Hour and see the result in Megamole/Second.
Manually converting Mole/Hour to Megamole/Second can be time-consuming,especially when you don’t have enough knowledge about Molar Flow units conversion. Since there is a lot of complexity and some sort of learning curve is involved, most of the users end up using an online Mole/Hour to Megamole/Second converter tool to get the job done as soon as possible.
We have so many online tools available to convert Mole/Hour to Megamole/Second, but not every online tool gives an accurate result and that is why we have created this online Mole/Hour to Megamole/Second converter tool. It is a very simple and easy-to-use tool. Most important thing is that it is beginner-friendly.
## How to Convert Mole/Hour to Megamole/Second (mol/h to Mmol/s)
By using our Mole/Hour to Megamole/Second conversion tool, you know that one Mole/Hour is equivalent to 2.7777777777778e-10 Megamole/Second. Hence, to convert Mole/Hour to Megamole/Second, we just need to multiply the number by 2.7777777777778e-10. We are going to use very simple Mole/Hour to Megamole/Second conversion formula for that. Pleas see the calculation example given below.
$$\text{1 Mole/Hour} = 1 \times 2.7777777777778e-10 = \text{2.7777777777778e-10 Megamole/Second}$$
## What Unit of Measure is Mole/Hour?
Mole per hour is a unit of measurement for molar flow. It describes flow rate of one mole of fluid which passes through a given surface each hour.
## What is the Symbol of Mole/Hour?
The symbol of Mole/Hour is mol/h. This means you can also write one Mole/Hour as 1 mol/h.
## What Unit of Measure is Megamole/Second?
Megamole per second is a unit of measurement for molar flow. It describes flow rate of one megamole of fluid which passes through a given surface each second.
## What is the Symbol of Megamole/Second?
The symbol of Megamole/Second is Mmol/s. This means you can also write one Megamole/Second as 1 Mmol/s.
## How to Use Mole/Hour to Megamole/Second Converter Tool
• As you can see, we have 2 input fields and 2 dropdowns.
• From the first dropdown, select Mole/Hour and in the first input field, enter a value.
• From the second dropdown, select Megamole/Second.
• Instantly, the tool will convert the value from Mole/Hour to Megamole/Second and display the result in the second input field.
## Example of Mole/Hour to Megamole/Second Converter Tool
Mole/Hour
1
Megamole/Second
2.7777777777778e-10
# Mole/Hour to Megamole/Second Conversion Table
Mole/Hour [mol/h]Megamole/Second [Mmol/s]Description
1 Mole/Hour2.7777777777778e-10 Megamole/Second1 Mole/Hour = 2.7777777777778e-10 Megamole/Second
2 Mole/Hour5.5555555555556e-10 Megamole/Second2 Mole/Hour = 5.5555555555556e-10 Megamole/Second
3 Mole/Hour8.3333333333333e-10 Megamole/Second3 Mole/Hour = 8.3333333333333e-10 Megamole/Second
4 Mole/Hour1.1111111111111e-9 Megamole/Second4 Mole/Hour = 1.1111111111111e-9 Megamole/Second
5 Mole/Hour1.3888888888889e-9 Megamole/Second5 Mole/Hour = 1.3888888888889e-9 Megamole/Second
6 Mole/Hour1.6666666666667e-9 Megamole/Second6 Mole/Hour = 1.6666666666667e-9 Megamole/Second
7 Mole/Hour1.9444444444444e-9 Megamole/Second7 Mole/Hour = 1.9444444444444e-9 Megamole/Second
8 Mole/Hour2.2222222222222e-9 Megamole/Second8 Mole/Hour = 2.2222222222222e-9 Megamole/Second
9 Mole/Hour2.5e-9 Megamole/Second9 Mole/Hour = 2.5e-9 Megamole/Second
10 Mole/Hour2.7777777777778e-9 Megamole/Second10 Mole/Hour = 2.7777777777778e-9 Megamole/Second
100 Mole/Hour2.7777777777778e-8 Megamole/Second100 Mole/Hour = 2.7777777777778e-8 Megamole/Second
1000 Mole/Hour2.7777777777778e-7 Megamole/Second1000 Mole/Hour = 2.7777777777778e-7 Megamole/Second
# Mole/Hour to Other Units Conversion Table
ConversionDescription
1 Mole/Hour = 0.00027777777777778 Mole/Second1 Mole/Hour in Mole/Second is equal to 0.00027777777777778
1 Mole/Hour = 0.016666666666667 Mole/Minute1 Mole/Hour in Mole/Minute is equal to 0.016666666666667
1 Mole/Hour = 24 Mole/Day1 Mole/Hour in Mole/Day is equal to 24
1 Mole/Hour = 0.27777777777778 Millimole/Second1 Mole/Hour in Millimole/Second is equal to 0.27777777777778
1 Mole/Hour = 16.67 Millimole/Minute1 Mole/Hour in Millimole/Minute is equal to 16.67
1 Mole/Hour = 1000 Millimole/Hour1 Mole/Hour in Millimole/Hour is equal to 1000
1 Mole/Hour = 24000 Millimole/Day1 Mole/Hour in Millimole/Day is equal to 24000
1 Mole/Hour = 277777.78 Nanomole/Second1 Mole/Hour in Nanomole/Second is equal to 277777.78
1 Mole/Hour = 16666666.67 Nanomole/Minute1 Mole/Hour in Nanomole/Minute is equal to 16666666.67
1 Mole/Hour = 1000000000 Nanomole/Hour1 Mole/Hour in Nanomole/Hour is equal to 1000000000
1 Mole/Hour = 24000000000 Nanomole/Day1 Mole/Hour in Nanomole/Day is equal to 24000000000
1 Mole/Hour = 2.7777777777778e-7 Kilomole/Second1 Mole/Hour in Kilomole/Second is equal to 2.7777777777778e-7
1 Mole/Hour = 0.000016666666666667 Kilomole/Minute1 Mole/Hour in Kilomole/Minute is equal to 0.000016666666666667
1 Mole/Hour = 0.001 Kilomole/Hour1 Mole/Hour in Kilomole/Hour is equal to 0.001
1 Mole/Hour = 0.024 Kilomole/Day1 Mole/Hour in Kilomole/Day is equal to 0.024
1 Mole/Hour = 277.78 Micromole/Second1 Mole/Hour in Micromole/Second is equal to 277.78
1 Mole/Hour = 16666.67 Micromole/Minute1 Mole/Hour in Micromole/Minute is equal to 16666.67
1 Mole/Hour = 2.7777777777778e-22 Examole/Second1 Mole/Hour in Examole/Second is equal to 2.7777777777778e-22
1 Mole/Hour = 2.7777777777778e-19 Petamole/Second1 Mole/Hour in Petamole/Second is equal to 2.7777777777778e-19
1 Mole/Hour = 2.7777777777778e-16 Teramole/Second1 Mole/Hour in Teramole/Second is equal to 2.7777777777778e-16
1 Mole/Hour = 2.7777777777778e-13 Gigamole/Second1 Mole/Hour in Gigamole/Second is equal to 2.7777777777778e-13
1 Mole/Hour = 2.7777777777778e-10 Megamole/Second1 Mole/Hour in Megamole/Second is equal to 2.7777777777778e-10
1 Mole/Hour = 0.0000027777777777778 Hectomole/Second1 Mole/Hour in Hectomole/Second is equal to 0.0000027777777777778
1 Mole/Hour = 0.000027777777777778 Dekamole/Second1 Mole/Hour in Dekamole/Second is equal to 0.000027777777777778
1 Mole/Hour = 0.0027777777777778 Decimole/Second1 Mole/Hour in Decimole/Second is equal to 0.0027777777777778
1 Mole/Hour = 0.027777777777778 Centimole/Second1 Mole/Hour in Centimole/Second is equal to 0.027777777777778
1 Mole/Hour = 277777777.78 Picomole/Second1 Mole/Hour in Picomole/Second is equal to 277777777.78
1 Mole/Hour = 277777777777.78 Femtomole/Second1 Mole/Hour in Femtomole/Second is equal to 277777777777.78
1 Mole/Hour = 277777777777780 Attomole/Second1 Mole/Hour in Attomole/Second is equal to 277777777777780
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# Electrical and magnetic fields are spherical right?
## Main Question or Discussion Point
When i think of electromagnetic waves i think of a fast moving sphere of expanding or contracting fields,either magnetic or electrical depending on where its at in its cycle. So i guess im picturing a single photon as a sphere. Is this a correct visualization?(i doubt it). If so, how does a sphere of electricity become perpendicular to sphere of magnetism? Haha
## Answers and Replies
Related Classical Physics News on Phys.org
It is most convenient to think of electromagnetic fields as arrows (as they are vector fields). So at every point in space there is an arrow of a certain length and direction.
When thinking of waves these arrows are perpendicular. To make a planewave just pick a crest and draw an arrow at that point (the field is greatest in this point) then draw arrows next to it in the direction the wave is moving that gradually (depending on the frequancy/wavelength of the wave) become smaller, then zero and then bigger in the opposite direction then you get something like this: http://hyperphysics.phy-astr.gsu.edu/hbase/electric/imgel2/emwavec.gif
There is also something like a spherical wave. In this case you still draw the arrows in much the same way but now froming a sphere like this: http://www.phy.uct.ac.za/demonline/virtual/images/23_Radiation3D.JPG [Broken]
this last image is one wavefront.
Last edited by a moderator:
Drakkith
Staff Emeritus
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# If all stars rotate, why was there a theory developed that requires non-rotating stars?
According to Penrose's research, a non-rotating star would end up, after gravitational collapse, as a perfectly spherical black hole. However, every star in the universe has some kind of angular momentum.
Why even bother doing that research if that won't ever happen in the universe and does it have any implications for the future of astrophysics?
• Frictionless spherical cows are useful abstractions too... Commented Aug 22, 2019 at 5:13
• I suppose it's the solution to a simplified model of reality as a first step? That's not unusual in science... Commented Aug 22, 2019 at 5:47
• "However, every star in the universe" You've checked them all have you? Commented Aug 22, 2019 at 13:03
• "All models are wrong, but some are useful" Commented Aug 22, 2019 at 22:14
In a similar way, we could ask...
No beams can be exactly 1 meter long. No beams can be exactly straight. The material making up a beam cannot be truly isotropic. So why should we bother calculating the stress in a 1 meter straight beam having isotropic material?
Because knowing how to perform this calculation is a building block for doing more complex calculations.
The non-rotating black hole calculation also provides a limiting solution. The solution for a spinning star's collapse will approach this solution as the spin approaches zero.
Similarly, Newton told us that as external forces approach zero, the path of a moving object will approach a straight line. This is useful to know even though there is no place in our universe that doesn't have gravitational influence.
• Assume a spherical cow... Commented Aug 22, 2019 at 3:23
• I'm not sure if the metre is still defined against a standard, but if so, there is one stick that is exactly 1 metre long (by definition). Perhaps not entirely relevant to your point though. Commented Aug 22, 2019 at 5:39
• @RolandHeath It hasn't been since 1960. Commented Aug 22, 2019 at 6:50
• +1, but is it obvious that the non-rotating case is a limiting solution? A priori there might be global (topological?) effects that come into play as the angular momentum density grows towards infinity just before a singularity forms. Commented Aug 22, 2019 at 12:54
• @James: My point is that a collapsing star with low but nonzero angular momentum has to go through a phase where its angular momentum density diverges to infinity during the collapse -- whereas a star with zero angular momentum can have zero angular momentum density during its entire collapse. That might (at least a priori) give rise to a qualitative difference that is not respected by the limiting process. Commented Aug 23, 2019 at 12:24
All models are approximations, we judge a model on how useful it is.
Understanding the collapse of a non-rotating star to a black hole gives insight into the nature of gravitational collapse. Much of the physics of collapse does not depend on spin. The formation of an event horizon, for example.
Models can be refined, and in this case, considering rotation leads to further insight, and a non-spherically symmetric structure with multiple singular horizons.
All models are necessarily simplifications. But the non-rotating model is still useful.
Another consideration is that the physics that describe a rotating black hole was much harder to develop.
The maths describing the Schwarzschild (uncharged, non-spinning) black hole was developed in 1916. This was expanded to charged, non-spinning black holes in 1918 (The Reissner–Nordström metric)
It wasn't until 1963 that the Kerr metric for uncharged spinning black holes was developed. Two years later, the most general form, the Kerr-Newman metric was found.
I wouldn't fancy waiting 47 years for a more accurate black hole model to be developed before doing any meaningful work in the field.
• Also note that the pure Schwarzschild solution is static: it's eternal, not formed by collapse, and it's the only object in an otherwise empty universe. But it's still a useful solution, despite these unnatural simplifications. Commented Aug 23, 2019 at 2:15
Our sun's rotation period is 24.47 days at the equator and almost 38 days at the poles, our planet's rotational period is 23h 56m 4.098,903,691s. Use of Schwarzschild equations for either case isn't exact.
If you used the equation for non-rotating objects to calculate the time at the altitude of GPS satellites (~ 20,200 km or 12,550 miles) then you would be off by 38,636 nanoseconds per day. A Julian year is defined as 365.25 days of exactly 86,400 seconds (SI base unit), totalling exactly 31,557,600 seconds in the Julian astronomical year. The Gregorian calendar year (400 year average) is 365.2425 days.
Multiplying 365.2425 x 38,636 = 14,111,509.23 nanoseconds, that's 0.0141 seconds per year. If being off by that amount isn't of any concern to you then you can use the easier equation, such as for calculations involving the star HR 1362 which has a rotational period that is 306.9 ± 0.4 days.
You're right: all stars rotate. The only reason I can think of why astrophysicists make calculations for a non-rotating star or black hole is that it makes their calculations a bit easier. Although all stars rotate, some rotate much faster than others, and their masses vary too, so there is a wide degree of uncertainty which is reduced by calculating for a star that does not rotate.
• How certain can we be that all stars rotate? There are a lot of stars and many many possible (theoretical) interactions that would slow rotation. Commented Aug 21, 2019 at 22:11
• @Valorum Yeah, I was thinking about a stellar collision where the stars are rotating in opposite directions. If the rotational energy is exactly opposite you'll get a non-rotating result. Very unlikely, not utterly impossible--thus it probably will happen somewhere, someday. Commented Aug 22, 2019 at 3:41
• @LorenPechtel The rotational momentum needs to be exactly equal. I think that counts as utterly impossible. Commented Aug 22, 2019 at 7:28
• @Valorum Because the chance for "zero" angular momentum approaches 0 much faster than the amount of stars grow with "sample size". Commented Aug 24, 2019 at 17:44
• @Valorum Nobody has any problem with approximately zero; that's obvious. For a star whose rotation is reversed: obviously "the system" (star+impactor) has angular momentum in the opposite direction to the star alone. The star + impactor will mix in complex ways, so I don't think it is helpful to talk about "the star" reversing direction. Commented Aug 26, 2019 at 8:29
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# A solution containing 0.400 g of a diprotic acid requires 40.00cm^3 of 0.100 mol dm^3 NaoH solution for complete neutralization.What is the approximate relative molecular mass of the acid? (1) 5.00...
A solution containing 0.400 g of a diprotic acid requires 40.00cm^3 of 0.100 mol dm^3 NaoH solution for complete neutralization.What is the approximate relative molecular mass of the acid?
(1) 5.00 × 10
(2) 1.00 × 10^2
(3) 2.00 × 10^2
(4) 3.00 × 10^2
(5) 4.00 × 10^2
jeew-m | College Teacher | (Level 1) Educator Emeritus
Posted on
A diprotic acid is a acid that have the capability to provide two `H^+ ` ions to the solution such as `H_2SO_4` . Let us take the acid as `H_2A`.
`2NaOH+H_2A+ rarr Na_2A+2H_2O`
Amount of NaOH used for reaction `= 0.1/1000xx40 = 0.004mol`
Since 0.004moles of NaOH reacted the amount of diprotic acid moles reacted will be half of NaOH reacted.
`NaOH:H^+ = 2:1`
Amount of diprotic acid reacted `= 0.004/2 = 0.002mol`
Relative molecular mass
= (mass)/(mol)
`= 0.4/0.002 `
`= 200 `
`= 2xx10^2`
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Design & Technology On The Web - Isometric, Planometric & Axonometric Drawings
ISOMETRIC DRAWINGS
Views use 30 degrees for lines showing
‘horizontals’ of the object:
This is the most widely used drawing convention and you should be very confident that you know what the term Isometricmeans.
It can sometimes be seen described as a ‘pictorial’ view. Examination questions asking for a ‘pictorial view’ mean an ‘isometric
This convention can be exceptionally useful as the drawing is ‘built-up’ from an actual plan. If a ground floor map of a room is completed and all the walls are shown as flat panels – vertical to the first ‘view’ – then a PLANOMETRIC view has been started. One big advantage of this drawing style is that all circles still appear as circles on the finished sketch,
~ Isometric, Planometric & Axonometric ~
Isometric Projection
X-axis ~ 30 degrees to the horizontal
Y-axis ~ 30 degrees to the horizontal
Z- axis ~ vertical
The same scale is used on all three axes
PLANOMETRIC DRAWINGS
Planometric
X-axis ~30 degrees to the horizontal
Y-axis ~ 60 degrees to the horizontal
Z- axis ~ vertical
The same scale is used on all three axes
This small line illustration should show how a ‘plan-view’ becomes an axonometric or a planometric dependent on the angle used to represent the horizontal lines.
Axonometric
(45 degree angle)
Planometric
(60/30 degree angles)
Check out the DESIGN-CYCLE section
http://en.wikipedia.org/wiki/Axonometric_projection
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Step by Step Group Development - A Trainer's Handbook
APPENDIX
GAMES
(introduction...) A - Games for Introducing/Games with Names: B - Songs: C - Games in a Circle D - Games for Trust-building in Groups
### D - Games for Trust-building in Groups
D1: ROBOTS
The players form pairs. One of them is the robot, the other the robot’s controller. The robot can only move when the controller helps him/her by tapping on his back.
But this only allows the robot to walk straight. If there are obstacles in the way or walls, the robot has to be moved to the side - by tapping on the right shoulder to go right, on the left shoulder for left -.
To stop the robot, touch it gently on the head.
Now try to move your robots through the room without touching the others.
After some time robots and controllers change roles.
Variations:
1. Try to move the robot to a special place in the room and back to the start.
2. One controller has to move two robots.
3. The robot is blind (closes the eyes) and therefore has to be moved very carefully.
Figure
D2: SWINGING
Small groups of up to eight form a close circle. One of the players (only if (s)he wants to!) goes into the center, and stiffens her/his pelvis, with the rest of the body loose, hands at sides.
The others hold their hands in front of their bodies. When all are ready, the one in the center lets him/herself fall in one direction (eyes open or closed as preferred). The standing people catch him/her and try to move him/her gently along the circle and across, when they get more experienced.
Figure
When the center one has had enough, they stop, discuss briefly how (s)he liked it and then the next one can try.
D3: HOLDING A CIRCLE
An even number of players form a circle, holding each other by the hands. Now they count “1-2-1-...” to divide the group into two sub-groups.
Figure
At a special command the 1’s lean into the circle while the 2’s lean to the outside. Feet are placed firmly on the floor, arms help to balance. When the circle is stable, at another command they change directions, the 1’s lean slowly outwards and the 2’s lean towards the center. Don’t let go your neighbours’ hands! When the players are tired, the circle is finished on a mutual command.
D4: THE GORDIAN KNOT
Figure
The players stand in a close circle, stretching their arms to the center. On a certain command they close their eyes and try to find two other hands they can hold. When each hand is linked to another one, the players open their eyes.
Now try to unravel this human knot! It is not allowed to let hands go, the players have to move up and down, just as needed.
The game is finished, when the players are standing in a circle, holding their neighbours by the hand.
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Community Profile
# Christopher Stapels
### MathWorks
33 total contributions since 2017
ThingSpeak Documentation Writer
and LCH enthusiast
#### Christopher Stapels's Badges
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Guru Meditation exception during publishing values to thing speak
You should also be able to update the value of a ThingSpeak field using the REST API. The MQTT connection is intended to be ve...
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matlab and thingspeak manipulate data before plotting
I would recommend setting up a new channel. Then you can set a react to call a MATLAB analysis to write the data to the new cha...
7 months ago | 0
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Battery Voltage
two six volt bettery packs in paralell in the basement
7 months ago
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WorkTime
8 months ago
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Soil Monitor Data Summary
Time aligned data
1 year ago
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Daily Stock Data
Google Stock values compiled daily.
1 year ago
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Home Temp Time
1 year ago
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A Simple Tide Gauge with MATLAB
*∿ ∿ ∿ ∿ ∿ ∿ ∿ ∿* You are standing in a few inches of sea water on a beach. You a...
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5 Prime Numbers
Your function will be given lower and upper integer bounds. Your task is to return a vector containing the first five prime numb...
1 year ago
I want to calculate the derivative of a channel. Can you suggest the best way to do so.
One way would be to use an instance of the React App to test if field 2 has a value in it, perhaps "if field2 < 0". Then have t...
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MyCar Count Channel
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Thingspeakwrite error when using string array
When you write string data to ThingSpeak, it generally wants to have a cell array. You should not usually need to do the Strin...
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Lights1
kitchen lights control
1 year ago
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Make an awesome ramp for a tiny motorcycle stuntman
Okay, given a vector, say v=[1 3 6 9 11], turn it into a matrix 'ramp' like so: m=[1 3 6 9 11; 3 6 9 11 0; 6 9 ...
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Back to basics 21 - Matrix replicating
Covering some basic topics I haven't seen elsewhere on Cody. Given an input matrix, generate an output matrix that consists o...
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Since Cody is 5 years old, it's pernicious. A <http://rosettacode.org/wiki/Pernicious_numbers Pernicious number> is an integer w...
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Energy of a photon
*⚛ ☢ ⚛ ☢ ⚛ ☢ ⚛* Given the frequency F of a photon in giga hertz. Find energy E of this...
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How to subtract?
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Significant Figures
• Jul 3rd 2012, 09:37 AM
karies4083
Significant Figures
I had this problem come up while studying for my Math Praxis II exam.
What can the difference between the following numbers that are rounded to 2 significant digits:
2.87 million
2.22 million
Thanks,
karies4083
• Jul 3rd 2012, 09:51 AM
emakarov
Re: Significant Figures
Hint: The first number belong to the interval $\displaystyle [2.865\cdot10^6,2.875\cdot10^6)$ and the second number belongs to the interval $\displaystyle [2.215\cdot10^6,2.225\cdot10^6)$.
• Jul 3rd 2012, 10:49 AM
karies4083
Re: Significant Figures
Thanks .. that helps . .ALOT!
karies4083
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Raw Score, x Population Mean, μ Standard Deviation, σ
## Z-score và Probability Converter
Please provide any one value to lớn convert between z-score & probability. This is the equivalent of referencing a z-table.
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Z-score, Z Probability, P(x Probability, P(x>Z) Probability, P(0 to lớn Z or Z lớn 0) Probability, P(-Z Probability, P(xZ)
## Probability between Two Z-scores
Use this hsnovini.com to lớn find the probability (area p. In the diagram) between two z-scores.
Left Bound, Z1 Right Bound, Z2
RelatedStandard Deviation hsnovini.com
### What is z-score?
The z-score, also referred khổng lồ as standard score, z-value, and normal score, among other things, is a dimensionless quantity that is used khổng lồ indicate the signed, fractional, number of standard deviations by which an event is above the mean value being measured. Values above the mean have positive z-scores, while values below the mean have negative z-scores.
The z-score can be calculated by subtracting the population mean from the raw score, or data point in question (a kiểm tra score, height, age, etc.), then dividing the difference by the population standard deviation:
z = x - μ σ
where x is the raw score, μ is the population mean, & σ is the population standard deviation.
The z-score has numerous applications & can be used lớn perform a z-test, calculate prediction intervals, process control applications, comparison of scores on different scales, & more.
### Z-table
A z-table, also known as a standard normal table or unit normal table, is a table that consists of standardized values that are used khổng lồ determine the probability that a given statistic is below, above, or between the standard normal distribution.
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The table below is a right-tail z-table. Although there are a number of types of z-tables, the right-tail z-table is commonly what is meant when a z-table is referenced. It is used lớn find the area between z = 0 và any positive value, & reference the area to the right-hand side of the standard deviation curve.
z 0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0 0 0.00399 0.00798 0.01197 0.01595 0.01994 0.02392 0.0279 0.03188 0.03586 0.1 0.03983 0.0438 0.04776 0.05172 0.05567 0.05962 0.06356 0.06749 0.07142 0.07535 0.2 0.07926 0.08317 0.08706 0.09095 0.09483 0.09871 0.10257 0.10642 0.11026 0.11409 0.3 0.11791 0.12172 0.12552 0.1293 0.13307 0.13683 0.14058 0.14431 0.14803 0.15173 0.4 0.15542 0.1591 0.16276 0.1664 0.17003 0.17364 0.17724 0.18082 0.18439 0.18793 0.5 0.19146 0.19497 0.19847 0.20194 0.2054 0.20884 0.21226 0.21566 0.21904 0.2224 0.6 0.22575 0.22907 0.23237 0.23565 0.23891 0.24215 0.24537 0.24857 0.25175 0.2549 0.7 0.25804 0.26115 0.26424 0.2673 0.27035 0.27337 0.27637 0.27935 0.2823 0.28524 0.8 0.28814 0.29103 0.29389 0.29673 0.29955 0.30234 0.30511 0.30785 0.31057 0.31327 0.9 0.31594 0.31859 0.32121 0.32381 0.32639 0.32894 0.33147 0.33398 0.33646 0.33891 1 0.34134 0.34375 0.34614 0.34849 0.35083 0.35314 0.35543 0.35769 0.35993 0.36214 1.1 0.36433 0.3665 0.36864 0.37076 0.37286 0.37493 0.37698 0.379 0.381 0.38298 1.2 0.38493 0.38686 0.38877 0.39065 0.39251 0.39435 0.39617 0.39796 0.39973 0.40147 1.3 0.4032 0.4049 0.40658 0.40824 0.40988 0.41149 0.41308 0.41466 0.41621 0.41774 1.4 0.41924 0.42073 0.4222 0.42364 0.42507 0.42647 0.42785 0.42922 0.43056 0.43189 1.5 0.43319 0.43448 0.43574 0.43699 0.43822 0.43943 0.44062 0.44179 0.44295 0.44408 1.6 0.4452 0.4463 0.44738 0.44845 0.4495 0.45053 0.45154 0.45254 0.45352 0.45449 1.7 0.45543 0.45637 0.45728 0.45818 0.45907 0.45994 0.4608 0.46164 0.46246 0.46327 1.8 0.46407 0.46485 0.46562 0.46638 0.46712 0.46784 0.46856 0.46926 0.46995 0.47062 1.9 0.47128 0.47193 0.47257 0.4732 0.47381 0.47441 0.475 0.47558 0.47615 0.4767 2 0.47725 0.47778 0.47831 0.47882 0.47932 0.47982 0.4803 0.48077 0.48124 0.48169 2.1 0.48214 0.48257 0.483 0.48341 0.48382 0.48422 0.48461 0.485 0.48537 0.48574 2.2 0.4861 0.48645 0.48679 0.48713 0.48745 0.48778 0.48809 0.4884 0.4887 0.48899 2.3 0.48928 0.48956 0.48983 0.4901 0.49036 0.49061 0.49086 0.49111 0.49134 0.49158 2.4 0.4918 0.49202 0.49224 0.49245 0.49266 0.49286 0.49305 0.49324 0.49343 0.49361 2.5 0.49379 0.49396 0.49413 0.4943 0.49446 0.49461 0.49477 0.49492 0.49506 0.4952 2.6 0.49534 0.49547 0.4956 0.49573 0.49585 0.49598 0.49609 0.49621 0.49632 0.49643 2.7 0.49653 0.49664 0.49674 0.49683 0.49693 0.49702 0.49711 0.4972 0.49728 0.49736 2.8 0.49744 0.49752 0.4976 0.49767 0.49774 0.49781 0.49788 0.49795 0.49801 0.49807 2.9 0.49813 0.49819 0.49825 0.49831 0.49836 0.49841 0.49846 0.49851 0.49856 0.49861 3 0.49865 0.49869 0.49874 0.49878 0.49882 0.49886 0.49889 0.49893 0.49896 0.499 3.1 0.49903 0.49906 0.4991 0.49913 0.49916 0.49918 0.49921 0.49924 0.49926 0.49929 3.2 0.49931 0.49934 0.49936 0.49938 0.4994 0.49942 0.49944 0.49946 0.49948 0.4995 3.3 0.49952 0.49953 0.49955 0.49957 0.49958 0.4996 0.49961 0.49962 0.49964 0.49965 3.4 0.49966 0.49968 0.49969 0.4997 0.49971 0.49972 0.49973 0.49974 0.49975 0.49976 3.5 0.49977 0.49978 0.49978 0.49979 0.4998 0.49981 0.49981 0.49982 0.49983 0.49983 3.6 0.49984 0.49985 0.49985 0.49986 0.49986 0.49987 0.49987 0.49988 0.49988 0.49989 3.7 0.49989 0.4999 0.4999 0.4999 0.49991 0.49991 0.49992 0.49992 0.49992 0.49992 3.8 0.49993 0.49993 0.49993 0.49994 0.49994 0.49994 0.49994 0.49995 0.49995 0.49995 3.9 0.49995 0.49995 0.49996 0.49996 0.49996 0.49996 0.49996 0.49996 0.49997 0.49997 4 0.49997 0.49997 0.49997 0.49997 0.49997 0.49997 0.49998 0.49998 0.49998 0.49998
ScientificFractionPercentageTriangleVolumeStandard DeviationRandom Number GeneratorMore Math hsnovini.coms
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# Different Doubles
## Introduction
The Different Doubles is craps side bet that has been seen at the Beau Rivage casino in Biloxi. It pays pays on the number of unique doubles the shooter rolls before rolling a total of seven. The pay table is as follow:
• 6 doubles pays 100 to 1
• 5 doubles pays 15 to 1
• 4 doubles pays 8 to 1
• 3 doubles pays 4 to 1
## Analysis
The table below shows the probability and contribution to the return of all possible outcomes. The lower right cell shows a house edge of 27.92% (ouch!).
### Different Doubles Return Table
Doubles Made Pays Combinations Probability Return
6 100 1 0.001082 0.108225
5 15 6 0.006494 0.097403
4 8 21 0.022727 0.181818
3 4 56 0.060606 0.242424
2 -1 126 0.136364 -0.136364
1 -1 252 0.272727 -0.272727
0 -1 462 0.500000 -0.500000
Total 924 1.000000 -0.279221
## Algebraic Analysis
The probability of any given double is 1/36. Thus, the probability of ANY double is 6/36 = 1/6. The probability of any seven is 1/6. We can ignore all rolls besides doubles and sevens. So, assuming there was a roll pertinent to the bet, below are the probabilities for that roll:
• Any double = 1/2
• Any seven = 1/2
The probability the first significant roll is a seven is 1/2. Thus there is a probability of zero doubles.
Otherwise, the player would have rolled a double. We can now ignore rolling that double again as a significant event. The probability of both types of significant events are now:
• Any significant double = 5/11
• Any seven = 6/11
The probability the player rolls a seven at this point, for one double, is 6/11. Thus, the overall probability of one double is (1/2)*(6/11) = 3/11 = apx. 27.27%.
Otherwise, the player would have rolled a second double. We can now ignore rolling two distinct doubles as a significant event. The probability of both types of significant events are now:
• Any significant double = 4/10
• Any seven = 6/10
The probability the player rolls a seven at this point, for two double, is 6/10. Thus, the overall probability of one double is (1/2)*(5/11)*(6/10) = 3/22 = apx. 13.6363636%.
Otherwise, the player would have rolled a third double. We can now ignore rolling three distinct doubles as a significant event. The probability of both types of significant events are now:
• Any significant double = 3/9
• Any seven = 6/9
The probability the player rolls a seven at this point, for two double, is 6/9. Thus, the overall probability of one double is (1/2)*(5/11)*(4/10)*(6/9) = 2/33 = apx. 6.060606%.
Otherwise, the player would have rolled a fourth double. We can now ignore rolling four distinct doubles as a significant event. The probability of both types of significant events are now:
• Any significant double = 2/8
• Any seven = 6/8
The probability the player rolls a seven at this point, for two double, is 6/8. Thus, the overall probability of one double is (1/2)*(5/11)*(4/10)*(3/9)*(6/8) = 1/44 = apx. 2.272727%.
Otherwise, the player would have rolled a fifth double. We can now ignore rolling five distinct doubles as a significant event. The probability of both types of significant events are now:
• Any significant double = 1/7
• Any seven = 6/7
The probability the player rolls a seven at this point, for two double, is 6/7. Thus, the overall probability of one double is (1/2)*(5/11)*(4/10)*(3/9)*(2/8)*(6/7) = 1/154 = apx. 0.649351%.
Otherwise, the player would have rolled that last remaining double. The probability of that is (1/2)*(5/11)*(4/10)*(3/9)*(2/8)*(1/7) = 1/924 = apx. 0.108225%.
We can now put all that together in the return table above. The number of combinations are out of 924.
## Integral Calculus Analysis
Imagine that instead of significant events being determined by the roll of dice, one at a time, consider them as an instant in time. Assume the time between events has a memory-less property, with an average time between events of one unit of time. In other words, the time between events follows an exponetial distribution with a mean of 1. This will not matter for purposes of adjudicating the bet, because event still happen one at a time.
The time between specific doubles will follow an exponential distribution with a mean of 12. The reason for the 12 is if there is a significant event, there is a 1/12 chance it was that specific double. So, the probability that a specific double has not happened in x units of time is exp(-x/12). The probability it has happened is thus 1-exp(-x/12).
Let x be the number of units of time since the bet started. The probability that every double has happened and a seven has not happened is (1-exp(-x/12))6 × exp(-x/2).
To close that off with a roll of a 7 at time x, multiply that by 1/2, the probability of a 7, to get a probability of a winning get at exactly time x of (1/2) ×(1-exp(-x/12))6 × exp(-x/2)
To find the probability of a winning bet over all time, integrate from 1 to infinity:
To integrate that, I recommend this integral calculator. In the integral field put, "(1-exp(-x/12))^6*(1/2)*exp(-x/2)." For the limits of integration, under options, put 0 and ∞. Then click "go." It will give you the integral below:
(-(1-e^(-x/12))^12+(60*(1-e^(-x/12))^11)/11-12*(1-e^(-x/12))^10+(40*(1-e^(-x/12))^9)/3-(15*(1-e^(-x/12))^8)/2+(12*(1-e^(-x/12))^7)/7)/2
However, we don't need to enter 0 into that. The calculator gives the answer as 1/924 = apx. 0.001082251082251082.
Here is the integral for five different doubles. The reason for the 6 is there are 6 possible doubles that were not made:
Here is the probability of winning within x units of time:
Putting in the bounds of integration from 0 to infinity, the probability of rolling exactly five distinct doubles is 1/154.
Here is the integral for four different doubles. The reason for the 15 is there are 6!/(4!*2!) = 15 possible combinations of 4 out of 6 doubles that were made:
Here is the probability of winning within x units of time:
Putting in the bounds of integration from 0 to infinity, the probability of rolling exactly four distinct doubles is 1/44.
Here is the integral for three distinct doubles. The reason for the 20 is there are 6!/(3!*3!) = 20 possible combinations of 3 out of 6 doubles that were made:
Here is the probability of winning within x units of time:
Putting in the bounds of integration from 0 to infinity, the probability of rolling exactly three distinct doubles is 2/33.
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https://pty.vanderbilt.edu/2018/01/spring-savy-2018-day-1-whats-the-matter-kindergarten/
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# Spring SAVY 2018, Day 1- What’s the Matter? (Kindergarten)
Posted by on Monday, January 29, 2018 in Grade K, SAVY.
We had a wonderful first day of SAVY discussing what it means to be a scientist and how we can use our skills of observations to study the world around us. As a class, we brainstormed all the different ways we could categorize the seemingly random objects in a box. We developed several ways that divided the objects by style, color, shape, utility, and size. We also applied a similar organizational strategy to divide our class into groups based on shirt color/style, gender, and age. We related these abilities to categorize and make observations to some of the skills that scientists use when studying matter. We learned that matter is anything that has mass (weight) and takes up space (has volume), and there are lots of things that are matter! We used water as an example of matter to conduct the rest of our experiments. First, we learned that the molecules that make up water are attracted to each other. When we put drops of water on wax paper, we saw that we could drag smaller drops together and all the little drops would combine to form larger drops that were difficult to separate. Students were able to practice their scientific investigation and reasoning as they recorded their experimental steps and observations through drawing and writing about the experiment. Some of the observations from the students included: “When I touch it, it expands and makes a giant one!” (Referring to moving one drop into another drop) and “If you make a drop in the middle of two small ones, it makes them together, bigger.”
After learning that water molecules are attracted to other water molecules, we did an experiment to find how many drops of water a coin can hold. First, we hypothesized as a class which coins would hold the most drops of water and which coins would hold the least. The students reasoned that quarters (because they were the largest coins) would hold the most drops of water and that dimes would hold the least drops of water. Each student was given their own coin and once again recorded their experimental procedure, observations, and results. Then we came together as a class and each student shared his or her results and we saw that our hypothesis was nearly correct. This gave us a chance to talk about why scientists repeat their experiments to make sure that their results are reproducible.
The last activity we did was to design an experiment to test whether or not the molecules that make up water are in motion. Our experiment consisted of putting drops of food coloring into a bottle of water and watching the food coloring and water mix without us stirring. Then we wanted to see what would happen if we added another color. We talked about the change that was occurring as the colors mixed together.
If you want to watch a video to see an example of water molecules being attracted to one another and have your student explain some of what they learned this week, check out this slow-motion video of a water balloon popping and watch how the water molecules stay together: https://www.youtube.com/watch?v=mn7K8SEWgYY
Additionally, a fun game to play with drops of water is Race Drop Raceway! The raceway paper is attached. Here are the instructions:
1. Take the “Race Drop Raceway” sheet onto a piece of cardboard to give it support.
2. Tape a piece of wax paper over the sheet.
3. Place 2-4 drops of water together to make one larger drop at the start.
4. As fast as you can, tilt the cardboard and guide your race drop around the track to the finish. Try not to touch the edge of the track!
I am looking forward to next weekend when we will learn about some of the physical properties of matter and take a field trip to a real chemistry lab to learn what kind of instruments they use to study matter.
-Dorothy
Race Drop Raceway
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http://www.teenxnxx.top/how-much-wood-is-a-quarter-cord/
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# how much wood is a quarter cord
4’x6’x16鈥?/div>
## People also ask
• ### How big is a quarter cord of firewood?
• Quarter Cord of Firewood. Clear. At Woodchuck a quarter cord of firewood is measured at 4 ft by 4 ft with a typical cut wood length of 16 inches to 18 inches. Most of our customers desire seasoned post oak for their firewood needs.
• ### How many cubic feet is a cord of wood?
• Then, the volume of the wood is taken. In the United States, the definition of a cord is typically a volume of 128 cubic feet鈥攐r a stack that is 4 feet wide, 4 feet high, and 8 feet long. Regulations about cord size can vary by state and country. There are several other terms used for firewood measurement.
• ### What is a quart of wood?
• Here is a guide to tell you what a quart of wood is. A quart of wood is how much? A quick answer to this question: a quart of wood is a quart or one-fourth of a cord of wood. A cord is the standard measurement for firewood and is 128 cubic feet. When a cord of wood is stacked, it will measure about 4 feet high, 4 feet wide and 8 feet long.
• ### How many cords of wood are in a stake?
• Since a cord is 128 cubic feet, divide the number of cubic feet in your stack by 128. That will tell you how many cords you have. The common dimension to get a cord of wood is 4鈥檟4鈥檟8, but there are many dimensions that will add up to 128 cubic feet. Examples are 2x4x16 and 1x4x32.
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http://datascience.sharerecipe.net/2019/06/21/an-introduction-to-bayesian-inference/
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An Introduction to Bayesian Inference
An Introduction to Bayesian Inferenceand Markov Chain Monte CarloNeel ParekhBlockedUnblockFollowFollowingJun 18Foreword: The following post is intended to be an introduction with some math.
Although it includes math, I’m by no means an expert.
I struggled to learn the basics of probabilistic programming, and hopefully this helps someone else make a little more sense out of the world.
If you find any errors, please drop a comment and help us learn.
Cheers!In data science we are often interested in understanding how data was generated, mainly because it allows us to answer questions about new and/or incomplete observations.
Given a (often noisy) dataset though, there are an infinite number of possible model structures that could have generated the observed data.
But not all model structures are made equally — that is, certain model structures are more realistic than others based on our assumptions of the world.
It is up to the modeler to choose the one that best describes the world they are observing.
GOAL: Find the model M that best explains how a dataset D was generatedFor example, we might believe that a model that predicts the probability of me eating gelato would be appropriate if we were to set it up so that the date and my location influence the temperature, which in turn influences the chances of me wolfing down dark chocolate gelato (the best kind).
But this only gives us the model structure.
We also need to know how much each of input influences each level of the model — these are the model parameters θ.
I love gelato.
Even after we have chosen which assumptions we’ve made about the world, the model structure itself so far is just abstract.
We would love to have a function that can assign probabilities to any possible gelato outcome for a given set of inputs, but with just an abstract model choice, we don’t yet know exactly which values to use for the parameters in the model we’ve chosen.
This is where the probability theory comes in.
REVISED GOAL: For a given model structure M that encompasses our assumptions about the structure of the world (i.
e.
how we believe the data were generated), what are the values of the model’s parameters θ that best explain the observed data D if it had been generated under this chosen model?In other words, we might be interested in finding the values of θ that maximize the likelihood of having observed the ys we did given the Xs we did (i.
e.
the likelihood of the observed data D).
This is called the maximum likelihood.
Back to the gelato: imagine you recorded every single day where I was and whether I ate gelato or not (weird, but maybe my doctor is concerned I’m eating too much gelato).
The maximum likelihood estimator would output the θs that maximize the likelihood function.
The likelihood is the probability that we would have observed the ys (my gelato consumption) and Xs (the date and my location) if those ys were actually generated with those θs and those Xs.
Maximizing the likelihood gives us the combination of θs that seemingly worked the best at explaining the data.
Often, we try to search through every single combination of values for model parameters θs and calculate the probability that the model with that specific set of model parameter values generated our observed data D.
Of course, it’s impossible to search through EVERY combination of θ in most practical scenarios, so in traditional machine learning we instead optimize this search with algorithms like gradient descent, which choose which values of θ to test next given how well this current combination of values did at explaining the observed data.
We then select and report the combination of θs that maximized the likelihood of our data.
One issue with this approach is that we are reporting a single combination of θs as our best estimate, but someone else who sees this report will have no idea how confident we are for each reported parameter value.
This is an issue because observed data is a sample of the true population and inherently noisy, and no matter how much data we collect, we should never be 100% confident in any point estimate of a model parameter.
Consider if we had observed a different or limited sample of the population (e.
g.
what if I only remembered to record my gelato consumption 75% of the time, or only when I was in NYC?), would our reported θs change drastically or minimally?.Knowing the credible interval for each θ value embeds information about how representative we think our estimate is for a given model parameter based on the data we’ve seen, which may often represent the effect of a true physical phenomenon.
We also know very little about how knowledge of my gelato consumption might effect the probability of it having been a certain temperature or me being in a certain city (i.
e.
inference in the reverse direction).
Bayesian InferenceIn order to address this issue, I’d like to introduce Bayesian Inference, where Bayes Rule is our mantra:Eqn 1: Bayes RuleLet’s see how it can be applied to our goal.
It would be ideal to know the probability distribution of a parameter value θ under the model structure we’ve chosen, given the data we’ve observed: P_M (θ | D).
Plugging this into Bayes’ Rule:Eqn 2Eqn 3We can now rewrite Eqn 2 as:Eqn 4If we choose values of θ to test for our pre-selected model, it’s really easy to calculate P_M (D | θ) because we know all the inputs/output/parameters to this modeling function (recall this is called the likelihood), and we also know the prior P(θ) because it represents our beliefs about a specific θ (e.
g.
for a seemingly fair coin the θ is likely a normal centered on 0.
5, and depending on our beliefs of how fair the coin is, we might increase or decrease the variance of that normal).
The numerator of EQN 4 is quite simple to calculate for a chosen θ .
Ask us for those quantities for every single possible θ and we quickly realize we run into the same problem as before in traditional machine learning: we can’t sample every single θ!.In other words, the denominator of EQN 4 is nearly impossible to compute.
MCMCMarkov Chain Monte Carlo (MCMC) techniques were developed in order to intelligently sample θs rather than directly sum the likelihood and prior for every possible θ .
The main idea behind these techniques is similar to the θ update techniques we saw in traditional machine learning: we “update” with a new set of θ values based on our evaluation of the likelihood of the current set of θ values.
The big difference here is that we are sampling rather than updating — in other words, we are interested in the history of values we have explored.
This is important because we are no longer simply interested in knowing the single best estimate of each of the parameters (which is what we do in gradient descent), but rather we are interested in knowing the collection of “good” estimates for each of the parameters and how likely they are.
(i.
e.
the probability distribution of θ given D).
Let’s delve into a high level overview of MCMC algorithms and why they work here.
MCMC cares about tracking two things:θ_current: a single θ we are currently interested intrace_θ: a list of all θ_currents in orderMCMC begins by choosing a random initial value of θ:θ_current = θ_0trace_θ = [θ_0]and calculates the likelihood and prior (i.
e.
the numerator of Eqn 4).
The point of creating MCMC was that although the denominator is constant across all choices of θ, we can’t calculate the sum in the denominator directly, so let’s put that aside for now.
At this point, we have no idea whether our chosen θ_0 is a good choice.
We instead choose a proposal θ_1 (at this point, consider it magically chosen, but we’ll get back to that really soon) and if it produces a larger numerator in EQN 4 , we can agree it’s a better choice for θ.
This is because in EQN 4 the denominator is constant, no matter our choice of θ to plug into EQN 4.
Since we know θ_1 is better, let’s add it to our trace, update our current θ, and calculate the likelihood and prior (i.
e.
numerator of EQN 4):θ_current = θ_1trace_θ = [θ_0, θ_1 ]We can now choose a new θ_2 to propose (still a magical process, for now) and if it’s not as good at explaining the data (i.
e.
it produces a smaller numerator of Eqn 4) then we don’t immediately accept it this time, instead we accept it into our θ_current and trace_θ with probability ????:Eqn 5The reason for this is because we can reason that θ_2 is only ????.as good as θ_current.
This means that in our sampling history, we should expect the number of times we sample θ_2 to be a factor of ????.of the number of times we sample θ_current:Eqn 6In this regime, after sufficient sampling, you should have an intuition that we will have explored a lot of θ values and have saved them in the trace variable with frequency proportional to how well the θ value explains the data relative to other θs.
That’s exactly the distribution we were interested in!And lastly we return to the question of how to generate good proposal θs.
If we have no method and choose θs to propose at random among all possible values of θ, we will reject far too many samples (considering the true density of θ is probably relatively narrowly distributed).
Instead we draw θs from a proposal distribution q(θ_proposed | θ_current).
For example we might choose to make q a Normal with a fixed variance and mean θ_current.
This is the essence of one of the most popular MCMC algorithms called Metropolis-Hastings (MH).
We incorporate this into our acceptance ratio ????.in the following way:Eqn 7If we choose a symmetric proposal distribution (e.
g.
the Normal distribution), then:Eqn 8and it follows that ????_r = ????.
This is called the Random Walk MH algorithm, and you’ll end up with plots like this:Image 1The left is a KDE plot (essentially a smoothed histogram) for an “intercept_mu” parameter, and the right is the trace over time for each sampling chain.
We would be able to infer that although the most likely value of the parameter (the mode of the trace, a.
k.
a.
the MAP) is 0.
16, the credible interval of the parameter is likely between 0.
1 and 0.
21.
Appendix:For an easy visual explanation of this process, check out this video.
Most packages that help you do MCMC will generally run 3 or more traces (a.
k.
a.
“chains”) with different random initializations to ensure the chains are “converging”.
This is important because converging chains indicate your Markov Chain has stabilized.
I generally factor in a burn-in period, which discards the first few thousand samples that may depend on the random initialization of thetaThe choice of prior has influence on the outcome of the sampling.
Don’t choose far too narrow priors (as your bias will prevent proper exploration of the parameter space and your chains will fail to converge) or far too wide priors (called an uninformative prior, your chances of converging decrease too as you’ll spend too much time rejecting useless samples).
I usually use the mode or mean of the trace to report a most likely value, and the HPD (Highest Posterior Density) Interval in order to establish the credible interval.
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Pair-Table Problem
Dear Lammps users,
I met one question when using Pair_style:“table” and “pair_write” commands. in this simulation, I am trying to give potential energy and force as a function of distance. we had the correct potentials from previous simulations and calculated force from cubic spline fit of MATLAB at each R. then, when I submitted the input file for table style and asked for the energy and force values at specific distances in the output file from Pair_write, the values of force in pair_write output file were surprisingly different from the ones in input file.
here are the input file for Pair_ style: table and output file from Pair_write respectively:
pair_style table linear 6251
input file for Pair_style table looks like:
#Pair potential for linear-polymer Melt
entry1
N 6251 RSQ 0.000000000001 625.0
1 1.0000000E-06 1.690000 -1.0514040E-03
2 0.3162293 1.657552 0.2030731
3 0.4472147 1.625750 0.2818211
4 0.5477235 1.594545 0.3385073
5 0.6324564 1.563940 0.3834179
6 0.7071075 1.533921 0.4203998
7 0.7745973 1.504482 0.4516596
8 0.8366606 1.475607 0.4785343
.
.
.
6244 24.98600 5.0875130E-03 0.0000000E+00
6245 24.98800 5.0875130E-03 0.0000000E+00
6246 24.99000 5.0875130E-03 0.0000000E+00
6247 24.99200 5.0875130E-03 0.0000000E+00
6248 24.99400 5.0875130E-03 0.0000000E+00
6249 24.99600 5.0875130E-03 0.0000000E+00
6250 24.99800 5.0875130E-03 0.0000000E+00
6251 25.00000 5.0875130E-03 0.0000000E+00
and the output file from pair_write command called table.txt is:
pair_write * * 6251 r 0.000001 25.0 table.txt entry1
Pair potential table for atom types 1 1: i,r,energy,force
entry1
N 6251 R 1e-06 25
1 1e-06 1.69 -1051.4
2 0.004001 1.69 -4.20599e+06
3 0.008001 1.69 -8.4069e+06
4 0.012001 1.69 -1.25997e+07
5 0.016001 1.69 -1.67804e+07
6 0.020001 1.69 -2.0945e+07
7 0.024001 1.69 -2.50894e+07
8 0.028001 1.69 -2.92095e+07
9 0.032001 1.69 -3.33014e+07
10 0.036001 1.69 -3.7361e+07
11 0.040001 1.69 -4.13843e+07
12 0.044001 1.69 -4.53671e+07
13 0.048001 1.69001 -4.93056e+07
I chose Ntable = Nfile and RSQ being evenly spaced between rlorlo and rhirhi to avoid preliminary spline interpolation,i.e., interpolated table matches exactly the tabulated file. apparently, this approach does not help.
can anyone help me with this problem?
Thanks,
Mouge
Dear Lammps users,
I met one question when using Pair_style:"table" and "pair_write" commands.
in this simulation, I am trying to give potential energy and force as a
function of distance. we had the correct potentials from previous
simulations and calculated force from cubic spline fit of MATLAB at each R.
then, when I submitted the input file for table style and asked for the
energy and force values at specific distances in the output file from
Pair_write, the values of force in pair_write output file were surprisingly
different from the ones in input file.
here are the input file for Pair_ style: table and output file from
Pair_write respectively:
pair_style table linear 6251
input file for Pair_style table looks like:
#Pair potential for linear-polymer Melt
entry1
N 6251 RSQ 0.000000000001 625.0
1 1.0000000E-06 1.690000 -1.0514040E-03
2 0.3162293 1.657552 0.2030731
3 0.4472147 1.625750 0.2818211
4 0.5477235 1.594545 0.3385073
5 0.6324564 1.563940 0.3834179
6 0.7071075 1.533921 0.4203998
7 0.7745973 1.504482 0.4516596
8 0.8366606 1.475607 0.4785343
.
.
.
6244 24.98600 5.0875130E-03 0.0000000E+00
6245 24.98800 5.0875130E-03 0.0000000E+00
6246 24.99000 5.0875130E-03 0.0000000E+00
6247 24.99200 5.0875130E-03 0.0000000E+00
6248 24.99400 5.0875130E-03 0.0000000E+00
6249 24.99600 5.0875130E-03 0.0000000E+00
6250 24.99800 5.0875130E-03 0.0000000E+00
6251 25.00000 5.0875130E-03 0.0000000E+00
and the output file from pair_write command called table.txt is:
pair_write * * 6251 r 0.000001 25.0 table.txt entry1
# Pair potential table for atom types 1 1: i,r,energy,force
entry1
N 6251 R 1e-06 25
1 1e-06 1.69 -1051.4
2 0.004001 1.69 -4.20599e+06
3 0.008001 1.69 -8.4069e+06
4 0.012001 1.69 -1.25997e+07
5 0.016001 1.69 -1.67804e+07
6 0.020001 1.69 -2.0945e+07
7 0.024001 1.69 -2.50894e+07
8 0.028001 1.69 -2.92095e+07
9 0.032001 1.69 -3.33014e+07
10 0.036001 1.69 -3.7361e+07
11 0.040001 1.69 -4.13843e+07
12 0.044001 1.69 -4.53671e+07
13 0.048001 1.69001 -4.93056e+07
I chose Ntable = Nfile and RSQ being evenly spaced between rlo*rlo and
rhi*rhi to avoid preliminary spline interpolation,i.e., interpolated table
matches exactly the tabulated file. apparently, this approach does not help.
can anyone help me with this problem?
you get what you ask for, where is the problem?
since you ask to write out a linear spaced R style table from an RSQ
table, you enforce that the potential is being resplined, and since
your potential is strongly curved at the low r end, you cannot avoid
that the spline function is behaving the way it does.
axel.
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latest
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http://www.ck12.org/book/CK-12-Math-Analysis/r19/section/8.2/
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# 8.2: Computing Limits
Difficulty Level: At Grade Created by: CK-12
## Learning objectives
• Demonstrate an understanding of the limit of a function from an algebraic perspective.
• Learn how to compute the limit for different kinds of functions.
## Introduction
In this lesson, we will present the limit of a function from an algebraic perspective. We will base the algebraic technique on the results obtained from the previous section. We first present the basic theorems that provide the tools necessary to calculate limits. These theorems are outlined in the two boxes below.
Box #1: Important Theorems of Limits
Let a be a real number and suppose that limxaf(x)=L1\begin{align*}\lim_{x\rightarrow a}f(x) = L_1 \end{align*} and limxag(x)=L2\begin{align*}\lim_{x\rightarrow a}g(x) = L_2 \end{align*}.
Then:
1. limxa[f(x)+g(x)]=limxaf(x)+limxag(x)=L1+L2\begin{align*} \lim_{x\rightarrow a} [f(x) + g(x)] = \lim_ {x\rightarrow a}f(x) + \lim_{x\rightarrow a}g(x) = L_1 + L_2\end{align*}, meaning the limit of the sum is the sum of the limits.
2. limxa[f(x)g(x)]=limxaf(x)limxag(x)=L1L2\begin{align*} \lim_{x\rightarrow a} [f(x) - g(x)] = \lim_ {x\rightarrow a}f(x) - \lim_{x\rightarrow a}g(x) = L_1 - L_2\end{align*}, meaning the limit of the difference is the difference of the limits
3. limxa[f(x)g(x)]=(limxaf(x))(limxag(x))=L1L2\begin{align*} \lim_{x\rightarrow a} [f(x)g(x)] = ( \lim_ {x\rightarrow a}f(x)) ( \lim_{x\rightarrow a}g(x)) = L_1L_2\end{align*}, meaning the limit of the product is the product of the limits.
4. limxaf(x)g(x)=limxaf(x)limxag(x)=L1L2L20\begin{align*} \lim_{x\rightarrow a} \frac{f(x)}{g(x)} =\frac{\lim_{x\rightarrow a} f(x)} {\lim_{x\rightarrow a} g(x)} = \frac {L_1} {L_2} L_2 \neq 0\end{align*}, meaning the limit of a quotient is the quotient of the limits (provided that the denominator does not equal zero.)
5. limxaf(x)n=limxaf(x)n=Ln1L1>0\begin{align*} \lim_{x\rightarrow a} \sqrt[n]{f(x)}= \sqrt[n] { \lim_{x\rightarrow a} f(x)} = \sqrt[n] L_1 L_1 > 0\end{align*} if n is even, meaning the limit of the nth root is the nth of the limit.
Other useful results follow from the above theorems:
Box #2
1. If a and k are real numbers, then limxak=k\begin{align*}\lim_{x\rightarrow a}k = k\end{align*}. That is, if f(x) = k, a constant function, then the values of f(x) do not change as x is varied, thus the limit of f(x) is k.
2. If a is a real number then limxax=a\begin{align*}\lim_{x\rightarrow a} x= a \end{align*}. That is, since f(x) = x is an identity function (its input equals its output), then as xa, f(x) = xa.
3. limxa(kf(x))=(limxak)(limxaf(x))=k(limxaf(x))\begin{align*}\lim_{x \rightarrow a} (k \cdot f(x)) = ( \lim_{x \rightarrow a} k) \cdot (\lim_{x \rightarrow a} f(x)) = k \cdot ( \lim_{x \rightarrow a} f(x))\end{align*}
4. limxaxn=(limxax)n=an\begin{align*}\lim_{x \rightarrow a} x^n = ( \lim_{x \rightarrow a} x)^n = a^n\end{align*}
## Limits of Polynomial Functions
Example 1: Find limx1(x23x+4)\begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4)\end{align*} and justify each step.
Solution: Using Equation (1) of Box #1 (the limit of the sum is the sum of the limits) we get
limx1(x23x+4)=limx1x2+limx1(3x)+limx14\begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4) = \lim_{x \rightarrow 1}x^2 + \lim_{x \rightarrow 1}(-3x) + \lim_{x \rightarrow 1} 4\end{align*}
From Equation (4) of Box #2, the first term becomes
limx1x2=(1)2=1\begin{align*}\lim_{x \rightarrow 1}x^2 = (1)^2 = 1\end{align*}
From Equations (2) and (3) of Box #2, the second term becomes
limx1(3x)=3limx1x=(3)(1)=3\begin{align*}\lim_{x \rightarrow 1}(-3x) = -3 \lim_{x \rightarrow 1}x = (-3)(1) = -3\end{align*}
Finally, from Equation (1) of Box #2, the third term becomes
limx14=4\begin{align*}\lim_{x \rightarrow 1}4 = 4\end{align*}
Thus the limit of the above polynomial is
limx1(x23x+4)=1+(3)+4=2\begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4) = 1 + (-3) + 4 = 2\end{align*}
However, for conciseness, we can compute the limit by simply substituting x = 1 directly into the polynomial,
limx1(x23x+4)=(1)23(1)+4=2\begin{align*}\lim_{x \rightarrow 1}(x^2 - 3x + 4) = (1)^2 - 3(1) + 4 = 2\end{align*}
Theorem: The limit of a polynomial
For any polynomial f(x) = cnxn + . . . + c1x + c0 and any real number a,\begin{align*}\lim_{x \rightarrow a}f(x) = c_n(a)^n + . . . + c_1(a) + c_0\end{align*}
\begin{align*}\lim_{x \rightarrow a}f(x) = f(a)\end{align*}
In other words, the limit of the polynomial is simply equal to f(a).
Example 2: Find \begin{align*}\lim_{x\rightarrow 3} (4x^3 - 4x - 5)\end{align*}
Solution: According to the theorem above, \begin{align*}\lim_{x\rightarrow 3}(4x^3 - 4x - 5) = 4(3)^3 - 4(3) - 5 =91\end{align*}
Example 3: Find \begin{align*}\lim_{x\rightarrow 5} \frac{x^2 - 4} {3x^2 - 2} \end{align*}
Solution: Using Equation (4) of Box #1 (the limit of the quotient is the quotient of the limit),
\begin{align*}\lim_{x \rightarrow 5} \frac{x^2 - 4} {3x^2 - 2} = \frac {\lim_{x \rightarrow 5} (x^2 - 4)} {\lim_{x \rightarrow 5}(3x^2 - 2)}\end{align*}
Making use of the limit of the polynomials theorem, we obtain,
\begin{align*}\lim_{x \rightarrow 5} \frac{x^2 - 4} {3x^2 - 2}\end{align*} \begin{align*}= \frac{(5)^2 - 4} {3(5)^2 - 2}\end{align*}
\begin{align*}\mathit = \frac{21} {73}\end{align*}
## Limits of Rational Functions
In Example 3 we found the limit of a rational function. Sometimes finding the limit of a rational function at a point a is difficult because evaluating the function at the point a leads to a denominator equal to zero. The box below describes finding the limit of a rational function.
Theorem: The limit of a Rational Function
For the rational function \begin{align*}f(x) = \frac{p(x)} {q(x)}\end{align*} and any real number a,
\begin{align*}\lim_{x \rightarrow a} f(x) = \frac{p(a)} {q(a)} \text{ if } q(a) \neq 0\end{align*}.
However, if \begin{align*} \,\! q(a) = 0 \end{align*} then the function may or may not exist. See Examples 5, 6, and 7 below.
Example 4: Find \begin{align*}\lim_{x \rightarrow 3} \frac{2 - x} {x - 2}\end{align*}.
Solution: Using the theorem above, we simply substitute x = 3: \begin{align*}\lim_{x \rightarrow 3} \frac{2 - x} {x - 2} = \frac{2 - 3} {3 - 2} = -1\end{align*}
Example 5: Find \begin{align*}\lim_{x \rightarrow 3} \frac{x + 1} {x - 3}\end{align*}.
Solution: Notice that the domain of the function is continuous (defined) at all real numbers except at x = 3. If we check the one-sided limits we see that \begin{align*}\lim_{x \rightarrow 3^{+}} \frac{x + 1} {x - 3}=\infty\end{align*} and \begin{align*}\lim_{x \rightarrow 3^{-}} \frac{x + 1} {x - 3}=-\infty\end{align*}. Because the one-sided limits are not equal, the limit does not exist.
Example 6: Find \begin{align*}\lim_{x \rightarrow 2} \frac{x^2 - 4} {x - 2}\end{align*}.
Solution: Notice that the function here is discontinuous at x = 2, that is, the denominator is zero at x = 2. However, it is possible to remove this discontinuity by canceling the factor x-2 from both the numerator and the denominator and then taking the limit:
\begin{align*}\lim_{x \rightarrow 2} \frac{x^2 -4} {x - 2} = \lim_{x \rightarrow 2} \frac{(x - 2)(x + 2)} {x - 2} = \lim_{x \rightarrow 2} (x + 2) = 4\end{align*}
This is a common technique used to find the limits of rational functions that are discontinuous at some points. When finding the limit of a rational function, always check to see if the function can be simplified. We will use this technique again in the next example.
Example 7: Find \begin{align*}\lim_{x \rightarrow 3} \frac{2x - 6} {x^2 + x - 12}\end{align*}.
Solution: The numerator and the denominator are both equal to zero at x = 3, but there is a common factor x - 3 that can be removed (that is, we can simplify the rational function):
\begin{align*}\lim_{x \rightarrow 3} \frac{2x - 6} {x^2 + x - 12}\end{align*} \begin{align*}= \lim_{x \rightarrow 3} \frac{2(x - 3)} {(x + 4) (x - 3)}\end{align*}
\begin{align*}= \lim_{x \rightarrow 3} \frac{2} {x + 4}\end{align*}
\begin{align*}= \frac{2} {7}\end{align*}
## Computing Limits Using One-Sided Limits
When we wish to find the limit of a function f(x) as it approaches a point a and we cannot evaluate f(x) at a because it is undefined at that point, we can compute the function's one-sided limits in order to find the desired limit. If its one-sided limits are the same, then the desired limit exists and is the value of the one-sided limits. If its one-sided limits are not the same, then the desired limit does not exist. This technique is used in the examples below.
Example 8: Find the limit f(x) as x approaches 1. That is, find \begin{align*}\lim_{x \rightarrow 1} f(x)\end{align*} if
\begin{align*}f(x) = \begin{cases}3 - x, & x < 1 \\ 3x - x^2, & x > 1 \end{cases}\end{align*}
Solution: Remember that we are not concerned about finding the value of f(x) at x but rather near x. So, for x < 1 (limit from the left),
\begin{align*}\lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (3 - x) = (3 - 1) = 2\end{align*}
and for x > 1 (limit from the right),
\begin{align*}\lim_{x \rightarrow 1^+} f(x) = \lim_{x \rightarrow 1^+} (3x - x^2) = 2\end{align*}
Now since the limit exists and is the same on both sides, it follows that
\begin{align*}\lim_{x \rightarrow 1} f(x) = 2\end{align*}
Example 9: Find \begin{align*}\lim_{x \rightarrow 2} \frac{3} {x - 2}\end{align*}.
Solution: From the figure below we see that \begin{align*}f(x)=\frac{3} {x - 2}\end{align*} decreases without bound as x approaches 2 from the left and \begin{align*}f(x)=\frac{3} {x - 2}\end{align*} increases without bound as x approaches 2 from the right.
This means that \begin{align*}\lim_{x \rightarrow 2^-} \frac{3} {x - 2} = -\infty\end{align*} and \begin{align*}\lim_{x \rightarrow 2^+} \frac{3} {x - 2} = +\infty\end{align*}. Since f(x) is unbounded (infinite) in either directions, the limit does not exist.
## Lesson Summary
In this lesson we learned many techniques for computing limits. In particular, we learned that we can find the limit of a polynomial function approaching a point a by evaluating the function at a. We also learned that the limits of some rational functions do not exist, but sometimes we can simplify the functions in order to compute their limits. Just because a function is discontinuous at a given point does not necessarily mean that it's limit does not exist at that point. Finally, we learned that limits can often be computed by finding both one-sided limits and checking to see if they are equal.
## Points to Consider
1. Can you prove that the theorems in the boxes at the beginning of this lesson are true?
2. Why does it make sense that a limit can exist for a function at a point of discontinuity (as in Example 6)?
## Review Questions
1. Find \begin{align*} \lim_{x \rightarrow -2} (2)\end{align*}.
2. Find \begin{align*} \lim_{x \rightarrow 0^+} (\pi)\end{align*}.
3. Find \begin{align*} \lim_{x \rightarrow 2} \frac{x^2 - 4} {x - 2}\end{align*}.
4. Find \begin{align*} \lim_{x \rightarrow 6} \frac{x - 6} {x^2 - 36}\end{align*}.
5. Find \begin{align*} \lim_{x \rightarrow 5} \sqrt{x^3 - 2x - 1}\end{align*}.
6. Find \begin{align*} \lim_{x \rightarrow 3^+} \frac{3} {x - 3}\end{align*}.
7. Show that \begin{align*}\lim_{x \rightarrow 0^+} \left (\frac{1} {x} - \frac{1} {x^2}\right ) = -\infty\end{align*}.
8. For an object in free fall, such as a stone falling off a cliff, the distance y(t) (in meters) that the object falls in t seconds is given by the kinematic equation y(t) = 4.9 t2. The object’s velocity after 2 seconds is given by \begin{align*}v(t) = \lim_{t \rightarrow 2} \frac{y(t) - y(2)} {t - 2}\end{align*}. What is the velocity of the object after 2 seconds?
1. 2
2. \begin{align*}\pi\end{align*}
3. 4
4. \begin{align*}\frac{1} {12}\end{align*}
5. \begin{align*}\sqrt{114}\end{align*}
6. \begin{align*}+\infty\end{align*}
7. Hint: Use a graph.
8. 19.6 m/sec
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Date Created:
Feb 23, 2012
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In studying Java development, the HasA and IsA relationships came up in the coursework. Is-A relationships are forms of inheritance. A car is a vehicle. So a vehicle is the base class and car would inherit from it. Alternatively, a table has a top, legs, maybe some drawers. This is an example of composition.
## Code: Inheritance
Suppose in an inheritance example we model a building as the base class:
``````public class Building {
private int walls;
private int height;
private int squareFeet;
private int numberOfDoors;
private boolean skylight;
private String flooringMaterial;
public Building(int walls, int height, int squareFeet, int numberOfDoors, boolean skylight, String flooringMaterial) {
this.walls = walls;
this.height = height;
this.squareFeet = squareFeet;
this.numberOfDoors = numberOfDoors;
this.skylight = skylight;
this.flooringMaterial = flooringMaterial;
}``````
As there are many types of buildings, each would extend from this (inherit) base class. For example, a condo Is-A building and could be defined like so:
``````public class Condo extends Building {
private boolean hoa;
private boolean sharedWall;
private int numberOfBedrooms;
private int numberOfBathrooms;
public Condo(int walls, int height, int squareFeet, int numberOfDoors, boolean skylight, String flooringMaterial,
boolean hoa, boolean sharedWall, int numberOfBathrooms, int numberOfBedrooms) {
super(walls, height, squareFeet, numberOfDoors, skylight, flooringMaterial);
this.hoa = hoa;
this.sharedWall = sharedWall;
this.numberOfBathrooms = numberOfBathrooms;
this.numberOfBedrooms = numberOfBedrooms;
}``````
## Code: Composition
Conversely, in composition, a relationship between code is specified in a Has-A relationship. Going back to the table analogy:
``````public class TableTop {
private String color;
private String material;
private Drawer drawer; // Composition, not inheritance.
public TableTop(String color, String material, Drawer drawer) {
this.color = color;
this.material = material;
this.drawer = drawer;
}
public String getColor() {
System.out.println("tabletop color: " + color);
return color;
}
public void setColor(String color) {
this.color = color;
}
public String getMaterial() {
System.out.println("tabletop material: "+ material);
return material;
}
public void setMaterial(String material) {
this.material = material;
}
public Drawer getDrawer() {
return drawer;
}
public void setDrawer(Drawer drawer) {
this.drawer = drawer;
}
}
``````
In the above, our TableTop class references another class called Drawer, so our table top might have a drawer hooked to the underside. I could also have chosen to use Legs… or something else.
The other class is defined:
``````public class Drawer {
private boolean sliding;
private int numberOfDrawers;
public Drawer(boolean sliding, int numberOfDrawers) {
this.sliding = sliding;
this.numberOfDrawers = numberOfDrawers;
}
public boolean isSliding() {
System.out.println("drawer sliding?: "+ sliding);
return sliding;
}
public void setSliding(boolean sliding) {
this.sliding = sliding;
}
public int getNumberOfDrawers() {
System.out.println("drawer number of: "+ numberOfDrawers);
return numberOfDrawers;
}
public void setNumberOfDrawers(int numberOfDrawers) {
this.numberOfDrawers = numberOfDrawers;
}
}``````
The table also has a Legs class:
``````public class Leg {
private int length;
private String material;
private Foot foot;
public Leg(int length, String material, Foot foot) {
this.length = length;
this.material = material;
this.foot = foot;
}
public int getLength() {
System.out.println("leg length: "+ length);
return length;
}
public void setLength(int length) {
this.length = length;
}
public String getMaterial() {
System.out.println("leg material: " + material);
return material;
}
public void setMaterial(String material) {
this.material = material;
}
public Foot getFoot() {
return foot;
}
public void setFoot(Foot foot) {
this.foot = foot;
}
}``````
Each leg has a foot, and that foot can be made of specific material, using a design pattern, and a color:
``````public class Foot {
private String color;
private String material;
private int designPattern;
public Foot(String color, String material, int designPattern) {
this.color = color;
this.material = material;
this.designPattern = designPattern;
}
public String getColor() {
System.out.println("foot color: "+color);
return color;
}
public void setColor(String color) {
this.color = color;
}
public String getMaterial() {
System.out.println("foot material: " + material);
return material;
}
public void setMaterial(String material) {
this.material = material;
}
public int getDesignPattern() {
System.out.println("foot pattern: "+designPattern);
return designPattern;
}
public void setDesignPattern(int designPattern) {
this.designPattern = designPattern;
}
}
``````
After each part of the composition is created, then they are assembled into an object (in this case a table) like so:
``````public class Table {
private TableTop tableTop;
private Leg leg;
public Table(TableTop tableTop, Leg leg) {
this.tableTop = tableTop;
this.leg = leg;
}
public void details() {
tableTop.getColor();
tableTop.getDrawer().getNumberOfDrawers();
tableTop.getDrawer().isSliding();
tableTop.getMaterial();
leg.getLength();
leg.getMaterial();
leg.getFoot().getColor();
leg.getFoot().getDesignPattern();
leg.getFoot().getMaterial();
}
public TableTop getTableTop() {
return tableTop;
}
public Leg getLeg() {
System.out.println(leg);
return leg;
}
}
``````
Instantiating the object is handled in each of the elements like so:
`````` TableTop theTableTop = new TableTop("redwood", "pressed wood", new Drawer(true, 1));
Leg theLeg = new Leg(36, "pressed wood", new Foot("brass", "brass", 22));
Table theTable = new Table(theTableTop, theLeg);
theTable.details();``````
Which would output something like:
tabletop color: redwood
drawer number of: 1
drawer sliding?: true
tabletop material: pressed wood
leg length: 36
leg material: pressed wood
foot color: brass
foot pattern: 22
foot material: brass
## Aggregation
A third type of OO methodology of code sharing is that of Aggregation. As described on this StackOverflow answer, aggregation is a Has-An type relationship.
The difference between a Has-A and Has-An is that the Has-A relationship is required. A table needs legs. The composition above assumes a requirement of each component. A house has walls, and walls are required.
A Has-An relationship is similar to Has-A, but one or more elements is not required. A car has a wheel(s), but it has an driver. The driver is not required for there to be a car.
#
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# Differentiate
Question:
Differentiate
$\frac{e^{x}}{\left(1+x^{2}\right)}$
Solution:
To find: Differentiation of $\frac{e^{x}}{\left(1+x^{2}\right)}$
Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)
(ii) $\frac{d e^{x}}{d x}=e^{x}$
(iii) $\frac{d x^{n}}{d x}=n x^{n-1}$
Let us take $u=e^{x}$ and $v=\left(1+x^{2}\right)$
$u^{\prime}=\frac{d u}{d x}=\frac{d\left(e^{x}\right)}{d x}=e^{x}$
$v^{\prime}=\frac{d v}{d x}=\frac{d\left(1+x^{2}\right)}{d x}=2 x$
Putting the above obtained values in the formula:-
$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)
$\left(\frac{e^{x}}{\left(1+x^{2}\right)}\right)^{\prime}=\frac{e^{x} \times\left(1+x^{2}\right)-e^{x} \times 2 x}{\left(1+x^{2}\right)^{2}}$
$=\frac{e^{x}\left(x^{2}-2 x+1\right)}{\left(1+x^{2}\right)^{2}}$
$=\frac{e^{x}(x-1)^{2}}{\left(1+x^{2}\right)^{2}}$
Ans $)=\frac{e^{x}(x-1)^{2}}{\left(1+x^{2}\right)^{2}}$
Leave a comment
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https://www.encyclopediaofmath.org/index.php/Hopf_manifold
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# Hopf manifold
A complex Hopf manifold is a quotient of by the infinite cyclic group of holomorphic transformations generated by , , . It is usually denoted by . As a differentiable manifold, any is diffeomorphic to . Consequently, when the first Betti number ; hence, such provide examples of compact complex manifolds (cf. Complex manifold) not admitting any Kähler metric (note that, for , is a -dimensional complex torus). However, these manifolds do carry a particularly interesting class of Hermitian metrics (cf. Hermitian metric), namely locally conformal Kähler metrics (a Hermitian metric is locally conformal Kähler if its fundamental -form satisfies the integrability condition with a closed -form , called the Lee form). An example of such a metric is the projection of the following metric on : , whose Lee form is . It is parallel with respect to the Levi-Civita connection of . This originated the study of the more general class of generalized Hopf manifolds: locally conformal Kähler manifolds with parallel Lee form. Their geometry is closely related to Sasakian and Kählerian geometries. Generically, a compact generalized Hopf manifold arises as the total space of a flat, principal bundle over a compact Sasakian orbifold and, on the other hand, fibres into -dimensional complex tori over a Kähler orbifold.
I. Vaisman conjectured that a compact locally conformal Kähler manifold that is not globally conformal Kähler must have an odd Betti number. To date (1996), this has only been proved for generalized Hopf manifolds (see [a5]).
It is rather difficult to characterize the Hopf manifolds among the locally conformal Kähler manifolds. However, in complex dimension several such characterizations are available; for instance, the only compact complex surface with and admitting conformally flat Hermitian metrics is (cf. [a4]; see also [a1]).
By analogy, "real Hopf manifoldreal Hopf manifolds" were defined as (compact) quotients of by an appropriate group of conformal transformations (see [a6] and [a2]). Similarly, "quaternion Hopf manifoldquaternion Hopf manifolds" are defined as quotients of (see [a3]).
#### References
[a1] C.P. Boyer, "Conformal duality and compact complex surfaces" Math. Ann. , 274 (1986) pp. 517–526 [a2] P. Gauduchon, "Structures de Weyl–Einstein, espaces de twisteurs et variétés de type " J. Reine Angew. Math. , 455 (1995) pp. 1–50 [a3] L. Ornea, P. Piccinni, "Locally conformal Kähler structures in quaternionic geometry" Trans. Amer. Math. Soc. , 349 (1997) pp. 641–655 [a4] M. Pontecorvo, "Uniformization of conformally flat Hermitian surfaces" Diff. Geom. Appl. , 3 (1992) pp. 295–305 [a5] I. Vaisman, "Generalized Hopf manifolds" Geom. Dedicata , 13 (1982) pp. 231–255 [a6] I. Vaisman, C. Reischer, "Local similarity manifolds" Ann. Mat. Pura Appl. , 35 (1983) pp. 279–292 [a7] S. Dragomir, L. Ornea, "Locally conformal Kähler geometry" , Birkhäuser (1997)
How to Cite This Entry:
Hopf manifold. L. Ornea (originator), Encyclopedia of Mathematics. URL: http://www.encyclopediaofmath.org/index.php?title=Hopf_manifold&oldid=18689
This text originally appeared in Encyclopedia of Mathematics - ISBN 1402006098
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https://stacks.math.columbia.edu/tag/0GIL
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Lemma 10.159.4. Let $A$ be a ring. Let $\kappa = \max (|A|, \aleph _0)$. Then every flat $A$-algebra $B$ is the filtered colimit of its flat $A$-subalgebras $B' \subset B$ of cardinality $|B'| \leq \kappa$. (Observe that $B'$ is faithfully flat over $A$ if $B$ is faithfully flat over $A$.)
Proof. If $B$ has cardinality $\leq \kappa$ then this is true. Let $E \subset B$ be an $A$-subalgebra with $|E| \leq \kappa$. We will show that $E$ is contained in a flat $A$-subalgebra $B'$ with $|B'| \leq \kappa$. The lemma follows because (a) every finite subset of $B$ is contained in an $A$-subalgebra of cardinality at most $\kappa$ and (b) every pair of $A$-subalgebras of $B$ of cardinality at most $\kappa$ is contained in an $A$-subalgebra of cardinality at most $\kappa$. Details omitted.
We will inductively construct a sequence of $A$-subalgebras
$E = E_0 \subset E_1 \subset E_2 \subset \ldots$
each having cardinality $\leq \kappa$ and we will show that $B' = \bigcup E_ k$ is flat over $A$ to finish the proof.
The construction is as follows. Set $E_0 = E$. Given $E_ k$ for $k \geq 0$ we consider the set $S_ k$ of relations between elements of $E_ k$ with coefficients in $A$. Thus an element $s \in S_ k$ is given by an integer $n \geq 1$ and $a_1, \ldots , a_ n \in A$, and $e_1, \ldots , e_ n \in E_ k$ such that $\sum a_ i e_ i = 0$ in $E_ k$. The flatness of $A \to B$ implies by Lemma 10.39.11 that for every $s = (n, a_1, \ldots , a_ n, e_1, \ldots , e_ n) \in S_ k$ we may choose
$(m_ s, b_{s, 1}, \ldots , b_{s, m_ s}, a_{s, 11}, \ldots , a_{s, nm_ s})$
where $m_ s \geq 0$ is an integer, $b_{s, j} \in B$, $a_{s, ij} \in A$, and
$e_ i = \sum \nolimits _ j a_{s, ij} b_{s, j}, \forall i, \quad \text{and}\quad 0 = \sum \nolimits _ i a_ i a_{s, ij}, \forall j.$
Given these choicse, we let $E_{k + 1} \subset B$ be the $A$-subalgebra generated by
1. $E_ k$ and
2. the elements $b_{s, 1}, \ldots , b_{s, m_ s}$ for every $s \in S_ k$.
Some set theory (omitted) shows that $E_{k + 1}$ has at most cardinality $\kappa$ (this uses that we inductively know $|E_ k| \leq \kappa$ and consequently the cardinality of $S_ k$ is also at most $\kappa$).
To show that $B' = \bigcup E_ k$ is flat over $A$ we consider a relation $\sum _{i = 1, \ldots , n} a_ i b'_ i = 0$ in $B'$ with coefficients in $A$. Choose $k$ large enough so that $b'_ i \in E_ k$ for $i = 1, \ldots , n$. Then $(n, a_1, \ldots , a_ n, b'_1, \ldots , b'_ n) \in S_ k$ and hence we see that the relation is trivial in $E_{k + 1}$ and a fortiori in $B'$. Thus $A \to B'$ is flat by Lemma 10.39.11. $\square$
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
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https://scicomp.stackexchange.com/questions/33539/lapack-symmetric-update-b-1ab-t?noredirect=1
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# Lapack symmetric update $B^{-1}AB^{-T}$
Does Lapack have a routine that, given symmetric $$A=A^T$$ and $$B$$, computes the symmetric matrix $$B^{-1}AB^{-T}$$ (while preserving symmetry exactly)?
It would be enough to have this routine for triangular $$B$$, because of course the first step here will be computing a factorization of $$B$$ (LU or LDL, according to whether $$B$$ is symmetric, too).
Motivation: I think it would be useful in this question, especially if one factorizes the expression as suggested in the second part of my answer.
Unfortunately I don't think there's a great way to do this, at least not without some effort.
In the event that $$\mathbf A$$ is nonsingular, it might be useful to point out that the desired matrix $$\mathbf C = \mathbf B^{-1} \mathbf A \mathbf B^{-T}$$ is the inverse of the symmetric Schur complement $$\mathbf S = \mathbf B^T \mathbf A^{-1} \mathbf B$$. I find that BLAS/LAPACK has much better support for forming a matrix like $$\mathbf S$$ than one like $$\mathbf C$$, so maybe it's profitable to compute $$\mathbf S$$ first, then apply a symmetry-preserving inversion routine? The utility of this idea depends on the conditioning of $$\mathbf A$$ and $$\mathbf B$$.
If you're lucky enough that $$\mathbf A$$ is positive definite in addition to symmetric, you can use [potrf] to Cholesky factor/overwrite $$\mathbf A = \mathbf L \mathbf L^T$$. Then $$\mathbf S = \mathbf B^T \left(\mathbf L \mathbf L^T \right)^{-1} \mathbf B = \left( \mathbf L^{-1} \mathbf B \right) ^T \left( \mathbf L^{-1} \mathbf B \right)$$. Using [trsm], you can overwrite $$\tilde{\mathbf B} = \mathbf L^{-1} \mathbf B$$, then use [syrk] to compute $$\mathbf S = \tilde{\mathbf B}^T \tilde{\mathbf B}$$ into a temporary (or just overwrite $$\mathbf A$$). If $$\mathbf B$$ has full column rank, then $$\mathbf S$$ is also positive definite, and the desired output $$\mathbf C = \mathbf S^{-1}$$ can be computed using [potrf] followed by [potri]. The symmetry of $$\mathbf A$$, $$\mathbf S$$ and $$\mathbf C$$ is enforced explicitly at the API level, because all of [syrk], [potrf] and [potri] operate only upon a single triangle (with the opposing triangle assumed to match).
When $$\mathbf A$$ is invertible but indefinite, you can follow the same basic idea but computing $$\mathbf S$$ is more involved due to pivoting considerations. If you'd like to see that procedure too, just leave a comment and I will add it as an edit.
• Thanks! Even if $A$ is invertible, I would be worried about its conditioning. However, I think the idea you suggested works also without the inversion part: factor $A=LL^T$ (or $LDL^T$ if indefinite), compute $C=B^{-1}L$, then the sought quantity is $CC^T$. This might even be cheaper than the other alternatives in terms of flops. Thanks for the good idea! Oct 10, 2019 at 16:46
• It's probably pretty close to a wash but only experimentation can show for certain. The answer proposes 2 [potrf]'s, 1 [trsm], 1 [syrk], 1 [potri]. The comment proposes 1 [getrf], 2 [trsm's], 1 [syrk]. Recalling that [getrf] is twice as expensive as [potrf], in the end it just boils down to 1 [potri] vs 1 [trsm]. It's worth noting that the [potri] might actually not be essential, because perhaps leaving the output $\mathbf C$ in factored form, as a solve-by-$\mathbf S$ operator, is sufficient? Might depend on context. Oct 10, 2019 at 17:38
• I will re-emphasize, though, that the indefinite case is messier. Unlike the positive case [potrf], the analogous indefinite factorization [sytrf] does not render $\mathbf L$ in a form that is suitable for other BLAS routines like [trsm], instead $\mathbf L$ is all tangled up with $\mathbf D$. Furthermore, the desired output is not $\mathbf C \mathbf C^T$ anymore, it's $\mathbf C \mathbf D \mathbf C^T$. This quantity cannot be computed readily using [syrk]/BLAS. Oct 10, 2019 at 17:46
• I agree with you that the indefinite case is messier. However, I am not sure about your complexity estimates for the posdef case. The method suggested in my comment needs a potrf for $LL^T$, a trsm for $B^{-1}L$ (and maybe there is still something to gain here because the RHS is triangular), and a syrk for $CC^T$. That's strictly less than the answer, am I missing anything? Oct 10, 2019 at 17:48
• I've assumed there's two trsms for $\mathbf B$ / for applying $\mathbf B^{-1}$, as it is factored into LU. Unless $\mathbf B$ itself is also symmetric? Oct 10, 2019 at 17:49
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## PROG0225 - Trilaterations
Trilateration is the process of determining (relative) locations or points by measurement of distances from this point to a few other, known points. It is based on the geometric properties of circles. Unlike with triangluation, there is no measuring of angles in trilateration. This technique has a couple of very practical applications, such as the GPS system, determining the epicenter of an earthquake…
The point where three circles intersect is the epicenter of an earthquake. This technique is called triangulation.
To determine the epicenter of an earthquake, for example, you can work as follows. You calculate the time between the first P-wave and the first S-wave for various weather stations. You can determine the distance from the epicenter to the weather station based on known tables. In the picture below you can see that a difference of 24 seconds between the first P-wave and the first S-wave corresponds with an epicenter in a 215 km radius. Now you can draw a circle around the area where the epicenter is situated. If you have multiple measuring points, you will find the epicenter in the intersection of multiple circles. In reality, there will always be a mistake in the measurement and the circles will never really intersect. You will however find an area of possibilities for the location of the epicenter.
The amplitude is used to determine both the strength of an earthquake and the distance between the weather station and the epicenter.
An important feature of trilateration is determining the position of two circles. In the table below you will find an overview of the mutual positions possible for two circles. $d$ stands for the distance between the centres of both circles. $r_1$ and $r_2$ respectively stand for the radius of the first and the second circle.
position condition
concentric $d=0$
touching inside $d=|r_1-r_2|$
touching outside $d=r_1+r_2$
enclosing $d < |r_1-r_2|$
intersecting $|r_1-r_2| < d < r_1+r_2$
separate $d > r_1+r_2$
The distance $d$ between two points $(x_1,y_1)$ and $(x_2,y_2)$ is calculated with the formula below:
$d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$.
Caution: If you work with real numbers on a computer, mistakes in rounding off occur very often. This is why 2 numbers that should be equal, often aren't. To fix this problem, we add a margin of error: we say that two numbers are equal if the difference between those two numbers is smaller than $10^{-2}$.
### Input
The input consists of six real numbers: one number per line. The first line represents the $x$ co-ordinate of the first circle, followed by a line with the $y$ co-ordinate of the circle and a line with the circle's radius. Then, this is repeated with the quantities of the second circle.
### Output
The output consists of a single line that contains one of six terms: concentric, touching inside, touching outside, enclosing, intersecting or separate. The term should state the mutual position of the two circles from the input.
### Example
Input:
3.5
2.7
7.86
6.5
6.7
2.86
Output:
touching inside
Trilateratie is het bepalen van de (relatieve) positie van een punt aan de hand van de afstanden van dit punt tot enkele gekende punten. Hierbij wordt gesteund op de meetkundige eigenschappen van cirkels. In tegenstelling tot triangulatie worden er bij trilateratie geen hoeken gemeten. Deze techniek heeft enkele zeer praktische toepassingen zoals het GPS-systeem, het bepalen van het epicentrum van een aardbeving, …
Het punt waar de drie cirkels elkaar snijden is het epicentrum van de aardbeving. Deze techniek word triangulatie genoemd.
Om bvb. het epicentrum van een aardbeving te bepalen kan je als volgt te werk gaan. Je bepaalt voor verschillende meetstation de tijd tussen de eerste P-golf en de eerste S-golf. Aan de hand van gekende tabellen kan je hieruit het afstand van het epicentrum tot het meetstation bepalen. Zo zie je op onderstaande afbeelding dat een verschil van 24 seconden tussen de eerste P-golf en de eerste S-golf correspondeert met een epicentrum op afstand 215 km. Dit geeft je dus een cirkel waarop het epicentrum kan liggen. Als je meerdere meetpunten hebt, dan wordt het epicentrum gegeven door het snijpunt van de verschillende cirkels. In werkelijkheid zal hier steeds een meetfout op zitten en zal je nooit echt een mooi snijpunt vinden, maar eerder een gebied van mogelijkheden.
De amplitude wordt gebruikt om sterkte van een aardbeving te bepalen, en de afstand van het waarnemingsstation tot het epicentrum.
Een belangrijk onderdeel van trilateratie is het bepalen van de onderlinge positie van twee cirkels. Hieronder zie je een overzicht van de mogelijke onderlinge posities van twee cirkels. In deze tabel staat $d$ voor de afstand tussen de middelpunten van de twee cirkels en $r_1$, respectievelijk $r_2$, voor de straal van de eerste, respectievelijk de tweede cirkel.
positie voorwaarde
concentrisch $d=0$
binnen rakend $d=|r_1-r_2|$
buiten rakend $d=r_1+r_2$
omsluitend $d < |r_1-r_2|$
snijdend $|r_1-r_2| < d < r_1+r_2$
gescheiden $d > r_1+r_2$
De afstand $d$ tussen twee punten $(x_1,y_1)$ en $(x_2,y_2)$ wordt hierbij berekend aan de hand van de volgende formule:
$d = \sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$.
Opgelet: Als je met reële getallen werkt op een computer, treden er al snel afrondingsfouten op. Hierdoor zullen twee getallen die gelijk zouden moeten zijn, vaak net niet gelijk zijn. Daarom voegen we een foutenmarge in: we zeggen dat twee getallen gelijk zijn, als het verschil tussen deze twee getallen kleiner is dan $10^{-2}$.
### Invoer
De invoer bestaat uit zes reële getallen: één getal per regel. Op de eerste regel staat de $x$-coördinaat van de eerste cirkel, gevolgd door een regel met de $y$-coördinaat van de cirkel en een regel met de straal van de cirkel. Dan herhalen deze drie grootheden zich voor de tweede cirkel.
### Uitvoer
De uitvoer bestaat uit één enkele regel die één van de volgende termen bevat: concentrisch, binnen rakend, buiten rakend, omsluitend, snijdend of gescheiden. De term die wordt uitgeschreven moet hierbij de onderlinge positie van de twee gegeven cirkels aangeven die via de invoer opgegeven werden.
### Voorbeeld
Invoer:
3.5
2.7
7.86
6.5
6.7
2.86
Uitvoer:
binnen rakend
Added by: Peter Dawyndt Date: 2012-02-16 Time limit: 10s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: PY_NBC Resource: None
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# The parts of a gable roof
the span. When you know the pitch of a roof plus the true span (measure the span from outside wall to outside wall with a long tape) or the run, you can determine the length of the common rafters. To do this, try using a book of rafter tables (see Sources on p. 198) or a pocket calculator.
A 24-ft.-wide building has a span of 24 ft. To find the rafter length of a 4-in — 12 pitch roof for this building, open your rafter-table book to the 4-in-12 page and look under the common-rafter table
at 24 ft. to see that the rafter length is 12 ft. 73/д in. If the span is 24 ft. 8 in., look under 8 in the inches column and add on an extra 41A in. That is the total length of the common rafter. It’s really that simple, so don’t make it difficult for yourself. After a few minutes with a rafter book, you can figure the length of almost any rafter for any pitch and any span. Subtract from this figure half the thickness of the ridge board (3A in. for a 2x ridge) and leave enough extra wood to cover the length of the tails in the overhang.
Updated: 21 ноября, 2015 — 6:26 пп
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# Parallel RLC Step Response with Voltage Source
I am learning about RLC circuits in my Circuits Analysis class and I had a question regarding Parallel RLC step responses.
In the book, they only provide examples of RLC circuits with a current source, such as the one below.
simulate this circuit – Schematic created using CircuitLab
The circuit and equations that go along with it make sense to me, but I started thinking about voltage sources, such as the one below.
simulate this circuit
This brings up a couple of questions:
1. Does the voltage change across the resistor, inductor or capacitor? According to KVL it would be 12V once the switch is closed. Does this produce a step response at all then?
2. When the switch is closed, the voltage across the capacitor goes from 0V(assuming no energy initially) to 12V because KVL would be violated otherwise. Wouldn't this be an instantaneous change, which cannot occur in capacitors?
3. What happens when you put a current source in a series RLC circuit? This raises similar questions.
Thanks for any help provided.
Assuming the context of ideal circuit theory, the voltage source current will not be finite at the instant the switch is closed; there will be an impulse of current to charge the capacitor instantly to 12V. After the initial impulse, the current will be a ramp, increasing linearly with time (without bound) due to the inductor.
However, ideal circuit theory is not physically relevant in this case since some of the assumptions of ideal circuit theory do not hold for the second circuit.
Physically, the current will be finite and the voltage across the capacitor will not be a step. This can be seen if the naive ideal circuit model is augmented with additional circuit elements to model the fact that, e.g.,
• All physical voltage sources have finite internal resistance (finite short circuit current)
• Physical capacitors have parasitic inductance and resistance which must be included in the ideal circuit model
• A physical circuit (the wires and circuit elements that form the closed path for current) have parasitic R, L, and C that must be modelled
• A physical circuit has radiation resistance, i.e., for large and fast current changes, the circuit will radiate energy into space and this must be modelled.
Regarding your perceptive question 3, replace current above with voltage, inductor with capacitor, short with open and then essentially the same argument holds.
• That clears up a bit, but I had another question. The second schematic would not produce a step response. Instead, adding a resistor in series to the voltage source would produce one. Am I correct in saying that? Feb 15, 2015 at 0:09
• @Addison, the 2nd schematic would produce a step output voltage at the expense of a current impulse (non-physical) and a current that increases without bound thereafter (non-physical). Adding a resistor in series with the voltage source prevents the current pathology (now a finite and bounded current) but the output voltage is no longer a step. In fact, the output voltage goes to zero as time increases. Feb 15, 2015 at 0:14
• What do you mean by step? My understanding is that a step response rises to a final voltage/current slowly, typically exponentially. In your answer, you said physically that the voltage across the capacitor will not be a step, but in the previous comment you said the 2nd schematic would produce a step output voltage. It seems contradictory. Also, by adding a resistor, you can do a source transformation to get a current source with a parallel resistor. This gives you the same circuit as the 1st schematic essentially. Therefore, the voltage should rise exponentially. Feb 15, 2015 at 0:34
• @Addison, I interpreted your first comment as asking if the output were a step function which is why I responded that the output would be a step (as in Heaviside step function). However, a step response is not necessarily an exponential rise to a final value. A step response is simply the output due to a step input and, in general, that can go to zero, non-zero, infinity, or oscillate. Feb 15, 2015 at 1:28
• @Addison, also, regarding the last sentence of your comment, in steady state, the voltage across an inductor is zero. If you add a resistor in series with the voltage source, the final (t goes to infinity) value of the output voltage is zero. This is (of course) true with the source transformation. Think about it. Feb 15, 2015 at 1:33
1. It would be 12V. The step response is an instantaneous rise from 0V to 12V.
2. The available current is infinite, so the capacitor can charge instantaneously.
3. The instantaneous current step would produce infinite voltage across the inductor.
This is that happens when you put theoretical ideal components into invalid configurations. A real voltage source has internal resistance that limits current, and a real current source has a maximum voltage that can be produced.
V is the source voltage, applied at t=0
Voltage across components must = source voltage at all values of time, therefore:
(1) Resistor current is V/R
(2) The inductor voltage is constant at V, so Ldi/dt = V; i = integral(V/L dt) = ramp with slope = V/L; therefore i = Vt/L
(3) Capacitor must charge instantaneously to V, so capacitor current is an impulse of strength VC (i.e. infinitesimal duration current to charge C to V)
The total current supplied by the source at time, t = V/R + Vt/L + (impulse of strength VC Coulomb at t=0)
When the impulse has finished, at t = 0+, total current is V/R + Vt/L, ie a ramp
• I'm new here, could someone explain why I've been awarded '-1'? Thank you. Chu
– Chu
Feb 16, 2015 at 12:41
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# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x)= (2x+4) /( x^2-3x-4)?
Sep 25, 2016
The vertical asymptotes are $x = 4$ and $x = - 1$. The horizontal asymptote is $y = 0$. There are no oblique asymptotes.
#### Explanation:
$f \left(x\right) = \frac{2 x + 4}{{x}^{2} - 3 x - 4}$
Factor the denominator.
$f \left(x\right) = \frac{2 x + 4}{\left(x - 4\right) \left(x + 1\right)}$
To find the vertical asymptotes, set the denominator equal to zero and solve.
$\left(x - 4\right) \left(x + 1\right) = 0$
$x - 4 = 0 \textcolor{w h i t e}{a a a} x + 1 = 0$
$x = 4 \textcolor{w h i t e}{a a a} x = - 1 \textcolor{w h i t e}{a a a}$These are the VA's.
To find the horizontal asymptotes, compare the degree of the numerator to the degree of the denominator.
• If the deg of the num > deg of den, there is an oblique asymptote.
• If the deg of the num = deg of den, the HA is the leading coefficient of the numerator divided by the leading coefficient of the denominator.
• If the deg of the num < deg of den, the HA is $y = 0$.
In this example, the degree of the numerator is $\textcolor{red}{1}$ and the degree of the denominator is $\textcolor{b l u e}{2}$
$f \left(x\right) = \frac{2 {x}^{\textcolor{red}{1}} + 4}{{x}^{\textcolor{b l u e}{2}} - 3 x - 4}$
The degree of the numerator < degree of the denominator, so the HA is $y = 0$.
There is no oblique asymptote, because the degree of the numerator is not greater than the degree of the denominator.
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# How to describe integer and floating point entry fields?
I'm going to be presenting university students with an entry field, into which they will have to enter a value. The value might be an integer, or it might be a floating point value. I'm trying to work out what the best way to label these is so they know what they have to enter.
Currently I have: `Enter an integer` and `enter a real number`. But I worry that some students are going to be confused by "real number" (Hmm, are the other types "unreal"?). We are pondering `Enter a whole number` and `Enter a number`, but the latter doesn't give an implicit clue that a decimal point is allowed. It is important that the student knows that the decimal is available, as it is most likely going to be needed (the question might be `What is 5 divided by 2?`).
What terms are good for a wide audience to refer to integers and real numbers in this manner?
There are some good suggestions for asking questions better, but in my case I have no idea what the question might be, and my prompt has to be suitable for whatever that question might be. And it is a single question at a time, so that might help. Another suggestion from someone here was `Enter a number` and `Enter a number (decimals allowed)` which lets people know that they can go floating point.
-
I think this brings up some more questions and potential issues:
1. Is it really necessary to have two separate kinds of fields for these types of numbers? The fact that you are having trouble properly labeling them means that it will likely result in a poor user experience.
2. What if the user thinks the answer contains a decimal, but is only allowed to put in a whole number? Should they round up or down?
3. Is there a way you can just provide them with one number field and take care of the rest programmatically? Maybe rounding or removing decimal places?
If you cannot combine them into one field type, how about being explicit on each label:
• Enter a number (decimals allowed)
• Enter a number (no decimals)
-
After consideration, I've decided that this really does bring up more questions, and while I like some of the other answers they can't apply in my situation. Accepted this one. – mj2008 May 11 '11 at 8:03
I think you're correct in going for:
Enter a whole number
for those fields that will only accept integers. This should be understandable by university students - what ever subject they are studying.
Enter a number
might be OK. Does this appear on the same page as the integer fields? This will make it stand out. If the first question makes it clear that a decimal is required (5 divided by 2 is a good example) then it should be obvious that they can enter e.g. `2.5`.
Is there room for adding instructions somewhere that state that these fields may require the number including any decimals?
-
If your example question is representative of the type of question that is going to be asked (I understand it is an oversimplified example), then I think the question is going to drive them to understanding of what type of number belongs. You shouldn't even really need the `Enter a number` labels because the answer is obvious. You might be confusing the matter more by doing so - adding text where you don't need it is rarely useful for your users.
-
I think this is a very good point, but there is a disconnect between the location of the question and the entry of the data (different screens) so some sort of prompt will help. Particularly as there are other types, like plain text (which I didn't mention). – mj2008 May 8 '11 at 10:25
@mj2008 - How are they going to answer the question if the question and answer form are on different screen? I think you've got a much bigger UX issue there. – Charles Boyung May 9 '11 at 14:49
An example might be where they are watching television, and the person on the telly asks a question. – mj2008 May 10 '11 at 10:29
Depending on the context, you could show default values (with precision) in the fields:
``````Enter an integer: [ 0]
Ender a real: [ 0.00]
``````
Or you could specify the precision/definiton as a hint:
``````What is 10 divided by 2: [ ] (a whole number)
What is 10 divided by 3: [ ] (with 2 decimals)
``````
-
Formatting it nicely that way isn't possible unfortunately. – mj2008 May 6 '11 at 13:45
Before the web era, there were quite smart input fields. The characters were fixed width, it was easy to recognize how many digits can fit about into the field. Started entering a number, e.g. 12.34, the following displayed (the '_' is the cursor):
``````get focused: [_ . ]
pressed '1': [1_ . ]
pressed '2': [12_ . ]
pressed '.': [ 12._ ]
pressed '3': [ 12.3_]
pressed '4': [ 12.34] - the cursor is blinking under the ']'
``````
It would be great to see such smart input fields again.
I'm not a native English, but I feel these two words appropiate to refer to integer and floating: "number of" foos, and "amount of" bar. For floating, I would add a remark: "(max. 2 decimals)", which makes the data type clear without explicit naming it on its frightening "scientific" name.
-
I don't think that masked edit fields are that uncommon? Eg: ui-patterns.com/users/1/collections/11/entry/614 – Jørn E. Angeltveit May 6 '11 at 14:03
Masked files went out of fashion because of HTML input field... err... primit... err... simplicity. – ern0 May 6 '11 at 17:15
There's a jQuery plugin for that: digitalbush.com/projects/masked-input-plugin – Matt Potts Nov 24 '11 at 13:07
Think in actions rather than in number classes. Number systems are only familiar to technical students. (I recall a conversation in a psychology class. Teacher asked 'do you know complex numbers?' Psychology student answered completely serious and with confidence 'yes, pi and e'.)
My suggestion: 'What is 10 divided by 3 (round to nearest whole number)?', as opposed to ''What is 10 divided by 3 (round to 2 decimal places)?'
An alternative would be: 'What is 10 divided by 3 (don't round off)?' -- here you could expect either a fraction for an answer, or a completely desparate student.
-
I think this is a good answer, but I don't know the context of the question. The prompt has to live completely separate to the question. – mj2008 May 9 '11 at 8:42
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12-Amethyst
## 3D curve with an helical sweep
Hi,
I have to make a curve (not solid or surface) with several rounds and helical portion.
It must follow other parts in an assembly.
Any idea on how to do this ?
(The picture is not a 3D model)
Thanks
Dominique
2 REPLIES 2
20-Turquoise
(To:Dom_CHENTRE)
Looking at your image, I would suggest you start with the helical sweep then sketch the two legs on planes and connect with curves through points. The legs can be sketched with the part in the assembly to follow the other parts. You can use references from the assembly to make a sketch, but I prefer to delete the external references before finishing the sketch.
Here is what I usually follow for a helical curve:
# HELICAL CURVE
1. Start curve from equation.
2. Select coordinate system.
3. Enter offset in “from” box
• e.g. .25 + 90°
4. Enter number of turns + offset in “To” box.
• e.g. 10 turns and .50 offset = 10.50
5. Create equation:
• r=”radius of curve”
• theta=t*360
• z= “pitch”*t
* Offset can also be added to theta calculation in degrees.
There is always more to learn in Creo.
22-Sapphire II
(To:Dom_CHENTRE)
A helical sweep or curve by equation as outlined below MAY or MAY NOT give the results you desire. If you are ok with the condition of the legs tangent to the helix, then either method WILL work. HOWEVER, if you are trying to get the legs perfectly normal to the axis of the helix, those methods will NOT work and you will have to use another method.
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Browse Scheme
Recent Scheme
# Coloring Pictures For Grade Page 3rd Christmas Pages Free Math
By Fleurette Auger at December 18 2018 05:25:59
However, with the creation of worksheets users can now calculate many simple and complex math and financial problems as well as display their stored data with many unique custom charts and graphs. Text data consists of alpha-numeric characters such as letters and words. Formulas are instructions that are included in a cell that allow you to manipulate and perform operations on other cells in the Workbook. When you put a formula in your cell, the calculated value is then displayed as a result.
Parents can go onto the various online portals to find suitable option to help their children learn math solving problem skills. Thorough research over the internet will give a wide range of websites that offers right set of math worksheets for children. 3rd grade math worksheets problems are experienced by many kids and their parents are usually frustrated. Not any more, there is information on how to solve the problem of your kids math homework. So what does the research show? The scientifically-based research shows that there are specific strategies that enhance reading for comprehension. Reading strategies are detailed steps students implement before reading the text, while reading the text and after reading the text. The strategies are purposeful in enhancing comprehension.
## Gallery of Coloring Pictures For Grade 3
So what kinds of worksheets should you get? Anything where you feel that your child needs further drill. We often have this notion that worksheets are just for math. This, of course, is not true. While they are excellent tools for reviewing math facts such as the multiplication tables and division facts, they are just as useful for reviewing parts of speech or the states in the union. When you're teaching your student to write, there are a whole host of worksheets online that you can use. Many of these include clipart that will help the students learn the sounds of letters and letter combinations. There are other sheets that help the student learn to write his or her numbers.
What are the Parts of a Worksheet? Worksheets consists of four primary parts. A cell is the most commonly used part within an Excel workbook. Cells are where users can enter data to be used within formulas and charts later on. Worksheets are the individual "pages" of an Excel file. A Worksheet is basically just a computer representation of a very large piece of paper. It is organized into columns and rows, with the columns denoted by alphabetical letters (A, B...AB, AC, AD,...etc) and rows denoted by numbers.
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# Define a function as a set of ordered pairs.
Question:
Define a function as a set of ordered pairs.
Solution:
A function is a set of ordered pairs with the property that no two ordered pairs have the same first component and different second components.
Sometimes we say that a function is a rule (correspondence) that assigns to each element of one set, X, only one element of another set, Y.
The elements of set X are often called inputs and the elements of set Y are called outputs.
The domain of a function is the set of all first components, x, in the ordered pairs.
The range of a function is the set of all second components, y, in the ordered pairs.
A function can be defined by a set of ordered pairs.
Example: {(1,a), (2, b), (3, c), (4,a)} is a function, since there are no two pairs with the same first component.
The domain is then the set {1,2,3,4} and the range is the set {a, b, c}.
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# How to Measure Loop Length of Knits Fabric?
The knitted fabric is made of loop formation. Some of the fabric properties depend on loop length. So, sometimes we need to know loop length of the fabric that we are going to use in garments. In this article, I have mentioned a method for measuring loop length of your sample.
To measure loop length of a knits fabric sample use following steps-
Step #1: Take your sample. Cut fabric swatch of 10 cm X 10 cm from the fabric sample. While cutting fabric swatch, consider cutting on the wales line. Count number of wales in the 10 cm of fabric swatch. For example see the right side image, that has 6 wales.
Image courtesy: bbc.co.uk
Step #2: Take out yarns by pulling the loop. Don't consider yarns those are not the full length of a swatch. Take five yarns of complete length and stretch yarns to remove curling on yarns.
Step #3: Measure yarn length. Use measuring tape or scale to measure yarn length. Measure all 5 sample yarns. Note yarn lengths in a paper or notebook. Calculate average length of the sample yarns.
Step #4: Calculate the loop length. Now divide the average length of the yarns by no. of loops on the fabric sample.
Suppose you count 'X' no. of loops (wales) in the swatch and average length of the stretched yarns are 'Y' cm. Therefore loop length of the sample fabric will be equal to Y/X centimeters.
[Note: The above method is not a standard method but a DIY guide to find approximate loop length]
### Prasanta Sarkar
Prasanta Sarkar is a textile engineer and a postgraduate in fashion technology from NIFT, New Delhi, India. He has authored 6 books in the field of garment manufacturing technology, garment business setup, and industrial engineering. He loves writing how-to guide articles in the fashion industry niche. He has been working in the apparel manufacturing industry since 2006. He has visited garment factories in many countries and implemented process improvement projects in numerous garment units in different continents including Asia, Europe, and South Africa. He is the founder and editor of the Online Clothing Study Blog.
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# Number 250844
Number 250,844 spell 🔊, write in words: two hundred and fifty thousand, eight hundred and forty-four . Ordinal number 250844th is said 🔊 and write: two hundred and fifty thousand, eight hundred and forty-fourth. Color #250844. The meaning of number 250844 in Maths: Is Prime? Factorization and prime factors tree. The square root and cube root of 250844. What is 250844 in computer science, numerology, codes and images, writing and naming in other languages. Other interesting facts related to 250844.
## What is 250,844 in other units
The decimal (Arabic) number 250844 converted to a Roman number is (C)(C)(L)DCCCXLIV. Roman and decimal number conversions.
#### Weight conversion
250844 kilograms (kg) = 553010.7 pounds (lbs)
250844 pounds (lbs) = 113782.1 kilograms (kg)
#### Length conversion
250844 kilometers (km) equals to 155868 miles (mi).
250844 miles (mi) equals to 403695 kilometers (km).
250844 meters (m) equals to 822969 feet (ft).
250844 feet (ft) equals 76459 meters (m).
250844 centimeters (cm) equals to 98757.5 inches (in).
250844 inches (in) equals to 637143.8 centimeters (cm).
#### Temperature conversion
250844° Fahrenheit (°F) equals to 139340° Celsius (°C)
250844° Celsius (°C) equals to 451551.2° Fahrenheit (°F)
#### Time conversion
(hours, minutes, seconds, days, weeks)
250844 seconds equals to 2 days, 21 hours, 40 minutes, 44 seconds
250844 minutes equals to 6 months, 6 days, 4 hours, 44 minutes
### Codes and images of the number 250844
Number 250844 morse code: ..--- ..... ----- ---.. ....- ....-
Sign language for number 250844:
Number 250844 in braille:
QR code Bar code, type 39
Images of the number Image (1) of the number Image (2) of the number More images, other sizes, codes and colors ...
## Mathematics of no. 250844
### Multiplications
#### Multiplication table of 250844
250844 multiplied by two equals 501688 (250844 x 2 = 501688).
250844 multiplied by three equals 752532 (250844 x 3 = 752532).
250844 multiplied by four equals 1003376 (250844 x 4 = 1003376).
250844 multiplied by five equals 1254220 (250844 x 5 = 1254220).
250844 multiplied by six equals 1505064 (250844 x 6 = 1505064).
250844 multiplied by seven equals 1755908 (250844 x 7 = 1755908).
250844 multiplied by eight equals 2006752 (250844 x 8 = 2006752).
250844 multiplied by nine equals 2257596 (250844 x 9 = 2257596).
show multiplications by 6, 7, 8, 9 ...
### Fractions: decimal fraction and common fraction
#### Fraction table of 250844
Half of 250844 is 125422 (250844 / 2 = 125422).
One third of 250844 is 83614,6667 (250844 / 3 = 83614,6667 = 83614 2/3).
One quarter of 250844 is 62711 (250844 / 4 = 62711).
One fifth of 250844 is 50168,8 (250844 / 5 = 50168,8 = 50168 4/5).
One sixth of 250844 is 41807,3333 (250844 / 6 = 41807,3333 = 41807 1/3).
One seventh of 250844 is 35834,8571 (250844 / 7 = 35834,8571 = 35834 6/7).
One eighth of 250844 is 31355,5 (250844 / 8 = 31355,5 = 31355 1/2).
One ninth of 250844 is 27871,5556 (250844 / 9 = 27871,5556 = 27871 5/9).
show fractions by 6, 7, 8, 9 ...
### Calculator
250844
#### Is Prime?
The number 250844 is not a prime number. The closest prime numbers are 250841, 250853.
#### Factorization and factors (dividers)
The prime factors of 250844 are 2 * 2 * 11 * 5701
The factors of 250844 are 1 , 2 , 4 , 11 , 22 , 44 , 5701 , 11402 , 22804 , 62711 , 125422 , 250844
Total factors 12.
Sum of factors 478968 (228124).
#### Powers
The second power of 2508442 is 62.922.712.336.
The third power of 2508443 is 15.783.784.853.211.584.
#### Roots
The square root √250844 is 500,843289.
The cube root of 3250844 is 63,066864.
#### Logarithms
The natural logarithm of No. ln 250844 = loge 250844 = 12,432587.
The logarithm to base 10 of No. log10 250844 = 5,399404.
The Napierian logarithm of No. log1/e 250844 = -12,432587.
### Trigonometric functions
The cosine of 250844 is 0,923771.
The sine of 250844 is 0,382944.
The tangent of 250844 is 0,414544.
### Properties of the number 250844
Is a Friedman number: No
Is a Fibonacci number: No
Is a Bell number: No
Is a palindromic number: No
Is a pentagonal number: No
Is a perfect number: No
## Number 250844 in Computer Science
Code typeCode value
PIN 250844 It's recommendable to use 250844 as a password or PIN.
250844 Number of bytes245.0KB
CSS Color
#250844 hexadecimal to red, green and blue (RGB) (37, 8, 68)
Unix timeUnix time 250844 is equal to Saturday Jan. 3, 1970, 9:40:44 p.m. GMT
IPv4, IPv6Number 250844 internet address in dotted format v4 0.3.211.220, v6 ::3:d3dc
250844 Decimal = 111101001111011100 Binary
250844 Decimal = 110202002112 Ternary
250844 Decimal = 751734 Octal
250844 Decimal = 3D3DC Hexadecimal (0x3d3dc hex)
250844 BASE64MjUwODQ0
250844 MD5b38cd986825e6e8de461d42f18dea378
250844 SHA19ac9f1cb6c6c692097808ac0a4d9f88ff513054a
250844 SHA224cf2d1f324fdef904a4cc1852fa433b85019c8cfd30455205585e115c
250844 SHA256452535f0f4109eed3ebc39687d75793cc48d4464232285a0676f112d97228226
250844 SHA3844c50373d1e7c28fab653f443c4e6d8da86f6af93d6a7fe1de3da1c2e29e7efe70b08cc8f787d726772297756380ab6b1
More SHA codes related to the number 250844 ...
If you know something interesting about the 250844 number that you did not find on this page, do not hesitate to write us here.
## Numerology 250844
### Character frequency in number 250844
Character (importance) frequency for numerology.
Character: Frequency: 2 1 5 1 0 1 8 1 4 2
### Classical numerology
According to classical numerology, to know what each number means, you have to reduce it to a single figure, with the number 250844, the numbers 2+5+0+8+4+4 = 2+3 = 5 are added and the meaning of the number 5 is sought.
## Interesting facts about the number 250844
### Asteroids
• (250844) 2005 UX182 is asteroid number 250844. It was discovered by Spacewatch from Obs. US National at Kitt Peak on 10/24/2005.
## № 250,844 in other languages
How to say or write the number two hundred and fifty thousand, eight hundred and forty-four in Spanish, German, French and other languages. The character used as the thousands separator.
Spanish: 🔊 (número 250.844) doscientos cincuenta mil ochocientos cuarenta y cuatro German: 🔊 (Nummer 250.844) zweihundertfünfzigtausendachthundertvierundvierzig French: 🔊 (nombre 250 844) deux cent cinquante mille huit cent quarante-quatre Portuguese: 🔊 (número 250 844) duzentos e cinquenta mil, oitocentos e quarenta e quatro Hindi: 🔊 (संख्या 250 844) दो लाख, पचास हज़ार, आठ सौ, चौंतालीस Chinese: 🔊 (数 250 844) 二十五万零八百四十四 Arabian: 🔊 (عدد 250,844) مئتانخمسون ألفاً و ثمانمائة و أربعة و أربعون Czech: 🔊 (číslo 250 844) dvěstě padesát tisíc osmset čtyřicet čtyři Korean: 🔊 (번호 250,844) 이십오만 팔백사십사 Danish: 🔊 (nummer 250 844) tohundrede og halvtredstusindottehundrede og fireogfyrre Dutch: 🔊 (nummer 250 844) tweehonderdvijftigduizendachthonderdvierenveertig Japanese: 🔊 (数 250,844) 二十五万八百四十四 Indonesian: 🔊 (jumlah 250.844) dua ratus lima puluh ribu delapan ratus empat puluh empat Italian: 🔊 (numero 250 844) duecentocinquantamilaottocentoquarantaquattro Norwegian: 🔊 (nummer 250 844) to hundre og femti tusen, åtte hundre og førti-fire Polish: 🔊 (liczba 250 844) dwieście pięćdziesiąt tysięcy osiemset czterdzieści cztery Russian: 🔊 (номер 250 844) двести пятьдесят тысяч восемьсот сорок четыре Turkish: 🔊 (numara 250,844) ikiyüzellibinsekizyüzkırkdört Thai: 🔊 (จำนวน 250 844) สองแสนห้าหมื่นแปดร้อยสี่สิบสี่ Ukrainian: 🔊 (номер 250 844) двiстi п'ятдесят тисяч вiсiмсот сорок чотири Vietnamese: 🔊 (con số 250.844) hai trăm năm mươi nghìn tám trăm bốn mươi bốn Other languages ...
## Comment
If you know something interesting about the number 250844 or any natural number (positive integer) please write us here or on facebook.
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Any views expressed within media held on this service are those of the contributors, should not be taken as approved or endorsed by the University, and do not necessarily reflect the views of the University in respect of any particular issue.
# School of Mathematics
Blog for the School of Mathematics community
# Samples of Mathematics in Data: DataFringe Festival 2023
Have you ever wondered about the role that mathematics plays in the world of data? On the morning of November 6, the School of Mathematics hosted a compelling series of talks at the Bayes Centre, aiming to shed light on this question. As part of the DataLab’s DataFringe 2023 festival, these mini-lectures brought together researchers from various specializations to explore the relationship between mathematics and data. The lectures covered diverse topics, unveiling unexpected ways in which mathematical data underpins our daily lives.
## How many islands are there?
### Dr Geoff Vasil
Dr Vasil kickstarted the event with this deceptively simple question – how do we quantify the number of islands on planet Earth? This led the audience down a philosophical rabbit hole, with wildly different guesses at the overall number. The presentation challenged conventional definitions of an island, the status of continents, and the complications that arise when attempting to determine the length of a coastline.
The potential for classifying even a single grain of sand as an island was raised! This thought experiment, supplemented with data as a means of illustrating the problem, highlighted how mathematics is used to investigate and redefine our ever-evolving understanding of the world.
## How statistics keep the lights on
### Dr Amanda Lenzi (joint work with Dr Mihai Anitescu)
The supply and demand of energy was the primary concern of the second talk. Statistical analysis can play a part in helping energy providers prepare for extreme events. Dr Lenzi gave context for the need of a constant, stable energy supply to enable societies worldwide to function. This resource is often taken for granted, and only when the absence of energy disrupts typical routines is the reality of losing power is the alarming reality recognized. A poignant example is the loss of life and destruction caused by severe weather in the blackouts that Texas experienced in February 2021.
Dr Lenzi introduced us to her data forecasting research, which aims at helping to prevent major power crises by observing the operations of the energy grid in maintaining an equal demand and generation by assuming fluctuating demands. This is increasingly important as renewable energy sources are utilized more – weather forecasts also need to be considered and are not as predictable in comparison to the traditional load forecast. Her insights into the FNET dataset were designed to increase the robustness of the system by issuing recommendations based on previous observations (10 per second in 70 participating homes across the state). Historical time series data, combined with variables of weather, can help companies decide whether to invest in back-up storage reserves in the event of a power outage.
## Bayesian Inverse Reinforcement Learning: Teaching Your Computer to Take Over Your Chores
### Dr Torben Sell
Dr Torben Sell asked attendees to imagine how much easier life would be if a robot that could handle all of your monotonous household chores existed. Off-the-shelf robots are currently unable to perform this function to the standard that we may desire, and Dr Sell argues that Bayesian decision-making theory could improve the situation.
Like self-driving cars, a Henry hoover that has been programmed with this data in mind, and subjected to extensive testing to mimic the actions of the average chore-doer, could eventually carry out the actions itself with a comparable degree of accuracy. The technology required to make this change operational is expensive and not widely used at present, but the talk provided an optimistic outlook for people who dislike vacuuming.
## Data-driven Optimization Under Uncertainty: Bridging Real-world Challenges with Mathematics
### Dr Merve Bodur
Dr Merve Bodur posited that mathematics is a powerful force in addressing real-world challenges. In an increasingly data-rich world, effective use of data is essential. Up to 90% of the world’s data was generated in the past two years alone, with this trend seeming set to continue. Dr Bodur highlighted the role of data-driven optimisation in uncertain circumstances, including in sectors susceptible to unpredictable conditions such as: transport, healthcare, and telecommunications.
AI tools can prove valuable even for presenting – ChatGPT helped outline the structure of the presentation itself! Research shows that leveraging data and incorporating mathematical models (such as stochastic programming) can lead to optimized solutions to problems through building and pre-empting future scenarios, giving those in charge methods of mitigating the adverse effects that a challenge may otherwise cause.
## Planning Outages in the Electricity System – Data and Models for Better Decisions
### Dr Lars Schewe
The morning was wrapped up by Dr Lars Schewe with a demonstration of how data-driven support for the outage planning for GB electricity grid leads to better outcomes. Dr Schewe shared his experience with a long-term project alongside National Grid ESO: the successes and difficulties encountered when working with real-world instances from the electricity grid. The talk illustrated the ways in which mathematics and data can compliment one another when making decisions in critical infrastructure.
In summary, the School of Mathematics’ event at DataLab’s DataFringe 2023 provided a multifaceted perspective on the interconnectedness of mathematics and data, prompting compelling discussions on seemingly unrelated topics. Attendees had the opportunity to ask questions and network before and after the talks took place. Whether an enthusiastic mathematician, or someone simply curious about the role that data plays, this event showcased the interplay between data and everyday problems.
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Solutions
Calculus
# Calculus (5th Edition) View more editions Solutions for Chapter 5.2
• 4704 step-by-step solutions
• Solved by publishers, professors & experts
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Over 90% of students who use Chegg Study report better grades.
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Chapter: Problem:
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SAMPLE SOLUTION
Chapter: Problem:
75% (4 ratings)
• Step 1 of 5
a)
Consider the following figure shows the Riemann sum approximation with subdivisions to
• Step 2 of 5
From the above approximation each rectangle length is equal to the function value at each left hand side end point of each sub interval
So this approximation is called left-hand approximation
Clearly observe that the area under the curve is smaller than this area given by left hand approximation.
• Step 3 of 5
The approximation is right hand side approximation that is each length of rectangle is the function value at right hand side end point of each sub interval
The following figure represents right hand approximation of given definite integral ‘
• Step 4 of 5
From the figure the area of right-approximation is lesser than area under the curve
That is the right approximation is lesser than left approximation
• Step 5 of 5
(b)
From the figures left end point of first interval is
The right end point of last interval is
The number of sub intervals is equal to number of rectangles is
The width of each rectangle is same in both approximations
Corresponding Textbook
Calculus | 5th Edition
9780470131596ISBN-13: 0470131594ISBN:
Alternate ISBN: 9780470420157, 9780470563724, 9780470564981, 9780470915028, 9780470921968, 9781118139486
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https://www.aqua-calc.com/one-to-one/density/long-ton-per-liter/tonne-per-cubic-centimeter/90
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# 90 long tons per liter [long tn/l] in tonnes per cubic centimeter
## long tons/liter to tonne/centimeter³ unit converter of density
90 long tons per liter [long tn/l] = 0.09 tonne per cubic centimeter [t/cm³]
### long tons per liter to tonnes per cubic centimeter density conversion cards
• 90
through
114
long tons per liter
• 90 long tn/l to t/cm³ = 0.09 t/cm³
• 91 long tn/l to t/cm³ = 0.09 t/cm³
• 92 long tn/l to t/cm³ = 0.09 t/cm³
• 93 long tn/l to t/cm³ = 0.09 t/cm³
• 94 long tn/l to t/cm³ = 0.1 t/cm³
• 95 long tn/l to t/cm³ = 0.1 t/cm³
• 96 long tn/l to t/cm³ = 0.1 t/cm³
• 97 long tn/l to t/cm³ = 0.1 t/cm³
• 98 long tn/l to t/cm³ = 0.1 t/cm³
• 99 long tn/l to t/cm³ = 0.1 t/cm³
• 100 long tn/l to t/cm³ = 0.1 t/cm³
• 101 long tn/l to t/cm³ = 0.1 t/cm³
• 102 long tn/l to t/cm³ = 0.1 t/cm³
• 103 long tn/l to t/cm³ = 0.1 t/cm³
• 104 long tn/l to t/cm³ = 0.11 t/cm³
• 105 long tn/l to t/cm³ = 0.11 t/cm³
• 106 long tn/l to t/cm³ = 0.11 t/cm³
• 107 long tn/l to t/cm³ = 0.11 t/cm³
• 108 long tn/l to t/cm³ = 0.11 t/cm³
• 109 long tn/l to t/cm³ = 0.11 t/cm³
• 110 long tn/l to t/cm³ = 0.11 t/cm³
• 111 long tn/l to t/cm³ = 0.11 t/cm³
• 112 long tn/l to t/cm³ = 0.11 t/cm³
• 113 long tn/l to t/cm³ = 0.11 t/cm³
• 114 long tn/l to t/cm³ = 0.12 t/cm³
• 115
through
139
long tons per liter
• 115 long tn/l to t/cm³ = 0.12 t/cm³
• 116 long tn/l to t/cm³ = 0.12 t/cm³
• 117 long tn/l to t/cm³ = 0.12 t/cm³
• 118 long tn/l to t/cm³ = 0.12 t/cm³
• 119 long tn/l to t/cm³ = 0.12 t/cm³
• 120 long tn/l to t/cm³ = 0.12 t/cm³
• 121 long tn/l to t/cm³ = 0.12 t/cm³
• 122 long tn/l to t/cm³ = 0.12 t/cm³
• 123 long tn/l to t/cm³ = 0.12 t/cm³
• 124 long tn/l to t/cm³ = 0.13 t/cm³
• 125 long tn/l to t/cm³ = 0.13 t/cm³
• 126 long tn/l to t/cm³ = 0.13 t/cm³
• 127 long tn/l to t/cm³ = 0.13 t/cm³
• 128 long tn/l to t/cm³ = 0.13 t/cm³
• 129 long tn/l to t/cm³ = 0.13 t/cm³
• 130 long tn/l to t/cm³ = 0.13 t/cm³
• 131 long tn/l to t/cm³ = 0.13 t/cm³
• 132 long tn/l to t/cm³ = 0.13 t/cm³
• 133 long tn/l to t/cm³ = 0.14 t/cm³
• 134 long tn/l to t/cm³ = 0.14 t/cm³
• 135 long tn/l to t/cm³ = 0.14 t/cm³
• 136 long tn/l to t/cm³ = 0.14 t/cm³
• 137 long tn/l to t/cm³ = 0.14 t/cm³
• 138 long tn/l to t/cm³ = 0.14 t/cm³
• 139 long tn/l to t/cm³ = 0.14 t/cm³
• 140
through
164
long tons per liter
• 140 long tn/l to t/cm³ = 0.14 t/cm³
• 141 long tn/l to t/cm³ = 0.14 t/cm³
• 142 long tn/l to t/cm³ = 0.14 t/cm³
• 143 long tn/l to t/cm³ = 0.15 t/cm³
• 144 long tn/l to t/cm³ = 0.15 t/cm³
• 145 long tn/l to t/cm³ = 0.15 t/cm³
• 146 long tn/l to t/cm³ = 0.15 t/cm³
• 147 long tn/l to t/cm³ = 0.15 t/cm³
• 148 long tn/l to t/cm³ = 0.15 t/cm³
• 149 long tn/l to t/cm³ = 0.15 t/cm³
• 150 long tn/l to t/cm³ = 0.15 t/cm³
• 151 long tn/l to t/cm³ = 0.15 t/cm³
• 152 long tn/l to t/cm³ = 0.15 t/cm³
• 153 long tn/l to t/cm³ = 0.16 t/cm³
• 154 long tn/l to t/cm³ = 0.16 t/cm³
• 155 long tn/l to t/cm³ = 0.16 t/cm³
• 156 long tn/l to t/cm³ = 0.16 t/cm³
• 157 long tn/l to t/cm³ = 0.16 t/cm³
• 158 long tn/l to t/cm³ = 0.16 t/cm³
• 159 long tn/l to t/cm³ = 0.16 t/cm³
• 160 long tn/l to t/cm³ = 0.16 t/cm³
• 161 long tn/l to t/cm³ = 0.16 t/cm³
• 162 long tn/l to t/cm³ = 0.16 t/cm³
• 163 long tn/l to t/cm³ = 0.17 t/cm³
• 164 long tn/l to t/cm³ = 0.17 t/cm³
• 165
through
189
long tons per liter
• 165 long tn/l to t/cm³ = 0.17 t/cm³
• 166 long tn/l to t/cm³ = 0.17 t/cm³
• 167 long tn/l to t/cm³ = 0.17 t/cm³
• 168 long tn/l to t/cm³ = 0.17 t/cm³
• 169 long tn/l to t/cm³ = 0.17 t/cm³
• 170 long tn/l to t/cm³ = 0.17 t/cm³
• 171 long tn/l to t/cm³ = 0.17 t/cm³
• 172 long tn/l to t/cm³ = 0.17 t/cm³
• 173 long tn/l to t/cm³ = 0.18 t/cm³
• 174 long tn/l to t/cm³ = 0.18 t/cm³
• 175 long tn/l to t/cm³ = 0.18 t/cm³
• 176 long tn/l to t/cm³ = 0.18 t/cm³
• 177 long tn/l to t/cm³ = 0.18 t/cm³
• 178 long tn/l to t/cm³ = 0.18 t/cm³
• 179 long tn/l to t/cm³ = 0.18 t/cm³
• 180 long tn/l to t/cm³ = 0.18 t/cm³
• 181 long tn/l to t/cm³ = 0.18 t/cm³
• 182 long tn/l to t/cm³ = 0.18 t/cm³
• 183 long tn/l to t/cm³ = 0.19 t/cm³
• 184 long tn/l to t/cm³ = 0.19 t/cm³
• 185 long tn/l to t/cm³ = 0.19 t/cm³
• 186 long tn/l to t/cm³ = 0.19 t/cm³
• 187 long tn/l to t/cm³ = 0.19 t/cm³
• 188 long tn/l to t/cm³ = 0.19 t/cm³
• 189 long tn/l to t/cm³ = 0.19 t/cm³
#### Foods, Nutrients and Calories
COOKED SLICED HAM, UPC: 050500191883 contain(s) 89 calories per 100 grams or ≈3.527 ounces [ price ]
#### Gravels, Substances and Oils
CaribSea, Marine, Aragonite, Florida Crushed Coral weighs 1 153.3 kg/m³ (71.99817 lb/ft³) with specific gravity of 1.1533 relative to pure water. Calculate how much of this gravel is required to attain a specific depth in a cylindricalquarter cylindrical or in a rectangular shaped aquarium or pond [ weight to volume | volume to weight | price ]
Cadmium hydroxide [Cd(OH)2] weighs 4 790 kg/m³ (299.02993 lb/ft³) [ weight to volume | volume to weight | price | mole to volume and weight | density ]
Volume to weightweight to volume and cost conversions for Refrigerant R-410A, liquid (R410A) with temperature in the range of -40°C (-40°F) to 60°C (140°F)
#### Weights and Measurements
A foot (ft) is a non-SI (non-System International) measurement unit of length.
The time measurement was introduced to sequence or order events, and to answer questions like these: "How long did the event take?" or "When did the event occur?"
nmi/min² to thou/s² conversion table, nmi/min² to thou/s² unit converter or convert between all units of acceleration measurement.
#### Calculators
Calculate gravel and sand coverage in a corner bow front aquarium
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en
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https://codereview.stackexchange.com/questions/206595/attempt-to-solve-mondrian-puzzle-with-c
| 1,563,233,423,000,000,000
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crawl-data/CC-MAIN-2019-30/segments/1563195524254.28/warc/CC-MAIN-20190715215144-20190716001144-00272.warc.gz
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# Attempt to solve Mondrian Puzzle with C++
I am looking for help in code proofreading. The program is trying to solve this question: Fit non-congruent rectangles into an array_size x array_size square grid. What is the smallest difference possible between the areas of the largest and the smallest rectangles?
I have a class of Rectangles, which I try to fit in a class called Boards. The function AutoInsert is basically the algorithm, where I have a linked list consisting of both Boards & Rectangles that adds and removes potential Rectangles from the Boards. This program works for small numbers of global variable array_size.
I appreciate any help at all: even if you don't understand the question or my programming, if you recognize any bad practices in my programming, please do tell.
#include<iostream>
#include<math.h>
#include<vector>
#include<time.h>
#include<stdlib.h>
#include <cstdio>
#include <ctime>
using namespace std;
const int array_size = 15; //array_size x array_size square
const int upperbound = ceil(3+array_size/log(array_size)); //given in reddit link
const int SQ = pow(array_size,2);
char getRandom(){
int n=rand()%78;
char c=(char)(n+49);
return c;
}
class Rect{
private:
int l,w,use; //int use: 0=haven't used, 1=using, 2=congruent use, 3=never use again ... should I add a possible use number w. respect to wnBounds?
int coords[4]; //top left x,tly,brx,bry
char random;
public:
Rect(){
l=0;
w=0;
use=0;
for(int i=0;i<4;i++){
coords[i]=-1;
}
}
void setL(int l){this->l=l;}
void setW(int w){this->w=w;}
void setUse(int use){this->use=use;}
void setCoords(int coords[]){
for(int i=0;i<4;i++){
this->coords[i]=coords[i];
}
}
int getL(){return l;}
int getW(){return w;}
int getArea(){return l*w;}
int getUse(){return use;}
int *getCoords(){return coords;}
int tlx(){return coords[0];}
int tly(){return coords[1];}
int brx(){return coords[2];}
int bry(){return coords[3];}
char setPiece(char c){random=c;}
char getPiece(){return random;}
};
class Rboard{
private:
bool inRange(int c[4]); //defensive functions
bool rectAtLoc(int c[4]); //defense
bool canUse(Rect n); //defense
Rect p[SQ]; //all possible rectangles, p[pow(array_size,2)-1] is the square so be careful
Rect first; //initial Rect
int difference;
public:
Rboard();
Rboard(Rect n); //initialize Rboard with first Rectangle, it will be up to Rsort to input good rectangles
bool coords[array_size][array_size]; //main coords, all begin as false, true is if occupied
void setFirst(Rect n){first=n;}
void editPoss(int c[4],int n);
Rect returnP(int n){return p[n];}
int getIndex(int l, int w);
int getDiff();
int spaceLeft();
vector<int> wnBound(); //returns rectangles within upperbound of initial rectangle
vector<int> pp(); //returns the Index of possible rectangles which can be placed
vector<int> pc(int i); //even index = top left x, odd index=y, can figure out rest of coords of rectangle from this information.
void display();
};
Rboard::Rboard(){
difference=upperbound;
int i=0;
for(int i=0;i<array_size;i++){
for(int j=0;j<array_size;j++){
coords[i][j]=false;
}
}
for(int j=0;j<array_size;j++){
for(int k=0;k<array_size;k++){
p[i].setL(j+1);
p[i].setW(k+1);
i++;
}
}
}
Rboard::Rboard(Rect n){
Rboard();
first=n;
}
int Rboard::getIndex(int l, int w){
for(int i=0;i<SQ;i++){
if((p[i].getL()==l)&&(p[i].getW()==w)){
return i;
}
}
}
void Rboard::editPoss(int c[4],int n){
Rect g=p[n];
g.setCoords(c);
g.setL(p[n].getL());
g.setW(p[n].getW());
if(canUse(g)){
p[n].setPiece(getRandom());
p[n].setCoords(c);
p[n].setUse(1);
if(g.getL()!=g.getW()){
p[getIndex(g.getW(),g.getL())].setUse(2);
}
for(int i=p[n].tly();i<=p[n].bry();i++){
for(int j=p[n].tlx();j<=p[n].brx();j++){
coords[j][i]=true;
}
}
}
}
bool Rboard::inRange(int c[4]){
for(int i=0;i<4;i++){
if((c[i]>=array_size)||(c[i]<0)){
return false;
}
}
return true;
}
bool Rboard::rectAtLoc(int c[4]){
for(int i=c[0];i<=c[2];i++){
for(int j=c[1];j<=c[3];j++){
if(coords[i][j]){
return true;
}
}
}
return false;
}
bool Rboard::canUse(Rect n){
if((n.getUse()==0)&&(inRange(n.getCoords()))&&(rectAtLoc(n.getCoords())==false)){
return true;
}
return false;
}
int Rboard::getDiff(){
int oriDiff=difference, counter=0, minN=SQ+1, maxN=-1; //biggest number, smallest number
for(int i=0;i<SQ;i++){
if(p[i].getUse()==1){
if(p[i].getArea()>maxN){
maxN=p[i].getArea();
}
if(p[i].getArea()<minN){
minN=p[i].getArea();
}
counter++;
}
}
if((counter<=1)&&(spaceLeft()==0)){
return oriDiff;
}else{
return maxN-minN;
}
}
int Rboard::spaceLeft(){
int sum=0;
for(int i=0;i<array_size;i++){
for(int j=0;j<array_size;j++){
if(coords[i][j]){
sum+=1;
}
}
}
return (SQ-sum);
}
vector<int> Rboard::wnBound(){
vector<int> output;
for(int i=0;i<SQ;i++){
if(abs(p[i].getArea()-first.getArea())<=upperbound){
output.push_back(i);
}
}
return output;
}
vector<int> Rboard::pc(int i){
vector<int> output;
int l=p[i].getL(),w=p[i].getW(),area=p[i].getArea(),counter=0;
for(int a=0;a<(array_size-l+1);a++){
for(int b=0;b<(array_size-w+1);b++){
for(int c=0;c<w;c++){
for(int d=0;d<l;d++){
if(!coords[b+c][a+d]){
counter++;
}
}
}
if(counter==area){
output.push_back(b);
output.push_back(a);
}
counter=0;
}
}
return output;
}
vector<int> Rboard::pp(){
vector<int> indexes;
for(int i=0;i<wnBound().size();i++){
int j=wnBound()[i];
if(p[j].getUse()==0){
if(spaceLeft()>=p[j].getArea()){
if(pc(j).size()!=0){
indexes.push_back(j);
}
}
}
}
return indexes;
}
void Rboard::display(){
for(int i=0;i<array_size;i++){
for(int j=0;j<array_size;j++){
if(coords[j][i]){
for(int a=0;a<SQ;a++){
if((p[a].tlx()<=j)&&(p[a].brx()>=j)&&(p[a].tly()<=i)&&(p[a].bry()>=i)){
cout<<p[a].getPiece()<<" ";
}
}
}else{
cout<<0<<" ";
}
}
cout<<endl;
}
}
class Rnode{
private:
Rect piece;
Rboard state;
Rnode *next;
public:
Rnode(Rect p, Rboard s, Rnode *n){piece=p; state=s; next=n;}
Rect getPiece(){return piece;}
Rboard getState(){return state;}
Rnode* getNext(){return next;}
void setPiece(Rect p){piece=p;}
void setState(Rboard s){state=s;}
void setNext(Rnode *next1){next=next1;}
};
class Rsort{
private:
Rnode *root;
vector<Rect> best;
int diff;
public:
Rsort(){
root=NULL;
diff=upperbound+1;
};
void setFirst(int i);
bool isLoser(Rboard b);
bool isDonut(Rboard b);
//to do: keep track of duplicates. e.j., 1x1+1x2+3x1 has been calculated... make sure don't branch to other 6!-1 combinations.
void autoInsert(Rnode *c,double duration,int maxDepth);
void autoFirst();
void display();
};
void Rsort::setFirst(int i){
Rboard board;
Rect f=board.returnP(i);
int coords[4]={0,0,f.getW()-1,f.getL()-1};
f.setCoords(coords);
board.editPoss(coords,i);
board.setFirst(f);
if(root!=NULL){
root=NULL;
}
root=new Rnode(f,board,NULL);
}
bool Rsort::isLoser(Rboard b){ //optimizes by 2 seconds
vector<int> pp=b.pp(),dup;
int sum=0,dupS=0;
for(int i=0;i<pp.size();i++){
if((b.returnP(pp[i]).getL())!=(b.returnP(pp[i]).getW())){
dup.push_back(b.returnP(pp[i]).getArea());
}
sum+=b.returnP(pp[i]).getArea();
}
for(int i=0;i<dup.size();i++){
dupS+=dup[i];
}
return ((sum-dupS/2)<b.spaceLeft()); //if sum of areas of possible rectangles < spaceLeft, this board will be a loser
}
bool Rsort::isDonut(Rboard b){ //check if there is an non-patchable "hole"
bool invBoard[array_size+2][array_size+2];
vector<int> p=b.wnBound();
for(int x=0;x<(array_size+2);x++){
invBoard[x][0]=true;
invBoard[x][array_size+1]=true;
}
for(int y=1;y<(array_size+1);y++){
invBoard[0][y]=true;
for(int x=1;x<(array_size+1);x++){
invBoard[x][y]=!b.coords[x-1][y-1]; //made coords public for this reason
}
invBoard[array_size+1][y]=true;
}
for(int i=0;i<p.size();i++){
int l=b.returnP(i).getL(),w=b.returnP(i).getW(),area=b.returnP(i).getArea(),counter=0;
for(int a=0;a<(array_size-l+3);a++){
for(int b=0;b<(array_size-w+3);b++){
for(int c=0;c<w;c++){
for(int d=0;d<l;d++){
if(invBoard[b+c][a+d]){
counter++;
}
}
}
if(counter==area){
return 0;
}
counter=0;
}
}
}
return 1;
}
void Rsort::autoInsert(Rnode *c,double duration,int maxDepth){ //add terminating counter for recursion?
if(isDonut(c->getState())||(maxDepth>=7)){ //max number of rectangles you want: change #. Need to define a function
//to calculate this based on array_size and upperbound. This variable
//alone has stopped my program from crashing at high values of array_size.
//I suspect this due to the stack overflow error.
return;
}
if(duration>=15.0){ //timer x.0 seconds => gives better solutions the longer the timer
root=NULL;
return;
}
clock_t start;
double d;
start = clock();
vector<int> pp=c->getState().pp(), pc;
c->setNext(NULL);
if(pp.size()==0){
if(c->getState().spaceLeft()==0){
if(c->getState().getDiff()<diff){
diff=c->getState().getDiff();
best.clear();
for(int a=0;a<SQ;a++){
if(c->getState().returnP(a).getUse()==1){
best.push_back(c->getState().returnP(a));
}
}
display();
c->getState().display();
}
}
}else if((root!=NULL)&&(!isLoser(c->getState()))){
for(int i=0;i<pp.size();i++){ //better restrictions here please
pc=c->getState().pc(pp[i]);
for(int j=0;j<pc.size();j+=2){
if(j==2){ //to cut down runtime
break;
}
Rect n;
Rboard n1=c->getState();
int coords[4]={pc[j],pc[j+1],pc[j]+c->getState().returnP(pp[i]).getW()-1,pc[j+1]+c->getState().returnP(pp[i]).getL()-1};
n.setL(c->getState().returnP(pp[i]).getL());
n.setW(c->getState().returnP(pp[i]).getW());
n.setCoords(coords);
n1.editPoss(coords,pp[i]);
Rnode *newnode=new Rnode(n,n1,NULL);
c->setNext(newnode);
d=(clock()-start)/(double)CLOCKS_PER_SEC;
autoInsert(c->getNext(),(duration+d),(maxDepth+1)); //remove +d if you have all the time in the world to wait for output
}
}
}
}
void Rsort::autoFirst(){
//to do: only iterate through non congruent rectangles, and interquartile range of that.
for(int i=SQ/4;i<3*SQ/4;i++){ //interquartile range
cout<<i<<endl;
setFirst(i);
autoInsert(root,0,0);
}
}
void Rsort::display(){
cout<<diff<<endl;
for(int i=0;i<best.size();i++){
cout<<best[i].getL()<<" x "<<best[i].getW()<<endl;
}
}
int main(){
srand(time(NULL));
Rsort r;
Rboard t;
r.isDonut(t);
r.autoFirst();
return 0;
}
• "The program is trying to solve this question" Yes, and it isn't entirely clear whether it succeeded to your satisfaction or not. Could you clarify this? And take a look at the help center while you're at it :-) – Mast Oct 30 '18 at 18:17
• Yes, my program works for array_size up til 15 x 15. Help center says "Questions about improving scalability are allowable, as long as your code works for small inputs." My program just takes too long after 15 x 15 (ie 20 x 20) for me to really say whether it works or not. I ran it for 10 minutes and gave up. Thanks for the link to the help center – Bo Work Oct 30 '18 at 18:53
## Look at your compiler's warnings
When I try to compile your code, the compiler gives two warnings: Rect::setPiece() doesn't have a return statement, and Rboard::getIndex() is missing a return statement after the loop.
The first warning can be fixed by simply changing the return type of Rect::setPiece() to void. The second warning might be harmless; the assumption is that Rboard::getIndex() will always be called with values for l and w that match one of the rectangles. But what if it doesn't? Then the for loop will end, and a bogus value will be returned. If this is never supposed to happen, just throw an exception there: the compiler warning will go away, and if your code ever does the wrong things, you will hopefully get a helpful error message.
## Try to write more C++
In general, your code looks very much like C with classes, and doesn't make good use of the features that the C++ language and its standard library provides. Try to find more C++-like ways to write your code. That doesn't mean "write templates, use inheritance, and overload every operator you possibly can", rather try to make better use of the STL, use features like range for, auto, and so on, to help you write more concise code.
## Use a proper random number generator
If you can use C++11 or later, use the functions from <random> to generate random numbers, instead of using the rather bad rand() function from C.
While it is not so important for this particular code, srand(time(NULL)) will only generate a new seed every second, which might be bad if your code is run multiple times per second, or multiple instances of the code are started in parallel. Also, rand() % N will, for most values of N, not give you a uniformly distributed random number. There are ways around both issues, but the C++11 RNG functions take care of this for you.
## Code style
Every programmer has his/her own favourite way of formatting their source code. While there is no right or wrong, you are using a very dense style, omitting spaces almost wherever possible. I would suggest you use spaces after punctuation (such as ;), and spaces around operators (such as =, <, and so on). For example this line:
if((b.returnP(pp[i]).getL())!=(b.returnP(pp[i]).getW())){
It's hard to see that this is comparing the result of two function called. Just adding some spaces (and removing some superfluous parentheses) results in:
if (b.returnP(pp[i]).getL() != b.returnP(pp[i]).getW()) {
And when you are initializing arrays, you can also put each element on its own line. So this line for example:
int coords[4]={pc[j],pc[j+1],pc[j]+c->getState().returnP(pp[i]).getW()-1,pc[j+1]+c->getState().returnP(pp[i]).getL()-1};
Will become:
int coords[4] = {
pc[j],
pc[j+1],
pc[j] + c->getState().returnP(pp[i]).getW() - 1,
pc[j+1] + c->getState().returnP(pp[i]).getL() - 1,
};
Also, declare one variable per line, so:
int l=p[i].getL(),w=p[i].getW(),area=p[i].getArea(),counter=0;
Becomes:
int l = p[i].getL();
int w = p[i].getW();
int area = p[i].getArea();
int counter = 0;
## Use descriptive variable and function names
It should be possible to determine what a variable or function does by looking at its name. There are some commonly used abbreviations, such as i for a loop index, x, y and z for coordinates, but otherwise you should not use abbreviations.
Instead of l and w, write length and width. Instead of SQ, write array_elements. Or better yet, array_size, and split the original array_size into array_length and array_width. This way, you'll be able to handle non-square boards.
Instead of Rboard, name your class either RectangleBoard or just Board. And what do Rboard::pc() and Rboard::pp() do? Even looking at the code I have no idea what those abbreviations mean.
## Move member variable initialization to the declaration
It's generally best to move initialization of variables as close as possible to their declaration. For example, in class Rect, instead of initializing the private member variables inside the constructor, just write:
class Rect {
private:
int l = 0;
int w = 0;
...
Here it is not too important, but if you have multiple constructors, or have a lot of member variables, it will become clear that this is better. Here, you might get rid of the constructor altogether this way.
## Don't store redundant information
Your class Rect stores the top-left and bottom-right coordinates of the rectangle, and its length and width. There is also nothing in that class that prevents these pieces of information from being in conflict with each other. Either store both coordinates, or one coordinate and the length and width. Your getters and setters should take care of calculating the required information if necessary.
## Avoid using an array to store coordinates
Unless you are going to store manydimensional coordinates, it's usually better to just name the coordinates x and y, or in this case if you don't want to store width and height, x1, y1, x2 and y2. The reason is that it's easy to make mistakes when you store the coordinates in an int[4]: did you store the coordinates in the aforementioned order or, was is x1, x2, y1, y2? Being explicit here avoids issues.
Even better is to define a struct coordinate {int x; int y;}, or use a library like GLM that provides you with various vector and matrix types, including all kinds of useful functions that operate on them.
## Use a single function to get/set multiple variables, if that is the typical use case
Instead of having separate functions setL(int l) and setW(int w), which you will always call in pairs, create a single function set_size(int l, int w). Of course, if you use a struct for coordinates, then you will automatically write code like that.
## Use std::vector instead of arrays where appropriate
In class Rboard, you declare an array Rect p[SQ]. You are not always using all elements. It makes much more sense to make this a std::vector<Rect> p. This way, you can add elements to the vector as needed, you don't have to have a member variable in class Rect to tell you whether the rectangle is in use or not.
## Avoid if (foo) return true; else return false;
Just directly return foo. For example Rboard::canUse() can be simplified to:
bool Rboard::canUse(Rect n) {
return n.getUse() == 0 && inRange(n.getCoords()) && !rectAtLoc(n.getCoords());
}
## Use const references where appropriate
Passing large classes by value might result in expensive copies, and might even trigger some undesired behaviour, depending on how these classes are implemented. Use const references to avoid that. For example, Rboard::canUse() can be rewritten as:
bool Rboard::canUse(const Rect &n) {
... // no need to change anything in the implementation
}
## Use std::list<> instead of writing your own linked lists
Your class Rnode implements a linked list. Let the STL do that for you! Remove the member variable Rnode *next, the function setNext(), and in class Rsort use std::list<Rnode> nodes instead of Rnode *root. Then instead of having to do thinks like Rnode *newnode = new Rnode(...) and setNext(newnode), you can just write nodes.push_back(...). As a bonus, this will take care of deleting the memory for you, which you forgot to do.
## Use nullptr instead of NULL
NULL is C, nullptr is C++.
## Use '\n' instead of std::endl
When you want to end a line, output a '\n' instead of using std::endl. The latter is the same but also flushes the output, which might slow down your program.
## Optimize your Rboard::display() function
Your function to display a board is quite inefficient: it has complexity O(array_size⁴). The reason is that for every position, you check every possible square if it is covering that position. It is better to create a 2D array that represents the board, and then for each square, draw it onto that representation, and at the end write out the whole array. This reduces the complexity to O(array_size²). For example:
void Rboard::display() {
char output[array_size][array_size];
for (auto &rect: p) {
char piece = getRandom();
for (int y = rect.tly(); y < rect.bry(); y++) {
for (int x = rect.tlx(); x < rect.brx(); x++) {
output[y][x] = piece;
}
}
}
for (int y = 0; y < array_size; y++) {
cout.write(output[y], array_size);
cout.put('\n');
}
}
Note that the above function also no longer needs the member variable char random in class Rect.
## Use an enum for Rect::use
Instead of using an int to represent different states, make them explicit by using an enum, preferrably even an enum class if you can use C++11 or later. For example:
class Rect {
public:
enum class UseType {
UNUSED,
USED,
CONGRUENT_USE
};
private:
UseType use = UseType::UNUSED;
...
public:
void setUse(UseType use) {
this->use = use;
}
UseType getUse() {
return use;
}
Then later in the code, you can for example write setUse(Rect::UseType::CONGRUENT_USE). That's very verbose, but it's clear from just that line of code what the intention is, whereas setUse(2) leaves the reader searching through the code to find out what 2 means. Also, you can now no longer accidentily set an invalid value, like setUse(9).
• Thank you for your insight. It must have taken you quite a while to write this, which I appreciate. – Bo Work Oct 31 '18 at 2:58
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https://stats.stackexchange.com/questions/59387/confusion-related-to-minimization-of-a-gaussian-likelihood-function
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# Confusion related to minimization of a gaussian likelihood function
I have this confusion related to minimization of gaussian likelihood function. The negative of the log likelihood of gaussian distribution is
$-\log \det(Q) + \text{tr}(SQ) + \lambda||Q||_{1}$ where $Q$ is the precision matrix to be estimated
I am following this paper
It's been mentioned that
Using sub-gradient notation, we can write the optimality conditions (aka “normal equations”) for a solution to the above equation as
$-Q^{-1} + S + \lambda\Gamma = 0$
where $\Gamma_{jk} = \text{sign}(Q_{jk})$ if $Q_{jk} \neq 0$ and
$\Gamma_{jk} \in [-1, 1]$ if $Q_{jk} = 0$
I didn't get how come
$\Gamma_{jk} \in [-1, 1]$ if $Q_{jk} = 0$
Can anyone please explain it should be 0 isn't it. I didn't get how the interval $[-1, 1]$ came
Suggestions?
The equation says that, at the minimum, one of the subgradients of $$-\log \det Q + \text{tr}( S Q ) + \lambda \| Q \|_1$$ is zero.
The last term, $\lambda \Gamma$ is a subgradient of $\lambda \left\| Q \right\|_1$. Since $\left\| Q \right\|_1 = \sum_{ij} \left| Q_{ij} \right|$, we just have to find the subgradients of the 1-dimensional function $f(q) = \left| q \right|$. If $q \neq 0$, $f$ is differentiable, and there is only one subgradient: $f'(q) = \text{sign}\,q$. For $q=0$, any number in $[-1,1]$ is a subgradient.
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http://www.convertaz.com/convert-18-dam-to-dm/
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# Convert 18 dam to dm
## Conversion details
To convert dam to dm use the following formula:
1 dam equals 100 dm
So, to convert 18 dam to dm, multiply 100 by 18 i.e.,
18 dam = 100 * 18 dm = 1800 dm
For conversion tables, definitions and more information on the dam and dm units scroll down or use the related dam and dm quick access menus located at the top left side of the page.
dam is the symbol for decameter
dm is the symbol for decimeter
## No conversion tables found for dam to dm
Click here for a list of all conversion tables of dam to other compatible units.
## decameter
Decameter is a multiple of the meter unit. The deca prefix stands for 10 therefore, 1 decameter = 10 meter units. Meter is a unit of measurement of length. The definition for meter is the following:
1 decameter is equal to 10 meters,
The symbol for decameter is dam
## decimeter
Decimeter is a subdivision of the meter unit. The deci prefix stands for 0.1 therefore, 1 decimeter = 0.1 meter units. Meter is a unit of measurement of length. The definition for meter is the following:
1 decimeter is one tenth (1/10, 0.1) of a meter
The symbol for decimeter is dm
## Other people are also searching for information on dam conversions.
Following are the most recent questions containing dam. Click on a link to see the corresponding answer.
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Home | Base units | Units | Conversion tables | Unit conversion calculator
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https://rdrr.io/cran/bootSVD/man/qrSVD.html
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# qrSVD: Wrapper for 'svd', which uses random preconditioning to... In bootSVD: Fast, Exact Bootstrap Principal Component Analysis for High Dimensional Data
## Description
In order to generate the SVD of the matrix `x`, `qrSVD` calls `genQ` to generate a random orthonormal matrix, and uses this random matrix to precondition `x`. The svd of the preconditioned matrix is calculated, and adjusted to account for the preconditioning process in order to find `svd(x)`.
## Usage
```1 2``` ```qrSVD(x, lim_attempts = 50, warning_type = "silent", warning_file = "qrSVD_warnings.txt", ...) ```
## Arguments
`x` a matrix to calculate the svd for `lim_attempts` the number of tries to randomly precondition x. We generally find that one preconditioning attempt is sufficient. `warning_type` controls whether the user should be told if an orthogonal preconditioning matrix is required, or if `svd` gives warnings. 'silent' ignores these warnings, 'print' prints the warning to the console, and 'file' saves the warnings in a text file. `warning_file` gives the location of a file to print warnings to, if `warning_type` is set to 'file'. `...` parameters passed to `svd`, such as `nv` and `nu`.
## Value
Solves svd(x)=UDV', where U is an matrix containing the left singular vectors of x, D is a diagonal matrix containing the singular values of x; and V is a matrix containing the right singular vectors of x (output follows the same notation convention as the `svd` function).
`qrSVD` will attempt the standard `svd` function before preconditioning the matrix x.
`fastSVD`
```1 2 3``` ```x <-matrix(rnorm(3*5),nrow=3,ncol=5) svdx <- qrSVD(x) svdx ```
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https://ftp.mcs.anl.gov/pub/fathom/moab-docs/GeomUtilTests_8cpp_source.html
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MOAB: Mesh Oriented datABase (version 5.3.0)
GeomUtilTests.cpp
Go to the documentation of this file.
00001 #include "moab/Core.hpp"
00002 #include "moab/GeomUtil.hpp"
00003
00004 using namespace moab;
00005 using namespace moab::GeomUtil;
00006
00007 #include <iostream>
00008
00009 #include "TestUtil.hpp"
00010 const double TOL = 1e-6;
00011 #define ASSERT_VECTORS_EQUAL( A, B ) assert_vectors_equal( ( A ), ( B ), #A, #B, __LINE__ )
00012 #define ASSERT_DOUBLES_EQUAL( A, B ) CHECK_REAL_EQUAL( A, B, TOL )
00013 #define ASSERT( B ) CHECK( B )
00014
00015 void assert_vectors_equal( const CartVect& a, const CartVect& b, const char* sa, const char* sb, int lineno )
00016 {
00017 if( fabs( a[0] - b[0] ) > TOL || fabs( a[1] - b[1] ) > TOL || fabs( a[2] - b[2] ) > TOL )
00018 {
00019 std::cerr << "Assertion failed at line " << lineno << std::endl
00020 << "\t" << sa << " == " << sb << std::endl
00021 << "\t[" << a[0] << ", " << a[1] << ", " << a[2] << "] == [" << b[0] << ", " << b[1] << ", " << b[2]
00022 << "]" << std::endl;
00023 FLAG_ERROR;
00024 }
00025 }
00026
00027 void test_box_plane_norm( CartVect norm, CartVect min, CartVect max )
00028 {
00029 CartVect c_lower = min;
00030 CartVect c_upper = max;
00031 for( int i = 0; i < 3; ++i )
00032 if( norm[i] < 0.0 ) std::swap( c_lower[i], c_upper[i] );
00033
00034 CartVect p_below = c_lower - norm;
00035 CartVect p_lower = c_lower + norm;
00036 CartVect p_upper = c_upper - norm;
00037 CartVect p_above = c_upper + norm;
00038
00039 double below = -( p_below % norm );
00040 double lower = -( p_lower % norm );
00041 double upper = -( p_upper % norm );
00042 double above = -( p_above % norm );
00043
00044 ASSERT( !box_plane_overlap( norm, below, min, max ) );
00045 ASSERT( box_plane_overlap( norm, lower, min, max ) );
00046 ASSERT( box_plane_overlap( norm, upper, min, max ) );
00047 ASSERT( !box_plane_overlap( norm, above, min, max ) );
00048 }
00049
00050 void test_box_plane_axis( int axis, double ns, const CartVect& min, const CartVect& max )
00051 {
00052 CartVect norm( 0.0 );
00053 norm[axis] = ns;
00054 test_box_plane_norm( norm, min, max );
00055 }
00056
00057 void test_box_plane_edge( int axis1, int axis2, bool flip_axis2, CartVect min, CartVect max )
00058 {
00059 CartVect norm( 0.0 );
00060 norm[axis1] = max[axis1] - min[axis1];
00061 if( flip_axis2 )
00062 norm[axis2] = min[axis2] - max[axis2];
00063 else
00064 norm[axis2] = max[axis2] - min[axis2];
00065 norm.normalize();
00066
00067 test_box_plane_norm( norm, min, max );
00068 }
00069
00070 void test_box_plane_corner( int xdir, int ydir, int zdir, CartVect min, CartVect max )
00071 {
00072 CartVect norm( max - min );
00073 norm[0] *= xdir;
00074 norm[1] *= ydir;
00075 norm[2] *= zdir;
00076 test_box_plane_norm( norm, min, max );
00077 }
00078
00079 void test_box_plane_overlap()
00080 {
00081 const CartVect min( -1, -2, -3 );
00082 const CartVect max( 6, 4, 2 );
00083
00084 // test with planes orthogonal to Z axis
00085 test_box_plane_axis( 2, 2.0, min, max );
00086 // test with planes orthogonal to X axis
00087 test_box_plane_axis( 1, -2.0, min, max );
00088 // test with planes orthogonal to Y axis
00089 test_box_plane_axis( 1, 1.0, min, max );
00090
00091 // test with plane orthogonal to face diagonals
00092 test_box_plane_edge( 0, 1, true, min, max );
00093 test_box_plane_edge( 0, 1, false, min, max );
00094 test_box_plane_edge( 0, 2, true, min, max );
00095 test_box_plane_edge( 0, 2, false, min, max );
00096 test_box_plane_edge( 2, 1, true, min, max );
00097 test_box_plane_edge( 2, 1, false, min, max );
00098
00099 // test with plane orthogonal to box diagonals
00100 test_box_plane_corner( 1, 1, 1, min, max );
00101 test_box_plane_corner( 1, 1, -1, min, max );
00102 test_box_plane_corner( 1, -1, -1, min, max );
00103 test_box_plane_corner( 1, -1, 1, min, max );
00104 }
00105
00106 class ElemOverlapTest
00107 {
00108 public:
00109 virtual bool operator()( const CartVect* coords, const CartVect& box_center, const CartVect& box_dims ) const = 0;
00110 };
00111 class LinearElemOverlapTest : public ElemOverlapTest
00112 {
00113 public:
00114 const EntityType type;
00115 LinearElemOverlapTest( EntityType t ) : type( t ) {}
00116 bool operator()( const CartVect* coords, const CartVect& box_center, const CartVect& box_dims ) const
00117 {
00118 return box_linear_elem_overlap( coords, type, box_center, box_dims );
00119 }
00120 };
00121 class TypeElemOverlapTest : public ElemOverlapTest
00122 {
00123 public:
00124 bool ( *func )( const CartVect*, const CartVect&, const CartVect& );
00125 TypeElemOverlapTest( bool ( *f )( const CartVect*, const CartVect&, const CartVect& ) ) : func( f ) {}
00126 bool operator()( const CartVect* coords, const CartVect& box_center, const CartVect& box_dims ) const
00127 {
00128 return ( *func )( coords, box_center, box_dims );
00129 }
00130 };
00131
00132 void general_box_tri_overlap_test( const ElemOverlapTest& overlap )
00133 {
00134 CartVect coords[3];
00135 CartVect center, dims;
00136
00137 // test box projection within triangle, z-plane
00138 coords[0] = CartVect( 0, 0, 0 );
00139 coords[1] = CartVect( 0, 4, 0 );
00140 coords[2] = CartVect( -4, 0, 0 );
00141 center = CartVect( -2, 1, 0 );
00142 dims = CartVect( 1, 0.5, 3 );
00143 ASSERT( overlap( coords, center, dims ) );
00144 // move box below plane of triangle
00145 center[2] = -4;
00146 ASSERT( !overlap( coords, center, dims ) );
00147 // move box above plane of triangle
00148 center[2] = 4;
00149 ASSERT( !overlap( coords, center, dims ) );
00150
00151 // test box projection within triangle, x-plane
00152 coords[0] = CartVect( 3, 3, 0 );
00153 coords[1] = CartVect( 3, 3, 1 );
00154 coords[2] = CartVect( 3, 0, 0 );
00155 center = CartVect( 3, 2.5, .25 );
00156 dims = CartVect( 0.001, 0.4, .2 );
00157 ASSERT( overlap( coords, center, dims ) );
00158 // move box below plane of triangle
00159 center[0] = 2;
00160 ASSERT( !overlap( coords, center, dims ) );
00161 // move box above plane of triangle
00162 center[0] = 4;
00163 ASSERT( !overlap( coords, center, dims ) );
00164
00165 // test tri slices corner at +x,+y,+z
00166 coords[0] = CartVect( 3, 1, 1 );
00167 coords[1] = CartVect( 1, 3, 1 );
00168 coords[2] = CartVect( 1, 1, 3 );
00169 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00170 // test with tri above the corner
00171 ASSERT( !overlap( coords, CartVect( 0, 0, 0 ), CartVect( 1, 1, 1 ) ) );
00172 // test tri slices corner at -x,-y,-z
00173 coords[0] = CartVect( -1, 1, 1 );
00174 coords[1] = CartVect( 1, -1, 1 );
00175 coords[2] = CartVect( 1, 1, -1 );
00176 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00177 // test with tri below the corner
00178 ASSERT( !overlap( coords, CartVect( 2, 2, 2 ), CartVect( 1, 1, 1 ) ) );
00179
00180 // test tri slices corner at -x,+y,+z
00181 coords[0] = CartVect( 0.5, 0.0, 2.5 );
00182 coords[1] = CartVect( 0.5, 2.5, 0.0 );
00183 coords[2] = CartVect( -0.5, 0.0, 0.0 );
00184 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00185 // test with tri above the corner
00186 ASSERT( !overlap( coords, CartVect( 2, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00187
00188 // test tri slices corner at +x,-y,-z
00189 coords[0] = CartVect( 0.5, 0.0, -1.5 );
00190 coords[1] = CartVect( 0.5, -1.5, 0.0 );
00191 coords[2] = CartVect( 1.5, 0.0, 0.0 );
00192 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00193 // test with tri above the corner
00194 ASSERT( !overlap( coords, CartVect( 0, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00195
00196 // test tri slices corner at +x,-y,+z
00197 coords[0] = CartVect( 1.0, 1.0, 2.5 );
00198 coords[1] = CartVect( 2.5, 1.0, 1.0 );
00199 coords[2] = CartVect( 1.0, -0.5, 1.0 );
00200 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00201 // test with tri above the corner
00202 ASSERT( !overlap( coords, CartVect( -1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00203
00204 // test tri slices corner at -x,+y,-z
00205 coords[0] = CartVect( 1.0, 1.0, -0.5 );
00206 coords[1] = CartVect( -0.5, 1.0, 1.0 );
00207 coords[2] = CartVect( 1.0, 2.5, 1.0 );
00208 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00209 // test with tri above the corner
00210 ASSERT( !overlap( coords, CartVect( 3, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00211
00212 // test tri slices corner at +x,+y,-z
00213 coords[0] = CartVect( -0.1, 1.0, 1.0 );
00214 coords[1] = CartVect( 1.0, -0.1, 1.0 );
00215 coords[2] = CartVect( 1.0, 1.0, -0.1 );
00216 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00217 // test with tri outside box
00218 ASSERT( !overlap( coords, CartVect( 1, 1, 3 ), CartVect( 1, 1, 1 ) ) );
00219
00220 // test tri slices corner at -x,-y,+z
00221 coords[0] = CartVect( 2.1, 1.0, 1.0 );
00222 coords[1] = CartVect( 1.0, 2.1, 1.0 );
00223 coords[2] = CartVect( 1.0, 1.0, 2.1 );
00224 ASSERT( box_tri_overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00225 // test with tri outside box
00226 ASSERT( !overlap( coords, CartVect( 1, 1, -1 ), CartVect( 1, 1, 1 ) ) );
00227
00228 // box edge parallel to x at +y,+z passes through triangle
00229 coords[0] = CartVect( 1.0, 1.0, 3.0 );
00230 coords[1] = CartVect( 1.0, 3.0, 3.0 );
00231 coords[2] = CartVect( 1.0, 3.0, 1.0 );
00232 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00233 // test with tri outside box
00234 ASSERT( !overlap( coords, CartVect( 1, 1, 0.3 ), CartVect( 1, 1, 1 ) ) );
00235
00236 // box edge parallel to x at +y,-z passes through triangle
00237 coords[0] = CartVect( 1.0, 3.0, 1.0 );
00238 coords[1] = CartVect( 1.0, 3.0, -1.0 );
00239 coords[2] = CartVect( 1.0, 1.0, -1.0 );
00240 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00241 // test with tri outside box
00242 ASSERT( !overlap( coords, CartVect( 1, 1, 1.7 ), CartVect( 1, 1, 1 ) ) );
00243
00244 // box edge parallel to x at -y,-z passes through triangle
00245 coords[0] = CartVect( 1.0, -1.0, 1.0 );
00246 coords[1] = CartVect( 1.0, -1.0, -1.0 );
00247 coords[2] = CartVect( 1.0, 1.0, -1.0 );
00248 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00249 // test with tri outside box
00250 ASSERT( !overlap( coords, CartVect( 1, 1, 1.7 ), CartVect( 1, 1, 1 ) ) );
00251
00252 // box edge parallel to x at -y,+z passes through triangle
00253 coords[0] = CartVect( 1.0, -1.0, 1.0 );
00254 coords[1] = CartVect( 1.0, -1.0, 3.0 );
00255 coords[2] = CartVect( 1.0, 1.0, 3.0 );
00256 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00257 // test with tri outside box
00258 ASSERT( !overlap( coords, CartVect( 1, 1, 0.3 ), CartVect( 1, 1, 1 ) ) );
00259
00260 // box edge parallel to y at +x,+z passes through triangle
00261 coords[0] = CartVect( 1.0, 1.0, 3.0 );
00262 coords[1] = CartVect( 3.0, 1.0, 3.0 );
00263 coords[2] = CartVect( 3.0, 1.0, 1.0 );
00264 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00265 // test with tri outside box
00266 ASSERT( !overlap( coords, CartVect( 1, 1, 0.3 ), CartVect( 1, 1, 1 ) ) );
00267
00268 // box edge parallel to y at -x,+z passes through triangle
00269 coords[0] = CartVect( 1.0, 1.0, 3.0 );
00270 coords[1] = CartVect( -1.0, 1.0, 3.0 );
00271 coords[2] = CartVect( -1.0, 1.0, 1.0 );
00272 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00273 // test with tri outside box
00274 ASSERT( !overlap( coords, CartVect( 1, 1, 0.3 ), CartVect( 1, 1, 1 ) ) );
00275
00276 // box edge parallel to y at +x,-z passes through triangle
00277 coords[0] = CartVect( 1.0, 1.0, -1.0 );
00278 coords[1] = CartVect( 3.0, 1.0, -1.0 );
00279 coords[2] = CartVect( 3.0, 1.0, 1.0 );
00280 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00281 // test with tri outside box
00282 ASSERT( !overlap( coords, CartVect( 1, 1, 1.7 ), CartVect( 1, 1, 1 ) ) );
00283
00284 // box edge parallel to y at -x,-z passes through triangle
00285 coords[0] = CartVect( 1.0, 1.0, -1.0 );
00286 coords[1] = CartVect( -1.0, 1.0, -1.0 );
00287 coords[2] = CartVect( -1.0, 1.0, 1.0 );
00288 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00289 // test with tri outside box
00290 ASSERT( !overlap( coords, CartVect( 1, 1, 1.7 ), CartVect( 1, 1, 1 ) ) );
00291
00292 // box edge parallel to z at +x,+y passes through triangle
00293 coords[0] = CartVect( 1.0, 3.0, 1.0 );
00294 coords[1] = CartVect( 3.0, 3.0, 1.0 );
00295 coords[2] = CartVect( 3.0, 1.0, 1.0 );
00296 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00297 // test with tri outside box
00298 ASSERT( !overlap( coords, CartVect( 0.3, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00299
00300 // box edge parallel to z at +x,-y passes through triangle
00301 coords[0] = CartVect( 1.0, -1.0, 1.0 );
00302 coords[1] = CartVect( 3.0, -1.0, 1.0 );
00303 coords[2] = CartVect( 3.0, 1.0, 1.0 );
00304 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00305 // test with tri outside box
00306 ASSERT( !overlap( coords, CartVect( 0.3, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00307
00308 // box edge parallel to z at -x,+y passes through triangle
00309 coords[0] = CartVect( 1.0, 3.0, 1.0 );
00310 coords[1] = CartVect( -1.0, 3.0, 1.0 );
00311 coords[2] = CartVect( -1.0, 1.0, 1.0 );
00312 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00313 // test with tri outside box
00314 ASSERT( !overlap( coords, CartVect( 1.7, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00315
00316 // box edge parallel to z at -x,-y passes through triangle
00317 coords[0] = CartVect( 1.0, -1.0, 1.0 );
00318 coords[1] = CartVect( -1.0, -1.0, 1.0 );
00319 coords[2] = CartVect( -1.0, 1.0, 1.0 );
00320 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00321 // test with tri outside box
00322 ASSERT( !overlap( coords, CartVect( 1.7, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00323
00324 // triangle penetrates +x face
00325 coords[0] = CartVect( 2.0, 2.0, 2.0 );
00326 coords[1] = CartVect( 5.0, 3.0, 2.0 );
00327 coords[2] = CartVect( 5.0, 1.0, 2.0 );
00328 ASSERT( overlap( coords, CartVect( 2, 2, 2 ), CartVect( 2, 2, 2 ) ) );
00329 // test with tri outside box
00330 ASSERT( !overlap( coords, CartVect( -1, 2, 2 ), CartVect( 2, 2, 2 ) ) );
00331
00332 // triangle penetrates -x face
00333 coords[0] = CartVect( 2.0, 2.0, 2.0 );
00334 coords[1] = CartVect( -1.0, 3.0, 2.0 );
00335 coords[2] = CartVect( -1.0, 1.0, 2.0 );
00336 ASSERT( overlap( coords, CartVect( 2, 2, 2 ), CartVect( 2, 2, 2 ) ) );
00337 // test with tri outside box
00338 ASSERT( !overlap( coords, CartVect( 5, 2, 2 ), CartVect( 2, 2, 2 ) ) );
00339
00340 // triangle penetrates +y face
00341 coords[0] = CartVect( 2.0, 2.0, 2.0 );
00342 coords[1] = CartVect( 3.0, 5.0, 2.0 );
00343 coords[2] = CartVect( 1.0, 5.0, 2.0 );
00344 ASSERT( overlap( coords, CartVect( 2, 2, 2 ), CartVect( 2, 2, 2 ) ) );
00345 // test with tri outside box
00346 ASSERT( !overlap( coords, CartVect( 2, -1, 2 ), CartVect( 2, 2, 2 ) ) );
00347
00348 // triangle penetrates -y face
00349 coords[0] = CartVect( 2.0, 2.0, 2.0 );
00350 coords[1] = CartVect( 3.0, -1.0, 2.0 );
00351 coords[2] = CartVect( 1.0, -1.0, 2.0 );
00352 ASSERT( overlap( coords, CartVect( 2, 2, 2 ), CartVect( 2, 2, 2 ) ) );
00353 // test with tri outside box
00354 ASSERT( !overlap( coords, CartVect( 2, 5, 2 ), CartVect( 2, 2, 2 ) ) );
00355
00356 // triangle penetrates +z face
00357 coords[0] = CartVect( 2.0, 2.0, 2.0 );
00358 coords[1] = CartVect( 2.0, 3.0, 5.0 );
00359 coords[2] = CartVect( 2.0, 1.0, 5.0 );
00360 ASSERT( overlap( coords, CartVect( 2, 2, 2 ), CartVect( 2, 2, 2 ) ) );
00361 // test with tri outside box
00362 ASSERT( !overlap( coords, CartVect( 2, 2, -1 ), CartVect( 2, 2, 2 ) ) );
00363
00364 // triangle penetrates -z face
00365 coords[0] = CartVect( 2.0, 2.0, 2.0 );
00366 coords[1] = CartVect( 2.0, 3.0, -1.0 );
00367 coords[2] = CartVect( 2.0, 1.0, -1.0 );
00368 ASSERT( overlap( coords, CartVect( 2, 2, 2 ), CartVect( 2, 2, 2 ) ) );
00369 // test with tri outside box
00370 ASSERT( !overlap( coords, CartVect( 2, 2, 5 ), CartVect( 2, 2, 2 ) ) );
00371 }
00372
00373 void general_box_hex_overlap_test( const ElemOverlapTest& overlap )
00374 {
00375 CartVect coords[8];
00376
00377 // test against axis-aligned rectilinear hex
00378 coords[0] = CartVect( -0.5, -0.5, -0.5 );
00379 coords[1] = CartVect( 0.5, -0.5, -0.5 );
00380 coords[2] = CartVect( 0.5, 0.5, -0.5 );
00381 coords[3] = CartVect( -0.5, 0.5, -0.5 );
00382 coords[4] = CartVect( -0.5, -0.5, 0.5 );
00383 coords[5] = CartVect( 0.5, -0.5, 0.5 );
00384 coords[6] = CartVect( 0.5, 0.5, 0.5 );
00385 coords[7] = CartVect( -0.5, 0.5, 0.5 );
00386
00387 ASSERT( overlap( coords, CartVect( 0, 0, 0 ), CartVect( 1, 1, 1 ) ) );
00388
00389 ASSERT( overlap( coords, CartVect( 1, 0, 0 ), CartVect( 1, 1, 1 ) ) );
00390 ASSERT( overlap( coords, CartVect( 0, 1, 0 ), CartVect( 1, 1, 1 ) ) );
00391 ASSERT( overlap( coords, CartVect( 0, 0, 1 ), CartVect( 1, 1, 1 ) ) );
00392 ASSERT( overlap( coords, CartVect( -1, 0, 0 ), CartVect( 1, 1, 1 ) ) );
00393 ASSERT( overlap( coords, CartVect( 0, -1, 0 ), CartVect( 1, 1, 1 ) ) );
00394 ASSERT( overlap( coords, CartVect( 0, 0, -1 ), CartVect( 1, 1, 1 ) ) );
00395
00396 ASSERT( overlap( coords, CartVect( 1, 1, 0 ), CartVect( 1, 1, 1 ) ) );
00397 ASSERT( overlap( coords, CartVect( -1, 1, 0 ), CartVect( 1, 1, 1 ) ) );
00398 ASSERT( overlap( coords, CartVect( -1, -1, 0 ), CartVect( 1, 1, 1 ) ) );
00399 ASSERT( overlap( coords, CartVect( 1, -1, 0 ), CartVect( 1, 1, 1 ) ) );
00400 ASSERT( overlap( coords, CartVect( 1, 0, 1 ), CartVect( 1, 1, 1 ) ) );
00401 ASSERT( overlap( coords, CartVect( -1, 0, 1 ), CartVect( 1, 1, 1 ) ) );
00402 ASSERT( overlap( coords, CartVect( -1, 0, -1 ), CartVect( 1, 1, 1 ) ) );
00403 ASSERT( overlap( coords, CartVect( 1, 0, -1 ), CartVect( 1, 1, 1 ) ) );
00404 ASSERT( overlap( coords, CartVect( 0, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00405 ASSERT( overlap( coords, CartVect( 0, -1, 1 ), CartVect( 1, 1, 1 ) ) );
00406 ASSERT( overlap( coords, CartVect( 0, -1, -1 ), CartVect( 1, 1, 1 ) ) );
00407 ASSERT( overlap( coords, CartVect( 0, 1, -1 ), CartVect( 1, 1, 1 ) ) );
00408
00409 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00410 ASSERT( overlap( coords, CartVect( -1, 1, 1 ), CartVect( 1, 1, 1 ) ) );
00411 ASSERT( overlap( coords, CartVect( -1, -1, 1 ), CartVect( 1, 1, 1 ) ) );
00412 ASSERT( overlap( coords, CartVect( 1, -1, 1 ), CartVect( 1, 1, 1 ) ) );
00413 ASSERT( overlap( coords, CartVect( 1, 1, -1 ), CartVect( 1, 1, 1 ) ) );
00414 ASSERT( overlap( coords, CartVect( -1, 1, -1 ), CartVect( 1, 1, 1 ) ) );
00415 ASSERT( overlap( coords, CartVect( -1, -1, -1 ), CartVect( 1, 1, 1 ) ) );
00416 ASSERT( overlap( coords, CartVect( 1, -1, -1 ), CartVect( 1, 1, 1 ) ) );
00417
00418 ASSERT( !overlap( coords, CartVect( 3, 0, 0 ), CartVect( 1, 1, 1 ) ) );
00419 ASSERT( !overlap( coords, CartVect( 0, 3, 0 ), CartVect( 1, 1, 1 ) ) );
00420 ASSERT( !overlap( coords, CartVect( 0, 0, 3 ), CartVect( 1, 1, 1 ) ) );
00421 ASSERT( !overlap( coords, CartVect( -3, 0, 0 ), CartVect( 1, 1, 1 ) ) );
00422 ASSERT( !overlap( coords, CartVect( 0, -3, 0 ), CartVect( 1, 1, 1 ) ) );
00423 ASSERT( !overlap( coords, CartVect( 0, 0, -3 ), CartVect( 1, 1, 1 ) ) );
00424
00425 ASSERT( !overlap( coords, CartVect( 3, 3, 0 ), CartVect( 1, 1, 1 ) ) );
00426 ASSERT( !overlap( coords, CartVect( -3, 3, 0 ), CartVect( 1, 1, 1 ) ) );
00427 ASSERT( !overlap( coords, CartVect( -3, -3, 0 ), CartVect( 1, 1, 1 ) ) );
00428 ASSERT( !overlap( coords, CartVect( 3, -3, 0 ), CartVect( 1, 1, 1 ) ) );
00429 ASSERT( !overlap( coords, CartVect( 3, 0, 3 ), CartVect( 1, 1, 1 ) ) );
00430 ASSERT( !overlap( coords, CartVect( -3, 0, 3 ), CartVect( 1, 1, 1 ) ) );
00431 ASSERT( !overlap( coords, CartVect( -3, 0, -3 ), CartVect( 1, 1, 1 ) ) );
00432 ASSERT( !overlap( coords, CartVect( 3, 0, -3 ), CartVect( 1, 1, 1 ) ) );
00433 ASSERT( !overlap( coords, CartVect( 0, 3, 3 ), CartVect( 1, 1, 1 ) ) );
00434 ASSERT( !overlap( coords, CartVect( 0, -3, 3 ), CartVect( 1, 1, 1 ) ) );
00435 ASSERT( !overlap( coords, CartVect( 0, -3, -3 ), CartVect( 1, 1, 1 ) ) );
00436 ASSERT( !overlap( coords, CartVect( 0, 3, -3 ), CartVect( 1, 1, 1 ) ) );
00437
00438 ASSERT( !overlap( coords, CartVect( 3, 3, 3 ), CartVect( 1, 1, 1 ) ) );
00439 ASSERT( !overlap( coords, CartVect( -3, 3, 3 ), CartVect( 1, 1, 1 ) ) );
00440 ASSERT( !overlap( coords, CartVect( -3, -3, 3 ), CartVect( 1, 1, 1 ) ) );
00441 ASSERT( !overlap( coords, CartVect( 3, -3, 3 ), CartVect( 1, 1, 1 ) ) );
00442 ASSERT( !overlap( coords, CartVect( 3, 3, -3 ), CartVect( 1, 1, 1 ) ) );
00443 ASSERT( !overlap( coords, CartVect( -3, 3, -3 ), CartVect( 1, 1, 1 ) ) );
00444 ASSERT( !overlap( coords, CartVect( -3, -3, -3 ), CartVect( 1, 1, 1 ) ) );
00445 ASSERT( !overlap( coords, CartVect( 3, -3, -3 ), CartVect( 1, 1, 1 ) ) );
00446
00447 // test against rectilinear hex rotated 45 degrees about z axis
00448 const double r = sqrt( 2.0 ) / 2.0;
00449 coords[0] = CartVect( r, 0, -0.5 );
00450 coords[1] = CartVect( 0, r, -0.5 );
00451 coords[2] = CartVect( -r, 0, -0.5 );
00452 coords[3] = CartVect( 0, -r, -0.5 );
00453 coords[4] = CartVect( r, 0, 0.5 );
00454 coords[5] = CartVect( 0, r, 0.5 );
00455 coords[6] = CartVect( -r, 0, 0.5 );
00456 coords[7] = CartVect( 0, -r, 0.5 );
00457
00458 ASSERT( overlap( coords, CartVect( 1, 0, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00459 ASSERT( overlap( coords, CartVect( -1, 0, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00460 ASSERT( overlap( coords, CartVect( 0, 1, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00461 ASSERT( overlap( coords, CartVect( 0, -1, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00462
00463 ASSERT( !overlap( coords, CartVect( 1, 0, 2 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00464 ASSERT( !overlap( coords, CartVect( -1, 0, 2 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00465 ASSERT( !overlap( coords, CartVect( 0, 1, 2 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00466 ASSERT( !overlap( coords, CartVect( 0, -1, 2 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00467
00468 ASSERT( !overlap( coords, CartVect( 2, 0, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00469 ASSERT( !overlap( coords, CartVect( -2, 0, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00470 ASSERT( !overlap( coords, CartVect( 0, 2, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00471 ASSERT( !overlap( coords, CartVect( 0, -2, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00472
00473 ASSERT( !overlap( coords, CartVect( 1, 1, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00474 ASSERT( !overlap( coords, CartVect( -1, 1, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00475 ASSERT( !overlap( coords, CartVect( -1, -1, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00476 ASSERT( !overlap( coords, CartVect( 1, -1, 0 ), CartVect( 0.5, 0.5, 0.5 ) ) );
00477
00478 ASSERT( overlap( coords, CartVect( 1, 1, 0 ), CartVect( 0.75, 0.75, 0.5 ) ) );
00479 ASSERT( overlap( coords, CartVect( -1, 1, 0 ), CartVect( 0.75, 0.75, 0.5 ) ) );
00480 ASSERT( overlap( coords, CartVect( -1, -1, 0 ), CartVect( 0.75, 0.75, 0.5 ) ) );
00481 ASSERT( overlap( coords, CartVect( 1, -1, 0 ), CartVect( 0.75, 0.75, 0.5 ) ) );
00482 }
00483
00484 void general_box_tet_overlap_test( const ElemOverlapTest& overlap )
00485 {
00486 CartVect coords[4];
00487
00488 // Octant I
00489 coords[0] = CartVect( 0, 0, 0 );
00490 coords[1] = CartVect( 1, 0, 0 );
00491 coords[2] = CartVect( 0, 1, 0 );
00492 coords[3] = CartVect( 0, 0, 1 );
00493 // tet entirely within box
00494 ASSERT( overlap( coords, CartVect( -1, -1, -1 ), CartVect( 3, 3, 3 ) ) );
00495 // box entirely within tet
00496 ASSERT( overlap( coords, CartVect( 0.2, 0.2, 0.2 ), CartVect( 0.1, 0.1, 0.1 ) ) );
00497 // box corner penetrates tet face
00498 ASSERT( overlap( coords, CartVect( 0.5, 0.5, 0.5 ), CartVect( 0.2, 0.2, 0.2 ) ) );
00499 // box corner does not penetrate face
00500 ASSERT( !overlap( coords, CartVect( 0.5, 0.5, 0.5 ), CartVect( 0.15, 0.15, 0.15 ) ) );
00501
00502 // Octant II
00503 coords[0] = CartVect( 0, 1, 0 );
00504 coords[1] = CartVect( -1, 0, 0 );
00505 coords[2] = CartVect( 0, 0, 0 );
00506 coords[3] = CartVect( 0, 0, 1 );
00507 // tet entirely within box
00508 ASSERT( overlap( coords, CartVect( 1, -1, -1 ), CartVect( 3, 3, 3 ) ) );
00509 // box entirely within tet
00510 ASSERT( overlap( coords, CartVect( -0.2, 0.2, 0.2 ), CartVect( 0.1, 0.1, 0.1 ) ) );
00511 // box corner penetrates tet face
00512 ASSERT( overlap( coords, CartVect( -0.5, 0.5, 0.5 ), CartVect( 0.2, 0.2, 0.2 ) ) );
00513 // box corner does not penetrate face
00514 ASSERT( !overlap( coords, CartVect( -0.5, 0.5, 0.5 ), CartVect( 0.15, 0.15, 0.15 ) ) );
00515
00516 // Octant III
00517 coords[0] = CartVect( 0, -1, 0 );
00518 coords[1] = CartVect( 0, 0, 0 );
00519 coords[2] = CartVect( -1, 0, 0 );
00520 coords[3] = CartVect( 0, 0, 1 );
00521 // tet entirely within box
00522 ASSERT( overlap( coords, CartVect( 1, 1, -1 ), CartVect( 3, 3, 3 ) ) );
00523 // box entirely within tet
00524 ASSERT( overlap( coords, CartVect( -0.2, -0.2, 0.2 ), CartVect( 0.1, 0.1, 0.1 ) ) );
00525 // box corner penetrates tet face
00526 ASSERT( overlap( coords, CartVect( -0.5, -0.5, 0.5 ), CartVect( 0.2, 0.2, 0.2 ) ) );
00527 // box corner does not penetrate face
00528 ASSERT( !overlap( coords, CartVect( -0.5, -0.5, 0.5 ), CartVect( 0.15, 0.15, 0.15 ) ) );
00529
00530 // Octant IV
00531 coords[0] = CartVect( 1, 0, 0 );
00532 coords[1] = CartVect( 0, -1, 0 );
00533 coords[2] = CartVect( 0, 0, 1 );
00534 coords[3] = CartVect( 0, 0, 0 );
00535 // tet entirely within box
00536 ASSERT( overlap( coords, CartVect( -1, 1, -1 ), CartVect( 3, 3, 3 ) ) );
00537 // box entirely within tet
00538 ASSERT( overlap( coords, CartVect( 0.2, -0.2, 0.2 ), CartVect( 0.1, 0.1, 0.1 ) ) );
00539 // box corner penetrates tet face
00540 ASSERT( overlap( coords, CartVect( 0.5, -0.5, 0.5 ), CartVect( 0.2, 0.2, 0.2 ) ) );
00541 // box corner does not penetrate face
00542 ASSERT( !overlap( coords, CartVect( 0.5, -0.5, 0.5 ), CartVect( 0.15, 0.15, 0.15 ) ) );
00543
00544 // Octant V
00545 coords[0] = CartVect( 0, 0, 0 );
00546 coords[1] = CartVect( 0, 1, 0 );
00547 coords[2] = CartVect( 1, 0, 0 );
00548 coords[3] = CartVect( 0, 0, -1 );
00549 // tet entirely within box
00550 ASSERT( overlap( coords, CartVect( -1, -1, 1 ), CartVect( 3, 3, 3 ) ) );
00551 // box entirely within tet
00552 ASSERT( overlap( coords, CartVect( 0.2, 0.2, -0.2 ), CartVect( 0.1, 0.1, 0.1 ) ) );
00553 // box corner penetrates tet face
00554 ASSERT( overlap( coords, CartVect( 0.5, 0.5, -0.5 ), CartVect( 0.2, 0.2, 0.2 ) ) );
00555 // box corner does not penetrate face
00556 ASSERT( !overlap( coords, CartVect( 0.5, 0.5, -0.5 ), CartVect( 0.15, 0.15, 0.15 ) ) );
00557
00558 // Octant VI
00559 coords[0] = CartVect( -1, 0, 0 );
00560 coords[1] = CartVect( 0, 1, 0 );
00561 coords[2] = CartVect( 0, 0, 0 );
00562 coords[3] = CartVect( 0, 0, -1 );
00563 // tet entirely within box
00564 ASSERT( overlap( coords, CartVect( 1, -1, 1 ), CartVect( 3, 3, 3 ) ) );
00565 // box entirely within tet
00566 ASSERT( overlap( coords, CartVect( -0.2, 0.2, -0.2 ), CartVect( 0.1, 0.1, 0.1 ) ) );
00567 // box corner penetrates tet face
00568 ASSERT( overlap( coords, CartVect( -0.5, 0.5, -0.5 ), CartVect( 0.2, 0.2, 0.2 ) ) );
00569 // box corner does not penetrate face
00570 ASSERT( !overlap( coords, CartVect( -0.5, 0.5, -0.5 ), CartVect( 0.15, 0.15, 0.15 ) ) );
00571
00572 // Octant VII
00573 coords[0] = CartVect( 0, 0, 0 );
00574 coords[1] = CartVect( 0, -1, 0 );
00575 coords[2] = CartVect( -1, 0, 0 );
00576 coords[3] = CartVect( 0, 0, -1 );
00577 // tet entirely within box
00578 ASSERT( overlap( coords, CartVect( 1, 1, 1 ), CartVect( 3, 3, 3 ) ) );
00579 // box entirely within tet
00580 ASSERT( overlap( coords, CartVect( -0.2, -0.2, -0.2 ), CartVect( 0.1, 0.1, 0.1 ) ) );
00581 // box corner penetrates tet face
00582 ASSERT( overlap( coords, CartVect( -0.5, -0.5, -0.5 ), CartVect( 0.2, 0.2, 0.2 ) ) );
00583 // box corner does not penetrate face
00584 ASSERT( !overlap( coords, CartVect( -0.5, -0.5, -0.5 ), CartVect( 0.15, 0.15, 0.15 ) ) );
00585
00586 // Octant VIII
00587 coords[0] = CartVect( 0, -1, 0 );
00588 coords[1] = CartVect( 1, 0, 0 );
00589 coords[2] = CartVect( 0, 0, -1 );
00590 coords[3] = CartVect( 0, 0, 0 );
00591 // tet entirely within box
00592 ASSERT( overlap( coords, CartVect( -1, 1, 1 ), CartVect( 3, 3, 3 ) ) );
00593 // box entirely within tet
00594 ASSERT( overlap( coords, CartVect( 0.2, -0.2, -0.2 ), CartVect( 0.1, 0.1, 0.1 ) ) );
00595 // box corner penetrates tet face
00596 ASSERT( overlap( coords, CartVect( 0.5, -0.5, -0.5 ), CartVect( 0.2, 0.2, 0.2 ) ) );
00597 // box corner does not penetrate face
00598 ASSERT( !overlap( coords, CartVect( 0.5, -0.5, -0.5 ), CartVect( 0.15, 0.15, 0.15 ) ) );
00599
00600 // Box edge -x,-z
00601 coords[0] = CartVect( 0, 0, 0 );
00602 coords[1] = CartVect( 2, -1, 0 );
00603 coords[2] = CartVect( 2, 1, 0 );
00604 coords[3] = CartVect( 0, 0, 2 );
00605 // box edge passes through tet
00606 ASSERT( overlap( coords, CartVect( 1.5, 0.0, 1.5 ), CartVect( 1, 1, 1 ) ) );
00607 // box edge does not pass through tet
00608 ASSERT( !overlap( coords, CartVect( 2.5, 0.0, 2.5 ), CartVect( 1, 1, 1 ) ) );
00609
00610 // Box edge -y,-z
00611 coords[0] = CartVect( 1, 2, 0 );
00612 coords[1] = CartVect( -1, 2, 0 );
00613 coords[2] = CartVect( 0, 0, 0 );
00614 coords[3] = CartVect( 0, 0, 2 );
00615 // box edge passes through tet
00616 ASSERT( overlap( coords, CartVect( 0.0, 1.5, 1.5 ), CartVect( 1, 1, 1 ) ) );
00617 // box edge does not pass through tet
00618 ASSERT( !overlap( coords, CartVect( 0.0, 2.5, 2.5 ), CartVect( 1, 1, 1 ) ) );
00619
00620 // Box edge +x,-z
00621 coords[0] = CartVect( -2, -1, 0 );
00622 coords[1] = CartVect( -2, 1, 0 );
00623 coords[2] = CartVect( 0, 0, 2 );
00624 coords[3] = CartVect( 0, 0, 0 );
00625 // box edge passes through tet
00626 ASSERT( overlap( coords, CartVect( -1.5, 0.0, 1.5 ), CartVect( 1, 1, 1 ) ) );
00627 // box edge does not pass through tet
00628 ASSERT( !overlap( coords, CartVect( -2.5, 0.0, 2.5 ), CartVect( 1, 1, 1 ) ) );
00629
00630 // Box edge +y,-z
00631 coords[0] = CartVect( 2, -1, 0 );
00632 coords[1] = CartVect( 0, 0, 0 );
00633 coords[2] = CartVect( -2, -1, 0 );
00634 coords[3] = CartVect( 0, 0, 2 );
00635 // box edge passes through tet
00636 ASSERT( overlap( coords, CartVect( 0.0, -1.5, 1.5 ), CartVect( 1, 1, 1 ) ) );
00637 // box edge does not pass through tet
00638 ASSERT( !overlap( coords, CartVect( 0.0, -2.5, 2.5 ), CartVect( 1, 1, 1 ) ) );
00639
00640 // Box edge -x,+z
00641 coords[0] = CartVect( 2, -1, 0 );
00642 coords[1] = CartVect( 0, 0, 0 );
00643 coords[2] = CartVect( 2, 1, 0 );
00644 coords[3] = CartVect( 0, 0, -2 );
00645 // box edge passes through tet
00646 ASSERT( overlap( coords, CartVect( 1.5, 0.0, -1.5 ), CartVect( 1, 1, 1 ) ) );
00647 // box edge does not pass through tet
00648 ASSERT( !overlap( coords, CartVect( 2.5, 0.0, -2.5 ), CartVect( 1, 1, 1 ) ) );
00649
00650 // Box edge -y,+z
00651 coords[0] = CartVect( -1, 2, 0 );
00652 coords[1] = CartVect( 1, 2, 0 );
00653 coords[2] = CartVect( 0, 0, 0 );
00654 coords[3] = CartVect( 0, 0, -2 );
00655 // box edge passes through tet
00656 ASSERT( overlap( coords, CartVect( 0.0, 1.5, -1.5 ), CartVect( 1, 1, 1 ) ) );
00657 // box edge does not pass through tet
00658 ASSERT( !overlap( coords, CartVect( 0.0, 2.5, -2.5 ), CartVect( 1, 1, 1 ) ) );
00659
00660 // Box edge +x,+z
00661 coords[0] = CartVect( -2, 1, 0 );
00662 coords[1] = CartVect( -2, -1, 0 );
00663 coords[2] = CartVect( 0, 0, -2 );
00664 coords[3] = CartVect( 0, 0, 0 );
00665 // box edge passes through tet
00666 ASSERT( overlap( coords, CartVect( -1.5, 0.0, -1.5 ), CartVect( 1, 1, 1 ) ) );
00667 // box edge does not pass through tet
00668 ASSERT( !overlap( coords, CartVect( -2.5, 0.0, -2.5 ), CartVect( 1, 1, 1 ) ) );
00669
00670 // Box edge +y,+z
00671 coords[0] = CartVect( 0, 0, 0 );
00672 coords[1] = CartVect( 2, -1, 0 );
00673 coords[2] = CartVect( -2, -1, 0 );
00674 coords[3] = CartVect( 0, 0, -2 );
00675 // box edge passes through tet
00676 ASSERT( overlap( coords, CartVect( 0.0, -1.5, -1.5 ), CartVect( 1, 1, 1 ) ) );
00677 // box edge does not pass through tet
00678 ASSERT( !overlap( coords, CartVect( 0.0, -2.5, -2.5 ), CartVect( 1, 1, 1 ) ) );
00679
00680 // Box edge -x,-y
00681 coords[0] = CartVect( 0, 0, 0 );
00682 coords[1] = CartVect( 0, 2, -1 );
00683 coords[2] = CartVect( 0, 2, 1 );
00684 coords[3] = CartVect( 2, 0, 0 );
00685 // box edge passes through tet
00686 ASSERT( overlap( coords, CartVect( 1.5, 1.5, 0.0 ), CartVect( 1, 1, 1 ) ) );
00687 // box edge does not pass through tet
00688 ASSERT( !overlap( coords, CartVect( 2.5, 2.5, 0.0 ), CartVect( 1, 1, 1 ) ) );
00689
00690 // Box edge +x,-y
00691 coords[0] = CartVect( 0, 2, -1 );
00692 coords[1] = CartVect( 0, 0, 0 );
00693 coords[2] = CartVect( 0, 2, 1 );
00694 coords[3] = CartVect( -2, 0, 0 );
00695 // box edge passes through tet
00696 ASSERT( overlap( coords, CartVect( -1.5, 1.5, 0.0 ), CartVect( 1, 1, 1 ) ) );
00697 // box edge does not pass through tet
00698 ASSERT( !overlap( coords, CartVect( -2.5, 2.5, 0.0 ), CartVect( 1, 1, 1 ) ) );
00699
00700 // Box edge -x,+y
00701 coords[0] = CartVect( 0, -2, 1 );
00702 coords[1] = CartVect( 0, -2, -1 );
00703 coords[2] = CartVect( 0, 0, 0 );
00704 coords[3] = CartVect( 2, 0, 0 );
00705 // box edge passes through tet
00706 ASSERT( overlap( coords, CartVect( 1.5, -1.5, 0.0 ), CartVect( 1, 1, 1 ) ) );
00707 // box edge does not pass through tet
00708 ASSERT( !overlap( coords, CartVect( 2.5, -2.5, 0.0 ), CartVect( 1, 1, 1 ) ) );
00709
00710 // Box edge +x,+y
00711 coords[0] = CartVect( 0, -2, -1 );
00712 coords[1] = CartVect( -2, 0, 0 );
00713 coords[2] = CartVect( 0, -2, 1 );
00714 coords[3] = CartVect( 0, 0, 0 );
00715 // box edge passes through tet
00716 ASSERT( overlap( coords, CartVect( -1.5, -1.5, 0.0 ), CartVect( 1, 1, 1 ) ) );
00717 // box edge does not pass through tet
00718 ASSERT( !overlap( coords, CartVect( -2.5, -2.5, 0.0 ), CartVect( 1, 1, 1 ) ) );
00719
00720 // Test tet edge through box
00721 coords[0] = CartVect( -0.13369421660900116, -2.9871494770050049, 0.0526076555252075 );
00722 coords[1] = CartVect( -0.00350524857640266, -3.3236153125762939, 0.2924639880657196 );
00723 coords[2] = CartVect( 0.16473215818405151, -2.9966945648193359, -0.1936169415712357 );
00724 coords[3] = CartVect( 0.26740345358848572, -2.8492588996887207, 0.1519143134355545 );
00725 ASSERT( overlap( coords, CartVect( -2.5, -2.8, -2.5 ), CartVect( 2.5, 0.31, 2.5 ) ) );
00726 }
00727
00728 void test_box_tri_overlap()
00729 {
00730 general_box_tri_overlap_test( TypeElemOverlapTest( &box_tri_overlap ) );
00731 }
00732
00733 void test_box_linear_elem_overlap_tri()
00734 {
00735 general_box_tri_overlap_test( LinearElemOverlapTest( MBTRI ) );
00736 }
00737
00738 void test_box_hex_overlap()
00739 {
00740 general_box_hex_overlap_test( TypeElemOverlapTest( &box_hex_overlap ) );
00741 }
00742
00743 void test_box_linear_elem_overlap_hex()
00744 {
00745 general_box_hex_overlap_test( LinearElemOverlapTest( MBHEX ) );
00746 }
00747
00748 void test_box_tet_overlap()
00749 {
00750 general_box_tet_overlap_test( TypeElemOverlapTest( &box_tet_overlap ) );
00751 }
00752
00753 void test_box_linear_elem_overlap_tet()
00754 {
00755 general_box_tet_overlap_test( LinearElemOverlapTest( MBTET ) );
00756 }
00757
00758 void test_ray_tri_intersect()
00759 {
00760 bool xsect;
00761 double t;
00762
00763 // define a triangle
00764 const CartVect tri[3] = { CartVect( 1.0, 0.0, 0.0 ), CartVect( 0.0, 1.0, 0.0 ), CartVect( 0.0, 0.0, 1.0 ) };
00765
00766 // try a ray through the center of the triangle
00767 xsect = ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( 1.0, 1.0, 1.0 ), t );
00768 ASSERT( xsect );
00769 ASSERT_DOUBLES_EQUAL( 1.0 / 3.0, t );
00770
00771 // try a same ray, but move base point above triangle
00772 xsect = ray_tri_intersect( tri, CartVect( 1.0, 1.0, 1.0 ), CartVect( 1.0, 1.0, 1.0 ), t );
00773 ASSERT( !xsect );
00774
00775 // try a same ray the other direction with base point below triangle
00776 xsect = ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( -1.0, -1.0, -1.0 ), t );
00777 ASSERT( !xsect );
00778
00779 // try a ray that passes above the triangle
00780 xsect = ray_tri_intersect( tri, CartVect( 1.0, 1.0, 1.0 ), CartVect( -1.0, -1.0, 1.0 ), t );
00781 ASSERT( !xsect );
00782
00783 // try a skew ray
00784 xsect = ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( 1.0, 1.0, -0.1 ), t );
00785 ASSERT( !xsect );
00786 }
00787
00788 void test_plucker_ray_tri_intersect()
00789 {
00790 bool xsect;
00791 double t;
00792
00793 // define a triangle
00794 const CartVect tri[3] = { CartVect( 1.0, 0.0, 0.0 ), CartVect( 0.0, 1.0, 0.0 ), CartVect( 0.0, 0.0, 1.0 ) };
00795
00796 // try a ray through the center of the triangle
00797 xsect = plucker_ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( 1.0, 1.0, 1.0 ), t );
00798 ASSERT( xsect );
00799 ASSERT_DOUBLES_EQUAL( 1.0 / 3.0, t );
00800
00801 // try a same ray, but move base point above triangle
00802 xsect = plucker_ray_tri_intersect( tri, CartVect( 1.0, 1.0, 1.0 ), CartVect( 1.0, 1.0, 1.0 ), t );
00803 ASSERT( !xsect );
00804
00805 // try a same ray the other direction with base point below triangle
00806 xsect = plucker_ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( -1.0, -1.0, -1.0 ), t );
00807 ASSERT( !xsect );
00808
00809 // try a ray that passes above the triangle
00810 xsect = plucker_ray_tri_intersect( tri, CartVect( 1.0, 1.0, 1.0 ), CartVect( -1.0, -1.0, 1.0 ), t );
00811 ASSERT( !xsect );
00812
00813 // try a skew ray
00814 xsect = plucker_ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( 1.0, 1.0, -0.1 ), t );
00815 ASSERT( !xsect );
00816
00817 // try a ray that intersects with wrong orientation
00818 const int orientation = -1.0;
00819 xsect = plucker_ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( 1.0, 1.0, 1.0 ), t, NULL, NULL,
00820 &orientation );
00821 ASSERT( !xsect );
00822
00823 // try a ray that intersects beyond the nonneg_ray_len
00824 const double nonneg_ray_len = 0.25;
00825 xsect = plucker_ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( 1.0, 1.0, 1.0 ), t, &nonneg_ray_len );
00826 ASSERT( !xsect );
00827
00828 // try a ray that intersects behind the origin
00829 const double neg_ray_len = -2.0;
00830 xsect = plucker_ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( -1.0, -1.0, -1.0 ), t, NULL,
00831 &neg_ray_len );
00832 ASSERT( xsect );
00833
00834 // try a ray that intersects a node
00835 intersection_type int_type;
00836 xsect = plucker_ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( 1.0, 0.0, 0.0 ), t, NULL, NULL, NULL,
00837 &int_type );
00838 ASSERT( xsect );
00839 ASSERT( NODE0 == int_type );
00840
00841 // try a ray that intersects an edge
00842 xsect = plucker_ray_tri_intersect( tri, CartVect( 0.0, 0.0, 0.0 ), CartVect( 1.0, 1.0, 0.0 ), t, NULL, NULL, NULL,
00843 &int_type );
00844 ASSERT( xsect );
00845 ASSERT( EDGE0 == int_type );
00846 }
00847
00848 void test_closest_location_on_tri()
00849 {
00850 CartVect result, input;
00851
00852 // define a triangle
00853 const CartVect tri[3] = { CartVect( 1.0, 0.0, 0.0 ), CartVect( 0.0, 1.0, 0.0 ), CartVect( 0.0, 0.0, 1.0 ) };
00854
00855 // try point at triangle centroid
00856 input = CartVect( 1.0 / 3.0, 1.0 / 3.0, 1.0 / 3.0 );
00857 closest_location_on_tri( input, tri, result );
00858 ASSERT_VECTORS_EQUAL( result, input );
00859
00860 // try point at each vertex
00861 closest_location_on_tri( tri[0], tri, result );
00862 ASSERT_VECTORS_EQUAL( result, tri[0] );
00863 closest_location_on_tri( tri[1], tri, result );
00864 ASSERT_VECTORS_EQUAL( result, tri[1] );
00865 closest_location_on_tri( tri[2], tri, result );
00866 ASSERT_VECTORS_EQUAL( result, tri[2] );
00867
00868 // try point at center of each edge
00869 input = 0.5 * ( tri[0] + tri[1] );
00870 closest_location_on_tri( input, tri, result );
00871 ASSERT_VECTORS_EQUAL( result, input );
00872 input = 0.5 * ( tri[0] + tri[2] );
00873 closest_location_on_tri( input, tri, result );
00874 ASSERT_VECTORS_EQUAL( result, input );
00875 input = 0.5 * ( tri[2] + tri[1] );
00876 closest_location_on_tri( input, tri, result );
00877 ASSERT_VECTORS_EQUAL( result, input );
00878
00879 // try a point above the center of the triangle
00880 input = CartVect( 1.0, 1.0, 1.0 );
00881 closest_location_on_tri( input, tri, result );
00882 ASSERT_VECTORS_EQUAL( result, CartVect( 1.0 / 3.0, 1.0 / 3.0, 1.0 / 3.0 ) );
00883
00884 // try a point below the center of the triangle
00885 input = CartVect( 0.0, 0.0, 0.0 );
00886 closest_location_on_tri( input, tri, result );
00887 ASSERT_VECTORS_EQUAL( result, CartVect( 1.0 / 3.0, 1.0 / 3.0, 1.0 / 3.0 ) );
00888
00889 // try a point closest to each vertex and 'outside' of both adjacent edges.
00890 input = 2 * tri[0];
00891 closest_location_on_tri( input, tri, result );
00892 ASSERT_VECTORS_EQUAL( result, tri[0] );
00893 input = 2 * tri[1];
00894 closest_location_on_tri( input, tri, result );
00895 ASSERT_VECTORS_EQUAL( result, tri[1] );
00896 input = 2 * tri[2];
00897 closest_location_on_tri( input, tri, result );
00898 ASSERT_VECTORS_EQUAL( result, tri[2] );
00899
00900 // try a point outside and closest to each edge
00901 input = tri[0] + tri[1];
00902 closest_location_on_tri( input, tri, result );
00903 ASSERT_VECTORS_EQUAL( result, 0.5 * input );
00904 input = tri[2] + tri[1];
00905 closest_location_on_tri( input, tri, result );
00906 ASSERT_VECTORS_EQUAL( result, 0.5 * input );
00907 input = tri[0] + tri[2];
00908 closest_location_on_tri( input, tri, result );
00909 ASSERT_VECTORS_EQUAL( result, 0.5 * input );
00910
00911 // define an equilateral triangle in the xy-plane
00912 const CartVect tri_xy[3] = { CartVect( 0.0, sqrt( 3.0 ) / 2.0, 0.0 ), CartVect( 0.5, 0.0, 0.0 ),
00913 CartVect( -0.5, 0.0, 0.0 ) };
00914
00915 // for each vertex, test point that is
00916 // - outside triangle
00917 // - closest to vertex
00918 // - 'inside' one of the adjacent edges
00919 // - 'outside' the other adjacent edge
00920 closest_location_on_tri( CartVect( -0.3, 1.2, 0.0 ), tri_xy, result );
00921 ASSERT_VECTORS_EQUAL( result, tri_xy[0] );
00922 closest_location_on_tri( CartVect( 0.3, 1.2, 0.0 ), tri_xy, result );
00923 ASSERT_VECTORS_EQUAL( result, tri_xy[0] );
00924 closest_location_on_tri( CartVect( 1.0, 0.1, 0.0 ), tri_xy, result );
00925 ASSERT_VECTORS_EQUAL( result, tri_xy[1] );
00926 closest_location_on_tri( CartVect( 0.6, -0.5, 0.0 ), tri_xy, result );
00927 ASSERT_VECTORS_EQUAL( result, tri_xy[1] );
00928 closest_location_on_tri( CartVect( -0.6, -0.5, 0.0 ), tri_xy, result );
00929 ASSERT_VECTORS_EQUAL( result, tri_xy[2] );
00930 closest_location_on_tri( CartVect( -1.0, 0.1, 0.0 ), tri_xy, result );
00931 ASSERT_VECTORS_EQUAL( result, tri_xy[2] );
00932 }
00933
00934 void test_closest_location_on_polygon()
00935 {
00936 CartVect result, input;
00937
00938 // define a unit square in xy plane
00939 const CartVect quad[4] = { CartVect( 0.0, 0.0, 0.0 ), CartVect( 1.0, 0.0, 0.0 ), CartVect( 1.0, 1.0, 0.0 ),
00940 CartVect( 0.0, 1.0, 0.0 ) };
00941
00942 // test input in center of square
00943 closest_location_on_polygon( CartVect( 0.5, 0.5, 0.0 ), quad, 4, result );
00944 ASSERT_VECTORS_EQUAL( result, CartVect( 0.5, 0.5, 0.0 ) );
00945 // test above center of square
00946 closest_location_on_polygon( CartVect( 0.5, 0.5, 1.0 ), quad, 4, result );
00947 ASSERT_VECTORS_EQUAL( result, CartVect( 0.5, 0.5, 0.0 ) );
00948 // test below center of square
00949 closest_location_on_polygon( CartVect( 0.5, 0.5, -1.0 ), quad, 4, result );
00950 ASSERT_VECTORS_EQUAL( result, CartVect( 0.5, 0.5, 0.0 ) );
00951
00952 // test points within square, but not at center
00953 input = CartVect( 0.25, 0.25, 0 );
00954 closest_location_on_polygon( input, quad, 4, result );
00955 ASSERT_VECTORS_EQUAL( result, input );
00956 input = CartVect( 0.75, 0.25, 0 );
00957 closest_location_on_polygon( input, quad, 4, result );
00958 ASSERT_VECTORS_EQUAL( result, input );
00959 input = CartVect( 0.75, 0.75, 0 );
00960 closest_location_on_polygon( input, quad, 4, result );
00961 ASSERT_VECTORS_EQUAL( result, input );
00962 input = CartVect( 0.25, 0.75, 0 );
00963 closest_location_on_polygon( input, quad, 4, result );
00964 ASSERT_VECTORS_EQUAL( result, input );
00965
00966 // test at each corner
00975
00976 // test at point on each edge
00978 closest_location_on_polygon( input, quad, 4, result );
00979 ASSERT_VECTORS_EQUAL( result, input );
00981 closest_location_on_polygon( input, quad, 4, result );
00982 ASSERT_VECTORS_EQUAL( result, input );
00984 closest_location_on_polygon( input, quad, 4, result );
00985 ASSERT_VECTORS_EQUAL( result, input );
00987 closest_location_on_polygon( input, quad, 4, result );
00988 ASSERT_VECTORS_EQUAL( result, input );
00989
00990 // test at point outside and closest to each corner
00991 closest_location_on_polygon( CartVect( -1.0, -1.0, 0.0 ), quad, 4, result );
00993 closest_location_on_polygon( CartVect( 2.0, -1.0, 0.0 ), quad, 4, result );
00995 closest_location_on_polygon( CartVect( 2.0, 2.0, 0.0 ), quad, 4, result );
00997 closest_location_on_polygon( CartVect( -1.0, 2.0, 0.0 ), quad, 4, result );
00999
01000 // test at point outside and closest to an edge
01001 CartVect x( 1.0, 0.0, 0.0 ), y( 0.0, 1.0, 0.0 );
01003 closest_location_on_polygon( input - y, quad, 4, result );
01004 ASSERT_VECTORS_EQUAL( result, input );
01006 closest_location_on_polygon( input + x, quad, 4, result );
01007 ASSERT_VECTORS_EQUAL( result, input );
01009 closest_location_on_polygon( input + y, quad, 4, result );
01010 ASSERT_VECTORS_EQUAL( result, input );
01012 closest_location_on_polygon( input - x, quad, 4, result );
01013 ASSERT_VECTORS_EQUAL( result, input );
01014 }
01015
01016 void test_segment_box_intersect()
01017 {
01018 const double box_min = 0.0;
01019 const double box_max = 2.0;
01020 const double box_wid = box_max - box_min;
01021 const double box_mid = 0.5 * ( box_min + box_max );
01022 const CartVect min( box_min );
01023 const CartVect max( box_max );
01024 const CartVect X( 1, 0, 0 ), Y( 0, 1, 0 ), Z( 0, 0, 1 );
01025 CartVect pt;
01026 double start, end;
01027 bool r;
01028
01029 // test line through box in +x direction
01030 double offset = 1;
01031 pt = CartVect( box_min - offset, box_mid, box_mid );
01032 start = -HUGE_VAL;
01033 end = HUGE_VAL;
01034 r = segment_box_intersect( min, max, pt, X, start, end );
01035 ASSERT( r );
01036 ASSERT_DOUBLES_EQUAL( start, box_min + offset );
01037 ASSERT_DOUBLES_EQUAL( end - start, box_wid );
01038
01039 // test with ray ending left of the box
01040 start = -HUGE_VAL;
01041 end = 0;
01042 r = segment_box_intersect( min, max, pt, X, start, end );
01043 ASSERT( !r );
01044
01045 // test with ray ending within box
01046 start = -HUGE_VAL;
01047 end = box_mid + offset;
01048 r = segment_box_intersect( min, max, pt, X, start, end );
01049 ASSERT( r );
01050 ASSERT_DOUBLES_EQUAL( start, box_min + offset );
01051 ASSERT_DOUBLES_EQUAL( end, box_mid + offset );
01052
01053 // test with ray beginning within box
01054 start = box_mid + offset;
01055 end = HUGE_VAL;
01056 r = segment_box_intersect( min, max, pt, X, start, end );
01057 ASSERT( r );
01058 ASSERT_DOUBLES_EQUAL( start, box_mid + offset );
01059 ASSERT_DOUBLES_EQUAL( end, box_max + offset );
01060
01061 // test with ray right of box
01062 start = offset + offset + box_max;
01063 end = HUGE_VAL;
01064 r = segment_box_intersect( min, max, pt, X, start, end );
01065 ASSERT( !r );
01066
01067 // test line through box in -y direction
01068 offset = 1;
01069 pt = CartVect( box_mid, box_min - offset, box_mid );
01070 start = -HUGE_VAL;
01071 end = HUGE_VAL;
01072 r = segment_box_intersect( min, max, pt, -Y, start, end );
01073 ASSERT( r );
01074 ASSERT_DOUBLES_EQUAL( end - start, box_wid );
01075 ASSERT_DOUBLES_EQUAL( end, box_min - offset );
01076
01077 // test with ray ending left of the box
01078 start = box_min;
01079 end = HUGE_VAL;
01080 r = segment_box_intersect( min, max, pt, -Y, start, end );
01081 ASSERT( !r );
01082
01083 // test with ray beginning within box
01084 start = -box_mid - offset;
01085 end = HUGE_VAL;
01086 r = segment_box_intersect( min, max, pt, -Y, start, end );
01087 ASSERT( r );
01088 ASSERT_DOUBLES_EQUAL( start, -box_mid - offset );
01089 ASSERT_DOUBLES_EQUAL( end, box_min - offset );
01090
01091 // test with ray ending within box
01092 start = -HUGE_VAL;
01093 end = -box_mid - offset;
01094 r = segment_box_intersect( min, max, pt, -Y, start, end );
01095 ASSERT( r );
01096 ASSERT_DOUBLES_EQUAL( start, -box_max - offset );
01097 ASSERT_DOUBLES_EQUAL( end, -box_mid - offset );
01098
01099 // test with ray right of box
01100 start = -HUGE_VAL;
01101 end = -box_max - offset - offset;
01102 r = segment_box_intersect( min, max, pt, -Y, start, end );
01103 ASSERT( !r );
01104
01105 // test ray outside in Z direction, parallel to Z plane, and
01106 // intersecting in projections into other planes
01107 pt = CartVect( box_mid, box_mid, box_max + 1 );
01108 start = 0;
01109 end = box_wid;
01110 r = segment_box_intersect( min, max, pt, X, start, end );
01111 ASSERT( !r );
01112 start = 0;
01113 end = box_wid;
01114 r = segment_box_intersect( min, max, pt, -X, start, end );
01115 ASSERT( !r );
01116 start = 0;
01117 end = box_wid;
01118 r = segment_box_intersect( min, max, pt, Y, start, end );
01119 ASSERT( !r );
01120 start = 0;
01121 end = box_wid;
01122 r = segment_box_intersect( min, max, pt, -Y, start, end );
01123 ASSERT( !r );
01124
01125 // try the other side (less than the min Z);
01126 pt = CartVect( box_mid, box_mid, box_min - 1 );
01127 start = 0;
01128 end = box_wid;
01129 r = segment_box_intersect( min, max, pt, X, start, end );
01130 ASSERT( !r );
01131 start = 0;
01132 end = box_wid;
01133 r = segment_box_intersect( min, max, pt, -X, start, end );
01134 ASSERT( !r );
01135 start = 0;
01136 end = box_wid;
01137 r = segment_box_intersect( min, max, pt, Y, start, end );
01138 ASSERT( !r );
01139 start = 0;
01140 end = box_wid;
01141 r = segment_box_intersect( min, max, pt, -Y, start, end );
01142 ASSERT( !r );
01143
01144 // now move the ray such that it lies exactly on the side of the box
01145 pt = CartVect( box_mid, box_mid, box_min );
01146 start = 0;
01147 end = box_wid;
01148 r = segment_box_intersect( min, max, pt, X, start, end );
01149 ASSERT( r );
01150 ASSERT_DOUBLES_EQUAL( start, 0 );
01151 ASSERT_DOUBLES_EQUAL( end, 0.5 * box_wid );
01152 start = 0;
01153 end = box_wid;
01154 r = segment_box_intersect( min, max, pt, -X, start, end );
01155 ASSERT( r );
01156 ASSERT_DOUBLES_EQUAL( start, 0 );
01157 ASSERT_DOUBLES_EQUAL( end, 0.5 * box_wid );
01158 start = 0;
01159 end = box_wid;
01160 r = segment_box_intersect( min, max, pt, Y, start, end );
01161 ASSERT( r );
01162 ASSERT_DOUBLES_EQUAL( start, 0 );
01163 ASSERT_DOUBLES_EQUAL( end, 0.5 * box_wid );
01164 start = 0;
01165 end = box_wid;
01166 r = segment_box_intersect( min, max, pt, -Y, start, end );
01167 ASSERT( r );
01168 ASSERT_DOUBLES_EQUAL( start, 0 );
01169 ASSERT_DOUBLES_EQUAL( end, 0.5 * box_wid );
01170
01171 // try a skew line segment
01172 pt = CartVect( box_min - 0.25 * box_wid, box_mid, box_mid );
01173 CartVect dir( 1.0 / sqrt( 2.0 ), 1.0 / sqrt( 2.0 ), 0 );
01174 start = 0;
01175 end = 1.5 / sqrt( 2.0 ) * box_wid;
01176 r = segment_box_intersect( min, max, pt, dir, start, end );
01177 ASSERT( r );
01178 ASSERT_DOUBLES_EQUAL( start, 0.5 / sqrt( 2.0 ) * box_wid );
01179 ASSERT_DOUBLES_EQUAL( end, box_wid / sqrt( 2.0 ) );
01180
01181 // try with skew line segment that just touches edge of box
01182 pt = CartVect( box_min - 0.5 * box_wid, box_mid, box_mid );
01183 start = 0;
01184 end = 3.0 / sqrt( 2.0 ) * box_wid;
01185 r = segment_box_intersect( min, max, pt, dir, start, end );
01186 ASSERT( r );
01187 ASSERT_DOUBLES_EQUAL( start, box_wid / sqrt( 2.0 ) );
01188 ASSERT_DOUBLES_EQUAL( end, box_wid / sqrt( 2.0 ) );
01189
01190 // try with skew line segment outside of box
01191 pt = CartVect( box_min - 0.75 * box_wid, box_mid, box_mid );
01192 start = 0;
01193 end = 3.0 / sqrt( 2.0 ) * box_wid;
01194 r = segment_box_intersect( min, max, pt, dir, start, end );
01195 ASSERT( !r );
01196 }
01197
01198 void test_closest_location_on_box()
01199 {
01200 const CartVect min( 0, 0, 0 ), max( 1, 2, 3 );
01201 CartVect pt;
01202
01203 // inside
01204 closest_location_on_box( min, max, CartVect( 0.5, 0.5, 0.5 ), pt );
01205 ASSERT_VECTORS_EQUAL( CartVect( 0.5, 0.5, 0.5 ), pt );
01206
01207 // closest to min x side
01208 closest_location_on_box( min, max, CartVect( -1.0, 0.5, 0.5 ), pt );
01209 ASSERT_VECTORS_EQUAL( CartVect( 0.0, 0.5, 0.5 ), pt );
01210
01211 // closest to max x side
01212 closest_location_on_box( min, max, CartVect( 2.0, 0.5, 0.5 ), pt );
01213 ASSERT_VECTORS_EQUAL( CartVect( 1.0, 0.5, 0.5 ), pt );
01214
01215 // closest to min y side
01216 closest_location_on_box( min, max, CartVect( 0.5, -1.0, 0.5 ), pt );
01217 ASSERT_VECTORS_EQUAL( CartVect( 0.5, 0.0, 0.5 ), pt );
01218
01219 // closest to max y side
01220 closest_location_on_box( min, max, CartVect( 0.5, 2.5, 0.5 ), pt );
01221 ASSERT_VECTORS_EQUAL( CartVect( 0.5, 2.0, 0.5 ), pt );
01222
01223 // closest to min z side
01224 closest_location_on_box( min, max, CartVect( 0.5, 0.5, -0.1 ), pt );
01225 ASSERT_VECTORS_EQUAL( CartVect( 0.5, 0.5, 0.0 ), pt );
01226
01227 // closest to max z side
01228 closest_location_on_box( min, max, CartVect( 0.5, 0.5, 100.0 ), pt );
01229 ASSERT_VECTORS_EQUAL( CartVect( 0.5, 0.5, 3.0 ), pt );
01230
01231 // closest to min corner
01232 closest_location_on_box( min, max, CartVect( -1, -1, -1 ), pt );
01233 ASSERT_VECTORS_EQUAL( min, pt );
01234
01235 // closest to max corner
01236 closest_location_on_box( min, max, CartVect( 2, 3, 4 ), pt );
01237 ASSERT_VECTORS_EQUAL( max, pt );
01238 }
01239
01240 int main()
01241 {
01242 int error_count = 0;
01243 error_count += RUN_TEST( test_box_plane_overlap );
01244 error_count += RUN_TEST( test_box_linear_elem_overlap_tri );
01245 error_count += RUN_TEST( test_box_linear_elem_overlap_tet );
01246 error_count += RUN_TEST( test_box_linear_elem_overlap_hex );
01247 error_count += RUN_TEST( test_box_tri_overlap );
01248 error_count += RUN_TEST( test_box_tet_overlap );
01249 error_count += RUN_TEST( test_box_hex_overlap );
01250 error_count += RUN_TEST( test_ray_tri_intersect );
01251 error_count += RUN_TEST( test_plucker_ray_tri_intersect );
01252 error_count += RUN_TEST( test_closest_location_on_tri );
01253 error_count += RUN_TEST( test_closest_location_on_polygon );
01254 error_count += RUN_TEST( test_segment_box_intersect );
01255 error_count += RUN_TEST( test_closest_location_on_box );
01256 return error_count;
01257 }
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| 2.578125
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https://www.iris.unina.it/handle/11588/593256
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Statistical modelling for ordinal data has received a considerable attention in the literature, and a consolidated theory relying on Generalized Linear Model approach has been developed. In this article, we present an innovative technique for modelling bivariate ordinal data. In particular, we consider the method introduced by Plackett for constructing a one-parameter bivariate distribution from given margins, and we apply it in order to represent correlated ordinal variables which individually follows a CUB model. This is a univariate mixture distribution defined by the convex Combination of a Uniform and a shifted Binomial distribution whose parameters may be related to rater's covariates. The article shows how the bivariate distribution can be defined and how its characterizing parameter, which describes the association between the component random variables, can be related to the subject's covariates. The proposed approach is applied to the study of two key drivers of extra virgin olive oil consumption in Italy. The technique allows a representation of the data whose meaning can be easily interpreted providing useful information for management support.
### Analyzing bivariate ordinal data with CUB margins
#### Abstract
Statistical modelling for ordinal data has received a considerable attention in the literature, and a consolidated theory relying on Generalized Linear Model approach has been developed. In this article, we present an innovative technique for modelling bivariate ordinal data. In particular, we consider the method introduced by Plackett for constructing a one-parameter bivariate distribution from given margins, and we apply it in order to represent correlated ordinal variables which individually follows a CUB model. This is a univariate mixture distribution defined by the convex Combination of a Uniform and a shifted Binomial distribution whose parameters may be related to rater's covariates. The article shows how the bivariate distribution can be defined and how its characterizing parameter, which describes the association between the component random variables, can be related to the subject's covariates. The proposed approach is applied to the study of two key drivers of extra virgin olive oil consumption in Italy. The technique allows a representation of the data whose meaning can be easily interpreted providing useful information for management support.
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https://quantumcomputing.stackexchange.com/questions/24490/redundant-parameters-in-2-times-2-unitary-operators
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# Redundant parameters in $2\times 2$ unitary operators?
A complex $$n \times n$$ unitary operator has $$n^2$$ free real parameters. For example, a $$2 \times 2$$ unitary matrix can be parametrized as $$$$\begin{pmatrix} e^{i(\alpha - \beta/2 - \delta/2)} \cos \frac{\gamma}{2} & -e^{i(\alpha - \beta/2 + \delta/2)} \sin \frac{\gamma}{2}\\ e^{i(\alpha + \beta/2 - \delta/2)} \sin \frac{\gamma}{2} & e^{i(\alpha + \beta/2 + \delta/2)} \cos \frac{\gamma}{2} \end{pmatrix}.$$$$
However, if a initial state is given as $$$$|\psi_i \rangle= \begin{pmatrix} \cos \frac{\theta_i}{2} \\ \sin \frac{\theta_i}{2} e^{i\phi_i} \end{pmatrix},$$$$ and the desired output is $$$$|\psi_f \rangle= \begin{pmatrix} \cos \frac{\theta_f}{2} \\ \sin \frac{\theta_f}{2} e^{i\phi_f} \end{pmatrix}.$$$$ For both input and output, the global phase is not of interest and therefore omitted.
I would like to find a unitary operator $$U$$ that converts $$|\psi_i \rangle$$ to $$|\psi_f \rangle$$. Since the initial state is known, there are only $$2$$ parameters underdetermined. This indicates that the $$U$$ such that $$|\psi_i \rangle \xrightarrow{U} |\psi_f \rangle$$ contains only two independent parameters.
It seems that two parameters in the general $$2\times 2$$ unitary matrix are redundant.
• – glS
Mar 11, 2022 at 23:25
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https://islamsight.org/hiiii9z/robust-standard-errors-in-r-381f17
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It takes a formula and data much in the same was as lm does, and all auxiliary variables, such as clusters and weights, can be passed either as quoted names of columns, as bare column names, or as a self-contained vector. when the assumed … To get heteroskadastic-robust standard errors in R–and to replicate the standard errors as they appear in Stata–is a bit more work. The robust standard errors are due to quasi maximum likelihood estimation (QMLE) as opposed to (the regular) maximum likelihood estimation (MLE). Active 4 months ago. This note deals with estimating cluster-robust standard errors on one and two dimensions using R (seeR Development Core Team[2007]). I have read a lot about the pain of replicate the easy robust option from STATA to R to use robust standard errors. Load in library, dataset, and recode. In R, robust standard errors are not “built in” to the base language. I replicated following approaches: StackExchange and Economic Theory Blog.They work but the problem I face is, if I … Examples of usage … They are robust against violations of the distributional assumption, e.g. This function performs linear regression and provides a variety of standard errors. I want to control for heteroscedasticity with robust standard errors. In practice, heteroskedasticity-robust and clustered standard errors are usually larger than standard errors from regular OLS — however, this is not always the case. Cluster-Robust Standard Errors 2 Replicating in R Molly Roberts Robust and Clustered Standard Errors March 6, 2013 3 / 35. An Introduction to Robust and Clustered Standard Errors Linear Regression with Non-constant Variance Review: Errors and Residuals Examples of usage … Details. To replicate the result in R takes a bit more work. First, we estimate the model and then we use vcovHC() from the {sandwich} package, along with coeftest() from {lmtest} to calculate and display the robust standard errors. This function performs linear regression and provides a variety of standard errors. Illustration showing different flavors of robust standard errors. None of them, unfortunately, are as simple as typing the letter r after a regression. Notice the third column indicates “Robust” Standard Errors. Using the High School & Beyond (hsb) dataset. Each has its ups and downs, but may serve different purposes. Ask Question Asked 4 months ago. Let's say that I have a panel dataset with the variables Y, ENTITY, TIME, V1. First we load the haven package to use the read_dta function that allows us to import Stata data sets. Do not really need to dummy code but may make making the X matrix easier. Details. For discussion of robust inference under within groups correlated errors, see There are a few ways that I’ve discovered to try to replicate Stata’s “robust” command. Now I want to have the same results with plm in R as when I use the lm function and Stata when I perform a heteroscedasticity robust and entity fixed regression. Viewed 123 times 1 \$\begingroup\$ I am looking for a way to implement (country) clustered standard errors on a panel regression with individual fixed effects. Cluster-robust stan-dard errors are an issue when the errors are correlated within groups of observa-tions. Hi! R plm cluster robust standard errors with multiple imputations. For further detail on when robust standard errors are smaller than OLS standard errors, see Jorn-Steffen Pische’s response on Mostly Harmless … There is a mention of robust standard errors in "rugarch" vignette on p. 25. Then we load two more packages: lmtest and sandwich.The lmtest package provides the coeftest function … It takes a formula and data much in the same was as lm does, and all auxiliary variables, such as clusters and weights, can be passed either as quoted names of columns, as bare column names, or as a self-contained vector. I get the same standard errors in R with this code
Is Stinking Chamomile Edible, Isilon A200 Spec Sheet, Thunderbird 68 Themes, La Paz, Bolivia Climate, Now Kidney Cleanse, Best Forever Roses, Freddy Fazbear's Pizzeria Location,
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# took GMAT 3 times still in the mid 600s....want 680+
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took GMAT 3 times still in the mid 600s....want 680+ [#permalink]
### Show Tags
27 Oct 2008, 22:10
here are my scores:
1st: 640 (V29, Q48) AWA 3.5 [taken 3 years ago]
2nd: 650 (V34, Q47) AWA 4.0 [Aug 08]
3rd: 630 (V32, Q44) AWA 4.0 [Sept 08]
I wanna apply for Stanford and Anderson so i need 700+ (i'll still apply if i get 680+)
I'm goin for Fall 09 so I have to acheive it by end of year. Any ideas to what approach i should take?
Thanks.
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Re: took GMAT 3 times still in the mid 600s....want 680+ [#permalink]
### Show Tags
28 Oct 2008, 08:56
Bfung
I suggest you do more practice from OG10 and OG11 along with their reviews and give the GMATprep and see where do you stand. You seem weak in verbal. So for the sentence correction part study the Manhattan SC guide and for CR part read the powerscore CR bible. If you diligently study from these two resources, it will help you to catapult the score above 38. And for quant practice from GMAT club challenges. It will help you to raise your score by 3-4 points.
Good luck
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Re: took GMAT 3 times still in the mid 600s....want 680+ [#permalink]
### Show Tags
28 Oct 2008, 10:14
Can't disagree with anything in this post. You also might look at your timing. Did you complete the entire verbal section? Did you have to guess at any of them? If so, how many? It sounds odd, but knowing your time limitations on each question will help you a lot. When a question stumps you, guess and move on.
subarao wrote:
Bfung
I suggest you do more practice from OG10 and OG11 along with their reviews and give the GMATprep and see where do you stand. You seem weak in verbal. So for the sentence correction part study the Manhattan SC guide and for CR part read the powerscore CR bible. If you diligently study from these two resources, it will help you to catapult the score above 38. And for quant practice from GMAT club challenges. It will help you to raise your score by 3-4 points.
Good luck
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Re: took GMAT 3 times still in the mid 600s....want 680+ [#permalink]
### Show Tags
28 Oct 2008, 12:34
thx guys...
yes my verbal seems week coz i'm coming from a non-english native background....altho it has got a bit better compared to 3 yrs ago w/ a lot of readings....and i got my AWA to 4 which should do
i'm as concerned about how i can't score better in quant coz they prolly expect more from me coming from an asian/engineering background...my quant has actually gotten worse compared to 3 yrs ago...seems like i'm doin as much math w/ my current job as i did in school which explains why....
i've finished the OG 11, did the manhattan free exam, did both GMATPrep exams twice....
i was very disappointed about my 3rd score coz i thought if i could maintain my best V(34) from my 2nd exam and my best Q(48) from the 1st exam i'd haf at least got a 680+...but it turned out to be the lowest score of the 3....
yes i had probs w/ timing in my 3rd quant...and i was rushing/guessing a bit for my last 10 questions....i guess that really hurt? btw, how big a penalty is it to not finish one question?
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took 3 times 630-650....want 680+ [#permalink]
### Show Tags
04 Nov 2008, 19:35
i'm plannin to take it again b4 xmas...i wonder if i should spend more time on quant or verbal...my verbal is lower which means more rooms for improvment, but i've been told that quant counts more...at least for my background
btw..subarao..wat's the powerscore CR bible? is it a book/CD/online test?
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Re: took GMAT 3 times still in the mid 600s....want 680+ [#permalink]
### Show Tags
08 Nov 2008, 14:58
I would suggest attacking your weakest area to improve your score. I am very weak in sentence correction and my practice scores have improved from focusing on my weakest area first.
I would also suggest typing in proper grammar and spelling. I'm not trying to sound like a jerk here, but I think it's beneficial to do so. Once I started studying for the exam, I noticed my weakness in not only with basic structure, but also with my everyday skills. I pushed myself to get away from typing "internet" spelling and emails and it's helping me improve in the grammar skills.
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Re: took GMAT 3 times still in the mid 600s....want 680+ [#permalink]
### Show Tags
12 Nov 2008, 20:04
Appreciate your advice. I see your point about staying away from "internet" style. I'm currently studying for another exam for late nov / early dec. My plan is, if I don't get 680+ I won't bother applying and i'm not gonna start the essays because it's not realistic. But if i do, i'll have a month to work on the apps.
hwiya320 wrote:
I would suggest attacking your weakest area to improve your score. I am very weak in sentence correction and my practice scores have improved from focusing on my weakest area first.
I would also suggest typing in proper grammar and spelling. I'm not trying to sound like a jerk here, but I think it's beneficial to do so. Once I started studying for the exam, I noticed my weakness in not only with basic structure, but also with my everyday skills. I pushed myself to get away from typing "internet" spelling and emails and it's helping me improve in the grammar skills.
Re: took GMAT 3 times still in the mid 600s....want 680+ [#permalink] 12 Nov 2008, 20:04
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0
# How do you write 19 over 6 as a decimal?
Updated: 9/20/2023
Wiki User
7y ago
The decimal form of nineteen over six is:
3.1666666666666666666666666666666666666666667
This is an approximation, since the sixes repeat forever, the 7 at the end is merely a rounding up .
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Related questions
### What is 19 over 6 in decimal notation?
19 over 6 in decimal notation is 3.167
### 19 over 6 written in a decimal is?
It's 3.16 and next time write the question correctly.
### How do you write 6 over 15 as a decimal?
It is: 6/15 = 0.4 as a decimal
### How do you write 19 over 6 in simplest form?
19/6 cannot be reduced further as the numerator and denominator do not have common factors. The result is 3 and 1/6 or 3.166666 in decimal
### How do you write six nineteenths as a decimal?
6/19 has no exact decimal equivalent. It is approximately .3157894.
### How do you write 5 over 6 in decimal?
By dividing the 5 by the 6. What is 5 over 6 as a decimal? 5/6 = 0.833... ≈ 0.833
### How can you write 6 over 8 as a decimal?
It is: 6/8 = 3/4 = 0.75 as a decimal
0.60
0.6
### How do you write 5 over 6 as a recurring decimal?
5/6 written as a recurring decimal is 0.83333...
3.1667
### What is 19 over 6 in decimal form?
3.16666 (repeating)
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# What can one tell a school kid about statistics and machine learning?
Next week we have an intern from a local school in the house. The concept behind his short internship is to get an idea how the real world works and what certain jobs deal with, how the daily work looks like etc.
Now I wondered, what one can tell/show/demonstrate such a young kid about Statistics and Machine Learning so that he/she
• gets the basic idea of this areas
• is getting enthusiastic (assuming that the kid's prior is not too heavily weighted in favor of other interests)
• wont forget about it the next day
I am primarily looking for sticking images, examples for demonstration etc..
The kid's background:
• 15-16 years old
• basic math concepts are known (what is a graph, rule of three, what is a variable (mathematically, not statistically speaking))
Since the hard part of this question is to explain your area to one without any background knowledge, this question might as well serve as reference for chats with relatives and friends.
Sidenote: I skipped the description of my job intentionally, so that this question is not too specific, this question is about the topic of this site in general.
I do research in computer vision and machine learning, when people ask me about machine learning, I like to mention how the 1 million Netflix challenge/prize is a great example of machine learning (stating the problem and input output), and the 3 million ongoing health prize challenge.
The Kinect uses random forests (very popular machine learning approach) and it's really simple to explain.
The website Kaggle.com hosts machine learning competitions, I WISH there was such website when I was a kid. Maybe you want to show it to him.
I would use some drawings to introduce concepts of classification and over fitting.
• the kinect example is great ... can create a link to the youngster ;). Can you recommend some documents explaining the application of RF in Kinect ? – steffen Feb 9 '12 at 20:27
• @steffen: the RF algorithm used in the Kinect is describe in details in this paper (received best paper award at CVPR 2011) - I couldn't find any simplified explanation of it online, but I'm willing to write one tonight and post it somewhere here on Stack Exchange. If you think it's a good idea, let me know an appropriate place in SE where I can post an illustration of the algorithm. – Roronoa Zoro Feb 9 '12 at 20:53
• thank you. I hope/think I can handle the paper, but thank you very much for the offer ! – steffen Feb 9 '12 at 21:03
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# independent study, conservation violation and dark energy
Discussion in 'Physics & Math' started by SimonsCat, Jan 24, 2017.
1. ### SimonsCatRegistered Member
Messages:
213
This is not just a post reporting a science news article, but I get into an informative discussion below, something that appears to be appreciated by members here.
An independent study that links energy conservation violation with dark energy, very similar to my own investigations. How do you define the leakage of energy in a universe? I claim energy doesn't leak from the universe, but a lot of it went towards a rotational property of the universe which has since decayed.
Conservation violation can also explain the problematics of a Wheeler de Witt equation and could solve the problem of time!
Let me explain
Does conservation exist in relativity?
The short answer, is yes and no. The long answer hits snags and complicates the discussion and whether there is a conservation or isn't, becomes tenebrous.
First of all, what does the physicist mean when they talk about energy conservation? In regards to general relativity, this may be discussed in quite a number of ways, one of the most popular explanations of conservation principles, comes from Noether's theorem.
Neother's theorem is about continuous translations over spacetime. To do this with energy, however, to see how it conserves, requires its conjugate of time. Now this is a big issue for cosmology, because... there is no global time in general relativity.
See, the way in which evolution happens in general relativity, is not actually generated by a time variable. Such an evolution, actually depends on difficult mathematics called diffeomorphism invariances which in a way, attempt to shuffle space coordinates freely. Thus, physicists say, motion is generated by a symmetry of the theory, it isn't actually a true time evolution. We will come back to this: The absence of time is taken on as a special feature of a quantum gravity equation known as the Wheeler deWitt equation.
Well that's interesting isn't it?
However, things only become complicated when I begin to tell you that general relativity contains two kinds of conservation equations. The most obvious one, is the de Sitter space solution of Friedmann's equations for cosmological expansion. Many physicists consider the general statement of the Friedmann equation as a conservation principle, which wouldn't be far wrong, because the equation does indeed conserve energy as the universe expands, which in previous work I have stated, was always an unfounded assumption.
Such a conservation happens because the universe is generally modeled as an adiabatic system (which means, it is a system which has no energy or heat that enters or leaves the system). This brings me back to the Wheeler deWitt equation (WDW) because in many ways, the resulting quantum gravity equation which Wheeler needed help quantizing back in the 60's, is a type of statement about conservation as well. It is no surprise, a universe with no internal changes would give rise to a static equation such as the WDW equation. You can only get change (and thus) perhaps a sense of time, if there was in fact some kind of conservation breakdown in a universe, such as an energy conservation breakdown.
So the jury may be out on this one, time may exist, energy conservation might exist, but equally, both could be a persistent problem for physics - general relativity doesn't actually make a statement in itself about conservation, only that it doesn't possess one. Whether one takes this to be the case of it being absent, is left to the reader.
REFERENCES
https://en.wikipedia.org/wiki/Friedmann_equations
http://vixra.org/pdf/1402.0145v1.pdf
http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/
3. ### SimonsCatRegistered Member
Messages:
213
And while I say it was unfounded by Friedmann to assume this, it probably wasn't all too unfounded, because it was generally believed that the universe is a closed system, so if the universe is adiabatic, how can energy enter or leave a system? Truth be told, we no longer need an outside to a universe to actually vary the energy inside of it.
karenmansker likes this.
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# Number 500850
### Properties of number 500850
Cross Sum:
Factorization:
2 * 3 * 3 * 3 * 5 * 5 * 7 * 53
Count of divisors:
Sum of divisors:
Prime number?
No
Fibonacci number?
No
Bell Number?
No
Catalan Number?
No
Base 2 (Binary):
Base 3 (Ternary):
Base 4 (Quaternary):
Base 5 (Quintal):
Base 8 (Octal):
7a472
Base 32:
f93i
sin(500850)
-0.99979182275372
cos(500850)
0.020403704437816
tan(500850)
-49.000505070085
ln(500850)
13.12406193404
lg(500850)
5.6997076781101
sqrt(500850)
707.70756672513
Square(500850)
### Number Look Up
Look Up
500850 (five hundred thousand eight hundred fifty) is a very special figure. The cross sum of 500850 is 18. If you factorisate 500850 you will get these result 2 * 3 * 3 * 3 * 5 * 5 * 7 * 53. 500850 has 96 divisors ( 1, 2, 3, 5, 6, 7, 9, 10, 14, 15, 18, 21, 25, 27, 30, 35, 42, 45, 50, 53, 54, 63, 70, 75, 90, 105, 106, 126, 135, 150, 159, 175, 189, 210, 225, 265, 270, 315, 318, 350, 371, 378, 450, 477, 525, 530, 630, 675, 742, 795, 945, 954, 1050, 1113, 1325, 1350, 1431, 1575, 1590, 1855, 1890, 2226, 2385, 2650, 2862, 3150, 3339, 3710, 3975, 4725, 4770, 5565, 6678, 7155, 7950, 9275, 9450, 10017, 11130, 11925, 14310, 16695, 18550, 20034, 23850, 27825, 33390, 35775, 50085, 55650, 71550, 83475, 100170, 166950, 250425, 500850 ) whith a sum of 1607040. The number 500850 is not a prime number. 500850 is not a fibonacci number. The figure 500850 is not a Bell Number. The number 500850 is not a Catalan Number. The convertion of 500850 to base 2 (Binary) is 1111010010001110010. The convertion of 500850 to base 3 (Ternary) is 221110001000. The convertion of 500850 to base 4 (Quaternary) is 1322101302. The convertion of 500850 to base 5 (Quintal) is 112011400. The convertion of 500850 to base 8 (Octal) is 1722162. The convertion of 500850 to base 16 (Hexadecimal) is 7a472. The convertion of 500850 to base 32 is f93i. The sine of the number 500850 is -0.99979182275372. The cosine of 500850 is 0.020403704437816. The tangent of the figure 500850 is -49.000505070085. The root of 500850 is 707.70756672513.
If you square 500850 you will get the following result 250850722500. The natural logarithm of 500850 is 13.12406193404 and the decimal logarithm is 5.6997076781101. I hope that you now know that 500850 is impressive figure!
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# Randomization Based Off a Set of Rules
Status
Not open for further replies.
#### Kawan
##### New Member
Hi everyone,
I have a rather tricky problem to solve.
For each row in my table (for each person) - refer to screenshot named "My Table", I would like excel to randomly generate a 9 digit number (in Column N) but based on a certain set of rules.
Here is how I currently generate them 1 by 1 manually:
View attachment 51533
1. Yellow cells: I only modify the numbers of this row.
• The first 3 digits of this row are based off the person's year of birth (refer to screenshot called "First 3 digits"). A 3 digit code is assigned based on the year of birth.
• Then I manually change the last 6 digits of this row one by one so that the green cell equals to an even number ending in 0, so: 10 or 20 or 30 or 40 or 50 or 60, etc.
2. Orange cells:
• This row is the factor row and never changes, each of the number above it is multiplied by this factor. So in the example above, 3 (in yellow) will be multiplied 1 (in red), 9 (in yellow) by 2 (in red), 4 (in yellow) by 1 (in red), and so fourth.
3. Red cells: The formula for each red cell is the same with the exception of the reference cells in the formula, so for example for cell F5 the formula is:
Excel Formula:
``=IF(SUM(F3*F4)>=10,SUM(F3*F4)-9,SUM(F3*F4))``
for cell G5 the formula is:
Excel Formula:
``=IF(SUM(G3*G4)>=10,SUM(G3*G4)-9,SUM(G3*G4))``
4. Green Cell: This cell is just the SUM of of the red cells, so
Excel Formula:
``=SUM(F5:N5)``
. It must always equal to an even number ending in 0, so: 10 or 20 or 30 or 40 or 50 or 60, etc...
To sum it up:
For each row in my table, so for each person (refer to screenshot called "My table" in the results worksheet), I would like excel to generate a random 9 digit number in the column called "9 Digit Number" (Column N) based on the set of rules explained above.
I have thousands of rows so you can understand why I would like this process to be automated.
Please do not hesitate to ask me should you require more clarification/information.
Thank you,
Kawan.
#### Attachments
• 9 Digits Worksheet.png
253.4 KB · Views: 4
• First 3 Digits.png
169.9 KB · Views: 4
• My Table - Results Worksheet.png
101.3 KB · Views: 4
### Excel Facts
What is the shortcut key for Format Selection?
Ctrl+1 (the number one) will open the Format dialog for whatever is selected.
#### Kawan
##### New Member
Hi everyone,
I have a rather tricky problem to solve.
For each row in my table (for each person) - refer to screenshot named "My Table", I would like excel to randomly generate a 9 digit number (in Column N) but based on a certain set of rules.
Here is how I currently generate them 1 by 1 manually:
View attachment 51533
1. Yellow cells: I only modify the numbers of this row.
• The first 3 digits of this row are based off the person's year of birth (refer to screenshot called "First 3 digits"). A 3 digit code is assigned based on the year of birth.
• Then I manually change the last 6 digits of this row one by one so that the green cell equals to an even number ending in 0, so: 10 or 20 or 30 or 40 or 50 or 60, etc.
2. Orange cells:
• This row is the factor row and never changes, each of the number above it is multiplied by this factor. So in the example above, 3 (in yellow) will be multiplied 1 (in red), 9 (in yellow) by 2 (in red), 4 (in yellow) by 1 (in red), and so fourth.
3. Red cells: The formula for each red cell is the same with the exception of the reference cells in the formula, so for example for cell F5 the formula is:
Excel Formula:
``=IF(SUM(F3*F4)>=10,SUM(F3*F4)-9,SUM(F3*F4))``
for cell G5 the formula is:
Excel Formula:
``=IF(SUM(G3*G4)>=10,SUM(G3*G4)-9,SUM(G3*G4))``
4. Green Cell: This cell is just the SUM of of the red cells, so
Excel Formula:
``=SUM(F5:N5)``
. It must always equal to an even number ending in 0, so: 10 or 20 or 30 or 40 or 50 or 60, etc...
To sum it up:
For each row in my table, so for each person (refer to screenshot called "My table" in the results worksheet), I would like excel to generate a random 9 digit number in the column called "9 Digit Number" (Column N) based on the set of rules explained above.
I have thousands of rows so you can understand why I would like this process to be automated.
Please do not hesitate to ask me should you require more clarification/information.
Thank you,
Kawan.
Also, I noticed I made a mistake here:
It should be:
2. Orange cells:
• This row is the factor row and never changes, each of the number above it is multiplied by this factor. So in the example above, 3 (in yellow) will be multiplied 1 (in orange), 9 (in yellow) by 2 (in orange), 4 (in yellow) by 1 (in orange), and so fourth.
#### Fluff
##### MrExcel MVP, Moderator
Duplicate to: Randomization Based Off a Set of Rules
In future, please do not post the same question multiple times. Per Forum Rules (#12), posts of a duplicate nature will be locked or deleted.
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Excel contains over 450 functions, with more added every year. That’s a huge number, so where should you start? Right here with this bundle.
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# Normal Tracing Worksheets Activities With Answers for Ages 4-6
Unleash the potential of effective learning with our curated Normal Tracing Worksheets Activities With Answers! Designed for educators, parents, and students alike, these worksheets serve as an unparalleled resource to enhance motor skills, focus, and understanding. Each activity is meticulously crafted to provide a balanced mix of challenge and fun, ensuring a rewarding experience for learners of all levels. With answers included, our worksheets not only offer immediate feedback but also empower users to learn from their mistakes and improve progressively. Dive into our collection and discover the perfect blend of education and engagement, promising a smoother learning journey for everyone involved.
Favorites
Interactive
• Tracing
• 4-6
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• Normal
## Trace Diagonal Lines Worksheet
This tracing diagonal lines worksheet offers preschoolers a great way to build fundamental skills in handwriting and drawing. Doing so helps them practice patterning, a key skill that will benefit them in writing, drawing, and math.
Trace Diagonal Lines Worksheet
Worksheet
## Letter O Tracing Page
Practice writing the letter "O" with our new worksheet. Trace and write the letter several times, starting at the red dot. First do the uppercase letter, then the lowercase. Use the pictures to complete the words "Ostrich" and "octopus". Kids Academy offers more fun alphabet worksheets.
Letter O Tracing Page
Worksheet
## Count and Trace 7 – 10 Worksheet
Let your child develop a lifelong love of math with Kids Academy's printable math worksheet based on the innovative Singapore Math method. It uses funny pictures and graphs to help solve math problems. Check out our amazing collection of free printable math worksheets now!
Count and Trace 7 – 10 Worksheet
Worksheet
## Using Triangles to Make Squares and Rhombuses Worksheet
Learning shapes is a fun way to boost spatial skills and critical thinking. This worksheet shows children how to use triangles to construct squares and rhombuses. Clear illustrations and a printout help children draw the new shape. A great activity for the math classroom! 80 words.
Using Triangles to Make Squares and Rhombuses Worksheet
Worksheet
## Cat Printable Sight Words Worksheet
Encourage your child's literacy with this fun sight words worksheet: cat PDF! It helps build a solid foundation for reading, featuring activities such as reading and tracing the word cat, and finding it amongst other sight words. With its cute cat, Kids Academy makes learning to read a delight!
Cat Printable Sight Words Worksheet
Worksheet
## Letter P Tracing Page
Trace the lines from the red dot to learn to write "P"! Then practice this letter with the fun activities: complete the word "Pig" and "Pumpkin". Check out Kids Academy to get more free ABC worksheets.
Letter P Tracing Page
Worksheet
## Letter I Tracing Page
Trace and write "I" uppercase and lowercase. An iguana's tail looks like the uppercase "I" and the lowercase "i" is as small as an inch. Make learning fun with Kids Academy worksheets.
Letter I Tracing Page
Worksheet
## Uppercase Letters J, K, and L Worksheet
Jewelry, kettle and lobster share a common trait: their names start with J, K and L, 3 letter neighbors found in the middle of the alphabet. This traceable worksheet helps kids learn letter sounds and practice writing. Arrows and tracing lines make it easy to form neat pencil strokes. Red dot shows where kids should start. Perfect for preschoolers and kindergarteners!
Uppercase Letters J, K, and L Worksheet
Worksheet
## Practice Drawing Hexahedrons And a Rhombus Worksheet
Trace the Robot's face, draw hexahedron and rhombus shapes, then trace again. Our tracing shapes worksheets make it easy for kids to learn geometry. Get more materials from Kids Academy to practice all the shapes.
Practice Drawing Hexahedrons And a Rhombus Worksheet
Worksheet
## Letter Q Tracing Page
Learn the letter "Q"! Trace and write it a few times in upper and lowercase. Then help the Queen by writing her initial letter. Finally, write "quarters" and practice with more alphabet worksheets.
Letter Q Tracing Page
Worksheet
## Cupid Maze: An Printable
Get a cute and creative way to boost their reading with this Valentine's Day-themed sight word worksheet. Perfect for kids, it features a fun maze and the article "an".
Cupid Maze: An Printable
Worksheet
## Write 8 Worksheet
Once your kids can count small numbers, it's time to learn how to write them. With your help and patience, they will soon be pros! This worksheet will make learning easier. Help them spot the 8s in the picture, then help them trace them.
Write 8 Worksheet
Worksheet
## Words That Start with "ch" Spelling Worksheet
With Kids Academy, learning phonics and spelling has never been easier. Kids Academy's "ch" tracing worksheet makes it easy for your little one to learn phonics and spelling. It features cute, brightly colored images to help them trace familiar words, making learning fun and enjoyable.
Words That Start with "ch" Spelling Worksheet
Worksheet
## Drawing a Triangle Worksheet
Help your child develop fine motor skills with this fun, traceable printable! It teaches them how to draw a triangle, with easy steps and cute picture motivation. They'll work on pre-writing skills, pencil position, and grip control without even realizing it. With this basic shape, they'll gain a foundation for strong writing and drawing skills.
Drawing a Triangle Worksheet
Worksheet
Ordinal Numbers: Let's Practice Numbers Printable
Worksheet
## Italian Word Tracing: Ciao Worksheet
Help kids learn to greet people in different languages with this fun worksheet! Featuring the Italian word 'Ciao', it helps kids trace and learn the pronunciation. Plus, it teaches hand-eye coordination and fine-motor skills. Global connectivity can start with this activity!
Italian Word Tracing: Ciao Worksheet
Worksheet
## A Pup, a Cap and a Pea Spelling Worksheet
Practice tracing and writing with three new words. Carefully trace each word and try to remember it, then move on to the next one. Head to Kids Academy for more tracing worksheets! Get your pencils ready! Let's practice tracing and writing three new words on Kids Academy's spelling worksheets. Carefully trace each line and remember the words. Pick more great worksheets on their website to continue practicing. Have fun!
A Pup, a Cap and a Pea Spelling Worksheet
Worksheet
## Yellow Tracing Color Words Worksheet
Tracing is a great way to start learning handwriting! Our printable worksheet helps your kindergartener practice and master the basics. They'll trace lines and fill in the word "yellow" with bright colors, sure to captivate and inspire them for more practice. Check out here for more tracing of color words.
Yellow Tracing Color Words Worksheet
Worksheet
## A Pen, a Hen and a Fox Spelling Worksheet
Practice spelling with this free worksheet: three lines, three new words. Say them, trace them, write them correctly. Keep learning with Kids Academy for more printable worksheets and more fun! (80 words)
A Pen, a Hen and a Fox Spelling Worksheet
Worksheet
## Tracing And Writing Number 6 Worksheet
Learn numbers in a fun way with our preschool number worksheets. Count six bananas, trace and write the number, find the monkey with six bananas, and trace the word "six." Explore more activities at Kids Academy.
Tracing And Writing Number 6 Worksheet
Worksheet
## Letter H Tracing Page
Trace the letter "H", then practice writing its lowercase form. Finally, choose your favorite form of transport and ride away! Get our worksheets to make learning fun and easy.
Letter H Tracing Page
Worksheet
## Triangle Rectangle Worksheet
This printable worksheet helps kids draw rectangles and triangles. Have your child trace the pizza and cracker shapes to become familiar, then try their own. Perfect for second graders! Coloring and creativity make this fun and educational.
Triangle Rectangle Worksheet
Worksheet
## Lowercase Letters y z Worksheet
Help your child get a good start in writing and reading by tracing and writing the lowercase letters y and z. With this activity, they will learn the alphabet, which is essential for literacy success. Guide them through the worksheet and make sure they form the letters correctly. This will help them learn the basics needed to write and read fluently.
Lowercase Letters y z Worksheet
Worksheet
## Tracing And Learning to Write Number 9 Worksheet
Explore our collection of learning numbers worksheets and make teaching numbers to preschoolers easier with Kids Academy. Trace numbers, circle houses with the specified number of windows, write the corresponding number words and revise previous numbers regularly to help kids learn.
Tracing And Learning to Write Number 9 Worksheet
Worksheet
Learning Skills
Normal Tracing Worksheets Activities with Answers are not just another set of exercises for young learners; they are a cornerstone of foundational education, especially in early childhood. These activities are crafted to help children develop a range of essential skills that are crucial for their academic and personal growth. Let's delve into why these tracing worksheets are exceedingly useful for children.
Firstly, normal tracing worksheets activities with answers enhance fine motor skills. As children trace over lines, shapes, letters, and numbers, they refine their hand-eye coordination and improve their pencil grip. This meticulous activity strengthens the small muscles in their hands, preparing them for more advanced tasks such as writing and drawing.
Secondly, these worksheets serve as an excellent introduction to letters and numbers. Through repetition and practice, children familiarize themselves with the basic building blocks of language and mathematics. This early exposure lays a solid foundation for reading, writing, and arithmetic skills that they will build upon in the years to come.
Moreover, tracing activities promote concentration and focus. Children learn to pay attention to detail as they follow specific patterns and shapes. This ability to concentrate on a task for an extended period is a skill that will benefit them in various aspects of their education and daily life.
Normal Tracing Worksheets Activities with Answers also provide children with instant feedback, which is crucial for effective learning. By checking their work against the answers, children can immediately see where they went wrong and learn from their mistakes. This self-assessment encourages a growth mindset, where children understand that practice leads to improvement.
In conclusion, normal tracing worksheets activities with answers are invaluable tools in early childhood education. They not only prepare children for academic success by developing fine motor skills, introducing foundational concepts, and enhancing focus but also foster a positive attitude towards learning through self-assessment and correction. These worksheets are indeed an essential part of a child's learning journey.
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If an event has probability *p* of occurring in some time interval, then the probability the event does not occur *q* is:
$$q=1-p$$
The probability of the event not occurring by time *t* will be:
$$P(q_1 \cap q_2 \cap...q_t) =q_1q_2...q_t = q^t$$
So the probability it *did occur* by is one minus this value ($q^t$), which is equal to the cumulative sum of *p* times the probability it had not occurred up to that point ($q^{t-1}$):
$$p_t=1-q^t=\sum\limits_{i=1}^t pq^{t-1}$$
If we are concerned with *n* independent events occurring with probabilities $p_1, p_2, ... p_n$ by time *t*, then:
$$P(p_{1t} \cap p_{2t} \cap...p_{nt}) =p_{1t}p_{2t}...p_{nt}$$
If $p_1=p_2= ... p_n$ then the above will be simply $p_t^n$. So the probability that all *n* events have occurred by time *t* will be:
$$P(t_{Allevents}\leq t)=(1-q^t)^n=\sum\limits_{i=1}^t (pq^{t-1})^n$$
If there is only one sequence of these events that results in the outcome of interest (e.g. $t_1<t_2<...t_n$, where $t_i$ refers to time of occurrence), the probability it is the observed sequence will be one over the total number of permutations ($1/n!$). So:
$$P(t_{Sequence}\leq t)=\frac{(1-q^t)^n}{n!}=\sum\limits_{i=1}^t \frac{(pq^{t-1})^n}{n!}$$
That gives us the CDF. To get the PDF we take the first derivative which is:
$$P(t\geq t_{Sequence}\leq t+1)=\frac{-nq^tln(q)(1-q^t)^{n-1}}{n!}$$
Given the above assumptions, we would expect the probability that the sequence of events completed at any given time interval to follow the PDF, shown in the lower row of plots:
t=1:100; p=.025; q=1-p
par(mfrow=c(2,4))
for(n in c(1,2,4,6)){
plot(t,(cumsum((p*q^(t-1)))^n)/factorial(n), xlab="Time",
ylab="P(t.Seq <= t)",main=paste(n, "Events"))
lines(t,((1-q^(t))^n)/factorial(n))
}
for(n in c(1,2,4,6)){
plot(t,(-n*(q^t)*log(q)*(1-q^t)^(n-1))/factorial(n), log="xy", xlab="Time",
ylab="P(t<= t.Seq<= t+1)",main=paste(n, "Events"))
lines(t[-1]-.5,diff(((1-q^(t))^n)/factorial(n)), col="Red")
}
![enter image description here][1]
I can find no flaw with the above reasoning. So my question is what did Armitage and Dodd calculate here: http://stats.stackexchange.com/questions/145214/i-have-an-epidemiology-question-with-logs ?
[1]: http://i.stack.imgur.com/wfArP.png
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# A Comprehensive Guide to Excelling in Your Probability Exam
August 30, 2023
Brandon Miles
Australia
Math
With a master’s in mathematics, Brandon Miles is a top-rates and experienced exam helper with hundreds of clients.
Probability is a fascinating branch of mathematics that allows us to quantify uncertainty and make informed decisions in a wide range of fields, from finance and science to everyday life. Whether you're a student preparing for a probability exam or someone looking to gain a deeper understanding of this topic, this comprehensive guide is here to help you master the subject. In this blog, we will explore the fundamental concepts of probability, key formulas, strategies for success, and practical tips to ensure you perform exceptionally well when you take your math exam.
## Understanding the Basics
Understanding the basics of probability, including events, outcomes, and sample spaces, is crucial for a strong foundation. These concepts form the building blocks for more complex probability calculations, enabling you to tackle a wide range of problems with confidence in your probability exam.
## Key Probability Concepts
Key probability concepts, such as the addition and multiplication rules, conditional probability, independence, and Bayes' Theorem, provide the framework for solving intricate probability problems. Proficiency in these concepts empowers you to analyze real-world scenarios and make informed decisions, making them pivotal in acing your probability exam. To excel in your probability exam, you need a strong grasp of these fundamental concepts:
### 1. What is Probability?
At its core, probability is the mathematical study of uncertainty. It provides a way to quantify the likelihood of events occurring. In probability, we work with events, outcomes, and sample spaces. Here's a brief breakdown:
Event: An event is a specific outcome or a combination of outcomes. For example, rolling a six on a fair six-sided die is an event.
Outcome: An outcome is a single result of an experiment. In the die example, getting a 6 is an outcome.
Sample Space: For a six-sided die, the sample space is {1, 2, 3, 4, 5, 6}.
### 2. Probability Notation
In probability, we use mathematical notation to describe the likelihood of events. For example, if we want to find the probability of rolling a 6 on a fair die, we write P(6) = 1/6, since there's one favorable outcome (rolling a 6) out of six possible outcomes.
### 3. Probability Rules
Probability rules like the addition and multiplication rules are fundamental tools for calculating probabilities of different events. These rules provide a systematic approach to handling complex probability scenarios, making them essential for excelling in your probability exam. They help you find the likelihood of combined events, a key skill for real-world applications.
The addition rule helps us find the probability of the union of two events A and B, denoted as P (A ∪ B). If the events are mutually exclusive (they cannot occur simultaneously), we use the simple addition rule:
P(A ∪ B) = P(A) + P(B)
If the events are not mutually exclusive (they can occur together), we use the general addition rule:
P(A ∪ B) = P(A) + P(B) - P(A ∩ B)
b. The Multiplication Rule
The multiplication rule helps us find the probability of the intersection of two events A and B, denoted as P(A ∩ B):
[P(A ∩ B) = P(A) * P(B|A)
Where P(B|A) represents the conditional probability of B occurring given that A has occurred.
### 4. Conditional Probability
Conditional probability deals with the likelihood of an event occurring under specific conditions. It's denoted as P(A|B), the probability of A occurring given that B has occurred. It's calculated as:
P(A|B) = (P(A∩B))/P(B)
### 5. Independence
Independence in probability is a fundamental concept that simplifies calculations. When two events are independent, the occurrence of one doesn't impact the other. This concept has broad applications, from statistical analysis to real-life scenarios like coin flips and dice rolls. Recognizing independence allows you to apply simplified probability rules, making complex problems more manageable. It's a key tool in your probability toolkit, enabling you to solve a wide range of problems with ease and confidence in your probability exam. Two events, A and B, are independent if the occurrence of one doesn't affect the occurrence of the other. In such cases, P(A ∩ B) = P(A) * P(B).
### 6. Complementary Events
Complementary events, often denoted as A' (pronounced "A prime"), are fundamental in probability theory. They represent the probability of an event not occurring. Understanding these events can simplify complex calculations. For instance, finding the probability of not rolling a 6 on a fair die can be easier than calculating the probability of rolling any of the other five numbers. Complementary events offer a strategic shortcut in probability problem-solving, making them a valuable concept to grasp for exam success. The probability of an event A not happening is denoted as P(A'). It's calculated as P(A') = 1 - P(A).
### 7. Bayes' Theorem
Bayes' Theorem is a powerful tool for updating probabilities based on new information. It's especially useful in situations involving conditional probability.
[P(A|B) = (P(B│A)*P(A))/(P(B))
## Key Probability Distributions
Key probability distributions like the uniform, binomial, normal, and Poisson are essential tools in probability theory. They provide powerful ways to model and analyze real-world situations, from random sampling to statistical analysis. Mastering these distributions equips you to handle diverse problems effectively in your probability exam.
### 1. The Uniform Distribution
The uniform distribution is a fundamental concept in probability, where each outcome in a sample space has an equal chance of occurring. This simple yet powerful distribution is often encountered in everyday scenarios, such as rolling a fair die or selecting a random card from a well-shuffled deck. Understanding the uniform distribution not only helps solve basic probability problems but also forms the basis for more complex probability models, making it a cornerstone of probability theory and its practical applications in various fields. For example, when rolling a fair six-sided die, each outcome has a probability of 1/6.
### 2. The Binomial Distribution
The binomial distribution is employed to model the number of successes, often denoted as 'x,' in a fixed number of independent trials, each with the same probability of success 'p.' Whether you're predicting the likelihood of a specific number of heads in multiple coin tosses or analyzing the success rate of a series of experiments, understanding the binomial distribution is essential for solving real-world problems and excelling in probability exams.
### 3. The Normal Distribution
The Normal Distribution, often referred to as the "bell curve," is a fundamental concept in probability and statistics. Understanding it is crucial because it describes the distribution of many real-world phenomena, from heights and test scores to errors in measurements. Comprehending the parameters, mean (μ) and standard deviation (σ), allows you to make accurate predictions and conduct hypothesis testing. Its ubiquity in statistics makes it a must-know topic for excelling in probability exams and applying statistical principles in various fields.
### 4. The Poisson Distribution
The Poisson Distribution is a powerful tool for modeling rare events that occur randomly over time or space. Its simplicity makes it useful in various fields, such as biology, finance, and telecommunications. The single parameter, λ (average event rate), allows you to predict the likelihood of a specific number of events happening in a fixed interval. Understanding and applying the Poisson Distribution effectively equips you to handle scenarios involving infrequent but significant occurrences, a valuable skill in real-world applications and probability exams.
## Strategies for Success in Probability Assignments
Strategies for success in probability assignments involve more than just calculations. It's about embracing a holistic approach. Consistent practice helps build confidence and mastery. Understanding underlying concepts, not just formulas, ensures you can tackle diverse problems effectively. Collaborating with peers for discussion and explanations enhances comprehension. Regular revision solidifies your knowledge. Seeking help when needed and using calculators for complex problems can be game-changers in excelling in your probability assignments.
### 1. Read the Instructions Carefully
Reading instructions carefully is a crucial exam strategy. It ensures you're on the right track from the beginning. Exam formats can vary, and understanding the specifics, like the number of questions, time allocation, and any special instructions, can prevent costly errors. Rushing without comprehension can lead to misunderstandings and lower scores. By taking a few extra moments to grasp the instructions, you set the stage for a smoother, more successful exam experience.
### 2. Practice, Practice, Practice
Practice is the cornerstone of success in probability. Repetition refines your skills and boosts your problem-solving confidence. By working through a variety of problems, from simple to challenging, you gain a deeper understanding of probability concepts and hone your analytical abilities. Textbook exercises, online resources, and past exams are valuable resources for practice. The more you practice, the better equipped you become to handle the diverse scenarios and questions that may arise in your probability assignments and exams.
### 3. Understand the Concepts
Understanding the concepts behind probability is vital for excelling in your probability exam. Beyond rote memorization, grasp the underlying principles. Visualize scenarios and real-world applications to make abstract concepts relatable. By truly comprehending the "why" behind formulas and rules, you can adapt and apply your knowledge to a variety of problems, rather than relying solely on memorized procedures. This deeper understanding will serve you well, not just in exams, but in real-life situations where probability plays a critical role in decision-making.
### 4. Use Probability Tables and Calculators
Leveraging probability tables and calculators can significantly simplify complex probability calculations. These tools provide quick access to cumulative probability values and can handle intricate statistical distributions. They not only save time during exams but also reduce the risk of computational errors. While understanding the underlying concepts is vital, these tables and calculators serve as valuable aids in performing accurate and efficient calculations, giving you a competitive edge in mastering probability-related assignments and exams.
### 5. Study in Groups
Leveraging probability tables and calculators can significantly simplify complex probability calculations. These tools provide quick access to cumulative probability values and can handle intricate statistical distributions. They not only save time during exams but also reduce the risk of computational errors. While understanding the underlying concepts is vital, these tables and calculators serve as valuable aids in performing accurate and efficient calculations, giving you a competitive edge in mastering probability-related assignments and exams.
### 6. Review and Revise
The process of reviewing and revising is a vital aspect of excelling in probability. Regularly revisiting the material helps reinforce your understanding, ensuring that the concepts stay fresh in your mind. Summarizing the key points, creating flashcards, or developing concise notes are effective revision techniques. Moreover, reviewing your work from previous assignments and exams can highlight areas where you might need additional practice, boosting your confidence and competence when tackling probability problems in the future.
### 7. Seek Help When Needed
Seeking help when needed is a smart strategy in mastering probability. It's a sign of maturity in learning, recognizing that everyone encounters challenges. Whether it's reaching out to a professor, seeking assistance from a tutor, or consulting online resources, these options offer valuable insights and clarify doubts. Don't hesitate to ask questions and address any uncertainties promptly. This proactive approach ensures a deeper understanding of complex concepts and ultimately leads to success in your probability endeavors.
## Conclusion
As you prepare to take your probability exam, remember that understanding this intriguing branch of mathematics goes beyond equations and calculations. It's about grasping the core concepts, recognizing patterns, and embracing effective study strategies. Armed with a solid foundation, key probability distributions, and a holistic approach to learning, you're well-equipped to excel. So, step into that exam room with confidence, knowing that your dedication and knowledge will guide you to success in your probability exam.
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# Dynes to Kilonewtons Converter
Enter the force in dynes below to get the value converted to kilonewtons.
## Result in Kilonewtons:
1 dyn = 1.0E-8 kN
Hint: use a scientific notation calculator to convert E notation to decimal
Do you want to convert kilonewtons to dynes?
## How to Convert Dynes to Kilonewtons
To convert a measurement in dynes to a measurement in kilonewtons, divide the force by the following conversion ratio: 100,000,000 dynes/kilonewton.
Since one kilonewton is equal to 100,000,000 dynes, you can use this simple formula to convert:
kilonewtons = dynes ÷ 100,000,000
The force in kilonewtons is equal to the force in dynes divided by 100,000,000.
For example, here's how to convert 50,000,000 dynes to kilonewtons using the formula above.
kilonewtons = (50,000,000 dyn ÷ 100,000,000) = 0.5 kN
## What Is a Dyne?
The dyne is the force needed to move one gram of mass at a rate of one centimeter per second squared.
The dyne is a centimeter-gram-second (CGS) unit of force. Dynes can be abbreviated as dyn; for example, 1 dyne can be written as 1 dyn.
Dynes can be expressed using the formula: 1 dyn = 1 gcm / s2
## What Is a Kilonewton?
One kilonewton is equal to 1,000 newtons, which are equal to the force needed to move one kilogram of mass at a rate of one meter per second squared.
The kilonewton is a multiple of the newton, which is the SI derived unit for force. In the metric system, "kilo" is the prefix for thousands, or 103. Kilonewtons can be abbreviated as kN; for example, 1 kilonewton can be written as 1 kN.
## Dyne to Kilonewton Conversion Table
Table showing various dyne measurements converted to kilonewtons.
Dynes Kilonewtons
1 dyn 0.00000001 kN
2 dyn 0.00000002 kN
3 dyn 0.00000003 kN
4 dyn 0.00000004 kN
5 dyn 0.00000005 kN
6 dyn 0.00000006 kN
7 dyn 0.00000007 kN
8 dyn 0.00000008 kN
9 dyn 0.00000009 kN
10 dyn 0.0000001 kN
100 dyn 0.000001 kN
1,000 dyn 0.00001 kN
10,000 dyn 0.0001 kN
100,000 dyn 0.001 kN
1,000,000 dyn 0.01 kN
10,000,000 dyn 0.1 kN
100,000,000 dyn 1 kN
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# Problem: Determine the molarity of each of the following solutions:(b) 0.515 g of H2SO4 in 1.00 L of solution
###### FREE Expert Solution
We’re being asked to calculate the molarity (M) of a solution of H2SO4
The molar mass of H2SO4 is:
H2SO4 2 H × 1 g/mol H = 2 g/mol
1 S × 32 g/mol S = 32 g/mol
4 O × 16 g/mol O = 64 g/mol
Total: 98 g/mol
99% (424 ratings)
###### Problem Details
Determine the molarity of each of the following solutions:
(b) 0.515 g of H2SO4 in 1.00 L of solution
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# [seqfan] nXk 1..Q matrices with each value adjacent to each other value
Ron Hardin rhhardin at att.net
Mon Oct 25 00:25:55 CEST 2010
```So far it seems like a difficult computational problem - T(n,k) = maximal number
Q in a 1..Q nXk array such that every value appears at least once adjacent to
every other value.
Some tiny n,k maximal Q solutions with Q=2,3,4,5,6
All.solutions.for.1X2..
..1..2....2..1..
Some.solutions.for.2X2..
..1..1....1..1....1..2....1..2....1..2....1..3....1..3....1..3....2..1....2..1..
..2..3....3..2....1..3....3..2....3..3....1..2....2..2....2..3....2..3....3..1..
Some.solutions.for.3X2..
..1..3....1..3....1..2....1..2....1..4....1..4....3..1....3..1....3..2....3..2..
..2..4....4..2....3..4....4..3....3..2....2..3....2..4....4..2....1..4....4..1..
..3..1....3..1....2..1....2..1....4..1....4..1....1..3....1..3....2..3....2..3..
Some.solutions.for.3X3..
..1..2..3....1..2..3....1..2..3....1..2..3....1..2..4....1..2..4....1..2..4..
..3..4..5....3..5..4....4..5..1....5..4..1....3..5..1....4..3..5....4..5..3..
..5..1..2....4..1..2....2..3..4....2..3..5....2..4..3....5..1..2....3..1..2..
Some.solutions.for.3X4..
..1..2..4..6....1..2..4..5....1..2..4..3....1..2..4..6....1..2..4..6..
..3..4..5..2....3..4..6..2....3..2..5..6....3..2..5..1....3..2..5..3..
..5..1..6..3....6..1..5..3....5..6..1..4....5..6..3..4....5..6..1..4..
rhhardin at mindspring.com
rhhardin at att.net (either)
```
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# Tag Info
3
The (in my opinion) simplest way to proceed about this is: First compute the square root $m:=\sqrt n$ of $n$ in $\mathbb N$; this can, for instance, be done in time $\mathcal O(\log^3n)$ using a binary search. The next step is to compute $\varphi(m)$ from $\varphi(n)$: by the properties of $\varphi$ we have \varphi(n) = (p-1)p(q-1)q=\varphi(m)\cdot m ...
4
When decrypting in lattice-based cryptosystems, one computes a value $v \in \mathbb{Z}_q$ that is guaranteed to be congruent to a "small" integer $e \in \mathbb{Z}$, where $e$ encodes the message (e.g., as the parity of $e$ modulo 2). By using the integer representatives between $-q/2$ and $q/2$, one can recover the small integer $e$ (and thereby recover ...
1
My understanding is that the coefficients of polynomials used in lattice crypto are often sampled from a discrete Gaussian distribution. A Gaussian is centered at 0, which would explain why the elements are represented as elements from the set $\{\frac{−(q−1)}{2},…,\frac{(q−1)}{2}\}$, as you mentioned.
1
There is no standard "multiply two group elements" operation in an additive group. So you first need to define what you mean by $P*Q$. From the comments I gather that you want $P*Q = q P = p Q = (p \cdot q) G$. The computational Diffie-Hellman (CDH) problem is: Given $P=pG$ and $Q=qG$ compute $(p\cdot q)G$. which is clearly equivalent to your problem. ...
0
Well the equation $R = P * Q$ simply isn't possible on an elliptic curve. The group of points on the EC is an additive group. Meaning it is only possible to compute $P + Q$ or $[m] P$ for some integer m. Taken $P=p \cdot G$ and $Q=q \cdot G$ you already got the answer yourself: $R=(p \cdot q)G$. Simply add the point $G$ to itself $(p \cdot q)$-times.
Top 50 recent answers are included
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# The Magic of… Bernoulli’s Principle. Aerodynamics is… The study of forces and the resulting motion of objects through the air.
## Presentation on theme: "The Magic of… Bernoulli’s Principle. Aerodynamics is… The study of forces and the resulting motion of objects through the air."— Presentation transcript:
The Magic of… Bernoulli’s Principle
Aerodynamics is… The study of forces and the resulting motion of objects through the air.
Airfoil - A part or surface, such as a wing, propeller blade, or rudder, whose shape and orientation control stability, direction, lift, thrust, or propulsion.
the force that directly opposes the weight of an airplane and holds the airplane in the air
Bernoulli’s Principle States… The pressure of a fluid decreases as the speed of the fluid increases.
Daniel Bernoulli first formulated this principle in the 1700’s.
Lift can be explained in part by the Bernoulli Principle. A Swiss mathematician, Bernoullli discovered that as the speed of a fluid increases, its pressure decreases. Air is a fluid so as it flows over a wing or airfoil, the wing’s shape and angle of attack cause the air to speed up above the wing’s surface. As the air speeds up, its pressure goes down, creating a low pressure area above the wing. This low pressure area creates lift, drawing the aircraft upward.
Bernoulli’s Principle Both streams must meet at the end of the wing at the same time. Stream A has farther to go; therefore, it must travel faster.
Factors that Affect Lift The Motion: Velocity and Angle of Attack The Object: Shape and Size The Air: Mass, Viscosity, Compressibility
Angle of Attack The acute angle between the direction of the relative wind and the chord of an airfoil.
Shape of the Wing Camber- A measure of the curvature of the airfoil Chord-The width of an airfoil or wing Leading Edge- The front, usually rounded, edge of an airplane wing or airfoil Trailing Edge- The rear edge of the wing
Chord Line Upper Camber Leading Edge Angle of Attack Lower Camber Trailing Edge
the force generated by the gravitational attraction of the earth on the aircraft; lift must be equal to weight in order to sustain flight.
a force applied to a body to propel it in a desired direction; the force which moves and aircraft through the air
Resistance of the air (technically a fluid) against the forward movement of an airplane
Download ppt "The Magic of… Bernoulli’s Principle. Aerodynamics is… The study of forces and the resulting motion of objects through the air."
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Limit of a composite function
• Dec 2nd 2012, 04:06 PM
Doubled144314
Limit of a composite function
I don't want an easy answer to this problem. However, I would be happy if you could provide me with theorems and/or techniques required to solve it.
$\displaystyle$\lim _{x\to \infty }x^2(ln({x+1\over x}) +ln({2x+3\over 2x}))$$I know that \displaystyle \lim _{x\to \infty }({x+1\over x})^x = \lim _{x\to \infty }(1+{1\over x})^x = e$$ But the natural logarithm is in the way and I think that you can't calculate the limit IN the logarithm first.
P.S. Oh, and by the way could anyone recommend a book on calculus with challenging problems, because most of the usual calculus textbooks aren't rigorous enough for my college course. (My lecturer always finds a way to give much more complicated problems than those in textbooks)
• Dec 2nd 2012, 06:55 PM
Soroban
Re: Limit of a composite function
Hello, Doubled144314!
We need these two theorems:
. . $\displaystyle \lim_{x\to\infty}\left(1 + \tfrac{1}{x}\right)^x \;=\;e$
. . $\displaystyle \lim_{x\to\infty}\left(1 + \tfrac{a}{x}\right)^x \;=\;e^a$
Quote:
$\displaystyle \displaystyle \lim _{x\to\infty}x^2\bigg[\ln\left(\tfrac{x+1}{x}\right) +\ln\left(\tfrac{2x+3}{2x}\right)\bigg]$
We have: .$\displaystyle \lim_{x\to\infty}x\cdot x\bigg[\ln\left(1+\tfrac{1}{x}\right) + \ln\left(1 + \tfrac{3}{2x}\right)\bigg]$
. . . . . . $\displaystyle =\;\lim_{x\to\infty}x\bigg[x\ln\left(1+\tfrac{1}{x}\right) + x\ln\left(1 + \tfrac{3}{2x}\right)^x\bigg]$
. . . . . . $\displaystyle =\;\lim_{x\to\infty}x\bigg[\ln\left(1 + \tfrac{1}{x}\right)^x + \ln\left(1 + \tfrac{\frac{3}{2}}{x}\right)^x\bigg]$
. . . . . . $\displaystyle =\;\lim_{x\to\infty}x\cdot \bigg[\ln\left(\lim_{x\to\infty}\left[1 + \tfrac{1}{x}\right]^x\right) + \ln\left(\lim_{x\to\infty}\left[1 + \tfrac{\frac{3}{2}}{x}\right]^x\right)\bigg]$
. . . . . . $\displaystyle =\;\infty\cdot \ln(e)\cdot \ln(e^{\frac{3}{2}}) \;=\;\infty\cdot1\cdot\tfrac{3}{2} \;=\;\infty$
• Dec 2nd 2012, 06:59 PM
Prove It
Re: Limit of a composite function
Quote:
Originally Posted by Doubled144314
I don't want an easy answer to this problem. However, I would be happy if you could provide me with theorems and/or techniques required to solve it.
$\displaystyle$\lim _{x\to \infty }x^2(ln({x+1\over x}) +ln({2x+3\over 2x}))$$I know that \displaystyle \lim _{x\to \infty }({x+1\over x}) = \lim _{x\to \infty }(1+{1\over x})^1 = e$$ But the natural logarithm is in the way and I think that you can't calculate the limit IN the logarithm first.
I would rewrite the function as \displaystyle \displaystyle \begin{align*} \lim_{x \to \infty} \frac{\ln{\left(\frac{x+1}{x}\right)} + \ln{\left(\frac{2x+3}{2x}\right)}}{\frac{1}{x^2}} \end{align*}, and since this goes to \displaystyle \displaystyle \begin{align*} \frac{0}{0} \end{align*} you can apply L'Hospital's Rule.
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You are previewing Applied Statistics and Probability for Engineers, 5th Edition.
1. Coverpage
2. Titlepage
4. Contents
5. Preface
6. INSIDE FRONT COVER Index of Applications in Examples and Exercises
7. CHAPTER 1 The Role of Statistics in Engineering
1. 1-1 The Engineering Method and Statistical Thinking
2. 1-2 Collecting Engineering Data
3. 1-3 Mechanistic and Empirical Models
4. 1-4 Probability and Probability Models
8. CHAPTER 2 Probability
1. 2-1 Sample Spaces and Events
2. 2-2 Interpretations and Axioms of Probability
4. 2-4 Conditional Probability
5. 2-5 Multiplication and Total Probability Rules
6. 2-6 Independence
7. 2-7 Bayes’ Theorem
8. 2-8 Random Variables
9. CHAPTER 3 Discrete Random Variables and Probability Distributions
10. CHAPTER 4 Continuous Random Variables and Probability Distributions
11. CHAPTER 5 Joint Probability Distributions
1. 5-1 Two or More Random Variables
2. 5-2 Covariance and Correlation
3. 5-3 Common Joint Distributions
4. 5-4 Linear Functions of Random Variables
5. 5-5 General Functions of Random Variables
12. CHAPTER 6 Descriptive Statistics
13. CHAPTER 7 Sampling Distributions and Point Estimation of Parameters
1. 7-1 Point Estimation
2. 7-2 Sampling Distributions and the Central Limit Theorem
3. 7-3 General Concepts of Point Estimation
4. 7-4 Methods of Point Estimation
14. CHAPTER 8 Statistical Intervals for a Single Sample
1. 8-1 Confidence Interval on the Mean of a Normal Distribution, Variance Known
2. 8-2 Confidence Interval on the Mean of a Normal Distribution, Variance Unknown
3. 8-3 Confidence Interval on the Variance and Standard Deviation of a Normal Distribution
4. 8-4 Large-Sample Confidence Interval for a Population Proportion
5. 8-5 Guidelines for Constructing Confidence Intervals
6. 8-6 Tolerance and Prediction Intervals
15. CHAPTER 9 Tests of Hypotheses for a Single Sample
1. 9-1 Hypothesis Testing
2. 9-2 Tests on the Mean of a Normal Distribution, Variance Known
3. 9-3 Tests on the Mean of a Normal Distribution, Variance Unknown
4. 9-4 Tests on the Variance and Standard Deviation of a Normal Distribution
5. 9-5 Tests on a Population Proportion
6. 9-6 Summary Table of Inference Procedures for a Single Sample
7. 9-7 Testing for Goodness of Fit
8. 9-8 Contingency Table Tests
9. 9-9 Nonparametric Procedures
16. CHAPTER 10 Statistical Inference for Two Samples
1. 10-1 Inference on the Difference in Means of Two Normal Distributions, Variances Known
2. 10-2 Inference on the Difference in Means of Two Normal Distributions, Variances Unknown
3. 10-3 A Nonparametric Test for the Difference in Two Means
4. 10-4 Paired t-Test
5. 10-5 Inference on the Variances of Two Normal Distributions
6. 10-6 Inference on Two Population Proportions
7. 10-7 Summary Table and Roadmap for Inference Procedures for Two Samples
17. CHAPTER 11 Simple Linear Regression and Correlation
1. 11-1 Empirical Models
2. 11-2 Simple Linear Regression
3. 11-3 Properties of the Least Squares Estimators
4. 11-4 Hypothesis Tests in Simple Linear Regression
5. 11-5 Confidence Intervals
6. 11-6 Prediction of New Observations
7. 11-7 Adequacy of the Regression Model
8. 11-8 Correlation
9. 11-9 Regression on Transformed Variables
10. 11-10 Logistic Regression
18. CHAPTER 12 Multiple Linear Regression
1. 12-1 Multiple Linear Regression Model
2. 12-2 Hypothesis Tests in Multiple Linear Regression
3. 12-3 Confidence Intervals in Multiple Linear Regression
4. 12-4 Prediction of New Observations
6. 12-6 Aspects of Multiple Regression Modeling
19. CHAPTER 13 Design and Analysis of Single-Factor Experiments: The Analysis of Variance
1. 13-1 Designing Engineering Experiments
2. 13-2 Completely Randomized Single-Factor Experiment
3. 13-3 The Random-Effects Model
4. 13-4 Randomized Complete Block Design
20. CHAPTER 14 Design of Experiments with Several Factors
1. 14-1 Introduction
2. 14-2 Factorial Experiments
3. 14-3 Two-Factor Factorial Experiments
4. 14-4 General Factorial Experiments
5. 14-5 2k Factorial Designs
6. 14-6 Blocking and Confounding in the 2k Design
7. 14-7 Fractional Replication of the 2k Design
8. 14-8 Response Surface Methods and Designs
21. CHAPTER 15 Statistical Quality Control
1. 15-1 Quality Improvement and Statistics
2. 15-2 Introduction to Control Charts
3. 15-3 X and R or S Control Charts
4. 15-4 Control Charts for Individual Measurements
5. 15-5 Process Capability
6. 15-6 Attribute Control Charts
7. 15-7 Control Chart Performance
8. 15-8 Time-Weighted Charts
9. 15-9 Other SPC Problem-Solving Tools
10. 15-10 Implementing SPC
22. APPENDICES
1. APPENDIX A: Statistical Tables and Charts
2. APPENDIX B: Answers to Selected Exercises
3. APPENDIX C: Bibliography
23. GLOSSARY
24. INDEX
25. INDEX OF APPLICATIONS IN EXAMPLES AND EXERCISES, CONTINUED
## INTRODUCTION
Engineers are often involved in estimating parameters. For example, there is an ASTM Standard E23 that defines a technique called the Charpy V-notch method for notched bar impact testing of metallic materials. The impact energy is often used to determine if the material experiences a ductile-to-brittle transition as the temperature decreases. Suppose that you have tested a sample of 10 specimens of a particular material with this procedure. You know that you can use the sample average to estimate the true mean impact energy μ. However, we also know that the true mean impact energy is unlikely to be exactly equal to your estimate. Reporting the results of your test as a single number is unappealing, because there is nothing inherent in that provides any information about how close it is to μ. Your estimate could be very close, or it could be considerably far from the true mean. A way to avoid this is to report the estimate in terms of a range of plausible values called a confidence interval. A confidence interval always specifies a confidence level, usually 90%, 95%, or 99%, which is a measure of the reliability of the procedure. So if a 95% confidence interval on the ...
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# Help with Date format
Studies that have been contributed to the community by other users. If you’ve got something useful to share, that’s great!
arnie
Posts: 1594
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### Help with Date format
Hi.
Let's see if someone can help me here with the following formula since rarely we see formulas being discussed here
My problem with it is drawing the trendlines from the beginning to the end of the selected input days.
How can I tranform the SessHi and SessLo variables into recognizable Date format, assuming that this is the only problem
Code: Select all
```Inputs: StDay (21), StMnth (11), StYear (108), EnDay (28), EnMnth (11), EnYear (108), PlotLines (True); Variables: SessST (true), SessEN (true), Hi (-999999), Lo (+999999), SessHi (0), SessLo (0), HiTL (-1), LoTL (-1); // Start time period SessST = Year(Date) > StYear or (Year(Date) = StYear and (Month(Date) > StMnth or Month(Date) = StMnth and DayOfMonth(Date) >= StDay)); // End time period SessEN = Year(Date) < EnYear or (Year(Date) = EnYear and (Month(Date) < EnMnth or Month(Date) = EnMnth and DayOfMonth(Date) <= EnDay)); // Highest High and Lowest Low inside time period if SessST and SessEN then begin If High > Hi then Hi = High; If Low < Lo then Lo = Low; SessHi = Hi; SessLo = Lo; // Plot TrendLines if PlotLines = false then begin HiTL = TL_New(date, Time, SessHi, Date, Time, SessHi); LoTL = TL_New(date, Time, SessLo, Date, Time, SessLo); end; if PlotLines = false then begin TL_SetBegin(HiTL, date, Time, SessHi); TL_SetEnd(HiTL, Date, Time, SessHi); TL_SetBegin(LoTL, date, Time, SessLo); TL_SetEnd(LoTL, Date, Time, SessLo); end; end; if PlotLines = true then begin if SessHi <> 0 then Plot1(SessHi); if SessLo <> 0 then Plot2(SessLo); end; ```
gregorio123456
Posts: 117
Joined: 08 Nov 2005
Been thanked: 3 times
well I don´t try you code but where you get value of SessHi and SessLo you put one line..... dateSessHi=date;timeSessHi=time;==> you have date and time of SessHi
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TJ
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### Re: Help with formula
Hi.
Let's see if someone can help me here with the following formula since rarely we see formulas being discussed here
My problem with it is drawing the trendlines from the beginning to the end of the selected input days.
can you post a mock up chart with arrows pointing to where you want your trendline and notes on how its done?
arnie
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OK.
See the images attached.
During the weekend I've been trying gregorio's way but with no success.
Attachments
ES image 01.png
ES image 02.png
SUPER
Posts: 638
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arnie,
I am not sure if you want to plot line from the end date, but this may plot lines similar to what you are looking for.
The last bit of your code will be written as follows:
if SessEN=false and PlotLines = true then begin
if SessHi <> 0 then
Plot1(SessHi);
if SessLo <> 0 then
Plot2(SessLo);
end;
TJ
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change from
Code: Select all
```Inputs: StDay (21), StMnth (11), StYear (108), ```
to
Inputs:
StDate (1081121),
if PlotLines = false then begin
TL_SetBegin(HiTL, stdate, Time, SessHi);
.
TJ
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or a more intuitive way of entering date:
input:
start.date(20081121);
var:
stdate(0);
stdate = start.date -19000000;
edit: corrected -19000000 order
sorry for the inconvenience.
Last edited by TJ on 15 Jun 2009, edited 2 times in total.
SUPER
Posts: 638
Joined: 03 Mar 2007
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Been thanked: 81 times
or a more intuitive way of entering date:
input:
start.date(20081121);
var:
stdate(0);
stdate = start.date -19000000;
TJ,
Thats very clever/interesting
2haerim
Posts: 502
Joined: 01 Sep 2006
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Can you explain a bit more detail how this works?
HR
gregorio123456
Posts: 117
Joined: 08 Nov 2005
Been thanked: 3 times
Can you explain a bit more detail how this works?
HR
try this
Code: Select all
```Inputs: StDay (1081121), // StMnth (11), // StYear (108), EnDay (1081228); // EnMnth (11), // EnYear (108); Variables: // SessST (true), // SessEN (true), Hi (-999999), Lo (+999999), SessHi (0), SessLo (0), HiTL (-1), LoTL (-1), PlotLines (True); // Start time period //SessST = Year(Date) > StYear // or (Year(Date) = StYear and (Month(Date) > StMnth // or Month(Date) = StMnth and DayOfMonth(Date) >= StDay)); // End time period //SessEN = Year(Date) < EnYear // or (Year(Date) = EnYear and (Month(Date) < EnMnth // or Month(Date) = EnMnth and DayOfMonth(Date) <= EnDay)); // Highest High and Lowest Low inside time period if StDay < date and EnDay>date then begin If High > Hi then Hi = High; If Low < Lo then Lo = Low; SessHi = Hi; SessLo = Lo; // Plot TrendLines if PlotLines = true then begin HiTL = TL_New(StDay, Time, SessHi, Date, Time, SessHi); LoTL = TL_New(StDay , Time, SessLo, Date, Time, SessLo); plotlines=false; end; if PlotLines = false then begin TL_SetBegin(HiTL, StDay , Time, SessHi); TL_SetEnd(HiTL, date , Time, SessHi); TL_SetBegin(LoTL, StDay , Time, SessLo); TL_SetEnd(LoTL, date , Time, SessLo); end; end; //if PlotLines = true then begin // if SessHi <> 0 then // Plot1(SessHi); // if SessLo <> 0 then // Plot2(SessLo); //end; ```
I post before but w error
jo
RobotMan
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Location: Los Altos, California, USA
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Contact:
Back before the turn of the century, back when my gran'pappy was still alive, back in the '80s. The 1980's. The Cruz brothers commissioned a small group of programmers to write a program called "SystemWriter". It was back in a time when phones had cords and email didn't have an "e" and was delivered to your house in small white envelopes called, well, envelopes.
No one thought much about the future because it was so far off. The only thing that mattered was right now, The Present. And right now, The Present always began with 19.
1980-Reagan as elected (yaa!), 1984-Apple Mac (wow, pricey), and 1987-Market Crash (lookout belowwwww). Who needs 20? We got 19 and that's all we will ever need.
But a long came the Y2k and elevators ceased to work, electricity went off in hospitals during surgery and ATMs zeroed out your accounts and planes fell from the sky. Remember? (10 years from now we will be looking back on Global Warming the same way).
"Time" for computers all started in the split second between 1899 and 1900. The month is known as "Month 00" and the day is known as "day 00" and the hour and minute is, well, you get the picture.
Computers also only had so many bits and bytes to play with so, everyone agreed time started, not at the big bang, but at 1900 and was represented as YYYMMDD past this magical demarcation.
Ie. Today is 20090616 (June, 16, 2009), but to Systemwriter, then Supercharts, then TS it would only look like:
20090616 - 19000000 = 1090616. Even to this day. The End.
Attachments
EL Date.jpg
arnie
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I want to apologise to you all for my late reply
I've been busy on family affairs and my free time has been shortened.
Thank you for all the answers given.
I'll test them this weekend.
Regards,
Fernando
arnie
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I've been testing the formula and I must say that TJ's way to input date is very "user friendly" Thanks for the tip
Gregorio, thanks for compiling it.
Just made 2 little modifications and now is 100% as I wanted.
Again, thank you all
Code: Select all
``` Inputs: StDay (20090507), EnDay (20090518); Variables: stDate (StDay - 19000000), enDate (EnDay - 19000000), Hi (-999999), Lo (+999999), SessHi (0), SessLo (0), HiTL (-1), LoTL (-1), PlotLines (True); // Highest High and Lowest Low inside time period If stDate <= date and enDate >= date then begin If High > Hi then Hi = High; If Low < Lo then Lo = Low; SessHi = Hi; SessLo = Lo; // Plot TrendLines if PlotLines = true then begin HiTL = TL_New(stDate, Time, SessHi, Date, Time, SessHi); LoTL = TL_New(stDate , Time, SessLo, Date, Time, SessLo); plotlines=false; end; if PlotLines = false then begin TL_SetBegin(HiTL, stDate , Time, SessHi); TL_SetEnd(HiTL, currentdate , Time, SessHi); TL_SetBegin(LoTL, stDate , Time, SessLo); TL_SetEnd(LoTL, currentdate , Time, SessLo); end; end; ```
TJ
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but this is not done yet.
see you soon.
gregorio123456
Posts: 117
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welcome
jo
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시간 제한메모리 제한제출정답맞힌 사람정답 비율
1 초 128 MB32266.667%
## 문제
Dr. Beverly has been experimenting with an interesting kind of organism. This organism’s DNA consists of a single string of k letters over some alphabet of size d. One hour after its birth, an individual gives birth to one offspring, after which it happily lives on for another 15 minutes or so. This process repeats until there is no cold pizza left to feed on.
A mutation is a replacement of a character in the string with any character from the alphabet, possibly with itself. Mutations may occur from one generation to the next. The probability of a mutation is the same for each of the characters of the alphabet and is denoted by p.
Unfortunately, after Dr. Beverly started an experiment with one such creature, she got so caught up in a computer game of some kind that she forgot all about her experiment. When she came back a while later, she found the remains of N creatures. She sampled their DNA, hoping she would be able to figure out which of the corpses belongs to the creature she started the experiment with. Oh if only she had made notes! Anyway, given N strings of DNA, can you compute the probability for each individual that it was the original creature? We may assume that each of the N strings is (a priori) equally likely to serve as the initial string. The order in which the N strings are given is random.
## 입력
The first line of the input contains a single number: the number of test cases to follow. Each test case has the following format:
• One line with three integers N, k and d, satisfying 1 ≤ N ≤ 15, 1 ≤ k ≤ 8, 1 ≤ d ≤ 4 and a real number p, satisfying 0.2 ≤ p ≤ 0.5. The numbers are separated by single spaces.
• N lines, each with a string of length k over some alphabet of size d. This alphabet is a subset of { A, B, C, . . . , Z } and is the same for all N strings. The string on the i-th line represents the DNA of the i-th creature.
## 출력
For every test case in the input, the output should contain N lines, each with a real number, rounded and displayed to six digits after the decimal point. The number on the i-th line should be the probability that the i-th creature in the input was the original creature.
## 예제 입력 1
2
3 1 2 0.25
A
C
C
4 4 4 0.50
GTTG
TGTG
TTTG
GTGT
## 예제 출력 1
0.466667
0.266667
0.266667
0.046602
0.393710
0.083333
0.476354
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Solve the question ; (i) x^4 − (x − z)^4 (ii) a^4 − 2a^2b^2 + b^4
42 views
Solve the question ;
(i) x4 − (x − z)4
(ii) a4 − 2a2b2 + b4
by (24.8k points)
selected
(i) x4 − (x − z)4 = (x2)2 − [(x − z)2]2
= [x2 − (x − z)2] [x2 + (x − z)2]
= [x − (x − z)] [x + (x − z)] [x2 + (x − z)2]
= z(2x − z) [x2 + x2 − 2xz + z2]
= z(2x − z) (2x2 − 2xz + z2)
(ii) a4 − 2a2b2 + b4
= (a2)2 − 2 (a2) (b2) + (b2)2
= (a2 − b2)2
= [(a − b) (a + b)]2
= (a − b)2 (a + b)2
| 281
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Reference documentation for deal.II version GIT relicensing-769-g16d39143b0 2024-05-28 17:50:02+00:00
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polynomials_barycentric.cc
Go to the documentation of this file.
1// ------------------------------------------------------------------------
2//
4// Copyright (C) 2021 - 2024 by the deal.II authors
5//
6// This file is part of the deal.II library.
7//
8// Part of the source code is dual licensed under Apache-2.0 WITH
9// LLVM-exception OR LGPL-2.1-or-later. Detailed license information
10// governing the source code and code contributions can be found in
11// LICENSE.md and CONTRIBUTING.md at the top level directory of deal.II.
12//
13// ------------------------------------------------------------------------
14
16
18
20
21namespace internal
22{
27 template <int dim>
28 unsigned int
30 const std::vector<typename BarycentricPolynomials<dim>::PolyType> &polys)
31 {
32 // Since the first variable in a simplex polynomial is, e.g., in 2d,
33 //
34 // t0 = 1 - x - y
35 //
36 // (that is, it depends on the Cartesian variables), we have to compute
37 // its degree separately. An example: t0*t1*t2 has degree 1 in the affine
38 // polynomial basis but is degree 2 in the Cartesian polynomial basis.
39 std::size_t max_degree = 0;
40 for (const auto &poly : polys)
41 {
42 const TableIndices<dim + 1> degrees = poly.degrees();
43
44 const auto degree_0 = degrees[0];
45 std::size_t degree_d = 0;
46 for (unsigned int d = 1; d < dim + 1; ++d)
47 degree_d = std::max(degree_d, degrees[d]);
48
49 max_degree = std::max(max_degree, degree_d + degree_0);
50 }
51
52 return max_degree;
53 }
54} // namespace internal
55
56
57template <int dim>
60{
61 std::vector<PolyType> polys;
62
63 const auto reference_cell = ReferenceCells::get_simplex<dim>();
64
65 switch (degree)
66 {
67 case 0:
69 1);
70 break;
71 case 1:
72 {
73 for (const unsigned int v : reference_cell.vertex_indices())
75 break;
76 }
77 case 2:
78 {
79 // vertices, then lines:
80 for (const unsigned int v : reference_cell.vertex_indices())
81 polys.push_back(
84 for (const unsigned int l : reference_cell.line_indices())
85 {
86 const auto v0 = reference_cell.line_to_cell_vertices(l, 0);
87 const auto v1 = reference_cell.line_to_cell_vertices(l, 1);
88 polys.push_back(4 *
91 }
92 break;
93 }
94 case 3:
95 {
96 // vertices, then lines, then quads:
97 for (const unsigned int v : reference_cell.vertex_indices())
98 polys.push_back(
102 for (unsigned int l : reference_cell.line_indices())
103 {
104 const auto v0 = reference_cell.line_to_cell_vertices(l, 0);
105 const auto v1 = reference_cell.line_to_cell_vertices(l, 1);
106 polys.push_back(
110 polys.push_back(
114 }
115
116 if (dim == 2)
117 {
118 polys.push_back(27 *
122 }
123 else if (dim == 3)
124 {
125 polys.push_back(27 *
129 polys.push_back(27 *
133 polys.push_back(27 *
137 polys.push_back(27 *
141 }
142
143 break;
144 }
145 default:
147 }
148
149 return BarycentricPolynomials<dim>(polys);
150}
151
152
153
154template <int dim>
156 const std::vector<PolyType> &polynomials)
157 : ScalarPolynomialsBase<dim>(internal::get_degree<dim>(polynomials),
158 polynomials.size())
159{
160 polys = polynomials;
161
163 poly_hessians.resize(polynomials.size());
164 poly_third_derivatives.resize(polynomials.size());
165 poly_fourth_derivatives.resize(polynomials.size());
166
167 for (std::size_t i = 0; i < polynomials.size(); ++i)
168 {
170 for (unsigned int d = 0; d < dim; ++d)
171 poly_grads[i][d] = polynomials[i].derivative(d);
172
173 // hessians
174 for (unsigned int d0 = 0; d0 < dim; ++d0)
175 for (unsigned int d1 = 0; d1 < dim; ++d1)
176 poly_hessians[i][d0][d1] = poly_grads[i][d0].derivative(d1);
177
178 // third derivatives
179 for (unsigned int d0 = 0; d0 < dim; ++d0)
180 for (unsigned int d1 = 0; d1 < dim; ++d1)
181 for (unsigned int d2 = 0; d2 < dim; ++d2)
182 poly_third_derivatives[i][d0][d1][d2] =
183 poly_hessians[i][d0][d1].derivative(d2);
184
185 // fourth derivatives
186 for (unsigned int d0 = 0; d0 < dim; ++d0)
187 for (unsigned int d1 = 0; d1 < dim; ++d1)
188 for (unsigned int d2 = 0; d2 < dim; ++d2)
189 for (unsigned int d3 = 0; d3 < dim; ++d3)
190 poly_fourth_derivatives[i][d0][d1][d2][d3] =
191 poly_third_derivatives[i][d0][d1][d2].derivative(d3);
192 }
193}
194
195
196
197template <int dim>
198void
200 const Point<dim> &unit_point,
201 std::vector<double> &values,
202 std::vector<Tensor<1, dim>> &grads,
204 std::vector<Tensor<3, dim>> &third_derivatives,
205 std::vector<Tensor<4, dim>> &fourth_derivatives) const
206{
207 Assert(values.size() == this->n() || values.empty(),
208 ExcDimensionMismatch2(values.size(), this->n(), 0));
210 ExcDimensionMismatch2(grads.size(), this->n(), 0));
213 Assert(third_derivatives.size() == this->n() || third_derivatives.empty(),
214 ExcDimensionMismatch2(third_derivatives.size(), this->n(), 0));
215 Assert(fourth_derivatives.size() == this->n() || fourth_derivatives.empty(),
216 ExcDimensionMismatch2(fourth_derivatives.size(), this->n(), 0));
217
218 for (std::size_t i = 0; i < polys.size(); ++i)
219 {
220 if (values.size() == this->n())
221 values[i] = polys[i].value(unit_point);
222
224 if (grads.size() == this->n())
225 for (unsigned int d = 0; d < dim; ++d)
227
228 // hessians
230 for (unsigned int d0 = 0; d0 < dim; ++d0)
231 for (unsigned int d1 = 0; d1 < dim; ++d1)
233
234 // third derivatives
235 if (third_derivatives.size() == this->n())
236 for (unsigned int d0 = 0; d0 < dim; ++d0)
237 for (unsigned int d1 = 0; d1 < dim; ++d1)
238 for (unsigned int d2 = 0; d2 < dim; ++d2)
239 third_derivatives[i][d0][d1][d2] =
240 poly_third_derivatives[i][d0][d1][d2].value(unit_point);
241
242 // fourth derivatives
243 if (fourth_derivatives.size() == this->n())
244 for (unsigned int d0 = 0; d0 < dim; ++d0)
245 for (unsigned int d1 = 0; d1 < dim; ++d1)
246 for (unsigned int d2 = 0; d2 < dim; ++d2)
247 for (unsigned int d3 = 0; d3 < dim; ++d3)
248 fourth_derivatives[i][d0][d1][d2][d3] =
249 poly_fourth_derivatives[i][d0][d1][d2][d3].value(unit_point);
250 }
251}
252
254
255template <int dim>
256double
258 const Point<dim> &p) const
259{
260 AssertIndexRange(i, this->n());
261 return polys[i].value(p);
262}
263
264
265
266template <int dim>
269 const Point<dim> &p) const
270{
272 for (unsigned int d = 0; d < dim; ++d)
273 result[d] = poly_grads[i][d].value(p);
274 return result;
275}
276
277
278
279template <int dim>
282 const Point<dim> &p) const
283{
284 Tensor<2, dim> result;
285 for (unsigned int d0 = 0; d0 < dim; ++d0)
286 for (unsigned int d1 = 0; d1 < dim; ++d1)
287 result[d0][d1] = poly_hessians[i][d0][d1].value(p);
289 return result;
290}
291
292
293
294template <int dim>
297 const Point<dim> &p) const
298{
299 Tensor<3, dim> result;
300 for (unsigned int d0 = 0; d0 < dim; ++d0)
301 for (unsigned int d1 = 0; d1 < dim; ++d1)
302 for (unsigned int d2 = 0; d2 < dim; ++d2)
303 result[d0][d1][d2] = poly_third_derivatives[i][d0][d1][d2].value(p);
304
305 return result;
306}
307
308
310template <int dim>
313 const Point<dim> &p) const
314{
315 Tensor<4, dim> result;
316 for (unsigned int d0 = 0; d0 < dim; ++d0)
317 for (unsigned int d1 = 0; d1 < dim; ++d1)
318 for (unsigned int d2 = 0; d2 < dim; ++d2)
319 for (unsigned int d3 = 0; d3 < dim; ++d3)
320 result[d0][d1][d2][d3] =
321 poly_fourth_derivatives[i][d0][d1][d2][d3].value(p);
323 return result;
324}
325
326
327
328template <int dim>
331 const Point<dim> &p) const
332{
333 return compute_1st_derivative(i, p);
335
336
337
338template <int dim>
341 const Point<dim> &p) const
342{
343 return compute_2nd_derivative(i, p);
344}
345
346
347
348template <int dim>
349std::unique_ptr<ScalarPolynomialsBase<dim>>
351{
352 return std::make_unique<BarycentricPolynomials<dim>>(*this);
353}
354
355
356
357template <int dim>
358std::string
360{
361 return "BarycentricPolynomials<" + std::to_string(dim) + ">";
362}
363
364
365
366template <int dim>
367std::size_t
369{
370 std::size_t poly_memory = 0;
371 for (const auto &poly : polys)
372 poly_memory += poly.memory_consumption();
376 MemoryConsumption::memory_consumption(poly_third_derivatives) +
377 MemoryConsumption::memory_consumption(poly_fourth_derivatives);
378}
379
380template class BarycentricPolynomials<1>;
381template class BarycentricPolynomials<2>;
382template class BarycentricPolynomials<3>;
383
static BarycentricPolynomial< dim, Number > monomial(const unsigned int d)
Tensor< 1, dim > compute_grad(const unsigned int i, const Point< dim > &p) const override
BarycentricPolynomials(const std::vector< BarycentricPolynomial< dim > > &polynomials)
virtual std::size_t memory_consumption() const override
Tensor< 2, dim > compute_grad_grad(const unsigned int i, const Point< dim > &p) const override
std::string name() const override
Tensor< 3, dim > compute_3rd_derivative(const unsigned int i, const Point< dim > &p) const override
Tensor< 1, dim > compute_1st_derivative(const unsigned int i, const Point< dim > &p) const override
Tensor< 2, dim > compute_2nd_derivative(const unsigned int i, const Point< dim > &p) const override
std::vector< PolyType > polys
void evaluate(const Point< dim > &unit_point, std::vector< double > &values, std::vector< Tensor< 1, dim > > &grads, std::vector< Tensor< 2, dim > > &grad_grads, std::vector< Tensor< 3, dim > > &third_derivatives, std::vector< Tensor< 4, dim > > &fourth_derivatives) const override
double compute_value(const unsigned int i, const Point< dim > &p) const override
std::vector< ThirdDerivativesType > poly_third_derivatives
virtual std::unique_ptr< ScalarPolynomialsBase< dim > > clone() const override
Tensor< 4, dim > compute_4th_derivative(const unsigned int i, const Point< dim > &p) const override
std::vector< HessianType > poly_hessians
static BarycentricPolynomials< dim > get_fe_p_basis(const unsigned int degree)
std::vector< FourthDerivativesType > poly_fourth_derivatives
Definition point.h:111
virtual std::size_t memory_consumption() const
#define DEAL_II_NAMESPACE_OPEN
Definition config.h:502
#define DEAL_II_NAMESPACE_CLOSE
Definition config.h:503
const unsigned int v0
const unsigned int v1
#define Assert(cond, exc)
static ::ExceptionBase & ExcDimensionMismatch2(std::size_t arg1, std::size_t arg2, std::size_t arg3)
#define AssertIndexRange(index, range)
#define DEAL_II_NOT_IMPLEMENTED()
std::enable_if_t< std::is_fundamental_v< T >, std::size_t > memory_consumption(const T &t)
unsigned int get_degree(const std::vector< typename BarycentricPolynomials< dim >::PolyType > &polys)
::VectorizedArray< Number, width > max(const ::VectorizedArray< Number, width > &, const ::VectorizedArray< Number, width > &)
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en
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https://outschool.com/classes/go-geometry-angles-perimeter-circumference-area-volume-more-NlYmtMY3
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English
### Go Geometry! Angles, Perimeter, Circumference, Area, Volume & More | Self-Paced
Ms. Ann (or Ms. W)
Star Educator
Average rating:5.0Number of reviews:(75)
This 6-part course reviews geometric measurement for angles and 2D and 3D objects, including perimeter, area, circumference, volume, and surface area.
#### Class experience
###### Aligned with Common Core State Standards (CCSS).css-1gf0lzu{-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;width:1em;height:1em;display:inline-block;fill:currentColor;-webkit-flex-shrink:0;-ms-flex-negative:0;flex-shrink:0;-webkit-transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;font-size:inherit;vertical-align:-0.125em;color:#380596;margin-left:0.4rem;}
3 units//8 lessons//6 Weeks
`.css-8atqhb{width:100%;}.css-9tf9ac{margin-bottom:1em;}.css-7yk5b4{padding:16px;border-radius:16px;background-color:#F7F7F7;margin-bottom:16px;}.css-1py0ztu{display:grid;grid-gap:0;grid-template-columns:96px 1fr 195px;padding-left:16px;padding-right:8px;margin-bottom:16px;}@media (max-width:767.95px){.css-1py0ztu{grid-template-columns:96px 1fr;}}.css-r0s4vb{margin:0;font-family:Ginto Normal,sans-serif;font-size:1.2rem;line-height:2rem;font-weight:500;letter-spacing:0.1rem;text-transform:uppercase;color:#7b2169;font-weight:700;}Unit 1.css-1kilzem{margin:0;font-family:Ginto Normal,sans-serif;font-size:1.6rem;line-height:1.5;font-weight:400;letter-spacing:0;text-transform:none;color:#1A1a1a;padding-left:16px;padding-right:16px;line-height:24px;font-weight:700;}@media (max-width:767.95px){.css-1kilzem{display:none;}}2D Objects.css-1iz5s76{padding-left:16px;padding-right:24px;line-height:1em;text-align:right;}@media (max-width:767.95px){.css-1iz5s76{padding-right:16px;width:100%;}}.css-zjrce1{margin:0;font-family:Ginto Normal,sans-serif;font-size:1.2rem;line-height:2rem;font-weight:500;letter-spacing:0.1rem;text-transform:uppercase;color:#004a77;margin-right:8px;font-weight:700;}4 lessons.css-1p5vpqy{margin:0;font-family:Ginto Normal,sans-serif;font-size:1.2rem;line-height:2rem;font-weight:500;letter-spacing:0.1rem;text-transform:none;color:#5C5C5C;font-weight:700;}.css-1neeyhn{-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;width:1em;height:1em;display:inline-block;fill:currentColor;-webkit-flex-shrink:0;-ms-flex-negative:0;flex-shrink:0;-webkit-transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;font-size:inherit;vertical-align:-0.125em;text-transform:uppercase;margin-right:5px;}3 Weeks.css-ddr7uv{margin:0;font-family:Ginto Normal,sans-serif;font-size:1.2rem;line-height:2rem;font-weight:500;letter-spacing:0.1rem;display:none;font-weight:700;}@media (max-width:767.95px){.css-ddr7uv{display:inherit;color:#1A1a1a;padding-left:16px;padding-right:16px;font-size:16px;}}2D Objects.css-1eswbhu{padding:16px;border-radius:16px;background-color:white;margin-top:16px;}.css-g8s3lw{display:grid;grid-gap:0;grid-template-columns:96px 1fr;margin-top:0rem;} .css-1ak6lo7{margin:0;font-family:Ginto Normal,sans-serif;font-size:1.2rem;line-height:2rem;font-weight:500;letter-spacing:0.1rem;text-transform:uppercase;border-bottom:1px solid;border-color:#E0E0E0;color:#5C5C5C;padding:0 0 5px 16px;}@media (max-width:767.95px){.css-1ak6lo7{padding:0 0 5px 8px;}}Week 1.css-199df33{display:grid;grid-template-columns:96px 1fr 40px;grid-gap:16px;padding-top:8px;}@media (max-width:767.95px){.css-199df33{grid-template-columns:96px 1fr 28px;grid-gap:8px;}}.css-12h69r9{margin:0;font-family:Ginto Normal,sans-serif;font-size:1.2rem;line-height:2rem;font-weight:500;letter-spacing:0.1rem;text-transform:uppercase;color:#004a77;line-height:24px;font-weight:700;}Lesson 1.css-t78n9j{color:#1A1a1a;}Introduction.css-kf9gkg{color:#5C5C5C;margin-top:12px;height:0;display:none;}This portion will outline material for the course..css-1ngtilx{cursor:pointer;padding:0 4px;}.css-v481rw{-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;width:1em;height:1em;display:inline-block;fill:currentColor;-webkit-flex-shrink:0;-ms-flex-negative:0;flex-shrink:0;-webkit-transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;font-size:inherit;vertical-align:-0.125em;height:24px;}Lesson 2AnglesThis lesson will review the 3 different types of angles. Week 2Lesson 3Triangles/Quadrilaterals: Area and PerimeterThis lesson will review the characteristics of triangles and quadrilaterals as well as area and perimeter. Week 3Lesson 4Circles: Area and CircumferenceThis lesson will review the characteristics of triangles and quadrilaterals as well as area and cicumference.Unit 23D Objects3 lessons3 Weeks3D Objects Week 4Lesson 5Rectangular Prisms: Volume and Surface AreaThis lesson will review the characteristics of rectangular prism as well as volume and surface area. Week 5Lesson 6Cylinders: Volume and Surface AreaThis lesson will review the characteristics of cylinder as well as volume and surface area. Week 6Lesson 7What are Net Figures?This lesson will review the characteristics of a net figure.Unit 3Review1 lesson1 WeekReview Week 6Lesson 8 Course ReviewThis lesson will allow students to review material from the entire course.`
`This class is taught in English.`
```Throughout this course, learners will get practice with:
- terminology and visual representations
- measuring angles
- measuring triangles, quadrilaterals, and circles
- measuring rectangular prisms, cylinders, and working with net objects```
`.css-1il00e6{display:-webkit-box;display:-webkit-flex;display:-ms-flexbox;display:flex;gap:1em;-webkit-flex-direction:column;-ms-flex-direction:column;flex-direction:column;}.css-mt1z2p{-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;width:1em;height:1em;display:inline-block;fill:currentColor;-webkit-flex-shrink:0;-ms-flex-negative:0;flex-shrink:0;-webkit-transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;font-size:inherit;vertical-align:-0.125em;text-align:center;width:1.25em;color:#A3A3A3;margin-right:0.5em;margin-top:0.2em;}.css-4j7l6r{margin:0;font-family:Ginto Normal,sans-serif;font-size:1.6rem;line-height:1.3;font-weight:500;letter-spacing:0.01rem;}Homework OfferedAssessments Offered.css-1do1xce{-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;width:1em;height:1em;display:inline-block;fill:currentColor;-webkit-flex-shrink:0;-ms-flex-negative:0;flex-shrink:0;-webkit-transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;transition:fill 200ms cubic-bezier(0.4, 0, 0.2, 1) 0ms;font-size:inherit;vertical-align:-0.125em;text-align:center;width:1.25em;color:#368139;margin-right:0.5em;margin-top:0.2em;}Grades Offered.css-l2z0vi{margin-top:0.5em;}`
`There are no prerequisites for this course.`
`Learners need paper, a pencil, and a calculator. A protractor for measuring angles is optional.`
`In addition to the Outschool classroom, this class uses:Google SlidesGoogle JamboardQuizletYoutube`
`.css-47miec{margin-top:0rem;}Third-party apps and websites may be used as part of this class. Students may need to log in and/or create accounts to use those applications/websites.`
Star Educator
Average rating:5.0Number of reviews:(75)
Profile
`Hello! My name is Ms. Ann and I have been a teacher for over 10 years in the beautiful state of VA. I have taught several grades in both Elementary and Middle School. I love to teach. I think the best part of teaching is sharing my knowledge in a... .css-gw2jo8{position:relative;display:inline-block;font-family:'Ginto Normal',sans-serif;font-style:normal;font-weight:500;font-size:1.6rem;text-align:center;text-transform:none;height:auto;max-width:100%;white-space:nowrap;cursor:pointer;-webkit-user-select:none;-moz-user-select:none;-ms-user-select:none;user-select:none;outline:none;border:none;background:none;padding:0;-webkit-transition:all ease-in-out 0.05s,outline 0s,;transition:all ease-in-out 0.05s,outline 0s,;line-height:1;color:#380596;}.css-gw2jo8:hover:not(:disabled),.css-gw2jo8:focus:not(:disabled){color:#380596;-webkit-text-decoration:underline;text-decoration:underline;}.css-gw2jo8:active:not(:disabled){color:#380596;}.css-gw2jo8:disabled{color:#C2C2C2;cursor:default;}.css-gw2jo8:focus-visible{outline-width:2px;outline-style:solid;outline-color:#4B01D4;outline-offset:2px;}`
#### \$10
weekly or \$60 for all content
8 pre-recorded lessons
6 weeks of teacher support
Choose your start date
Completed by 2 learners
Ages: 9-13
Enroll Now, Start Anytime
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http://idevgames.com/forums/thread-3971-post-36308.html#pid36308
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## Closest Point on Polygon Boundry
Sage
Posts: 1,403
Joined: 2005.07
Post: #1
Im trying to get some collision response now, but it requires some information I dont have or know how to calculate,
I can detect the collision fine, but I need to get a point of collision and a normal. I've thought about various ways to do this but most of them would be quite likely to produce bad results or very slow.
I have two points in time, just before the collision occurred and just after. Im thinking now of going to the time just before the collision like this:
and then attempting to find two points, one on the boundary of each polygon, such that the two points are the closest possible, I can use the vector from one point to the other as the normal (blue), and the midpoint (red) as the point the collision occurred.
Just now I imagine I would have to go through each point on polygon A and find the minimum distance to each line on polygon B and taking the minimum one. I guess that would get really slow when a polygon with lots of edges collides multiple times per frame.
Does anyone know of a better way of finding the point and normal (given two intersecting or non intersection polygons composed of triangles) or a fast algorithm to find the two points like I was describing?
Sir, e^iπ + 1 = 0, hence God exists; reply!
Moderator
Posts: 529
Joined: 2003.03
Post: #2
unknown Wrote:Just now I imagine I would have to go through each point on polygon A and find the minimum distance to each line on polygon B and taking the minimum one. I guess that would get really slow when a polygon with lots of edges collides multiple times per frame.
Would you really have to compare to every line? You already know which ones collided.
"Yes, well, that's the sort of blinkered, Philistine pig-ignorance I've come to expect from you non-creative garbage."
Sage
Posts: 1,487
Joined: 2002.09
Post: #3
I don't remember exactly what they are called, but when creating your poly regions, you can create a bitmap with the normal and distance information in it. It doesn't even need to be full resolution as you can interpolate to get a decent approximation.
When looking for the collision point, you simply poll the vertices in one poly against the map of the other. I've never used this myself as I only handle convex polys, but it sounds interesting.
What I do to approximate collision points is simply to use the current positions of the predicted penetrating vertices in the next time step. If you want a good non-elastic collision, you may need more than one. That works as a good approximation as long as things aren't moving to fast, otherwise you can get some screwy rotation.
Scott Lembcke - Howling Moon Software
Author of Chipmunk Physics - A fast and simple rigid body physics library in C.
Member
Posts: 26
Joined: 2006.09
Post: #4
The easiest way to detect the collision between a circle and edge is to calculate the distance between the edge and the point. That will handle all cases nicely, even the corners. The direction from the center of the collider to the nearest point on edge is the normal of the collision.
Since you can operate on the level of edges, it does not matter if you have holes in the polygons or not. Just make sure your polygons are winded correctly, that allows you to do "backface culling". If the dot product of the direction from your collider to the edge with the normal of the edge is negative, you can skip the edge. The normal of the edge is perproduct of the direction of the edge (that is, either (y, -x) or (-y, x) depending on the winding order). Another way to speed it up is to fist check if the point is inside the AABB of the shape (of course you have to pad the bbox with the radius of the collider, or check AABB vs. AABB),
Here's code to calculte the nearest point. I'm not 100% sure it works, I was unable to test it.
Vec2 NearestPointOnSegment(const Vec2& pt, const Vec2& p0, const Vec2& p1)
{
Vec2 dir = p1 - p0;
float len = Dot(dir, dir);
if(len < EPSILON) return p0; // Zero len segment.
Vec2 diff = pt - p0;
float u = Dot(dir, diff);
u /= len;
if (u > 0.0f)
{
if (u < 1.0f)
return p0 + u * dir;
else
return p1;
}
else
return p0;
}
If you want to test a (fast) moving object, you can use line-line distance checks. It gets a bit trickier, but is doable
This is a good source for and tutorial for 2D collision detection stuff:
http://www.harveycartel.org/metanet/tuto...rialA.html
Possibly Related Threads...
Thread: Author Replies: Views: Last Post Finding the closest point to (x1,y1) in array of [x1,y1,z1,x2... mikey 17 13,147 Aug 28, 2009 08:12 AM Last Post: mikey 2D Polygon intersection testing unknown 6 9,385 Aug 15, 2006 04:17 PM Last Post: unknown
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| 3.15625
| 3
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CC-MAIN-2016-50
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latest
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en
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https://questioncove.com/updates/53c41d8ee4b05c273e9cfe04
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OpenStudy (anonymous):
what do these numbers mean: |23| -|23| |-23|
3 years ago
OpenStudy (abmon98):
The modulus of a number is the magnitude of that number. For example, the modulus of -1 ( |-1| ) is 1. The modulus of x, |x|, is x for values of x which are positive and -x for values of x which are negative.
3 years ago
OpenStudy (cwrw238):
the 2 vertical lines around the number means the 'absolute value of' that means you always take the positive value so the third one = 23 the first one = 23 and 2nd = -23 because the negative sign is outside the vertical lines
3 years ago
OpenStudy (cwrw238):
absolute value or magnitude (as Abmon said)
3 years ago
OpenStudy (cwrw238):
or modulus
3 years ago
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CC-MAIN-2018-13
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https://solvedlib.com/n/x-and-y-are-real-numbers-what-is-the-result-of-x-y-3-5-4,21039922
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# X and y are real numbers, what is the result of x*y?3.5* +4.2y = 107 4.5* + 5.2Y = 140
###### Question:
x and y are real numbers, what is the result of x*y? 3.5* +4.2y = 107 4.5* + 5.2Y = 140
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# Divide polygone by calculated area ratio
Discussion created by annetudd on Feb 24, 2014
Latest reply on Feb 25, 2014 by annetudd
Hello there,
I am pretty new to ArcMap10.1 but so far I haven´t found a proper solution to my problem in the Forum.
The main topic is tree crown delineation. I am using airborne data to create a shape file for tree crowns. It works really well for single trees but if the corwns are touching I only get a big "chunky" polygon. What I did so far after I created the polygon: By calculating area, perimter and radius of each polygone I then calculated an index (Rindex=perimeter*radius/area). For a nomral circle this ratio is 2. So if I have a big polygone with Rindex = 6, I now know, that there are approx. 3 trees inside (--> Nr_Trees = Rindex/2).
Now my actual question: How can I divide my polygons by the calculated Nr_Trees (see attached screen shot)?
If you need more data to help me solve my problem please let me know! Thanks in advance for any suggestions.
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## Hungarian Method
The Hungarian method is a computational optimization technique that addresses the assignment problem in polynomial time and foreshadows following primal-dual alternatives. In 1955, Harold Kuhn used the term “Hungarian method” to honour two Hungarian mathematicians, Dénes Kőnig and Jenő Egerváry. Let’s go through the steps of the Hungarian method with the help of a solved example.
## Hungarian Method to Solve Assignment Problems
The Hungarian method is a simple way to solve assignment problems. Let us first discuss the assignment problems before moving on to learning the Hungarian method.
## What is an Assignment Problem?
A transportation problem is a type of assignment problem. The goal is to allocate an equal amount of resources to the same number of activities. As a result, the overall cost of allocation is minimised or the total profit is maximised.
Because available resources such as workers, machines, and other resources have varying degrees of efficiency for executing different activities, and hence the cost, profit, or loss of conducting such activities varies.
Assume we have ‘n’ jobs to do on ‘m’ machines (i.e., one job to one machine). Our goal is to assign jobs to machines for the least amount of money possible (or maximum profit). Based on the notion that each machine can accomplish each task, but at variable levels of efficiency.
## Hungarian Method Steps
Check to see if the number of rows and columns are equal; if they are, the assignment problem is considered to be balanced. Then go to step 1. If it is not balanced, it should be balanced before the algorithm is applied.
Step 1 – In the given cost matrix, subtract the least cost element of each row from all the entries in that row. Make sure that each row has at least one zero.
Step 2 – In the resultant cost matrix produced in step 1, subtract the least cost element in each column from all the components in that column, ensuring that each column contains at least one zero.
Step 3 – Assign zeros
• Analyse the rows one by one until you find a row with precisely one unmarked zero. Encircle this lonely unmarked zero and assign it a task. All other zeros in the column of this circular zero should be crossed out because they will not be used in any future assignments. Continue in this manner until you’ve gone through all of the rows.
• Examine the columns one by one until you find one with precisely one unmarked zero. Encircle this single unmarked zero and cross any other zero in its row to make an assignment to it. Continue until you’ve gone through all of the columns.
Step 4 – Perform the Optimal Test
• The present assignment is optimal if each row and column has exactly one encircled zero.
• The present assignment is not optimal if at least one row or column is missing an assignment (i.e., if at least one row or column is missing one encircled zero). Continue to step 5. Subtract the least cost element from all the entries in each column of the final cost matrix created in step 1 and ensure that each column has at least one zero.
Step 5 – Draw the least number of straight lines to cover all of the zeros as follows:
(a) Highlight the rows that aren’t assigned.
(b) Label the columns with zeros in marked rows (if they haven’t already been marked).
(c) Highlight the rows that have assignments in indicated columns (if they haven’t previously been marked).
(d) Continue with (b) and (c) until no further marking is needed.
(f) Simply draw the lines through all rows and columns that are not marked. If the number of these lines equals the order of the matrix, then the solution is optimal; otherwise, it is not.
Step 6 – Find the lowest cost factor that is not covered by the straight lines. Subtract this least-cost component from all the uncovered elements and add it to all the elements that are at the intersection of these straight lines, but leave the rest of the elements alone.
Step 7 – Continue with steps 1 – 6 until you’ve found the highest suitable assignment.
## Hungarian Method Example
Use the Hungarian method to solve the given assignment problem stated in the table. The entries in the matrix represent each man’s processing time in hours.
$$\begin{array}{l}\begin{bmatrix} & I & II & III & IV & V \\1 & 20 & 15 & 18 & 20 & 25 \\2 & 18 & 20 & 12 & 14 & 15 \\3 & 21 & 23 & 25 & 27 & 25 \\4 & 17 & 18 & 21 & 23 & 20 \\5 & 18 & 18 & 16 & 19 & 20 \\\end{bmatrix}\end{array}$$
With 5 jobs and 5 men, the stated problem is balanced.
$$\begin{array}{l}A = \begin{bmatrix}20 & 15 & 18 & 20 & 25 \\18 & 20 & 12 & 14 & 15 \\21 & 23 & 25 & 27 & 25 \\17 & 18 & 21 & 23 & 20 \\18 & 18 & 16 & 19 & 20 \\\end{bmatrix}\end{array}$$
Subtract the lowest cost element in each row from all of the elements in the given cost matrix’s row. Make sure that each row has at least one zero.
$$\begin{array}{l}A = \begin{bmatrix}5 & 0 & 3 & 5 & 10 \\6 & 8 & 0 & 2 & 3 \\0 & 2 & 4 & 6 & 4 \\0 & 1 & 4 & 6 & 3 \\2 & 2 & 0 & 3 & 4 \\\end{bmatrix}\end{array}$$
Subtract the least cost element in each Column from all of the components in the given cost matrix’s Column. Check to see if each column has at least one zero.
$$\begin{array}{l}A = \begin{bmatrix}5 & 0 & 3 & 3 & 7 \\6 & 8 & 0 & 0 & 0 \\0 & 2 & 4 & 4 & 1 \\0 & 1 & 4 & 4 & 0 \\2 & 2 & 0 & 1 & 1 \\\end{bmatrix}\end{array}$$
When the zeros are assigned, we get the following:
The present assignment is optimal because each row and column contain precisely one encircled zero.
Where 1 to II, 2 to IV, 3 to I, 4 to V, and 5 to III are the best assignments.
Hence, z = 15 + 14 + 21 + 20 + 16 = 86 hours is the optimal time.
## Practice Question on Hungarian Method
Use the Hungarian method to solve the following assignment problem shown in table. The matrix entries represent the time it takes for each job to be processed by each machine in hours.
$$\begin{array}{l}\begin{bmatrix}J/M & I & II & III & IV & V \\1 & 9 & 22 & 58 & 11 & 19 \\2 & 43 & 78 & 72 & 50 & 63 \\3 & 41 & 28 & 91 & 37 & 45 \\4 & 74 & 42 & 27 & 49 & 39 \\5 & 36 & 11 & 57 & 22 & 25 \\\end{bmatrix}\end{array}$$
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## Frequently Asked Questions on Hungarian Method
What is hungarian method.
The Hungarian method is defined as a combinatorial optimization technique that solves the assignment problems in polynomial time and foreshadowed subsequent primal–dual approaches.
## What are the steps involved in Hungarian method?
The following is a quick overview of the Hungarian method: Step 1: Subtract the row minima. Step 2: Subtract the column minimums. Step 3: Use a limited number of lines to cover all zeros. Step 4: Add some more zeros to the equation.
## What is the purpose of the Hungarian method?
When workers are assigned to certain activities based on cost, the Hungarian method is beneficial for identifying minimum costs.
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## Hungarian Method Examples
Now we will examine a few highly simplified illustrations of Hungarian Method for solving an assignment problem .
Later in the chapter, you will find more practical versions of assignment models like Crew assignment problem , Travelling salesman problem , etc.
Example-1, Example-2
## Example 1: Hungarian Method
The Funny Toys Company has four men available for work on four separate jobs. Only one man can work on any one job. The cost of assigning each man to each job is given in the following table. The objective is to assign men to jobs in such a way that the total cost of assignment is minimum.
This is a minimization example of assignment problem . We will use the Hungarian Algorithm to solve this problem.
Identify the minimum element in each row and subtract it from every element of that row. The result is shown in the following table.
"A man has one hundred dollars and you leave him with two dollars, that's subtraction." -Mae West
On small screens, scroll horizontally to view full calculation
Identify the minimum element in each column and subtract it from every element of that column.
Make the assignments for the reduced matrix obtained from steps 1 and 2 in the following way:
• For every zero that becomes assigned, cross out (X) all other zeros in the same row and the same column.
• If for a row and a column, there are two or more zeros and one cannot be chosen by inspection, choose the cell arbitrarily for assignment.
An optimal assignment is found, if the number of assigned cells equals the number of rows (and columns). In case you have chosen a zero cell arbitrarily, there may be alternate optimal solutions. If no optimal solution is found, go to step 5.
Use Horizontal Scrollbar to View Full Table Calculation
Draw the minimum number of vertical and horizontal lines necessary to cover all the zeros in the reduced matrix obtained from step 3 by adopting the following procedure:
• Mark all the rows that do not have assignments.
• Mark all the columns (not already marked) which have zeros in the marked rows.
• Mark all the rows (not already marked) that have assignments in marked columns.
• Repeat steps 5 (ii) and (iii) until no more rows or columns can be marked.
• Draw straight lines through all unmarked rows and marked columns.
You can also draw the minimum number of lines by inspection.
Select the smallest element (i.e., 1) from all the uncovered elements. Subtract this smallest element from all the uncovered elements and add it to the elements, which lie at the intersection of two lines. Thus, we obtain another reduced matrix for fresh assignment.
Now again make the assignments for the reduced matrix.
## Final Table: Hungarian Method
Since the number of assignments is equal to the number of rows (& columns), this is the optimal solution.
The total cost of assignment = A1 + B4 + C2 + D3
Substituting values from original table: 20 + 17 + 17 + 24 = Rs. 78.
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## Hungarian Algorithm for Assignment Problem | Set 2 (Implementation)
Given a 2D array , arr of size N*N where arr[i][j] denotes the cost to complete the j th job by the i th worker. Any worker can be assigned to perform any job. The task is to assign the jobs such that exactly one worker can perform exactly one job in such a way that the total cost of the assignment is minimized.
Input: arr[][] = {{3, 5}, {10, 1}} Output: 4 Explanation: The optimal assignment is to assign job 1 to the 1st worker, job 2 to the 2nd worker. Hence, the optimal cost is 3 + 1 = 4. Input: arr[][] = {{2500, 4000, 3500}, {4000, 6000, 3500}, {2000, 4000, 2500}} Output: 4 Explanation: The optimal assignment is to assign job 2 to the 1st worker, job 3 to the 2nd worker and job 1 to the 3rd worker. Hence, the optimal cost is 4000 + 3500 + 2000 = 9500.
Different approaches to solve this problem are discussed in this article .
Approach: The idea is to use the Hungarian Algorithm to solve this problem. The algorithm is as follows:
• For each row of the matrix, find the smallest element and subtract it from every element in its row.
• Repeat the step 1 for all columns.
• Cover all zeros in the matrix using the minimum number of horizontal and vertical lines.
• Test for Optimality : If the minimum number of covering lines is N , an optimal assignment is possible. Else if lines are lesser than N , an optimal assignment is not found and must proceed to step 5.
• Determine the smallest entry not covered by any line. Subtract this entry from each uncovered row, and then add it to each covered column. Return to step 3.
Consider an example to understand the approach:
Let the 2D array be: 2500 4000 3500 4000 6000 3500 2000 4000 2500 Step 1: Subtract minimum of every row. 2500, 3500 and 2000 are subtracted from rows 1, 2 and 3 respectively. 0 1500 1000 500 2500 0 0 2000 500 Step 2: Subtract minimum of every column. 0, 1500 and 0 are subtracted from columns 1, 2 and 3 respectively. 0 0 1000 500 1000 0 0 500 500 Step 3: Cover all zeroes with minimum number of horizontal and vertical lines. Step 4: Since we need 3 lines to cover all zeroes, the optimal assignment is found. 2500 4000 3500 4000 6000 3500 2000 4000 2500 So the optimal cost is 4000 + 3500 + 2000 = 9500
For implementing the above algorithm, the idea is to use the max_cost_assignment() function defined in the dlib library . This function is an implementation of the Hungarian algorithm (also known as the Kuhn-Munkres algorithm) which runs in O(N 3 ) time. It solves the optimal assignment problem.
Below is the implementation of the above approach:
Time Complexity: O(N 3 ) Auxiliary Space: O(N 2 )
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## Procedure, Example Solved Problem | Operations Research - Solution of assignment problems (Hungarian Method) | 12th Business Maths and Statistics : Chapter 10 : Operations Research
Chapter: 12th business maths and statistics : chapter 10 : operations research.
Solution of assignment problems (Hungarian Method)
First check whether the number of rows is equal to the numbers of columns, if it is so, the assignment problem is said to be balanced.
Step :1 Choose the least element in each row and subtract it from all the elements of that row.
Step :2 Choose the least element in each column and subtract it from all the elements of that column. Step 2 has to be performed from the table obtained in step 1.
Step:3 Check whether there is atleast one zero in each row and each column and make an assignment as follows.
Step :4 If each row and each column contains exactly one assignment, then the solution is optimal.
Example 10.7
Solve the following assignment problem. Cell values represent cost of assigning job A, B, C and D to the machines I, II, III and IV.
Here the number of rows and columns are equal.
∴ The given assignment problem is balanced. Now let us find the solution.
Step 1: Select a smallest element in each row and subtract this from all the elements in its row.
Look for atleast one zero in each row and each column.Otherwise go to step 2.
Step 2: Select the smallest element in each column and subtract this from all the elements in its column.
Since each row and column contains atleast one zero, assignments can be made.
Step 3 (Assignment):
Thus all the four assignments have been made. The optimal assignment schedule and total cost is
The optimal assignment (minimum) cost
Example 10.8
Consider the problem of assigning five jobs to five persons. The assignment costs are given as follows. Determine the optimum assignment schedule.
∴ The given assignment problem is balanced.
Now let us find the solution.
The cost matrix of the given assignment problem is
Column 3 contains no zero. Go to Step 2.
Thus all the five assignments have been made. The Optimal assignment schedule and total cost is
The optimal assignment (minimum) cost = ` 9
Example 10.9
Solve the following assignment problem.
Since the number of columns is less than the number of rows, given assignment problem is unbalanced one. To balance it , introduce a dummy column with all the entries zero. The revised assignment problem is
Here only 3 tasks can be assigned to 3 men.
Step 1: is not necessary, since each row contains zero entry. Go to Step 2.
Step 3 (Assignment) :
Since each row and each columncontains exactly one assignment,all the three men have been assigned a task. But task S is not assigned to any Man. The optimal assignment schedule and total cost is
The optimal assignment (minimum) cost = ₹ 35
Related Topics
Index Assignment problem Hungarian algorithm Solve online
## The Hungarian algorithm: An example
We consider an example where four jobs (J1, J2, J3, and J4) need to be executed by four workers (W1, W2, W3, and W4), one job per worker. The matrix below shows the cost of assigning a certain worker to a certain job. The objective is to minimize the total cost of the assignment.
Below we will explain the Hungarian algorithm using this example. Note that a general description of the algorithm can be found here .
Step 1: Subtract row minima
We start with subtracting the row minimum from each row. The smallest element in the first row is, for example, 69. Therefore, we substract 69 from each element in the first row. The resulting matrix is:
Step 2: Subtract column minima
Similarly, we subtract the column minimum from each column, giving the following matrix:
Step 3: Cover all zeros with a minimum number of lines
We will now determine the minimum number of lines (horizontal or vertical) that are required to cover all zeros in the matrix. All zeros can be covered using 3 lines:
First, we find that the smallest uncovered number is 6. We subtract this number from all uncovered elements and add it to all elements that are covered twice. This results in the following matrix:
Again, We determine the minimum number of lines required to cover all zeros in the matrix. Now there are 4 lines required:
Because the number of lines required (4) equals the size of the matrix ( n =4), an optimal assignment exists among the zeros in the matrix. Therefore, the algorithm stops.
The optimal assignment
The following zeros cover an optimal assignment:
This corresponds to the following optimal assignment in the original cost matrix:
Thus, worker 1 should perform job 3, worker 2 job 2, worker 3 job 1, and worker 4 should perform job 4. The total cost of this optimal assignment is to 69 + 37 + 11 + 23 = 140.
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## The Hungarian method for solving an assignment problem can also be used to solve.
A. A transportation problem
B. A travelling salesman problem
C. A LP problem
D. Both a & b
## Solution(By Examveda Team)
This Question Belongs to Management >> Operations Research
## Join The Discussion
Related Questions on Operations Research
The use of decision models.
A. Is possible when the variables value is known
B. Reduces the scope of judgement & intuition known with certainty in decision-making
C. Require the use of computer software
D. None of the above
Every mathematical model.
A. Must be deterministic
B. Requires computer aid for its solution
C. Represents data in numerical form
D. All of the above
A physical model is example of.
A. An iconic model
B. An analogue model
C. A verbal model
D. A mathematical model
The qualitative approach to decision analysis relies on.
A. Experience
B. Judgement
C. Intuition
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#### IMAGES
1. How to Solve an Assignment Problem Using the Hungarian Method
2. Assignment Problem (Part-2) Introduction to Hungarian Method to Solve
3. How to Solve Assignment Problem Using Hungarian Method (Maximization
4. Assignment problem Hungarian method
5. Unbalanced Assignment Problem
6. Assignment Problem
#### VIDEO
1. Assignment Hungarian method operations management CMA INTER
2. 2. Minimal Assignment problem {Hungarian Method}
3. Assignment problem, simple probl
4. Assignment in Ms Excel
5. Assignment Problems-Hungarian Method Operation Research-04
6. Assignment problem = Hungarian method = Reduced matrix method
1. Hungarian Method
The Hungarian method is a simple way to solve assignment problems. Let us first discuss the assignment problems before moving on to learning the Hungarian method. What is an Assignment Problem? A transportation problem is a type of assignment problem. The goal is to allocate an equal amount of resources to the same number of activities.
2. Hungarian method calculator
Hungarian method calculator . Operation Research - Assignment problem calculator - Find solution of Assignment Problem Hungarian method, step-by-step online
3. Hungarian algorithm
The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in polynomial time and which anticipated later primal-dual methods.
4. Hungarian Method Examples, Assignment Problem
Solution. This is a minimization example of assignment problem. We will use the Hungarian Algorithm to solve this problem. Step 1 Identify the minimum element in each row and subtract it from every element of that row. The result is shown in the following table. "A man has one hundred dollars and you leave him with two dollars, that's subtraction."
5. How to Solve an Assignment Problem Using the Hungarian Method
How to Solve an Assignment Problem Using the Hungarian Method Shokoufeh Mirzaei 16.5K subscribers 217K views 5 years ago Linear Programming In this lesson we learn what is an assignment...
6. Hungarian Algorithm for Assignment Problem
The Hungarian algorithm, aka Munkres assignment algorithm, utilizes the following theorem for polynomial runtime complexity ( worst case O (n3)) and guaranteed optimality: If a number is added to or subtracted from all of the entries of any one row or column of a cost matrix, then an optimal assignment for the resulting cost matrix is also an op...
7. Solve the assignment problem online
Fill in the cost matrix of an assignment problem and click on 'Solve'. The optimal assignment will be determined and a step by step explanation of the hungarian algorithm will be given. Fill in the cost matrix ( random cost matrix ): Don't show the steps of the Hungarian algorithm Maximize the total cost
8. Using the Hungarian Algorithm to Solve Assignment Problems
The Hungarian algorithm is useful to identify minimum costs when people are assigned to specific activities based on cost. Practice using this algorithm in example equations of real-world...
9. Hungarian Algorithm for Assignment Problem
Different approaches to solve this problem are discussed in this article.. Approach: The idea is to use the Hungarian Algorithm to solve this problem. The algorithm is as follows: For each row of the matrix, find the smallest element and subtract it from every element in its row. Repeat the step 1 for all columns.
10. linear programming
Assignment problem using Hungarian method Asked 4 years, 4 months ago Modified 4 years, 4 months ago Viewed 525 times 8 There are five jobs to be assigned to five machines and associated cost matrix is as follows
11. Solution of assignment problems (Hungarian Method)
Solution: Here the number of rows and columns are equal. ∴ The given assignment problem is balanced. Now let us find the solution. Step 1: Select a smallest element in each row and subtract this from all the elements in its row. Look for atleast one zero in each row and each column.Otherwise go to step 2.
12. Second best solution to an assignmentproblem using the Hungarian Algorithm
4. For finding the best solution in the assignment problem it's easy to use the Hungarian Algorithm. For example: A | 3 4 2 B | 8 9 1 C | 7 9 5. When using the Hungarian Algorithm on this you become: A | 0 0 1 B | 5 5 0 C | 0 1 0. Which means A gets assigned to 'job' 2, B to job 3 and C to job 1. However, I want to find the second best solution ...
13. An Assignment Problem solved using the Hungarian Algorithm
An Assignment Problem solved using the Hungarian Algorithm - HungarianAlgorithm.com The Hungarian algorithm: An example We consider an example where four jobs (J1, J2, J3, and J4) need to be executed by four workers (W1, W2, W3, and W4), one job per worker. The matrix below shows the cost of assigning a certain worker to a certain job.
14. PDF The Hungarian Method for The Assignment Problem, With Generalized
In this section an algorithm to solve assignment problem with generalized interval arithmetic using Hungarian method: Step 1: Find out the mid values of each interval in the cost matrix.
15. The Hungarian Method for the Assignment Problem, With Generalized
Although the assignment problem can be solved as an ordinary transportation problem or as Linear programming problem, its unique structure can be exploited, resulting in special purpose...
16. Difference between solving Assignment Problem using the Hungarian
1 Answer Sorted by: 7 The main differences probably are that there is a somewhat large overhead you have to pay when solving the AP as a linear program: You have to build an LP model and ship it to a solver. In addition, an LP solver is a generalist.
17. Using Interval Operations in the Hungarian Method to Solve ...
Assignment problem (AP) is an entrenched tool for solving engineering and management problems. The Hungarian method is always used to fathom the AP in crisp cases. This paper presents an algorithm of finding the optimum solution of the fuzzy AP by using the modified Hungarian method. This method is utilized to get a minimum assignment cost in the fuzzy environment for a fuzzy AP. Firstly, we ...
18. [#1]Assignment Problem[Easy Steps to solve
Here is the video about assignment problem - Hungarian method with algorithm.NOTE: After row and column scanning, If you stuck with more than one zero in th...
19. PDF Using of Hungarian Method to Solve the Transportation Problems
dummy row contains the amount equal 50, and then use the VAM method as usual. First: the solution using the VAM method. The total cost of transportation is equal to 65500. Second: the solution using the Hungarian method. Now, the Hungarian method is applied to solve this problem and find the total cost matrix after adding dummy row in
20. Solved Qs: solve the following assignment problem using
Question: Qs: solve the following assignment problem using Hungarian method. The matrix entries represent the processing time in hours. A B C D E 5918 19 12 18 11 9 6 ...
21. PDF Assessment of Assignment Problem using Hungarian Method
Many practitioners and researchers used the Hungarian method in the past to solve assignment problems (Kuhn 1955) ; (Chopra et al. 2017). The existing Hungarian method for solving unbalanced assignment problems is based on the assumption that some jobs should be assigned to dummy or pseudo machines, but those jobs are left unexecuted by
22. Hungarian Method for Solving Assignment Problem
We examine a numerical example by using Hungarian method .The Hungarian method is a systematic procedure, easy to apply for solving assignment problem. Discover the world's research 25+ million ...
23. The Hungarian method for solving an assignment problem can ...
The Hungarian method for solving an assignment problem can also be used to solve a travelling salesman problem. The Hungarian method is a combinatorial optimization algorithm that solves the assignment problem in polynomial time and which anticipated later primal-dual methods. This Question Belongs to Management >> Operations Research.
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# Finance
You purchase XYZ company's perpetual preferred stock, which pays a perpetual annual dividend of \$8 per share. If the appropriate discount rate for this investment is 14%, what is the price of one share of stock?
Thank You!
8/.14 = 57.14
1. 👍
2. 👎
3. 👁
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NESA is regularly updating its advice as the coronavirus outbreak unfolds. Get our latest COVID-19 advice
This webpage has been archived to prepare for transfer to the new NESA website. Reference to syllabus outcomes and content on this webpage may not be current. Teachers are encouraged to visit the Key Learning Area page for recent student work samples on the NESA website.
Assessment Resource Centre (ARC)
Home
1. Stage 2
2. Mathematics
3. Activities
4. Is it fair?
## Is it fair?
End of Stage 2 (end of Year 4)
New Work Samples
Kerry
## Foundation Statement strands
The following strands are covered in this activity:
## Description of activity
Students work in teams to predict possible outcomes in an experimental context.
1. The class is organised into four teams. Each team is allocated a colour name: red, blue, green or yellow.
2. The teacher has a bag containing 1 yellow, 10 red, 5 blue, and 4 green counters. Students are told that there are twenty counters and that each colour is represented in the bag. The number of each coloured counter is not revealed to the students.
3. The teacher draws a counter from the bag and a point is given to the team with the corresponding colour.
4. The counter is then returned to the bag and the process is repeated for twenty draws.
• their prediction of the composition of coloured counters in the bag
• the reasons for their prediction
• whether the game is fair.
The suggested time allocation for this activity is 60 minutes.
## Suggested materials
10 red counters, 5 blue counters, 4 green counters, 1 yellow counter, a cloth bag, paper, pens.
## Prior learning
Students have encountered a variety of chance concepts through games – including randomness, prediction and likelihood. They are familiar with the language of chance.
Board of Studies NSW, Mathematics K–6 Syllabus, p 69
Board of Studies NSW, Mathematics K–6 Sample Units of Work, p 104
## Outcomes
Chance (NS2.5)
Describes and compares chance events in social and experimental contexts
Applying Strategies (WMS 2.2)
Selects and uses appropriate mental or written strategies, or technology, to solve problems
Communicating (WMS2.3)
Uses appropriate terminology to describe, and symbols to represent, mathematical ideas
Reasoning (WMS2.4)
Checks the accuracy of a statement and explains the reasoning used
Reflecting (WMS2.5)
Links mathematical ideas and makes connections with, and generalisations about, existing knowledge and understanding in relation to Stage 2 content
## Criteria for assessing learning
Students will be assessed on their ability to:
• describe chance events in an experimental context
• use appropriate mental or written strategies to solve problems
• communicate mathematical ideas and reasoning using appropriate terminology
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Primoz
• Member for 11 years, 8 months
• Last seen this week
• Slovenia
## 13 Answers
13 votes
Lie algebras over rings (Lie rings) are important in group theory. For instance, to every group $G$ one can associate a Lie ring $$L(G)=\bigoplus _{i=1}^\infty \gamma _i(G)/\gamma _{i+1}(G),$$ where ...
View answer
Accepted answer
11 votes
Tarski monsters provide examples of 2-generator noetherian groups that is not finitely presented. Edit (YCor): Tarski monsters, as defined in the link (infinite groups of prime exponent $p$ in which ...
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8 votes
There is another description of $H_2(G)$ due to Miller: Miller, Clair The second homology group of a group; relations among commutators. Proc. Amer. Math. Soc. 3, (1952). 588--595. In modern ...
View answer
5 votes
There is a whole theory of functional equations (or functional identities) in algebras. It was used for instance to obtain solutions to some of Herstein's problems on Lie homomorphisms. An overview of ...
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5 votes
Every group of exponent 3 is nilpotent of class at most 3, and this bound is best possible. The question whether finitely generated groups of exponent $n$ are finite is also known as the Burnside ...
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Accepted answer
3 votes
The answer is yes, $S$ is an inverse limit of its lower central quotients. As these have bounded derived length, the same goes for the Cartesian product of these groups. By the way, all pro-$p$ groups ...
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Accepted answer
3 votes
This should follow along the lines of Robinson's book "Finiteness conditions and generalized soluble groups", Part 2, Section 9.2. I will try to sketch an argument using Robinson's terminology. Since $... View answer 3 votes The negative solution of Burnside's problem by Novikov and Adjan is an outstanding example of use of complicated induction. Quoting from Wikipedia's entry on Adjan: "The solution of the Burnside ... View answer 3 votes The 1967 book Varieties of Groups by Hanna Neumann has the following footnote (p. 21): The term metabelian will always mean solvable of length two in agreement with current English usage; note however ... View answer 2 votes Classification of$p$-groups by nilpotency class is hard in general. As pointed above, coclass seems to be a better invariant. On the other hand, Ahmad, Magidin and Morse recently finished a ... View answer 2 votes Given an arbitrary countable group$H$containing an element of large enough order but no involutions, there exists a two-generator simple group$G$such that$H$is a proper subgroup of$G$and$G=\...
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1 votes
L.-C. Kappe and M. Meriano dealt with a similar problem, see Meriano's Groups St. Andrews 2013 slides They defined certain subgroups ${}^*w_2(g)$ for a two-variable word $w$ which seem to be what ...
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Accepted answer
1 votes
Try the book "Functional identities" by Bresar, Chebotar, and Martindale. It deals with functional equations in the realm of associative algebras, Lie algebras and Jordan algebras. As I mentioned in ...
View answer
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# Several torques are applied to a shaft as shown. Prepare a torque diagram and find the torque carried by each segment of the shaft. Assume the torque carried in segment EG is positive and indicate the direction ofother torques with positive values if they are the same as segment EG; negative if they are opposite.ТвTcTpTG10.9 kN - m2.9 kN - m24.4 kN · m5 kN - m10.4 kN · m2014 Michael SwanbomTEGGGG:To
Question
248 views
P1
check_circle
Step 1
Draw the schematic diagram of the shaft and show all action and the reaction torques that are acting on it at different locations. In the diagram, TA is the reaction torque that is acting on the shaft at point A.
Step 2
Consider, the above diagram and apply the torque equilibrium condition by considering the counterclockwise direction as positive and opposite to it as negative.
Step 3
Substitute the relevant values in the equation (1) to determine ...
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Cody
# Problem 624. Get the length of a given vector
Solution 201374
Submitted on 6 Feb 2013 by Jeremy
This solution is locked. To view this solution, you need to provide a solution of the same size or smaller.
### Test Suite
Test Status Code Input and Output
1 Pass
%% x = [1 2 3]; y_correct = 3; assert(isequal(VectorLength(x),y_correct))
ans = 3
2 Pass
%% x = 1:10; y_correct = 10; assert(isequal(VectorLength(x),y_correct))
ans = 10
3 Pass
%% x = rand(1,928); y_correct = 928; assert(isequal(VectorLength(x),y_correct))
ans = 928
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# Balance Mechanics 3.1: More on income
We had previously defined income and consumption as the transactions that change a unit’s net worth, i.e.:
$y-c=\Delta nw \$
Now I want to explain in more detail what income is because income can be different things that have to be kept apart. Also what income is is often not sufficiently well explained in normal textbooks.
First we should define income. Income is defined as a unit’s consumption plus its change in net worth:
$y=c+\Delta nw \$
Further, changes in net worth can be separated into changes in tangible assets ($\Delta ta \$) and changes in net financial assets ($\Delta nfa=\Delta ofa -\Delta l \$ ) so that:
$y=c+\Delta ta +\Delta nfa \$
The change in tangible assets is also called investment ($\Delta ta \equiv=i \$ ); and you can change your net financial assets (as we had defined in the previous post) by having an expenditure surplus or deficit ($r-e=\Delta nfa \$). Putting all that together we can write income thus:
$y=c+i + r-e \$
Using this definition we can distinguish between three kinds of income: production as income, revenues as income and capital gains as income.
Production as income
The things that you produce and then either consume or invest are also your income. For instance, if you have a garden where you grow potatoes, the potato is your production and as such your income. When you eat the potato, you consume your product and your income is equal to your consumption:
$y=c \$
When you do not eat the potato but just keep it, you increase your stock of tangible assets — because you can plant the potato and have more potatoes later on:
$y=i\$
You can also produce services: if you cook your dinner yourself (having found and not planted the potato) this service is also your income and your consumption. While this kind of service is not counted in the national accounts (whereas going out for dinner is), it still is income and consumption. There is no reason one should not count it in the national accounts because other strange but comparable things are.
Take the implicit rent of people who own their house and do not rent it. The housing service which in value is equivalent to the rent you would pay the landlady to live in the house is counted in the national accounts. Otherwise economies with a high rate of home owners would have a much lower GDP than economies where most people rent.
As far as investment as income is concerned, in the real world this is mostly important for companies. Companies which build their own machines and other tangible assets increase their income (which is their profit) one-by-one with their investment — if no expenditures are associated with the investment.
Income as revenue
For most of us, the main source of our income are our revenues, mostly from wages, but for some also from interest and dividends (capital income). Companies receive revenues from their sales (but this is not yet companies‘ income).
There are three things that you can do with those revenues: you can use them for consumption, investment or to increase your net financial assets if you do not spend as much as you receive in revenues.
If you receive some revenues (for instance wages) and consume part of its, the consumption occurs on two places in the income equation:
$y=c+r_{wage}-e_{consumption} \$
Consumption occurs both as $c$ as a positive entry and as an expenditure $e_{consumption}$ as a negative entry. This is different from the consumption goods that you produce and consume yourself and which directly increase your income. Here, consumption does not change your income because you buy your consumption good somewhere. Your income is thus equal to your wage. If we did not add the term $c$, income would be equal to the expenditure surplus only and not the full amount of your wage income. To the extent that you buy all your consumpion goods, $c - e_{consumption}$ is always zero. When you also produce some of your consumption goods, consumption will be higher than consumption expenditures. This is why it is important to keep consumption and consumption expenditures apart.
The same principle applies if you buy tangible assets with your income:
$y=i+r_{wage}-e_{investment} \$
Lastly, whatever you buy, you change your net financial assets by the amount of your revenue supluses / deficits so that:
$y=c+i+r-e=c+i+\Delta nfa \$
Income as capital gains
The third kind of income is if your net worth changes due to capital gains. Capital gains change the value of your assets. We can write:
$y=\Delta nw=\Delta gfa_{capgains}+ \Delta ta_{capgains}-\Delta l \$
Since we had defined income as consumption plus the change in capital gains, capital gains are also income. Normally liabilities do not change in value so that capital gains only apply to changes in the value of assets.
# Balance Mechanics 3: Flows that change the balance sheet
We had already defined the different items on a balance sheet that we are interested in here. Now the question is how one can change any of those items. In German standard accounting there are three kinds of flows that change the items on the balance sheet:
• Income$y \$, increases and consumption$c\$, decreases an economic unit’s net worth: $y-c=\Delta nw \$ (Remember, net worth is the difference between assets (both tangible and financial) and liabilities: $nw=ta+fa-l \$,).
• Revenues, $r \$, increase and expenditures$r \$, decrease net financial assets: $r-e=\Delta nfa \$ (Remember, net financial assets are the difference between financial assets and liabilities: $nfa=fa-l \$).
• Receipts increase and payments decrease a unit’s means of payment (stock of money): , $receipts-payments=\Delta m \$,
Now, it is very important to keep those items clearly apart. Too often, receipts, income and revenues are mixed up with each other as are consumption, expenditures or payments. While there might sometimes be an overlap between those different flows (a revenue might also be income and come in the form of a receipt), they have to be kept strictly apart.
Here are some examples:
• Income but no revenue: Here only tangible assets increase and net financial assets do not. The simplest example would be a non-monetary gift.But it also contains all production of tangible assets to which no expenditures correspond (this will be very important when we will talk about profit). Another example is the appreciation of an asset (financial or tangible) you already own. An important note: Although net financial assets can increase through an appreciation because after one they are worth more, this does not count as a revenue since revenues are transactions and a change in an asset’s value is not a transaction.
• Consumption but no expenditure: Those are all transactions that leave your net financial assets unchanged. Imagine you cook for yourself and eat what you cooked. This is a consumption but not an expenditure. For companies, depreciation of your tangible assets (sometimes also called consumption of fixed assets) is a consumption that is no expenditure.
• Revenue but no income: This is an increase in net financial assets that leaves your net worth unchanged. This then has to be a transaction in which an increase of your net financial assets is compensated for by a decrease in your tangible assets. This could be the sale of a machine — it increases your net financial assets (more money) and decreases your stock of tangible assets (less machines).
• Receipt but no revenue: The stock of means of payments increase without an increase in net financial assets. This can be due to two things: a) liabilities increase by the same amount as your means of payment, i.e. you take out a loan. This is a balance sheet lengthening (more on that later). b) somebody pays you back money she owes you. Then your other financial assets decrease (your claim) and your stock of money incrases. This is an asset exchange.
And so on. You can make up other examples but the bottom line is: in order to understand what is going on in the economy, you have to keep those different transactions apart (and as an appetizer: I think much of academic economics does not properly keep those different transactions apart with grave consequences…).
# Balance Mechanics 2.1: Assets and liabilities for groups and the aggregate economy
Having defined the balance sheet and its composition, we can now employ the framework of partial, relational and global statements to see what we can say about assets and liabilities in the aggregate and for groups of economic units. While basic and sometimes trivial, we can derive very important conclusions for economic policy from this exercise.
To recap, we had separated assets into tangible assets (machines, houses etc.) and financial assets (financial claims); liabilities are units’ debts and net worth is the difference between all assets and liabilities:
$ta+fa-l=nw \$
Net financial assets are financial assets minus liabilities, where financial assets can be separated into means of payment and other financial assets so that:
$m+ofa-l=nfa \$
Now it is essential to keep in mind that every financial assets is always someone’s liability; and every financial liability is someone else’s asset so that:
$fa\equiv l \$
Net financial assets – how much you can hold
With this insight we can make an important statement about the economy, namely that you can only hold a positive (negative) stock of net financial assets if the rest of the economy (what we had called the complementary group) has an negative (positive) stock of net financial assets of the same absolute amount. To see that, we can separate all holdings of financial assets and liabilities in an economy into those of some group, $g \$, and its complementary group, $cg \$:
$fa_g+fa_{cg}= l_g+l_{cg} \$
Bringing the group’s liabilities to the left and the complementary group’s liabilities to the right yields the net financial asset holdings of both the group and the complementary group:
$fa_g-l_g=-(fa_{cg}-l_{cg}) \$
That means that whatever amount of net financial assets you hold – the rest of the world will by necessity always hold the exact same amount but with the reverse sign. Expressed in terms of the three statements we can write:
• Partial statement (valid for individuals or groups): Individuals or groups of economic units can have a positive (or negative) stock of net financial assets.
• Relational statement (which tells us under which conditions the partial statement is valid): Individuals or groups can only have a positive (negative) stock of net financial assets if its complementary group has the exact same amount of negative (positive) net financial assets.
• Global statement (valid for all units): The aggregate economy’s (a closed economy or the world economy) net financial assets are zero.
Why is that important? A first application would be pension systems based on the accumulation of financial assets. If households wanted to save in the form of financial assets, they could only do so if the rest of the economy (i.e. companies, the government, foreigners) issued the corresponding liabilities (among which we also count stocks). No liabilities, no financial assets.
But this is the risk of pension systems based on an accumulation of financial assets: somebody will have to increase liabilities by the same amount that pensioners increase their financial assets. And liabilities are payment commitments. The higher payment commitments are, the more likely – all else being equal – are defaults which in turn hurt pension savers etc. The problem is exacerbated if the government would not be allowed to issue debts. In most industrialised countries, government bonds are a safe asset, i.e. a default free assets because in the extreme case in which the private sector is not willing to refinance government debt, the central bank might step in an (more on that in a later post); also governments can coerce (at least to some degree) the population to pay more taxes in order to service its debts which private companies can (normally) not do. This ability of governments to get taxes gives holders of government debts the assurance that the government will keep its promise and pay. If you save for your pension or are already a pensioner you certainly want to receive your pension payouts and thus need some safe asset.
A second – and related – question is the one about whether government debts are a liability to “future generations”. They certainly are a liability of future governments which future tax payers will have to shoulder. For those tax payers, government debt then really will be a burden. But in those “future generations” are also those holding the corresponding net financial assets who will receive interest. There certainly is a problem (if you think it is a problem) of distribution between those who pay taxes in order that the governments pays interest and debt holders who receive this interest. But future generations can by definition not be burdened in their entirety by government debt.
A third application are net financial assets of entire countries. The US today is the world’s biggest net debtor – i.e. it has a huge stock of negative net financial assets. This means that the rest of the world holds the corresponding positive financial assets, among them claims vis-à-vis Americans. Some see that as a huge problem that needs to be corrected (one can have a debate about whether this is a problem for the US or even an advantage reflecting the fact that the US issues the world currency, i.e. the dollar. But we leave that to another post).
Assets in the aggregate
A final point I want to make is to ask what kind of assets the aggregate economy can hold? That is fairly easy if you keep in mind that the aggregate economy’s net financial assets are by definition zero because (gross) financial assets and liabilities sum to zero in the aggregate. Since tangible assets are also assets, this means that an aggregate economy’s only assets are tangible assets, i.e. the houses machines etc. This finding is important: National economies that are open to the world (which today are mostly all economies) have both positive and negative net financial assets, i.e. they are either net creditors or net debtors. But they mostly hold huge amounts of tangible assets.
Coming back to the US example: While the US is the world’s biggest net debtor, it also holds huge amounts of tangible assets that are by far worth more than their net debts (= negative net financial assets). It is in this sense that the US is probably the richest country in the world even with high net debts.
But that in the aggregate (which means in a closed economy or the world as a whole) there are only tangible assets has important implications for saving to which we will have to say more later on. Here is only a short teaser: Since saving is the change in net worth, and net worth consists of tangible assets and net financial assets, saving is both a change in machines and houses (=investment) and a change in net financial assets. Since the aggregate economy’s net financial assets are always zero, the change in net financial assets is also zero. This means that an aggregate economy can only save by increasing its tangible assets, i.e. invest. We’ll come to what that means later on in more detail.
# Balance Mechanics 2: Balance Sheets
Now we come to one of the fundamental building blocks of balance mechanics, the balance sheet. A proper understanding of what a balance sheet is and the items it contains is the basis for all of balance mechanics (since balance mechanics has the word “balance” in it for a reason…).
Each economic unit has a balance sheet, be it individuals (you and me), households (your wife / husband, kids etc.), firms, governments, countries etc. This balance sheet contains assets, $a \$, liabilities, $l \$, and the difference between assets and liabilities: net worth, $nw \$.
Thus, net worth is:
$nw_t \equiv ta_t+fa_t-l_t \$
Assets can be both tangible assets, $ta \$, and financial assets, $fa \$. Tangible assets are: machines, houses, cars etc. Financial assets are financial claims like money on the bank, a loan, a bond, a stock etc. The difference between those two kinds of assets will be very important in balance mechanics. A unit’s liabilities are the debt it owes and also the part of equity that are stocks. To include stocks as a liability is somewhat strange. We do it however anyway in order to be consistent (more on that later).
Now we can define another important concept that will be widely used later one, namely net financial assets. This is the difference between financial assets and liabilities:
$nfa_t \equiv gfa_t-l_t \$
It is very, very, very important to keep financial assets – more exactly: gross financial assets – and net financial assets apart. (Gross) financial assets are assets that can be traded on financial markets and are – more or less – concrete things. Net financial assets cannot be traded on any market but are just an accounting concept where you subtract two accounting items from each other. Many economists tend to confuse the two (which is a stark claim which will be substantiated in a later post).
The last important definition we will use is that financial assets contain two kind of financial assets: means of payment, $m \$ – i.e. money – and all other financial assets, $ofa \$ – i.e. financial claims that are not means of payment:
$gfa_t\equiv m_t+ofa_t \$
The latter distinction is crucial for all of economics, and especially if we want to analyze financial crises, monetary policy, what banks do etc. So what are other financial assets — $ofa$? Those are all financial claims except for money. They are are promises to receive means of payment, $m$, but tend to be no means of payment themselves. Since any unit’s financial asset is another unit’s liability, liabilities are promises to make payments. And here comes the important point: A payment is the act of servicing a contractual debt.
However, there is some difficulty to exactly define the financial assets that are means of payment and that are not. Why is that the case? Take a euro coin or a euro banknote. Those are obviously not accepted to service dollar debts. In the US, a euro banknote is generally not a means of payment. But since it is a financial asset, it then becomes an “other financial assets”. Also, a sight-deposit at a commercial bank is normally accepted by non-banks (us) as a means of payment. When we see a higher balance on our bank account, we accept this as payment. We do normally not go to our employer and demand our wages in cash (At least not not any longer. Our grand parents used to do exactly that). But commercial banks among each other do not accept their respective sight-deposits as means of payment. This is an important fact that determines much of what banks and monetary policy can and can’t do. We will come back to this later on (however not in this post).
While this context-dependence of what constitutes a means of payment makes it hard to exactly define money independent of context, the distinction between means of payment and other financial assets is at the heart of every financial crisis: in a crisis, debtors have difficulties to make good on their promises to deliver the contractually promised means of payment. Even if they held other financial assets they may well not be able to convert them into means of payment, or only at far lower prices than anticipated. They may therefore be forced into default owing to a lack of liquidity. This is why you should always keep apart other financial assets and means of payments – both for your personal financial health as well as analytically.
All balance sheet items are shown in the table below. All assets are listed on the left hand side, all liabilities and net worth are listed on the right hand side. As I already wrote: every economic unit (whether implicit or explicit) has such a balance sheet. However, what kinds of assets, liabilities and what amount of those assets and liabilities each unit holds is quite different: For instance, non-financial firms normally hold mainly tangible assets like machines and only relatively few financial assets. They often have a high net worth and low debt. Private households typically hold both tangible assets (mainly houses) and financial assets (deposits, bonds, stocks) and have high net worth. Banks’ tangible assets are mostly negligible. They mainly hold loans, bonds, derivatives and other financial assets, have very high debts and very low net worth.
So, this is basically what is there to know about balance sheets, at least for our purposes. In the next post I will show you how those different balance sheet items can be changed.
# Balance Mechanics 1: Groups, the Aggregate Economy and Fallacies of Composition
This is the first official lecture on balance mechanics and it will be about a fundamental issue: the difference between what you can say about individual economic units, groups of economic units and the sum of all economic units, i.e. the aggregate economy.
One of the most interesting aspects of balance mechanics is that it teaches you to always keep different kinds of statements for different levels of analysis apart – and to both detect and avoid fallacies of composition, i.e. the application of statements valid for individual economic units or groups of economic units to the aggregate economy.
What do I mean by the term “economic unit”? I mean by that everyone and everything that is relevant in the economy, i.e. individuals, firms, governments, countries etc. Groups of economic units can be defined at will: It can be all individuals, all firms, or – if you want – some kinds of households, some kind of firms etc.
Also important is that to each group in the economy corresponds a complementary group which are all economic units in the economy that are not members of that group. This means that the group plus the complementary group are the sum of economic units in the economy:
$group + \text{\textit{complementary group}}=\text{\textit{all economic units}}$
We can also – depending on our interest and field of analysis – define groups and complementary groups for certain types of economic units. For instance, we might be interested in the relation of banks between each other. Then we might define the aggregate of all banks and divide this aggregate into a group of banks and all other banks which would then constitute the complementary group.
Standing up while watching a play
It is best to illustrate this with a little example. Imagine a theater filled with the audience. Now one man sits up to improve his view (I take a man. They tend to be ruder than women). However, he will only be successful in improving his view if all other members of the audience stay seated. Or, expressed with the terms defined above: the group (here a group with just one member) will be successful in improving its view if the complementary group (all other members of the audience) stay seated. If everybody stood up, nobody would be able to improve their view.
This means that the statement “one can improve one’s view in the theater by standing up” cannot be applied to all of the members in the audience. To do so would constitute a “fallacy of composition“.
Now we can use three kinds of statements to sort out what is valid for individuals and groups of units, under what conditions it is valid and what is valid for the sum of all economic units:
1. Partial statement (valid for individuals or groups): an individual or a group of members of the audience can stand up to improve its view.
2. Relational statement (which tells us under which conditions the partial statement is valid): an individual or a group of members of the audience can only improve its view if the rest of the audience stays seated.
3. Global statement (valid for all units): If all members of the audience stood up, nobody could improve their view.
Now, in this theater example the global statement might not be 100 % correct for each and every individual because some will certainly be able to improve their view while others won’t, depending on where you sit, the height of different audience members etc. It will most likely hold only on average. There will however be cases to be discussed later where the aggregate statement will hold for everybody and not only on average.
We can now define a fallacy of composition in terms of those statements: obviously, a fallacy of composition is when you apply a partial statement to the aggregate economy. There might be cases and conditions under which what is true for a single group is also valid for the aggregate. But those are normally special cases and not general cases. We will come to those cases later on.
You could of course also have the reverse, i.e. that you falsely apply a global statement to a group or an individual. I don’t know whether there is a term for that but that would evidently also be a problem.
Fallacy of composition vs. the rationality trap
The term “fallacy of composition” is used when you make an ex ante analytical statement about a certain situation. It does not depend on the concrete behavior of individuals. If behavior is concerned, i.e. if people actually tried to stand up to improve their view and found out that they could not actually improve their view because everybody did the same, it would constitute a “rationality trap”. Such a trap is that people individually behave rationally but their intention will be thwarted in the aggregate.
This is not to say that those people are dumb or do not understand that their standing up might be based on the fallacy of composition and might lead them into a rationality trap. Take the example of a fire breaking out in the theater. Everybody will be absolutely rational and right to stand up and run to the emergency exit. However, given that emergency exits are often very small compared to big audiences and fire tends to expand quickly, some will not get out, will be trampled down and burn. But it would hardly have been wise to stay seated and wait to be burnt alive. They had to take their chances.
Many economic situations (and other situations as well) are of this sort that one might know that some collective solution might be better but will have no chance of being realized so that people have to act rationally even if they get – on aggregate – into a rationality trap.
While balance mechanics is more about detecting fallacies of composition, one needs to find those fallacies to avoid rationality traps if possible. This is one of the appeals of balance mechanics.
While the above example is trivial, I will later show that quite a number of economics problems will be easier to analyze with the framework of groups, complementary groups and the three statements in your mind.
# An Easy Introduction to Balance Mechanics and its Inventor
Who was Wolfgang Stützel, the inventor of “balance mechanics”? Hardly known in the anglo-saxon world and largely forgotten in Germany today, he was an economics professor and published his landmark book “balance mechanics” (in German: “Volkswirtschaftliche Saldenmechanik“) in 1958.
In the 1960s and 1970s he was widely known beyond economics because he was a member of the famed German Council of Economic Experts which is still influential today. In his work, he shows that no abstract models and fantasy assumptions about human behavior are necessary to draw rigorous and logically necessary conclusions about the actual economy. Essentially, you only need a profound understanding of something as trivial as accounting. Thus Stützel wrote in his book:
Apart from things that depend on human behavior, […] there are many economic relations […] about which one can make strictly generalisable statements, relations that do not depend on human behavior but which would still be unchanged if people behaved in absolutely strange way.
[In German: [Es gibt] neben Zusammenhängen, die vom menschlichen Verhalten abhängen, […] viele Größenbeziehungen in der Wirtschaft […], über die sich streng Allgemeingültiges aussagen läßt, Zusammenhänge, die nicht vom menschlichen Verhalten abhängen, sondern auch dann unverändert bestehen bleiben würden, wenn die Menschen sich noch so ungewöhnlich verhielten.]
Those relations are very trivial things, like: Someone’s purchases are another one’s sales because nobody can sell something without someone else buying it; or that someone’s debts are someone else’s claims because nobody can have financial claims without someone who holds the corresponding liabilities.
One would have to assume that economists know those “trival arithmetic relationships” and their implications. However, this is mostly not so. Stützel always showed with great relish how his colleagues stubbornly held beliefs which one would never hold if one only knew the basics of accounting and balance mechanics.
Here are some examples: The trivial but always true statement that somebody’s claims are someone else’s debts is often ignored by economists and politicians alike if they claim that today’s government debt would burden future generations. The underlying assumptions of this statement is that in the future there are debts but no corresponding financial claims — a logical impossibility. If there are debts in the future, holders have to pay interest and somebody will earn this interest. People do not only inherit debts but also the corresponding claims.
Or another example: Savers increase their (net) financial claims. But this in only possible if others increase their (net) debts by the same amount. However, if nobody is willing to increase her net debts, nobody will be able to save. While this is trivial, it becomes politically salient if applied to the privatization of pensions. Those willing to increase privatization of pensions by logical necessity also have to be in favor of higher debts.
But who should increase their debts? In many parts of the world (Germany, for instance), companies do not increase their net debts at all but finance their investment out of their current earnings; all over the world the government is held not to increase or even to decrease its debts. But then only foreign countries could increase their debts – i.e. those countries like the European crisis countries that lived through a debt crisis… However, if nobody is willing to increase their debts, there cannot be higher private pension saving (and other saving as well) — and one cannot ask persons to do so anyway.
Austerity in Europe after 2010 can also be understood by the trivial accounting rules and balance mechanics: in order to pay back their debts, governments in the crisis countries had to radically reduce their expenditures. However, since someone’s expenditures are another one’s revenues, this cut in government spending also made life hard for the rest of Europe which had to deal with a drop of exports — its revenues — to the crisis countries.
All this very much sounds like simple Keynesianism — but Stützel shows that it is just the consequence of pure accounting: If everybody is asked to reduce their expenditures, it would be unreasonable to expect an increase in revenues; if everybody is expected to save more and increase financial assets, one can not demand that nobody is allowed to increase debts.
But Stützel himself was no left leaning Keynesian. In the introduction to the works of his teacher Wilhelm Lautenbach, he wrote:
The claim that deficit spending — the creation of money by the state through higher debts — could fulfill all economic needs, has not been made by serious employment theorists. It is an invention of demagogues who have taken money creation from modern credit theory and increasing debts from Keynesianism to make the masses believe illusions and to poison the atmosphere for sober thinking. Perhaps it is just a bugaboo which was invented by the enemies of employment theory in order not to think too much when arguing against employment theory.
[In German: „Die Behauptung, daß mit der Druckknopftherapie des deficit spending, der staatlichen Geldschöpfung durch Verschuldung, alle wirtschaftlichen Nöte gelindert werden könnten, stammt nicht von ernsthaften Beschäftigungstheoretikern. Sie ist eine Erfindung von Demagogen, die sich aus den modernen Kredittheorien das Geldschöpfen und aus dem Keynesianismus das Schuldenmachen geholt haben, um damit breiten Massen Illusionen vorzugaukeln und die Atmosphäre für nüchterne Überlegungen zu vergiften. Vielleicht ist es auch ein Popanz, den sich manche Gegner der Beschäftigungstheorie zusammengezimmert haben, damit sie in ihrer Argumentation gegen die Beschäftigungstheorie nicht so viel zu denken gezwungen sind.]
But while Stützel saw that much can go wrong if one relied too much on the market, he believed in capitalism and the market economy, was an active member of the economically (neo-)liberal free democratic party (FDP) and part of the neoliberal “Kronberger” circle which propagated an extension of markets in the 1970s and beyond.
Since the early 1970s Stützel was against the financing of the welfare state by contributions levied on wages and explained the increasing unemployment at the time not with the lack of aggregate demand but of too high wage costs. Also, when capital controls were still in place almost everywhere, he was a strong defender of the free movement of capital because he trusted market more than bureaucrats.
On the other hand, in his book “Market price and human dignity” (“Marktpreis und Menschenwürde”) he demanded a lower limit for wages (and favored a labour union that encompassed all workers to enforce this limit). By this he wanted to prevent a “wage paradoxon” according to which each individual, in order to save her job, had to accept wage reductions and a lengthening of working time which would — if applied to everybody — only result in wage reductions for all with at an unchanged overall employment level. He took that thought from Karl Marx. And while he rejected high social contributions on wages, he was not concerned to decrease the welfare state but to finance it by taxes in order not to disturb prices on the labour market.
Thus, Stützel was an economic liberal, even a neoliberal. However, he knew when the market reached its limits and when the pursuit of self interest by each individual could result in collective disaster.
This text is a translation (with slight changes) of an article originally published in German on the Blog “Herdentrieb” of the German weekly “Die Zeit”.
# Saving does not finance investment!
Why do neoclassical economists want the budget to be balanced and people to save more? Because they believe that “saving finances investment”. They believe that in order to finance investment, somebody has to have saved beforehand. This might sound intuitive, but it is one of the biggest (and unfortunately oldest) fallacies in economics. Saving never finances anything – money and credit do; and nobody has to save for anybody else to have credit and thus to invest.
According to the “saving-finances-investment”-theory (often called the loanable funds theory), households have to save first, bring their money to the bank so that firms can borrow and invest – as long as government deficits have not taken away the precious savings from firms. From this it follows that more household saving and lower government deficits are the best way to promote investment: More saving leads to a higher supply of credit and thus more investment. Zero government deficits avoid crowding out private investment. And so if investment is not high enough, both households and the government should save more by cutting their spending and thus allow banks to increase credit.
### Credit in the real world – a simple booking procedure
At first sight, the theory seems to be intuitively appealing. But once you look how credit is actually created in the real world, the theory quickly becomes utter nonsense (find more on this in my recent working paper). Nobody has to save before a credit can be created. A bank (either a commercial bank or the central bank) creates credit and money out of thin air by a stroke on a computer keyboard: a bank creates a deposit for its debtor which the debtor can withdraw to pay for something. Later on, the debtor has to repay its debt – if he can’t, the bank has a problem. Giving credit is a simple booking procedure. No bank has ever denied a credit to a potential borrower because of a lack of prior saving.
But if banks can create money out of thin air, why do they need deposits? Banks create bank money, but not central bank money – which only the central bank can do. But central bank money – coins, banknotes, deposits at the central bank – is what people need to make payments. So banks somehow have to get hold of central bank money, either directly from the central bank, from other banks via the interbank market or via deposits.
But if banks want to get money from depositors, this does not necessitate in any way that depositors save. Money does not disappear when it is spent and not saved – it is not “used up” by government deficits or investment. If everybody would always spend all of their income and not save at all, the money would not disappear but flow to firms and back to households: it is transferred from the customer’s bank account to the firm’s bank account; from the firm’s bank account in the form of wages and interest to the household’s bank account and so on. The money is not lost at all to the banking system whatever it is spent on and whatever amount people save. When people do not spend all of their income but save, this is money that the business sector does not get in the form of sales.
In general, as long as people keep their money at the banks – whatever their amount of saving – the banking system has no problem refinancing itself with deposits. And if people were less willing to keep their money at the bank (which today happens for instance in Europe’s crisis countries), banks can refinance themselves at the central bank. So, no saving has to take place either for deposits to be a source of bank credit or for credit to be created.
Financial saving is a zero-sum game
Anybody with sufficient collateral who wants to finance a physical investment (produce or buy a machine or a house) can get this credit, use the money so obtained and invest (with the caveat that banking regulation can of course restrict credit creation, but again, this has nothing to do with saving). Most adherents of the view that saving finances investment seem not to know that investment itself is saving. Normally, we tend to identify saving with financial saving – i.e. to have more money, more bonds or more equity. However, accountants have defined saving as increases in all kinds of wealth, both financial and non-financial.
While the production of new capital goods always increases saving in an economy, financial saving is a zero-sum game. One can only increase net financial assets (financial assets minus financial liabilities) by spending less than one earns. The problem is that your spending is my earning. So if anybody cuts his spending to increase his net financial assets, somebody else will see his earnings cut by the same amount – and thus his ability to save money. Since earning and spending are necessarily equal in the whole economy, the whole economy cannot save financially – net financial assets are zero in the aggregate.
On the other hand, investment creates spending – if a firm builds new machines and houses, it spends money on wages which are income for wage earners; firms that buy machines and households that buy houses spend money – which sellers earn. In contrast to financial saving, increasing production of physical capital goods – investment – is not a zero-sum game.
So what would really happen if people save more money believing that this leads to a higher availability of credit and more investment? Since households mostly spend their money on goods sold by firms, cutting spending in order to increase saving automatically reduces firms’ revenues. Will that incite firms to build more machines?
This seems unlikely. Even if they take out more debts when they see their revenues fall, they would probably not use it to increase their investment. If they would like to continue spending the same amount of money on their employees which they did before households had cut their spending in order to save, they would have to borrow the money which they had earned before by their sales.
But is it even likely that they will take out more credit when they see their earnings fall? That depends on their expectations of future household spending. If households firmly believe in loanable funds theory, they will try to keep their saving up in the long-term – and in consequence firms’ revenues will be permanently reduced. If firms still kept up their former level of spending, their higher credits would lead to higher interest commitments in the future. Thus, rational businesspeople are more likely to cut spending themselves than to increase future spending by taking out new credits. What kind of spending are they likely to cut? Probably they will reduce the wage bill by either reducing wages or firing people. Unfortunately, those are the private households’ revenues, out of which households had planned to save.
Thus, if firms reduced their payrolls households would have shot themselves in the foot with their decision to save: they wanted to save but what they actually achieved is to cut their own revenues. If they would then again reduce their expenditures, firmly believing in the merits of thrift, firms would again face lower revenues etc.
This is the paradox of thrift: households’ plans to save more leads to a decrease in aggregate revenues and expenditures – and no financial saving has actually taken place. Since firms are also likely to cut back their investment, overall saving will have fallen. The government could of course counteract the whole process by spending more where households spend less – but if the government also believes in loanable funds theory, it will cut spending itself and consequently the private sector’s revenues – welcome in recession-land. Or rather: welcome in euro-land.
Those are the economic costs if people who firmly believe in “saving finances investment” rule or advise the rulers. This is not just theory, the fallacies of loanable funds theory are what makes millions of unemployed in the euro zone suffer every day. Economic theory has real world consequences – sometimes for the better, often for the worse.
The text was first published in 2012 on the Social Europe Blog.
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Worksheet # Topics/Curriculum Unit-1 Numeration Review Unit-2 Subtracting Numbers Ending in 0 or with a 0 in the Middle Unit-3 Multiplying 2 Digit Numbers by 2 Digit Number Unit-4 Multiplying 2 or 3 Digit Numbers by 2 Digit Numbers Unit-5 Number Concepts Unit-6 Dividing 2 or 3 Digit Numbers Ending at 0 or 5 Unit-7 Dividing 3 Digit Numbers by 2 Digit Numbers part I Unit-8 Dividing 3 Digit Numbers by 2 Digit Numbers part II Unit-9 Dividing 3 or 4 Digit Numbers by 2 Digit Numbers Unit-10 Introducing Fractions Unit-11 Reducing Fractions Unit-12 Reducing Fractions Using GCF Unit-13 More Problems on Reducing Fractions Using GCF Unit-14 Conversion of Fractions Unit-15 Adding Fractions with Like Denominators Unit-16 More Problems on Adding Fractions with Like Denominators Unit-17 Addition of Unlike Fractions Unit-18 Subtraction of Fractions, Mixed Numbers, and Whole Numbers Unit-19 Subtraction of Unlike Fractions and Mixed Numbers Unit-20 Addition of Decimals Unit-21 Subtraction of Decimals Unit-22 Introducing Integers (Z) Unit-23 Evaluating Algebraic Expressions Unit-24 Adding and Subtracting Algebraic Expressions Unit-25 More Problems on Adding and Subtracting Algebraic Expressions Unit-26 Ratio, Percent, and Proportion Unit-27 Shapes Unit-28 Perimeter and Area
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# [R] user defined covariance structure
From: Jonathan Smith <jon3smith_at_hotmail.com>
Date: Tue 01 Aug 2006 - 00:08:22 EST
I am writing as I am still having trouble trying to define my own covariance matrix. My code is displayed below. I am defining the covariance matrix in the form of an AR1 process so it can be easily checked if working correctly. Another question I have is if it is possible to define the matrix without giving p a specific value and leaving it in as a coefficient. For the reason that it can be estimated when model is run. I figure that is the way AR1 works in GLS.
Thank you
Jon Smith
I would appreciate any help s I am stuck here and need to figure this out prior to continuing my research.
function ()
{
library(nlme)
tim<-c(1,2,3,4,1,2,3,4,1,2,3,4,1,2,3,4)
peep<-c(1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4) y<-c(11.78,9.53,11.03,9.89,10.80,8.74,10.25,10.69,5.60,7.27,6.81,4.56,7.01,5.64,6.30,8.31)
#This y data was created from and AR1 model with correlation coefficient
equaling 0.7.
timMat<-matrix(c(tim),ncol=1,nrow=16)
peepMat<-matrix(c(peep),ncol=1,nrow=16)
yMat<-matrix(c(y),ncol=1,nrow=16)
dataframe<-data.frame(yMat,timMat,peepMat) p=0.7
g=1
tester2<-corSymm(value = c(p^(1),p^(2),p^(3),p^(1),p^(2),p^(1)),form = ~ timMat|peepMat)
tester2<-Initialize(tester2, data = dataframe) testMat2<-corMatrix(tester2)
print(testMat2)
# this appears to be working correctly
#smanGls<-gls(yMat~timMat,data = dataframe, corr = corAR1(form =
~timMat|peepMat))
# works perfectly
smanGls<-gls(yMat~timMat,data = dataframe, corr = corSymm(tester2))
arsum<-summary(smanGls)
print(arsum)
#this is what message I get when I try to use the covarince matrix I
defined.
#Error in gls(yMat ~ timMat, data = dataframe, corr = corSymm(tester2)) :
# false convergence (8)
}
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Received on Tue Aug 01 00:16:38 2006
Archive maintained by Robert King, hosted by the discipline of statistics at the University of Newcastle, Australia.
Archive generated by hypermail 2.1.8, at Tue 01 Aug 2006 - 02:16:53 EST.
Mailing list information is available at https://stat.ethz.ch/mailman/listinfo/r-help. Please read the posting guide before posting to the list.
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## Monday, October 26, 2009
### Soda Springs power estimate
The Soda Springs Low-Key went rather well in most regards. As a participant,I was pleased with my finishing place. I usually train with a PowerTap, but for most races (especially hillclimbs!) I leave the powertap wheel home and instead use a carbon race-only tubular (either my Reynolds MV-32T or my Mt Washington).
On hillclimbs, though, I can generally estimate the power. The key parameters are total weight, rolling resistance coefficient, wind drag coefficient, and air density. For PowerTap-equivalent power I do not need drivetrain efficiency, since drivetrain losses aren't measured by the PowerTap, anyway. Given these, it's simple enough to calculate power from speed.
So, some numbers:
• weight: I was 125 lb in the morning upon wakening. Given input - output, I was probably close to this at the start. Curiously, I'd been almost 3 lb lighter the last time I weighed myself two days before. A lot depends on what I've been eating and how hydrated I am. Additionally, my bike weighed out at 11 lb 6 ounces before the start. This was one ounce more than I expected, but I'd accidentally mounted the wrong tire on my front wheel. Hard to comprehend how I could have gone through an entire tire-glueing procedure without noticing this.... I only noticed it, in fact, when I was looking at a photo of myself after the race! Add in the weight of shoes, clothing, and my water bottle and I figure around 4 extra pounds is pretty close: should be within 1% of actual total weight.
The pre-climb weigh-in (Rich Hill photo)
• Rolling Resistance: My rear tire has a value reported by Al Morrison to be 0.234%. I accidentally had a Veloflex Carbon on my front wheel, with a value of 0.312%. Given the slope, the rear wheel should be responsible for around 2/3 of the total, the front wheel around 1/3, putting the average around 0.26%. The road on the climb is rough in sections, so I'll assume an actual value of around 0.4%.
• Air density: At Redwood Estates in the hills over Los Gatos (1690 ft elevation), air pressure was 29.98 inches of Hg, while the temperature was 67.2 F with 70% humidity at 10:32 pm, around the time of the climb. These values yield an air density of 1.13 kg/meter³.
• Wind drag coefficient: I'll use the value I extracted from Old La Honda data, CdA = 0.36 meters².
Okay, then. I climbed 2368 feet over 5.35 miles in 32:04. For these sorts of calculations, with mixed imperial and metric units, the Unix "units" program is great, as it handles the unit conversions.
1. Climbing power = weight × feet gained / time = 234 watts
2. Rolling resistance power = weight × Crr × speed = 11 watts
3. Wind resistance power = ½ ρ CdA × speed³ = 18 watts
4. Acceleration power: I went from 0 to 10 mph over 32:04, which works out to 0.3 watts, or zero with the 1 W precision used here. I never braked during the climb, so there were no other accelerations which were not canceled by other decelerations.
5. Total = 263 watts
If I wanted to determine SRM or Quarq Cinqo watts (or MetriGear Vector), I'd divide by an efficiency term, for example 0.97, and thereby estimate maybe 271 watts.
I was probably around 290 watts (PowerTap equivalent) when I set my Old La Honda PR in July. Sure, Soda was almost twice the duration, but that's only responsible for a few %. Specifically, assuming a critical power model where AWC = CP × 60 seconds, which generally fits my data, I'd expect a 2.7% loss in power going from OLH to SS: around 8 watts. This at least brings my effective OLH power above 270 watts PT-equivalent, not bad for the Noon Ride, but not up to competition standards.
I think it shows a difference between off-season (lifting, running, generally not pushing things too hard) and in-season fitness. The same applies for everyone in the Low-Keys, however. None of the top guys are in mid-summer fitness.
D'oh! (Rich Hill photo)
So what was the penalty of my bad tire choice? Let's use Al's numbers of 0.08% difference for the wheel, blow that up to 0.11% for the rougher road, and assume 1/3 of weight on the wheel, yielding 1.0 watts lost to propulsion out of 263. Using a formula I derived earlier, with 88% of the power mass-proportional, this yields a speed penalty of 0.30%, or 5.9 seconds. There's also a mass penalty, as the Record is slightly heavier, around 1 ounce. That penalty is an extra 0.6 seconds, bringing the total to 6.5 seconds. I was 22 seconds down on Eric, however, so this mistake didn't affect my placing.
For error analysis, a large source of uncertainty is the total climbing. For every 10 feet of error in the total climbing feet that's a difference of 1.0 watt. But the climbing feet agree with those recorded by two other riders on their barometric altimeters, each recording slightly less, not more climbing than the Garmin data I used as a reference. I'll stick with the Garmin numbers, since they're less susceptible to barometic drift due to the GPS signal used as a calibration correction to the barometric signal, the latter better for detail.
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# 24 Summer Time Math Activities which can be done at home!
I realize many of you (teachers and parents) may be searching for ways to link every day activities to math so that children can learn in a practical way while at home during this surrealistic period. Happy Fourth of July and . . . .Here’s a list of things you might like to try:
Outdoors
1. While bouncing a ball, skip count by any number. See how high you get before missing the ball. Good to keep your multiplication facts current.
2. How high can you bounce a ball? Tape a yardstick or tape measure to a vertical surface (tree, side of house, basketball goal). While one person bounces, one or two others try to gauge the height. Try with different balls. Figure an average of heights after 3-4 bounces.
3. Play basketball, but instead of 2 points per basket, assign certain shots specific points and keep a mental tally.
4. Get out the old Hot Wheels. Measure the distance after pushing them. Determine ways to increase or decrease the distance. Compete with a sibling or friend to see who has the highest total after 3-4 pushes. Depending on the age of your child, you may want to measure to the nearest foot, inch, half-inch or cm.
5. Measure the stopping distance of your bicycle.
6. Practice broad jumps in the lawn. Measure the distance you can jump. Older students can compute an average of their best 3-4 jumps. Make it a competition with siblings or friends.
7. Some good uses for a water squirt gun:
• Aim at a target with points for how close you come. The closer to the center is more points.
• Measure the distance of your squirts. What is your average distance?
• How many squirts needed to fill up a bucket? This would be a good competition.
8. Competitive sponge race (like at school game days): Place a bucket of water at the starting line. Each player dips their sponge in and runs to the opposite side of the yard and squeezes their sponge into their own cup or bowl. Keep going back and forth. The winner is the one who fills up their container first. Find out the volume of the cup and the volume of water a sponge can hold.
9. Build a fort with scrap pieces of wood. Make a drawing to plan it. Measure the pieces to see what fits. Use glue or nails (with adult supervision).
10. Take walks around the neighborhood. Estimate the perimeter distance of the walk.
11. In the pool:
• Utilize a pool-safe squirt gun (as in #6 above).
• Estimate the height of splashes after jumping in.
• Measure the volume of the pool (l x w x h). The result will be in cubic feet. Convert using several online conversion calculators such as this one: https://www.metric-conversions.org/volume/
• Measure the perimeter of the pool. If it is rectangular, does your child realize the opposite sides are equal. This is a very important concept for students regarding geometry (opposite sides of rectangles are equal).
• What if you want to cover the pool? What would the area of the cover be?
• Measure how far you can swim. Time the laps. What is the average time?
12. Watch the shadows during the day. Notice the direction and the length. Many kids don’t realize the connection between clocks and the sun. Make your own sun dial. Here are a few different resources for getting that done, some easier than others:
Indoors
1. Keep track of time needed (or allowed) for indoor activities: 30 minutes ipad, 1 hour tv, 30 minutes fixing lunch, 30 minutes for chores, etc. This helps children get a good concept of time passage. Even many 4th and 5th graders have difficulty realizing how long a minute is. This is also helpful as a practical application of determining elapsed time. Examples:
• It’s 11:30 now. I’ll fix lunch in 45 minutes. What time will that be?
• I need you to be cleaned up and ready for bed at 8:30. It’s 6:30 now. How much time do you have?
• You can use your ipad for games for 1 hour and 20 minutes. It is 2:30 now. What time will you need to stop?
2. Explore various recipes and practice using measuring tools. What if the recipe calls for 3/4 cup flour and you want to double it?
3. In the bathtub, use plastic measuring cups to notice how many 1/4 cups equal a whole cup. How many 1/3 cups in a cup? How many cups in a gallon (using a gallon bucket or clean, empty milk carton)?
4. While reading, do some text analysis regarding frequency of letter usage.
• Select a passage (short paragraph). Count the number of letters.
• Keep track of how often each letter appears in that passage. Are there letters of the alphabet never used?
• Compare with other similar length passages.
• After analyzing a few, can you make predictions about the frequency of letters in any given passage?
• How does this relate to letters requested on shows such as “Wheel of Fortune” or letters used in Scrabble?
5. Fluency in reading is a measure of several different aspects: speed, accuracy, expression, phrasing, intonation.
• To work on the speed aspect, have your child read a selected passage (this can vary depending on the age of the child). Keep track of the time down to number of seconds. This is a baseline.
• Have the child repeat the passage to see if the time is less. Don’t really focus on total speed because that it not helpful for a child to think good reading is super fast reading. Focus more on smoothness, accuracy and phrasing.
• Another way is to have a child read a passage and stop at 1 minute. How many words per minute were read? Can the child increase the # of words per minute (but still keep accuracy, smoothness, and expression at a normal pace)?
6. Play Yahtzee! Great for addition and multiplication. Lots of other board games help with number concepts (Monopoly, etc.)
7. Lots of card games using a standard deck of cards have math links. See my last post for ideas.
8. Measure the temperature of the water in the bathtub (pool thermometers which float would be great for that). How fast does the temperature decrease. Maybe make a line graph to show the decline over time.
9. Gather up all of the coins around the house. Read or listen to “Pigs Will be Pigs” for motivation. Keep track of how much money the pigs find around the house. Count up what was found. Use the menu in the back of the book (or use another favorite menu) to plan a meal. Be sure to check out Amy Axelrod’s other Pig books which have a math theme Amy Axelrod Pig Stories – Amazon Here is a link to “Pigs Will be Pigs”: Pigs Will Be Pigs – Youtube version
10. Help kids plan a take-out meal that fits within the family’s budget. Pull up Door Dash for a variety of menus or get them online from your favorite eateries. This gives great practical experience in use of the dollar to budget.
11. Look at the local weekly newspaper food advertisements. Given a certain amount of \$, can your child pick items to help with your shopping list? If they accompany you to the store, make use of the weighing stations in the produce section to check out the weights and cost per pound.
12. Visit your favorite online educational programs for math games or creative activities. See a previous post regarding “Math Learning Centers.” The pattern blocks and Geoboard apps allow for a lot of creativity while experiencing the concept of “trial and error” and perseverance. These can be viewed at the website or as an app. Here’s a link to it to save you time. Virtual math tools (cindyelkins.edublogs.org)
Please share other activities you recommend!! Just click on the speech bubble at the top of this post or complete the comments section below. I miss you all!
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# Lorentz Factor
1. Jul 13, 2012
### Vorde
Is it fair to call the lorentz factor the derivative of measured time relative to proper time? I've seen the lorentz factor equated twice now to $\frac{dt}{d\tau}$ and I wanted to know whether that was a legitimate way to look at it.
Thank you
2. Jul 14, 2012
### ghwellsjr
That's the way it is shown in the wikipedia article on Lorentz Factor, however, the article doesn't explain what t is. It is not a measured time but rather the arbitrarily defined coordinate time of a frame. As such, we usually use the reciprocal of gamma to show the time dilation of a clock moving at some speed v according to a particular previously selected frame. So it isn't generally the Proper Time on a moving clock (unless the clock is inertial) but rather the rate of ticking of the moving clock compared to the rate of ticking of the coordinate time of the frame.
3. Jul 14, 2012
### Vorde
Sort of what I meant.
For the latter part though, aren't the two statements equal? Assuming by 'compared to' you mean something equivalent to the rate of change between the two. If I measure object A as moving from point x to point x+dx in time t, and the object measures the time difference as $\tau$, the derivative dt/d$\tau$ would be equal to the lorentz factor performed on the object's instantaneous velocity during that period, no?
4. Jul 14, 2012
### ghwellsjr
When you're talking about the motion of the clock in a particular frame, don't you always refer to dx/dt as the speed? Would you ever try to figure out what dt/dx meant? In the same way, wouldn't it make more sense to talk about dτ/dt rather than dt/dτ?
5. Jul 14, 2012
### Vorde
I don't think so, to the latter part, not the first part.
$\tau$ is a constant from all points of reference, but t is frame-dependent. Wouldn't it make sense to be deriving with respect to the invariant?
6. Jul 14, 2012
### ghwellsjr
Wow, that's a novel way to mix up a definition. According to the wikipedia article on "invariant", it "is a property of a system which remains unchanged under some transformation". Since the tick rate of a clock is what changes when transformed to different frames, we cannot call it an invariant.
7. Jul 14, 2012
### Vorde
But proper time is an invariant, thats why it's a useful tool. If an object follows a world-line that passes through two events, the time separation between two events that it measures is the proper time.
It sort of makes sense to me why the derivative of measured time with respect to proper time is useful, it tells you (in a somewhat roundabout way) the instantaneous velocity of the object, but the roundaboutness might be useful in calculations.
8. Jul 16, 2012
### ghwellsjr
Yes, proper time is an invariant but I wasn't talking about proper time. I was talking about the tick rate of a clock moving in a frame compared to the tick rate of the coordinate time of the frame which is not an invariant. When you integrate the tick rate with respect to coordinate time over the coordinate time separation between the two events, then you get the invariant proper time. You cannot get the invariant proper time just from the coordinate time separation between the two events unless the clock is inertial but I assume it is not since in your next sentence, you mention instantaneous velocity, implying that it's speed is changing.
It doesn't make any sense to me, can you explain exactly what you mean by this? You can start by explaining what you mean by "measured time" and how you measure it.
9. Jul 16, 2012
### Vorde
That confusion was my mistake, I thought you were talking about something different.
What I meant was that (and I still haven't taken the time to do any of this even semi-rigorously, so this is all just a mental image) the Lorentz factor is a function of velocity (I'm going to drop the relatives because I'll always be talking about an object and a single observer). Time dilation is obviously directly connected to the Lorentz Factor. It makes sense (non-mathematically, so far at least) to me that at any point you could measure the elapsed time and also measure the proper time (this is assuming you know the equations of motion) of the object. From that, as time elapsed becomes infinitesimal, it seems that you could define a ratio that is somewhat akin to saying 'how much faster is my time flowing than his'.
In my purely mental image, it seems like this ratio would be the derivative $\frac{dt}{d\tau}$
Assuming everything I have said so far is correct, which I am prepared to be wrong about, it would then be a matter of reverse-working the dt/d$\tau$ in order to get the instantaneous velocity of the object at that point.
I guess it's silly to be arguing for this method's usefulness because the confusion is whether this is a legitimate way to represent something that's known to be useful. My only real interest at the beginning was curiosity towards the origins of the lorentz factor in a non historical setting.
10. Jul 16, 2012
### ghwellsjr
I'm assuming that the object is moving in some yet-to-be-determined way and that it is (or will be) far removed from the observer.
I'm assuming that the object has an observable clock displaying proper time riding along with it.
I'm assuming that the observer has an observable clock displaying coordinate time next to him.
I'm assuming that the observer has to wait for the image of the object's clock to propagate to him at the speed of light.
I'm assuming that the observer's calculation involves the delayed observation of the proper time on the object's clock and the non-delayed observation of the coordinate time on his own clock.
I'm not sure if you are envisioning the object moving around in 3D space or just 2D space or just 1D space--can you please clarify?
Once we get the scenario ironed out, what calculation does the observer do to determine the instantaneous speed of the object?
11. Jul 16, 2012
### Vorde
I was envisioning a textbook like scenario where you were given the equations of motion. Remember I was thinking this originally totally as an interesting educational point.
I think we've gotten a little too literal. My original question was simply to verify the lorentz factor's legitimacy as a derivative, for my piece of mind as well as to further my intuitiveness with the equations of special relativity. I can't think of any reason why you would ever use it for calculations: as it requires derivatives of variables you can avoid entirely using the classical methods.
I also just realized something else.
If $\gamma$ is defined as $\frac{dt}{d\tau}$
Then the time dilation equation really just says $t=\frac{dt}{d\tau}Δ\tau$
Changing the time to infitesimal quantities and making it an integral of all the little changes, the time dilation equation just says $t=\int\frac{dt}{d\tau}d\tau$ from t1 to t2.
Last edited: Jul 16, 2012
12. Jul 16, 2012
### ghwellsjr
This seems to be the exact opposite of what you are now saying.
If you want to "verify the lorentz factor's legitimacy as a derivative" you should take it as dτ/dt=1/γ=√(1-β2). This means the instantaneous tick rate of a clock moving at the instantaneous speed β compared to the coordinate tick rate is equal to √(1-β2).
So at β = 0.6, the moving clock is ticking at 80% of normal.
You could also say that if a clock is ticking at 80% of normal, it has a speed of 0.6c.
But it seems rather unusual, even weird, to say that if normal time is ticking at 1.25% of a moving clock, then that means the clock is moving at 0.6c.
13. Jul 16, 2012
### ghwellsjr
γ is not defined as dt/dτ. It's defined as 1/√(1-β2). See the wikipedia article on Special Relativity, under the section called "Metric and transformations of coordinates".
The rest of your post doesn't make sense to me. Maybe you could show some examples of what you mean.
14. Jul 16, 2012
### Vorde
I think we are arguing about different ways of saying the same thing. Directly, whether to say $t'= \gamma t0 or \frac{t'}{\gamma} = t0$
I hate LaTeX sometimes.
15. Jul 16, 2012
### ghwellsjr
I hate LaTex all the time. It doesn't work on my very old mobile device so I only use it when I quote someone else.
I think you are saying that t'=γt0 is equivalent to t'/γ=t0 which is true since it is simple algebra (not that I know what it means, since you haven't defined your variables). But up until now you have been talking about a derivative and you can't do the same thing with derivatives. I tried to point this out in post #4.
But since I can't tell what you are trying to say and you don't want to get precise, I have no way of knowing whether what you are saying is the same as the standard way of presenting and understanding time dilation and the Lorentz factor.
But why are you pursuing this?
16. Jul 16, 2012
### Vorde
I completely agree with you. I still have a math related question lingering but its small and rather tangential so I think I'll be able to answer it myself.
You answered my original question but I got carried away in the details, so I'm sorry for that.
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# Guessing the number game
## Main Question or Discussion Point
Hello Forum,
I just learned about the "guess the number" game. There are 100 numbers, from 1 to 100 and the computer must ask questions to find the correct number.
The computer could start from 1 and go up until it finds the right number. That would involve a certain number of questions.
Or the computer could ask more intelligent questions, like "is the number smaller or larger than 50", etc....
Most of the codes I have seen don't ask that "is the number smaller or larger than 50?" but simply ask some binary questions that are all similar (is this number smaller or larger than the previously guessed number? etc....)
Why?
thanks
fisico30
Related Programming and Computer Science News on Phys.org
uart
Hi fisico30. This game usually uses a simple binary search, in which it always halves the interval in which it currently knows where the solution lies. So the "computer" would typically start with a guess of 50 and then simply require to be told if it was either correct or high or low.
If correct the game stops with a score of "one guess". If it was told 50 was high then it would next guess 25, if told 50 was low then it would next guess 75 and so on. Searching by continually halving the interval is easy to implement and fairly efficient.
BTW. This game is sometimes given as a exercise for students learning an introductory programming course. It's about the simplest example the you can think of that in some sense constitutes an "AI" game.
Thanks uart,
great help. Ideally, the least number of guesses, questions, to get to the right answer, the better...
What would be a slightly more sophisticated version of this game? Instead of asking questions based on halving the interval in which it currently knows where the solution lies, would could be asked?
thanks
fisico30
Increase the range, say to 0-1,000, and ask if *any* (not which) digit is correct.
Thanks, I see.
So, if we get the number 3456, we would ask if any of the digits forming this number are correct...
that still seems far from being able to correctly guess the right number though....
thanks
fisico30
You know it has to be 3xxx or x4xx or xx5x or xxx6.
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ScaLAPACK 2.1 2.1 ScaLAPACK: Scalable Linear Algebra PACKage
pdgsepsubtst.f
Go to the documentation of this file.
1 *
2 *
3 SUBROUTINE pdgsepsubtst( WKNOWN, IBTYPE, JOBZ, RANGE, UPLO, N, VL,
4 \$ VU, IL, IU, THRESH, ABSTOL, A, COPYA, B,
5 \$ COPYB, Z, IA, JA, DESCA, WIN, WNEW,
7 \$ WORK, LWORK, LWORK1, IWORK, LIWORK,
8 \$ RESULT, TSTNRM, QTQNRM, NOUT )
9 *
10 * -- ScaLAPACK routine (version 1.7) --
11 * University of Tennessee, Knoxville, Oak Ridge National Laboratory,
12 * and University of California, Berkeley.
13 * May 1, 1997
14 *
15 * .. Scalar Arguments ..
16 LOGICAL WKNOWN
17 CHARACTER JOBZ, RANGE, UPLO
19 \$ LIWORK, LWORK, LWORK1, N, NOUT, RESULT
20 DOUBLE PRECISION ABSTOL, QTQNRM, THRESH, TSTNRM, VL, VU
21 * ..
22 * .. Array Arguments ..
23 INTEGER DESCA( * ), ICLUSTR( * ), IFAIL( * ),
24 \$ IWORK( * )
25 DOUBLE PRECISION A( * ), B( * ), COPYA( * ), COPYB( * ),
26 \$ GAP( * ), WIN( * ), WNEW( * ), WORK( * ),
27 \$ Z( * )
28 * ..
29 *
30 * Purpose
31 * =======
32 *
33 * PDGSEPSUBTST calls PDSYGVX and then tests the output of
34 * PDSYGVX
35 * If JOBZ = 'V' then the following two tests are performed:
36 * |AQ -QL| / (abstol + eps * norm(A) ) < THRESH
37 * |QT * Q - I| / eps * norm(A) < THRESH
38 * If WKNOWN then
39 * we check to make sure that the eigenvalues match expectations
40 * i.e. |WIN - WNEW(1+IPREPAD)| / (eps * |WIN|) < THRESH
41 * where WIN is the array of eigenvalues as computed by
42 * PDSYGVX when eigenvectors are requested
43 *
44 * Arguments
45 * =========
46 *
47 * NP = the number of rows local to a given process.
48 * NQ = the number of columns local to a given process.
49 *
50 * WKNOWN (global input) INTEGER
51 * .FALSE.: WIN does not contain the eigenvalues
52 * .TRUE.: WIN does contain the eigenvalues
53 *
54 * IBTYPE (global input) INTEGER
55 * Specifies the problem type to be solved:
56 * = 1: sub( A )*x = (lambda)*sub( B )*x
57 * = 2: sub( A )*sub( B )*x = (lambda)*x
58 * = 3: sub( B )*sub( A )*x = (lambda)*x
59 *
60 *
61 * JOBZ (global input) CHARACTER*1
62 * Specifies whether or not to compute the eigenvectors:
63 * = 'N': Compute eigenvalues only.
64 * = 'V': Compute eigenvalues and eigenvectors.
65 * Must be 'V' on first call to PDGSEPSUBTST
66 *
67 * RANGE (global input) CHARACTER*1
68 * = 'A': all eigenvalues will be found.
69 * = 'V': all eigenvalues in the interval [VL,VU]
70 * will be found.
71 * = 'I': the IL-th through IU-th eigenvalues will be found.
72 * Must be 'A' on first call to PDGSEPSUBTST
73 *
74 * UPLO (global input) CHARACTER*1
75 * Specifies whether the upper or lower triangular part of the
76 * symmetric matrix A is stored:
77 * = 'U': Upper triangular
78 * = 'L': Lower triangular
79 *
80 * N (global input) INTEGER
81 * Size of the matrix to be tested. (global size)
82 *
83 * VL (global input) DOUBLE PRECISION
84 * If RANGE='V', the lower bound of the interval to be searched
85 * for eigenvalues. Not referenced if RANGE = 'A' or 'I'.
86 *
87 * VU (global input) DOUBLE PRECISION
88 * If RANGE='V', the upper bound of the interval to be searched
89 * for eigenvalues. Not referenced if RANGE = 'A' or 'I'.
90 *
91 * IL (global input) INTEGER
92 * If RANGE='I', the index (from smallest to largest) of the
93 * smallest eigenvalue to be returned. IL >= 1.
94 * Not referenced if RANGE = 'A' or 'V'.
95 *
96 * IU (global input) INTEGER
97 * If RANGE='I', the index (from smallest to largest) of the
98 * largest eigenvalue to be returned. min(IL,N) <= IU <= N.
99 * Not referenced if RANGE = 'A' or 'V'.
100 *
101 * THRESH (global input) DOUBLE PRECISION
102 * A test will count as "failed" if the "error", computed as
103 * described below, exceeds THRESH. Note that the error
104 * is scaled to be O(1), so THRESH should be a reasonably
105 * small multiple of 1, e.g., 10 or 100. In particular,
106 * it should not depend on the precision (single vs. double)
107 * or the size of the matrix. It must be at least zero.
108 *
109 * ABSTOL (global input) DOUBLE PRECISION
110 * The absolute tolerance for the eigenvalues. An
111 * eigenvalue is considered to be located if it has
112 * been determined to lie in an interval whose width
113 * is "abstol" or less. If "abstol" is less than or equal
114 * to zero, then ulp*|T| will be used, where |T| is
115 * the 1-norm of the matrix. If eigenvectors are
116 * desired later by inverse iteration ("PDSTEIN"),
117 * "abstol" MUST NOT be bigger than ulp*|T|.
118 *
119 * A (local workspace) DOUBLE PRECISION array
120 * global dimension (N, N), local dimension (DESCA(DLEN_), NQ)
121 * A is distributed in a block cyclic manner over both rows
122 * and columns.
123 * See PDSYGVX for a description of block cyclic layout.
124 * The test matrix, which is then modified by PDSYGVX
126 *
127 * COPYA (local input) DOUBLE PRECISION array, dimension(N*N)
128 * COPYA holds a copy of the original matrix A
129 * identical in both form and content to A
130 *
131 * B (local workspace) DOUBLE PRECISION array, dim (N*N)
132 * global dimension (N, N), local dimension (LDA, NQ)
133 * A is distributed in a block cyclic manner over both rows
134 * and columns.
135 * The B test matrix, which is then modified by PDSYGVX
136 *
137 * COPYB (local input) DOUBLE PRECISION array, dim (N, N)
138 * COPYB is used to hold an identical copy of the array B
139 * identical in both form and content to B
140 *
141 * Z (local workspace) DOUBLE PRECISION array, dim (N*N)
142 * Z is distributed in the same manner as A
143 * Z contains the eigenvector matrix
144 * Z is used as workspace by the test routines
145 * PDGSEPCHK and PDSEPQTQ.
147 *
148 * IA (global input) INTEGER
149 * On entry, IA specifies the global row index of the submatrix
150 * of the global matrix A, COPYA and Z to operate on.
151 *
152 * JA (global input) INTEGER
153 * On entry, IA specifies the global column index of the submat
154 * of the global matrix A, COPYA and Z to operate on.
155 *
156 * DESCA (global/local input) INTEGER array of dimension 8
157 * The array descriptor for the matrix A, COPYA and Z.
158 *
159 * WIN (global input) DOUBLE PRECISION array, dimension (N)
160 * If .not. WKNOWN, WIN is ignored on input
161 * Otherwise, WIN() is taken as the standard by which the
162 * eigenvalues are to be compared against.
163 *
164 * WNEW (global workspace) DOUBLE PRECISION array, dimension (N)
165 * The eigenvalues as copmuted by this call to PDSYGVX
166 * If JOBZ <> 'V' or RANGE <> 'A' these eigenvalues are
167 * compared against those in WIN().
170 *
171 * IFAIL (global output) INTEGER array, dimension (N)
172 * If JOBZ = 'V', then on normal exit, the first M elements of
173 * IFAIL are zero. If INFO > 0 on exit, then IFAIL contains the
174 * indices of the eigenvectors that failed to converge.
175 * If JOBZ = 'N', then IFAIL is not referenced.
178 *
179 * ICLUSTR (global workspace) integer array, dimension (2*NPROW*NPCOL)
180 *
181 * GAP (global workspace) DOUBLE PRECISION array,
182 * dimension (NPROW*NPCOL)
183 *
184 * WORK (local workspace) DOUBLE PRECISION array, dimension (LWORK)
187 *
188 * LWORK (local input) INTEGER
189 * The actual length of the array WORK after padding.
190 *
191 *
192 * LWORK1 (local input) INTEGER
193 * The amount of real workspace to pass to PDSYGVX
194 *
195 * IWORK (local workspace) INTEGER array, dimension (LIWORK)
198 *
199 * LIWORK (local input) INTEGER
200 * The length of the array IWORK after padding.
201 *
202 * RESULT (global output) INTEGER
203 * The result of this call to PDSYGVX
204 * RESULT = -3 => This process did not participate
205 * RESULT = 0 => All tests passed
206 * RESULT = 1 => ONe or more tests failed
207 *
208 * TSTNRM (global output) DOUBLE PRECISION
209 * |AQ- QL| / |A|*N*EPS
210 *
211 * QTQNRM (global output) DOUBLE PRECISION
212 * |QTQ -I| / N*EPS
213 *
214 * .. Parameters ..
215 *
216 INTEGER BLOCK_CYCLIC_2D, DLEN_, DTYPE_, CTXT_, M_, N_,
217 \$ MB_, NB_, RSRC_, CSRC_, LLD_
218 PARAMETER ( BLOCK_CYCLIC_2D = 1, dlen_ = 9, dtype_ = 1,
219 \$ ctxt_ = 2, m_ = 3, n_ = 4, mb_ = 5, nb_ = 6,
220 \$ rsrc_ = 7, csrc_ = 8, lld_ = 9 )
221 DOUBLE PRECISION PADVAL, FIVE, NEGONE
222 PARAMETER ( PADVAL = 13.5285d+0, five = 5.0d+0,
223 \$ negone = -1.0d+0 )
225 PARAMETER ( IPADVAL = 927 )
226 * ..
227 * .. Local Scalars ..
228 LOGICAL MISSLARGEST, MISSSMALLEST
229 INTEGER I, IAM, INDIWRK, INFO, ISIZESUBTST, ISIZESYEVX,
230 \$ isizetst, j, m, maxeigs, maxil, maxiu, maxsize,
231 \$ minil, mq, mycol, myil, myrow, nclusters, np,
232 \$ npcol, nprow, nq, nz, oldil, oldiu, oldnz, res,
233 \$ sizechk, sizemqrleft, sizemqrright, sizeqrf,
234 \$ sizeqtq, sizesubtst, sizesyevx, sizetms,
235 \$ sizetst, valsize, vecsize
236 DOUBLE PRECISION EPS, ERROR, MAXERROR, MAXVU, MINERROR, MINVL,
237 \$ NORMWIN, OLDVL, OLDVU, ORFAC, SAFMIN
238 * ..
239 * .. Local Arrays ..
240 INTEGER DESCZ( DLEN_ ), DSEED( 4 ), ITMP( 2 )
241 * ..
242 * .. External Functions ..
243 *
244 LOGICAL LSAME
245 INTEGER NUMROC
246 DOUBLE PRECISION PDLAMCH
247 EXTERNAL LSAME, NUMROC, PDLAMCH
248 * ..
249 * .. External Subroutines ..
250 EXTERNAL blacs_gridinfo, descinit, dgamn2d, dgamx2d,
251 \$ dlacpy, igamn2d, igamx2d, pdchekpad, pdelset,
254 \$ slboot, sltimer
255 * ..
256 * .. Intrinsic Functions ..
257 INTRINSIC abs, max, min, mod
258 * ..
259 * .. Executable Statements ..
260 * This is just to keep ftnchek happy
261 IF( block_cyclic_2d*csrc_*ctxt_*dlen_*dtype_*lld_*mb_*m_*nb_*n_*
262 \$ rsrc_.LT.0 )RETURN
264 \$ sizemqrright, sizeqrf, sizetms, sizeqtq,
265 \$ sizechk, sizesyevx, isizesyevx, sizesubtst,
266 \$ isizesubtst, sizetst, isizetst )
267 *
268 tstnrm = negone
269 qtqnrm = negone
270 eps = pdlamch( desca( ctxt_ ), 'Eps' )
271 safmin = pdlamch( desca( ctxt_ ), 'Safe min' )
272 *
273 normwin = safmin / eps
274 IF( n.GE.1 )
275 \$ normwin = max( abs( win( 1 ) ), abs( win( n ) ), normwin )
276 *
277 * Make sure that we aren't using information from previous calls
278 *
279 nz = -13
280 oldnz = nz
281 oldil = il
282 oldiu = iu
283 oldvl = vl
284 oldvu = vu
285 *
286 DO 10 i = 1, lwork1, 1
287 work( i+iprepad ) = 14.3d+0
288 10 CONTINUE
289 DO 20 i = 1, liwork, 1
290 iwork( i+iprepad ) = 14
291 20 CONTINUE
292 *
293 DO 30 i = 1, n
294 wnew( i+iprepad ) = 3.14159d+0
295 30 CONTINUE
296 *
297 iclustr( 1+iprepad ) = 139
298 *
299 IF( lsame( jobz, 'N' ) ) THEN
300 maxeigs = 0
301 ELSE
302 IF( lsame( range, 'A' ) ) THEN
303 maxeigs = n
304 ELSE IF( lsame( range, 'I' ) ) THEN
305 maxeigs = iu - il + 1
306 ELSE
307 minvl = vl - normwin*five*eps - abstol
308 maxvu = vu + normwin*five*eps + abstol
309 minil = 1
310 maxiu = 0
311 DO 40 i = 1, n
312 IF( win( i ).LT.minvl )
313 \$ minil = minil + 1
314 IF( win( i ).LE.maxvu )
315 \$ maxiu = maxiu + 1
316 40 CONTINUE
317 *
318 maxeigs = maxiu - minil + 1
319 END IF
320 END IF
321 *
322 *
323 CALL descinit( descz, desca( m_ ), desca( n_ ), desca( mb_ ),
324 \$ desca( nb_ ), desca( rsrc_ ), desca( csrc_ ),
325 \$ desca( ctxt_ ), desca( lld_ ), info )
326 *
327 CALL blacs_gridinfo( desca( ctxt_ ), nprow, npcol, myrow, mycol )
328 indiwrk = 1 + iprepad + nprow*npcol + 1
329 *
330 iam = 1
331 IF( myrow.EQ.0 .AND. mycol.EQ.0 )
332 \$ iam = 0
333 *
334 * If this process is not involved in this test, bail out now
335 *
336 result = -3
337 IF( myrow.GE.nprow .OR. myrow.LT.0 )
338 \$ GO TO 150
339 result = 0
340 *
341 *
342 * DSEED is not used in this call to PDLASIZESYEVX, the
343 * following line just makes ftnchek happy.
344 *
345 dseed( 1 ) = 1
346 *
347 CALL pdlasizesyevx( wknown, range, n, desca, vl, vu, il, iu,
348 \$ dseed, win, maxsize, vecsize, valsize )
349 *
350 np = numroc( n, desca( mb_ ), myrow, 0, nprow )
351 nq = numroc( n, desca( nb_ ), mycol, 0, npcol )
352 mq = numroc( maxeigs, desca( nb_ ), mycol, 0, npcol )
353 *
354 CALL dlacpy( 'A', np, nq, copya, desca( lld_ ), a( 1+iprepad ),
355 \$ desca( lld_ ) )
356 *
357 CALL dlacpy( 'A', np, nq, copyb, desca( lld_ ), b( 1+iprepad ),
358 \$ desca( lld_ ) )
359 *
360 CALL pdfillpad( desca( ctxt_ ), np, nq, b, desca( lld_ ), iprepad,
362 *
363 CALL pdfillpad( desca( ctxt_ ), np, nq, a, desca( lld_ ), iprepad,
365 *
366 CALL pdfillpad( descz( ctxt_ ), np, mq, z, descz( lld_ ), iprepad,
368 *
371 *
372 CALL pdfillpad( desca( ctxt_ ), nprow*npcol, 1, gap, nprow*npcol,
374 *
377 *
380 *
383 *
384 CALL pifillpad( desca( ctxt_ ), 2*nprow*npcol, 1, iclustr,
386 *
387 * Make sure that PDSYGVX does not cheat (i.e. use answers
389 *
390 DO 60 i = 1, n, 1
391 DO 50 j = 1, maxeigs, 1
392 CALL pdelset( z( 1+iprepad ), i, j, desca, 13.0d+0 )
393 50 CONTINUE
394 60 CONTINUE
395 *
396 orfac = -1.0d+0
397 *
398 CALL slboot
399 CALL sltimer( 1 )
400 CALL sltimer( 6 )
401 CALL pdsygvx( ibtype, jobz, range, uplo, n, a( 1+iprepad ), ia,
402 \$ ja, desca, b( 1+iprepad ), ia, ja, desca, vl, vu,
403 \$ il, iu, abstol, m, nz, wnew( 1+iprepad ), orfac,
405 \$ lwork1, iwork( 1+iprepad ), liwork,
407 \$ gap( 1+iprepad ), info )
408 CALL sltimer( 6 )
409 CALL sltimer( 1 )
410 *
411 IF( thresh.LE.0 ) THEN
412 result = 0
413 ELSE
414 CALL pdchekpad( desca( ctxt_ ), 'PDSYGVX-B', np, nq, b,
417 *
418 CALL pdchekpad( desca( ctxt_ ), 'PDSYGVX-A', np, nq, a,
420 *
421 CALL pdchekpad( descz( ctxt_ ), 'PDSYGVX-Z', np, mq, z,
424 *
425 CALL pdchekpad( desca( ctxt_ ), 'PDSYGVX-WNEW', n, 1, wnew, n,
427 *
428 CALL pdchekpad( desca( ctxt_ ), 'PDSYGVX-GAP', nprow*npcol, 1,
431 *
432 CALL pdchekpad( desca( ctxt_ ), 'PDSYGVX-WORK', lwork1, 1,
435 *
436 CALL pichekpad( desca( ctxt_ ), 'PDSYGVX-IWORK', liwork, 1,
438 *
439 CALL pichekpad( desca( ctxt_ ), 'PDSYGVX-IFAIL', n, 1, ifail,
441 *
442 CALL pichekpad( desca( ctxt_ ), 'PDSYGVX-ICLUSTR',
443 \$ 2*nprow*npcol, 1, iclustr, 2*nprow*npcol,
445 *
446 *
447 * Since we now know the spectrum, we can potentially reduce MAXSIZE.
448 *
449 IF( lsame( range, 'A' ) ) THEN
450 CALL pdlasizesyevx( .true., range, n, desca, vl, vu, il, iu,
451 \$ dseed, wnew( 1+iprepad ), maxsize,
452 \$ vecsize, valsize )
453 END IF
454 *
455 *
456 * Check INFO
457 *
458 *
459 * Make sure that all processes return the same value of INFO
460 *
461 itmp( 1 ) = info
462 itmp( 2 ) = info
463 *
464 CALL igamn2d( desca( ctxt_ ), 'a', ' ', 1, 1, itmp, 1, 1, 1,
465 \$ -1, -1, 0 )
466 CALL igamx2d( desca( ctxt_ ), 'a', ' ', 1, 1, itmp( 2 ), 1, 1,
467 \$ 1, -1, -1, 0 )
468 *
469 *
470 IF( itmp( 1 ).NE.itmp( 2 ) ) THEN
471 IF( iam.EQ.0 )
472 \$ WRITE( nout, fmt = * )
473 \$ 'Different processes return different INFO'
474 result = 1
475 ELSE IF( mod( info, 2 ).EQ.1 .OR. info.GT.7 .OR. info.LT.0 )
476 \$ THEN
477 IF( iam.EQ.0 )
478 \$ WRITE( nout, fmt = 9999 )info
479 result = 1
480 ELSE IF( mod( info / 2, 2 ).EQ.1 .AND. lwork1.GE.maxsize ) THEN
481 IF( iam.EQ.0 )
482 \$ WRITE( nout, fmt = 9996 )info
483 result = 1
484 ELSE IF( mod( info / 4, 2 ).EQ.1 .AND. lwork1.GE.vecsize ) THEN
485 IF( iam.EQ.0 )
486 \$ WRITE( nout, fmt = 9996 )info
487 result = 1
488 END IF
489 *
490 *
491 IF( lsame( jobz, 'V' ) .AND. ( iclustr( 1+iprepad ).NE.
492 \$ 0 ) .AND. ( mod( info / 2, 2 ).NE.1 ) ) THEN
493 IF( iam.EQ.0 )
494 \$ WRITE( nout, fmt = 9995 )
495 result = 1
496 END IF
497 *
498 * Check M
499 *
500 IF( ( m.LT.0 ) .OR. ( m.GT.n ) ) THEN
501 IF( iam.EQ.0 )
502 \$ WRITE( nout, fmt = 9994 )
503 result = 1
504 ELSE IF( lsame( range, 'A' ) .AND. ( m.NE.n ) ) THEN
505 IF( iam.EQ.0 )
506 \$ WRITE( nout, fmt = 9993 )
507 result = 1
508 ELSE IF( lsame( range, 'I' ) .AND. ( m.NE.iu-il+1 ) ) THEN
509 IF( iam.EQ.0 )
510 \$ WRITE( nout, fmt = 9992 )
511 result = 1
512 ELSE IF( lsame( jobz, 'V' ) .AND.
513 \$ ( .NOT.( lsame( range, 'V' ) ) ) .AND. ( m.NE.nz ) )
514 \$ THEN
515 IF( iam.EQ.0 )
516 \$ WRITE( nout, fmt = 9991 )
517 result = 1
518 END IF
519 *
520 * Check NZ
521 *
522 IF( lsame( jobz, 'V' ) ) THEN
523 IF( lsame( range, 'V' ) ) THEN
524 IF( nz.GT.m ) THEN
525 IF( iam.EQ.0 )
526 \$ WRITE( nout, fmt = 9990 )
527 result = 1
528 END IF
529 IF( nz.LT.m .AND. mod( info / 4, 2 ).NE.1 ) THEN
530 IF( iam.EQ.0 )
531 \$ WRITE( nout, fmt = 9989 )
532 result = 1
533 END IF
534 ELSE
535 IF( nz.NE.m ) THEN
536 IF( iam.EQ.0 )
537 \$ WRITE( nout, fmt = 9988 )
538 result = 1
539 END IF
540 END IF
541 END IF
542 IF( result.EQ.0 ) THEN
543 *
544 * Make sure that all processes return the same # of eigenvalues
545 *
546 itmp( 1 ) = m
547 itmp( 2 ) = m
548 *
549 CALL igamn2d( desca( ctxt_ ), 'a', ' ', 1, 1, itmp, 1, 1, 1,
550 \$ -1, -1, 0 )
551 CALL igamx2d( desca( ctxt_ ), 'a', ' ', 1, 1, itmp( 2 ), 1,
552 \$ 1, 1, -1, -1, 0 )
553 *
554 IF( itmp( 1 ).NE.itmp( 2 ) ) THEN
555 IF( iam.EQ.0 )
556 \$ WRITE( nout, fmt = 9987 )
557 result = 1
558 ELSE
559 *
560 * Make sure that different processes return the same eigenvalues
561 *
562 DO 70 i = 1, m
563 work( i ) = wnew( i+iprepad )
564 work( i+m ) = wnew( i+iprepad )
565 70 CONTINUE
566 *
567 CALL dgamn2d( desca( ctxt_ ), 'a', ' ', m, 1, work, m, 1,
568 \$ 1, -1, -1, 0 )
569 CALL dgamx2d( desca( ctxt_ ), 'a', ' ', m, 1,
570 \$ work( 1+m ), m, 1, 1, -1, -1, 0 )
571 *
572 DO 80 i = 1, m
573 *
574 IF( result.EQ.0 .AND. ( abs( work( i )-work( m+
575 \$ i ) ).GT.five*eps*abs( work( i ) ) ) ) THEN
576 IF( iam.EQ.0 )
577 \$ WRITE( nout, fmt = 9986 )
578 result = 1
579 END IF
580 80 CONTINUE
581 END IF
582 END IF
583 *
584 * Make sure that all processes return the same # of clusters
585 *
586 IF( lsame( jobz, 'V' ) ) THEN
587 nclusters = 0
588 DO 90 i = 0, nprow*npcol - 1
589 IF( iclustr( 1+iprepad+2*i ).EQ.0 )
590 \$ GO TO 100
591 nclusters = nclusters + 1
592 90 CONTINUE
593 100 CONTINUE
594 itmp( 1 ) = nclusters
595 itmp( 2 ) = nclusters
596 *
597 CALL igamn2d( desca( ctxt_ ), 'a', ' ', 1, 1, itmp, 1, 1, 1,
598 \$ -1, -1, 0 )
599 CALL igamx2d( desca( ctxt_ ), 'a', ' ', 1, 1, itmp( 2 ), 1,
600 \$ 1, 1, -1, -1, 0 )
601 *
602 IF( itmp( 1 ).NE.itmp( 2 ) ) THEN
603 IF( iam.EQ.0 )
604 \$ WRITE( nout, fmt = 9985 )
605 result = 1
606 ELSE
607 *
608 * Make sure that different processes return the same clusters
609 *
610 DO 110 i = 1, nclusters
611 iwork( indiwrk+i ) = iclustr( i+iprepad )
612 iwork( indiwrk+i+nclusters ) = iclustr( i+iprepad )
613 110 CONTINUE
614 CALL igamn2d( desca( ctxt_ ), 'a', ' ', nclusters*2+1, 1,
615 \$ iwork( indiwrk+1 ), nclusters*2+1, 1, 1,
616 \$ -1, -1, 0 )
617 CALL igamx2d( desca( ctxt_ ), 'a', ' ', nclusters*2+1, 1,
618 \$ iwork( indiwrk+1+nclusters ),
619 \$ nclusters*2+1, 1, 1, -1, -1, 0 )
620 *
621 *
622 DO 120 i = 1, nclusters
623 IF( result.EQ.0 .AND. iwork( indiwrk+i ).NE.
624 \$ iwork( indiwrk+nclusters+i ) ) THEN
625 IF( iam.EQ.0 )
626 \$ WRITE( nout, fmt = 9984 )
627 result = 1
628 END IF
629 120 CONTINUE
630 *
631 IF( iclustr( 1+iprepad+nclusters*2 ).NE.0 ) THEN
632 IF( iam.EQ.0 )
633 \$ WRITE( nout, fmt = 9983 )
634 result = 1
635 END IF
636 END IF
637 END IF
638 *
639 *
640 CALL igamx2d( desca( ctxt_ ), 'a', ' ', 1, 1, result, 1, 1, 1,
641 \$ -1, -1, 0 )
642 IF( result.NE.0 )
643 \$ GO TO 150
644 *
645 * Note that a couple key variables get redefined in PDGSEPCHK
646 * as described by this table:
647 *
648 * PDGSEPTST name PDGSEPCHK name
649 * ------------- -------------
650 * COPYA A
651 * Z Q
652 * B B
653 * A C
654 *
655 *
656 IF( lsame( jobz, 'V' ) ) THEN
657 *
658 * Perform the residual check
659 *
660 CALL pdfillpad( desca( ctxt_ ), sizechk, 1, work, sizechk,
662 *
663 CALL pdgsepchk( ibtype, n, nz, copya, ia, ja, desca, copyb,
664 \$ ia, ja, desca, thresh, z( 1+iprepad ), ia,
665 \$ ja, descz, a( 1+iprepad ), ia, ja, desca,
667 \$ sizechk, tstnrm, res )
668 *
669 CALL pdchekpad( desca( ctxt_ ), 'PDGSEPCHK-WORK', sizechk,
671 \$ 4.3d+0 )
672 *
673 IF( res.NE.0 )
674 \$ result = 1
675 END IF
676 *
677 * Check to make sure that we have the right eigenvalues
678 *
679 IF( wknown ) THEN
680 *
681 * Set up MYIL if necessary
682 *
683 myil = il
684 *
685 IF( lsame( range, 'V' ) ) THEN
686 myil = 1
687 minil = 1
688 maxil = n - m + 1
689 ELSE
690 IF( lsame( range, 'A' ) ) THEN
691 myil = 1
692 END IF
693 minil = myil
694 maxil = myil
695 END IF
696 *
697 * Find the largest difference between the computed
698 * and expected eigenvalues
699 *
700 minerror = normwin
701 *
702 DO 140 myil = minil, maxil
703 maxerror = 0
704 *
705 * Make sure that we aren't skipping any important eigenvalues
706 *
707 misssmallest = .true.
708 IF( .NOT.lsame( range, 'V' ) .OR. ( myil.EQ.1 ) )
709 \$ misssmallest = .false.
710 IF( misssmallest .AND. ( win( myil-1 ).LT.vl+normwin*
711 \$ five*thresh*eps ) )misssmallest = .false.
712 misslargest = .true.
713 IF( .NOT.lsame( range, 'V' ) .OR. ( myil.EQ.maxil ) )
714 \$ misslargest = .false.
715 IF( misslargest .AND. ( win( myil+m ).GT.vu-normwin*five*
716 \$ thresh*eps ) )misslargest = .false.
717 IF( .NOT.misssmallest ) THEN
718 IF( .NOT.misslargest ) THEN
719 *
720 * Make sure that the eigenvalues that we report are OK
721 *
722 DO 130 i = 1, m
723 error = abs( win( i+myil-1 )-wnew( i+iprepad ) )
724 maxerror = max( maxerror, error )
725 130 CONTINUE
726 *
727 minerror = min( maxerror, minerror )
728 END IF
729 END IF
730 140 CONTINUE
731 *
732 *
733 * If JOBZ = 'V' and RANGE='A', we might be comparing
734 * against our estimate of what the eigenvalues ought to
735 * be, rather than comparing against what PxSYGVX computed
736 * last time around, so we have to be more generous.
737 *
738 IF( lsame( jobz, 'V' ) .AND. lsame( range, 'A' ) ) THEN
739 IF( minerror.GT.normwin*five*five*thresh*eps ) THEN
740 IF( iam.EQ.0 )
741 \$ WRITE( nout, fmt = 9997 )minerror, normwin
742 result = 1
743 END IF
744 ELSE
745 IF( minerror.GT.normwin*five*thresh*eps ) THEN
746 IF( iam.EQ.0 )
747 \$ WRITE( nout, fmt = 9997 )minerror, normwin
748 result = 1
749 END IF
750 END IF
751 END IF
752 *
753 *
754 * Make sure that the IL, IU, VL and VU were not altered
755 *
756 IF( il.NE.oldil .OR. iu.NE.oldiu .OR. vl.NE.oldvl .OR. vu.NE.
757 \$ oldvu ) THEN
758 IF( iam.EQ.0 )
759 \$ WRITE( nout, fmt = 9982 )
760 result = 1
761 END IF
762 *
763 IF( lsame( jobz, 'N' ) .AND. ( nz.NE.oldnz ) ) THEN
764 IF( iam.EQ.0 )
765 \$ WRITE( nout, fmt = 9981 )
766 result = 1
767 END IF
768 *
769 END IF
770 *
771 * All processes should report the same result
772 *
773 CALL igamx2d( desca( ctxt_ ), 'a', ' ', 1, 1, result, 1, 1, 1, -1,
774 \$ -1, 0 )
775 *
776 150 CONTINUE
777 *
778 *
779 RETURN
780 *
781 9999 FORMAT( 'PDSYGVX returned INFO=', i7 )
782 9998 FORMAT( 'PDSEPQTQ returned INFO=', i7 )
783 9997 FORMAT( 'PDGSEPSUBTST minerror =', d11.2, ' normwin=', d11.2 )
784 9996 FORMAT( 'PDSYGVX returned INFO=', i7,
785 \$ ' despite adequate workspace' )
786 9995 FORMAT( .NE..NE.'ICLUSTR(1)0 but mod(INFO/2,2)1' )
787 9994 FORMAT( 'M not in the range 0 to N' )
788 9993 FORMAT( 'M not equal to N' )
789 9992 FORMAT( 'M not equal to IU-IL+1' )
790 9991 FORMAT( 'M not equal to NZ' )
791 9990 FORMAT( 'NZ > M' )
792 9989 FORMAT( 'NZ < M' )
793 9988 FORMAT( 'NZ not equal to M' )
794 9987 FORMAT( 'Different processes return different values for M' )
795 9986 FORMAT( 'Different processes return different eigenvalues' )
796 9985 FORMAT( 'Different processes return ',
797 \$ 'different numbers of clusters' )
798 9984 FORMAT( 'Different processes return different clusters' )
799 9983 FORMAT( 'ICLUSTR not zero terminated' )
800 9982 FORMAT( 'IL, IU, VL or VU altered by PDSYGVX' )
801 9981 FORMAT( 'NZ altered by PDSYGVX with JOBZ=N' )
802 *
803 * End of PDGSEPSUBTST
804 *
805 END
subroutine pichekpad(ICTXT, MESS, M, N, A, LDA, IPRE, IPOST, CHKVAL)
max
#define max(A, B)
Definition: pcgemr.c:180
pdlasizesyevx
subroutine pdlasizesyevx(WKNOWN, RANGE, N, DESCA, VL, VU, IL, IU, ISEED, WIN, MAXSIZE, VECSIZE, VALSIZE)
Definition: pdlasizesyevx.f:5
pdgsepsubtst
subroutine pdgsepsubtst(WKNOWN, IBTYPE, JOBZ, RANGE, UPLO, N, VL, VU, IL, IU, THRESH, ABSTOL, A, COPYA, B, COPYB, Z, IA, JA, DESCA, WIN, WNEW, IFAIL, ICLUSTR, GAP, IPREPAD, IPOSTPAD, WORK, LWORK, LWORK1, IWORK, LIWORK, RESULT, TSTNRM, QTQNRM, NOUT)
Definition: pdgsepsubtst.f:9
subroutine pdchekpad(ICTXT, MESS, M, N, A, LDA, IPRE, IPOST, CHKVAL)
sltimer
subroutine sltimer(I)
Definition: sltimer.f:47
pdgsepchk
subroutine pdgsepchk(IBTYPE, MS, NV, A, IA, JA, DESCA, B, IB, JB, DESCB, THRESH, Q, IQ, JQ, DESCQ, C, IC, JC, DESCC, W, WORK, LWORK, TSTNRM, RESULT)
Definition: pdgsepchk.f:6
descinit
subroutine descinit(DESC, M, N, MB, NB, IRSRC, ICSRC, ICTXT, LLD, INFO)
Definition: descinit.f:3
slboot
subroutine slboot()
Definition: sltimer.f:2
subroutine pdfillpad(ICTXT, M, N, A, LDA, IPRE, IPOST, CHKVAL)
pdlasizegsep
subroutine pdlasizegsep(DESCA, IPREPAD, IPOSTPAD, SIZEMQRLEFT, SIZEMQRRIGHT, SIZEQRF, SIZETMS, SIZEQTQ, SIZECHK, SIZESYEVX, ISIZESYEVX, SIZESUBTST, ISIZESUBTST, SIZETST, ISIZETST)
Definition: pdlasizegsep.f:8
pdelset
subroutine pdelset(A, IA, JA, DESCA, ALPHA)
Definition: pdelset.f:2
pdsygvx
subroutine pdsygvx(IBTYPE, JOBZ, RANGE, UPLO, N, A, IA, JA, DESCA, B, IB, JB, DESCB, VL, VU, IL, IU, ABSTOL, M, NZ, W, ORFAC, Z, IZ, JZ, DESCZ, WORK, LWORK, IWORK, LIWORK, IFAIL, ICLUSTR, GAP, INFO)
Definition: pdsygvx.f:6
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| 2.546875
| 3
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CC-MAIN-2024-18
|
latest
|
en
| 0.38709
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http://www.pzzls.com/all_puzzles_and_riddles/most_popular_on_top/40.html
| 1,547,644,781,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2019-04/segments/1547583657470.23/warc/CC-MAIN-20190116113941-20190116135941-00604.warc.gz
| 366,749,486
| 4,438
|
Logic puzzles, riddles, math puzzles and brainteasers - pzzls.com
Vandaag is het 16 January 2019
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# Elevator woman - puzzling puzzle
Difficulty:
Rating: 2.6/5.0
A woman lived in an apartment building on the seventh floor. Every morning she took the lift to the ground floor and went to her work. But when she came home, she took the lift to the fourth floor and then continued by walking the stairs up. Why would she do that?
# Apple - math puzzle
Difficulty:
Rating: 2.6/5.0
Eric has a perfect spherical shaped apple. With a cilindrically shaped drill bit he removes the core of the apple. A cilindrically shaped hole remains, the center of this hole is exactly at the center of the apple. Eric measures the length of the hole with a piece of rope. It turns out to be exactly 8 cm. What is the volume of the remaining apple?
# Who me? - puzzling puzzle
Difficulty:
Rating: 2.6/5.0
| 251
| 949
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| 2.609375
| 3
|
CC-MAIN-2019-04
|
latest
|
en
| 0.956938
|
https://marinegyaan.com/exercise-28-azimuth-sun/
| 1,718,595,438,000,000,000
|
text/html
|
crawl-data/CC-MAIN-2024-26/segments/1718198861696.51/warc/CC-MAIN-20240617024959-20240617054959-00676.warc.gz
| 355,398,386
| 29,930
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# EXERCISE 28 — AZIMUTH SUN (Numerical Solution)
###### NOTE:
1. If LHA<180, P = LHA and If LHA>180, P = 360-LHA
2. Naming of A, B, C: A is named opposite to the latitude when LHA is between 270 & 90 and same as latitude when LHA is between 90 & 270. B is named same as declination. For C, if A & B are of same names, add and retain names. If of contrary names then subtract and retain name of larger one.
We know that:
| 131
| 422
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| 3.125
| 3
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CC-MAIN-2024-26
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latest
|
en
| 0.827925
|
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