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https://stats.stackexchange.com/questions/179882/is-the-posterior-distribution-on-means-in-a-bayesian-gaussian-mixture-model-with | 1,718,472,309,000,000,000 | text/html | crawl-data/CC-MAIN-2024-26/segments/1718198861605.77/warc/CC-MAIN-20240615155712-20240615185712-00399.warc.gz | 489,481,828 | 40,017 | # Is the posterior distribution on means in a Bayesian Gaussian mixture model with symmetric priors Gaussian?
I am reading through a document on learning Gaussian mixture models in Infer.NET. They assume the data is generated from 2 Gaussians where the prior distribution on means is Gaussian and the prior distribution on precisions is a Whishart distribution. The prior distribution on the mixture is a Dirichlet distribution. All of these priors are symmetric in the two Gaussians.
They do some inference on some data, and they get back that the posterior distribution on each of the two means is the same Gaussian. They then go on to talk about how to break the symmetry in the model so that the means can converge to different Gaussians.
How can it possibly be that the posteriors on the means are Gaussian? If I observe a million samples from a Gaussian Mixture Model (say unbeknownst to me the data is created by choosing with equal probability a normal distribution of mean 0 and variance 1 or a normal distribution with mean 100 and variance 1) it should be ABSOLUTELY CLEAR what the two means and standard deviations are. The symmetry of course means that the model doesn't know whether the first or the second Gaussian has mean 0 or mean 100, so shouldn't the posterior have two peaks, one near 0 and one near 100? If so, it's obviously not Gaussian.
I would appreciate any help in this matter.
• TLDR; Could this be possibly related: stats.stackexchange.com/questions/178321/… ?
– Tim
Commented Nov 3, 2015 at 14:56
• @time : Yes. It is exactly the label-switching problem I am talking about. Reading through the comments and papers in the link you gave makes it clear that the posteriors in this situation can have multiple modes and so, in general are not Gaussian. In my opinion, this exposes a bug in Infer.NET. Commented Nov 3, 2015 at 17:18
• TLDR: there is probably a misreading in here. It is well known that the posteriors on individual mixture components can have conjugate priors (including gaussians). Commented Jun 27, 2017 at 11:31 | 466 | 2,062 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2024-26 | latest | en | 0.939202 |
https://picaxeforum.co.uk/threads/spot-the-possible-optimisation.9276/ | 1,652,930,624,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522741.25/warc/CC-MAIN-20220519010618-20220519040618-00464.warc.gz | 537,465,846 | 15,686 | # Spot the possible optimisation
#### beny1949
##### Senior Member
I dont know if anyone remembers, but last year about may/june time, i was working on a micromouse. things didn't go so well at the competition, but i have been having another go and things are looking much better now in the fact that my mouse now works!
The main issue with it now is that it runs very slowely when trying to do the bellman flood fill algorithm.
I was wondering if anyone can see any way that I could shorten this loop whilst performing the same function. this loop may have to be run over 100 times when it gets to the middle of a maze, so even shaving a few microseconds here and there will make a huge difference to the overal time taken
The loop goes through an array of 256 and checks that each cell has a value which is one more than its lowest neibouring cell. there is alot more information about this at: http://micromouse.cannock.ac.uk/maze/solving.htm
Here is the code:
Code:
`````` for FloodP = \$00 to \$FF
'get flood data
hi2cin FloodP,(FloodD)
Temp2 = FloodD
'get wall data for the current cell
Temp1 = WallD & \$01 'North
if Temp1 = 0 then
let WallP = FloodP + 16 'point north
hi2cin WallP,(Temp1) 'read the flood value
if Temp1 < FloodD then
let FloodD = Temp1 + 1
endif
endif
Temp1 = WallD & \$04 'East
if Temp1 = 0 then
let WallP = FloodP + 1
hi2cin WallP,(Temp1) 'read the flood value
if Temp1 < FloodD then
let FloodD = Temp1 + 1
endif
endif
Temp1 = WallD & \$10 'South
if Temp1 = 0 then
let WallP = FloodP - 16
hi2cin WallP,(Temp1) 'read the flood value
if Temp1 < FloodD then
let FloodD = Temp1 + 1
endif
endif
Temp1 = WallD & \$40 'west
if Temp1 = 0 then
let WallP = FloodP - 1
hi2cin WallP,(Temp1) 'read the flood value
if Temp1 < FloodD then
let FloodD = Temp1 + 1
endif
endif
hi2cout FloodP,(FloodD)'write the number which is one bigger than the nearest cell
if Temp2 != FloodD then: Let Change = 1: endif
next 'go and look a``````
If anyone needs anymore details let me know. I am not sure really how much background anyone would need to be able to suggest improvements.
#### hippy
##### Technical Support
Staff member
I don't see any obvious scope for optimisation so it might have to be algorithm / technique design which changes.
Just realised the X1's only have 128 bytes of scratchpad, so the italicised here isn't particularly useful at the moment ...
If you can use the 256 byte scratchpad rather than external Eeprom that would speed things up considerably.
It may be possible to dump current scratchpad to Eeprom, load it with flood data, process and save that and reload later and still save time as there is a lot of processing to be done. Premature Eeprom burnout is the price you'd have to accept, or use 256x8 I2C RAM for the temporary copy.
The 40X2 can work as an I2C Slave so one running a "Do:Loop", do nothing program should act as a 256x8 I2C RAM as actual I2C RAM seems hard to find these days.
You could also use the 40X2 as the flood fill processor. Dump the initial data to that, tell it to process, and when you come to read later it will have the results you want ready and waiting. You can run that at 20MHz and I'd be tempted to try overclocking it to 24MHz/32MHz.
I'd also consider over-clocking all PICAXE's in the system.
If I recall right, any PICmicro/PICAXE set for running from an external crystal/resonator can be driven by an external oscillator module instead, so you'd only need one module for the entire board ( and some careful tracking ). As X1's / X2's have selectable clocking they'll all revert to 4MHz for programming.
If someone has a frequency generator or some oscillator modules it would be interesting to see how far the 28X1 and 40X1 can be over-clocked. Sparkfun pushed a 16F873A to 32MHz without even bothering with the XTAL capacitors. 32MHz is handy because it means serial comms will still work with other PICAXE's, but no reason not to strive higher for fastest operation. I2C shouldn't really care no matter what peculiar value is used, 33.8688MHz crystals are a common slavage from PC cards.
http://www.sparkfun.com/commerce/present.php?p=Overclocking a PIC
Another possibility is to use a non-PICAXE slave for the flood-fill part of the operation.
#### beny1949
##### Senior Member
Thanks very much Hippy!
I am actually using I2c Ram, i have a couple of PCF8750P's which i got from Rev-ed some time ago, kind of handy seen as they dont have them at the moment!
I have spent some time tring to think of another way to do this, which a picaxe can handle, but i think i am going to just have to wait for the X2's to go any faster. I am currently running at 16mhtz. Overclocking to a non standard/supported speed might be a bit of an issue for me currently as I am using the servo command. switching to an interupt and pulse out isn't much hassle having said that.
A quick question about the RAM, in the data sheet, it says that the maximum clock speed is 100Khtz. I was running it at that speed, but i have started to run it at 400Khtz. Is it safe to push it further, and is there anything i can do to make it more reliable at high speeds?
Ben
#### Dippy
##### Moderator
I was doing something using FRAM to hold my 'arrays'. The faster clock speed had only a sligt effect in my application as most of the delay was overhead involved in the actual command.
Also had to reduce the value of the pullups.
#### hippy
##### Technical Support
Staff member
At least with I2C RAM you cannot damage it so I'd simply try writing and reading changing bit patterns to see when it falls over then back-off. I don't think there's a lot to be done to make it more reliable at high speed other than keep the capacitance low and tweak the pull-ups. Watching the bus and signal slew on a scope may be the best trial and error method.
Code:
``````Do
b0 = b0 * 2 : If b0 = 0 Then : b0 = 1 : End If
For b1 = 0 to 255
Hi2cOut b1,(b0)
b0 = b0 + 1
Next
For b1 = 0 to 255
Hi2cIn b1,(b2)
If b2 <> b0 Then : SerTxd( "Oops - Failed !" ) : End If
b0 = b0 + 1
Next
Loop``````
I can appreciate the issue with non-standard / over-clocked speeds and Servo. One approach may be another 28X1 I2C slave handling the Servo commands and perhaps other I/O ? That can run at 16MHz, the main controller at anything providing I2C works. I2C handily sidesteps the mismatched SerIn/SerOut baud rate issues.
The X1's auto-adjust to keep Servo running at the right speed regardless of crystal so there may be somewhere in SFR to poke to allow non-standard speeds and correct servo refresh operation.
Massive parallelism using multiple PICAXE's is the key IMO for maximimising throughput in a PICAXE system but not always cheap and it does require considerable design effort at times. It's often easy and inexpensive to throw in another PICAXE-08M to work as a front-end or back-end for wireless, serial or servo handling and to overcome delays like reading DS18B20 temperatures, but more complicated sub-systems take more effort, require higher-end PICAXE's and there's a limit to scaleability.
Maybe a dual-PICAXE system where the slave starts to calculates the flood-fill as a square is entered and by the time the master needs to use that information it will have the data ready ? It only needs to pass back a single byte saying which way to go.
Last edited:
#### beny1949
##### Senior Member
I will experiment!
Thanks very much,
Ben
#### Brietech
##### Senior Member
I know it is a bit blasphemous on this forum, but if all you genuinely care about is speed, and you aren't limited to a Picaxe, I have a Basic-X 24p chip you could have for the cost of shipping (which is probably pretty low, even though I'm in NYC). It has the following specs:
Speed 83,000 Basic instructions per second
EEPROM 32K bytes (User program and data storage)
Max program length 8000+ lines of Basic code
RAM 400 bytes
Available I/O pins 21 (16 standard + 2 serial only + 3 accessed outside standard dip pin area)
Analog Inputs (ADCs) 8 (8 of the 16 standard I/O pins can individually function as 10bit ADCs or standard digital I/Os or a mixture of both)
Serial I/O speed 1200 - 460.8K Baud
Floating point math 32bit x 32bit floating point math built-in
Programming interface High speed Serial
Physical Package 24 pin DIP module
I think they normally cost about \$50 (+ shipping from the states), but I've had this lying around forever and could never think of anything fun for it. It should chew through your calculations in a blink, without being too much more complicated to program.
#### Brietech
##### Senior Member
Here is the website: http://www.basicx.com/
I'd have to make sure it was still in working order, but it probably is =)
#### beny1949
##### Senior Member
That would be very kind of you indeed!!
The next competition for this mousy is minos 08 which is this weekend, so i guess its no good for then, however i am very keen to get this mouse working to its full potential!
If you are sure that you do not have a use for your basic X, I am sure that would make a great co-processor!
Ben
#### hippy
##### Technical Support
Staff member
If you're not sticking entirely with the PICAXE there are many more options available. The Parallax Propeller will give you eight processors on one chip, equivalent of 130,000 basic instructions per second ( plus 24MIPS Assembler ) per processor. So 50 times the oomph of a 16MHz 40X1 in total and as all data is in shared 32KB Ram almost zero comms overhead. \$13+I2c Eeprom.
Last edited:
#### BCJKiwi
##### Senior Member
You've not indicated what i2c bus speed you are using nor which PE version.
There are issues with the i2c commands in earlier versions of PE which did not in fact run the i2c bus at the speed selected but always in slow but this was fixed with PE 5.2.0
If you are are only running the bus at 100 (slow) then there will be room for improvement by running the bus faster (provided the external eeprom you have supports it of course).
#### Dippy
##### Moderator
Going to another chip or compiler is pretty bloomin' obvious. I didn't mention it as this is a PICAXE forum. So keep the free advertising to a minimum eh?
#### beny1949
##### Senior Member
BCJKiwi,
I am running a PCF8570P RAM chip, which according to the data sheet is only upto 100Khtz, But i am running it at 400K at the moment without having any problems. I think I am now past the point of being able to make any changes for this weekend, after all i do have a worker!
thanks very much,
Ben
#### Dippy
##### Moderator
beny: "...after all i do have a worker!"
- if you've got got a worker, then get him to do some work and you can do your project.
Do you have a butler as well?
#### beny1949
##### Senior Member
no unfortuately not, although a buttler could be classed a worker.
Thanks all the same
Dippy:" this is a PICAXE forum." | 2,810 | 10,852 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2022-21 | latest | en | 0.89458 |
http://cran.r-project.org/web/packages/aghq/vignettes/aghq.html | 1,618,612,244,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038092961.47/warc/CC-MAIN-20210416221552-20210417011552-00319.warc.gz | 21,071,269 | 51,221 | # aghq: Adaptive Gauss-Hermite Quadrature for Bayesian inference
library(aghq)
library(dplyr)
#>
#> Attaching package: 'dplyr'
#> The following objects are masked from 'package:stats':
#>
#> filter, lag
#> The following objects are masked from 'package:base':
#>
#> intersect, setdiff, setequal, union
library(ggplot2)
This vignette describes the aghq package for Bayesian inference using Adaptive Gauss Hermite Quadrature as described by Bilodeau, Stringer, and Tang (2020). It is taken almost verbatim from Section 2 (Basic Use) of Stringer (2020). It shows how to use the basic functions of the package to implement a very simple example of Bayesian inference using adaptive quadrature. For six challenging and illuminating examples, consult Stringer (2020).
# A simple example
The following conjugate model is used by Bilodeau, Stringer, and Tang (2020): \begin{aligned} Y_i | \lambda &\overset{ind}{\sim} \text{Poisson}(\lambda), i\in[n], \\ \lambda &\sim \text{Exponential}(1), \lambda > 0, \\ \implies \lambda | Y &\sim \text{Gamma}\left(1 + \sum_{i=1}^{n}y_{i},n+1\right), \\ \pi(\lambda,Y) &= \frac{\lambda^{\sum_{i=1}^{n}y_{i}}e^{-n\lambda}}{\prod_{i=1}^{n}y_{i}!}\times e^{-\lambda}, \\ \pi(Y) &= \frac{\Gamma\left(1 + \sum_{i=1}^{n}y_{i}\right)}{(n+1)^{\left(1 + \sum_{i=1}^{n}y_{i}\right)}\prod_{i=1}^{n}y_{i}!} \end{aligned} We apply a transformation: \begin{aligned} \eta &= \log\lambda, \\ \implies \log\pi(\eta,Y) &= \eta\sum_{i=1}^{n}y_{i} - (n+1)e^{\eta} - \sum_{i=1}^{n}\log(y_{i}!) + \eta, \\ \end{aligned} which, because the jacobian $$|d\lambda/d\eta|$$ is included, does not change the value of $$\pi(Y)$$.
To approximate $$\pi(Y)$$ and then make Bayesian inferences based on $$\pi(\lambda|Y)$$, first install and load the package:
# Stable version:
# install.packages("aghq")
# Or devel version:
# devtools::install_github("awstringer1/aghq")
We will simulate some data from the model:
set.seed(84343124)
y <- rpois(10,5) # True lambda = 5, n = 10
The main function is the aghq::aghq function. The user supplies a list containing the log-posterior and its first two derivatives:
# Define the log posterior, log(pi(eta,y)) here
logpietay <- function(eta,y) {
sum(y) * eta - (length(y) + 1) * exp(eta) - sum(lgamma(y+1)) + eta
}
# The objective function has to take a single argument, eta, so pass
# in the data:
objfunc <- function(x) logpietay(x,y)
# Simple numerical derivatives suffice here:
objfunchess <- function(x) numDeriv::hessian(objfunc,x)
# Now create the list to pass to aghq()
funlist <- list(
fn = objfunc,
he = objfunchess
)
Now perform the quadrature. The only other required input is the number of quadrature points, $$k$$, and a starting value for the optimization:
# AGHQ with k = 3
# Use eta = 0 as a starting value
thequadrature <- aghq::aghq(ff = funlist,k = 3,startingvalue = 0)
The object thequadrature has class aghq, with summary and plot methods:
summary(thequadrature)
#> AGHQ on a 1 dimensional posterior with 3 quadrature points
#>
#> The posterior mode is: 1.493925
#>
#> The log of the normalizing constant/marginal likelihood is: -23.32123
#>
#> The posterior Hessian at the mode is:
#> [,1]
#> [1,] 49
#>
#> The covariance matrix used for the quadrature is...
#> [,1]
#> [1,] 0.02040816
#>
#> ...and its Cholesky is:
#> [,1]
#> [1,] 0.1428571
#>
#> Here are some moments and quantiles for theta:
#>
#> mean median mode sd 2.5% 97.5%
#> theta1 1.483742 1.482532 1.493925 0.1424943 1.204135 1.762909
plot(thequadrature)
The actual object is a list with elements (see for notation):
• normalized_posterior: the output of the aghq::normalize_posterior function, which is itself a list with elements:
• nodesandweights: a dataframe containing the nodes and weights from the quadrature and the normalized and un-normalized log posterior at the nodes,
• thegrid: a NIGrid object from the mvQuad package, see ?mvQuad::createNIGrid,
• lognormconst: the numeric value of the approximation of the log of the normalizing constant $$\pi(Y)$$.
• marginals: a list of the same length as startingvalue of which element j is the result of calling aghq::marginal_posterior with that j. This is a tbl_df/tbl/data.frame containing the normalized marginal posterior for the $$j^{th}$$ element of $$\eta$$. In this one-dimensional example, this is exactly the same information that is present in thequadrature$normalized_logpost$nodesandweights.
• optresults: information and results from the optimization of $$\log\pi^{*}(\eta,Y)$$. This is itself a list with elements:
• ff: the function list passed to aghq,
• mode: the mode of the log-posterior,
• hessian: the negative Hessian of the log-posterior, at the mode, and
• convergence: the convergence code, specific to the optimizer used.
We can observe the structure of the object and compare the results to the truth for this simple example:
# The normalized posterior:
thequadrature$normalized_posterior$nodesandweights
#> theta1 weights logpost logpost_normalized
#> 1 1.246489 0.2674745 -23.67784 -0.3566038
#> 2 1.493925 0.2387265 -22.29426 1.0269677
#> 3 1.741361 0.2674745 -23.92603 -0.6047982
# The log normalization constant:
options(digits = 6)
thequadrature$normalized_posterior$lognormconst
#> [1] -23.3212
# Compare to the truth:
lgamma(1 + sum(y)) - (1 + sum(y)) * log(length(y) + 1) - sum(lgamma(y+1))
#> [1] -23.3195
# Quite accurate with only n = 10 and k = 3; this example is very simple.
# The mode found by the optimization:
thequadrature$optresults$mode
#> [1] 1.49393
# The true mode:
log((sum(y) + 1)/(length(y) + 1))
#> [1] 1.49393
options(digits = 3)
Of course, in practical problems, the true mode and normalizing constant won’t be known.
The package provides further routines for computing moments, quantiles, and distributions/densities. These are especially useful when working with transformations and we are in the example, since interest here is in the original parameter $$\lambda = e^{\eta}$$. The main functions are:
• compute_pdf_and_cdf: compute the density and cumulative distribution function for a marginal posterior distribution of a variable and (optionally) a smooth monotone transformation of that variable,
• compute_moment: compute the posterior moment of any function,
• compute_quantiles: compute posterior quantiles for a marginal posterior distribution.
We will consider several examples. To compute the approximate marginal posterior for $$\lambda$$, $$\widetilde{\pi}(\lambda|Y)$$, first compute the marginal posterior for $$\eta$$ and then transform: $$$\widetilde{\pi}(\lambda|Y) = \widetilde{\pi}(\eta|Y)|\frac{d\eta}{d\lambda}|$$$
The aghq::compute_pdf_and_cdf function has an option, transformation, which allows the user to specify a parameter transformation that they would like the marginal density of. The user specifies functions fromtheta and totheta which convert from and to the transformed scale (on which quadrature was done), and the function returns the marginal density on both scales, making use of a numerically differentiated jacobian. This is all done as follows:
# Compute the pdf for eta
pdfwithlambda <- compute_pdf_and_cdf(
thequadrature$marginals[[1]], transformation = list( totheta = function(x) log(x), fromtheta = function(x) exp(x) ) ) # Plot along with the true posterior lambdapostplot <- pdfwithlambda %>% ggplot(aes(x = transparam,y = pdf_transparam)) + theme_classic() + geom_line() + stat_function(fun = dgamma, args = list(shape = 1+sum(y),rate = 1 + length(y)), linetype = 'dashed') + labs(x = expression(lambda),y = "Density") lambdapostplot We may compute the posterior mean of $$\lambda = e^{\eta}$$, $$\EE(\lambda|Y)$$, using the compute_moment function. This can also be used to confirm that the posterior is, in fact, properly normalized: options(digits = 6) # Check if the posterior integrates to 1, by computing the "moment" of "1": compute_moment(thequadrature$normalized_posterior,
ff = function(x) 1)
#> [1] 1
# Posterior mean for eta:
compute_moment(thequadrature$normalized_posterior, ff = function(x) x) #> [1] 1.48374 # Posterior mean for lambda = exp(eta) compute_moment(thequadrature$normalized_posterior,
ff = function(x) exp(x))
#> [1] 4.45441
# Compare to the truth:
(sum(y) + 1)/(length(y) + 1)
#> [1] 4.45455
options(digits = 3)
Quantiles are computed using the compute_quantiles function. For example, to get the first and third percentiles along with median and first and third quartiles:
# Get the quantiles for eta:
etaquants <- compute_quantiles(
q = c(.01,.25,.50,.75,.99)
)
# The quantiles for lambda = exp(eta) are obtained directly:
exp(etaquants)
#> 1% 25% 50% 75% 99%
#> 3.17 4.00 4.40 4.85 6.15
# and compared with the truth:
qgamma(c(.01,.25,.50,.75,.99),shape = 1+sum(y),rate = 1+length(y))
#> [1] 3.11 4.01 4.42 4.87 6.07
The estimation of quantiles, especially extreme quantiles, is much more difficult than that of moments, and this is reflected in the small differences in accuracy observed between the two in this example.
For more details, see Bilodeau, Stringer, and Tang (2020) and Stringer (2020).
# References
Bilodeau, Blair, Alex Stringer, and Yanbo Tang. 2020. “Higher Order Accurate Approximate Bayesian Inference Using Adaptive Quadrature.” In Preparation.
Stringer, Alex. 2020. “Implementing Adaptive Quadrature for Bayesian Inference: The Aghq Package.” In Preparation. | 2,779 | 9,451 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 2, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 3, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2021-17 | latest | en | 0.644405 |
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Categories: Miscellaneous | 701 | 3,110 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2024-30 | latest | en | 0.965042 |
https://www.numbersaplenty.com/6868 | 1,696,116,843,000,000,000 | text/html | crawl-data/CC-MAIN-2023-40/segments/1695233510730.6/warc/CC-MAIN-20230930213821-20231001003821-00836.warc.gz | 986,373,697 | 3,229 | Search a number
6868 = 2217101
BaseRepresentation
bin1101011010100
3100102101
41223110
5204433
651444
726011
oct15324
910371
106868
115184
123b84
133184
142708
15207d
6868 has 12 divisors (see below), whose sum is σ = 12852. Its totient is φ = 3200.
The previous prime is 6863. The next prime is 6869. The reversal of 6868 is 8686.
It can be written as a sum of positive squares in 2 ways, for example, as 6084 + 784 = 78^2 + 28^2 .
6868 is an undulating number in base 10.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (6863) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (7) of ones.
It is a polite number, since it can be written in 3 ways as a sum of consecutive naturals, for example, 18 + ... + 118.
It is an arithmetic number, because the mean of its divisors is an integer number (1071).
26868 is an apocalyptic number.
6868 is a gapful number since it is divisible by the number (68) formed by its first and last digit.
It is an amenable number.
6868 is a deficient number, since it is larger than the sum of its proper divisors (5984).
6868 is a wasteful number, since it uses less digits than its factorization.
With its predecessor (6867) it forms a Ruth-Aaron pair, since the sum of their prime factors is the same (122).
6868 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 122 (or 120 counting only the distinct ones).
The product of its digits is 2304, while the sum is 28.
The square root of 6868 is about 82.8733974204. The cubic root of 6868 is about 19.0083066172.
It can be divided in two parts, 68 and 68, that added together give a triangular number (136 = T16).
The spelling of 6868 in words is "six thousand, eight hundred sixty-eight".
Divisors: 1 2 4 17 34 68 101 202 404 1717 3434 6868 | 556 | 1,897 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2023-40 | latest | en | 0.916603 |
http://www.clipsnation.com/2012/2/13/2796911/adjusted-point-differential-and-pythagorean-wins | 1,406,365,719,000,000,000 | text/html | crawl-data/CC-MAIN-2014-23/segments/1405997900573.25/warc/CC-MAIN-20140722025820-00166-ip-10-33-131-23.ec2.internal.warc.gz | 594,241,691 | 23,061 | ## Adjusted Point Differential and Pythagorean Wins
More terrific statistical work from Citizen BelgianClipper, who apparently has a little too much time on his hands and some truly mad Excel skillz. Dude, step away from the keyboard. Maybe go see the sights. I hear the canals in Bruges are nice. Steve
Fellow citizens,
This is a follow-up fanpost on this fanshot which Steve so graciously promoted to the front page (thanks) - (you're welcome).
First off, I'm adding the Adjusted Point Differential for all NBA teams. I have also added this to the original fanshot but some of you might not have seen so:
the 76ers have a actual point difference of 9.11 but the average team would be expected to have a point diff of 1.41 against their opponents. Thus their adjusted point difference is 7.70. The last 2 stats are the difference between their actual points and points expected for and against. This has the nice effect of dividing the Adjusted Point differential between offence and defence. Thus the Sixers have an Adjusted Point Diff of 7.70 which is composed of 2.11 points of offence and 5.59 on defence.
Secondly, Citizen Bandany wondered how this would effect Pythagorean Wins.
I'm using Basketball-reference.com method to expand this topic
W PythPythagorean Wins; the formula is G * (Tm PTS14 / (Tm PTS14 + Opp PTS14)). The formula was obtained by fitting a logistic regression model with log(Tm PTS / Opp PTS) as the explanatory variable.
this leads to the next table
Act pyth is the Pythagorean Wins calculated using the actual point difference. Act/exp pyth is the Pythagorean Wins calculated by dividing the the actual point per game for and against by the respective expected values. Using this measure you get a corrected Pythagorean Wins which takes in account the strength of the opposition. If we compare both calculations you see that the Clips would be expected to have won 0.807 games more if we use this last calculation. the last 2 columns compare these calculations with the actual number of wins. it is a bit weird that the Clips get harped because of their point differential whilst OKC, even without adjustments, have by far the greatest difference between their actual wins and Pythagorean Wins.
Well now I went completely mad: using the expected points values I calculated the Pythagorean Wins the average team would have against the opposition each team faced (exp pyth). In golf terms: par for the course. Then I compared them to the actual wins and the adjusted Pythagorean Wins. To counteract the difference in number of games I divided those comparisons by the number of games each team has played. "W vs exp pyth" per game essentially gives an indication by how much a team exceeds their expected Pythagorean Wins. "Act/exp vs exp" compares the adjusted Pythagorean Wins against the expected Pythagorean Wins. This could be seen as some sort of "objective" evaluation of a team results. The "mean" in the regression to the mean idea.
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We have our own Community Guidelines at Clips Nation. You should read them. | 759 | 3,511 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2014-23 | longest | en | 0.955425 |
https://goprep.co/ex-20.5-q10-integrate-0-infinity-x-1x-1x-2-dx-evaluate-the-i-1nl3wk | 1,621,204,152,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243989914.60/warc/CC-MAIN-20210516201947-20210516231947-00394.warc.gz | 309,862,667 | 43,832 | # <span lang="EN-US
Let us assume
Adding – 1 and + 1
Let
Thus I = I1 – I2 …….equation 1
Solving for I1
since
I1 = [tan – 1(∞) – tan – 1(0)]
I1 = π/2 ……….equation 2
Solving for I2
Let .....…..equation 3
a + b = 0; a + c = 1; b + c = 0
solving we get
a = c = 1/2
b = – 1/2
substituting the values in equation 3
Thus substituting the values in I2, thus
Solving :
Let 1 + x2 = y
2xdx = dy
For x = ∞
y = ∞
For x = 0
y = 0
substituting values
Thus
……….equation 4
Substituting values equation 2 and equation 4 in equation 1
Thus
I = I1 – I2
I = π/2 – π/4
I = π/4
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https://ai.stackexchange.com/questions/7178/multiple-sets-of-input-in-neural-network-or-other-form-of-ml | 1,623,705,722,000,000,000 | text/html | crawl-data/CC-MAIN-2021-25/segments/1623487613453.9/warc/CC-MAIN-20210614201339-20210614231339-00479.warc.gz | 100,920,426 | 37,365 | # Multiple sets of input in Neural network (or other form of ML)
I'm currently working on a research project where I try to apply different kinds of Machine Learning on some existing software I wrote a few years ago.
This software will scan for people in the room continuously. Some of these detections are either True or False. However, this is not known, so I cannot use supervised learning to train a network to make a distinction. I do however have a number that is correlated to the number of detections that should be True in a given period of time (let's say 30 seconds - 2 minutes), which can be used as an output feature to train a regression model. But the problem is... How can I give these multiple "detections" as an input? The way I see it now, would be something like this:
+--------------------------------------------------------------+-----------+------------+------------+----------------+--+
| Detections | Variable1 | Variable 2 | Variable n | Output Feature | |
+--------------------------------------------------------------+-----------+------------+------------+----------------+--+
| {person a, person b, person h, person z} | 132 | 189 | 5 | 50 | |
| {person a, person b, person c, person d, person k, person m} | 1 | 50 | 147 | 80 | |
| {person c, person e, person g, person f} | 875 | 325 | 3 | 20 | |
+--------------------------------------------------------------+-----------+------------+------------+----------------+--+
Each of these persons would be a tuple of values: var_1, var_2, var_3, var_4. These values are not constant however! They do change between observations.
Different approach to explain it: there's multiple observations (variable amount) in each time segment (duration of time segment is a fixed integer to be chosen). These observations have a few variables that would indicate whether the observation is true or false. However, the threshold for it being true or false, is very much dependant on the other variables, that are not tied to the information of the persons. (These variables are the same for all of them, but vary in between time segments. Let's call'm "environment features") Lastly, the output feature is the product of the count of persons that resulted in "True" and a (varying) factor that is correlated to the environment features.
So I've been thinking about probabilistic AI, but the problem is that there isn't a known distribution between True/False.
• Is there any technique I can apply to be able to use this kind of data as an input of a Neural Network (or other forms of ML)? Or is there a specific form of ML that is used for this kind of problems? | 594 | 2,833 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2021-25 | latest | en | 0.844603 |
http://stackoverflow.com/questions/7727502/how-to-get-table-size-232-in-java?answertab=active | 1,432,730,211,000,000,000 | text/html | crawl-data/CC-MAIN-2015-22/segments/1432207928965.67/warc/CC-MAIN-20150521113208-00104-ip-10-180-206-219.ec2.internal.warc.gz | 222,340,576 | 17,962 | # How to get table size 2^32 in java
I have to get in java `protected final static int [] SIEVE = new int [ 1 << 32 ];` But i cant force java to that.
Max sieve what i get is 2^26 i need 2^32 to end my homework. I tried with mask but i need to have SIEVE[n] = k where min{k: k|n & k >2}.
EDIT I need to find Factor numbers from 2 to 2^63-1 using Sieve and sieve must have information that P[n]= is smallest prime with divide n. I know that with sieve i can Factorise number to 2^52. But how do that exercises with holding on to the content.
EDIT x2 problem solved
-
What do you really need to do? Allocating a huge array is not the correct way to do what it. – unholysampler Oct 11 '11 at 14:24
Only the ints would occupy 16 GB, so get yourself a high end server. – Mister Smith Oct 11 '11 at 14:25
I don't think you're supposed to do this for your homework. – Ishtar Oct 11 '11 at 14:26
If you really need to store that much data in memory, try using java.util.LinkedList collection instead.
However, there's a fundamental flaw in your algorithm if you need to store 16GB of data in memory. If you're talking about Sieve of Eratosthenes and you need to store all primes < 2^32 in an array, you still wouldn't need an array of size 2^32. I'd suggest you use `java.util.BitSet` to find the primes and either iterate and print or store them in a LinkedList as required.
-
A LinkedList is by far less efficient than an array, both in RAM and performance. – Mister Smith Oct 11 '11 at 14:33
I just gave a way to work around his problem, irrespective of the consequences. It works. I've however qualified my answer by pointing a possible flaw in his approach. – Jagat Oct 11 '11 at 14:35
Yes, I totally agree with what you said and I've read about its performance impact as well. Again, the answer I gave isn't a clean one and is just something that "works" for him. – Jagat Oct 11 '11 at 14:47
You can't. A Java array can have at most `2^31 - 1` elements because the `size` of an array has to fit in a signed 32-bit integer.
This applies whether you run on a 32 bit or 64 bit JVM.
I suspect that you are missing something in your homework. Is the requirement to be able to find all primes less than `2^32` or something? If that is the case, they expect you to treat each `int` of the `int[]` as an array of 32 bits. And you need an array of only `2^25` ints to do that ... if my arithmetic is right.
A `BitSet` is another good alternative.
A `LinkedList<Integer>` is a poor alternative. It uses roughly 8 times the memory that an array of the same size would, and the performance of `get(int)` is going to be horribly slow for a long list ... assuming that you use it in the obvious fashion.
If you want something that can efficiently use as much memory as you can configure your JVM to use, then you should use an `int[][]` i.e. an array of arrays of integers, with the `int[]` instances being as large as you can make them.
I need to find Factor numbers from 2 to 2^63-1 using Sieve and sieve must have information that P[n]= is smallest prime with divide n. I know that with sieve i can Factorise number to 2^52. But how do that exercises with holding on to the content.
I'm not really sure I understand you. To factorize a number in the region of 2^64, you only need prime numbers up to 2^32 ... not 2^52. (The square root of 2^64 is 2^32 and a non-prime number must have a prime factor that is less than or equal to its square root.)
It sounds like you are trying to sieve more numbers than you need to.
-
I need to find Factor numbers from 2 to 2^63-1 using Sieve and sieve must have information that P[n]= is smallest prime with divide n. I know that with sieve i can Factorise number to 2^52. But how do that exercises with holding on to the content. – Bla bla Oct 11 '11 at 15:05 | 1,017 | 3,809 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.703125 | 3 | CC-MAIN-2015-22 | latest | en | 0.95908 |
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ACC2233 Hand-In Assignment 6v2
Problem One
PLANT-WIDE RATE AND ACTIVITY-BASED COSTING
The comptroller for Douglas Inc. has established the following activity cost pools and
cost drivers.
Budgeted
Overhead
Activity Cost Pool Cost Cost Driver Budgeted
Level for
Cost Driver Machine setups
Material handling
Waste control
Quality control
Other overhead costs Number of setups
160
Weight of material 50,000 kilograms
Weight of chemicals10,000 kilograms
Number of inspections 300
Machine hours
20,000 \$200,000
100,000
50,000
75,000
200,000 Each order must be a minimum of 1,000 boxes of film chemicals.
The production requirements for each minimum order are as follows:
Machine setups
Material handling
Waste Control
Quality Control
Other Overhead Costs 5 setups
1,500 kilograms
200 kilograms
5 inspections
500 machine hours Required:
1 Calculate the overhead that should be assigned to each minimum order of film
chemicals using the above data.
2 Calculate the overhead that should be assigned to each minimum order of film
chemicals if Douglas decides to use a single predetermined overhead rate based on
machine hours. Problem Two
PLANT-WIDE RATE AND ACTIVITY-BASED COSTING
Hood Manufacturing Inc. has four categories of overhead. The expected overhead costs
for each of the categories for the next year are as follows: Maintenance
Material handling
Setups
Inspection Costs Cost Driver Expected
Activity \$510,000
250,000
60,000
21,000 Machine hours
Material moves
Setups
Inspections 60,000
20,000
3,000
12,000 Currently, the company applies overhead using a predetermined overhead rate based
upon budgeted direct labour hours of 100,000.
The company has been asked to submit a bid on a proposed job. Usually bids are
based upon full manufacturing costs plus a percentage of 10 percent.
Estimates for the proposed job are as follows:
Direct materials
Direct labour
Number of direct labour hours
Number of material moves
Number of inspections
Number of setups
Number of machine hours \$30,000
\$24,000
8,000
100
120
24
4,000 Required:
1. If the company used activity-based cost drivers to assign overhead, calculate the bid
price of the proposed job.
2. If the company used direct labour hours as the cost driver, calculate the bid price of
the proposed job. Problem Three
NON-VALUE-ADDED COSTS, ACTIVITY COSTS, ACTIVITY COST REDUCTION
Bob Daly, vice president of Dwan Company (a producer of a variety of plastic products),
has been supervising the implementation of an ABC management system. One of Bob’s
objectives is to improve process efficiency by improving the activities that define the
processes. To illustrate the potential of the new system to the president, Bob has
decided to focus on two processes: production and customer service. Within each process, one activity will be selected for improvement: materials usage for
production and sustaining engineering for customer service (sustaining engineers are
responsible for redesigning products based on customer needs and feedback). Valueadded standards are identified for each activity (the level of efficiency so that no waste
exists). For materials usage, the value-added standard calls for six kilograms per unit of
output (although the plastic products differ in shape and function, their size—as
measured by weight—is uniform). The value-added standard is based on the elimination
of all waste due to defective moulds. The standard price of materials is \$5 per kilogram.
For sustaining engineering, the standard is 58 percent of current practical activity
capacity. This standard is based on the fact that about 42 percent of the complaints
have to do with design features that could have been avoided or anticipated by the
company.
Current practical capacity (at the end of 2013) is defined by the following requirements:
6,000 engineering hours for each product group that has been on the market or in
development for five years or less and 2,400 hours per product group of more than five
years. Four product groups have less than five years’ experience, and 10 product
groups have more. Each of the 24 engineers is paid a salary of \$60,000. Each engineer
can provide 2,000 hours of service per year. No other significant costs are incurred for
the engineering activity.
Actual materials usage for 2014 was 25 percent above the level called for by the Valueadded standard; engineering usage was 46,000 hours. A total of 80,000 units of output
were produced. Bob and the operational managers have selected some improvement
measures that promise to reduce non-value-added activity usage by 40 percent in 2015.
Selected actual results achieved for 2015 are as follows:
Units produced
80,000
Materials used
584,800
Engineering hours 35,400
The actual prices paid for materials and engineering hours are identical to the standard
or budgeted prices.
Required:
1. For 2014, calculate the non-value-added usage and costs for materials usage and
sustaining engineering.
2. Using the budgeted improvements, calculate the expected activity usage levels for
2015. Now, calculate the 2015 usage variances (the difference between the
expected and actual values), expressed in both physical and financial measures, for
materials and engineering. Comment on the company’s ability to achieve its targeted
reductions. In particular, discuss what measures the company must take to capture
any realized reductions in resource usage. | 1,190 | 5,395 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-33 | latest | en | 0.866348 |
http://www.slidesearchengine.com/slide/ssc-cgl-numerical-aptitude-basic-algebra | 1,516,617,676,000,000,000 | text/html | crawl-data/CC-MAIN-2018-05/segments/1516084891277.94/warc/CC-MAIN-20180122093724-20180122113724-00429.warc.gz | 564,221,802 | 6,347 | # Ssc cgl-numerical-aptitude-basic-algebra
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Published on March 4, 2014
Author: upscportal
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Numerical Aptitude http://sscportal.in Basic Algebra A Civil Servant should be well-versed in basics of Algebra. In the Civil Services Aptitude Test Paper 2, in Basic Numeracy, certainly there will be asked some questions based on equations and their roots. © 2014 SSCPORTAL .IN Join Online Coaching Click Here
Numerical Aptitude http://sscportal.in Basic Algebra Polynomials An expression in term of some variable(s) is called a polynomial. For example f(x) = 2x – 5 is a polynomial in variable x g(y) = 5y2 – 3y + 4 is a polynomial in variable y © 2014 SSCPORTAL .IN Join Online Coaching Click Here
Numerical Aptitude http://sscportal.in Basic Algebra Degree of a Polynomial The exponent of the highest degree term in a polynomial is known as its degree. Linear Polynomial A polynomial of degree one is called a linear polynomials. In general f(x) = ax + b, where a ¹ 0 is a linear polynomial. © 2014 SSCPORTAL .IN Join Online Coaching Click Here
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Numerical Aptitude http://sscportal.in Basic Algebra Useful Formulae (i) (x + y)2 = x2 + y2 + 2xy (ii) (x – y)2 = x2 + y2 – 2xy (iii) (x2 – y2) = (x + y) (x – y) (iv) (x + y)3 = x3 + y3 + 3xy(x + y) (v) (x – y)3 = x3 – y3 – 3xy(x – y) (vi) (x3 + y3) = (x + y) (x2 + y2 – xy) (vii) (x3 – y3) = (x – y) (x2 + y2 + xy) (viii) (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) © 2014 SSCPORTAL .IN Join Online Coaching Click Here
Numerical Aptitude http://sscportal.in Basic Algebra (ix) (x3 + y3 + z3 – 3xyz) = (x + y + z) (x2 + y2 + z2 – xy – yz – zx) (x) If x + y + z = 0, then x3 + y3 + z3 = 3xyz Also, (i) (xn – an ) is divisible by (x – a) for all values of n. (ii) (xn + an ) is divisible by (x + a) only when n is odd. (iii) (xn– an ) is divisible by (x + a) only for even values of n. (iv) (xn + an) is never divisible by (x – a). © 2014 SSCPORTAL .IN Join Online Coaching Click Here
Numerical Aptitude http://sscportal.in Basic Algebra Linear Equations Equations Linear equations in one variable Linear Equations in two variables Consistency of a System of Linear Equations © 2014 SSCPORTAL .IN Join Online Coaching Click Here
Numerical Aptitude http://sscportal.in Basic Algebra Theory of Equation Quadratic Equation To Find the Roots of Quadratic Equation Method 1: By Factorisation Method 2 : Using Formula Sum and Product of Roots of a Quadratic Equation Constructing a Quadratic Equation © 2014 SSCPORTAL .IN Join Online Coaching Click Here
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1. According to Dalton's law, the total pressure of the mixture of gases is equal to
(A) Greater of the partial pressures of all
(B) Average of the partial pressures of all
(C) Sum of the partial pressures of all
(D) Sum of the partial pressures of all divided by average molecular weight
2. Which of the following can be regarded as gas so that gas laws could be applicable, within the commonly encountered temperature limits?
(A) O₂, N₂, steam, CO₂
(B) O₂, N₂, water vapour
(C) SO₂, NH₃, CO₂, moisture
(D) O₂, N₂, H₂, air
3. The measurement of a thermodynamic property known as temperature is based on
(A) First law of thermodynamics
(B) Second law of thermodynamics
(C) Zeroth law of thermodynamics
(D) None of these
4. For a perfect gas, according to Boyle’s law (where P = Absolute pressure, V = Volume and T = Absolute temperature)
(A) V/T = constant, if p is kept constant
(B) P v = constant, if T is kept constant
(C) T/P = constant, if v is kept constant
(D) P/T = constant, if v is kept constant
5. The gas turbine cycle with regenerator improves
(A) Work ratio
(B) Thermal efficiency
(C) Avoid pollution
(D) None of these
6. The compression ratio is the ratio of
(A) Total volume to swept volume
(B) Swept volume to clearance volume
(C) Swept volume to total volume
(D) Total volume to clearance volume
7. The unit of pressure in S.I. units is
(A) kg/cm²
(B) mm of water column
(C) Pascal
(D) Dyne per square cm
8. A closed system is one in which
(A) Mass does not cross boundaries of the system, though energy may do so
(B) Mass crosses the boundary but not the energy
(C) Neither mass nor energy crosses the boundaries of the system
(D) Both energy and mass cross the boundaries of the system
9. Temperature of a gas is produced due to
(A) Its heating value
(B) Kinetic energy of molecules
(C) Repulsion of molecules
(D) Surface tension of molecules
10. According to kinetic theory of gases, the absolute zero temperature is attained when
(A) Volume of the gas is zero
(B) Pressure of the gas is zero
(C) Kinetic energy of the molecules is zero
(D) Specific heat of gas is zero
11. Those substances which have so far not been resolved by any means into other substances of simpler form are called
(A) Atoms
(B) Compounds
(C) Elements
(D) Molecules
12. Coke is produced
(A) When coal is strongly heated continuously for 42 to 48 hours in the absence of air in a closed vessel
(B) From the finely ground coal by moulding under pressure with or without a binding material
(C) When coal is first dried and then crushed to a fine powder by pulverizing machine
(D) By heating wood with a limited supply of air to a temperature not less than 280°C
13. The hard coke is obtained when carbonization of coal is carried out at
(A) 500° to 700°C
(B) 700° to 900°C
(C) 300° to 500°C
(D) 900° to 1100°C
14. Which of the following is an intensive property of a thermodynamic system?
(A) Mass
(B) Temperature
(C) Energy
(D) Volume
15. Otto cycle is also known as
(A) Constant pressure cycle
(B) Constant volume cycle
(C) Constant temperature cycle
(D) Constant temperature and pressure cycle
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• you will enter the following information in MySQL Workbench when connecting to the “Instructor’s Online Server” :
• hostname: hult-mysql.cmsi20r5frst.us-west-1.rds.amazonaws.com
• port: 3306You will be given access (read-only) to an existing database – the schema is called “accounting” in the Instructor’s Online Server. You can take a look at the EER diagram here: Accounting ER.mwb Download Accounting ER.mwb (Just don’t use the EER diagram to create your tables – it won’t work – all data is in the Instructor’s Online Server)
You will need to build queries and insert results into tmp views to create the Profit and Loss (P&L) and Balance Sheet (B/S) statements
Your queries must achieve the following:
• Receive the year in which the user wants to produce the P&L and the B/S for.
• Do the necessary calculations to obtain the net profit or loss for that year.
• Use that value, to produce the B/S.
• Demonstrarte that A = L + E
• Print both the P&L and B/S as clear as possible on the Result Grids
• Show the % change vs. the previous year for every major line item on the P&L and B/S
• Give some headings to each major account line item. And to each section of the P&L and B/S
• Add any additional financial metrics and ratios that could be useful to analyze this start-up.
• Wanna ensure an “A”? Produce a cash flow statement as an added challenge
Students will submit the following:
1. (60 points) SQL script (if you save it as txt, be sure to keep the adequate indentations – you will receive points for correct syntax, following best practices, comments in the code, correctness, efficiency ):
• While there are multiple ways to approach the building of their P&L and B/S statements, most likely you will need need analytical queries
• be sure to include the comprehensive list of queries in ONE single .sql script that if the instructor runs on their computer, they will be able to obtain exactly the same results as per your P&L and B/S screenshots from #2.
2. (20 points) Screenshot of their resulting P&L and B/S outputs. You will receive points for correct results, 10 points for correct B/S and 10 points for correct P&L.
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https://slideplayer.com/slide/5930585/ | 1,580,011,430,000,000,000 | text/html | crawl-data/CC-MAIN-2020-05/segments/1579251684146.65/warc/CC-MAIN-20200126013015-20200126043015-00150.warc.gz | 658,548,394 | 21,290 | # INHERENT LIMITATIONS OF COMPUTER PROGRAMS CSci 4011.
## Presentation on theme: "INHERENT LIMITATIONS OF COMPUTER PROGRAMS CSci 4011."— Presentation transcript:
INHERENT LIMITATIONS OF COMPUTER PROGRAMS CSci 4011
TIME COMPLEXITY Definition: Let M be a TM that halts on all inputs. The running time or time-complexity of M is the function f : N N, where f(n) is the maximum number of steps that M uses on any input of length n. Definition: TIME(t(n)) = { L | L is a language decided by a O(t(n)) time Turing Machine }
P = TIME(n c ) c N IMPORTANT COMPLEXITY CLASSES Problems in P can be efficiently solved. Problems in NP can be efficiently verified. NP = NTIME(n c ) c N L ∈ P iff there is an n c -time TM that decides L. L ∈ NP iff there is an n c -time NTM that decides L iff there is an n c -time “verifier” for L
NP = { } Languages that have “nifty proofs” (certificates) of membership. 3SAT = { ϕ | ∃ y so that ϕ (y)=1 and ϕ is in 3cnf } certificate: the satisfying assignment y HAMPATH = { 〈 G,s,t 〉 | G has a hamiltonian path from s to t} certificate: the hamiltonian path CLIQUE= { 〈 G,k 〉 | G has a clique of size k} certificate: the clique (set of k nodes) FACTOR = { 〈 N,k 〉 | N has a prime factor ≥ k} certificate: the factor p ≥ k
If P = NP… Recognition is the same as Generation. We could design optimal: cars, airplanes, fusion generators, circuits, medicines… Cryptography as we know it is impossible. In 40+ years, no one has found a proof that P NP. Mathematicians are out of a job.
POLY-TIME REDUCIBILITY A language A is polynomial time reducible to language B, written A P B, if there is a polynomial time computable function ƒ: Σ* Σ*, where for every w, w A ƒ(w) B ƒ is called a polynomial time reduction of A to B
Theorem. If A ≤ P B and B ∈ P, then A ∈ P. (so if A ≤ P B and A P, then B P.) Proof. B P ⇒ TIME COMPLEXITY = O(|w| ab ) |s| b -time program test_b(s) to decide B A ≤ P B ⇒ def test_a(w): s = map(w) return test_b(s) |w| a -time function map s.t. map(w) B iff w A.
Definition: A language B is NP-complete if: 1. B NP 2. Every A in NP is poly-time reducible to B (i.e. B is NP-hard) HARDEST PROBLEMS IN NP
B is NP-Complete P NP B
Theorem. If B is NP-Complete, C ∈ NP, and B ≤ P C, then C is NP-Complete. P NP B C
Theorem. If B is NP-Complete and B ∈ P, then P=NP. Corollary. If B is NP-Complete, and P NP, there is no fast algorithm for B.
Let BA NTM = { 〈 M,x,t 〉 | M is an NTM that accepts x in at most t steps} Theorem. BA NTM is NP-Complete. 1. BA NTM NP: The list of guesses M makes to accept x in t steps is the proof that 〈 M,x,t 〉 BA NTM. 2. For all A NP, A ≤ P BA NTM. A NP iff there is an NTM for A that runs in time O(n c ). Let ƒ A (w) = 〈 N,w,|w| c 〉. 〈 N,w,|w| c 〉 BA NTM iff N accepts w, iff w A. Hyper-technical detail: we let n denote 1 n.
Let C DTM = { 〈 M,x,t 〉 | ∃ y, |y| ≤ t and M(x,y) accepts in at most t steps} Theorem. C DTM is NP-Complete. 1. C DTM ∈ NP: The string y proves 〈 M,x,t 〉 ∈ C DTM.
Let C DTM = { 〈 M,x,t 〉 | ∃ y, |y| ≤ t and M(x,y) accepts in at most t steps} Theorem. C DTM is NP-Complete. 2. For all A ∈ NP, A ≤ P C DTM. A ∈ NP ⇒ There is a n a -time TM V A and c where A = { x | ∃ y. |y| ≤ |x| c and V A (x,y) accepts } Define ƒ A (w) = 〈 V, w, t 〉, where t=(|w| c + |w|) a
Let EA DTM = { 〈 M,n,t 〉 | ∃ y ∈ {0,1} n : M accepts y in at most t steps} Theorem. EA DTM is NP-Complete. 2. C DTM ≤ P EA DTM. Define ƒ(M,x,t) = 〈 M x, t, t 〉, where M x (y) = “run M(x,y).” 1. EA DTM ∈ NP: The string y proves 〈 M,n,t 〉 EA DTM.
A CIRCUIT … is a collection of (boolean) gates and inputs connected by wires. ⋁ ∧ ¬ ∧ x0x0 x1x1 x2x2 … is satisfiable if some setting of inputs makes it output 1. … has arbitrary fan-in and fan-out CIRCUIT-SAT = { 〈 C 〉 | C is a satisfiable circuit } ∧
Theorem. CIRCUIT-SAT is NP-Complete. Proof. 1. CIRCUIT-SAT ∈ NP. 2. EA DTM ≤ P CIRCUIT-SAT. Recall that the language EA DTM = { 〈 M, n, t 〉 | M is a TM and ∃ y {0,1} n such that M accepts y in ≤ t steps.} is NP-Complete. We prove EA DTM ≤ P C-SAT.
WARMUP Given 〈 M,t 〉, there is a function C : {0,1} t → {0,1} such that C(x) = 1 iff M(x) accepts in t steps. Any boolean function in k variables can be written as a CNF with at most 2 k clauses: x0x0 x1x1 F 001 010 100 111 ⇔ (x 0 ⋁ ¬x 1 ) ∧ (¬x 0 ⋁ x 1 ) Why doesn’t this prove the theorem?
Given TM M and time bound t, we create a circuit that takes n input bits and runs up to t steps of M. The circuit will have t “rows”, where the i th row represents the configuration of M after i steps: q00q0010101 0q11q110101 0q10q10101 0q01q0101 01q00q001 010q11q11 010q1q1 010qaqa t t a “tableau”
q0q0 Rows are made up of cells. Each cell has a “light” for every state and every tape symbol. q1q1 qaqa 01 q0q0 q1q1 qaqa 01 q0q0 q1q1 qaqa 01 q0q0 q1q1 qaqa 01 Each light has a circuit that turns it on or off based on the previous row. q0q0 ∧ ∧ ⋁
q0q0 EXAMPLE q0q0 q1q1 qaqa 0 → 0, R 0 → □, R 1 → 1, R 1 → □, R □ → □, L q1q1 qaqa 01 q0q0 q1q1 qaqa 01 q0q0 q1q1 qaqa 01 q0q0 q1q1 qaqa 01 q0q0 ∧ ∧ ⋁ q1q1 ∧ ∧ ⋁ ∧ ⋁ qaqa 0 ∧ ⋁ ⋁ ∙ ∧ 1 ∧ ⋁ ⋁ ∙ ∧ ∧ ⋁ ⋁ ∙ ∧ ∧ ∧
The lights in the first row are connected to the circuit inputs and the tape head is hardwired in: x1x1 x2x2 x3x3 … x n-1 xnxn q0q0 The circuit should output 1 iff M ends in q accept. qaqa qaqa qaqa qaqa qaqa qaqa qaqa qaqa ⋁
How big is the circuit if M has m states+symbols? O(m 2 t 2 ) How long does it take to build the circuit? O(m 2 t 2 ) When is the circuit satisfiable? Iff there is a n-bit input that makes M accept in at most t steps. | 2,090 | 5,586 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.375 | 3 | CC-MAIN-2020-05 | latest | en | 0.835626 |
http://gamedev.stackexchange.com/questions/24957/doing-an-snes-mode-7-affine-transform-effect-in-pygame?answertab=votes | 1,448,483,080,000,000,000 | text/html | crawl-data/CC-MAIN-2015-48/segments/1448398445679.7/warc/CC-MAIN-20151124205405-00106-ip-10-71-132-137.ec2.internal.warc.gz | 94,337,748 | 18,446 | # Doing an SNES Mode 7 (affine transform) effect in pygame
Is there such a thing as a short answer on how to do a Mode 7 / mario kart type effect in pygame?
I have googled extensively, all the docs I can come up with are dozens of pages in other languages (asm, c) with lots of strange-looking equations and such.
Ideally, I would like to find something explained more in English than in mathematical terms.
I can use PIL or pygame to manipulate the image/texture, or whatever else is necessary.
I would really like to achieve a mode 7 effect in pygame, but I seem close to my wit's end. Help would be greatly appreciated. Any and all resources or explanations you can provide would be fantastic, even if they're not as simple as I'd like them to be.
If I can figure it out, I'll write a definitive how to do mode 7 for newbies page.
edit: mode 7 doc: http://www.coranac.com/tonc/text/mode7.htm
-
Please post a link to such a doc, so we can guide you through the math. – Maik Semder Mar 4 '12 at 12:16
there seems to be equations here: en.wikipedia.org/wiki/Mode_7 Although, these days we have 3D acceleration, things like Mode 7, or the whacky way doom worked are more of a curiosity than a solution. – salmonmoose Mar 4 '12 at 12:27
@2D_Guy this page explain the algorithm very well for me. You want to know how to do it, or you want it already implemented for you? – Gustavo Maciel Mar 5 '12 at 3:53
@stephelton On the SNES systems, the only layer that could be distorted, rotated..(applied affine transformations with matrices) is the seventh layer. The Background layer. All other layers were used to simple sprites, So if you wanted a 3D effect, you had to use this layer, this is where the name came from :) – Gustavo Maciel May 8 '12 at 11:02
@GustavoMaciel: That's a bit inaccurate. The SNES had 8 different modes (0-7), in which up to 4 background layers had different functionality, but only one mode (mode 7, hence the name) supported rotation and scaling (and also restricted you to a single layer). You couldn't really combine the modes. – Michael Madsen May 8 '12 at 15:30
Mode 7 is a very simple effect. It projects a 2D x/y texture (or tiles) to some floor/ceiling. Old SNES use hardware to do this, but modern computers are so powerful that you can do this realtime (and no need of ASM as you mention).
Basic 3D math formula to project a 3D point (x, y, z) to a 2D point (x, y) is :
``````x' = x / z;
y' = y / z;
``````
When you think about it, it makes sense. Objects that are far in distance are smaller than objects near you. Think about railroad tracks going to nowhere :
If we look back at the formula input values : `x` and `y` will be the current pixel we are processing, and `z` will be distance information about how far the point is. To understand what `z` should be, look at that picture, it shows `z` values for image above :
purple = near distance, red = far away
So in this example, `z` value is `y - horizon` (assuming `(x:0, y:0)` is at the center of screen)
If we put everything together, it becomes : (pseudocode)
``````for (y = -yres/2 ; y < yres/2 ; y++)
for (x = -xres/2 ; x < xres/2 ; x++)
{
horizon = 20; //adjust if needed
fov = 200;
px = x;
py = fov;
pz = y + horizon;
//projection
sx = px / pz;
sy = py / pz;
scaling = 100; //adjust if needed, depends of texture size
color = get2DTexture(sx * scaling, sy * scaling);
//put (color) at (x, y) on screen
...
}
``````
One last thing : if you want to make a mario kart game, I suppose you also want to rotate the map. Well its also very easy : rotate `sx` and `sy` before getting texture value. Here is formula :
`````` x' = x * cos(angle) - y * sin(angle);
y' = x * sin(angle) + y * cos(angle);
``````
and if you want to move trough the map, just add some offset before getting texture value :
`````` get2DTexture(sx * scaling + xOffset, sy * scaling + yOffset);
``````
NOTE : i tested the algorithm (almost copy-paste) and it works. Here is the example : http://glslsandbox.com/e#26532.3 (require recent browser and WebGL enabled)
NOTE2 : i use simple math because you said you want something simple (and dont seems familiar with vector math). You can achieve same things using wikipedia formula or tutorials you give. The way they did it is much more complex but you have much more possibilities for configuring the effect (in the end it works the same...). | 1,187 | 4,379 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.96875 | 3 | CC-MAIN-2015-48 | longest | en | 0.944335 |
https://exercism.io/tracks/elixir/exercises/accumulate/solutions/cb2c2c9358ec434280a5d7ceb18e0d5c | 1,601,237,844,000,000,000 | text/html | crawl-data/CC-MAIN-2020-40/segments/1600401578485.67/warc/CC-MAIN-20200927183616-20200927213616-00690.warc.gz | 364,353,227 | 6,909 | 🎉 Exercism Research is now launched. Help Exercism, help science and have some fun at research.exercism.io 🎉
# thekeele's solution
## to Accumulate in the Elixir Track
Published at Jul 13 2018 · 0 comments
Instructions
Test suite
Solution
#### Note:
This solution was written on an old version of Exercism. The tests below might not correspond to the solution code, and the exercise may have changed since this code was written.
Implement the `accumulate` operation, which, given a collection and an operation to perform on each element of the collection, returns a new collection containing the result of applying that operation to each element of the input collection.
Given the collection of numbers:
• 1, 2, 3, 4, 5
And the operation:
• square a number (`x => x * x`)
Your code should be able to produce the collection of squares:
• 1, 4, 9, 16, 25
Check out the test suite to see the expected function signature.
## Restrictions
Keep your hands off that collect/map/fmap/whatchamacallit functionality provided by your standard library! Solve this one yourself using other basic tools instead.
## Running tests
Execute the tests with:
``````\$ elixir accumulate_test.exs
``````
### Pending tests
In the test suites, all but the first test have been skipped.
Once you get a test passing, you can unskip the next one by commenting out the relevant `@tag :pending` with a `#` symbol.
For example:
``````# @tag :pending
test "shouting" do
assert Bob.hey("WATCH OUT!") == "Whoa, chill out!"
end
``````
Or, you can enable all the tests by commenting out the `ExUnit.configure` line in the test suite.
``````# ExUnit.configure exclude: :pending, trace: true
``````
For more detailed information about the Elixir track, please see the help page.
## Source
Conversation with James Edward Gray II https://twitter.com/jeg2
## Submitting Incomplete Solutions
It's possible to submit an incomplete solution so you can see how others have completed the exercise.
### accumulate_test.exs
``````if !System.get_env("EXERCISM_TEST_EXAMPLES") do
Code.load_file("accumulate.exs", __DIR__)
end
ExUnit.start()
ExUnit.configure(exclude: :pending, trace: true)
defmodule AccumulateTest do
use ExUnit.Case
test "accumulate empty list" do
assert Accumulate.accumulate([], fn n -> n * n end) == []
end
@tag :pending
test "accumulate square numbers" do
assert Accumulate.accumulate([1, 2, 3], fn n -> n * n end) == [1, 4, 9]
end
@tag :pending
test "accumulate upcased strings" do
fun = fn w -> String.upcase(w) end
assert Accumulate.accumulate(["hello", "world"], fun) == ["HELLO", "WORLD"]
end
@tag :pending
test "accumulate reversed strings" do
fun = fn w -> String.reverse(w) end
words = ~w(the quick brown fox etc)
expected = ["eht", "kciuq", "nworb", "xof", "cte"]
assert Accumulate.accumulate(words, fun) == expected
end
@tag :pending
test "nested accumulate" do
chars = ~w(a b c)
nums = ~w(1 2 3)
fun = fn c -> Accumulate.accumulate(nums, &(c <> &1)) end
expected = [["a1", "a2", "a3"], ["b1", "b2", "b3"], ["c1", "c2", "c3"]]
assert Accumulate.accumulate(chars, fun) == expected
end
end``````
``````defmodule Accumulate do
@doc """
Given a list and a function, apply the function to each list item and
replace it with the function's return value.
Returns a list.
## Examples
iex> Accumulate.accumulate([], fn(x) -> x * 2 end)
[]
iex> Accumulate.accumulate([1, 2, 3], fn(x) -> x * 2 end)
[2, 4, 6]
"""
@spec accumulate(list, (any -> any)) :: list
def accumulate([], _), do: []
def accumulate([car | cdr], fun) do
[fun.(car) | accumulate(cdr, fun)]
end
end``````
## Community comments
Find this solution interesting? Ask the author a question to learn more.
### What can you learn from this solution?
A huge amount can be learned from reading other people’s code. This is why we wanted to give exercism users the option of making their solutions public.
Here are some questions to help you reflect on this solution and learn the most from it.
• What compromises have been made?
• Are there new concepts here that you could read more about to improve your understanding? | 1,108 | 4,109 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.515625 | 3 | CC-MAIN-2020-40 | longest | en | 0.84828 |
https://wanderin.dev/coding-challenges/binary-search-leetcode-704/ | 1,679,544,629,000,000,000 | text/html | crawl-data/CC-MAIN-2023-14/segments/1679296944996.49/warc/CC-MAIN-20230323034459-20230323064459-00785.warc.gz | 702,070,479 | 20,826 | ## Challenge Statement
• Given an array of integers `nums` which is sorted in ascending order, and an integer `target`, write a function to search `target` in `nums`. If `target` exists, then return its index. Otherwise, return `-1`.
• You must write an algorithm with `O(log n)` runtime complexity.
• This challenge corresponds to LeetCode #704.
### Constraints
• `1 <= nums.length <= 104`
• `-104 < nums[i], target < 104`
• All the integers in `nums` are unique.
• `nums` is sorted in ascending order.
### Example 1:
``````Input: nums = [-1, 0, 3, 5, 9, 12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4``````
### Example 2:
``````Input: nums = [-1, 0, 3, 5, 9, 12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1``````
### Example 3:
``````Input: head = []
Output: []``````
## Solution
Below is my solution and some test cases. This solution has a logarithmic time complexity O(log n) and a constant space complexity O(1), where n is the length of the input list. | 307 | 1,028 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2023-14 | longest | en | 0.740501 |
http://stackoverflow.com/questions/13648318/calculating-pearson-correlation?answertab=active | 1,419,699,247,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1419447552343.99/warc/CC-MAIN-20141224185912-00044-ip-10-231-17-201.ec2.internal.warc.gz | 88,018,445 | 17,130 | # Calculating Pearson correlation
I'm trying to calculate the Pearson correlation coefficient of two variables. These variables are to determine if there is a relationship between number of postal codes to a range of distances. So I want to see if the number of postal codes increases/decreases as the distance ranges changes.
I'll have one list which will count the number of postal codes within a distance range and the other list will have the actual ranges.
Is it ok to have a list that contain a range of distances? Or would it be better to have a list like this [50, 100, 500, 1000] where each element would then contain ranges up that amount. So for example the list represents up to 50km, then from 50km to 100km and so on.
-
@Krab Removed unnecessary information inline with SO policy, SO is a question and answer site so saying I would appreciate help is redundant, to say thanks you upvote and accept answer.. if you want more information on this read the faq and dig around on meta stackoverflow – iiSeymour Nov 30 '12 at 16:08
Use scipy :
``````scipy.stats.pearsonr(x, y)
``````
Calculates a Pearson correlation coefficient and the p-value for testing non-correlation.
The Pearson correlation coefficient measures the linear relationship between two datasets. Strictly speaking, Pearson’s correlation requires that each dataset be normally distributed. Like other correlation coefficients, this one varies between -1 and +1 with 0 implying no correlation. Correlations of -1 or +1 imply an exact linear relationship. Positive correlations imply that as x increases, so does y. Negative correlations imply that as x increases, y decreases.
The p-value roughly indicates the probability of an uncorrelated system producing datasets that have a Pearson correlation at least as extreme as the one computed from these datasets. The p-values are not entirely reliable but are probably reasonable for datasets larger than 500 or so.
Parameters :
``````x : 1D array
y : 1D array the same length as x
Returns :
(Pearson’s correlation coefficient, :
2-tailed p-value)
``````
-
Ok, so what matters more is that both the x and y arrays are of the same length. Then you are comparing elements x[i] with element y[i]? – user94628 Nov 30 '12 at 16:43
yep. In your case, x should be equal to the distances considered, and y[i] should return the number of postal code at distances[i]. To see the actual computation for the Pearson : stackoverflow.com/questions/3949226/… – georgesl Nov 30 '12 at 16:49
Cool, so x[i] could mean up to that distance? – user94628 Nov 30 '12 at 16:52 | 597 | 2,593 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2014-52 | latest | en | 0.914179 |
http://infosciphi.info/poker/craps-strategy-bet-odds.php | 1,550,919,997,000,000,000 | text/html | crawl-data/CC-MAIN-2019-09/segments/1550249500704.80/warc/CC-MAIN-20190223102155-20190223124155-00012.warc.gz | 131,595,432 | 12,509 | 4692
# Craps strategy bet odds
What is the free odds bet in craps? Here we summarize to give you an overall understanding and we link to our other articles that cover it in more detail. Craps Strategy - Learn How to Win at Craps with the Best Craps Strategy and Tips. A Complete Guide for the Best Way to Play Craps and Win | INFOSCIPHI.INFO CRAPS STRATEGY TIPS When playing craps at casinos, get comfortable with the pass, come, and free odds wagers. The only good wagers are the pass, don’t pass, come, don’t come, free odds, and placing the 6 and 8 wagers.
## CrapsPit.org
Keep off from Proposition Bets The worse strategy for online craps is without contradiction placing proposition bets. This is compensated after the point number is thrown as it is the casino who has the edge. It informs other players and dealers that betting needs to stop to usher in the next toss of dice. Craps is considered the most exciting game in the casinos, it is said you can find your way blindfolded to a craps table led by the screaming and yelling of the players. To win with any rolled number besides a seven. How to Play Craps. It is important to remember the chart as it is used in basic craps strategy.
• Croupier's clothes are given out by the casino. There are no pockets there, so you can not hide or steal chips.
• The longest game of poker in the casino is 90 hours. The record belongs to the Irishman Phil Laak. Phil not only set a record, but also won 7 thousand dollars.
## Craps Strategy: All You Want To Know
To understand why the Free Odds bet is so important you first have to understandthe house edge. In craps, the house edge on the Pass Line bet is 1. The house edge is the mathematical average for thelong run. Different bets carry a different house edge. The Free Odds bet carries no house edge.
The casino makes no profit on thisbet. You make a Pass Line bet, and the shooterrolls a number to establish a point 4,5,6,8,9, or Now if the shooter rolls the point again,you not only win your Pass Line bet, you also win the Odds bet.
The way the casino makes its profiton all other bets in the casino is by paying less than true odds. The true odds varies according to the point, and so the payoff varies as well. The amount you can bet on the Free Odds varies from casino to casino, and is postedon a sign on the table. Since the Oddsbet carries no house edge, it pays to seek out casinos that offer the maximum oddspossible. In Vegas you can get better odds downtown vs the strip; many downtowncasinos offer 10x Odds, or even better.
The exception on the strip is the Casino Royale , which offers an amazing x Odds on craps. The Free Odds bethas the effect of diluting the edge of the flat bet you have to make before youcan make the Odds bet. With Full Double Odds means the player can take 2. If the player takes the maximum oddsthen the payoff will conveniently be seven times the pass or come bet. See ourseparate article about odds. With the Pass or Come you bet a smaller amount to wina bigger amount. Now you have the opportunityto make a Free Odds bet.
Голубые стены хорошо гармонировали с сисечками бестии и ее лобком, на котором была небольшая щетина. Она разделась до трусиков и стала перед ним, заслонив море своей красивой и упругой попкой. "As long as this was seen as a gay disease. She then looked up at me, smiled wickedly, and said, "I think you had better get out of these young man!". Да, если желток растечётся, получишь пять плетей. Прекрасный прошмандовки с удовольствием показывают упругие ягодицы не прикрытые трусами.
### View Details
Play to Win Real Money! Looking for a little more excitement? Try a free cash bonus offer at a safe, regulated online casino? A good strategy in craps is to start with a pass line bet. Once you have established a point number then take the maximum odds on that number. This will ensure you can paid at a better rate. We can now take odds of 4 times that amount on most tables. The chance of a 5 hitting before a 7 is always the same, so really we want the maximum percentage return on our bet if it does hit.
On the next roll of the nice you should make a Come bet. Once a number gets established via the Come you should take the full odds on that. Again this is for the same reason as our example above. And then we should repeat that again to establish a 2nd number via the Come.
This will give us 3 numbers in total. Lets say they were 8 on the come out roll the true point number , and 5 and 9 on the Come bets the pseudo point numbers. This puts us in the situation of winning on 5,8 and 9. Nothing happening on 2,3,4,6,10,11, and And losing on 7. Stopping at 3 numbers is recommended because a 7 is going to be rolled once every 6 times on average, so you want to minimise you losses when the dice is unkind to you. If you hit the 5,8, or 9, the number you hit gets paid off and cleared so it is good to re-establish 3 numbers again all with max odds either via the pass line or the come box depending which number was cleared.
After that you might want to consider not recharging the numbers, as a seven has to appear eventually. If we manage to get say get 5,8,9 cleared, establishing replacements of 4,5,10 as we go, which also get cleared. And then keep playing a single number say 6 , hitting it once, and then losing, our outcome would be like this: We want to make enough profit by hitting numbers to more than cover the loss when a 7 hits.
Another good way to get more numbers in action after the initial point number is established is to make some direct place bets.
## Craps on the hop bet
Although you can play craps knowing only a couple of bets, you can have more fun and change things up a bit when you know all the betting possibilities available to you and how and when to place them. Two six-sided dice are used in the game.
The dice are passed counter clockwise after each new round. When a new round of craps begins the following takes place:. Other players at the table also make bets. The shooter's first roll is known as the come-out roll. If the come-out roll is a 7 or 11, then the Pass bet wins and the Don't Pass bets lose. This scenario ends the round. This also ends the round.
## Video
### Free Roulette
The thrill of watching the spinning red and black Roulette wheel has long served to grip many avid gamblers around the g... | 1,527 | 6,359 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2019-09 | latest | en | 0.926549 |
https://mycalcu.com/20-ms-to-mph | 1,653,031,930,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662531762.30/warc/CC-MAIN-20220520061824-20220520091824-00707.warc.gz | 456,782,654 | 6,509 | # 20 Ms To Mph
20 Ms to Mph calculator converts 20 meters per second (m/s) into miles per hour (mph) quickly .
## How many miles per hour made up 20 meters per second?
In order to carry out the conversion quickly, you can simply multiply 20 Ms by 2.23694 to convert it into Mph.
## What is the value of 20 Ms in Mph?
20 M/s is equal to 44.74 Mph.
## 20 Miles Per Second Conversion
Mile Per Hour 44.74 Meter Per Second 44.74 Kilometers Per Hour 72 Knot 38.88
20 M/s to Mph calculator converts 20 meters per second into miles per hour and also converts 20 Ms into other units such as knot, Fps, and more simultaneously.
ms mph
20.00 ms 44.74 mph
20.01 ms 44.76237 mph
20.02 ms 44.78474 mph
20.03 ms 44.80711 mph
20.04 ms 44.82948 mph
20.05 ms 44.85185 mph
20.06 ms 44.87422 mph
20.07 ms 44.89659 mph
20.08 ms 44.91896 mph
20.09 ms 44.94133 mph
20.10 ms 44.9637 mph
20.11 ms 44.98607 mph
20.12 ms 45.00844 mph
20.13 ms 45.03081 mph
20.14 ms 45.05318 mph
20.15 ms 45.07555 mph
20.16 ms 45.09792 mph
20.17 ms 45.12029 mph
20.18 ms 45.14266 mph
20.19 ms 45.16503 mph
20.20 ms 45.1874 mph
20.21 ms 45.20977 mph
20.22 ms 45.23214 mph
20.23 ms 45.25451 mph
20.24 ms 45.27688 mph
20.25 ms 45.29925 mph
20.26 ms 45.32162 mph
20.27 ms 45.34399 mph
20.28 ms 45.36636 mph
20.29 ms 45.38873 mph
20.30 ms 45.4111 mph
20.31 ms 45.43347 mph
20.32 ms 45.45584 mph
20.33 ms 45.47821 mph
20.34 ms 45.50058 mph
20.35 ms 45.52295 mph
20.36 ms 45.54532 mph
20.37 ms 45.56769 mph
20.38 ms 45.59006 mph
20.39 ms 45.61243 mph
20.40 ms 45.6348 mph
20.41 ms 45.65717 mph
20.42 ms 45.67954 mph
20.43 ms 45.70191 mph
20.44 ms 45.72428 mph
20.45 ms 45.74665 mph
20.46 ms 45.76902 mph
20.47 ms 45.79139 mph
20.48 ms 45.81376 mph
20.49 ms 45.83613 mph
20.50 ms 45.8585 mph
20.51 ms 45.88087 mph
20.52 ms 45.90324 mph
20.53 ms 45.92561 mph
20.54 ms 45.94798 mph
20.55 ms 45.97035 mph
20.56 ms 45.99272 mph
20.57 ms 46.01509 mph
20.58 ms 46.03746 mph
20.59 ms 46.05983 mph
20.60 ms 46.0822 mph
20.61 ms 46.10457 mph
20.62 ms 46.12694 mph
20.63 ms 46.14931 mph
20.64 ms 46.17168 mph
20.65 ms 46.19405 mph
20.66 ms 46.21642 mph
20.67 ms 46.23879 mph
20.68 ms 46.26116 mph
20.69 ms 46.28353 mph
20.70 ms 46.3059 mph
20.71 ms 46.32827 mph
20.72 ms 46.35064 mph
20.73 ms 46.37301 mph
20.74 ms 46.39538 mph
20.75 ms 46.41775 mph
20.76 ms 46.44012 mph
20.77 ms 46.46249 mph
20.78 ms 46.48486 mph
20.79 ms 46.50723 mph
20.80 ms 46.5296 mph
20.81 ms 46.55197 mph
20.82 ms 46.57434 mph
20.83 ms 46.59671 mph
20.84 ms 46.61908 mph
20.85 ms 46.64145 mph
20.86 ms 46.66382 mph
20.87 ms 46.68619 mph
20.88 ms 46.70856 mph
20.89 ms 46.73093 mph
20.90 ms 46.7533 mph
20.91 ms 46.77567 mph
20.92 ms 46.79804 mph
20.93 ms 46.82041 mph
20.94 ms 46.84278 mph
20.95 ms 46.86515 mph
20.96 ms 46.88752 mph
20.97 ms 46.90989 mph
20.98 ms 46.93226 mph
20.99 ms 46.95463 mph
## More Calculations
21 ms to mph 25 ms to mph 20.1 ms to mph 20.2 ms to mph 20.3 ms to mph 20.4 ms to mph 20.5 ms to mph 20.6 ms to mph 20.7 ms to mph 20.8 ms to mph 20.9 ms to mph | 1,354 | 3,007 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.078125 | 3 | CC-MAIN-2022-21 | latest | en | 0.449519 |
http://hvac-talk.com/vbb/showthread.php?178699-indirect-heat-and-solar-thermal&goto=nextnewest | 1,495,582,690,000,000,000 | text/html | crawl-data/CC-MAIN-2017-22/segments/1495463607704.68/warc/CC-MAIN-20170523221821-20170524001821-00005.warc.gz | 160,539,251 | 21,187 | # Thread: Economic balance point cut over question…
1. Regular Guest
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## Economic balance point cut over question…
Economic balance point cut over question…
Just went DF. New unit keeping the house cool without humidity this summer. Thanks to all!
New 14 SEER 4 ton Amana AZ 410A heat pump with coil.
I have set the cut over to change over to oil heat at 20* outdoor temp.
Based on my calculations, with a COP of 2.65 at 20* - oil at 3.85 a gallon and .21 KWH
#2 Oil needs to plummet to roughly 2.65 a gallon to begin considering bring the cut over upwards to 25* or 30*
Is my math on?
I used many of the online calculators on this site so I think I am close to reasonable number.
Thanks again…
2. Your math is correct for economical balance point.
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By cut over I assume you mean locking out the HP to not run at all. Without knowing the efficiency of your oil furnace I can't confirm the calculations, but it sounds like you have the COP numbers and therefore are on the right track.
I wouldn't be surprised if your oil furnace has to run at least part time at those lower temps.
Its been too long since I did COP and EER calculations on test units, so maybe others on this forum can tell us if defrost is accounted for in those COP numbers. If not, that COP number will be lower (negative?) on days when defrost occurs often. If you're in a very dry climate this isn't much of an issue.
4. COP NEVER includes defrost.
HSPF includes defrost.
COP is just BTU output per WATT consumed in heat mode. Defrost is not heat mode.
At 82% efficiecny, oil is still \$1.03 per 100,000 BTU output more expensive then the heat pump at a COP of 2.65.
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That sat will cut out the heat pump for 90 sec. before it kicks on the oil furnace.
The furnace is a 5 ton 80% armstrong.
How often might my heat pump be in defrost mode when it is say 25* outside?
Unit is on the unshadded South side of the house.
Long story short, unless #2 goes below \$3 I donot even have to think about it... yes?
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are you considering the difference in power usage for the blower? The blower will be running considerably longer for a heat pump at 20 degrees than it would be for your furnace. That might bump the balance point up a bit.
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Blower is a 5 ton 3/4 horse PCS motor.
How woud I handy-cap the added elcecitcal use of the blower into my cut over numbers?
Sounds like tall order.
9. A ¾HP PSC blower motor, will consume around 746 watts.(give or take a couple)
That 746 watts will also add 2,546 BTUs per hour of run time to the air.
So its a semi wash. It used more electric running longer, but it also added heat to the home.
It would have a COP of 1. So it would be the same as an electric strip heater.
You will get 3413 BTUs for every kilowatt it consumes.
10. Professional Member
Join Date
Jun 2008
Posts
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Originally Posted by jrongo
How often might my heat pump be in defrost mode when it is say 25* outside?
In part it depends on humidity, but I know it can occur at 40 degrees on down. Would love to know from the pros on its perceived or observed frequency at a given temp and humidity.
Originally Posted by beenthere
At 82% efficiecny, oil is still \$1.03 per 100,000 BTU output more expensive then the heat pump at a COP of 2.65.
With that kind of BTU cost spread, is there any reason to lock out the HP at 20 degree temp?
11. Thats his economic balanc point. Not his thermal.
I have some dual fuels in, that teh HP isn't locked out. Its first stage heat no matter what OD temp. If it can't keep up, the stat calls for second.
Every once in awhile. A customer calls and wants is locked out before the discharge air temp gets cool for them.
So, its what ever the customer wants. High comfort, or high efficiency.
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Have faith. But just because it's possible, doesn't mean it will be easy. Know that whatever life you want, the grades you want, the job you want, the reputation you want, friends you want, that it's possible.
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Discussion: All Topics Topic: HELP!!!!!!!!
Post a new topic to the General Discussion in Algebra for Rational Equations discussion
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Subject: HELP!!!!!!!! Author: Frank Date: Sep 26 2003
y-5= -10/11(x-(-7)).
(x+7)
y-5= -10/11x - 70/11
y= -10/11x - 70/11 + 5/1
????????????????????????
Im stuck here, Plz help!I need to sovle this in standard from as
Ax-By-C=0
Answer that my professor gave me was 8x-11y-23=0
Plz explain how he got this answer.
Reply to this message Quote this message when replying? yes no
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#### Bending Moment & Shear Force Calculator for point load on Cantilever beam
Point Load on cantilever
This calculator provides the result for bending moment and shear force at a section (distance "x") from the fixed support of cantilever beam subjected to a point load. It also calculates maximum bending moment value which occurs at the fixed end. This calculator gives shear force values on both sides (left and right) of the section. Please refer to the figure and enter the values of load and distances in the form given below and then press "Calculate". All upward acting loads are taken as positive and downward as negative. All the distances should be measured with reference to the origin taken at the fixed end. This calculator can also be used to find the ordinates of Influence line diagram for structures just by putting P=1 at different locations on the beam.
You can visit instructions for bending moment calculator if you need more help.
INPUT VALUES
Span of cantilever (m):
Load P (kN):
Distance 'a' (m):
Distance 'x' (m):
Please make sure that
all the distances are Positive.
OUTPUT RESULTS
Fx-Left (kN):
Fx-Right (kN):
Bending Moment Mx (kNm):
Max. Bending Moment (kNm):
Maximum Bending Moment occurs
at fixed support of cantilever
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# Question A metal rod of resistance 20 Ω is fixed along a diameter of a conducting ring of radius 0.1 m and lies on x-y plane. There is a magnetic field = (50 T) . The ring rotates with an angular velocity ω = 20 rad/sec about its axis. An external resistance of 10Ω is connected across the centre of the ring and rim. Find the current through external resistance.
State two uses of a convex mirror
Solution:
Explanation:
The convex mirror reflects the light outwards and is therefore it is used to spread the light.
The two uses of convex mirror.
It is used as a rear view mirror in the vehicles.
It is used in the street light reflectors.
It is used as a sunglass.
The convex mirror is used as vehicles mirror,magnifying glasses.
Name three bad conductors of heat
Solution:
Explanation:
The object is bad conductors of heat means it is not conducted the heat easily.
The bad conductor heats are Plastic,rubber,wood.
The bad conductors of heat are plastic,rubber,wood,glass.
What is the force between two small charged spheres having charges
of 2×10^-7C and 3× 10^-7 C placed 30 cm apart in air?
Solution:
Explanation:
Charge on the first sphere, q1=2×10-7C
Charge on the second sphere, q2=3×10-7C
Distance between the spheres, r=30 cm=0.3m
Hence, force between the two small charged spheres is 6×10-3N .
The charges are of same nature. Hence, force between them will be repulsive
The ground state energy of hydrogen atom
is -13.6 eV. What are the kinetic and
potential energies of the electron in this state?
Ground state energy of hydrogen atom, E = - 13.6 eV
This is the total energy of a hydrogen atom. Kinetic energy is equal to the negative of
the total energy.
Kinetic energy = - E= - (- 13.6) = 13.6 eV
Potential energy is equal to the negative of two times of kinetic energy.
Potential energy = - 2 x (13.6) = - 27.2 eV
Solution:
Explanation:
Ground state energy of hydrogen atom is E=−13.67eV
The kinetic energy of the electron is given as: KE=−E =+13.67eV
The potential energy of the electron is given as: PE=−2×KE =−27.2eV
Hence,
The kinetic energy of the electron is +13.67eV
The potential energy of the electron is −27.2eV
The following configuration of gate equivalent to
A: NAND
B: XOR
C: OR
D: None of these
Solution:
Explanation:
The given output can be written as
Y=(A+B)(¯A+¯B)
Y=AA+A¯B+A¯B +B¯B
Y=0+A¯B+A¯B +0
Y=A¯B+A¯B
This the output of XOR gate. | 669 | 2,442 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.1875 | 4 | CC-MAIN-2022-49 | latest | en | 0.91777 |
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Hey everyone!
I was wondering if someone could help me fix two of my algorithms. I am trying to factorize semi-prime numbers by using a brute force method and a square root method.
The brute force method should start by multiplying all odd numbers between 1 and the semi-prime number. Meaning that 1 should multiply 3 then 5 then 7, etc. Then three should multiply 5 then 7 then 9 and so forth. The first number would never multiply a number equal to or less than itself.
Any help at all would be greatly appreciated!
Here is what I have:
``123456789101112`` ``````bruteForce(int x){ f1 = 1; f2 = f1 + 2; while (f1 < x){ for (f2 = x; f2 <= x; f2-2){ f1 * f2; } if (x%f1 == f2 || x%f2 == f1) printFactors(f1,f2); } f1 = f1 + 2; }; ``````
For my other algorithm, I want it to get the square root of the semi-prime and then start multiplying odd numbers around it. I made it so that the square root would have to be set to the next even number (in case it was a factor). If the product of the two odd numbers (a * b) is greater than the original semi-prime, I would multiply b by the next smaller number than a. Conversely, if the product of a and b is less than the original semi-prime, it will multiply a by the next odd number larger than b. This will repeat until the two correct factors of the original semi-prime are found.
Algorithm:
``123456789101112131415161718`` ``````SquareRootFactorize(int x){ int N = x; sqrtN = sqrt(N); if (isOdd(sqrtN) == true) sqrtN = sqrtN + 1; f1 = sqrtN - 1; f2 = sqrtN + 1; cout << sqrtN; while (N != n){ n = multFacts(f1,f2); if (N < (f1 * f2)) f1 = f1 + 2; else if (N > (f1 * f2)) f2 = f2 + 2; else if (N = n) printFactors(f1,f2); } }``````
The fun about algorithms is that you sometimes think your brain is bleeding trying to get a glance of what the hell is going on
Giving you the answer wouldn't help you at all (even if you were to understand what's happening)
The key of learning how-to is to come up with a solution yourself and learn the process how to think to get there
Maybe I can point you in a direction and show you that whatever you think the code below is doing isn't what it acutally does.
``1234567891011121314`` ``````bruteForce(int x){ f1 = 1; f2 = f1 + 2; // here you set f2 to 3 while (f1 < x){ // looks okay for (f2 = x; f2 <= x; f2-2){ // and here you set f2 to x which overwrites the 3 f1 * f2; // I think your heart is in the right place (is x an odd number?) // if not then you are about to multiply all even numbers from x down to infinity // you have to save the result of f1 * f2 somewhere here so you use the result for your next iteration // On top of all that it looks like an infinite loop (since your f2 is getting smaller it will always be <= x ( btw you wrote f2-2 it should be f2 -= 2) } if (x%f1 == f2 || x%f2 == f1) // I don't get that printFactors(f1,f2); } f1 = f1 + 2; // this seems okay };``````
If you figured this out come back for the second one but don't try to multitask while both Tasks include algorithms ;)
Well I hope this helps you somehow
Last edited on
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### Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code
Apple Division CSES Problem Set Solution | CSES Problem Set Solution Apple division with code - Apple Division CSES Problem Solution Easy Explanation. Apple division is problem is taken form cses introductory problem set.Let's Read Problem statement first. Problem Statement- Time limit: 1.00 s Memory limit: 512 MB There are n n apples with known weights. Your task is to divide the apples into two groups so that the difference between the weights of the groups is minimal. Input The first input line has an integer n n : the number of apples. The next line has n n integers p 1 , p 2 , … , p n p 1 , p 2 , … , p n : the weight of each apple. Output Print one integer: the minimum difference between the weights of the groups. Constraints 1 ≤ n ≤ 20 1 ≤ n ≤ 20 1 ≤ p i ≤ 10 9 1 ≤ p i ≤ 10 9 Example Input: 5 3 2 7 4 1 Output: 1 Explanation: Group 1 has weights 2, 3 and 4 (total weight 9), and group 2 has weights 1 and 7 (total weight 8). Join Telegram channel for code discussi
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Movie Festival CSES problem set solution - Problem statement- Time limit: 1.00 s Memory limit: 512 MB In a movie festival n n movies will be shown. You know the starting and ending time of each movie. What is the maximum number of movies you can watch entirely? Input The first input line has an integer n n : the number of movies. After this, there are n n lines that describe the movies. Each line has two integers a a and b b : the starting and ending times of a movie. Output Print one integer: the maximum number of movies. Constraints 1 ≤ n ≤ 2 ⋅ 10 5 1 ≤ n ≤ 2 ⋅ 10 5 1 ≤ a < b ≤ 10 9 1 ≤ a < b ≤ 10 9 Example Input: 3 3 5 4 9 5 8 Output: 2 Solution - Step -1 take input in a vector as a pair first element is ending time ans second element is starting time Step -2 sort the vector in increasing order because we will watch that movie which is ending first. Step -3 iterate over vector and calculate ans . follow code below- | 909 | 3,780 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.046875 | 3 | CC-MAIN-2024-33 | latest | en | 0.868442 |
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Solutions to Quiz 4
# Solutions to Quiz 4 - Solutions to Quiz 4 Q1 A Keeping In...
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Solutions to Quiz 4 Q1) A) Keeping In order to calculate, we need and Thus, Now we have, Here, Thus we have, B) Using the above derived equation: C) In that case, we will have a 4×4 martrix. In order to identify our equations, we need to put restrictions on all the elements above principle diagonal, i.e., 6 restrictions. D) For a shock in t 0 1 0 1 0.2 0.8 2 0.68 0.32 3 0.392 0.608 4 0.5648 0.4352 5 0.46112 0.53888 6 0.523328 0.476672 7 0.486002 0.513997 8 0.508398 0.491602
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.2 .4 .6 .8 1 yt 0 5 10 15 20 25 t For a shock in t 0 1 0 1 0.8 0.2 2 0.32 0.68 3 0.608 0.392 4 0.4352 0.5648 5 0.53888 0.46112 6 0.476672 0.523328 7 0.513997 0.486003 8 0.491602 0.508398
0 .2 .4 .6 .8 yt2 0 5 10 15 20 25 t Q2) ACF or correlogram is a correlation coefficient between pairs of values of , separated by an interval of length k. First ACF depict a white noise series where all the coefficients are zero Second ACF depict a non-stationary time series where coefficients are slowly decreasing Third ACF depict a AR(1) process where coefficients are fatly decreasing B) PACF is the correlation between the current value and the value k period ago, after controlling for observation at the intermediate lags, i.e., all lags < k
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Jill Tulane University ‘16, Course Hero Intern | 766 | 2,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.875 | 4 | CC-MAIN-2018-22 | latest | en | 0.815655 |
https://www.askiitians.com/forums/Magical-Mathematics%5BInteresting-Approach%5D/29/7149/plzz-solve-this-sum.htm | 1,723,496,868,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641048885.76/warc/CC-MAIN-20240812190307-20240812220307-00425.warc.gz | 501,003,567 | 43,071 | # x6 - 10x3 + 27:
148 Points
14 years ago
Dear ajinkya
x6 - 10x3 + 27=0
let x3 =t
so t2 -10t +27 =0
discrimnent of this equation is negative so no real value of t
so this equation has no real value of x
Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed
solution very quickly. | 126 | 419 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2024-33 | latest | en | 0.920916 |
https://math.stackexchange.com/questions/1285412/closed-forms-for-definite-integrals-involving-error-functions | 1,576,150,362,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540543252.46/warc/CC-MAIN-20191212102302-20191212130302-00370.warc.gz | 457,344,043 | 32,465 | # Closed forms for definite integrals involving error functions
I have been working for a while with these kinds of integrals
$$\int_0^\infty dx\,\text{erfc}\left(c +i x\right)\exp \left(-\frac{1}{2}d^2x^2+i cx\right)$$ $$\int_\Lambda^\infty dx\,\frac{1}{x}\text{erfc}\left(c +i x\right) \exp \left(-\frac{1}{2}d^2x^2+i cx\right)$$ where $c$ and $d$ are just real constants and $\Lambda>0$.
I have also been working with other similar integrals that have a closed-form expression, but I can't figure out the form of these ones. Does anyone know if these integrals have a closed-form solution?
• Hint: Replace $\text{erfc}(\color{red}c+ix)$ with $\text{erfc}(\color{red}u+ix)$, then differentiate both expressions with regard to u, and the second one also with regard to c. – Lucian May 17 '15 at 2:19
First of all for both integrals to exist we need $$d> \sqrt{2}$$ or otherwise the integral is divergent around plus infinity. Now the first integral has a closed form in terms of the Owen's T function as shown below: $$\begin{eqnarray} &&\int\limits_0^\infty \operatorname{erfc}(c+ \imath x) \exp(-\frac{1}{2} d^2 x^2+\imath c x) dx=\\ && \exp(-\frac{1}{2} \frac{c^2}{d^2} ) \frac{\sqrt{2\pi}}{d} \int\limits_{-\imath \frac{c}{d}}^\infty \operatorname{erfc}(c-\frac{c}{d^2}+\imath \frac{x}{d}) \frac{\exp(-1/2 x^2)}{\sqrt{2 \pi}} dx=\\ && \exp(-\frac{1}{2} \frac{c^2}{d^2} ) \frac{\sqrt{2\pi}}{d} \left( \frac{1}{2} \operatorname{erfc}(-\imath \frac{c}{\sqrt{2} d}) - 2 T(-\imath \frac{c}{d}, \imath\frac{\sqrt{2}}{d},(c-\frac{c}{d^2})\sqrt{2})\right)=\\ &&- \frac{\exp(-\frac{1}{2} \frac{c^2}{d^2} )}{\sqrt{2 \pi}d} \left( 4 \pi T\left(\frac{c-\frac{c}{d^2}}{\sqrt{\frac{1}{2}-\frac{1}{d^2}}},\frac{i d}{\sqrt{2} \left(d^2-1\right)}\right)+4 \pi T\left(\frac{i c}{d},i \sqrt{2} d\right)+\pi \operatorname{erf}\left(\frac{c-\frac{c}{d^2}}{\sqrt{1-\frac{2}{d^2}}}\right)-\pi \operatorname{erfc}\left(-\frac{i c}{\sqrt{2} d}\right)-2 i \tanh^{-1}\left(\frac{d}{\sqrt{2} \left(d^2-1\right)}\right)+2 i \tanh ^{-1}\left(\frac{\sqrt{2}}{d}\right)-2 i \tanh ^{-1}\left(\sqrt{2} d\right) \right) \end{eqnarray}$$ where $$T(h,a)$$ is the Owen's T function and $$T(h,a,b)$$ is the generalized Owen's T function (see Generalized Owen's T function).
In[1201]:= {d} = RandomReal[{Sqrt[2], 2}, 1, WorkingPrecision -> 50];
{c} = RandomReal[{0, 2}, 1, WorkingPrecision -> 50];
NIntegrate[ Erfc[c + I x] Exp[-1/2 d^2 x^2 + I c x], {x, 0, Infinity}]
NIntegrate[
Erfc[c + I Sqrt[2]/d x] Exp[- x^2 + I c Sqrt[2]/d x], {x, 0,
Infinity}] Sqrt[2]/d
Exp[-1/2 c^2/d^2] 1/d NIntegrate[
Erfc[c - c/d^2 + (I x)/d] Exp[-1/2 (x)^2], {x, -I c /(d) ,
Infinity}]
Exp[-1/2 c^2/d^2] 1/d Sqrt[
2 Pi] (1/2 Erfc[-I c /(d) 1/Sqrt[2]] -
2 T[-I c /(d), I/d Sqrt[2], (c - c/d^2) Sqrt[2]])
-Exp[-1/2 c^2/d^2] 1/(d Sqrt[2 Pi]) (2 I ArcTanh[Sqrt[2]/d] -
2 I ArcTanh[Sqrt[2] d] -
2 I ArcTanh[d/(Sqrt[2] (-1 + d^2))] + \[Pi] Erf[(c - c/d^2)/Sqrt[
1 - 2/d^2]] - \[Pi] Erfc[-((I c)/(Sqrt[2] d))] +
4 \[Pi] OwenT[(c - c/d^2)/Sqrt[1/2 - 1/d^2], (I d)/(
Sqrt[2] (-1 + d^2))] + 4 \[Pi] OwenT[(I c)/d, I Sqrt[2] d])
Out[1203]= 0.0185396 - 0.0478548 I
Out[1204]= 0.0185396 - 0.0478548 I
Out[1205]= 0.0185396 - 0.0478548 I
Out[1206]= 0.018539649231816650876610218020057720732515452423 -
0.047854753638600728058690767430930302821895582282 I
Out[1207]= 0.018539649231816650876610218020057720732515452423 -
0.0478547536386007280586907674309303028218955822824 I
If $u =\operatorname{erfc}(c+ix)$ then
$$du = \dfrac {-2} {\sqrt\pi} e^{-(c+ix)^2} i \, dx = \dfrac {-2} {\sqrt\pi} e^{-(c^2-2icx - x^2)^2} i \, dx. \tag 1$$
We need to work with $$\exp\left( \frac {-1}2 d^2x^2 + icx\right).$$ The exponent is $$\frac{-d^2} 2 \left( x^2 - \frac{2icx}{d^2} - \frac{c^2}{d^4} \right) - \frac{c^2}{2d^2} = \frac{-d^2}2\left( x - \frac{ic}{d^2} \right)^2 - \frac{c^2}{2d^2} = \frac{-d^2}2 w^2 - \frac{c^2}{2d^2}.$$
Where we find $x$ in $(1)$, replace it with $$x = w + \frac{ic}{d^2}.$$ Then do routine algebra and go on from there.
• i think there is a typo in yout dirst line... – tired May 16 '15 at 21:48
• @tired : Thank you. I've fixed it now, I how. – Michael Hardy May 16 '15 at 22:47
• Thank you for your answer. However, I don't follow how you obtain $x=u+i\frac{c}{d^2}$ from $u=\text{erfc}(c+ix)$, or maybe you are saying to perform two different changes of variables? Additionally there is still a typo in $(1)$, the last exponential should be $e^{-(c^2+2icx-x^2)}$ right? – Alex May 16 '15 at 23:27
• Sorry -- I did intend two different ones. I've changed ti. – Michael Hardy May 17 '15 at 1:25
• @MichaelHardy, I have been trying to follow your steps, and I have arrived to the following expression for the first one $$\frac{i\sqrt{\pi}}{2}e^{(d^2/2-1)c^2}\int_{\text{erfc}(c)}^0 du\,u\, e^{(d^2/2+1)[\text{erfc}^{-1}(u)]^2-c(d^2+1)\text{erfc}^{-1}(u)}.$$I have tried integrating by parts since the integral of the whole exponential has an analytic expression, but I haven't succeeded in getting a final result. Am I in the right track? – Alex May 20 '15 at 4:04 | 2,121 | 5,028 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 4, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.671875 | 4 | CC-MAIN-2019-51 | latest | en | 0.687755 |
https://www.numbersaplenty.com/28358197 | 1,675,525,945,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764500140.36/warc/CC-MAIN-20230204142302-20230204172302-00547.warc.gz | 919,959,265 | 3,358 | Search a number
28358197 = 723317387
BaseRepresentation
bin110110000101…
…1011000110101
31222100202010121
41230023120311
524224430242
62451451541
7463016650
oct154133065
958322117
1028358197
1115009a2a
1295b6bb1
135b4b8ca
143aa2897
152752667
hex1b0b635
28358197 has 8 divisors (see below), whose sum is σ = 32550336. Its totient is φ = 24201312.
The previous prime is 28358191. The next prime is 28358257. The reversal of 28358197 is 79185382.
It is a sphenic number, since it is the product of 3 distinct primes.
It is a cyclic number.
It is not a de Polignac number, because 28358197 - 211 = 28356149 is a prime.
It is a congruent number.
It is not an unprimeable number, because it can be changed into a prime (28358191) by changing a digit.
It is a pernicious number, because its binary representation contains a prime number (13) of ones.
It is a polite number, since it can be written in 7 ways as a sum of consecutive naturals, for example, 7063 + ... + 10324.
It is an arithmetic number, because the mean of its divisors is an integer number (4068792).
Almost surely, 228358197 is an apocalyptic number.
It is an amenable number.
28358197 is a deficient number, since it is larger than the sum of its proper divisors (4192139).
28358197 is a wasteful number, since it uses less digits than its factorization.
28358197 is an odious number, because the sum of its binary digits is odd.
The sum of its prime factors is 17627.
The product of its digits is 120960, while the sum is 43.
The square root of 28358197 is about 5325.2414968713. The cubic root of 28358197 is about 304.9482918702.
The spelling of 28358197 in words is "twenty-eight million, three hundred fifty-eight thousand, one hundred ninety-seven".
Divisors: 1 7 233 1631 17387 121709 4051171 28358197 | 539 | 1,790 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.296875 | 3 | CC-MAIN-2023-06 | latest | en | 0.849016 |
https://physics-network.org/what-is-the-meaning-of-universal-constant/ | 1,669,844,288,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446710771.39/warc/CC-MAIN-20221130192708-20221130222708-00506.warc.gz | 500,980,526 | 17,589 | # What is the meaning of universal constant?
Definition of universal constant : a physical constant of wide application and frequent occurrence in physical formulas the speed of light, c, the electronic charge, e, and the Planck constant, h, are universal constants.
## What is universal gravitational constant in physics?
The universal gravitational constant is the gravitational force acting between two bodies of unit mass, kept at a unit distance from each other. The value of G is a universal constant and doesn’t change. Its value is 6.67×10−11 Nm2/kg2.
## What is a gravitational constant simple definition?
gravitational constant. A constant relating the force of the gravitational attraction between two bodies to their masses and their distance from each other in Newton’s law of gravitation. The gravitational constant equals approximately 6.67259 X 10-11 newton square meters per square kilogram. Its symbol is G.
## What is universal law of gravitation short answer?
Newton’s Law of Universal Gravitation states that every particle attracts every other particle in the universe with force directly proportional to the product of the masses and inversely proportional to the square of the distance between them.
## Why is the gravitational constant important?
‘ The constant, G, determines the strength of Newton’s inverse square law in a particular system of physical units and is, not surprisingly, known as Newton’s constant of gravitation. It is considered to be a fundamental constant of nature.
## Where does the universal gravitational constant come from?
Although the gravitational constant was first introduced by Isaac Newton as part of his popular publication in 1687, the Philosophiae Naturalis Principia Mathematica, it was not until 1798 that the constant was observed in an actual experiment.
## What is the value of universal constant G and its unit?
In SI units, G has the value 6.67 × 10-11 Newtons kg-2 m2.
## How is universal gravitational constant calculated?
Derivation – Gravitational Constant Now, the SI unit of gravitational constant is given here below: 6.67 x 10-11 Newton meters square per kilogram square (N x m2 x kg-2). Throughout our solar system and galaxy, also the galaxy within the vicinity, the value of the constant is uniform.
## What is the value of G gravitational constant on Earth?
The gravitational constant denoted by G=6.67×10−11Nm2kg2 and it is constant throughout, so its value remains same all over the universe.
## Why is Newton’s law of gravitation universal?
Newton’s law of gravity is considered “universal” because it is believed to be applicable to the entire Universe. It is called so because it is applicable on all bodies having mass(like the sun, moon, earth or an apple) and the bodies will be governed by the same law, that is newton’s law of gravitation.
## What do you mean by universal law?
In law and ethics, universal law or universal principle refers as concepts of legal legitimacy actions, whereby those principles and rules for governing human beings’ conduct which are most universal in their acceptability, their applicability, translation, and philosophical basis, are therefore considered to be most …
## What are the examples of universal law of gravitation?
The moon revolving around the Earth: The moon revolves around the Earth due to Earth’s gravity. The gravitational pull of the Earth on the moon and the outward force of the moon due to its revolution around the Earth (centrifugal force) are balanced, resulting in the fixed orbital motion of the moon around the Earth.
## Who discovered universal gravitational constant?
Hint: The value of gravitational constant, G first determined was 6.75×10−11Nm2kg−2. Complete step by step answer: After nearly a century in 1798, Lord Henry Cavendish came to determine the value for that constant G by the experiment of torsional balance.
## What is the importance of G Why is it difficult to calculate?
G is quite difficult to measure because gravity is much weaker than other fundamental forces, and an experimental apparatus cannot be separated from the gravitational influence of other bodies.
## What does gravitational constant depend on?
The value of Gravitational constant is a universal constant and does not depend on any factor. It remains constant anywhere in universe and its value is equal to 6.67×10−11Nm2kg−2.
## What is Newton’s law of gravitational constant?
In symbols, the magnitude of the attractive force F is equal to G (the gravitational constant, a number the size of which depends on the system of units used and which is a universal constant) multiplied by the product of the masses (m1 and m2) and divided by the square of the distance R: F = G(m1m2)/R2.
## Is gravity constant everywhere in the universe?
Yep, the Force of Gravity is the Same Throughout the Universe.
## Is value of G same everywhere?
The value of g at all place is not the same, it varies. The value of g is more at the poles and less at the equator.
## Is 9.8 the force of gravity?
The numerical value for the acceleration of gravity is most accurately known as 9.8 m/s/s. There are slight variations in this numerical value (to the second decimal place) that are dependent primarily upon on altitude.
## At what distance from the Earth is gravity zero?
Gravity can never become zero except maybe at infinity. As we move away from the surface of the Earth the gravitational force becomes weaker but it will never become zero.
## Is gravity a law or a theory?
Universal Gravity is a theory, not a fact, regarding the natural law of attraction. This material should be approached with an open mind, studied carefully, and critically considered. The Universal Theory of Gravity is often taught in schools as a fact, when in fact it is not even a good theory.
## Is Ohm’s law a universal law?
No. Ohm’s law is not a universal law. This is because Ohm’s law is only applicable to ohmic conductors such as iron and copper but is not applicable to non-ohmic conductors such as semiconductors.
## What are the 7 Universal laws?
There are seven Universal Laws or Principles by which everyone and everything in the Universe is governed. To name them, they are the Laws of Mentalism, Correspondence, Vibration, Polarity, Rhythm, Cause and Effect and Gender. The Universe exists in perfect harmony and order by virtue of these Laws.
## What is the SI unit of universal law of gravitation?
Detailed Solution Option 3 is the correct answer: N m2 kg-2 is the SI unit of G-the universal gravitation constant. Key Points. According to the law of gravitation, the gravitational force of attraction between two bodies of mass ‘m1’ and ‘m2’, separated by a distance ‘r’ is given by. F = G m 1 m 2 r 2.
## Is gravity a force or acceleration?
Gravity is measured by the acceleration that it gives to freely falling objects. At Earth’s surface the acceleration of gravity is about 9.8 metres (32 feet) per second per second. Thus, for every second an object is in free fall, its speed increases by about 9.8 metres per second. | 1,510 | 7,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.6875 | 4 | CC-MAIN-2022-49 | latest | en | 0.920989 |
http://www.davideaversa.it/2016/01/fast-approximated-moving-average-computation/ | 1,534,760,555,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221216333.66/warc/CC-MAIN-20180820101554-20180820121554-00602.warc.gz | 491,491,525 | 28,355 | # Fast (Approximated) Moving Average Computation
Computing the Moving Average of a sequence is a common task in games (and other applications). The motivation is simple: if something happened too far in the past, probably it does not matter anymore.
One of the main problems with the standard average is that it “slows” over time. This can be a serious problem. Suppose that you are designing a game in which your player have to quickly press a button to keep a value (e.g., player speed) above a certain average. If you use the global average, after a certain amount of time, your player can stop pressing the button and still keep the average above the threshold.
The demonstration of this is quite intuitive. If you have an average a(t) at frame t and the player will not press anything in the current frame, the average at frame t+1 will be
$a(t+1) = a(t) \frac{t}{t+1}$
This factor depends on the elapsed time and becomes “almost 1” very quickly. You don’t want that. You want to keep your player on the narrow land between “boredom” and “frustration”. You cannot give to your player the possibility to win without doing nothing for 30 seconds.
The solution to this problem is simple. Use a Moving Average. The player will have to push the button faster than the threshold, but the average is computed only using the data from the last 5 second (or any other time window you want).
Unfortunately, unlike the global average, computing the moving average require additional information. For the “global average” you just need to keep in memory the current average and the elapsed time. To compute a moving average, instead, you have to add a timestamp to every sample. You have to keep the timestamp because you must know which sample is old enough to be removed from the average computation when the time window shifts.
That’s BORING. Can we do something to avoid this unnecessary complexity? Yes, we can. Let’s write the theoretical optimal formula to update the moving average.
$a(t+\Delta t) = \frac{(a(t) T - X) + S_{\Delta t}}{T}$
In the previous formula T is the time window in which we are computing the average, S is the amount of new samples we have measured in the interval between the current time and the last computation of the average and X is the “amount of average” we have lost in the last time windows shifting (that is the sum of the samples we are going to remove). We know everything but the value of X. As we have seen before, we cannot know X without storing each sample with the corresponding timestamp.
However, if we don’t need the exact value of the moving average – and in games this is usually the case – we can compute the expected value of it way faster.
$E[a(t+\Delta t]) = \frac{(a(t) T - E[X]) + S_{\Delta t}}{T}$
We are just moving the problem to compute the expected value of X. However, we can go completely blind and assume a uniform distribution. If we compute the average frequently enough, this is “almost” true in any case. In this setting, the expected value is just the current average multiplied for the time increment:
$E[X]) = a(t) \Delta t$
Therefore, we can put this in the original formula:
$E[a(t+\Delta t]) = a(t) \left( 1 - \frac{\Delta t}{T} \right) + \frac{S_{\Delta t}}{T}$
Look at the formula. You don’t need to know how many samples are gone, so you can avoid implementing timestamp for them. The main “problem” with this formula is that it never goes to zero. Fortunately, you can workaround this assuming that the moving average is zero if it goes below a certain threshold. In general, this formula is a “smoother” (filtered) version of the real moving average. Fortunately, in games nobody will never have a problem with that and, in the meanwhile, you will be spared a lot of effort and additional complexity.
## Demo
Update: Here is a small demo on how this thing works! :D
### Average: 0
This small applet computes the moving average of clicks in the last 5 seconds. Note that, if you stop clicking, it slowly decays to 0 (while the true moving average should be exactly 0 after 5 seconds of inactivity). However, in the average case, the average is good (oh, silly words). | 962 | 4,163 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 5, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.984375 | 4 | CC-MAIN-2018-34 | latest | en | 0.941788 |
http://slideplayer.com/slide/4271995/ | 1,511,610,942,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809778.95/warc/CC-MAIN-20171125105437-20171125125437-00011.warc.gz | 276,228,047 | 22,377 | # Image Processing A brief introduction (by Edgar Alejandro Guerrero Arroyo)
## Presentation on theme: "Image Processing A brief introduction (by Edgar Alejandro Guerrero Arroyo)"— Presentation transcript:
Image Processing A brief introduction (by Edgar Alejandro Guerrero Arroyo)
What is an image? We can think in an image as a two dimensional array of pixels. We can think in an image as a two dimensional array of pixels.
What is Image Processing? Image processing is any form of signal processing for which the input is an image. Image processing is any form of signal processing for which the input is an image. So...is the image a signal? So...is the image a signal? Yes it is!! Yes it is!! A image can be tought as a two dimensional signal. A image can be tought as a two dimensional signal.
Brief Motivation Quantization: It’s a signal process of approximating a continuous range of values by a relatively small discrete set of integer values. Quantization: It’s a signal process of approximating a continuous range of values by a relatively small discrete set of integer values.
Brief Motivation Image Editing: Alterating the image. Image Editing: Alterating the image.
Brief Motivation High dynamic range image: changing the range of luminances. High dynamic range image: changing the range of luminances.
Why in Parallel? It is expensive in time. It is expensive in time. Suppose a pix max of 1024x1024 8-bit pixels. Suppose a pix max of 1024x1024 8-bit pixels. In order to operate every pixel we need to do operations! In order to operate every pixel we need to do operations! We can remember the example of project 1 of class. We can remember the example of project 1 of class.
Low Level Image Processing It is natural to think about doing the operations in a image in a pixel level. It is natural to think about doing the operations in a image in a pixel level. There are several low level operations that can be applied in images. There are several low level operations that can be applied in images. Such operations can be divided depending of in what is the output based. Such operations can be divided depending of in what is the output based. A single pixel: point processing A single pixel: point processing A goup of pixels: local operations A goup of pixels: local operations All the pixels in the image: global operations All the pixels in the image: global operations
Low Level Image Operations Thresholding The idea is to define a threshold value. The idea is to define a threshold value. Later on, all the pixels with values above the predetermined threshold value are kept. Later on, all the pixels with values above the predetermined threshold value are kept. The rest of the pixels are reduce to 0. The rest of the pixels are reduce to 0.
Low Level Image Operations Contrast Stretching The range of the gray-level values is extended. The range of the gray-level values is extended. Therefore the details are more visible. Therefore the details are more visible. Original range Original range Contrast range Contrast range
Low Level Image Operations Histograms It’s a function that shows the # of pixels of an image at each gray level. It’s a function that shows the # of pixels of an image at each gray level. It is useful to find the variations of gray levels in an image. It is useful to find the variations of gray levels in an image.
Low Level Image Operations Smooothing It suppresses large fluctuations in intensity over the image area. It suppresses large fluctuations in intensity over the image area. It could be achieved by reducing the high frequency contend. It could be achieved by reducing the high frequency contend. It reduces the noise in the image but blurs it. It reduces the noise in the image but blurs it. A simple technique is to take the mean of a group of pixels as the new value of the central pixel (project 1). A simple technique is to take the mean of a group of pixels as the new value of the central pixel (project 1).
Low Level Image Operations Computing the Mean
Low Level Image Operations Sharpening It accentuates the transitions enhancing the detail. It accentuates the transitions enhancing the detail. One approach is to reduce the low frequence content One approach is to reduce the low frequence content Another one is to accentuate changes through differentiation. Another one is to accentuate changes through differentiation.
Low Level Image Operations Sharpening
Noise Reduction It suppresses a noise signal present in the image. It suppresses a noise signal present in the image. The noise signal itself may be a random signal completely uncorrelated with the image signal. The noise signal itself may be a random signal completely uncorrelated with the image signal. One way to reduce this last is capturating the image several times and taking the averange of each pixel. One way to reduce this last is capturating the image several times and taking the averange of each pixel.
Low Level Image Operations Noise Reduction Luminance fluctuations along thin blue and red strips of pixels in the top and the bottom images respectively.
Low Level Image Operations Noise Reduction
Edge Detection The idea of identify objects from other is clearly quite important. The idea of identify objects from other is clearly quite important. What is an edge? It’s a significant change in the gray level intensity. It’s a significant change in the gray level intensity.
Edge Detection Gradient and Magnitude Let f(x) be a one-dimentional gray level function. Let f(x) be a one-dimentional gray level function. f '(x) measures the gradient of the transition. f '(x) measures the gradient of the transition. f ''(x) helps to identify the exact position of transition. f ''(x) helps to identify the exact position of transition.
Edge Detection But an image is a two dimensional discretized gray level function f(x,y). But an image is a two dimensional discretized gray level function f(x,y). The norm of the gradient can be aproximated to reduce computations. The norm of the gradient can be aproximated to reduce computations.
Edge Detection Edge Detection Using Masks The idea is to use the partial derivates of the function, (discrete). The idea is to use the partial derivates of the function, (discrete). In this way we can know the difference between neighbouring pixel gray levels in a row and in a column. In this way we can know the difference between neighbouring pixel gray levels in a row and in a column.
Edge Detection Example (Done in class)
Edge Detection Prewitt Operator As always, using more points we get better results. (Example) As always, using more points we get better results. (Example)
Edge Detection Sobel Operator Both edge detection and Smoothing. Both edge detection and Smoothing.
Edge Detection Sobel Operator
Edge Detection Sobel Operator
Edge Detection Laplace Operator If the first derivate is good, the second is better. If the first derivate is good, the second is better. After it, it’s good to apply an edge detection operator, (threshold is prefered), to get either black or white pixels. After it, it’s good to apply an edge detection operator, (threshold is prefered), to get either black or white pixels.
Edge Detection Laplace Operator
The Hough Transform It ‘s useful to find the parameters of equations of lines that most likely fit the sets of pixels in an image. It ‘s useful to find the parameters of equations of lines that most likely fit the sets of pixels in an image. The proporse of use this is because is cheaper than using a edge detector in simple lines, (straight lines, circles, etc). The proporse of use this is because is cheaper than using a edge detector in simple lines, (straight lines, circles, etc).
The Hough Transform Example (Done in class) Consider the problem of find the line that contains 3 given points. For each data point, a number of lines are plotted going through it, all at different angles. For each data point, a number of lines are plotted going through it, all at different angles. For each solid line a line is plotted which is perpendicular to it and which intersects the origin. For each solid line a line is plotted which is perpendicular to it and which intersects the origin. The length and angle of each dashed line is measured. The length and angle of each dashed line is measured. This is repeated for each data point. This is repeated for each data point.
The Hough Transform Example cont (Done in class)
The Hough Transform Example cont. (Done in class)
Fourier Series Joseph Fourier, (1768-1830), found that every function can be expressed as an infinite sum of Sines and Cosines. Joseph Fourier, (1768-1830), found that every function can be expressed as an infinite sum of Sines and Cosines.
Fourier Transformation The Fourier transform is an application that sends a complex function f into an “scary” function g as follow: The Fourier transform is an application that sends a complex function f into an “scary” function g as follow: Brief explanation. (Done in class) Brief explanation. (Done in class)
Discrete Fourier Transform The sequence of N complex numbers X 0,..., X N−1 is transformed into the sequence of N complex numbers X 0,..., X N−1 by the DFT according to the formula: The sequence of N complex numbers X 0,..., X N−1 is transformed into the sequence of N complex numbers X 0,..., X N−1 by the DFT according to the formula: where is a primitive N'th root of unity. Calculate sequentialy this serie will cost N xN products and N sums => O(N^2) Calculate sequentialy this serie will cost N xN products and N sums => O(N^2)
Discrete Fourier Transform The Fourier Transform is used if we want to access the geometric characteristics of a spatial domain image. The Fourier Transform is used if we want to access the geometric characteristics of a spatial domain image. In the image we can calculate a two dimensional Fourier Transformation (!!!). In the image we can calculate a two dimensional Fourier Transformation (!!!). But instead, we can calculate the Fourier transformation of rows and columns separately (clever). But instead, we can calculate the Fourier transformation of rows and columns separately (clever).
Discrete Fourier Transform The kth element of the DFT is given by: The kth element of the DFT is given by: Then, the vector result can be seen as the dot product of a matrix and a vector. Then, the vector result can be seen as the dot product of a matrix and a vector.
Discrete Fourier Transform Therefore, since the DFT can be computed as a matrix product, we can apply a numerical algorithm to achieve this, (maybe with our project 3). Therefore, since the DFT can be computed as a matrix product, we can apply a numerical algorithm to achieve this, (maybe with our project 3).
Fast Fourier Transform This method allows us to find the DFT in O(NLogN), in sequential time, instead of (N^2). This method allows us to find the DFT in O(NLogN), in sequential time, instead of (N^2).
Fast Fourier Transform Binary Algorithm
Fast Fourier Transform Binary Algorithm Analisis Computation: Given P processors and N points, at each steap each processor will calculate N/P points, (1 sum and 1 product). That’s it O(NLogN). Computation: Given P processors and N points, at each steap each processor will calculate N/P points, (1 sum and 1 product). That’s it O(NLogN). Communication: if P = N, then communication occurs at each step, one data. That’s it O(N). If P < N, then communication occurs only the first Log P steaps. That’s it O(LogP). Communication: if P = N, then communication occurs at each step, one data. That’s it O(N). If P < N, then communication occurs only the first Log P steaps. That’s it O(LogP).
Fast Fourier Transform The image before and after Fourier Transformation. We take the logarithm after FT, otherwise we will only see a point. The image before and after Fourier Transformation. We take the logarithm after FT, otherwise we will only see a point.
Bibliography Wikipedia.com Wikipedia.com Parallel Programing, Wilkinson & Allen. Parallel Programing, Wilkinson & Allen. Internet Internet
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Similar presentations | 2,605 | 12,258 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.234375 | 3 | CC-MAIN-2017-47 | longest | en | 0.916766 |
https://math.stackexchange.com/questions/647818/norm-of-differential-operator-on-pn0-1 | 1,716,544,277,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058709.9/warc/CC-MAIN-20240524091115-20240524121115-00052.warc.gz | 327,187,571 | 35,621 | # norm of differential operator on $P^n[0,1]$
Consider the space $P^n[0,1]$ of polynomials of degree $\leq n$ on $[0,1]$, equipped with the sup norm. Now, this is a finite dimensional space, so all linear operators have to be continuous, hence bounded.
My question is: what is the norm of the differential operator $d/dx : P^n[0,1] \rightarrow P^{n-1}[0,1]$? I can't see the relation between the supremum of a polynomial $p(x)=a_0+\dots+a_nx^n$ and the supremum of $p'(x)=a_1 + 2a_2 x +\dots+n a_n x^{n-1}$.
• Do you want the exact norm or some bound? Jan 22, 2014 at 18:35
• Such results are called Bernstein inequalities. For trigonometric polyomials the norm is $n$, and the only inequality that I remember is $|P'(x)| \leq ||P|| n /(\sqrt{x(1-x)})$. That does not give any bounds, but, maybe, it would be helpfull. Bernstein inequalities (and thus concrete norms) are known in a large number of cases. and I can not google the one you need :) Jan 22, 2014 at 18:58
## 1 Answer
The answer is given here for the interval $[-1,1]$ by Markov's inequality for polynomials: The norm is $n^2$ and $||P'||_\infty=n^2||P||_\infty$ for the $n$-th Chebyshev polynomial $T_n$. For the interval $[0,1]$, we have to use scaling: If $p$ is a polynomial considered on $[0,1]$, then $q(x)=p(\frac{x+1}2)$ is a polynomial considered on $[-1,1]$. Hence $||q'||_{\infty,[-1,1]}\leq n^2||q||_{\infty,[-1,1]}$. Since $q'(x)=\frac12 p'(\frac{x+1}2)$, we find $||p'||_{\infty,[0,1]}\leq 2n^2||p||_{\infty,[0,1]}$ and equality is attained for $t_n(x)=T_n(2x-1)$. | 539 | 1,546 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.65625 | 4 | CC-MAIN-2024-22 | latest | en | 0.839489 |
http://www.cardtrick.ca/m/trick/Magic-Spelling-Card-Trick-Secret.html | 1,534,359,777,000,000,000 | text/html | crawl-data/CC-MAIN-2018-34/segments/1534221210249.41/warc/CC-MAIN-20180815181003-20180815201003-00265.warc.gz | 478,452,224 | 4,479 | # Magic Spelling Card Trick Secret
A very clever self working trick secret which is guaranteed to baffle your audience
Effect: The magician shuffles the deck and takes the top thirteen cards. Holding the cards face down, he proceeds to spell the first card name, Ace. "A-C-E," and for each letter, he puts one card under the packet of thirteen cards. He then flips over the next card (the fourth,) and it is an Ace. He repeats this process for each card number, Ace through King. At the end, he has all thirteen cards face up on the table, in sequential order.
Preparation: First, take out 13 cards from a deck, the cards do not have to be the same suit, but you need the thirteen cards from Ace to King. After you have them out, place them in this order, begin by placing the three face-up on the table. On top of that, place the eight. Then, place the Seven, Ace, Queen, Six, Four, Two, Jack, King, Ten, Nine, and last of all Five. And then put these on top of the deck face down.
Performance: The trick almost works itself. To start, pretend to shuffle the cards, leaving the top thirteen untouched. You can do some false shuffles, or you can do real shuffles but keep the top 13 cards on top. After you shuffle, you can fan through the deck, and show every card, for the pre-arranged 13 cards don't look like they are arranged at all. Remove the top thirteen cards as a group and arrange them like a fan, so that your audience can see their faces. Square up the cards, and hold them face down.
When you spell out each card, do it as follows: let's say you're spelling the word ACE. Spell A, remove the top card and place it on the bottom. Then spell C, and remove the top card and place that on the bottom. Next spell E, remove this top card and place it on the bottom. Flip the new top card and show that it's an Ace, and place it on the table.
Continue in this manner until all the cards are face up on the table. ( You spell the cards in order: Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K) Your audience may realize that the cards must have been set up beforehand, but this only adds to the mystery - and you can treat it as a puzzle for them to try to figure out to add to the mystery. Don't use cards all of the same suit. A mixed group of suits makes it seem less like a "stacked" deck
Magic Slapping Magic Tooth Pick
Feedback | 578 | 2,341 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2018-34 | longest | en | 0.940501 |
http://mathforum.org/kb/plaintext.jspa?messageID=8400510 | 1,511,596,017,000,000,000 | text/html | crawl-data/CC-MAIN-2017-47/segments/1510934809695.97/warc/CC-MAIN-20171125071427-20171125091427-00310.warc.gz | 194,959,929 | 2,138 | ```Date: Feb 23, 2013 4:43 PM
Author: mueckenh@rz.fh-augsburg.de
Subject: Re: Matheology § 222 Back to the roots
On 23 Feb., 22:15, Virgil <vir...@ligriv.com> wrote:> In article> <62781b70-dff9-4093-85d0-ff6e5bfcb...@u20g2000yqj.googlegroups.com>,>>>>>> WM <mueck...@rz.fh-augsburg.de> wrote:> > On 22 Feb., 23:39, Virgil <vir...@ligriv.com> wrote:> > > In article> > > <6cfcca98-d4e5-475d-a4bf-168639050...@n2g2000yqg.googlegroups.com>,>> > > WM <mueck...@rz.fh-augsburg.de> wrote:> > > > On 22 Feb., 22:29, Virgil <vir...@ligriv.com> wrote:> > > > > In article> > > > > <c3c197e4-2161-4ecf-a84e-d479adb05...@k4g2000yqn.googlegroups.com>,>> > > > > WM <mueck...@rz.fh-augsburg.de> wrote:> > > > > > On 21 Feb., 21:51, Virgil <vir...@ligriv.com> wrote:>> > > > > > > > Or consider the union of natural numbers in a set B while there> > > > > > > > remains always one number in the intermediate reservoir A.>> > > > > > > > A B> > > > > > > > --> 1 -->{ }> > > > > > > > --> 2,1 -->{ }> > > > > > > > --> 2 -->1> > > > > > > > --> 3, 2 -->1> > > > > > > > --> 3 -->1, 2> > > > > > > > --> 4, 3 -->1, 2> > > > > > > > --> 4 -->1, 2, 3> > > > > > > > ...> > > > > > > > --> n -->1, 2, 3, ..., n-1> > > > > > > > --> n+1, n -->1, 2, 3, ..., n-1> > > > > > > > --> n+1 -->1, 2, 3, ..., n-1, n> > > > > > > > ...>> > > > > > > > One would think that never all naturals can be collected in B,> > > > > > > > since a> > > > > > > > number n can leave A not before n+1 has arrived.>> > > > > > > > Of course this shows that ZF with its set of all natural numbers> > > > > > > > is> > > > > > > > contradicted.>> > > > > > > WM's A and B are not sets but sequences of sets, so if WM wants to> > > > > > > consider a limit to any such sequences, he must first define what> > > > > > > he> > > > > > > means by such a limit, as there is no universal definition for> > > > > > > "the"> > > > > > > limit of a sequence of sets.>> > > > > > By definition of A we know it is never empty.>> > > > > There is no such thing as an "A" but only an infinite sequence of> > > > > differing A's, indexable by the infinite set of natural numbers,>> > > > In any case there is never an A = { }.> > > > Therefore similarly there is never a B = |N.>> > > There is never an A or a B which is a subset of |N either, though their> > > members are subsets of |N.>> > You are in error.> > The set A_1 = {1} is a subset of |N.>> But A_1 is merely a member of the sequence A, and is not A itself.A_1 is A in the first step.>> > The set B_1 = { } is a subset of |N>> But B_1 is merely a term of sequence B and not equal to B.B_1 is B in the first setp.Regards, WM
``` | 1,005 | 2,633 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2017-47 | longest | en | 0.621222 |
https://socialsci.libretexts.org/Bookshelves/Economics/International_Trade_-_Theory_and_Policy/05%3A_The_Heckscher-Ohlin_(Factor_Proportions)_Model | 1,723,782,461,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641333615.45/warc/CC-MAIN-20240816030812-20240816060812-00548.warc.gz | 419,466,714 | 31,926 | # 5: The Heckscher-Ohlin (Factor Proportions) Model
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The Heckscher-Ohlin (H-O; aka the factor proportions) model is one of the most important models of international trade. It expands upon the Ricardian model largely by introducing a second factor of production. In its two-by-two-by-two variant, meaning two goods, two factors, and two countries, it represents one of the simplest general equilibrium models that allows for interactions across factor markets, goods markets, and national markets simultaneously.
These interactions across markets are one of the important economics lessons displayed in the results of this model. With the H-O model, we learn how changes in supply or demand in one market can feed their way through the factor markets and, with trade, the national markets and influence both goods and factor markets at home and abroad. In other words, all markets are everywhere interconnected.
Among the important results are that international trade can improve economic efficiency but that trade will also cause a redistribution of income between different factors of production. In other words, some will gain from trade, some will lose, but the net effects are still likely to be positive.
The end of the chapter discusses the specific factor model, which represents a cross between the H-O model and the immobile factor model. The implications for income distribution and trade are highlighted.
This page titled 5: The Heckscher-Ohlin (Factor Proportions) Model is shared under a CC BY-NC-SA 3.0 license and was authored, remixed, and/or curated by Anonymous via source content that was edited to the style and standards of the LibreTexts platform. | 2,099 | 6,054 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2024-33 | latest | en | 0.198457 |
https://www.physicsforums.com/threads/car-door-shutting-at-a-certain-acceleration.380369/ | 1,508,519,909,000,000,000 | text/html | crawl-data/CC-MAIN-2017-43/segments/1508187824226.31/warc/CC-MAIN-20171020154441-20171020174441-00550.warc.gz | 963,973,000 | 14,563 | # Car door shutting at a certain acceleration
1. Feb 21, 2010
1. The problem statement, all variables and given/known data
A car has its door open at 90 degrees. The door is considered a uniform square sheet of steel of side .8 m and mass 20kg. The hinges on the door are frictionless. At time t = 0 the car accelerates with constant acceleration a = 3 m/sec^2. How long does it take the door to close?
2. Relevant equations
Torque = Mass * radius * acceleration
Torque = Mass * radius^2 * alpha
3. The attempt at a solution
T = 20 (Mass) * (.8)(Radius) * 3 (Acceleration)
T = 48
alpha = 48(Torque) / (.8^2)(Radius) * (20)(Mass) = 3.75
Now that I have angular acceleration, I need only to integrate twice to find the time.
omega = 3.75t
theta = 3.75t^2 / 2 theta = 90 degrees or pi/2 in radians so,
pi/2 = 3.75t^2 / 2 pi/3.75 = t^2 t = sqr root (pi/3.75)
t = .92 seconds
Hopefully these are the correct formulas, but I could use someone to check if these are right. Thanks in advance.
Last edited: Feb 22, 2010
2. Feb 22, 2010 | 325 | 1,033 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.75 | 4 | CC-MAIN-2017-43 | longest | en | 0.892649 |
https://structbio.vanderbilt.edu/archives/amber-archive/2004/2406.php | 1,723,713,160,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641278776.95/warc/CC-MAIN-20240815075414-20240815105414-00757.warc.gz | 447,081,200 | 4,323 | # AMBER Archive (2004)Subject: Re: AMBER: torsion scan
From: Furse, Kristina Elisabet (kristina.e.furse_at_vanderbilt.edu)
Date: Tue Jul 13 2004 - 12:37:35 CDT
Don't know about your iat section--I typically use:
iat=1,2,5,8,
with all 4 atom numbers sequentially instead of listing individually, but I use
amber7, not 8 (yet), so your way might work, too. The force looks a little high,
so maybe that's why you're not seeing a whole lot of resistance. Have you looked
at the trajectory (vmd or mddisplay) to see if the torsion is really moving?
I've rotated dihedrals before in order to do umbrella sampling, but I typically
break out the restraints in the DISANG file as follows:
#
# Drive to 75deg
&rst
iat=1,2,5,8, nstep1= 500001, nstep2= 600000,
iresid= 0, ifvari= 1, ninc= 0, imult= 0, ir6= 0,
r1= 30.0, r2= 60.0, r3= 60.0, r4= 90.0,
rk2=1.0, rk3=1.0,
r1a= 45.0, r2a= 75.0, r3a= 75.0, r4a= 105.0,
rk2a=5.0, rk3a=5.0,
&end
#
# Equilbration at 75deg
&rst
iat=1,2,5,8, nstep1= 600001, nstep2= 700000,
iresid= 0, ifvari= 0, ninc= 0, imult= 0, ir6= 0,
r1= 45.0, r2= 75.0, r3= 75.0, r4= 105.0,
rk2=5.0, rk3=5.0,
r1a= 45.0, r2a= 75.0, r3a= 75.0, r4a= 105.0,
rk2a=5.0, rk3a=5.0,
&end
#
# Data collection at 75deg
&rst
iat=1,2,5,8, nstep1= 700000, nstep2= 1000000,
iresid= 0, ifvari= 0, ninc= 0, imult= 0, ir6= 0,
r1= 45.0, r2= 75.0, r3= 75.0, r4= 105.0,
rk2=5.0, rk3=5.0,
r1a= 45.0, r2a= 75.0, r3a= 75.0, r4a= 105.0,
rk2a=5.0, rk3a=5.0,
&end
#
# Drive to 90deg
etc.
Good luck!
Kristina
Quoting Guanglei Cui <cuigl_at_csb.sunysb.edu>:
> Dear all,
>
> I'm trying to rotate along a dihedral angle by 360 degrees with nmropt =
> 1 in amber8. Here is the DISANG file
>
> &rst
> iat(1) = 11565, iat(2) = 11567, iat(3) = 11568, iat(4) = 11569,
> nstep1 = 1, nstep2 = 360000, irstyp = 0, ifvari = 1,
> r1 = 165.021, r2 = 170.021, r3 = 170.021, r4 = 175.021,
> r1a = 525.021, r2a = 530.021, r3a = 530.021, r4a = 535.021,
> rk2 = 5000, rk3 = 5000, rk2a = 5000, rk3a = 5000,
> &end
> &rst
> iat(1) = 0,
> &end
>
> From the output, I can see this restraint is read and applied. But from
> the value of restraint energy (fairly small), it doesn't seem to me the
> r2 and r3 are being varied during the calculation. Is there a mistake in
> the input or did I misunderstand the manual? Any input is very appreciated.
>
> Guanglei
>
-----------------------------------------------------------------
Kristina E. Furse
Department of Chemistry
Center for Structural Biology
Vanderbilt University
Email: kristina.e.furse_at_Vanderbilt.Edu
-----------------------------------------------------------------------
The AMBER Mail Reflector
To post, send mail to amber_at_scripps.edu
To unsubscribe, send "unsubscribe amber" to majordomo_at_scripps.edu | 1,135 | 2,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.671875 | 3 | CC-MAIN-2024-33 | latest | en | 0.749759 |
https://rdrr.io/cran/randtoolbox/man/pokertest.html | 1,527,392,784,000,000,000 | text/html | crawl-data/CC-MAIN-2018-22/segments/1526794867995.55/warc/CC-MAIN-20180527024953-20180527044953-00410.warc.gz | 629,050,641 | 14,859 | # pokertest: the Poker test In randtoolbox: Toolbox for Pseudo and Quasi Random Number Generation and RNG Tests
## Description
The Poker test for testing random number generators.
## Usage
`1` ```poker.test(u , nbcard = 5, echo = TRUE) ```
## Arguments
`u` sample of random numbers in ]0,1[. `echo` logical to plot detailed results, default `TRUE` `nbcard` a numeric for the number of cards, we assume that the length of `u` is a multiple of `nbcard`.
## Details
We consider a vector `u`, realisation of i.i.d. uniform random variables U1... Un.
Let us note k the card number (i.e. `nbcard`). The poker test computes a serie of 'hands' in {0, ..., k-1} from the sample u_i = floor(u_i k) (`u` must have a length dividable by k). Let n_j be the number of 'hands' with (exactly) j different cards. The probability is
p_j = 1/k^k * k! / (k-j)!) * S_k^j,
where S_k^j denotes the Stirling numbers of the second kind. Finally the chi-squared statistic is
S = ∑_{j=1}^k [n_j - np_j/k ]^2/[np_j/k].
## Value
a list with the following components :
`statistic` the value of the chi-squared statistic.
`p.value` the p-value of the test.
`observed` the observed counts.
`expected` the expected counts under the null hypothesis.
`residuals` the Pearson residuals, (observed - expected) / sqrt(expected).
## Author(s)
Christophe Dutang.
## References
Planchet F., Jacquemin J. (2003), L'utilisation de methodes de simulation en assurance. Bulletin Francais d'Actuariat, vol. 6, 11, 3-69. (available online)
L'Ecuyer P. (2001), Software for uniform random number generation distinguishing the good and the bad. Proceedings of the 2001 Winter Simulation Conference. (available online)
L'Ecuyer P. (2007), Test U01: a C library for empirical testing of random number generators. ACM Trans. on Mathematical Software 33(4), 22.
other tests of this package `freq.test`, `serial.test`, `gap.test`, `order.test` and `coll.test`
`ks.test` for the Kolmogorov Smirnov test and `acf` for the autocorrelation function.
``` 1 2 3 4 5 6 7 8 9 10 11``` ```# (1) hands of 5 'cards' # poker.test(runif(50000)) # (2) hands of 4 'cards' # poker.test(runif(40000), 4) # (3) hands of 42 'cards' # poker.test(runif(420000), 42) ``` | 654 | 2,223 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2018-22 | latest | en | 0.646421 |
https://www.bills.com/resources/personal-finance/compound-interest-calculator | 1,723,554,221,000,000,000 | text/html | crawl-data/CC-MAIN-2024-33/segments/1722641076695.81/warc/CC-MAIN-20240813110333-20240813140333-00424.warc.gz | 511,734,015 | 18,399 | # Compound Interest Calculator - Boost Your Savings
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## Compound Interest - Increase Your Savings
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So, for example, if you make monthly deposits of \$100 at 4% interest, your balance at the end of year one would be \$1200. If your account is compounded annually, then you will not earn any interest in the first year. At the end of the second year, you saved \$2,448, which is comprised of \$2,400 in deposits and \$48 in interest. However, if your savings account is compounded monthly, then your balance at the end of the first year is \$1222 and the end of the second year, \$2,494. If you continue to save for 30 years, you will have 3% more through the monthly compounding calculation.
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## Set Your Saving Plan - Let Compound Interest Work For You
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However, it may not be so obvious, but saving money helps you build your financial health. Are you saving enough? Check out Bills.com Monthly Saving Calculator to see if you are putting away enough money in your emergency fund, retirement account, and investment accounts. Keeping money in a liquid account helps you cover emergency expenses. Also, putting money away in special savings accounts enables you to build a down payment for a house, or pay for college.
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Take the first step towards a debt-free life with personalized debt reduction strategies. | 512 | 2,407 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2024-33 | latest | en | 0.933249 |
http://mathhelpforum.com/advanced-statistics/24278-binomial-distribution-proof-mean.html | 1,481,410,821,000,000,000 | text/html | crawl-data/CC-MAIN-2016-50/segments/1480698543577.51/warc/CC-MAIN-20161202170903-00302-ip-10-31-129-80.ec2.internal.warc.gz | 172,617,499 | 11,074 | # Thread: Binomial distribution proof of mean
1. ## Binomial distribution proof of mean
Can anyone provide a proof for the variance of binomial distribution?
Show that it is npq without using the Bernoulli distribution and independence way..( which is the typical way of summations or expectations)
thank you so much..
2. Originally Posted by eugene1687
Can anyone provide a proof for the variance of binomial distribution?
Show that it is npq without using the Bernoulli distribution and independence way..( which is the typical way of summations or expectations)
thank you so much..
Here is the solution from my probability book.
$E[X^k] = \sum_{i=0}^n i^k {n\choose i} p^i(1-p)^{n-i} = \sum_{i=1}^n i^k {n\choose i}p^i (1-p)^{n-i}$
Use the identity,
$i{n\choose i} = n{{n-1}\choose {i-1}}$
Thus, letting $j=i-1$,
$E[X^k] = np\sum_{j=0}^{n-1}(j+1)^{k-1}{{n-1}\choose j}p^j(1-p)^{n-1-j}=$ $npE[(Y+1)^{k-1}] \mbox{ where }Y \sim \mbox{bino}(n-1,p)$.
So that means,
$E[X] = np E[(Y+1)^0] = np$
$E[X^2] = np E[(Y+1)^1] = np[(n-1)p+1]$
That means,
$\mbox{Var}[X] = E[X^2] - (E[X])^2 = np(1-p)$
3. Originally Posted by eugene1687
Can anyone provide a proof for the variance of binomial distribution?
Show that it is npq without using the Bernoulli distribution and independence way.
Actually not knowing what text material you have been given, this could be seen as an odd question.
Therefore what I do is just a guess. I expect that your text would have shown the if X is the binomial variable then $E\left( {X^2 } \right) = np\left[ {\left( {n - 1} \right)p + 1} \right]$.
Using the standard definition for variance:
$V(X) = E\left( {X^2 } \right) - E^2 \left( X \right) = np\left[ {\left( {n - 1} \right)p + 1} \right] - \left( {np} \right)^2$.
From which the result follows.
4. ## thank you very much
hey there,
Thank you sooo much
I got the idea on how to show the proof from the help u gave..
Appreciate it..
thanks | 643 | 1,925 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 10, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2016-50 | longest | en | 0.840919 |
https://business-accounting.net/understanding-bond-yield-and-return/ | 1,708,937,038,000,000,000 | text/html | crawl-data/CC-MAIN-2024-10/segments/1707947474653.81/warc/CC-MAIN-20240226062606-20240226092606-00058.warc.gz | 147,275,547 | 8,701 | Understanding Bond Yield and Return
As a result, the bond’s yield to maturity will fluctuate, while the coupon rate for a previously existing bond will remain the same. In comparison, the current yield on a bond is the annual coupon income divided by the current price of the bond security. The Yield to Maturity (YTM) represents the expected annual rate of return earned on a bond under the assumption that the debt security is held until maturity. For bonds with multiple coupons, it is not generally possible to solve for yield in terms of price algebraically. A numerical root-finding technique such as Newton’s method must be used to approximate the yield, which renders the present value of future cash flows equal to the bond price.
This is done by using a variety of rates that are substituted into the current value slot of the formula. The true YTM is determined once the price matches that of the security’s actual current market price. Many brokerages will automatically calculate yield to maturity for you, which makes it easy to compare potential investments.
If you have an interest in corporate bonds then you will need a brokerage account. Whether or not a higher YTM is positive depends on the specific circumstances. On the one hand, a higher YTM might indicate that a bargain opportunity is available since the bond in question is available for less than its par value.
• If an investor purchases a bond at par or face value, the yield to maturity is equal to its coupon rate.
• The higher the yield to maturity, the less susceptible a bond is to interest rate risk.
• Calculating the YTM on each can give you an idea of what they’re getting for their money.
• The coupon rate is contractually fixed, whereas the YTM changes based on the price paid for the bond as well as the interest rates available elsewhere in the marketplace.
• To the bond trader, the potential for gains or losses is generated by variations in the bond’s market price.
If you can determine that a premium is worth the risk, you may have found a great fixed-income investment for your portfolio that can pay consistently and appreciate in value. Therefore, the price of bonds will fall, naturally resulting in a rise in the yield to maturity rate. Alternatively, as interest rates fall, the bonds become more attractive due to their fixed rates, their prices increase due to demand, and their yield falls. If an investor purchases a bond at par or face value, the yield to maturity is equal to its coupon rate.
What is the Yield to Maturity (YTM)?
The coupon rate is the stated periodic interest payment due to the bondholder at specified times. The bond’s yield is the anticipated rate of return from the coupon payments alone, calculated by dividing the annual coupon payment by the bond’s current market price. If the bond’s price changes and is no longer offered at par value, the coupon rate and the yield will no longer be the same. This is because the coupon rate is fixed, and yield is a derivative calculation based on the bond price.
• A bond’s yield will often stray from the original yield at the time of issue.
• Yield to maturity (YTM) is one of the most frequently used returns metrics for evaluating potential bond and fixed-income investments by investors.
• In practice, the rates that will actually be earned on reinvested interest payments are a critical component of a bond’s investment return.[9] Yet they are unknown at the time of purchase.
First and foremost is the impact of interest rate changes on realized returns. Yield to maturity assumes an investor can reinvest all of the coupon payments at the same rate as the calculated yield to maturity. This is typically not the case, as market fluctuations make it difficult to get the exact same yield from one day to the next.
Calculating yield to maturity (YTM)
However, the benefits related to comparability tend to outweigh the drawbacks, which explains the widespread usage of YTM across the debt markets and fixed-income investors. Alternatively, this process can be sped up by utilizing the SOLVER function in Excel, which determines a value based on conditions that can be set. This means that an analyst can set the present value (price) of the security and solve for the YTM which acts as the interest rate for the PV calculation. The YTM of a discount bond that does not pay a coupon is a good starting place in order to understand some of the more complex issues with coupon bonds. Yield to maturity can be useful for investors trying to decide between multiple investment options. Calculating the YTM on each can give you an idea of what they’re getting for their money.
Yield to maturity is similar to a discounted cash flow model for valuing stocks. In the case of calculating yield to maturity, all future cash flows can be known. Bonds have fixed coupon rates that pay on fixed schedules, and they have a certain face value paid at maturity on a certain date. Unlike current yield, YTM accounts for the present value of a bond’s future coupon payments. In other words, it factors in the time value of money, whereas a simple current yield calculation does not.
Figuring Bond Return
The YTM is merely a snapshot of the return on a bond because coupon payments cannot always be reinvested at the same interest rate. As interest rates rise, the YTM will increase; as interest rates fall, the YTM will decrease. The term yield to maturity (YTM) refers to the total return anticipated on a bond if the bond is held until it matures. Yield to maturity is considered a long-term bond yield but is expressed as an annual rate. In other words, it is the internal rate of return (IRR) of an investment in a bond if the investor holds the bond until maturity, with all payments made as scheduled and reinvested at the same rate.
Understanding Yield to Maturity (YTM)
A bond’s yield to maturity rises or falls depending on its market value and how many payments remain. Solving the equation by hand requires an understanding of the relationship between a bond’s price and its yield, as well as the different types of bond prices. When the bond is priced at par, the bond’s interest rate is equal to its coupon rate. A bond priced above par, called a premium bond, has a coupon rate higher than the realized interest rate, and a bond priced below par, called a discount bond, has a coupon rate lower than the realized interest rate.
Example of a YTM Calculation
Interest rates regularly fluctuate, making each reinvestment at the same rate virtually impossible. While helpful, it’s important to realize that YTM and YTC may not be the same as a bond’s total return. Such a figure is only accurately computed when you sell a bond or when it matures. The relationship between the current YTM and interest rate risk is inversely proportional, which means the higher the YTM, the less sensitive the bond prices are to interest rate changes. The relationship between the yield to maturity and coupon rate (and current yield) are as follows.
That last assumption makes yield to maturity purely theoretical since market conditions are constantly in flux. Yield to Maturity (YTM) – otherwise referred to as redemption or book yield – is the speculative rate of return or interest rate of a fixed-rate security, such as a bond. In practice, the rates that will actually be earned on reinvested interest payments are a critical component of a bond’s investment return.[9] Yet they are unknown at the time of purchase. | 1,520 | 7,498 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.03125 | 3 | CC-MAIN-2024-10 | latest | en | 0.931604 |
http://topcoder.bgcoder.com/print.php?id=2046 | 1,726,088,536,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651400.96/warc/CC-MAIN-20240911183926-20240911213926-00425.warc.gz | 28,403,415 | 2,062 | ### Problem Statement
A serial-parallel resistor circuit is either
1. a single resistor. The resistance of such circuit is equal to the resistance of the resistor. Or
2. several circuits R1,R2,...,Rn combined in serial. The resistance is equal to R1+R2+...+Rn. Or
3. several circuits R1,R2,...,Rn combined in parallel. The resistance is equal to 1/((1/R1)+(1/R2)+...+(1/Rn)).
Given two positive integers a and b, your task is to build a serial-parallel resistor circuit that has resistance equal to a/b. You are only allowed to use two kinds of resistors: R=1 and R=2. Return the minimal number of resistors needed. If the circuit cannot be built with 16 or less resistors, return -1.
### Definition
Class: BuildCircuit Method: minimalCount Parameters: int, int Returns: int Method signature: int minimalCount(int a, int b) (be sure your method is public)
### Constraints
-a and b will each be between 1 and 50000, inclusive.
### Examples
0)
`1` `1`
`Returns: 1`
One unit resistor is enough.
1)
`2` `3`
`Returns: 2`
Combine R=1 and R=2 in parallel.
2)
`6` `5`
`Returns: 3`
Combine R=1 and R=2 in serial, then with R=2 in parallel.
3)
`42` `47`
`Returns: 7`
4)
`1` `20`
`Returns: -1`
5)
`756` `874`
`Returns: 10`
#### Problem url:
http://www.topcoder.com/stat?c=problem_statement&pm=8510
#### Problem stats url:
http://www.topcoder.com/tc?module=ProblemDetail&rd=11124&pm=8510
srbga
#### Testers:
PabloGilberto , Olexiy , Andrew_Lazarev , ivan_metelsky
#### Problem categories:
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# Problem Set 6sol - 1 ECONOMICS 100 NOTES ON PROBLEM SET #6...
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ECONOMICS 100 NOTES ON PROBLEM SET #6 Short and Long Run Competition, Monopoly and Other Markets For diagrams, see attachment at the end of this document/separate file. 1. Constant Cost Perfectly Competitive Industry Consider a cabbage growing industry which is perfectly competitive and which has a large number of firms with identical cost curves. Assume each firm uses two factors of production: land, which is fixed in the short run and variable in the long run; and labour, which is variable in both the short and the long run. Use diagrams to determine the impact of an increase in the price of land, assuming that the long-run supply curve of the cabbage industry exhibits constant costs. In particular, show the impact on: * the price, output, and profit of the individual firm in the short and long run, * the price and output of the industry in the short and long run. Only SAC shifts in short run. Since SMC has not shifted, there is no change in Industry SRS. Consequently equilibrium P and Q are unchanged. Each firm suffers losses. In the long run, some firms exit. As they do, SRS shifts to the left, causing equilibrium price to rise, along the D schedule. This process of firms exiting continues until losses are eliminated i.e. until P = Min SAC (on the new higher SAC schedule). Here are some elements of the Final Result: Fewer firms; higher P; Lower Q; Each firm’s Profits = 0. 2. Short and Long Run Cumquats are produced in a perfectly competitive industry consisting of a large number of identical and independently managed firms. All inputs, including labour, are available to the cumquat industry at unchanged input prices. The industry is initially in long run equilibrium with each firm earning normal profits. With the aid of carefully labelled diagrams, examine the full impact on the
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## This note was uploaded on 01/16/2011 for the course ECO ECO100 taught by Professor Inheart during the Fall '09 term at University of Toronto- Toronto.
### Page1 / 8
Problem Set 6sol - 1 ECONOMICS 100 NOTES ON PROBLEM SET #6...
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Ask a homework question - tutors are online | 563 | 2,515 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.921875 | 3 | CC-MAIN-2017-13 | longest | en | 0.925634 |
https://dorigo.wordpress.com/2007/10/06/a-shot-in-the-dark/ | 1,561,044,261,000,000,000 | text/html | crawl-data/CC-MAIN-2019-26/segments/1560627999261.43/warc/CC-MAIN-20190620145650-20190620171650-00512.warc.gz | 413,871,969 | 25,660 | ## A shot in the dark ? October 6, 2007
Posted by dorigo in personal, physics, science.
Yesterday I was in Bari, where I followed a meeting with some members of the CMS-tracker italian group. At the meeting we discussed a research project aimed at developing new technologies which will be useful for the upgrade of the detector whose construction our group heavily contributed: the silicon tracker.
The silicon tracker of CMS (of which you can see in the picture the inner layers, called TIB -for Tracker Inner Barrel- being assembled last year) is a fantastic device, and we confidently expect it to deliver what it was designed to, so why are we thinking about upgrading it if it has not seen more than cosmic rays so far ?
Well, experimental physicists specialized in the design of particle experiments have learned to think ahead – ten, fifteen years ahead: that is the time scale of today’s experiments. The Large Hadron Collider (LHC), which will provide proton-proton collisions at 14 TeV to the CMS and ATLAS experiments from next year onwards, cannot be easily upgraded to increase the beam energy it provides, but its instantaneous luminosity can be pushed up by a full order of magnitude by increasing the number of protons and the interaction rate.
Instantaneous luminosity is a quantity basically obtained by multiplying the number of circulating protons in each beam $N_1, N_2$ by the revolution frequency $f$, and dividing by the area $A$ of the beams at the interaction point: $L = N_1 N_2 f / A$. The larger the number of particles in each beam and their crossing rate, and the smaller the area where they interact, the higher is the resulting luminosity. Luminosity is thus directly related to the production rate of a rare process: if we call $\sigma$ the cross section for the process, we have simply that the rate is $R = \sigma L$.
If LHC works and finds new physics -that is a big if, but let me finish the sentence- an increase of the beam luminosity of LHC is useful in many scenarios: more luminosity means more data, and a larger discovery potential, plus smaller statistical errors on all measurements. Of course, after twenty years spent designing and building the LHC and the experiments, running for only a few years and then selling the pieces as scrap metal is not the best option! Better to think of an upgrade which can extend the lifetime of the experiments and increase their discovery potential, if possible.
While in principle an increase in beam luminosity is a good thing, the detectors that benefit from it must come prepared, because of several issues.
First, a high luminosity causes larger fluences of particles through the detectors, which thus have to withstand a larger radiation damage. Of course, radiation dose goes inversely with the squared distance from the interaction point, so the closest devices must be the toughest. The silicon microstrip and pixel detectors of CMS are radiation hard, but tolerances are not allowing for an increase by an order of magnitude in doses.
Second, the 100 interactions occurring every 12.5 nanoseconds in the core of the detectors during SLHC operation will cause a large occupancy of the detector components closest to the interaction region: this imposes constraints on a design capable of maintaining the current tracking performances. One would not like to collect ten times more data if the deal involved a detector performing much worse!
Third, muons – particles which are crucial for several searches of new physics – require a very good momentum resolution at trigger level in order to allow an efficient filtering of the few interesting events among the 80 MHz of interactions. This is because of the steeply falling probability of finding a true muon as its transverse momentum $P_T$ increases. We are interested in the high-momentum ones (derived from the decay of W and Z bosons, or Higgs bosons too), and so we select them with a cut $P_T> X$ GeV, a threshold dictated by our limited ability to write events to tape. Now, if there is even a slight chance -because of insufficiently good momentum resolution- that we mistake a low momentum muon for one passing the threshold we are dead, because there are so many more muons with the lower momentum that whatever threshold X we set, we will collect mostly the low momentum ones – filling our tapes with uninteresting stuff.
The jury will soon be out to decide on a design for the upgrade of the CMS tracker that addresses all these issues. There are quite a few subtleties, many issues involving cost effectiveness, redundancy, triggering capabilities by the tracker alone, and so on. However, what I feel is most urgently needed is a preliminary assessment of the physics the upgrade will address. Nobody’s fault: it is simply impossible to make a really meaningful case for a LHC upgrade right now, if you ask me. That is: an upgrade is certainly a good idea, but which one, well…
Indeed, deciding now on an upgrade of LHC experiment seems to me a shot in the dark. We can only guess whether CMS and ATLAS will discover the Higgs, and whether they will find SUSY. We can argue on the likelihood of new unpredicted discoveries. But what one would need in order to decide whether design A is better or worse than design B, C, D…, in a scenario where LHC runs for a few years with a tenfold increase in luminosity, would be a clear idea of what we would gain in the study of a few flagship signals. And knowing whether SUSY is there or not makes the hell of a difference.
Sure, we could make a physics case with stuff we are almost certain we will find, such as the Higgs. To me that would be more meaningful than trying to sell a better determination of the full spectrum of SUSY particles or large extra dimensions. An example: Would we be able to see exclusive higgs production, and thus measure that particle’s quantum numbers, only with design A ?
Knowing that would be something… But it looks unimpressive to somebody who has to decide whether to fund a multi-million-dollar project!
1. Plato - October 6, 2007
Very Interesting
2. anomalous cowherd - October 6, 2007
Dear Prof. Dorigo
Congratulations on the CMS silicon tracker. Having had tours of both CMS and ATLAS, I can report that it is really impossible to understand the scope of the accomplishment that construction of these detectors represents, until you have seen them with your own eyes.
Good luck on the commissioning, the running, and the upgrades!
We all eagerly await LHC data.
Cowherd
3. Paolo - October 7, 2007
Very interesting indeed. And many thanks for , of course! Hopefully, I will be able to contribute something to the discussion, but in the meanwhile at least a link I didn’t know about until a few days ago:
that is, this year VIDEO material too! Nima Arkani-Hamed is rather enjoyable also for now experts, IMHO. Highly recommended!
4. Paolo - October 7, 2007
Of course I meant “… many thanks for -Instantaneous luminosity- …”
5. dorigo - October 7, 2007
Thank you all. Indeed, the CMS and ATLAS experiments are among the most complex things ever built by humanity. I am serious…. One only understands it by visiting the experiments, which is a reason for visiting Geneva alone.
Paolo, Nima is a string theorist… I have no doubts he can give entertaining lectures, but I would advise you a very good book: “Not Even Wrong”, by Peter Woit. See the link to his site in my blogroll.
Cheers,
T.
6. anomalous cowherd - October 7, 2007
5. dorigo – October 7, 2007 writes:
“Paolo, Nima is a string theorist…”
I’m not sure that either Nima, or mainstream string theorists, would characterize him as such. He was a Ph.D. student of Berkeley phenomenologist Lawrence Hall [himself a student of Howard Georgi]. He is a gifted and incredibly productive creator of phenomenological models of physics beyond the standard model; in my view he is the most influential phenomenologist of his generation. He has worked on models of: fermion masses and the flavour problem, hypercolour, supercomposite models, large extra dimensions, neutrino mass, supersymmetry, orbifold GUTs, deconstruction, little Higgs, inflation, ghost modifications of gravity, split supersymmetry, and signature driven Monte-Carlo analysis tools for LHC physics [with experimentalist Bruce Knutson, who had already implemented systems like this for the TeVatron experiments. Of his 75 papers, only a handful have had any string theory collaborators. Like his Harvard colleague Lisa Randall, he has frequently adopted the approach of taking ideas that had originated in string theory, and incorporating those ideas in phenomenological models [she too is one of the most influential phenomenologists of her generation].
To get a sense of the difference in interests and style between string theory and beyond the standard model phenomenology and model building, you could do a SPIRES search to compare the publications of the string theory group at Harvard [Strominger and Vafa] to those of the phenomenology group [Arkani-Hamed, Georgi, and Randall]. Or try the same for Stanford [Shenker, Susskind, Kallosh, Kachru and Silverstein, (strings), compared to Dimopoulos, Peskin, Quinn, Wacker, Hewitt, and Rizzo (phenomenology)]. You see some collaboration between members of the groups [which is a very positive thing], but on the whole their approaches and problems are different. This difference is often ignored in popular books whose publishers will put anything on the cover that they think will enhance sales.
P.S. And yes, Nima is an exceptionally charismatic lecturer, which is why he is in so much in demand for physics summer schools.
7. dorigo - October 8, 2007
Hi anomalous,
I will take your word for it – I must admit I was not aware of the breadth of his works. However, Nima’s position on the anthropic landscape of ST is enough for me to warn any student about his views.
Cheers,
T.
8. Paolo - October 9, 2007
Tommaso, is it easy to visit the experiments? I’d like to follow your advice as soon as possible, Geneva isn’t a big deal of a travel for us Italians, anyway…
Otherwise, well, many thanks to AC for the details… what can I add, definitely from the actual lectures neither the landscape neither string theory appear as a dogma. I would say in the lectures there is something for everyone, a lot of “philosophy”, a lot of mathematics, a lot of genuine “physical intuition”. Personally, I had a lot of fun… and his walking incessantly from left to right and way back reminded me a lion in a cage, a LOT of energy
9. anomalous cowherd - October 9, 2007
8. Paolo – October 9, 2007 writes:
” is it easy to visit the experiments? I’d like to follow your advice as soon as possible”
– Go to: http://www.cern.ch
– Click on “For CERN Users”
– Click on “Media and Public Corner”
– Scroll Down
– Click on “Visits Service”
This will give you information on, and booking forms for, visits.
10. Paolo - October 9, 2007
Thanks!
11. Plato - October 9, 2007
Tammaso:However, Nima’s position on the anthropic landscape of ST is enough for me to warn any student about his views.
So is this support for what you think is relevant. I have followed the discussions between Lee Smolin, Jacques Distler, Clifford of Asymptotia and Peter Woit of “Not Event Wrong.”
I wonder if you had “more information,” if this might change your statement above, and conclusion, you may have drawn from seeing another view ? One you might not of seen before?
Is mathematical consistency of value to you when it is developing?
12. dorigo - October 9, 2007
Paolo, it should not be too hard to get a tour of CMS or ATLAS. I have no access to the caverns – I could ask for it but am too lazy to do it, since there is no reason for me to be downstairs. If you decide to go, let me know and I’ll see if I can find a contact for you on site.
Cheers,
T.
13. dorigo - October 9, 2007
Hi Plato,
I think there are scientific theories, and one has a chance to draw one’s own conclusion on those based on proven and unproven fact and experimental hints; and then, there is scientific method, on which one has an opinion regardless of any input. I think that “solving” the problem of foldings of the extra dimensions by calling it a “feature” of the theory, is not science.
Cheers,
T.
14. Paolo - October 9, 2007
Tommaso, thanks a lot again about the visit. To tell you the truth now I’m a bit confused, because never considered visiting before and now, all of a sudden, I have only to fill a form, too good to be true! Probably, I will try to make a reservation for next spring, ideally I would find something else to do in Switzerland (e.g., the famous kunstmuseum in Basel?) and spend overall ~1 week. Since you are so kind, I will probably contact you privately later for some hints (about lodging too, probably)… In the meanwhile, if you think there is something I should know immediately (like, no-no times of the year, for some reason), I’m all ears… Ciao, Paolo.
15. anomalous cowherd - October 10, 2007
14. Paolo – October 9, 2007
If you want to visit ATLAS or CMS you should plan your visit soon! Once the experiments close up for beam commissioning there will be no more visits to them. [You will be able to do CERN visits, but you won’t be able to see ATLAS or CMS during the visit].
Check with Prof. Dorigo that my information is accurate, but that is what I had been told by my experimental colleagues.
Cowherd
16. Plato - October 10, 2007
Tammasobased on proven and unproven fact and experimental hints; and then, there is scientific method, on which one has an opinion regardless of any input.
Then there is a division amongst your colleques? Ones who have scientiifc theories and those that practise scientific method?
17. Plato - October 11, 2007
Jacques Distler:
It’s not a matter of liking or disliking the alternatives. It’s a matter of whether they are viable candidates. As I’ve tried to explain, they are not.
That doesn’t mean I don’t think they’re worth thinking about. There’s a long and noble history of “wrong” ideas having useful (and, sometimes, unexpected) spinoffs. The same may well turn out to be true of String Theory, which continues to have a rich set of spinoffs for Quantum Field Theory and for Mathematics but which could, ultimately, turn out to be wrong.
Quote is direct link.
If one did not extend the methods of “genus figures” then mathematically the issue would have been dead, and Lee Smolin’s point would have been right. But he was shown to be wrong in his new book, and Peter Woit had to have recognize this?
That’s a simple question of Peter Woit in face of the link Tammaso you supplied on this matter. I have nothing to gain other then be corrected on the insights that people point too, to help me decide what is where and what is the state of things.
As you have corrected on dual blackholes.
Sometimes recognizing “the differences” you have pointed out can hurt others who continue to research by stating explicitly that the individual character could have been flaw in any scientist, by association?
That’s all I have right now
18. dorigo - October 11, 2007
Plato,
as always I have some trouble decyphering you. Can you try writing simpler sentences, like explaining what are “genus figures”, what is Smolin’s point, where he was shown wrong, why peter should know, when did I discuss dual blackholes… My memory is at fault, help me answer you meaningfully!
I even find it hard to understand your last sentence. I apologize, it must be too late – I think I’ll just go to sleep 🙂
Sorry for being useless,
Cheers,
T.
19. Plato - October 13, 2007
Tammaso,
That is my difficulty in being clear. You have every right to ask for it.
1.What are “genus figures”
We can also consider donuts with more handles attached. The number of handles in a donut is its most important topological information. It is called the genus.
2.What is Smolin’s point?
His view on Finiteness of String theory. and the development to two loop genus figure. Saying, that this is all that has been done.
The Trouble With Physics,” by Lee Smolin, Index page 382, Mandelstam, Stanley, and string theory finiteness, pages 117,187, 278-79, 280, 281, 367n14,15
3.Why Peter should know?
In regards to any arguments Peter Woit has on string theory, this particular development is explicitly clear in the mathematical progress.
Jacques Distler :
This is false. The proof of finiteness, to all orders, is in quite solid shape. Explicit formulæ are currently known only up to 3-loop order, and the methods used to write down those formulæ clearly don’t generalize beyond 3 loops.
4.When did I discuss dual blackholes?
Tammaso:
Plato, the paper you link to appears a bit outdated, I think. The RHIC has called off early claims of being able of producing black holes, IIRC.
20. Plato - October 13, 2007
Tammaso,
Mouse over the quotes. They are directly linked to their source. Click on them.
21. dorigo - October 14, 2007
Hi Plato,
since you are by now a regular here, you should learn to spell my name right 😉 Tommaso.
Anyway, thank you for the information – things look a whole lot clearer to me now! Ok, genus figures I think I should have known. I have read about them, and indeed in English. But some “new words” stick less than others…
As for dual black holes, ok… I did not “discuss” them but you are right, I mentioned that.
As for Jacques Distler vs Peter Woit, I take the latter 🙂 Distler is a bit too partisan. I have a bet out against him actually, on finding SUSY at LHC.
Cheers,
T.
Sorry comments are closed for this entry | 4,085 | 17,620 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 8, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.609375 | 3 | CC-MAIN-2019-26 | latest | en | 0.92275 |
https://socratic.org/questions/how-do-you-graph-and-solve-2x-abs-x-4-8 | 1,586,024,076,000,000,000 | text/html | crawl-data/CC-MAIN-2020-16/segments/1585370524604.46/warc/CC-MAIN-20200404165658-20200404195658-00171.warc.gz | 704,314,042 | 5,973 | # How do you graph and solve 2x-abs[x+4]> 8?
Oct 27, 2017
$x > 12$
#### Explanation:
$2 x - \left\mid x + 4 \right\mid = 8 + {\epsilon}^{2}$ with $\epsilon \ne 0$ or
$2 x - 8 - {\epsilon}^{2} = \left\mid x + 4 \right\mid$ or
${\left(2 x - 8 - {\epsilon}^{2}\right)}^{2} - {\left(x + 4\right)}^{2} = 0$ or
$\left(- {\epsilon}^{2} - 12 + x\right) \left(- {\epsilon}^{2} - 4 + 3 x\right) = 0$ or
$\left\{\begin{matrix}- {\epsilon}^{2} - 12 + x = 0 \\ - {\epsilon}^{2} - 4 + 3 x = 0\end{matrix}\right.$
or
$\left\{\begin{matrix}x > 12 \\ x > \frac{4}{3}\end{matrix}\right.$
and the solution is $x > 12$ | 271 | 609 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 9, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.53125 | 5 | CC-MAIN-2020-16 | longest | en | 0.444613 |
https://programmingpraxis.com/2010/03/12/traveling-salesman-brute-force/?like=1&source=post_flair&_wpnonce=744caefc44 | 1,524,221,856,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125937440.13/warc/CC-MAIN-20180420100911-20180420120911-00551.warc.gz | 685,571,777 | 18,576 | ## Traveling Salesman: Brute Force
### March 12, 2010
The traveling salesman problem is classic: Find the minimum-length tour that visits each city on a map exactly once, returning to the origin. It is an important problem in practice; consider, for instance, that the cities are soldering points on a large circuit board, each of which must be visited by a soldering robot. It is also an important problem in mathematics, and is known to be in NP, meaning that optimal solutions for non-trivial problems are impossible. Nonetheless, there exist heuristic algorithms that do a good job at minimizing the tour length.
We will examine the traveling salesman problem in an occasional series of exercise, beginning with today’s exercise that looks at the brute-force solution of examining all possible tours. The algorithm is simple: First, select a tour at random, then determine all possible permutations of the tour. Second, determine the length of each tour. Third, report the tour with minimum length as the solution.
Your task is to write a program that solves the traveling salesman problem using the brute-force algorithm. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.
Pages: 1 2
### 9 Responses to “Traveling Salesman: Brute Force”
1. […] Praxis – Traveling Salesman: Brute Force By Remco Niemeijer In today’s Programming Praxis exercise we have to implement a brute-force algorithm for solving the well-known […]
2. Remco Niemeijer said
My Haskell solution (see http://bonsaicode.wordpress.com/2010/03/12/programming-praxis-traveling-salesman-brute-force/ for a version with comments):
import Data.List
import qualified Data.List.Key as K
dist :: Floating a => (a, a) -> (a, a) -> a
dist (x1, y1) (x2, y2) = sqrt ((x1 - x2) ** 2 + (y1 - y2) ** 2)
tours :: [b] -> [[(Int, b)]]
tours = map (\(x:xs) -> x:xs ++ [x]) . permutations . zip [0..]
cost :: Floating a => [(b, (a, a))] -> a
cost xs = sum \$ zipWith dist xs (tail xs)
shortestPath :: [(Double, Double)] -> [Int]
shortestPath = init . map fst . K.minimum cost . tours
3. Jebb said
A C solution. Quite verbose obviously, and I did manage to fall into a couple of pitfalls along the way. It is fast, though (1 second for the 10-second problem).
#include <stdio.h>
#include <math.h>
#define NUMCITY 10
#define LANDSIZE 100
#define square(A) ((A) * (A))
typedef int City[2];
void generate(City cities[]);
void print_cities(City cities[]);
float distance(City city1, City city2);
void copy_tour(City citiesDest[], City citiesSource[]);
void copy_City(City dest, City source);
void swap_cities(City city1, City city2);
void circ_perm(City cities[], int numCities);
void scramble(City cities[], City *pivot, int numCities);
void target_function(City cities[]);
float tour_length(City cities[]);
float shortestTourLength;
City shortestTour[NUMCITY];
int main()
{
City cities[NUMCITY];
generate(cities);
shortestTourLength = tour_length(cities);
copy_tour(shortestTour, cities);
scramble(cities, cities, NUMCITY);
printf("found the shortest tour:\n");
print_cities(shortestTour);
printf("Length is: %f\n", shortestTourLength);
return 0;
}
float tour_length(City cities[])
{
int i;
float length = 0.0;
for(i = 0; i < NUMCITY - 1; i++)
length += distance(cities[i], cities[i+1]);
length += distance(cities[NUMCITY - 1], cities[0]);
return length;
}
void target_function(City cities[])
{
float length;
length = tour_length(cities);
if (length < shortestTourLength) {
shortestTourLength = length;
copy_tour(shortestTour, cities);
}
}
/*pivot is the base address in the cities[NUMCITY] array
*for the recursive scrambling; the only reason we also
*pass the unchanged cities address is because we need it
*to call the target function (which does *something* to
*the scrambled array) at each recursive call to scramble
*/
void scramble(City cities[], City *pivot, int numCities)
{
int i;
City *newPivot;
if (numCities <= 1) { //Scrambled! Call the target function
target_function(cities);
return;
}
for (i = 0; i < numCities; i++) {
newPivot = &pivot[1];
scramble(cities, newPivot, numCities - 1);
circ_perm(pivot, numCities);
}
}
void circ_perm(City cities[], int numCities)
{
int i;
City tmp;
copy_City(tmp, cities[0]);
for (i = 0; i < numCities - 1; i++)
copy_City(cities[i], cities[i + 1]);
copy_City(cities[numCities - 1], tmp);
}
void copy_tour(City citiesDest[], City citiesSource[])
{
int i;
for (i = 0; i < NUMCITY; i++)
copy_City(citiesDest[i], citiesSource[i]);
}
void copy_City(City dest, City source)
{
dest[0] = source[0];
dest[1] = source[1];
}
void swap_cities(City city1, City city2)
{
City tmp;
copy_City(tmp, city1);
copy_City(city1, city2);
copy_City(city2, tmp);
}
float distance(City city1, City city2)
{
float result;
result = sqrtf((float)square(city2[0] - city1[0]) +
(float)square(city2[1] - city1[1]));
return result;
}
void generate(City cities[])
{
int i, j;
for (i = 0; i < NUMCITY; i++)
for (j = 0; j < 2; j++)
cities[i][j] = rand() % LANDSIZE;
}
void print_cities(City cities[])
{
int i;
for (i = 0; i < NUMCITY; i++) {
printf("(%d,%d)", cities[i][0], cities[i][1]);
if (i < NUMCITY - 1)
printf(" , ");
}
printf("\n");
}
4. […] with the length of the data. On Programming Praxis they have proposed to resolve the problem using brute force, and using the closest neighbor (a simplification of the […]
5. Mike said
One observation is that most of the permutations of the cities are mere rotations of another permutation. For example, the tour [city2, … cityN, city1] is a rotation of the tour [city1, city2, … cityN]. Indeed, any tour starting with any of city2 .. cityN is a rotation of a tour starting with city1.
A smarter brute force solution is to only consider permutations that start with city1. For 10 cities this optimisation eliminates 90% of the calculations.
In python:
def brute_force(cities):
other_cities = cities.copy()
start_city = tuple(other_cities.pop())
return min((start_city+p for p in permutations(other_cities)), key=path_cost)
6. harsha said
On executing this programme on C-free5.0 version, it showed an error for “rand()” at line no.:121
so i overcomed it by including “stdlib.h” header file for this “rand()” function
7. Anusha said
In the c program there is an error inthe distance function:
in sqrtf()
8. mumus said
Why use brute force when you can do it easily ? Just visit this link and try the actual program.
http://stackoverflow.com/questions/28702426/no-matching-function-for-call-to-mapflatfunctionname/29174891#29174891 | 1,781 | 6,584 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.625 | 4 | CC-MAIN-2018-17 | latest | en | 0.888895 |
https://www.brainkart.com/article/Tools-of-financial-statement-analysis_38855/ | 1,726,040,899,000,000,000 | text/html | crawl-data/CC-MAIN-2024-38/segments/1725700651344.44/warc/CC-MAIN-20240911052223-20240911082223-00135.warc.gz | 623,758,637 | 6,940 | Home | | Accountancy 12th Std | Tools of financial statement analysis
# Tools of financial statement analysis
Different tools are used for analysing the financial statements. The tool is selected based on the purpose of analysis. Following are the commonly used tools of financial statement analysis:
Tools of financial statement analysis
Different tools are used for analysing the financial statements. The tool is selected based on the purpose of analysis. Following are the commonly used tools of financial statement analysis:
### (i) Comparative statement
A statement giving comparison of net increase or decrease in the individual items of financial statements of two or more years of a business concern is called comparative statement. It shows the actual figures at different periods of time, the increase or decrease in these figures in absolute terms and the percentages of such increase or decrease. The two basic comparative statements prepared are comparative income statement and comparative balance sheet.
### (ii) Common-size statements
The common–size statements show the relationship of various items with some common base, expressed as percentage of the common base. The common bases are total of assets or total of equity and liabilities or revenue from operations (net sales). The common size statements include common-size income statement and common-size balance sheet.
In the common–size income statement, revenue from operations is taken as 100 and various expenses and incomes are expressed as a percentage to the revenue from operations. In the common-size balance sheet, the total of balance sheet, that is, the total of assets or total of equity and liabilities is taken as 100 and various assets and liabilities are expressed as a percentage of the total of assets or total of equity and liabilities.
The common-size statements can be compared with those of previous years. They can also be compared with those of other similar businesses with similar accounting policies.
### (iii) Trend analysis
Trend refers to the tendency of movement. Trend analysis refers to the study of movement of figures over a period. The trend may be increasing trend or decreasing trend or irregular. When data for more than two years are to be analysed, it may be difficult to use comparative statement. For this purpose, trend analysis may be used. One year, generally, the first year is taken as the base year. The figures of the base year are taken as 100. The figures for the other years are expressed as a percentage to the base year and the trend is determined.
### (iv) Funds flow analysis
The term ‘fund’ refers to working capital. Working capital refers to the excess of current assets over current liabilities. The term ‘flow’ means movement and includes both ‘inflow’ and ‘outflow’. Funds flow analysis is concerned with preparation of funds flow statement which shows the inflow (sources) and outflow (applications) of funds in a given period of time. Funds flow analysis is useful in judging the credit worthiness, financial planning and preparation of budgets.
### (v) Cash flow analysis
Cash flow analysis is concerned with preparation of cash flow statement which shows the inflow and outflow of cash and cash equivalents in a given period of time. Cash includes cash in hand and demand deposits with banks. Cash equivalents denote short term investments which can be realised easily within a short period of time, without much loss in value. Cash flow analysis helps in assessing the liquidity and solvency of a business concern.
Tags : Accountancy , 12th Accountancy : Chapter 8 : Financial Statement Analysis
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
12th Accountancy : Chapter 8 : Financial Statement Analysis : Tools of financial statement analysis | Accountancy | 733 | 3,867 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-38 | latest | en | 0.930701 |
https://www.coursehero.com/file/6641246/908-Mechanics-SolutionInstructors-SolManual-Mechanics-Materials-7ebook-Gere-light1/ | 1,493,564,912,000,000,000 | text/html | crawl-data/CC-MAIN-2017-17/segments/1492917125654.80/warc/CC-MAIN-20170423031205-00132-ip-10-145-167-34.ec2.internal.warc.gz | 889,410,950 | 111,113 | 908_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1
# 908_Mechanics SolutionInstructors_Sol.Manual-Mechanics_Materials_7e.book_Gere_light.1
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Problem 11.9-19 Find the required outside diameter d for a steel pipe column (see figure) of length L ± 11.5 ft that is pinned at both ends and must support an axial load P ± 80 k. Assume that the wall thickness t is 0.30 in. (Use E ± 29,000 ksi and .) s Y ± 42 ksi 902 CHAPTER 11 Columns Solution 11.9-19 Pipe column Pinned ends . Seltect various values of diameter d until we obtain . P allow ± P a L r b c ± A 2 p 2 E s Y ± 116.7 L c ± (116.7) r I ± p 64 [ d 4 ² ( d ² 2 t ) 4 ] r ± A I A A ± p 4 [ d 2 ² ( d ² 2 t ) 2 ] E ± 29,000 ksi s Y ± 42 ksi d ± outside diameter t ± 0.30 in. L ± 11.5 ft ± 138 in. P ± 80 k ( K ± 1) For P ± 80 k, d ± 5.23 in. ; d (in.) 5.20 5.25 5.30 A (in. 2 ) 4.618 4.665 4.712 I (in. 4 ) 13.91 14.34 14.78 r (in.) 1.736 1.753 1.771 L c (in.) 203 205 207 L/r 79.49 78.72 77.92 n 1 (Eq. 11-79) 1.883 1.881 1.880 (Eq. 11-81) s allow / s Y 0.4079 0.4107 0.4133 s allow (ksi) 17.13 17.25 17.36 P allow ± A s allow 79.1 k 80.5 k 81.8 k Problem 11.9-20 Find the required outside diameter d for a steel pipe column (see figure) of length L ± 3.0 m that is pinned at both ends and must support an axial load P ± 800 kN. Assume that the wall thickness t is 9 mm. (Use E ± 200 GPa and .) s Y ± 300 MPa Solution 11.9-20 Pipe column
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Unformatted text preview: Pinned ends . Seltect various values of diameter d until we obtain . If , Use Eqs. (11-79) and (11-81). If , Use Eqs. (11-80) and (11-82). L Ú L c L … L c P allow ± P a L r b c ± A 2 p 2 E s Y ± 114.7 L c ± (114.7) r I ± p 64 [ d 4 ² ( d ² 2 t ) 4 ] r ± A I A A ± p 4 [ d 2 ² ( d ² 2 t ) 2 ] E ± 200 GPa s Y ± 300 MPa d ± outside diameter t ± 9.0 mm L ± 3.0 m P ± 800 kN ( K ± 1) For P ± 800 kN, d ± 194 mm ; d (mm) 193 194 195 A (mm 2 ) 5202 5231 5259 I (mm 4 ) 20.08 * 10 6 22.43 * 10 6 22.80 * 10 6 r (mm) 65.13 65.48 65.84 L c (mm) 7470 7510 7550 L/r 46.06 45.82 45.57 n 1 (Eq. 11-79) 1.809 1.809 1.808 (Eq. 11-81) s allow / s Y 0.5082 0.5087 0.5094 s allow (MPa) 152.5 152.6 152.8 P allow ± A s allow 793.1 kN 798.3 kN 803.8 kN If , Use Eqs. (11-79) and (11-81). If , Use Eqs. (11-80) and (11-82). L Ú L c L … L c 11Ch11.qxd 9/27/08 2:27 PM Page 902...
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## This note was uploaded on 12/22/2011 for the course MEEG 310 taught by Professor Staff during the Fall '11 term at University of Delaware.
Ask a homework question - tutors are online | 1,128 | 2,655 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.390625 | 3 | CC-MAIN-2017-17 | longest | en | 0.657645 |
https://calculator.academy/return-on-profit-calculator/ | 1,686,444,862,000,000,000 | text/html | crawl-data/CC-MAIN-2023-23/segments/1685224646652.16/warc/CC-MAIN-20230610233020-20230611023020-00148.warc.gz | 186,188,155 | 54,023 | Enter the total profit ($) and the total cost ($) into the Return on Profit Calculator. The calculator will evaluate and display the Return on Profit.
## Return on Profit Formula
The following formula is used to calculate the Return on Profit.
ROProfit = TP / TC * 100
• Where ROProfit is the Return on Profit (%)
• TP is the total profit ($) • TC is the total cost ($)
## How to Calculate Return on Profit?
The following example problems outline how to calculate Return on Profit.
Example Problem #1:
1. First, determine the total profit ($). • The total profit ($) is given as: 500.
2. Next, determine the total cost ($). • The total cost ($) is provided as: 375.
3. Finally, calculate the Return on Profit using the equation above:
ROProfit = TP / TC * 100
The values given above are inserted into the equation below and the solution is calculated:
ROProfit = 500 / 375 * 100 = 133.33 (%)
Example Problem #2:
For this problem, the variables needed are provided below:
total profit ($) = 800 total cost ($) = 600
This example problem is a test of your knowledge on the subject. Use the calculator above to check your answer.
ROProfit = TP / TC * 100 = ? | 272 | 1,171 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.734375 | 4 | CC-MAIN-2023-23 | latest | en | 0.763651 |
http://library.thinkquest.org/3493/noframes/glossary.html | 1,386,561,464,000,000,000 | text/html | crawl-data/CC-MAIN-2013-48/segments/1386163870408/warc/CC-MAIN-20131204133110-00090-ip-10-33-133-15.ec2.internal.warc.gz | 103,468,232 | 4,015 | # Glossary Of Terms And Phrases
This is a comprehensive glossary of terms and phrases relating to chaos theory, dynamic systems, and fractal geometry. Most of the difficult words used in these pages, and others relating to the subject are listed below. The words are in alphabetical order for ease of location. To search for a word or phrase, enter it in the form below. If no match is found, the closest match will be shown. To add a word, or buil upon the definition of a word already in the glossary, click below on the link which says Add A Word.
Search Word:
# A
A super fractal
# B
butterfly effect- Submitted By The Authors
See sensitive dependance.
# C
cantor set- Submitted By The Authors
The Cantor Set is a fractal generated by repeatedly removing the center third from a line of unit length.
chaos theory- Submitted By The Authors
complex- Submitted By The Authors
The complexity of a system is defined by the comlexity of the model necessary to effectively predict the behavior of the system
converge- Submitted By The Athors
A point converges if it is stays withing a set boundry throughout iteration.
critical points- Submitted By The Authors
The critical points of an equation is defined as the point where the graph of a function changes from increasing to decreasing
# D
diverge- Submitted By The Authors
A point diverges if it escapes from a set boundry, or goes to infinity, when the equation is iterated.
# E
euclidean geometry- Submitted By The Authors
Euclidean geometry is geometry based on Euclid's axioms and is the geometry of Euclidean space. Throughout these documents, it is used in reference with objects of an integer dimension.
# F
feigenbaun's constant- Submitted By The Authors
In a period doubling situation, such as the logisitc equation, the ratio of distances between the consecutive doubling parameter values of the equation is iterated to infity is called Feignbaun's constant. It is aproximately 4.669201609102990671853
fractal- Submitted By The Authors
fractal attractor- Submitted By The Authors
See Strange Attractor.
fractal dimension- Submitted By The Authors
fractal geometry- Submitted By The Authors
# J
julia set- Submitted By The Authors
The Julia Set, closely related to the Mandelbrot Set, is the set of converging points in a plane when tested with equation z=z2+c, using varying z values.
# K
koch curve- Submitted By The Authors
The Koch Curve is a fractal generated by removing the center third from a line segment of unit length and replacing it with an equilateral triangle, which has its base removed.
koch snowflake- Submitted By The Authors
The Koch Snowflake is a fractal, similiar to the Koch Curve, which is generated by removing the center third from each line segment of and equilateral triangle and replacing it with an equilateral triangle which is missing its base.
# L
logistic equation- Submitted By The Authors
The Logistic Equation, t =c(1-t), is a model for animal population
lorenz model- Submitted By The Authors
The Lorenz Model is a model for the convection of thermal engergy.
# M
mandelbrot set- Submitted By The Authors
The Mandelbrot Set is the set of converging points of a plane when tested with the equation z=z2+c, with varying c values.
# N
nonlinear- Submitted By The Authors
Nonlinear implies that a set of data does not fit linear model
# S
self similiar- Submitted By The Authors
Self-similiarity is the property of invariance under a change of scale. In other words, a magnified image is virtually indistinguisahble from the whole.
sensitive dependance- Submitted By The Authors
Sensitive dependance, otherwise known as the butterfly effect, is one of the key elements in a complex system, stating that complex systems have extreme sensitivity upon initial conditions
sierpinski triangle- Submitted By The Authors
The Sierpinski Triangle, or the Sierpinski Sieve is a fractal which is generated in the following manner: A triangle is repeatedly divided into four conguent triangles and the center triangle is removed.
strange attractor- Submitted By The Authors
A Strange Attractor is the limit set of a chaotic trajectory. A Strange Attractor is an attractor that is topologically from a periodic orbit or a limit cycle. It can also be considered a fractal attractor.
system- Submitted By The Authors
The understanding of the relationship between things which interact. For example, a pile of stones is a system based upon how they are piled. This site deals mainly with chaotic systems.
# Z
[ Main Page | Chaos Theory | Fractal Geometry | Glossary Of Terms And Phrases | Appendix I - Documentation | Appendix II - Source Code ] | 1,043 | 4,665 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.890625 | 3 | CC-MAIN-2013-48 | latest | en | 0.899472 |
http://love2d.org/forums/search.php?author_id=134799&sr=posts | 1,576,101,469,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540533401.22/warc/CC-MAIN-20191211212657-20191212000657-00484.warc.gz | 91,356,315 | 7,828 | ## Search found 13 matches
Mon Apr 23, 2018 6:29 pm
Forum: Support and Development
Topic: How would I go about drawing simple 3D shapes in 11.0? [Solved!]
Replies: 4
Views: 1131
### Re: How would I go about drawing simple 3D shapes in 11.0? [Solved!]
pgimeno wrote:
Mon Apr 23, 2018 9:28 am
There are two problems you may have run into; one is https://github.com/excessive/cpml/issues/33 and the other is viewtopic.php?p=219678#p219678
Yeah, I did run into that. The makeProper function reorders the matrix.
Mon Apr 23, 2018 1:40 am
Forum: Support and Development
Topic: How would I go about drawing simple 3D shapes in 11.0? [Solved!]
Replies: 4
Views: 1131
### Re: How would I go about drawing simple 3D shapes in 11.0?
After fiddling with my matrix function, I got it to work! I'm not sure what the issue was, but now I've got a 3D triangle. Also, turns out my calculations seem to have their Y coordinate flipped. Screenshot from 2018-04-22 18-37-54.png If you want to see how it works, you can visit the GitHub repo: ...
Sun Apr 22, 2018 9:40 pm
Forum: Support and Development
Topic: How would I go about drawing simple 3D shapes in 11.0? [Solved!]
Replies: 4
Views: 1131
### Re: How would I go about drawing simple 3D shapes in 11.0?
I've been working on trying to get it to work, and I've been making progress. I've been trying to adapt the love3d demos' code to work in my environment. Doing the calculations outside of the shader code seems to suggest that it should be drawing something, but it's not. Screenshot from 2018-04-22 1...
Fri Apr 20, 2018 4:56 pm
Forum: Support and Development
Topic: How would I go about drawing simple 3D shapes in 11.0? [Solved!]
Replies: 4
Views: 1131
### How would I go about drawing simple 3D shapes in 11.0? [Solved!]
11.0 adds a bunch of new features to be used from shaders for supporting depth buffers, but I don't have any clue how I'd actually use them. Are there any examples out there or something to get me started?
Tue Mar 06, 2018 10:22 pm
Forum: Libraries and Tools
Topic: love-joycon - Switch controller support for LÖVE
Replies: 3
Views: 2847
### Re: love-joycon - Switch controller support for LÖVE
Can you rewrite it so it works exactly like normal joystick API? So that, you know, joycons are automatically supported, without you having to explicitly write its own code for them? Imagine you had to write its own code for XBox controller, PS4 controller, all the third party controllers, etc. The...
Tue Mar 06, 2018 4:59 am
Forum: Libraries and Tools
Topic: love-joycon - Switch controller support for LÖVE
Replies: 3
Views: 2847
### love-joycon - Switch controller support for LÖVE
love-joycon love-joycon is a smallish library that adds Joy-Con and Pro Controller support. The library is fairly simple to use (unless you're supporting Windows, which you probably should be). Just joycon = require("joycon") and call the joycon.joystick* functions. If you support Windows, you'll h...
Sun Dec 14, 2014 4:59 am
Forum: Libraries and Tools
Topic: Locale - A simple localization system for LOVE
Replies: 1
Views: 1174
### Locale - A simple localization system for LOVE
Hello everyone! While writing my game (that hasn't been released yet), I added in a localization system, and I thought I may as well post it here so that others can use it also. Usage is pretty easy: locale = require "locale" function love.load() locale.setLocalization("en_US") end function love.dra...
Fri Dec 05, 2014 3:47 pm
Forum: Games and Creations
Topic: Pacturn (Turn-Based Pacman)
Replies: 2
Views: 1269
### Re: Pacturn (Turn-Based Pacman)
A word of an advice, but I would highly consider changing name pacman. NAMCO at from what I've seen in the internet, has a bit of a reputation of not really liking random people using their trade marked name on other projects. Yeah, I just wasn't really sure what to call it yet. But thanks for the ...
Fri Dec 05, 2014 4:26 am
Forum: Games and Creations
Topic: Pacturn (Turn-Based Pacman)
Replies: 2
Views: 1269
### Pacturn (Turn-Based Pacman)
Hello all, Pacturn is my first "completed" LÖVE game. The basic premise is that it's a turn-based Pacman game. The game is fairly simple: One player plays Pacman, and the other player(s) control the ghosts (Blinky, Speedy, Inky, and Clyde). Although there is no win condition (yet), Pacman can lose b...
Fri Nov 28, 2014 2:28 am
Forum: Support and Development
Topic: Why does this code not work?
Replies: 4
Views: 1582
### Re: Why does this code not work?
Or, if you really, really, want to:
Code: Select all
function love.draw()
function mainBackground()
love.graphics.setBackgroundColor(50, 50, 75)
love.graphics.rectangle('fill', 0, 520, 1280, 200)
end
mainBackground()
end | 1,344 | 4,723 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.546875 | 3 | CC-MAIN-2019-51 | longest | en | 0.945106 |
https://www.jntuworldupdates.org/antenna-and-wave-propagation-notes-pdf/ | 1,618,399,549,000,000,000 | text/html | crawl-data/CC-MAIN-2021-17/segments/1618038077810.20/warc/CC-MAIN-20210414095300-20210414125300-00169.warc.gz | 951,322,168 | 18,888 | # [Pdf] #1: AWP Notes – Antenna and Wave Propagation Notes Pdf Free Download
AWP Pdf notes -Here you can get lecture notes of Antenna and Wave Propagation pdf notes with unit wise topics. Here we have listed different units wise downloadable links of Antenna and Wave Propagation notes where you can click to download respectively.
## AWP Notes – Antenna and Wave Propagation Notes Pdf Free Download
Please find the Notes below according to the New and Old Syllabus:
### AWP Notes – Antenna and Wave Propagation Notes Pdf Free Download
• #### UNIT I Antenna Basics:
Introduction. Basic Antenna Parameters – Patterns, Beam Area, Radiation Intensity. Beam Efficiency, Directivity-Gain-Resolution. Antenna Apertures, Effective Height. Illustrative Problems. Fields from Oscillating Dipole, Field Zones, Shape-Impedance Considerations, Antenna Temperature, Front – to-back Ratio, Antenna Theorems, Radiation- Basic Maxwell’s Equations, Retarded Potentials – Helmholtz Theorem
Unit – 1 :N/A
• #### UNIT II Thin Linear Wire Antennas – Antenna and Wave Propagation Notes
Radiation from Small Electric Dipole, Quarter Wave Monopole and Half Wave Dipole – Current Distributions, Field Components, Radiated Power, Radiation Resistance, Beam Width, Directivity, Effective Area and Effective Height, Natural Current Distributions, Far Fields and Patterns of Thin Linear Centre-fed Antennas of Different Lengths, Illustrative Problems. Loop Antennas — Introduction, Small Loop. Comparison of Far Fields of Small Loop and Short Dipole, Radiation Resistances and Directivities of Small and Large Loops (Qualitative Treatment).
Unit – 2 : N/A
• #### UNIT III Antenna Arrays: Point Sources
Definition, Patterns, arrays of 2 Isotropic Sources – Different Cases, Principle of Pattern Multiplication, Uniform Linear Arrays – Broadside Arrays, Endfire Arrays, EFA with Increased Directivity. Derivation of their Characteristics and Comparison, BSAs with Non-uniform Amplitude Distributions — General Considerations and Binomial Arrays, Illustrative Problems.
Unit – 3 : N/A
• #### UNIT IV VHF. UHF and Microwave Antennas – I : Antenna and Wave Propagation Pdf
Arrays with Parasitic Elements. Yagi-Uda Array. Folded Dipoles and their Characteristics, Helical Antennas – Helical Geometry. Helix Modes. Practical Design Considerations for Monofilar Helical Antenna in Axial and Normal Modes. Horn Antennas -Types, Fermat’s Principle, Optimum Horns. Design Considerations of Pyramidal Horns, Illustrative Problems.
Unit – 4 : N/A
• #### UNIT V VHF, UHF and Microwave Antennas – II: Antenna and Wave Propagation Pdf
Microstrip Antennas ~- Introduction. Features, Advantages and Limitations. Rectangular Patch Antennas — Geometry and Parameters, Characteristics of Microstrip Antennas. Impact of Different Parameters on Characteristics, Reflector Antennas — Introduction, Flar Sheet and Corner Reflectors, Paraholoidal Reflectors — Geometry. Pattern Characteristics, Feed Methods. Reflector Types – Related Features. Illustrative Problems.
Unit – 5 : N/A
• #### UNIT VI Lens Antennas – Antenna and Wave Propagation Notes Pdf
Introduction, Geometry of Non–metallic Dielectric Lenses, Zoning, Tolerances, Applications. Antenna Measurements: Introduction. Concepts – Reciprocity, Near and Far Fields, Coordinate System, Sources of Errors. Patterns to be Measured, Pattern Measurement Arrangement, Directivity Measurement. Gain Measurements (by Comparison, Absolute and 3-Antenna Methods)
Unit – 6 : N/A
• #### UNIT VII Wave Propagation – I: Antenna and Wave Propagation Notes Pdf
Introduction, Definitions, Categorizations and General Classifications, Different Modes of Wave Propagation, Ray/Mode Concepts. Ground Wave Propagation (Qualitative Treatment) — introduction, Platte Earth Reflections, Space and Surface Waves, Wave Tilt, Curved Earth Reflections. Space Wave Propagation – Introduction. Field Strength Variation with Distance and Height, Effect ofEarth‘s Curvature. Absorption. Super Refraction. M-Curves and Duct Propagation. Scattering Phenomena. Tropospheric Propagation, Fading and Path Loss Calculations.
Unit – 7 : N/A
• #### UNIT VIII Wave Propagation – II: Antenna and wave propagation pdf
Sky Wave Propagation — Introduction. Structure of ionosphere, Refraction and Reflection of Sky Waves by ionosphere, Ray Path, Critical Frequency, MUF, LUF, OF, Virtual Hlght and Skip Distance. Relation between and Skip Distance, Multi-hop Propagation. Energy Loss in ionosphere. Summary of Wave Characteristics in Different Frequency Ranges.
Unit – 8 : N/A
TEXT BOOKS : 1. Antennas for All Applications – John D. Kraus and Ronald J. Marhefka, TMHl, 3rd Edn., 2003. 2. Electromagnetic Waves and Radiating Systems – E.C. Jordan and K.G. Balmain, PHI, 2nd ed., 2000.
Note :- These notes are according to the r09 Syllabus book of JNTUH.In R13 ,8-units of R09 syllabus are combined into 5-units in r13 syllabus. | 1,180 | 4,908 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2021-17 | longest | en | 0.803259 |
http://www.dreamincode.net/forums/topic/327514-looping-question/page__pid__1891802__st__0 | 1,464,373,495,000,000,000 | text/html | crawl-data/CC-MAIN-2016-22/segments/1464049276964.77/warc/CC-MAIN-20160524002116-00068-ip-10-185-217-139.ec2.internal.warc.gz | 468,961,075 | 18,734 | # Looping Question
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### #1 kyle_denney
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• Posts: 171
• Joined: 10-August 12
# Looping Question
Posted 22 August 2013 - 02:42 PM
I am trying to find the easiest way to do the following:
I have pseudo-code but think it needs some optimization.
The pseudo-code I think one could use is something like this
```Dim resultNeeded = 123
Dim x = 0
Dim y = 0
For x as int16 = 0 to 1000
For y as int16 = 0 to 1000
Perform some quick calculation in here to get resultNeeded
Next
Next
```
What is inside the loop does not matter. What I am wanting to know is to see if the above loop is the best way to iterate through the two combinations of 1000 or is there a quicker way in VB.Net or C#. Thanks!
Basically set up two variables x, y such that each number is iterated through a loop all the way up to 1000 for both x and y running an algorithm on the result to see if it matches a certain criteria.
I am trying to run an algorithm that when using two numbers between 0-1000 will give a certain output - lets say a key like ACD123-45232.
All I am wanting to know is how to iterate through the two sets of 0-1000 the fastest way and run the algorithm through each iteration to test if it matches the value like above. The algorithm doesn't matter and I know in VB.Net you can use a For Each/Next scenario taking each number and looping it against the other number but I feel there has to be a better way. Any advice is appreciated. Thanks! - Remember any actual code is VB.Net or C#. Thanks!
Is This A Good Question/Topic? 0
## Replies To: Looping Question
### #2 modi123_1
• Suitor #2
Reputation: 11617
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## Re: Looping Question
Posted 22 August 2013 - 02:49 PM
Yes.. that would be the most straight forward. So.. trying to brute force a key? Office product keys? | 644 | 2,239 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.015625 | 3 | CC-MAIN-2016-22 | latest | en | 0.873723 |
http://examcrazy.com/signal-flow-graph-solved-examples | 1,540,118,642,000,000,000 | text/html | crawl-data/CC-MAIN-2018-43/segments/1539583513844.48/warc/CC-MAIN-20181021094247-20181021115747-00525.warc.gz | 126,863,056 | 30,796 | # Signal flow graphs
Signal flow graphs solved examples
Consider the following two-port network
A signal flow graph is a diagram depicting the relationships between signals in a network. It can also be used to solve for ratios of these signals. Signal flow graphs are used in control systems, power systems and other fields besides microwave engineering. Key elements of a signal flow graph are:
1. The network must be linear,
2. Nodes represent the system variables,
3. Branches represent paths for signal flow.
For example, referring to the two-port above, the nodes and branches are
4. A signal yk traveling along a branch between nodes ak and bjis multiplied by the gain of that branch:
That is, bj = Sjk ak
5. Signals travel along branches only in the direction of the arrows.
This restriction exists so that a branch from ak to bj denotes a proportional dependence of bj on ak , but not the reverse.
Solving Signal Flow Graphs
Signal flow graphs (SFGs) can form an intuitive picture of the signal flow in a network. As an application, we will develop SFGs in the next lecture to help us calibrate out systematic errors present when we make measurements with a VNA. Another useful characteristic is that we can solve for ratios of signals directly from a SFG using a simple algebra. There are four rules that form the algebra of SFGs:
1. Series Rule. Given the two proportional relations then
V2 = S21V1 and V3 = S32 V2
V3= (S32S21)V1
In a SFG, this is represented as
In other words, two series paths are equivalent to a single path with a transmission factor equal to a product of the two original transmission factors.
2. Parallel Rule. Consider the relation:
V2 = Sa V1 + Sb V1 =(Sa +Sb) V1
In a SFG, this is represented as
In other words, two parallel paths are equivalent to a single path with a transmission factor equal to the sum of the original transmission coefficients.
3. Self-Loop Rule. Consider the relations
V2 = S21 V1 + S22 V2
and V3 = S32V2 We will choose to eliminate V2 . From
V2(1-S22)=S11V1 ÞV2=(S21/1-S22)V1
Substituting this into (4) gives
V3=(S32S21/1-S22)V1
In a SFG, this is represented as
In other words, a feedback loop may be eliminated by dividing the input transmission factor by one minus the transmission factor around the loop.
4. Splitting Rule. Consider the relationships:
V4 = S42 V2 V2 = S21 V1 and V3=S32V2
The SFG is
From (6) and (7) we find that V4 = S42S21V1. Hence, if we use the Series Rule “in reverse” we can define:
V4 S42 V4= and V4= S21 V1
In a SFG, this is represented as
In other words, a node can be split such that the product of transmission factors from input to output is unchanged.
Example N21.1. Construct a signal flow graph for the network shown below. Determine tin and VL using only SFG algebra.
The signal flow graph is:
Notice the arrow directions for Ts and TL . These are the correct orientations since
And
We will systematically apply the four rules above to reduce this diagram to a form that will directly allow us to determine both Tin and VL .
Starting this solution process is probably the most difficult part. The rest is fairly systematic.
Step 1. Start by splitting node a2 using Rule 4:
Why split TL and not S22 ? Because the TL arrow is in the same direction as S12 .
Step 2. Eliminate nodes a2 and a2 using Rule 1:
Step 3. Eliminate the self loop at b2 using Rule 3:
Step 4. Eliminate node b2 using Rule 1:
Step 5. Apply Rule 2:
From this last diagram we can directly solve for Tin :
Next, we will determine the load voltage VL . The voltage on this
TL can be expressed as V2( z) = b2 (e-jbz + TL e-jbz) . At the terminal plane z = 0, then
V2(0) =VL = b2(1+TL)
In this expression, V2(0) =VL because the TL is very short. So, we see from (9) that to find VL we need to determine b2 . In Step 4, however, we eliminated that node. Let’s start again from Step 3, but now split node b1 :
Step 4?. Split node b1 using Rule 4:
Step 5?. Next, we can use Rule 1, then Rule 3 (Self-Loop Rule) to all branches feeding node a1 :
Step 6?. Split node b2 using Rule 4:
Step 7?. Apply Rule 3 one last time to remove the self loop:
Step 8?. Using the Series Rule, we can now find b2 as:
Or
Using this b2 , we can determine VL from (9) to be | 1,113 | 4,233 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2018-43 | longest | en | 0.924803 |
https://www.rocknets.com/converter/length/feet-to-cm-converter | 1,721,664,753,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763517890.5/warc/CC-MAIN-20240722160043-20240722190043-00448.warc.gz | 839,800,103 | 16,492 | # Feet to Centimeters Converter (ft to cm)
Example: 1 ft = 30.48 cm feet: ft cm: cm
You may also interested in: cm to feet Converter
The online feet to cm Converter is used to convert the length from feet to cm.
#### The feet to cm Conversion Formula
You can use the following formula to convert from cm to feet :
X(cm) = y(feet) × 30.48
Example: How to convert 2 Feet to cm?
X(cm) = 2(feet) × 30.48
Anwser: 60.96 cm
#### feet to cm conversion table
Feet (ft) Centimeters (cm)
1 ft 30.48 cm
2 ft 60.96 cm
3 ft 91.44 cm
4 ft 121.92 cm
5 ft 152.4 cm
6 ft 182.88 cm
7 ft 213.36 cm
8 ft 243.84 cm
9 ft 274.32 cm
10 ft 304.8 cm
20 ft 609.6 cm
30 ft 914.4 cm
40 ft 1219.2 cm
50 ft 1524 cm
60 ft 1828.8 cm
70 ft 2133.6 cm
80 ft 2438.4 cm
90 ft 2743.2 cm
100 ft 3048 cm
500 ft 15240 cm
1000 ft 30480 cm
Full feet to cm conversion table
#### Frequently asked questions to convert feet to cm
What is 4 feet in cm ?
What is 5 feet in cm ?
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#### Online Converter to convert feet to cm
We provide the online converter for free. You can use the feet to cm converter to convert the feet to cm.
#### References
More references for Feet/Foot unit and Centimetre | 419 | 1,234 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.71875 | 3 | CC-MAIN-2024-30 | latest | en | 0.67446 |
https://socratic.org/questions/what-is-the-slope-of-the-tangent-line-of-r-theta-sin-10theta-3-pi-8-at-theta-pi- | 1,652,881,240,000,000,000 | text/html | crawl-data/CC-MAIN-2022-21/segments/1652662522270.37/warc/CC-MAIN-20220518115411-20220518145411-00065.warc.gz | 609,885,658 | 6,016 | # What is the slope of the tangent line of r=theta-sin((10theta)/3-(pi)/8) at theta=(pi)/4?
Slope of the tangent line of
$r = \theta - \sin \left(\frac{10 \theta}{3} - \frac{\pi}{8}\right)$
at $\theta = \frac{\pi}{4}$ is
$\frac{\mathrm{dy}}{\mathrm{dx}} = 0.995$
#### Explanation:
Given:
$r = \theta - \sin \left(\frac{10 \theta}{3} - \frac{\pi}{8}\right)$
$\frac{\mathrm{dr}}{d \left(\theta\right)} = 1 - \left(\frac{10}{3}\right) \cos \left(\frac{10 \theta}{3} - \frac{\pi}{8}\right)$
wkt
$x = r \cos \theta$
$\frac{\mathrm{dx}}{d \left(\theta\right)} = r \left(- \sin \theta\right) + \cos \theta \left(\frac{\mathrm{dr}}{d \left(\theta\right)}\right)$
at $\theta = \frac{\pi}{4}$
$r = \frac{\pi}{4} - \sin \left(\frac{10 \frac{\pi}{4}}{3} - \frac{\pi}{8}\right)$
$= - 0.008$
$\frac{\mathrm{dr}}{d \left(\theta\right)} = 1 - \left(\frac{10}{3}\right) \cos \left(\frac{10 \frac{\pi}{4}}{3} - \frac{\pi}{8}\right)$
$= 3.029$
$\frac{\mathrm{dx}}{d \left(\theta\right)} = - 0.008 \left(- \sin \frac{\pi}{4}\right) + \cos \left(\frac{\pi}{4}\right) \left(3.029\right)$
$= 0.008 \left(0.707\right) + 0.707 \left(3.029\right)$
$\frac{\mathrm{dx}}{d \left(\theta\right)} = 2.147$
$y = r \sin \theta$
$\frac{\mathrm{dy}}{d \left(\theta\right)} = r \left(\cos \theta\right) + \sin \theta \left(\frac{\mathrm{dr}}{d \left(\theta\right)}\right)$
At $\theta = \frac{\pi}{4}$
$\frac{\mathrm{dy}}{d \left(\theta\right)} = - 0.008 \left(\cos \frac{\pi}{4}\right) + \sin \left(\frac{\pi}{4}\right) \left(3.029\right)$
$\frac{\mathrm{dy}}{d \left(\theta\right)} = 2.136$
Slope of the tangent line of
$r = \theta - \sin \left(\frac{10 \theta}{3} - \frac{\pi}{8}\right)$
at $\theta = \frac{\pi}{4}$ is
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\frac{\mathrm{dy}}{d \left(\theta\right)}}{\frac{\mathrm{dx}}{d \left(\theta\right)}} = \frac{2.136}{2.147}$
Thus,
$\frac{\mathrm{dy}}{\mathrm{dx}} = 0.995$ | 780 | 1,886 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 24, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.5 | 4 | CC-MAIN-2022-21 | latest | en | 0.335181 |
https://math.stackexchange.com/questions/1983722/why-is-determinant-called-determinant | 1,716,265,542,000,000,000 | text/html | crawl-data/CC-MAIN-2024-22/segments/1715971058383.61/warc/CC-MAIN-20240521025434-20240521055434-00159.warc.gz | 333,670,167 | 41,499 | Why is determinant called determinant?
Can someone explain to me why do we call the determinant of a matrix "determinant"? Does it have any meaning? Like it determines something for example!
• It also determines whether the corresponding system of linear equations has a solution.
– J126
Oct 25, 2016 at 1:14
• Systems of equations can be called "overdetermined" or "underdetermined" if they have too many or too few number of equations versus unknowns respectively. When there are the same number of equations as unknowns, the determinant will be nonzero precisely when there is a unique solution and will be zero when there is the possibility of infinitely many or no solutions. Oct 25, 2016 at 1:14
• Can anyone deterministically determine what it is that the determinant determines? :) Oct 25, 2016 at 1:18
• (I'm determined to find out...) Oct 25, 2016 at 1:19
• en.wikipedia.org/wiki/Determinant#History Oct 25, 2016 at 3:58
Here is some information about the origin of the term determinant. This term was introduced the first time $1801$ by C.F. Gauss in his Disquisitiones arithmeticae, XV, p. 2 in connection with a form of second degree.
• The following is from The Theory of Determinants in the historical order of development (1905) by Thomas Muir.
[Muir, p. 64]: Gauss writes the form as \begin{align*} axx+2bxy+cyy \end{align*} and for shortness speaks of it as the form $(a,b,c)$.
The function of the coefficients $a,b,c$, which was found by Lagrange to be of notable importance in the discussion of the form, Gauss calls the determinant of the form, the exact words being
• [Gauss, 1801] Numerum $bb-ac$, a cuius indole preprietates formae $(a,b,c)$ imprimis pendere in sequentibus decebimus, determinantem huius formae uocabimus.
and Muir continues:
• [Muir, p.64] ... Here then we have the first use of the term which with an extended signification has in our day come to be so familiar. It must be carefully noted that the more general functions, to which the name came afterwards to be given, also repeatedly occur in the course of Gauss' work, ...
Besides the historic reasons, which are covered in previous answers, here's another take at why we would say the determinant "determines" something. This is clearly not the origin of the word, I think, but it gives you another curious answer to the question.
I believe that an interesting way of looking at it is beginning with alternating multilinear forms, which are mappings such $$t: V \times \overset{n}{\dots}\times V \to \mathbb{R}$$ (where $$V$$ is a vector space and $$\mathbb{R}$$ can be substituted with any other scalar field) which are linear in every entry and evaluate to $$0$$ if two or more entries are equal eg. Suppose $$V = \mathbb{R}$$, then $$t(1,2, \dots, n-1, 2) = 0$$ (that is what 'alternating' means). There are plenty of places where you can read about multilinear mappings: see Wikipedia for example, read it first and then come back, I will focus on the fact that interests us.
Suppose that you have a basis of $$V$$, say $$\lbrace \vec{u}_1, \dots, \vec{u}_n \rbrace$$. Now take any family of vectors you like in the input space $$\lbrace \vec{x}_1, \dots, \vec{x}_n \rbrace$$ and note that those vectors can be expressed in terms of the basis we chose
$$$$(\vec{x}_1, \dots, \vec{x}_n) = (\vec{u}_1, \dots, \vec{u}_n) \begin{pmatrix} a_1^1 & a_2^1 & \dots & a_n^1 \\ a_1^2 & a_2^2 & \dots & a_n^2 \\ \vdots & \vdots & \vdots &\vdots \\ a_1^n & a_2^n & \dots & a_n^n \end{pmatrix}$$$$
Now we are ready to expand $$t(\vec{x}_1, \dots, \vec{x}_n)$$ using the mutilinearity of $$t$$:
\begin{align*} &t(\vec{x}_1, \dots, \vec{x}_n) = t(\sum_{i_1=1}^n a^{i_1}_1 \vec{u}_{i_1}, \dots, \sum_{i_n=1}^n a^{i_n}_n \vec{u}_{i_n}) =\\ &= \sum_{i_1=1}^n a^{i_1}_1 \dots \sum_{i_n=1}^n a^{i_n}_n\cdot t(\vec{u}_{i_1}, \dots,\vec{u}_{i_n}) = \sum_{i_1,\dots, i_n=1}^n a^{i_1}_1 \dots a^{i_n}_n \cdot t(\vec{u}_{i_1}, \dots,\vec{u}_{i_n}) \end{align*}
In this sum we do not put any restrictions to the indexes and so there will be terms where we will find repeated $$i_j$$, and then that term is equal to $$0$$ because $$t$$ is alternating eg. $$t(\vec{u}_{i_1}, \dots,\vec{u}_{i_j}, \dots, \vec{u}_{i_j}, \dots,\vec{u}_{i_n}) = 0$$. Therefore, the only terms in this sum which are not zero are those such $$(i_1, \dots, i_n)$$ are all different; in other words, the indexes are a permutation of $$(1,\dots,n)$$.
\begin{align*} t(\vec{x}_1, \dots, \vec{x}_n) &= \sum_{\sigma \in \mathcal{S}_n} a^{\sigma(1)}_1 \dots a^{\sigma(n)}_n \cdot t(\vec{u}_{\sigma(1)}, \dots,\vec{u}_{\sigma(n)}) = \\ &= \sum_{\sigma \in \mathcal{S}_n} a^{\sigma(1)}_1 \dots a^{\sigma(n)}_n \cdot \text{sig}(\sigma) \cdot t(\vec{u}_1, \dots,\vec{u}_n) \end{align*}
Finally, we see that the way $$t$$ acts upon any arbitrary family of vectors has a part which will be common to every family of vectors ($$t$$ acting on the basis), and a differentiating part which will be particular to that set of vectors (because their coordinates with respect to that basis identify them uniquely). We call that differentiating part the determinant of that set of vectors, because indeed is what determines the value of $$t(\vec{x}_1, \dots, \vec{x}_n)$$.
$$$$\operatorname{det}_{\lbrace \vec{u}_i \rbrace}(\vec{x}_1, \dots, \vec{x}_n) = \sum_{\sigma \in \mathcal{S}_n} a^{\sigma(1)}_1 \dots a^{\sigma(n)}_n \cdot \operatorname{sig}(\sigma)$$$$
You see that the definition of the determinant of a matrix is exactly the same. Indeed it can be seen as the determinant of the vectors that form the columns of the matrix. The final part of this reasoning would be to show that there is a single multilinear form $$d$$ that verifies $$d(\vec{u}_1, \dots, \vec{u}_n) = 1$$ for a given basis, and thus $$d(\vec{x}_1, \dots, \vec{x}_n) = \operatorname{det}_{\lbrace \vec{u}_i \rbrace}(\vec{x}_1, \dots, \vec{x}_n)$$, and so determinant is itself an alternating multilinear form. Then we get all the nice properties of multilinear forms for our concept of determinant.
• I would say so, @EpsilonDelta. The second to last paragraph closes that argument. My point is that the concept of determinant of a matrix can be derived this way (my last paragraph tries to link multilinear forms with determinants of matrices briefly) and in this derivation we see why it makes sense to call it determinant. The historical reason might be different as it is pointed out in one of the other comments, but disregarding it, this is another view as to why we would say it determines something. I hope you agree on this. If not, should I delete this answer? Jul 1, 2019 at 11:27
• I interpreted the question as the historic origin of the determinant. Now I see your point, but maybe make this somewhat clearer in your answer. +1 Jul 1, 2019 at 13:03
• Cool, and thanks for the feedback! Jul 9, 2019 at 9:57
It determines whether a linear system of equations has a solution.
A linear system of $$m$$ equations in $$n$$ variables, denoted $$A\mathbf{x}=\mathbf{b}$$, has a unique solution if the rank of its augmented matrix equals $$n$$, denoted $$\operatorname{rank}(A^{\#}) = n$$. (The rank of a matrix is the number of nonzero rows in its row-echelon form and the augmented matrix, $$A^{\#}$$, is $$A$$ extended with $$\mathbf{b}$$ in the last column.)
For $$1\times1$$ $$A$$ ($$[a_{11}]$$), the determinant is $$a_{11}$$ and the linear system it represents has a solution if $$a_{11} \ne 0$$ (this is the trivial case of $$a_{11}x = b$$).
For $$2\times2$$ $$A$$, represented
$$A=\left[ \matrix{ a_{11} & a_{12} \\ a_{21} & a_{22} } \right]$$
the row-echelon form is
$$\left[ \matrix{ a_{11} & a_{12} \\ 0 & a_{22} - \frac{a_{21}a_{12}}{a_{11}} } \right]$$
so as long as $$a_{22} - \frac{a_{21}a_{12}}{a_{11}} \ne 0$$, or equivalently $$a_{11}a_{22} - a_{12}a_{21} \ne 0$$, the system has a unique solution.
Extending this to larger 2D matrices,
$$\det(A) = \sum \sigma \left( p_1, p_2, \ldots, p_n \right)a_{1 p_1} a_{2 p_2} \cdots a_{n p_n}$$
where the summation is over the $$n!$$ distinct permutations $$\left( p_1, p_2, \ldots, p_n \right)$$ of the integers $$1, 2, 3, \ldots, n$$ and
$$\sigma \left( p_1, p_2, \ldots, p_n \right) = \begin{cases} +1 & \text{if } \left( p_1, p_2, \ldots, p_n \right) \text{ has even parity}, \\ -1 & \text{if } \left( p_1, p_2, \ldots, p_n \right) \text{ has odd parity.} \end{cases}$$
This is explained well in Chapter 3 of "Linear Analysis and Differential Equations" by Goode and Annin (Amazon Link).
Determinants can be used to calculate areas and volumes in a geometric sense, but the term itself originated from its use in determining whether or not systems of equations have solutions.
NOTE: NOT to be confused with the discriminant, which "discriminates" (i.e. deciphers) the types and numbers of solutions to a polynomial equation. | 2,734 | 8,834 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 52, "wp-katex-eq": 0, "align": 0, "equation": 2, "x-ck12": 0, "texerror": 0} | 3.609375 | 4 | CC-MAIN-2024-22 | latest | en | 0.92209 |
https://visp-doc.inria.fr/doxygen/visp-3.3.0/classvpAdaptiveGain.html | 1,721,404,001,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514908.1/warc/CC-MAIN-20240719135636-20240719165636-00022.warc.gz | 525,572,483 | 7,089 | Visual Servoing Platform version 3.3.0 under development (2020-02-17)
#include <visp3/vs/vpAdaptiveGain.h>
## Public Member Functions
vpAdaptiveGain (double gain_at_zero, double gain_at_infinity, double slope_at_zero)
void initFromConstant (double c)
void initFromVoid (void)
void initStandard (double gain_at_zero, double gain_at_infinity, double slope_at_zero)
double setConstant (void)
double value_const (double x) const
double value (double x) const
double limitValue_const (void) const
double limitValue (void) const
double getLastValue (void) const
double operator() (double x) const
double operator() (const vpColVector &x) const
double operator() (void) const
## Static Public Attributes
static const double DEFAULT_LAMBDA_ZERO = 1.666
static const double DEFAULT_LAMBDA_INFINITY = 0.1666
static const double DEFAULT_LAMBDA_SLOPE = 1.666
## Friends
VISP_EXPORT std::ostream & operator<< (std::ostream &os, const vpAdaptiveGain &lambda)
## Detailed Description
As described in [20], a varying gain could be used in the visual servoing control law
with
where:
• is the gain in 0, that is for very small values of
• is the gain to infinity, that is for very high values of
• is the slope of at
As described in Tutorial: How to boost your visual servo control law, the interest of Using an adaptive gain is to reduce the time to convergence in order to speed up the servo.
The following example shows how to use this class in order to use an adaptive gain with the following parameters , and .
#include <visp3/vs/vpServo.h>
int main()
{
vpAdaptiveGain lambda(4, 0.4, 30); // lambda(0)=4, lambda(oo)=0.4 and lambda'(0)=30
vpServo servo;
servo.setLambda(lambda);
while(1) {
}
}
This other example shows how to use this class in order to set a constant gain that will ensure an exponential decrease of the task error.
#include <visp3/vs/vpServo.h>
int main()
{
vpServo servo;
servo.setLambda(lambda);
while(1) {
}
}
Examples:
mbot-apriltag-2D-half-vs.cpp, mbot-apriltag-ibvs.cpp, mbot-apriltag-pbvs.cpp, servoAfma4Point2DCamVelocityKalman.cpp, servoBebop2.cpp, servoFlirPtuIBVS.cpp, servoFrankaIBVS.cpp, servoFrankaPBVS.cpp, servoViper850Point2DCamVelocityKalman.cpp, tutorial-ibvs-4pts-plotter-continuous-gain-adaptive.cpp, and tutorial-ibvs-4pts-plotter-gain-adaptive.cpp.
Definition at line 121 of file vpAdaptiveGain.h.
## Constructor & Destructor Documentation
Basic constructor which initializes all the parameters with their default value:
Definition at line 71 of file vpAdaptiveGain.cpp.
References initFromVoid().
explicit
Constructor that initializes the gain as constant. In that case .
Parameters
c : Value of the constant gain. A typical value is 0.5.
Definition at line 84 of file vpAdaptiveGain.cpp.
References initFromConstant().
Constructor that initializes the gain as adaptive.
Parameters
gain_at_zero : the expected gain when : . gain_at_infinity : the expected gain when : . slope_at_zero : the expected slope of when : .
Definition at line 96 of file vpAdaptiveGain.cpp.
References initStandard().
## ◆ getLastValue()
double vpAdaptiveGain::getLastValue ( void ) const
inline
Gets the last adaptive gain value which was stored in the class.
Returns
It returns the last adaptive gain value which was stored in the class.
Definition at line 195 of file vpAdaptiveGain.h.
## ◆ initFromConstant()
void vpAdaptiveGain::initFromConstant ( double c )
Initializes the parameters to have a constant gain. In that case .
Parameters
c : Value of the constant gain. A typical value is 0.5.
Examples:
servoAfma4Point2DCamVelocityKalman.cpp.
Definition at line 115 of file vpAdaptiveGain.cpp.
## ◆ initFromVoid()
Initializes the parameters with the default value :
Definition at line 130 of file vpAdaptiveGain.cpp.
## ◆ initStandard()
void vpAdaptiveGain::initStandard ( double gain_at_zero, double gain_at_infinity, double slope_at_zero )
Set the parameters used to compute .
Parameters
gain_at_zero : the expected gain when : . gain_at_infinity : the expected gain when : . slope_at_zero : the expected slope of when : .
Examples:
mbot-apriltag-2D-half-vs.cpp, mbot-apriltag-ibvs.cpp, mbot-apriltag-pbvs.cpp, servoAfma4Point2DCamVelocityKalman.cpp, and servoViper850Point2DCamVelocityKalman.cpp.
Definition at line 147 of file vpAdaptiveGain.cpp.
## ◆ limitValue()
double vpAdaptiveGain::limitValue ( void ) const
Gets the value of the gain at infinity (ie the value of ) and stores it as a parameter of the class.
Returns
It returns the value of the gain at infinity.
Definition at line 252 of file vpAdaptiveGain.cpp.
References limitValue_const().
Referenced by operator()().
## ◆ limitValue_const()
double vpAdaptiveGain::limitValue_const ( void ) const
Gets the value of the gain at infinity (ie the value of ). This function is similar to limitValue() except that here the value is not stored as a parameter of the class.
Returns
It returns the value of the gain at infinity.
Definition at line 218 of file vpAdaptiveGain.cpp.
Referenced by limitValue().
## ◆ operator()() [1/3]
double vpAdaptiveGain::operator() ( double x ) const
Operator that computes where
Parameters
x : Input value to consider. During a visual servo this value can be the euclidian norm or the infinity norm of the task function.
Returns
It returns the value of the computed gain.
value()
Definition at line 280 of file vpAdaptiveGain.cpp.
References value().
## ◆ operator()() [2/3]
double vpAdaptiveGain::operator() ( const vpColVector & x ) const
Operator which computes where
Parameters
x : Input vector to consider.
Returns
It returns the value of the computed gain.
Definition at line 303 of file vpAdaptiveGain.cpp.
References vpColVector::infinityNorm(), and value().
## ◆ operator()() [3/3]
double vpAdaptiveGain::operator() ( void ) const
Gets the value of the gain at infinity (ie the value of ).
Returns
It returns the value of the gain at infinity.
limitValue()
Definition at line 290 of file vpAdaptiveGain.cpp.
References limitValue().
## ◆ setConstant()
Sets the internal parameters in order to obtain a constant gain equal to the gain in 0 set through the parameter .
Returns
It returns the value of the constant gain .
Definition at line 174 of file vpAdaptiveGain.cpp.
## ◆ value()
double vpAdaptiveGain::value ( double x ) const
Computes the value of the adaptive gain using:
This value is stored as a parameter of the class.
Parameters
x : Input value to consider. During a visual servo this value can be the euclidian norm or the infinity norm of the task function.
Returns
It returns the value of the computed gain.
Definition at line 239 of file vpAdaptiveGain.cpp.
References value_const().
Referenced by operator()().
## ◆ value_const()
double vpAdaptiveGain::value_const ( double x ) const
Computes the value of the adaptive gain using:
Parameters
x : Input value to consider. During a visual servo this value can be the euclidian norm or the infinity norm of the task function.
Returns
It returns the value of the computed gain.
Definition at line 204 of file vpAdaptiveGain.cpp.
Referenced by value().
## ◆ operator<<
VISP_EXPORT std::ostream& operator<< ( std::ostream & os, const vpAdaptiveGain & lambda )
friend
Prints the adaptive gain parameters .
Parameters
os : The stream where to print the adaptive gain parameters. lambda : The adaptive gain containing the parameters to print.
Definition at line 319 of file vpAdaptiveGain.cpp.
## ◆ DEFAULT_LAMBDA_INFINITY
static
Definition at line 125 of file vpAdaptiveGain.h.
Referenced by initFromVoid().
## ◆ DEFAULT_LAMBDA_SLOPE
static
Definition at line 126 of file vpAdaptiveGain.h.
Referenced by initFromVoid(). | 1,947 | 7,733 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2024-30 | latest | en | 0.543566 |
http://ncatlab.org/nlab/show/crossing+number | 1,448,432,152,000,000,000 | application/xhtml+xml | crawl-data/CC-MAIN-2015-48/segments/1448398444974.3/warc/CC-MAIN-20151124205404-00122-ip-10-71-132-137.ec2.internal.warc.gz | 170,118,620 | 4,964 | # nLab crossing number
Let $K$ be a knot.
###### Definition
The crossing number, $c(K)$, of $K$ is the minimum number of crossings in a diagram in the isotopy class of $K$.
The crossing number is thus the number of crossings in the simplest picture of a knot. A diagram of a knot $K$ with exactly $c(K)$ crossings is called a minimal diagram.
##### Examples
• $c(unknot) = 0$;
• $c(trefoil) = 3$;
• $c(figure-8) = 4$.
• if a diagram has $1$ or $2$ crossings it represents the unknot, so there are no knots with $c(K) = 1$ or $2$.
The crossing number is related to the unknotting number, but in quite a subtle way.
In the books by Burde and Zeischang (1985) and Kauffman (1987), the tables of knots are arranged according to crossing number. (Choices have been made of one mirror image? or the other.) Given some arbitrary diagram, the crossing number of the knot that it respresents may be hard to determine.
Created on October 15, 2010 17:07:09 by Tim Porter (95.147.236.87) | 279 | 986 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 13, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2015-48 | latest | en | 0.890185 |
https://senselab.med.yale.edu/modeldb/getModelFile?model=155565&AttrID=23&s=yes&file=/%2FWebPublication%2Flib%2FGaussian.m | 1,695,748,853,000,000,000 | text/plain | crawl-data/CC-MAIN-2023-40/segments/1695233510214.81/warc/CC-MAIN-20230926143354-20230926173354-00565.warc.gz | 570,993,488 | 883 | function z=Gaussian(x,y,mu,sigma) x=x-mu(1); y=y-mu(2); z=1/(2*pi*sigma^2) * exp(-1*(x.^2 + y.^2)/(2*sigma^2)); end | 57 | 115 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2023-40 | latest | en | 0.188539 |
https://denisegaskins.com/tag/mtap-playful-math-carnival/ | 1,638,229,766,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358847.80/warc/CC-MAIN-20211129225145-20211130015145-00384.warc.gz | 294,737,528 | 54,806 | ## Carnival 150: Keeping Playful Math Alive
Welcome to the 150th edition of the Playful Math Education Blog Carnival — a smorgasbord of delectable tidbits of mathy fun. It’s like a free online magazine devoted to learning, teaching, and playing around with math from preschool to high school.
Bookmark this post, so you can take your time browsing.
There’s so much playful math to enjoy!
We didn’t have a volunteer host for October, so I’m squeezing this in between extended-family situations that require my time. If you’d like to help keep the Playful Math Carnival alive, we desperately need hosts for 2022!
## Try These Sum-Difference Puzzles
By tradition, we start the carnival with a puzzle/activity in honor of our 150th edition. This puzzle comes from the archives of Iva Sallay’s delightful Find the Factors blog.
Copy the puzzle diagrams below. Write numbers in the boxes to make true statements going uphill, and use the same numbers to make true statements going downhill:
Six has two factor pairs. One of those factor pairs adds up to 5, and the other one subtracts to 5. Can you place those factors in the proper boxes to complete the first puzzle?
150 has six factor pairs. One of those factor pairs adds up to 25, and another one subtracts to 25. If you can identify those factor pairs, then you can complete the second puzzle!
How are the two puzzles related?
Iva also posts one of my favorite multiplication puzzles, her scrambled times tables. Check out puzzle #150 in this post (it’s Level 3, or medium difficulty) and then browser her site for plenty more.
## Playful Math Carnival 149 via Nature Study Australia
Check out the latest carnival of playful math:
Each monthly Playful Math Education Blog Carnival brings you a great new collection of puzzles, math conversations, teaching tips, and all sorts of mathy fun. It’s like a free online magazine of mathematical adventures, helpful and inspiring no matter when you read them.
Johanna put together a beautiful collection of mathematical resources and activities for studying math in nature, biometric data, journaling, math poetry, and other delights.
### Help Us Keep the Carnival Going
The Playful Math Blog Carnival wants you!
The carnival is a joint effort. We depend on our volunteer hosts to collect blog posts and write the carnival each month.
Putting together a blog carnival can be a lot of work, but it’s a great opportunity to share the work of bloggers you admire and to discover new math-friends online. I love that part of being a host!
Classroom teachers, homeschoolers, college professors, unschoolers, or anyone who likes to play around with math — if you would like to take a turn hosting the carnival, please speak up.
## Playful Math Carnival 148 via Math Book Magic
Check out the latest carnival of playful math:
Each monthly Playful Math Education Blog Carnival brings you a great new collection of puzzles, math conversations, teaching tips, and all sorts of mathy fun. It’s like a free online magazine of mathematical adventures, helpful and inspiring no matter when you read them.
Kelly put together this magical collection of mathematical games, activities, art projects, hands-on fun, math storybooks, and inspiration.
### Help Us Keep the Carnival Going
The Playful Math Blog Carnival wants you!
The carnival is a joint effort. We depend on our volunteer hosts to collect blog posts and write the carnival each month.
Putting together a blog carnival can be a lot of work, but it’s a great opportunity to share the work of bloggers you admire and to discover new math-friends online. I love that part of being a host!
Classroom teachers, homeschoolers, college professors, unschoolers, or anyone who likes to play around with math — if you would like to take a turn hosting the carnival, please speak up.
## Playful Math Education Carnival 147
Welcome to the 147th edition of the Playful Math Education Blog Carnival — a smorgasbord of delectable tidbits of mathy fun. It’s like a free online magazine devoted to learning, teaching, and playing around with math from preschool to high school.
Bookmark this post, so you can take your time browsing. There’s so much playful math to enjoy!
By tradition, we start the carnival with a puzzle in honor of our 147th edition. But if you’d rather jump straight to our featured blog posts, click here to see the Table of Contents.
## Playful Math Carnival 146 via Find the Factors
Check out the latest carnival of playful math:
Each monthly Playful Math Education Blog Carnival brings you a great new collection of puzzles, math conversations, teaching tips, and all sorts of mathy fun. It’s like a free online magazine of mathematical adventures, helpful and inspiring no matter when you read them.
Iva put together this huge and amazing collection of mathematical games, activities, art projects, hands-on fun, math storybooks, poetry, and more.
### Help Us Keep the Carnival Going
The Playful Math Blog Carnival wants you!
The carnival is a joint effort. We depend on our volunteer hosts to collect blog posts and write the carnival each month.
Putting together a blog carnival can be a lot of work, but it’s a great opportunity to share the work of bloggers you admire and to discover new math-friends online. I love that part of being a host!
Classroom teachers, homeschoolers, college professors, unschoolers, or anyone who likes to play around with math — if you would like to take a turn hosting the carnival, please speak up.
## Playful Math Carnival 145 via Math Hombre
Check out the latest carnival of playful math:
Each monthly Playful Math Education Blog Carnival brings you a great new collection of puzzles, math conversations, teaching tips, and all sorts of mathy fun. It’s like a free online magazine of mathematical adventures, helpful and inspiring no matter when you read them.
John put together this wonderful collection of mathematical games, art projects, books, math essays, puzzles, and more.
Continue reading Playful Math Carnival 145 via Math Hombre
## Playful Math Carnival 144: Anniversary Edition
Welcome to the 144th edition of the Playful Math Education Blog Carnival — a smorgasbord of delectable tidbits of mathy fun. It’s like a free online magazine devoted to learning, teaching, and playing around with math from preschool to high school.
Bookmark this post, so you can take your time browsing.
There’s so much playful math to enjoy!
By tradition, we would start the carnival with a puzzle/activity in honor of our 144th edition. But this time, I want to take a peek back at the history of our carnival.
## Playful Math Carnival 143 via Learning Well at Home
Check out the latest carnival of playful math:
Each monthly Playful Math Education Blog Carnival brings you a great new collection of puzzles, math conversations, teaching tips, and all sorts of mathy fun. It’s like a free online magazine of mathematical adventures, helpful and inspiring no matter when you read them.
Sonya put together this wonderful collection of mathematical games, art projects, holiday activities, paper crafts (LOTS of snowflakes!), and more.
She writes:
“It doesn’t matter where you look, there are always things to count and lots of things to wonder about. How many snowflakes in a snowman? How many points are on a star? How many turkeys do we eat every year? Does anyone actually eat all the fruitcake that is sold each year?
“Here is Winter and her holidays by the numbers…
“This was probably my favorite Playful Math Education Blog Carnival to write. There is so much out there to explore that I’m sure I’ve barely scratched the surface.
“Come back because I’m going to keep adding to this post…”
—Sonya Post, Carnival #143
## Playful Math Education 142
Welcome to the 142nd edition of the Playful Math Education Blog Carnival — a smorgasbord of delectable tidbits of mathy fun. It’s like a free online magazine devoted to learning, teaching, and playing around with math from preschool to high school.
Bookmark this post, so you can take your time browsing.
Seriously, plan on coming back to this post several times. There’s so much playful math to enjoy!
By tradition, we start the carnival with a puzzle/activity in honor of our 142nd edition. But if you’d rather jump straight to our featured blog posts, click here to see the Table of Contents.
## Activity: Planar Graphs
According to the OEIS Wiki, 142 is “the number of planar graphs with six vertices.”
What does that mean?
And how can our students play with it?
A planar graph is a set of vertices connected (or not) by edges. Each edge links two vertices, and the edges cannot intersect each other. The graph doesn’t have to be fully connected, and individual vertices may float free.
Children can model planar graphs with three-dimensional constructions using small balls of playdough (vertices) connected by toothpicks (edges).
Let’s start with something smaller than 142. If you roll four balls of playdough, how many different ways can you connect them? The picture shows five possibilities. How many more can you find?
Sort your planar graphs into categories. How are they similar? How are they different?
A wise mathematician once said, “Learning is having new questions to ask.” How many different questions can you think of to ask about planar graphs?
Play the Planarity game to untangle connected planar graphs (or check your phone store for a similar app).
Or play Sprouts, a pencil-and-paper planar-graph game.
For deeper study, elementary and middle-school students will enjoy Joel David Hamkins’s Graph coloring & chromatic numbers and Graph theory for kids. Older students can dive into Oscar Levin’s Discrete Mathematics: An Open Introduction. Here’s the section on planar graphs.
## Working on the Playful Math Carnival
Every time I put together a Playful Math Education Blog Carnival, it becomes my favorite blog post of all time.
At least until the next edition.
I’m always delighted by the posts I discover. There’s so much richness in the math blogging community. This month’s carnival is no different.
I think you’re going to love it!
### Hint of Things to Come
The Paul Klee painting above is from the carnival. Isn’t it beautiful?
In the carnival post, I use the image to complement a math activity about graphs.
But I think it’s also a wonderful reminder of how connections (between individual bloggers or between math topics in a carnival) make a richer whole than any of us could create on our own.
### Would You Like to Help?
We need volunteers! Classroom teachers, homeschoolers, unschoolers, or anyone who likes to play around with math (even if the only person you “teach” is yourself) — if you would like to take a turn hosting the Playful Math Carnival, please speak up!
CREDITS: “Geöffneter Berg” by Paul Klee, 1914 via John Golden’s (Mathhombre) Miscellanea. | 2,339 | 10,914 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.40625 | 3 | CC-MAIN-2021-49 | longest | en | 0.859026 |
https://deepai.org/publication/monotone-drawings-of-k-inner-planar-graphs | 1,675,173,878,000,000,000 | text/html | crawl-data/CC-MAIN-2023-06/segments/1674764499871.68/warc/CC-MAIN-20230131122916-20230131152916-00609.warc.gz | 213,058,128 | 69,259 | DeepAI
# Monotone Drawings of k-Inner Planar Graphs
A k-inner planar graph is a planar graph that has a plane drawing with at most k internal vertices, i.e., vertices that do not lie on the boundary of the outer face of its drawing. An outerplanar graph is a 0-inner planar graph. In this paper, we show how to construct a monotone drawing of a k-inner planar graph on a 2(k+1)n × 2(k+1)n grid. In the special case of an outerplanar graph, we can produce a planar monotone drawing on a n × n grid, improving previously known results.
• 2 publications
• 12 publications
03/20/2019
### Drawing planar graphs with few segments on a polynomial grid
The visual complexity of a plane graph drawing is defined to be the numb...
02/19/2018
### The Complexity of Drawing a Graph in a Polygonal Region
We prove that the following problem is complete for the existential theo...
08/14/2019
### On Strict (Outer-)Confluent Graphs
A strict confluent (SC) graph drawing is a drawing of a graph with verti...
02/19/2018
### Convexity-Increasing Morphs of Planar Graphs
We study the problem of convexifying drawings of planar graphs. Given an...
08/24/2017
### Ordered Level Planarity, Geodesic Planarity and Bi-Monotonicity
We introduce and study the problem Ordered Level Planarity which asks fo...
05/27/2021
### On Morphing 1-Planar Drawings
Computing a morph between two drawings of a graph is a classical problem...
08/15/2019
### Homotopy height, grid-major height and graph-drawing height
It is well-known that both the pathwidth and the outer-planarity of a gr...
## 1 Introduction
A straight-line drawing of a graph G is a mapping of each vertex to a distinct point on the plane and of each edge to a straight-line segment between the vertices. A path is monotone if there exists a line such that the projections of the vertices of on appear on in the same order as on . A straight-line drawing of a graph G is monotone, if a monotone path connects every pair of vertices. We say that an angle is convex if . A tree is called ordered if the order of edges incident to any vertex is fixed. We call a drawing of an ordered tree rooted at near-convex monotone if it is monotone and any pair of consecutive edges incident to a vertex, with the exception of a single pair of consecutive edges incident to , form a convex angle.
Monotone graph drawing has been lately a very active research area and several interesting results have appeared since its introduction by Angelini et al. [1]. In the case of trees, Angelini et al. [1] provided two algorithms that used ideas from number theory (Stern-Brocot trees [3, 14] [6, Sect. 4.5]) to produce monotone tree drawings. Their BFS-based algorithm used an grid while their DFS-based algorithm used an grid. Later, Kindermann et al. [12] provided an algorithm based on Farey sequence (see [6, Sect. 4.5]) that used an grid. He and He in a series of papers [7, 8, 9] gave algorithms, also based on Farey sequences, that eventually reduced the required grid size to . Their monotone tree drawing uses a grid which is asymptotically optimal as there exist trees which require at least area [8]. In a recent paper, Oikonomou and Symvonis [13] followed a different approach from number theory and gave an algorithm based on a simple weighting method and some simple facts from geometry, that draws a tree on an grid.
Hossain and Rahman [11] showed that by modifying the embedding of a connected planar graph we can produce a planar monotone drawing of that graph on an grid. To achieve that, they reinvented the notion of an orderly spanning tree introduced by Chiang et al. [4] and referred to it as good spanning tree. Finally, He and He [10] proved that Felsner’s algorithm [5] builds convex monotone drawings of 3-connected planar graphs on grids in linear time.
A -inner planar graph is a planar graph that has a plane drawing with at most inner vertices, i.e., vertices that do not lie on the boundary of the outer face of its drawing. An outerplanar graph is a -inner planar graph. In this paper, we show how to construct a monotone drawing of a -inner planar graph on a grid. This yields monotone drawings for outerplanar graphs on a grid, improving the results in [2, 11]. Proofs of lemmata/theorems can be found in the Appendix.
## 2 Preliminaries
Let be a tree rooted at a vertex . Denote by the sub-tree of rooted at vertex . By we denote the number of vertices of . In the rest of the paper, we assume that all tree edges are directed away from the root.
Our algorithm for obtaining a monotone plane drawing of a -inner planar graph in based on our ability: i) to produce for any rooted tree a compact monotone drawing that satisfies specific properties and ii) to identify for any planar graph a good spanning tree.
###### Theorem 2.1 (Oikonomou, Symvonis [13])
Given an -vertex tree rooted at vertex , we can produce in time a monotone drawing of where: i) the root is drawn at , ii) the drawing in non-strictly slope-disjoint111See [13] for the definition of non-strictly slope-disjoint drawings of trees., iii) the drawing is contained in the first quadrant, and iv) it fits in an grid.
By utilizing and slightly modifying the algorithm that supports Thm. 2.1, we can obtain a specific monotone tree drawing that we later use in our algorithm for monotone -inner planar graphs.
###### Theorem 2.2
Given an -vertex tree rooted at vertex , we can produce in time a monotone drawing of where: i) the root is drawn at , ii) the drawing in non-strictly slope-disjoint, iii) the drawing is near-convex, iv) the drawing is contained in the second octant (defined by two half-lines with a common origin and slope and , resp.), and v) it fits in a grid.
The following definition of a good spanning tree is due to Hossain and Rahman [11]. Let be a connected embedded plane graph and let be a vertex of lying on the boundary of its outer face. Let be an ordered spanning tree of rooted in that respects the embedding of . Let be the unique path in from the root to a vertex . The path divides the children of , except , into two groups; the left group and the right group . A child of is in group and, more specifically in a subgroup of denoted by , if the edge appears before edge in clockwise order of the edges incident to when the ordering starts from edge . Similarly, a child of is in group and more specifically in a subgroup of denoted by if the edge appears after edge in clockwise order of the edges incident to when the ordering starts from edge . A tree is called a good spanning tree of if every vertex of satisfies the following two conditions with respect to (see Figure 2 for an example of a good spanning tree (solid edges)):
1. does not have a non-tree edge ; and
2. The edges of incident to vertex , excluding , can be partitioned into three disjoint (and possibly empty) sets and satisfying the following conditions:
1. Each of and is a set of consecutive non-tree edges and is a set of consecutive tree edges.
2. Edges of and appear clockwise in this order after edge .
3. For each edge , for some , , and for each edge , for some , .
###### Theorem 2.3 ([4, 11])
Let be a connected planar graph of vertices. Then has a planar embedding that contains a good spanning tree. Furthermore, and a good spanning tree of can be found in time.
Consider an embedded plane graph for which a good spanning tree exists. We say that a non-tree edge of covers vertex if lies in the inner face delimited by the simple cycle formed by tree-edges of and . Vertices on the cycle are not covered by edge .
###### Lemma 1
If a planar graph has a -inner embedding, then has a -inner embedding which contains a good spanning tree , where . Moreover, each non-tree edge covers at most of ’s leaves.
## 3 k-inner Monotone Drawings
The general idea for producing a monotone drawing of plane -inner graph which has a good spanning tree is to first obtain a monotone drawing of satisfying the properties of Thm 2.2 and then to insert the remaining non-tree edges in a way that the drawing remains planar. The insertion of a non-tree edge may require to slightly adjust the drawing obtained up to that point, resulting in a slightly larger drawing. As it turns out, the insertion of a subset of the non-tree edges may violate the planarity of the drawing and, moreover, if these edges are considered in a proper order, the increase on the size of the drawing can be kept small (up to a factor of for each dimension).
Consider a plane graph that has a good spanning tree . For any non-tree edge of , we denote by the set of leaf-vertices of covered by . A non-tree edge is called a leader edge if and there doesn’t exist another edge such that with lying in the inner of the cycle induced by the edges of and . In Fig. 2, leader edges are drawn as dashed edges. We also have that , , , and .
###### Lemma 2
Let be a -inner plane graph that has a good spanning tree. Then, there exist at most leader edges in .
###### Proof (Sketch)
Firstly observe that a boundary vertex of that is a leaf-vertex in cannot be covered by any edge of . That is, the set , for any non-tree edge , contains only inner-vertices of , and thus, . The proof then easily follows by observing that for any two distinct leader edges , exactly one of the following statements holds: i) , ii) , iii) . ∎
###### Lemma 3
Let be a plane graph that has a good spanning tree and let be a monotone drawing of that is near-convex and non-strictly slope-disjoint. Let be the drawing produced if we add all leader edges of to and be the drawing produced if we add all non-tree edges of to . Then, is planar if and only if is planar.
Lemma 3 indicates that we only have to adjust the original drawing of so that after the addition of all leader edges it is still planar. Then, the remaining non-tree edges can be drawn without violating planarity. Note that, for the proof of Lemma 3, it is crucial that the original drawing of is near-convex.
Consider the near-convex drawing of a good spanning tree rooted at that is non-strictly slope-disjoint. Assume that is drawn in the first quadrant with its root at . Let be a vertex of and be its parent. Vertices and are drawn at grid points and , resp. We define the reference vector of with respect to its parent in , denoted by
, to be vector
. We emphasize that the reference vector of a tree vertex is defined wrt the original drawing of and does not change even if the drawing of is modified by our drawing algorithm.
The elongation of edge by a factor of , (also referred to as a -elongation) translates the drawing of subtree along the direction of edge so that the new length of increases by times the length of the reference vector of . Since the elongation factor is a natural number, the new drawing is still a grid drawing. Let . After a -elongation of , vertex is drawn at point . A 0-elongation leaves the drawing unchanged. Note that, by appropriately selecting the elongation factor for an upward tree edge , we can reposition so that it is placed above any given point .
If we insert the leader edges in the drawing of the good spanning tree in an arbitrary order, we may have to adjust the drawing more than one time for each inserted edge. This is due to dependencies between leader edges. Fig. 2 describes the two types of possible dependencies. In the case of Fig. 2(a), the leader edge must by inserted first since . Inserting leader so that it is not intersected by any tree edge can be achieved by elongating edges and by appropriate factors so that vertices and are both placed at grid points above (i.e., with larger coordinate) vertex .
In the case of Fig. 2(b), we have that , , and . However, leader must be inserted first since one of its endpoints () is an ancestor of an endpoint () of . Again, inserting leader so that it is not intersected by any tree edge can be achieved by elongating edges and by appropriate factors so that vertices and are both placed at grid points above vertex .
###### Lemma 4
Let be a plane graph that has a good spanning tree and let be a drawing of that satisfies the properties of Thm 2.2. Then, there exists an ordering of the leader edges, such that if they are inserted into (with the appropriate elongations) in that order they need to be examined exactly once.
Our method for producing monotone drawings of -inner planar graphs is summarized in Algorithm 1. A proof that the produced drawing is actually a monotone plane drawing follows from the facts that i) there is always a good spanning tree with at most inner vertices (Lemma 1), ii) there is a monotone tree drawing satisfying the properties of Thm 2.2, iii) the operation of edge elongation on the vertices of a near-convex monotone non-strictly slope-disjoint tree drawing maintains these properties, iv) the ability to always insert the leader edges into the drawing without violating planarity (through elongation), and v) the ability to insert the remaining non-tree edges (Lemma 3).
Let and be the reference vectors of and , resp. When we process leader edge (lines 9-15 of Algorithm 1), factors and are used for the elongation of edges and , resp. The use of these elongation factors ensures that both and are placed above vertex , and thus the insertion of edge leaves the drawing planar. When the leader edges are processed in the order dictated by their dependencies (line 6 of Algorithm 1), we can show that:
###### Lemma 5
Let be the drawing of the good spanning tree satisfying the properties of Thm 2.2 and let be the drawing of immediately after the insertion of leader edge by Algorithm 1. Let and be drawn in at points and , resp. Then, in the drawings of and are contained in and grids, resp.
Based on Lemma 5, we can easily show the main result of our paper.
###### Theorem 3.1
Let be an -vertex -inner planar graph. Algorithm 1 produces a planar monotone drawing of on a grid.
A corollary of Thm 3.1 is that for an -vertex outerplanar graph Algorithm 1 produces a planar monotone drawing of on a grid. However, we can further reduce the grid size down to .
###### Theorem 3.2
Let be an -vertex outerplanar graph. Then, there exists an planar monotone grid drawing of .
###### Proof
Simply observe that since an outerplanar graph has no leader edges, the drawing produced by Algorithm 1 is identical to that of the original drawing of the good spanning tree . In Algorithm 1 we used a drawing of in the second octant in order to simplify the elongation operation. Since outerplanar graphs have no leader edges, they require no elongations and we can use instead the (first quadrant) monotone tree drawing of [13], appropriately modified so that it yields a near-convex drawing. ∎
## 4 Conclusion
We defined the class of -inner planar graphs which bridges the gap between outerplanar and planar graphs. For an -vertex -inner planar graph , we provided an algorithm that produces a monotone grid drawing of . Building algorithms for -inner graphs that incorporate into their time complexity or into the quality of their solution is an interesting open problem.
## References
• [1] Angelini, P., Colasante, E., Di Battista, G., Frati, F., Patrignani, M.: Monotone drawings of graphs. Journal of Graph Algorithms and Applications 16(1), 5–35 (2012). https://doi.org/10.7155/jgaa.00249
• [2] Angelini, P., Didimo, W., Kobourov, S., Mchedlidze, T., Roselli, V., Symvonis, A., Wismath, S.: Monotone drawings of graphs with fixed embedding. Algorithmica 71(2), 233–257 (2015). https://doi.org/10.1007/s00453-013-9790-3
• [3] Brocot, A.: Calcul des rouages par approximation, nouvelle methode. Revue Chronometrique 6, 186–194 (1860)
• [4] Chiang, Y., Lin, C., Lu, H.: Orderly spanning trees with applications. SIAM Journal on Computing 34(4), 924–945 (2005). https://doi.org/10.1137/S0097539702411381
• [5] Felsner, S.: Convex drawings of planar graphs and the order dimension of 3-polytopes. Order 18(1), 19–37 (Mar 2001). https://doi.org/10.1023/A:1010604726900
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• [7] He, D., He, X.: Nearly optimal monotone drawing of trees. Theoretical Computer Science 654, 26–32 (2016). https://doi.org/10.1016/j.tcs.2016.01.009
• [8] He, D., He, X.: Optimal monotone drawings of trees. SIAM Journal on Discrete Mathematics 31(3), 1867–1877 (2017). https://doi.org/10.1137/16M1080045, https://doi.org/10.1137/16M1080045
• [9] He, X., He, D.: Compact monotone drawing of trees. In: Xu, D., Du, D., Du, D. (eds.) Computing and Combinatorics: 21st International Conference, COCOON 2015, Beijing, China, August 4-6, 2015, Proceedings. pp. 457–468. Springer International Publishing, Cham (2015). https://doi.org/10.1007/978-3-319-21398-9_36
• [10] He, X., He, D.: Monotone drawings of 3-connected plane graphs. In: Bansal, N., Finocchi, I. (eds.) Algorithms - ESA 2015. pp. 729–741. Springer Berlin Heidelberg, Berlin, Heidelberg (2015)
• [11] Hossain, M.I., Rahman, M.S.: Good spanning trees in graph drawing. Theoretical Computer Science 607, 149–165 (2015). https://doi.org/10.1016/j.tcs.2015.09.004
• [12] Kindermann, P., Schulz, A., Spoerhase, J., Wolff, A.: On monotone drawings of trees. In: Duncan, C., Symvonis, A. (eds.) Graph Drawing: 22nd International Symposium, GD 2014, Würzburg, Germany, September 24-26, 2014, Revised Selected Papers. pp. 488–500. Springer Berlin Heidelberg, Berlin, Heidelberg (2014). https://doi.org/10.1007/978-3-662-45803-7_41
• [13] Oikonomou, A., Symvonis, A.: Simple compact monotone tree drawings. In: Frati, F., Ma, K. (eds.) Graph Drawing and Network Visualization - 25th International Symposium, GD 2017, Boston, MA, USA, September 25-27, 2017, Revised Selected Papers. Lecture Notes in Computer Science, vol. 10692, pp. 326–333. Springer (2017). https://doi.org/10.1007/978-3-319-73915-1_26
• [14] Stern, M.: Ueber eine zahlentheoretische funktion. Journal fur die reine und angewandte Mathematik 55, 193–220 (1858)
## Appendix 0.A Additional material for Section 2
### 0.a.1 Some more definitions
Let be a drawing of a graph and be an edge from vertex to vertex in . The slope of edge , denoted by , is the angle spanned by a counter-clockwise rotation that brings a horizontal half-line starting at and directed towards increasing -coordinates to coincide with the half-line starting at and passing through .
Angelini et al. [1] defined the notion of slope-disjoint tree drawings and proved that a every such drawing is monotone. In order to simplify the presentation of our compact tree drawing algorithm presented in [13], we slightly extended the definition of slope-disjoint tree drawings given by Angelini et al. [1]. More specifically, we called a tree drawing of a rooted tree a non-strictly slope-disjoint drawing if the following conditions hold:
1. For every vertex , there exist two angles and , with such that for every edge that is either in or enters from its parent, it holds that .
2. For every two vertices such that is a child of , it holds that .
3. For every two vertices with the same parent, it holds that either or .
The idea behind the original definition of slope-disjoint tree drawings is that all edges in the subtree as well as the edge entering from its parent will have slopes that strictly fall within the angle range defined for vertex . is called the angle range of with and being its boundaries. In our extended definition, we allowed for angle ranges of adjacent vertices (parent-child relationship) or sibling vertices (children of the same parent) to share angle range boundaries. Note that replacing the “” symbols in our definition by the “” symbol gives us the original definition of Angelini et al. [1] for the slope-disjoint tree drawings. Non-strictly slope-disjoint tree drawings are also monotone.
### 0.a.2 Proof of Theorem 2.2
Theorem 2. Given an -vertex tree rooted at vertex , we can produce in time a monotone drawing of where: i) the root is drawn at , ii) the drawing in non-strictly slope-disjoint, iii) the drawing is near-convex, iv) the drawing is contained in the second octant (defined by two half-lines with a common origin and slope and , resp.), and v) it fits in a grid.
###### Proof
We show how to extend/modify the algorithm presented in [13] so that it satisfies all properties specified in the theorem. For brevity, we refer to the algorithm of [13] as Algorithm MQuadTD (abbreviation for “Monotone Quadrant Tree Drawing”).
Algorithm MQuadTD constructs a non-strictly slope-disjoint tree drawing with the root of the tree drawn at . So, the first two conditions are satisfied. In addition, the tree is drawn at the first quadrant of its root. We first show how Algorithm MQuadTD can be extended so that it produces near-convex drawings. Since Algorithm MQuadTD produces a non-strictly slope-disjoint tree drawing, it splits the angle-range of every internal tree vertex into as many slices as ’s children and assigns these slices to the children of .
We denote by the parent of internal tree vertex . Note that, in a non-strictly slope-disjoint tree drawing that is drawn is an circular sector of size at most , i.e., the angle-range of the root is , a non-convex angle can be only formed between the edge entering an internal vertex and the edge connecting with its first or last child, say and , resp. Moreover, in this case a convex angle may be formed only if the slope of edge falls within the angle-range of either or . W.l.o.g., say that the slope of falls within the angle range of ; the other case is symmetric. It turns out that we can easily avoid creating a non-convex angle by simply drawing point on the first available grid point on the extension of . If we place in this way, then the length of edge will be equal to that of , which guarantees that this modification will not increase the required grid size.
Having modified Algorithm MQuadTD as described above, yields a drawing of that satisfies conditions i), ii) and iii). In order get a drawing that satisfies all five conditions we work as follows. We construct a new tree by getting two copies of and making them subtrees of a new root . Since has vertices, we can get a drawing of that lies in the first quadrant, satisfies conditions i), ii) and iii) and, moreover, it fits on a grid with its root being the only vertex being place on the or the axis. By the construction of and its symmetry, we observe that its sub-drawing that lies in the second octant corresponds to a drawing of the initial tree . In this way, we obtain a drawing that satisfies all 5 conditions. To complete the proof, simply note that all described modification of Algorithm MQuadTD can be easily accommodated in time. ∎
The following Lemma in not included in the main part of the paper.
###### Lemma 6
Let be a drawing for tree which satisfies all properties of Thm-2.2 (and can be produced by the modification of algorithm MQuadTD as described in the proof of Thm-2.2). Then, if we substitute in the edges entering the leaves of by semi-infinite lines (i.e., rays), then partitions the first quadrant into unbounded convex regions.
###### Proof
See Figure 4 for an example drawing. Note that since is monotone, it is also planar. Let be the drawing we obtain by substituting each leaf edge in by a ray. In addition, we add to two extra rays, the first coinciding with the positive -axis and the second with the positive -axis. The lemma states that all unbounded inner faces in are unbounded convex regions.
Since the original drawing of is near-convex and by the fact that the two extra rays lie on the and axes, all angles (in the first quadrant) between consecutive edges incident to any vertex are convex. Moreover, since the drawing of is also (non-strictly) slope-disjoint, each pair of leaf edges is drawn completely inside non-overlapping (but possibly touching) angular sectors. Thus, the rays which substitute these edges diverge. Then, the proof follows by the planarity of the original tree drawing of . ∎
### 0.a.3 Proof of Lemma 1
Lemma 1. If a planar graph has a -inner embedding, then has a -inner embedding which contains a good spanning tree, where . Moreover, each non-tree edge covers at most of ’s leaves.
###### Proof
The algorithm that supports Theorem 2.3 (see [11]), when given a -inner embedding, it produces a planar embedding which contains a good spanning tree by changing the original embedding. It does so by only moving -components and -split components out of a cycle induced by the constructed good spanning tree222For the definitions of a -components and -split components ans well as details of the good spanning tree construction algorithm see [11].. Therefore, the number of vertices in the outer face of is equal or greater than the number of vertices in the original embedding of , which implies that is a -inner planar graph, where . Moreover there are at most leaf vertices that are covered by any non-tree edge since there are at most inner vertices. ∎
## Appendix 0.B Additional material for Section 3
Let be a plane graph and being a good spanning tree of . For any vertex of , the leftmost path (resp. rightmost path) of , denoted by (resp. ), starts at , traverses edges of by following the last (resp. first) tree edge of each vertex in the counter-clockwise order, and terminates at a leaf vertex. Recall that denotes the set of leaf vertices of covered by then non-tree edge . We will call the non-tree vertices of that are not leader edges, ordinary edges. Finally, by we denote the convex hull of a point-set .
The following technical lemma is not included in the main part of the paper.
###### Lemma 7
Let be a plane graph, be a good spanning tree of , a monotone drawing of which in near-convex and non-strictly slope-disjoint, and a leader edge of . W.l.o.g., let the inner face of the cycle induced by and edges of lie to the right of edge when we traverse it from to . Finally, let be the set of vertices contained in paths and . Then, all vertices of are located on the the convex hull . Moreover, any ordinary edge that covers exactly the same leaf vertices as , (i.e., ) has one of its endpoints on and the other on .
###### Proof
Refer to Figure 4. First, we prove that all vertices of are located on the convex hull . Let and be the leaves at the end of and , resp., and also let (resp. ) be the first (resp. last) edge of and (resp. ) be the first (resp. last) edge of .
We observe that if we substitute and with rays toward the decreasing -coordinates, then the two rays intersect. Moreover, if we substitute and with rays toward the increasing -coordinates, then the two rays diverge. Both of these two facts are due to that drawing is non-strictly slope-disjoint. Combined with the fact that is near convex, proves that contains all vertices of .
Now we prove that every ordinary edge that covers exactly the same leaf vertices as (i.e., set ), has one of its endpoints in and the other in .
For a contradiction, assume that there is an ordinary edge that covers exactly set of leaf vertices and does not have its endpoints in . Edge cannot lie inside the cycle induced by and because it would violate the fact that is a leader edge. Since do not have its endpoints in , then covers at least one of leaf vertices or , a contradiction. Thus, has its endpoints on . By taking also into account that is a good spanning tree, and thus, no non-tree edge connects two vertices on the same path from the root to a leaf, we conclude that one of the endpoints of lies on and the other on . ∎
### 0.b.1 Proof of Lemma 3
Lemma 3. Let be a plane graph that has a good spanning tree and let be a monotone drawing of that is near-convex and non-strictly slope-disjoint. Let be the drawing produced if we add all leader edges of to and be the drawing produced if we add all non-tree edges of to . Then, is planar if and only if is planar.
###### Proof
First recall that when we add edges to an existing drawing we only add them as straight line segments.
It trivially holds. is a subgraph of by definition. Any subgraph of a planar graph is planar.
We simply have to show that we can insert all ordinary edges into without violating planarity. Consider and arbitrary ordinary edge . Then, there exists a leader edge such that . If not, would be a leader edge. W.l.o.g., let the inner face of the cycle induced by and edges of lie to the right of edge when we traverse it from to . Consider paths and and let and be their leaf endpoints, resp. Let be the set of vertices of paths and .
Since drawing is planar and, by Lemma7, we know that contains all points of , the internal face induced by the cycle formed by edge , , and the line segment is empty and convex. Thus, since one endpoint of lies on and the other on (Lemma7), can be inserted to the drawing without violating planarity. In the same way, we can insert all ordinary edges. The proof is completed by observing that since is a plane graph, no two ordinary edges intersect.
### 0.b.2 Proof of Lemma 4
Lemma 4. Let be a plane graph that has a good spanning tree and let be a drawing of that satisfies the properties of Thm 2.2. Then, there exists an ordering of the leader edges, such that if they are inserted into (with the appropriate elongations) in that order they need to be examined exactly once.
###### Proof (Sketch)
We create a directed dependency graph that has the leader edges as its vertices. The edges of correspond to the dependencies between leader edges of . Let and be two leader edges of . If we insert a directed edge from to in . If we insert a directed edge from to in . These dependency edges correspond to the dependency depicted in Figure 2(a). In addition, there may be a dependency between and even though . This is depicted in Figure 2(b) and occurs when an endpoint of is an ancestor of or vice versa. In this case we insert an edge in from the “ancestor” to the “descendant” edge. The fact that is planar and is a good spanning tree imply that is actually a directed acyclic graph (DAG). Then, the ordering of the leader edges is obtained by topologically sorting DAG .
We note that the dependency graph is not necessarily connected. Since we study -inner planar graphs, there are at most leaders and, thus, has at most vertices, and consequently . So, we need time for the topological ordering of which, together with the time required for the recognition of the leader edges, yield an time algorithm. However, we can improve it to time by avoiding to insert transitive edges into . ∎
### 0.b.3 Proof of Lemma 5
Lemma 5. Let be the drawing of the good spanning tree satisfying the properties of Thm 2.2 and let be the drawing of immediately after the insertion of leader edge by Algorithm 1. Let and be drawn in at points and , resp. Then, in the drawings of and are contained in and grids, resp.
###### Proof
We only prove the lemma for since the case for is symmetric.
Let be the parent of in the good spanning tree used in Algorithm 1. The insertion of leader edge in the drawing causes the elongation of edge and, as a result, translates the drawing of along the direction of .
Observe that no edge in has been elongated in a previous step. In order for an edge in to be elongated, a leader edge , such that or is in , must be selected for insertion. So, must be an ancestor of or . Then, leader depends on leader and has to follow in the sorted leaders list (line 6 of Algorithm 1). Thus, no edge in has been elongated prior to the elongation of .
The lemma follows from the facts that i) the initial drawing of the good spanning tree fits on a grid (Thm-2.2) and ii) in , vertex is placed at point . Therefore, is contained in a grid. ∎
### 0.b.4 Proof of Theorem 3.1
Theorem 4. Let be an -vertex -inner planar graph. Algorithm 1 produces a planar monotone drawing of on a grid.
###### Proof
The proof that the drawing produced by Algorithm 1 is actually a monotone plane drawing follows from the facts that i) there is always a good spanning tree with at most inner vertices (Lemma 1), ii) there is a monotone tree drawing satisfying the properties of Thm 2.2, iii) the operation of edge elongation on the vertices of a near-convex monotone non-strictly slope-disjoint tree drawing maintains these properties, iv) the ability to always insert the leader edges into the drawing without violating planarity (through elongation), and v) the ability to insert the remaining ordinary edges (Lemma 3).
Now we turn our attention to the grid size of the drawing produced by Algorithm 1. W.l.o.g., we assume that we deal with plane graphs that have only tree and leader edges. We can safely do so since, by Lemma 3, all ordinary edges can be inserted in the drawing without violating planarity.
We use induction on the number of leader edges. For the base case where , the grid bound holds by Theorem 2.2. For the induction step, we assume that the Theorem holds for plane graphs with less than leader edges, and we show that it also holds for plane graphs with leader edges.
Firstly note that, since all edges in the drawing have slopes greater than , the grid side length is dictated by the maximum -coordinate of a leaf vertex.
Consider a plane graph that has a good spanning tree and of its non-tree edges are leader edges. Let be the last leader edge considered by Algorithm 1. If we remove from we get a plane graph with leader edges.
By the induction hypothesis, the produced drawing of fits in a grid. Thus, the path from the root of to is contained in a grid.
Consider now the case where Algorithm 1 inserts leader edge into the drawing it has already produced for graph . During the iteration of the while-loop (lines 9-16 of Algorithm 1) for edge , at most two edges of are elongated. This means that the grid size of the drawing for , which is dictated as the maximum coordinate of a leaf vertex, is the maximum between the coordinate of a leaf vertex in either or , or it does not change. By Lemma 5 and the fact that during the elongation of edge , vertex may be translated at most grid points333Recall that denotes the reference vector of . above the highest leaf of the drawing of of , it follows that, after the insertion edge , the path from the root of to fits in a grid. So, the path from the root to any leaf of fits in a grid. However, we have that:
2kn+udy+2n−uy = 2(k+1)n−(uy−udy) ≤ 2(k+1)n
since . Thus, the path from the root to any leaf of fits in a grid. Similarly, the same holds for the path from the root of to any leaf of . This completes the proof. ∎ | 8,296 | 34,666 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.78125 | 3 | CC-MAIN-2023-06 | latest | en | 0.898632 |
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# Seek help on Create new var with Do Loop
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``````I want to create new variables (newna1-newna5) from na1 -na5, if there is missing value of na then newna will be equal to the previous nonmissing value of na.But I have problems running the program below, error with array length.Please help me to figure it out.Thank you very muchdata try;
input id na1 na2 na3 na4 na5;
datalines;
1 2 3 . . 4
2 3 5 4 1 .
3 6 . 7 5 4
4 5 3 . 4 .
5 4 . . . 6
;
run;
data try1;
set try;
array na(*) na1-na5;
array newna(*) newna1-newna5;
do i=1 to 5;
newna(i)=na(i);
do j=1 to 4;
if na(i)=. then newna(i)=na(i-j);
end;
end;
run;
output;``````
Accepted Solutions
Solution
04-19-2016 06:40 PM
Super User
Posts: 6,641
## Re: Seek help on Create new var with Do Loop
You could simplify the process and get the correct result most of the time:
``````data try1;
set try;
array na(*) na1-na5;
array newna(*) newna1-newna5;
do i=1 to 5;
if na(i) > . then newna(i)=na(i);
else newna(i)=newna(i-1);
end;
run;``````
When I say "most of the time" I'm looking at the possibility that na1 might be missing. What should happen in that case? The program will generate an error as it stands now. So if it's possible that na1 might be missing, you would need to specify what the result should be so the program can be modified.
All Replies
Solution
04-19-2016 06:40 PM
Super User
Posts: 6,641
## Re: Seek help on Create new var with Do Loop
You could simplify the process and get the correct result most of the time:
``````data try1;
set try;
array na(*) na1-na5;
array newna(*) newna1-newna5;
do i=1 to 5;
if na(i) > . then newna(i)=na(i);
else newna(i)=newna(i-1);
end;
run;``````
When I say "most of the time" I'm looking at the possibility that na1 might be missing. What should happen in that case? The program will generate an error as it stands now. So if it's possible that na1 might be missing, you would need to specify what the result should be so the program can be modified.
Contributor
Posts: 34
## Re: Seek help on Create new var with Do Loop
Dear Astouding,
Thanks a lot for your help.
It's so simple!
Best,
Trang
🔒 This topic is solved and locked. | 693 | 2,200 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.6875 | 3 | CC-MAIN-2018-26 | latest | en | 0.77659 |
https://gist.github.com/lfreeman/4196181 | 1,524,433,104,000,000,000 | text/html | crawl-data/CC-MAIN-2018-17/segments/1524125945660.53/warc/CC-MAIN-20180422212935-20180422232935-00097.warc.gz | 609,566,000 | 12,481 | Instantly share code, notes, and snippets.
# lfreeman/merge_sort.py Created Dec 3, 2012
What would you like to do?
Merge Sort
import numpy.random as nprnd import time def timeit(method): def timed(*args, **kwargs): ts = time.time() result = method(*args, **kwargs) te = time.time() print '%s (%s) %2.2f sec' % \ (method.__name__, len(args[0]), te - ts) return result return timed @timeit def merge_sort(list,merge): return _merge_sort(list, merge) def merge1(left, right): i = j = inv = 0 merged = [] while i < len(left) and j < len(right): if left[i] <= right[j]: merged.append(left[i]) i += 1 else: merged.append(right[j]) j += 1 inv += len(left[i:]) merged += left[i:] merged += right[j:] return merged, inv def merge2(array1, array2): inv = 0 merged_array = [] while array1 or array2: if not array1: merged_array.append(array2.pop()) elif (not array2) or array1[-1] > array2[-1]: merged_array.append(array1.pop()) inv += len(array2) else: merged_array.append(array2.pop()) merged_array.reverse() return merged_array, inv def _merge_sort(list, merge): len_list = len(list) if len_list < 2: return list, 0 middle = len_list / 2 left, left_inv = _merge_sort(list[:middle], merge) right, right_inv = _merge_sort(list[middle:], merge) l, merge_inv = merge(left, right) inv = left_inv + right_inv + merge_inv return l, inv def main(): test_list = nprnd.randint(1000, size=50000).tolist() test_list_tmp = list(test_list) merge_sort(test_list_tmp, merge1) test_list_tmp = list(test_list) merge_sort(test_list_tmp, merge2) if __name__ == '__main__': main() | 444 | 1,554 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.796875 | 3 | CC-MAIN-2018-17 | latest | en | 0.408339 |
https://chargersgame.org/find-the-position-of-the-center-of-mass-of-the-bar-x-measured-from-the-bars-left-end-2/ | 1,670,225,852,000,000,000 | text/html | crawl-data/CC-MAIN-2022-49/segments/1669446711013.11/warc/CC-MAIN-20221205064509-20221205094509-00274.warc.gz | 186,951,573 | 16,558 | # Find the position of the center of mass of the bar, x, measured from the bar’s left end.?
A nonuniform, horizontal bar of mass m is supported by two massless wires against gravity. The left wire makes an angle phi_1 with the horizontal, and the right wire makes an angle phi_2. The bar has length L.
Let F₁ and F₂ be the forces exerted by lift and right wire on the bar. Both forces act upward in direction of the wires. So you can split the forces into x- and y-component.
F₁x = -F₁ · sin(φ₁)
F₁y = F₁ · cos(φ₁)
F₂x = F₂ · sin(φ₂)
F₂y = F₂ · cos(φ₂)
As always for such a plane system in equilibrium you can balances horizontal forces(x) , vertical forces (y) and the torques about an arbitrary fulcrum.
Set up balance of forces acting on the bar in x-direction
F₁x + F₂x = 0
<=>
-F₁ · sin(φ₁) + F₂ · sin(φ₂) = 0
<=>
F₁/F₂ = sin(φ₂)/sin(φ₁)
From the balance in y-direction you could calculate the magnitude of F₁ and F₂ but this is not the focus, so skip this step.
As fulcrum for torque balance choose the point directly above the center of mass at the upper edge of the bar. By this choice there are no torques exerted by the F₁x and F₂x nor the weight of the bar.
– F₁y · x + F₂y · (L-x) = 0
=>
x = L · F₂y / (F₁y + F₂y)
<=>
x = L / ( (F₁y/F₂y ) + 1)
express forces in terms of magnitude an angle
x = L / ( (F₁/F₂)·(cos(φ₁)/cos(φ₂)) + 1)
use ratio derive from x-direction balance
x = L / ( (sin(φ₂)/sin(φ₁))·(cos(φ₁)/cos(φ₂)) + 1)
<=>
x = L / ( (tan(φ₂)/tan(φ₁) + 1)
The answer is actually supposed to be
L tan(phi_2)/tan(phi_2)+tan(phi_1) | 521 | 1,552 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.25 | 4 | CC-MAIN-2022-49 | latest | en | 0.844652 |
https://brainly.com/question/86531 | 1,484,870,209,000,000,000 | text/html | crawl-data/CC-MAIN-2017-04/segments/1484560280761.39/warc/CC-MAIN-20170116095120-00257-ip-10-171-10-70.ec2.internal.warc.gz | 792,273,742 | 9,321 | 2014-07-22T18:25:46-04:00
### This Is a Certified Answer
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
There are two laws named for Kirchhoff. The both concern electrical circuits.
Here they are in my own words:
1). The sum of the voltage drops around any closed loop in a circuit is zero.
2). The sum of the currents at any single point in a circuit is zero.
• Brainly User
2014-07-23T11:36:21-04:00
there are two
1 Kirchhoff's current law:
it states that the sum of all currents flowing in all conductors that meet at single point is zero..
2 Kirchhoff's voltage law:
it states that the algebraic sum of all voltages around any closed network is zero... | 219 | 891 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.984375 | 3 | CC-MAIN-2017-04 | latest | en | 0.929714 |
https://lambdapage.org/28382/verbs-worksheets-for-grade-6/verbs-worksheets-have-fun-teaching-with-verbs-worksheets-for-grade-6/ | 1,620,293,383,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988753.91/warc/CC-MAIN-20210506083716-20210506113716-00112.warc.gz | 394,720,600 | 7,335 | verbs worksheets o have fun teaching with verbs worksheets for grade 6
verbs worksheets o have fun teaching with verbs worksheets for grade 6.
You should give an effective time management before doing so. The worksheet also works nicely in a color to attract kids. It also has a feature by using the representative picture for the letter. For example, the “A” letter is represented by the apple picture. This is very good for introducing your kids for the letter.
These preschool worksheets are simple. It’s also didn’t have any complex solution. You just need to explore your kids’ creativity. Try to get a worksheet with the picture that have a colorful theme.
It helps to understand that the number they will write has a sequence. If your kid start recognizing the number, then you can start to write the number. The simplest number is 1.
It’s another activity that will stimulate their memory and recognize the colors. Use the crayon to finish this project. As a reward, kids are allowed to hang their creation on their wall. It will make they enjoy and want to learn more using the other worksheets.
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The instruction is simple. You can ask your kid to put the color by saying the letter. For example, instruct them to color the “A” letter using the blue. Don’t point where is the A, let them pick the letter. If they pick the right letter, then they can continue to color the alphabet.
As usual, you have to give an instruction for your kid. Teach them about the difference between straight line and spiral line. The spiral line usually comes with the complex shape.
Another fun activity that can be done during the learning in home is by tracing a picture with kids. The picture tracing worksheets are awesome tool that can be used to achieve this activity.
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3Rd Grade Math Worksheets Multiplication
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Start by teaching your kid about the word that they want to practice. You also can teach them to spell the word. After that, make an example on how to trace the word.
Similar with the alphabet tracing, the alphabet coloring also can be fun activity to do. However, the main purpose of this exercise is to introducing the alphabet.
This can be done step by step. After a quarter, continue to the half and finish until the perfect spiral. It’s better to start the spiral from the center.
The letter set is also important. You have to make a proper amount of the letter. Don’t expect your kid to color all the 26 alphabets at one exercise. Start it by introducing them 5-7 letters.
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March 7, 2021 | 759 | 3,099 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2021-21 | latest | en | 0.964347 |
https://www.jiskha.com/questions/279422/The-graph-of-the-cosine-function-is-the-same-as-the-graph-of-the-sine-function-but | 1,563,467,566,000,000,000 | text/html | crawl-data/CC-MAIN-2019-30/segments/1563195525659.27/warc/CC-MAIN-20190718145614-20190718171614-00024.warc.gz | 733,619,690 | 5,161 | # Math
The graph of the cosine function is the same as the graph of the sine function but translated:
a) 90 degrees to the right
b) 90 degrees to the left
c) 180 degrees to the right
d) 180 degrees to the left
I was thinking it could a) and b) but there can only be one answer. Could someone help please? Thanks :)
1. 👍 0
2. 👎 0
3. 👁 83
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compare the parent function f(x)=x^2 to the quadractic function f(x)=-2x^2-6. the 6 in the function does which of the following? a.it makes the graph narrower than the parent function. b.it makes the graph wider than the parent | 880 | 3,076 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.53125 | 4 | CC-MAIN-2019-30 | latest | en | 0.910339 |
https://www.statisticshowto.com/find-t-critical-value-ti-83/ | 1,721,249,550,000,000,000 | text/html | crawl-data/CC-MAIN-2024-30/segments/1720763514801.32/warc/CC-MAIN-20240717182340-20240717212340-00560.warc.gz | 872,517,063 | 14,687 | # How to Find t Critical Value on TI 83
Probability and Statistics > TI-83 for Statistics > how to find t critical value on ti 83
IMPORTANT: The TI-83 doesn’t have a function to calculate the t critical value directly. In other words, you won’t find it listed on a menu as something like “invT”. However, you can calculate it; it just takes a couple of extra steps.
• if you want a t critical value for a two tailed test, follow the steps below. The confidence interval limits are the same as the t critical values.
• if you want the t critical value for a one tailed test, you can’t use the TI83 unless you program it (see this video for instructions ).
## How to Find T Critical Value on TI 83: Steps
Sample Question: Calculate the two-talied t-critical value for 36 degrees of freedom (df = 36) for an alpha level of 0.05 (α = 0.05).
Step 1: Press the “STAT” key, then press the left arrow key to arrive at the Tests menu.
Step 2: Press “8” for TInterval.
Step 3: Arrow right to choose STATS. Press ENTER.
Step 3: Change the values in the list. For this sample question:
• Set the mean (Xbar) to 0
• Set Sx to the square root of your sample size. The sample size is df+1, so for this example press √37.
• Set n as your sample size. For this example, it’s 37.
• Set your confidence level, which is 1 – α For this example, set it to .95.
Step 4: Highlight CALCULATE and then press Enter. The t-critical value is The same as the given confidence level limits: 2.028. | 388 | 1,477 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.265625 | 3 | CC-MAIN-2024-30 | latest | en | 0.774173 |
https://www.nmstpk.com/science/question-science-fair-variable.html | 1,620,896,884,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243990584.33/warc/CC-MAIN-20210513080742-20210513110742-00639.warc.gz | 865,961,720 | 10,906 | # Question: Science Fair Variable?
## What are the 3 variables in science?
There are three main variables: independent variable, dependent variable and controlled variables.
## What are the 4 variables in science?
Variables are the factors, traits, and conditions you can modify and measure. You’ll find different variables in all types of subjects. But, the most common variables found in a science experiment include dependent, independent, and controlled. Check out what each is through examples.
## What are the two variables in science?
There must be at least two variables in any experiment: a manipulated variable and a responding variable. A manipulated variable is a variable that is changed by the researcher. A manipulated variable is also called an independent variable.
## How do you identify variables in a science experiment?
An easy way to think of independent and dependent variables is, when you’re conducting an experiment, the independent variable is what you change, and the dependent variable is what changes because of that. You can also think of the independent variable as the cause and the dependent variable as the effect.
## What are the 5 types of variables?
There are six common variable types:
• DEPENDENT VARIABLES.
• INDEPENDENT VARIABLES.
• INTERVENING VARIABLES.
• MODERATOR VARIABLES.
• CONTROL VARIABLES.
• EXTRANEOUS VARIABLES.
## What are 3 control variables?
An experiment usually has three kinds of variables: independent, dependent, and controlled.
## What are variables in science experiments?
These changing quantities are called variables. A variable is any factor, trait, or condition that can exist in differing amounts or types. An experiment usually has three kinds of variables: independent, dependent, and controlled. The independent variable is the one that is changed by the scientist.
## What is a science variable?
Learn about the different types of variables in an experiment. Variables are an important part of an eye tracking experiment. A variable is anything that can change or be changed. In other words, it is any factor that can be manipulated, controlled for, or measured in an experiment.
## What are the 4 variables?
There are four variables you have to deal with: resources, time, quality, and scope.
## What are constants in science?
Updated August 08, 2018. A constant is a quantity that does not change. Although you can measure a constant, you either cannot alter it during an experiment or else you choose not to change it. Contrast this with an experimental variable, which is the part of an experiment that is affected by the experiment.
## What’s a independent variable in science?
Answer: An independent variable is exactly what it sounds like. It is a variable that stands alone and isn’t changed by the other variables you are trying to measure. For example, someone’s age might be an independent variable.
You might be interested: What Does A Spatula Do In Science?
## Why do we standardize variables?
Variables are standardized for a variety of reasons, for example, to make sure all variables contribute evenly to a scale when items are added together, or to make it easier to interpret results of a regression or other analysis.
## How do you identify variables?
A variable in research simply refers to a person, place, thing, or phenomenon that you are trying to measure in some way. The best way to understand the difference between a dependent and independent variable is that the meaning of each is implied by what the words tell us about the variable you are using.
## How do you identify a control group?
The control group and experimental group are compared against each other in an experiment. The only difference between the two groups is that the independent variable is changed in the experimental group. The independent variable is ” controlled ” or held constant in the control group.
## How do you identify dependent and independent variables?
Independent and dependent variables
1. The independent variable is the cause. Its value is independent of other variables in your study.
2. The dependent variable is the effect. Its value depends on changes in the independent variable. | 814 | 4,209 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.8125 | 3 | CC-MAIN-2021-21 | latest | en | 0.919919 |
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# is not an error excel Harwich, Massachusetts
The resulting array looks like this: {0;1;0;1;1} SUMPRODUCT then sums the items in this array and returns the total, which in the example is the number 3. This allows your formulas to evaluate properly without your intervention. For example, if cell A1 contains the formula =B1/C1, and the value of C1 is 0, the formula in A1 returns the #DIV/0! TRUE =ISNA(A4) Checks whether the value in cell A4, #REF!, is the #N/A error.
by neonstorm / February 28, 2005 4:40 AM PST In reply to: Help with Excel / =IF(ISERROR) function Alright, I am still having problems. Copyright © 2003-2016 TechOnTheNet.com. Also please exercise your best judgment when posting in the forums--revealing personal information such as your e-mail address, telephone number, and address is not recommended. If a cell contains an error, the value 5 is returned.
ISREF Value refers to a reference. Here's how to do it. Answer: Often times your spreadsheet contains a large amount of formulas which will not properly calculate when an error is encountered. After copying the example to a blank worksheet, select the range C2:C4, press F2, and then press CTRL+SHIFT+ENTER.
Workbooks.Open Filename:=nFileName nNewBook = ActiveWorkbook.Name ' Copy Data file to main file Range("A1").CurrentRegion.Copy Windows(nWBook).Activate Worksheets(nSheet).Range("A1").PasteSpecial Paste:=xlValues ' Close Data file. More... If used with an expression of TRUE, then FALSE is returned. Select A1, and press F2 to edit the formula.
Advertisement About Us Contact Us Testimonials Donate Follow us Home MS Excel Formulas / Functions TechOnTheNet.com requires javascript to work properly. While not necessarily harmful, these errors will be displayed in your spreadsheet until corrected or until the required data is entered, which can make the overall table less attractive and more The IS functions are useful in formulas for testing the outcome of a calculation. Description Each of these functions, referred to collectively as the IS functions, checks the specified value and returns TRUE or FALSE depending on the outcome.
Bad cut and paste job. Skip to main content Login Cart Exceljet Quick, clean, and to the point Training Videos Functions Formulas Shortcuts Blog Search form Search Count cells that do not contain errors =SUMPRODUCT(--NOT(ISERR(rng))) To Format error values by applying a white font color to the text Use the following procedure to format cells that contain errors so that the text in those cells is displayed All rights reserved.
You can prevent these indicators from being displayed by using the following procedure. In the example below, select the cell that contains the text “Data” and then drag the cursor to select through the last cell in the “Description (Result)” column. Applies To Excel 2016, Excel 2013, Excel 2011 for Mac, Excel 2010, Excel 2007, Excel 2003, Excel XP, Excel 2000 Type of Function Worksheet function (WS) VBA function (VBA) Example (as a sheet where data is imported.I need to make a cell that, when the imported data is refreshed, simply displays a field from page 3.
Ryan,Here a portion of code that will help you modify your macro. FALSE =ISNUMBER(A5) Checks whether the value in cell A5, 330.92, is a number. All rights reserved. If not, the value 100 is returned.
Related functions Excel ISERROR Function Excel NOT Function Excel SUMPRODUCT Function Author Dave Bruns Excel Formula Training Bite-sized videos in plain English. Learn nested IF, VLOOKUP, INDEX & MATCH, COUNTIFS, RANK, SUMIFS, Databases SQL Oracle / PLSQL SQL Server MySQL MariaDB PostgreSQL SQLite MS Office Excel Access Word Web Development HTML CSS Color Picker Languages C Language More ASCII Table Linux UNIX Java Databases SQL Oracle / PLSQL SQL Server MySQL MariaDB PostgreSQL SQLite MS Office Excel Access Word Web Development HTML CSS Color Picker Languages C Language More ASCII Table Linux UNIX Java Everthing else seems ok other than this.
Copyright © 2003-2016 TechOnTheNet.com. and replaces the C332 refernce in my formula with #REF!. NA Use this function to return the string #N/A in a cell. On the Home tab, in the Styles group, click the arrow next to Conditional Formatting and then click Manage Rules.The Conditional Formatting Rules Manager dialog box appears.
View the discussion thread. The function name is followed by a pair of empty parentheses, like this: =NA(). It can be used as a worksheet function (WS) in Excel. Select the range of cells that contain the error value.
Remarks If Value or Value_if_error is an empty cell, IFERROR treats it as an empty string value (""). Under Error Checking, clear the Enable background error checking check box. To do this, you can use the IFERROR and NA functions, as the following example shows. We're using a small weight loss tracking spreadsheet as an example of the kind of table that would produce a calculation error (weight lost percentage calculation) while waiting for new data
error, I'm trying to get it to pull from a corresponding cell on page 2 that the user can key in (in the case that it is missing from page 3).After error. by neonstorm / February 28, 2005 12:13 AM PST In reply to: Help with Excel / =IF(ISERROR) function You guys are awesome! For example, the ISBLANK function returns the logical value TRUE if the value argument is a reference to an empty cell; otherwise it returns FALSE.
If I manually deleted the data in C332, A16 displays the data in B16. Please try again now or at a later time. Insted of getting the #REF! All rights reserved.
The NOT function is a built-in function in Excel that is categorized as a Logical Function. excel 2010 tutorial | how to use excel | microsoft excel 2010 | vba in excel Connect with us facebook twitter CNET Reviews Top Categories CNET 100 Appliances Audio Cameras Cars FALSE =ISNA(A6) Checks whether the value in cell A6, #N/A, is the #N/A error. If you need to, you can adjust the column widths to see all the data.
error. Frequently Asked Questions Question: Can you give me specific examples of when and how the ISERROR function is used. Remarks The value arguments of the IS functions are not converted. You will need to add validation, etc. ' Allow the user to select Data file. | 1,424 | 6,317 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.5625 | 3 | CC-MAIN-2018-51 | latest | en | 0.801594 |
https://plainmath.net/28188/coefficient-determine-behavior-determine-algebraically-polynomial | 1,638,144,398,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358673.74/warc/CC-MAIN-20211128224316-20211129014316-00456.warc.gz | 540,578,183 | 7,658 | # use the leading coefficient to determine the graph’s end behavior.Determine algebraically whether the polynomial is even,odd,or neither. f(x)=3x−x^3
For the following exercise, for each polynomial, a. find the degree. b. find the zeros, if any. c. find the y-intercept(s), if any. d. use the leading coefficient to determine the graph’s end behavior. and e. determine algebraically whether the polynomial is even, odd, or neither. $$\displaystyle{f{{\left({x}\right)}}}={3}{x}−{x}^{{3}}$$
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a) Degree is 3.
b) $$\displaystyle{x}=\pm\sqrt{{3}}$$, x=0
c) y-intercept
d) Graph rises to the left and falls to the right
e) Function is odd. | 237 | 855 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.703125 | 4 | CC-MAIN-2021-49 | latest | en | 0.789295 |
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1 General Finite State General Finite State Machines: Machines: Mealy and Moore Mealy and Moore ECSE-2610 Co mputer C perations (CoCO) Fall 2006 11/02/06
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2 Recap: Vending Machine Recap: Vending Machine Delivers package of gum after 15 cents is deposited Single coin slot for dimes, nickels No change given Reset N N N, D [ open ] 15¢ 10¢ D D Present State 10¢ 15¢ D 0 0 1 1 0 0 1 1 0 0 1 1 X N 0 1 0 1 0 1 0 1 0 1 0 1 X Inputs Next State 10¢ X 10¢ 15¢ X 10¢ 15¢ 15¢ X 15¢ Output Open 0 0 0 X 0 0 0 X 0 0 0 X 1
3 Combinational logic for outputs Recap: Vending Machine FSM Recap: Vending Machine FSM D 1 = Q 1 + D + Q 0 N D 0 = N Q 0 + Q 0 N + Q 1 N + Q 1 D OPEN = Q 1 Q 0 Step 6. Implementation: D FFs Step 6. Implementation: D FFs Combinational logic for next state inputs Flip-Flops (memory) CLK OPEN CLK Q 0 D R Q Q D R Q Q \ Q 1 \reset \reset \ Q 0 \ Q 0 Q 0 Q 0 Q 1 Q 1 Q 1 Q 1 D D N N N \ N D 1 D 0
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4 General Circuit Structure General Circuit Structure
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This homework help was uploaded on 04/22/2008 for the course ECSE 2610 taught by Professor Ji during the Spring '08 term at Rensselaer Polytechnic Institute.
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Ask a homework question - tutors are online | 552 | 1,730 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2017-17 | longest | en | 0.782682 |
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Question: if f is analytical and $|f(z)| < M$ for $|z| =< R$ find an upper bound for $|f('n)(z)|$ in $|z| =< \rho < R$ (where $f('n)$ means the nth derivative of $f$).
So I cited Cauchy's Inequality, to find the bound $\frac{n!M_R} {R^n}$ , but since we have the additional limit of $\rho$ I think we should be able to find a better bound. The geometric interpretation is that the the circle of radius $\rho$ limits the radius of $|z|$. Is there further geometric intuition that I should see to solve this problem?
-
## 2 Answers
No, there is no better bound. Consider the function $f(z)=M \frac{z^n}{R^n}$ which fulfills the requirement (ok, that $|f(z)|$ is really smaller than $M$ one should replace $M$ by $M-\epsilon$ and in the end let $\epsilon\to0$). It is easy to check that $$f^{(n)}(z) = \frac{n! M}{R^n},$$ i.e., it exhausts the Cauchy bound. Note however that the $n$-th derivative is independent on $z$. So there is no way you get a better bound by restricting $|z|<\rho$.
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This might not be correct: See math.stackexchange.com/questions/555820/… – bryanj Nov 7 '13 at 17:38
This is a question of Lars V. Ahlfors, Complex Analysis, Third Edition, McGraw-Hill International Editions, 1979, at the end of the subsection "Higher Derivatives" of the section "Cauchy's Integral Formula". In this context, the answer should take this into account. My suggestion is to apply the formula $f^{\left(n\right)}\left(z\right)=\frac{n!}{2\pi i}\int_{C}\frac{f\left(\zeta\right)d\zeta}{\left(\zeta-z\right)^{n+1}}$. Thus we have
$\begin{split}\left|f^{\left(n\right)}\left(z\right)\right| & =\left|\frac{n!}{2\pi i}\int_{C}\frac{f\left(\zeta\right)d\zeta}{\left(\zeta-z\right)^{n+1}}\right|\\ & =\frac{n!}{2\pi}\left|\int_{C}\frac{f\left(\zeta\right)d\zeta}{\left(\zeta-z\right)^{n+1}}\right|\\ & \leq\frac{n!}{2\pi}\int_{C}\frac{\left|f\left(\zeta\right)\right|d\zeta}{\left|\zeta-z\right|^{n+1}}\\ & \leq\frac{n!}{2\pi}M\int_{C}\frac{d\zeta}{R^{n+1}}\\ & =\frac{n!M}{2\pi R^{n+1}}\int_{C}d\zeta\\ & \leq\frac{n!M}{2\pi R^{n+1}}2\pi R\\ & =\frac{n!M}{R^{n}}.\end{split}$
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# Land Nav Task 14 - Determine Direction without a Compass
Standards: Identified north and east within 15 degrees.
Conditions:
During daylight and at night (with a clear
view of the Big Dipper), given a wristwatch
(not digital), the soldier must determine
direction in a field environment with
natural vegetation available.
Standards:
Identified north and east within 15 degrees.
Note.
All of the procedures given in this task
give approximate directions. For accurate
directions, a compass must be used.
Evaluation
Preparation:
Setup:
Directionally orient yourself to an area
that is unfamiliar to the soldier to be
tested.
Brief
Soldier: Accompany the soldier to the
area and tell the soldier to use the
field-expedient methods to determine which
direction is north and east. Use a compass
direction of your choice. Do not tell the
soldier how he or she did on performance
measure 1 until he or she completes
performance measure 2.
Note.
Before the soldier is scored a GO for this
task, he or she must display proficiency in
all three field-expedient methods of
determining direction without a compass.
However, performance measure 3 must be
tested in a different location.
Performance Measures GO NO GO 1. Determined direction using the shadow-tip field-expedient method — — a. Placed a stick vertically into the ground at a desired spot. b. Marked the tip of the stick’s shadow to represent west. c. Waited 10 to 15 minutes. d. Marked a new position of the tip of the stick’s shadow to represent east. e. Drew an east-west line through the two shadow-tip marks. f. Drew a north-south line at a right angle to the east-west line. g. Pointed in the required direction within 15 degrees. 2. Determined direction using the watch field-expedient method. — — a. In the Northern Hemisphere, pointed the hour hand of the watch at the sun; in the Southern Hemisphere, pointed the 12 o’clock position of the watch at the sun. b. Pointed in the required direction within 15 degrees. 3. Determined direction using the North Star field-expedient method. — — a. Located the Big Dipper. b. Located Polaris, the North Star. c. Pointed in the required direction within 15 degrees.
Evaluation
Guidance:
Score the soldier GO if all
performance measures are passed. Score the
soldier NO GO if any performance measure is
failed. If the soldier scores NO GO, show
the soldier what was done wrong and how to
do it correctly.
References Required Related FM 3-25.26
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As an example, what is wrong with the concern is incorrect suggestions, such as obtaining a partial value when you do not have enough working components to complete the entire task. There is a great reason that this was wrong, and it is a matter of logic, not intuition. Logic enables you to follow a detailed procedure that makes good sense, as well as when you make a wrong action, you are generally required to either attempt to go forward as well as remedy the blunder, or attempt to step and do a backwards action.
An additional instance is when the trainee does not recognize an action of a procedure. These are both rational failures, as well as there is no way around them. Also when you are stuck in a location that does not permit you to make any kind of type of action, such as a triangle, it is still vital to recognize why you are stuck, to ensure that you can make a much better step and go from the action you are stuck at to the following location.
With this in mind, the most effective way to address a stuck circumstance is to simply take the progression, instead of attempting to step. Both processes are different in their technique, yet they have some fundamental resemblances. Nevertheless, when they are attempted with each other, you can quickly tell which one is better at fixing the issue, and also you can likewise tell which one is a lot more effective.
Let's speak about the very first example, which relates to the Sphere Packing mathematics lab. This is not too complicated, so allow's initial go over just how to begin. Take the following process of attaching a component to a panel to be made use of as a body. This would certainly need 3 measurements, and would be something you would need to connect as part of the panel.
Now, you would have an extra dimension, however that does not suggest that you can just maintain that measurement as well as go from there. When you made your initial step, you can easily ignore the measurement, and then you would have to go back as well as backtrack your steps.
Nevertheless, instead of bearing in mind the extra measurement, you can utilize what is called a "mental faster way" to aid you bear in mind that extra measurement. As you make your initial step, envision yourself taking the dimension and also attaching it to the part you wish to affix to, and then see just how that makes you really feel when you duplicate the procedure.
Visualisation is a really powerful strategy, as well as is something that you need to not skip over. Visualize what it would certainly seem like to actually attach the part as well as be able to go from there, without the measurement.
Currently, allow's consider the 2nd instance. Let's take the exact same process as in the past, but now the student has to keep in mind that they are mosting likely to return one step. If you tell them that they have to return one action, but after that you remove the idea of having to return one action, after that they will not know how to wage the issue, they will not recognize where to try to find that action, and the procedure will certainly be a mess.
Instead, use a psychological faster way like the mental diagram to mentally show them that they are going to return one action. and also put them in a setting where they can move forward from there. without needing to consider the missing out on an action.
## Hire Someone To Do Your Sphere Packing Lab
" Sphere Packing - Required Assist With a Mathematics lab?" Unfortunately, lots of trainees have actually had an issue comprehending the ideas of linear Sphere Packing. The good news is, there is a new style for linear Sphere Packing that can be used to instruct straight Sphere Packing to students who have problem with this concept. Students can use the lab Aid Solution to help them discover new techniques in straight Sphere Packing without facing a mountain of troubles and also without needing to take an examination on their principles.
The lab Assist Service was created in order to assist battling students as they move from university and secondary school to the university as well as work market. Numerous pupils are unable to handle the anxiety of the understanding process as well as can have really little success in comprehending the principles of direct Sphere Packing.
The lab Assist Solution was developed by the Educational Screening Service, who uses a range of different online examinations that pupils can take and also exercise. The Test Assist Service has helped several trainees improve their ratings as well as can assist you boost your ratings also. As students move from college and high school to the university and also work market, the TTS will certainly help make your trainees' shift simpler.
There are a few different ways that you can capitalize on the lab Aid Solution. The major manner in which pupils make use of the lab Help Service is with the Answer Managers, which can aid students discover strategies in direct Sphere Packing, which they can utilize to help them prosper in their training courses.
There are a number of issues that students experience when they initially make use of the lab Assist Solution. Trainees are often overwhelmed and do not recognize just how much time they will certainly require to commit to the Solution. The Response Supervisors can help the trainees examine their principle learning and also help them to review all of the product that they have actually already found out in order to be prepared for their next course work.
The lab Aid Solution works similarly that a teacher does in terms of aiding pupils understand the ideas of direct Sphere Packing. By providing your pupils with the tools that they require to learn the crucial principles of linear Sphere Packing, you can make your trainees much more effective throughout their research studies. Actually, the lab Aid Service is so efficient that lots of students have switched from standard mathematics class to the lab Assist Service.
The Job Supervisor is created to help students manage their homework. The Task Supervisor can be set up to schedule just how much time the pupil has readily available to complete their designated research. You can also set up a customized amount of time, which is a great attribute for trainees that have a busy timetable or an extremely busy secondary school. This feature can assist students stay clear of feeling overwhelmed with math tasks.
One more useful function of the lab Help Solution is the Trainee Assistant. The Trainee Aide aids students manage their job as well as provides a location to post their homework. The Trainee Aide is useful for students who don't intend to get bewildered with addressing multiple inquiries.
As trainees get more comfy with their projects, they are motivated to connect with the Job Manager and the Pupil Aide to obtain an online support group. The on-line support system can help trainees maintain their focus as they answer their tasks.
All of the assignments for the lab Help Service are consisted of in the package. Trainees can login and also finish their designated job while having the student aid offered behind-the-scenes to help them. The lab Aid Service can be a great assistance for your trainees as they begin to navigate the tough college admissions and job hunting waters.
Students have to be prepared to obtain used to their jobs as swiftly as possible in order to reach their primary goal of getting into the university. They have to strive sufficient to see outcomes that will allow them to stroll on at the next degree of their research studies. Obtaining made use of to the process of finishing their tasks is really essential.
Pupils are able to discover different ways to help them find out how to use the lab Assist Service. Learning exactly how to make use of the lab Aid Solution is important to pupils' success in college and task application.
## Hire Someone To Take My Sphere Packing Lab
Sphere Packing is utilized in a lot of institutions. Some teachers, nonetheless, do not use it really effectively or utilize it inaccurately. This can have a negative effect on the pupil's learning.
So, when assigning assignments, utilize a good Sphere Packing assistance service to help you with each lab. These solutions supply a variety of practical services, consisting of:
Assignments may require a lot of assessing and also searching on the computer system. This is when utilizing an aid service can be a great benefit. It permits you to get more work done, increase your comprehension, and also avoid a lot of stress and anxiety.
These kinds of research services are a superb means to start working with the very best sort of assistance for your demands. Sphere Packing is among the most tough based on grasp for students. Working with a service, you can see to it that your demands are fulfilled, you are instructed appropriately, and you recognize the material appropriately.
There are a lot of manner ins which you can show on your own to function well with the class as well as be successful. Use a proper Sphere Packing help solution to lead you as well as get the job done. Sphere Packing is just one of the hardest courses to find out but it can be easily grasped with the best help.
Having a homework service additionally aids to enhance the student's qualities. It allows you to include additional credit report as well as boost your GPA. Obtaining additional credit is frequently a massive benefit in several universities.
Pupils who don't maximize their Sphere Packing class will certainly wind up continuing of the remainder of the class. The bright side is that you can do it with a fast as well as very easy service. So, if you wish to move ahead in your class, make use of a good assistance solution. One point to keep in mind is that if you really intend to increase your quality level, your course work needs to get done. As much as feasible, you need to comprehend as well as collaborate with all your problems. You can do this with a good assistance solution.
One benefit of having a research solution is that you can help on your own. If you don't feel confident in your ability to do so, after that a good tutor will certainly have the ability to help you. They will be able to solve the troubles you face and also help you recognize them to get a better grade.
When you graduate from secondary school and go into university, you will certainly require to work hard in order to remain ahead of the various other pupils. That suggests that you will require to strive on your homework. Making use of an Sphere Packing service can aid you get it done.
Maintaining your grades up can be tough because you typically require to study a great deal and also take a lot of examinations. You do not have time to service your grades alone. Having an excellent tutor can be an excellent assistance since they can aid you as well as your research out.
A help service can make it much easier for you to handle your Sphere Packing class. In addition, you can learn more concerning yourself and aid you prosper. Discover the very best tutoring solution and you will be able to take your research skills to the following level. | 2,457 | 12,545 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.640625 | 3 | CC-MAIN-2021-04 | latest | en | 0.980181 |
https://www.it610.com/article/1201355984786792448.htm | 1,575,926,985,000,000,000 | text/html | crawl-data/CC-MAIN-2019-51/segments/1575540523790.58/warc/CC-MAIN-20191209201914-20191209225914-00010.warc.gz | 746,636,182 | 4,621 | # LeetCode - 23. Merge k Sorted Lists
### 题目描述
Merge k sorted linked lists and return it as one sorted list. Analyze and describe its complexity.
### Java代码实现
``````/**
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeKLists(ListNode[] lists) {
if (lists == null || lists.length == 0) {
return null;
}
// new PriorityQueue<>(lists.length, Comparator.comparingInt(o -> o.val));
PriorityQueue pq = new PriorityQueue<>(lists.length, (o1, o2) -> {
return Integer.compare(o1.val, o2.val);
});
ListNode dummy = new ListNode(0);
ListNode walk = dummy;
for (ListNode node : lists) {
if (node != null) {
pq.offer(node);
}
}
while (!pq.isEmpty()) {
walk.next = pq.poll();
walk = walk.next;
if (walk.next != null) { | 214 | 806 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.859375 | 3 | CC-MAIN-2019-51 | longest | en | 0.417588 |
https://jajawriters.org/2021/04/19/1-calculate-the-molar-mass-of-cocl2-6h2o-2-what-mass-in-g-is-needed-to-make-100ml-of-a-0-080m-cocl2-6h2o-solution-3-after-the-0-080m-cocl2-6h2o-solution-was-made-its-absorbance-at-510-n/ | 1,620,368,067,000,000,000 | text/html | crawl-data/CC-MAIN-2021-21/segments/1620243988775.25/warc/CC-MAIN-20210507060253-20210507090253-00634.warc.gz | 348,209,945 | 10,630 | 1. Calculate the molar mass of CoCl2 6H2O. 2 . What mass (in g) is needed to make 100ml of a 0.080M CoCl2 6H2O solution? 3 . After the 0.080M CoCl2 6H2O solution was made, it’s absorbance at 510 n
1. Calculate the molar mass of CoCl2 6H2O. 2 . What mass (in g) is needed to make 100ml of a 0.080M CoCl2 6H2O solution? 3 . After the 0.080M CoCl2 6H2O solution was made, it’s absorbance at 510 nm was measured. It was found to be .370. what is the molarity of the solution based on the absorbance? 4 . What kind of Molarity is the value calculated above? 5 . What is the percent error for the solution made in part A? 6 . What volume (in mL) of the 0.080M CoCl2 6H2O solution in part A is needed to make 100.00mL of a 0.060M CoCl2 6H2O solution? 7 . After the 0.060M CoCl2 6H2O solution was made, it’s absorbance at 510nm was measured to be 0.270. What is the Molarity of the solution based on the absorbance? 8 . What is the percent error for the solution made in part B?
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1. Calculate the molar mass of CoCl2 6H2O. 2 . What mass (in g) is needed to make 100ml of a 0.080M CoCl2 6H2O solution? 3 . After the 0.080M CoCl2 6H2O solution was made, it’s absorbance at 510 n
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Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result. | 757 | 2,625 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.90625 | 3 | CC-MAIN-2021-21 | latest | en | 0.897117 |
https://poitevinpm.medium.com/leetcode-2131-longest-palindrome-by-concatenating-two-letter-words-c76d28ba336a | 1,713,033,963,000,000,000 | text/html | crawl-data/CC-MAIN-2024-18/segments/1712296816832.57/warc/CC-MAIN-20240413180040-20240413210040-00890.warc.gz | 445,918,325 | 25,310 | # Leetcode 2131: Longest Palindrome by Concatenating Two Letter Words
In this problem, we want to form the longest possible palindrome by concatenating two letter words from a list:
You are given an array of strings `words`. Each element of `words` consists of two lowercase English letters.
Create the longest possible palindrome by selecting some elements from `words` and concatenating them in any order. Each element can be selected at most once.
Return the length of the longest palindrome that you can create. If it is impossible to create any palindrome, return `0`.
A palindrome is a string that reads the same forward and backward.
We are also given a few examples:
To solve the problem, we want to match strings with there “reverse”, for instance, match `lc` with `cl`. Each time we find such a pair, we are able to add 4 characters to the maximum palindrome.
Also, we are able to use use one “self palindrome”, that is a string with 2 characters that are the same, in the middle of the palindrome, and add 2 characters to the maximum palindrome. For instance in the first example: “lcggcl”.
That’s it, there is no other way to extend the palindrome. We want to solve the problem in linear time with the number of words `n` as much as possible. to do that, we will keep track of all the words we encountered, or more precisely, we will store their inverse (`lc` to `cl`), along with the number of time we encountered it. When we find a matching string, we will decrease the count, as there is one less string to match. We will use a map to keep track of all the words encountered.
We will also keep track of the number of self palindromes we encountered. Each time we find an unmatched self palindrome, we add one to the count. Each time we find a palindrome that can be matched to an existing one, we remove one to the count. If the count is greater than 0 at the end of the loop, we add 2 to the result as we are able to put one of the unmatched self palindrome in the middle of the string.
In Java, I obtained the following code:
`class Solution { public int…` | 471 | 2,085 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.1875 | 3 | CC-MAIN-2024-18 | latest | en | 0.91846 |
http://stackoverflow.com/questions/7144791/how-to-find-a-vector-using-angle?answertab=oldest | 1,417,010,891,000,000,000 | text/html | crawl-data/CC-MAIN-2014-49/segments/1416931006885.62/warc/CC-MAIN-20141125155646-00038-ip-10-235-23-156.ec2.internal.warc.gz | 268,408,169 | 16,160 | # How to find a vector using angle?
In 3 dimensional space i have an ine object between x and y axis.If the angle between x and y axis given,how can we find a vector value(i,j,k) ?please help to solve this problem.
-
I think this website might not be the right place for this question... Try here maybe: math.stackexchange.com – Eran Zimmerman Aug 22 '11 at 8:30
The x and y axis are in the same plane so you would actually need the z axis angle as well to do this. (at least a combination of x, y and z (not just x and y)). – ExtremeCoder Aug 22 '11 at 8:31
Yes,the z axis angle alse given. – user519675 Aug 22 '11 at 9:15 | 175 | 628 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.625 | 3 | CC-MAIN-2014-49 | longest | en | 0.885237 |
http://hvac-talk.com/vbb/showthread.php?467112-Questions-regarding-HVAC-Calc | 1,501,118,790,000,000,000 | text/html | crawl-data/CC-MAIN-2017-30/segments/1500549426693.21/warc/CC-MAIN-20170727002123-20170727022123-00585.warc.gz | 145,315,533 | 19,808 | 1. New Guest
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## Questions regarding HVAC-Calc
I have a couple of questions regarding the HVAC-Calc program and admit I am on a steep learning curve regarding heat loss calculations. I downloaded the program and understand the basic operation, required input data is fairly straight forward. The design that I am trying to enter is a three level home with a finished basement, main living floor and a conditioned attic. The home will be SIP construction with 6” walls and 8” SIP roof. Five of the rooms on the main level will have vaulted ceilings, so there is not attic deck above. The remainder of the main level will have flat ceilings with the attic deck above.
1. Since the program asks for a ceiling height, should I take the average ceiling height for the rooms with vaulted ceilings?
2. The program does not have a SIP roof, should I just use the R-30 roof choice?
3. Should the attic be included as a second floor or should I just include the attic as a vaulted ceiling over that portion of the home and adjust the wall height to account for the gable ends? The attic deck as well as the deck under the main level will be un-insulated.
4. I entered the total length of each exterior wall and a room for each section of exterior wall, since the remainder of the walls are interior, I assume that I should be unconcerned with them since correct? The square footage I entered is the total for each level and not just for the rooms with exterior walls
5. Finally, I have an attached garage that will also be constructed from SIPS with a SIP roof. The garage will be separated from the living areas by a SIP wall. The garage will be partially conditioned (abt 40 degrees in the winter) to prevent the freezing of water lines in our RV. How can this situation be handled by the HVAC-Calc program? I had originally modeled the living areas without the garage. Should I continue this way so that I have a more conservative model?
Thanks!
2. Being a bit more clear and specific will help with your problem - what program your using ? what kind of construction your dealing with ..,2 story, 3 story , in ground ..All of these are in my program and have done many .. Heat Gain & Loss are much easier than they use to be.. If your program has a training tool use it a few times.. There's probably areas in your program your missing ..
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The Program that I have been using is HVAC-Calc. I have played around a little more with it and used the tutorial videos. I believe the results of the various model runs I have done are in the ballpark. It appears that I have a heat loss of about 52 kBTUh. The partially conditioned garage adds another 13 kBTUh for a total loss of about 65 kBTUh.
In reading product specifications from the various GHP manufacturers, specifically the “Hydronic Heating Capacities”, how is the heat loss related to determining the size of the GHP needed? I know that I need to replace the calculated loss to maintain the interior temperature specified in the model and I know that the model is based on conditions that seldom occur in reality. Referring to the product performance data sheets provided the the various manufactures, which reflect the system performance under various conditions, how do I use this data to size an appropriate system for my home?
General questions:
Where can I find groundwater temperatures for a specific area?
It’s the EWT listed on the specification sheets that groundwater temperature?
Is Total Heating Capacity (HC) the number used to size the system or is Total Heat Extracted (HE) the number used or neither?
How is the hot water storage tank sized?
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Don Sleeth, author of HVAC-Calc, provides tech support last I checked. Google ground water or deep earth temps for a national map with temperature contour lines.
EWT is only same as ground water temp on an open loop system
System output (HC) is HE + power it uses in heating mode. In cooling power used is subtracted from Heat Rejected to determine nest cooling
Best hot water tank arrangement is two tanks. Size the upstream buffer tank to a day's household use. Size the downstream finishing tank to the house's worst hour of use
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# Problem 2911. Matlab Basics II - squares
Solution 1119060
Submitted on 9 Feb 2017 by Jihye Sofia Seo
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 = [1 4 9]; assert(isequal(square_elements(x),y_correct))
2 Pass
x = [33 44 5; 56 20 12]; y_correct = [1089 1936 25;3136 400 144]; assert(isequal(square_elements(x),y_correct))
3 Pass
x=[3.3 4.4 5; 5.6 2.0 1.2]; y_correct = [10.889999999999999 19.360000000000003 25; 31.359999999999996 4 1.440000000000000]; assert(isequal(square_elements(x),y_correct)) | 237 | 660 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.75 | 3 | CC-MAIN-2020-34 | latest | en | 0.578839 |
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### Dti2143 chapter 5
1. 1. CHAPTER 6<br />FUNCTION<br />1<br />
2. 2. What is function?<br />A function is a section of a program that performs a specific task .<br />Solving a problem using different functions makes programming much simpler with fewer defects .<br />It’s a solution for a big project that split into small sub project. <br />
3. 3. Overview<br />Same book published in several volumes.<br />Easily manageable<br />Huge Book of 3000 pages<br />
4. 4. Advantages of Functions<br />Problem can be viewed in a smaller scope<br />Program development are much faster compared to the common structure<br />Program becomes easier to maintain<br />
5. 5. Classification of Functions<br />Library functions<br /><ul><li>defined in the language
6. 6. provided along with the compiler</li></ul>Example:printf(), scanf() etc.<br />User Defined functions<br /><ul><li> written by the user</li></ul>Example:main() or any other user-defined function<br />
7. 7. More about Function..<br />Functions are used to perform a specific task on a set of values<br />Values can be passed to functions so that the function performs the task on these values<br />Values passed to the function are called arguments<br />After the function performs the task, it can send back the results to the calling function.<br />The value sent back by the function is called return value<br />A function can return back only one valueto the calling function<br />
8. 8. Writing User-Defined Functions<br />intfnAdd(int iNumber1, int iNumber2)<br />{<br /> /* Variable declaration*/<br />intiSum;<br /> /* Find the sum */<br />iSum = iNumber1 + iNumber2;<br /> /* Return the result */<br /> return (iSum);<br />}<br />Return data type<br />Arguments (Parameters)<br />Function header<br />Function Body<br />Can also be written as return isum;<br />
9. 9. Example1: Writing User-Defined Functions<br />void fnDisplayPattern(unsigned intiCount)<br />{<br /> unsigned intiLoopIndex;<br /> for (iLoopIndex = 1; iLoopIndex <= iCount; iLoopIndex++) {<br />printf(“*”);<br /> }<br /> /* return is optional */<br /> return;<br />} <br />
10. 10. Example: Writing User-Defined Functions<br />Example2 <br />intfnAdd(int iNumber1, int iNumber2)<br />{<br /> /* Return the result */Can also be written as <br /> return (iNumber1 + iNumber2);<br />} <br />=======================================================<br />Example3<br />/* Function to display “UTHM.” */<br />void fnCompanyNameDisplay()<br />{<br />printf(“UTHM.”);<br />} <br />
11. 11. Returning values<br /><ul><li>The result of the function can be given back to the calling functions
12. 12. Return statement is used to return a value to the calling function
13. 13. Syntax:</li></ul>return (expression) ;<br /><ul><li>Example:</li></ul>return(iNumber * iNumber); <br /> return 0;<br /> return (3);<br /> return;<br /> return (10 * i);<br />
14. 14. Function Terminologies<br />Function Prototype<br />void fnDisplay() ;<br />int main(int argc, char **argv) {<br /> fnDisplay();<br /> return 0;<br />}<br />void fnDisplay() {<br /> printf(“Hello World”);<br />}<br />Calling Function<br />Function Call Statement<br />Function Definition<br />Called Function<br />
15. 15. Formal and Actual Parameters<br /><ul><li>The variables declared in the function header are called as formal parameters
16. 16. The variables or constants that are passed in the function call are called as actualparameters
17. 17. The formal parameter names and actual parameters names can be the same or different </li></li></ul><li>Functions – Example<br />intfnAdd(int iNumber1, int iNumber2) ;<br />int main(intargc, char **argv) {<br />int iResult,iValue1, iValue2;<br /> /* Function is called here */<br />iResult = fnAdd(iValue1, iValue2);<br />printf(“Sum of %d and %d is %dn”,iValue1, iValue2, iResult);<br /> return 0;<br />}<br />/* Function to add two integers */<br />intfnAdd(int iNumber1, int iNumber2)<br />{<br /> /* Variable declaration*/<br />intiSum;<br />iSum = iNumber1 + iNumber2; /* Find the sum */<br /> return (iSum); /* Return the result */<br />}<br />Actual Argument<br />Formal Argument<br />Return value<br />
18. 18. Types of Function in C Language<br />Function Definition<br />Function Calls<br />Function Prototypes<br />
19. 19. Element Of Functions<br />Function definitions<br />The first line<br /> A function type<br />A function name<br />An optional list of formal parameters enclosed in parenthesis<br />Eg: <br />function_type function_name(formal parameters)<br />The body of the function<br />The function body is the expression of the algorithm for the <br />module in C.<br />The function body consist of variable declarations and statements<br />
20. 20. Example of Function Definition<br />void print_menu(void)<br />/* example of function definition. The first line specifies the type of the function as void. This type of function will not return a value under its name. If a function is designed such that it does not return any value under its name, its type must be Void.*/<br />{<br />printf(“THIS PROGRAM DRAWS A RECTANGLE OR A TRIANGLE ON THE”);<br />printf(“SCREEN.n”);<br />printf(“Enter 1 to draw a rectangle.n”);<br />printf(“Enter 2 to draw a triangle.”);<br />} /*end function print_menu*/<br />
21. 21. Function Calls<br />A function call requires the name of the function followed by a list of actual parameters (or arguments), if any enclosed in parentheses.<br />If a function has no formal parameters in its definition, it cannot have any actual parameters in calls to it.<br />In this case, in a function call, the name of the function must be followed by the function call operator, (). To indicate that it has no parameters<br /> Eg1 : Function that has no parameters<br /> polynomial ()<br />
22. 22. The actual parameters may be expressed as constants, single variables or more complex expressions.<br />Eg2: Function that return value to y<br />y=polynomial(x);<br />Eg3: Function that does not returns anything<br />polynomial (a,b,c)<br />
23. 23. Example Function that Return Value<br />/* determine the largest of three integer quantities*/<br />#include <stdio.h><br />#include <conio.h><br />int maximum (intx,int y)<br />{<br />int z;<br /> z=(x>=y)? x :y ;<br /> return(z);<br />}<br />main()<br />{<br />inta,b,c,d;<br /> /* read the integer quantities*/<br />printf("na=");<br />scanf("%d",&a);<br />printf("nb=");<br />scanf("%d",&b);<br />printf("nc = ");<br />scanf("%d",&c);<br /> /*calculate and display the maximum value*/<br /> d=maximum(a,b);<br />printf("nnmaximum =%d",maximum(c,d));<br />getch();<br /> }<br />
24. 24. Function Prototypes<br />In general, all function in C must be declared<br />But function main, must not be declared.<br />Function prototype consist of<br />A function type<br />A function name<br />A list of function parameters, the list of function parameter types is written as (void) or (). If the function has more than one formal parameter, the parameter types in the list must be saparated by commas.<br />Eg format:<br />function_typefunction_name(parameters); <br />
25. 25. Example of Function Prototype <br />Format: <br />function_typefunction_name(parameters);<br />Example:<br />void print_menu(void);<br />double squared (double number);<br />intget_menu_choice(void);<br />
26. 26. Function prototype can be placed in the source file outside function definition and in a function definition.<br />
27. 27. Outside Function Definition<br />Example of outside function definition:Global prototype<br />If a function prototype is global, any function in the program may use it because of this flexibility. <br />Global prototype will be place after the processor directives and before the definition or function main.<br />
28. 28. Inside Function Definition<br />Example of inside function definition: Local prototype<br />The variables that are declared inside a function are called as local variables<br />Their scope is only within the function in which they are declared <br />These variables cannot be accessed outside the function <br />Local variables exist only till the function terminates its execution<br />The initial values of local variables are garbage values<br />
29. 29. Do and Don’t in Function<br />
30. 30. Passing Arguments to a Function<br />List them in parentheses following the function name.<br />The number of arguments and the type of each arguments must match the parameter in the function header and prototype.<br />Example<br />if a function is defined to take two type intarguments, you must pass it exactly two intarguments.<br />
31. 31. Each argument cab be any valid C expression such as:<br />A constant<br />A variable<br />A mathematical or logical expression or event another function( one with a return value)<br />
32. 32. <ul><li>Example:</li></ul>X=half (third(square(half(y))));<br />How to solve it?<br />The program first calls half(), passing it y as an argument.<br />When execution returns from half(), the program calls square(), passing half()’s return values as the argument.<br />Then, half() is called again, this time with third()’s return values as an argument<br />Finally, half()’s return value is assigned to the variable x.<br />
33. 33. The following is an equivalent piece of code:<br />a= half(y);<br />b=square(a);<br />c= third(b);<br />x= half(c);<br />
34. 34. Recursion <br />The term recursion refers to a situation in which a function calls itself either directly or indirectly.<br />Indirectly recursion:<br />Occurs when one functions and they can be useful in some situations.<br />This type of recursion can be used to calculated the factorial of a number and others situation.<br />
35. 35. /*Demonstrates function recursion. Calculate the factorial of a number*/<br />#include <stdio.h><br />unsigned intf,x;<br />unsigned int factorial(unsigned int a);<br />main()<br />{<br /> puts ("Enter an integer value between 1 and 8:");<br />scanf("%d",&x); <br />if(x>8||x<1) {<br />printf("Only values from 1 to 8 are acceptable!");}<br />else{<br />f=factorial(x);<br />printf("%u factorial equals %un",x,f);} <br /> return 0;<br />}<br />unsigned int factorial (unsigned int a){<br /> if (a==1) <br /> return 1; <br /> else{<br /> a *=factorial(a-1);<br /> return a;<br />} }<br />
36. 36. Others examples..<br />
37. 37. Example – Finding the sum of two numbers using functions ( No parameter passing and no return)<br />#include <stdio.h><br />#include <conio.h><br />void fnSum();<br />int main( intargc, char **argv ) {<br />fnSum();<br />getch();<br /> return 0;<br />}<br />void fnSum() {<br />int iNum1,iNum2,iSum;<br />printf("nEnter the two numbers:");<br />scanf("%d%d",&iNum1,&iNum2);<br />iSum = iNum1 + iNum2;<br />printf("nThe sum is %dn",iSum); <br />} <br />
38. 38. Example – Finding the sum of two numbers using functions ( parameter passing )<br />#include <stdio.h><br />#include <conio.h><br />void fnSum( int iNumber1, int iNumber2);<br />int main( intargc, char **argv ) {<br />int iNumber1,iNumber2;<br />printf("nEnter the two numbers:");<br />scanf("%d%d",&iNumber1,&iNumber2);<br />fnSum(iNumber1,iNumber2);<br />getch();<br /> return 0;<br />}<br />void fnSum(int iNum1,int iNum2){ <br />intiSum;<br />iSum=iNum1 + iNum2;<br />printf("nThe sum is %dn",iSum);<br />}<br />
39. 39. Example – Finding the sum of two numbers using functions ( parameter passing and returning value)<br />#include <stdio.h><br />#include <conio.h><br />intfnSum( int iNumber1, int iNumber2);<br />int main( intargc, char **argv ){<br />int iNumber1,iNumber2,iSum;<br />printf("nEnter the two numbers:");<br />scanf("%d%d",&iNumber1,&iNumber2);<br />iSum = fnSum(iNumber1,iNumber2);<br />printf("nThe sum is %dn",iSum);<br />getch();<br /> return 0;<br />}<br />intfnSum(int iNum1,int iNum2){<br />intiTempSum;<br />iTempSum=iNum1 + iNum2;<br /> return iTempSum;<br />}<br />
40. 40. Simple Example 1<br />#include<stdio.h><br />#include<conio.h><br />int addition (int a, int b)<br />{<br /> int r;<br /> r=a+b;<br /> return (r);<br />}<br />int main ()<br />{<br /> int z;<br /> z = addition (5,3);<br /> printf("The result is %d",z);<br /> getch();<br /> return 0;<br />}<br />
41. 41. #include<stdio.h><br />#include<conio.h><br />int subtraction (int a, int b)<br />{<br /> int r;<br /> r=a-b;<br /> return (r);<br />}<br />int main ()<br />{<br /> int x=5, y=3, z;<br /> z = subtraction (7,2);<br /> printf("The first result is %dn",z);<br /> printf("The second result is %dn",subtraction (7,2));<br /> printf("The third result is %dn",subtraction (x,y));<br /> z= 4 + subtraction (x,y);<br /> printf("The fourth result is %dn",z);<br /> getch();<br /> return 0;<br />}<br />
42. 42. Thank you !<br />38<br />38<br /> | 3,690 | 13,206 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2017-43 | latest | en | 0.740386 |
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Hi! I'm implementing Catmull-Clark subdivision for my terrain as described in the paper "Rapid evaluation of Catmull-Clark surfaces". I created the basis function tables with the program "Subdivide 2.0" exactly as in the paper. It works very well and the tessellation works fine. However, I'm stuck with creating of the basis function tables of derivatives. I need them to compute vertex normals. How do they generate the tables? Is it just sampling the derivatives on the tessellated unit pulse base mesh? I would appreciate some clarification! Thanks very much
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I'm not quite sure what you mean by sampling the derivatives.
It's the same principle as the limit-surface mask which brings, in one final averaging step, the control mesh to the limit surface. The limit tangent and binormal each have their own limit masks which are applied in the same way when generating the basis tables. So I think the answer to your question is yes.
BTW: Because I didn't use subdivide2.0 to generate my tables I noticed that the limit tangent/binormal mask mentioned in "Piecewise Smooth Subdivision Surfaces with Normal Control" in the appendix is incorrect and the correct mask can be found in the source code of subdivide2.0 (quadrule.cpp). I found this out because for valency 4 vertices you're supposed to get...
-1 -4 -1
0 0 0
1 4 1
...or some rotation of that.
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Quote:
Original post by SoiledI'm not quite sure what you mean by sampling the derivatives.
I'll try to explain it in more detail...
What I do now:
First, I create the tesselated base mesh(unit pulse) by subdivide2.0. Then I sample the y-values of the vertices of the patch. These sampled values form the basis function table I later use at run-time.
The idea of sampling the derivatives came from the paper (Rapid evaluation of CC surfaces):
Quote:
Page 4, just under the picture of the base mesh...For all these cases, limit positions and partial derivatives in the two parametric directions were sampled on a 33x33 grid....
I thought I would evaluate the partial derivatives for each of the 33x33 points. Now I don't know if this will work and also what should I sample from the derivatives? The y-component of their vectors?
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Quote:
I thought I would evaluate the partial derivatives for each of the 33x33 points. Now I don't know if this will work and also what should I sample from the derivatives? The y-component of their vectors?
I'd say so. Provide a unit pulse to subdivide2.0 (x=0,y=1,z=0) and ask it to produce a limit tangent and a limit binormal (assuming it can do that?) and take the y-component of results and put in your basis table (33x33 for tangent and 33x33 for binormal).
At run-time when subdividing a real control mesh you need to calculate limit-surface normal from cross product of limit tangent and limit binormal (both calculated using basis tables) - I don't think you can generate a table directly for the limit normal.
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Quote:
Original post by SoiledI'd say so. Provide a unit pulse to subdivide2.0 (x=0,y=1,z=0) and ask it to produce a limit tangent and a limit binormal (assuming it can do that?)
No, subdivide2.0 can't produce the tangent and binormal directly. I thought I would take the unit pulse tesselated by Subdivide2.0 and estabilish the tangents and binormals at each point with something like u=(2, y(x+1, z)-y(x-1, z), 0) v=(0, y(x, z+1)-y(x, z-1), 2) where y(x, z) is the y-coordinate of vertex in the grid at x, z. The y-values of u and v vectors would form the basis table. I don't really know if this would work.
I'd like to produce all basis tables from the unit pulse mesh generated with Subdivide2.0 if possible.
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That's a shame about subdivide2.0 :(
I know it's got the masks for tangent and binormal in there (I looked at the source code) in quadrule.cpp - I guess it just doesn't expose the limit tangent/binormal in the library interface.
What you're suggesting is effectively a limit mask for tangent and binormal that's applied directly to the limit surface (instead of subdivided control mesh). Something like...
0 0 0
-1 0 1
0 0 0
...for tangent and...
0 -1 0
0 0 0
0 1 0
...for the binormal (or vice-versa or some rotation of). Whereas the correct solution is to apply
-1 -4 -1
0 0 0
1 4 1
...to the subdivided control mesh (instead of limit surface) and that's only for valency=4 vertices.
I guess it would work however there's two issues I see, (1) you also need masks for the extraordinary vertices (you've got them for the regular vertices only), (2) the real masks are designed to make the tangent space continuous so by using your own your tangent-space might look non-smooth in places where it's supposed to be smooth. Although it might be good enough for what you need (ie, might look good enough).
If you're after correctness though, since you've already invested in using subdivide2.0 perhaps you might want to look at the source code and try to figure out how to get access to the limit tangent/binormal (even if you have to hack the code a bit). Just a suggestion since the alternative of coding up your own with all the different combinations of crease edges, etc seems daunting. My case was easier because I dissallowed crease edges, etc so the rules become simple enough for me to code them myself.
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The extraordinary vertices are no problem for me. The mesh I'm working on is made of quads only (there is no vertex with other valence than 4). No crease edges. And I really don't care whether the subdivision is correct on the boundary. That all comes from the fact that I will use this algorithm only for my terrain, where I may set these "limitations".
What would you suggest now?
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So you have a regular heightfield terrain ? I thought you were trying to avoid heightmaps ? You can have extraordinary vertices with a quad-only mesh (eg, a cube has valency 3 vertices).
Having said that, it is possible to create overhangs with a regular (valency 4) catmull-clark surface but it's much easier when you allow arbitrary valency because you can then have tunnels, protrusions, etc, and also your cliffs/overhangs don't stretch too much (because you extrude quads where there's an overhang, ie a quad becomes a cube popping out of the surface, instead of just moving vertices).
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Quote:
Original post by SoiledSo you have a regular heightfield terrain ? I thought you were trying to avoid heightmaps ? You can have extraordinary vertices with a quad-only mesh (eg, a cube has valency 3 vertices).Having said that, it is possible to create overhangs with a regular (valency 4) catmull-clark surface but it's much easier when you allow arbitrary valency because you can then have tunnels, protrusions, etc, and also your cliffs/overhangs don't stretch too much (because you extrude quads where there's an overhang, ie a quad becomes a cube popping out of the surface, instead of just moving vertices).
OK, I will perhaps try to implement the basis table generation myself with the help of the subdivide2.0 source code. At first I'll perhaps restrict the vertices to valence 4 as I do now, but later, if all goes well, I'll allow other valences(well, I must admit that you have convinced me of the benefits of this [smile]).
I'll start coding and I will see...
And also, be prepared for another load of questions as soon as I get stuck somewhere in the implementation [grin]
And thanks for your input, the whole thing is much clearer to me now!
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Starting with valency 4 is a good idea.
Are you planning on subdividing on-the-fly as the camera moves around or doing it at pre-compile and paging on-the-fly (or loading whole thing at load-time) ?
Also are you planning on using some form of level-of-detail ?
Reason I ask is because it'll determine how you handle things like hairline cracks appearing between adjacent control quads subdivided to same level due to float imprecision ((a+b)+c not always equal to a+(b+c) for floats) and obvious cracks due to differing subdivision levels.
Quote:
And also, be prepared for another load of questions as soon as I get stuck somewhere in the implementation [grin]
No probs.
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### 4march
Dare to guess what day it is Problem
In one small town, there is a liar who lies on six days of the week. But on the seventh day he always tells the truth. He made the following statements on three successive days:
Day 1: 'I lie on Monday and Tuesday.'
Day 2: 'Today, it's Thursday, Saturday, or Sunday.'
Day 3: 'I lie on Wednesday and Friday.'
What day does the guy tell the truth?
1. Tuesday :)
2. I got this table to work on :
M L
T L
W T
T N
F L
S N
S N
------
LLLT
LLTL
LTLL
The first column is the first letters of days.
L is lie
T is do not lie
N is this day is not today
He tells the truth on Wednesday and we are Wednesday.
I lie on Monday and Thursday is TRUE
Today, it's Thursday, Saturday, or Sunday is FALSE and we are Wednesday, so he is lying.
I lie on Wednesday and Friday is FALSE since he does not lie on Wednesday. | 241 | 871 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.515625 | 4 | CC-MAIN-2016-44 | longest | en | 0.963847 |
http://kineticvisuals.com/eoir-closures-seszpm/a5dedf-how-to-find-identity-element-and-inverse | 1,639,046,109,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964363791.16/warc/CC-MAIN-20211209091917-20211209121917-00514.warc.gz | 37,725,459 | 6,510 | A n × n square matrix with a main diagonal of 1's and all other elements 0's is called the identity matrix I n. If A is a m × n matrix, thenI m A = A and AI n = A. Thus, there can only be one element in Rsatisfying the requirements for the multiplicative identity of the ring R. Problem 16.13, part (b) Suppose that Ris a ring with unity and that a2Ris a unit Make a note that while there exists only one identity for every single element in the group, each element in the group has a different inverse. Every element has an inverse: for every member a of S, there exists a member a −1 such that a ∗ a −1 and a −1 ∗ a are both identical to the identity element. It is called the identity element of the group. Whenever the identity element for an operation is the answer to a problem, then the two items operated on to get that answer are inverses of each other.. More explicitly, let S S S be a set and ∗ * ∗ be a binary operation on S. S. S. Then For example, the operation o on m defined by a o b = a(a2 - 1) + b has three left identity elements 0, 1 and -1, but there exists no right identity element. The operation is associative: if a, b and c are members of S, then (a ∗ b) ∗ c is identical to a ∗ (b ∗ c). You can't name any other number x, such that 5 + x = 0 besides -5. Actually that is what you are looking for to be satisfied when trying to find the inverse of an element, nothing else. The definition in the previous section generalizes the notion of inverse in group relative to the notion of identity. -5 + 5 = 0, so the inverse of -5 is 5. Writing $1/3$ is just a shorthand notation for $1 \cdot 3^{-1}$ in the rational (or real or complex) group; of course you can define it in any other group (like the one you have provided) but using it can be confusing. It's also possible, albeit less obvious, to generalize the notion of an inverse by dropping the identity element but keeping associativity, i.e., in a semigroup.. The identity matrix for is because . One can show that the identity element is unique, and that every element ahas a unique inverse. An identity element in a set is an element that is special with respect to a binary operation on the set: when an identity element is paired with any element via the operation, it returns that element. ... inverse or simply an inverse element. The Identity Matrix This video introduces the identity matrix and illustrates the properties of the identity matrix. The inverse of ais usually denoted a−1, but it depend on the context | for example, if we use the symbol ’+’ as group operator, then −ais used to denote the inverse of a. Such an element is unique (see below). Identity element There exists an element e in G such that, for every a in G, one has e ⋅ a = a and a ⋅ e = a. In fact, if a is the inverse of b, then it must be that b is the inverse of a. Inverses are unique. Example 3.12 Consider the operation ∗ on the set of integers defined by a ∗ b = a + b − 1. Similarly, an element v is a left identity element if v * a = a for all a E A. Is A is a n × n square matrix, then Whenever a set has an identity element with respect to a binary operation on the set, it is then in order to raise the question of inverses. Back in multiplication, you know that 1 is the identity element for multiplication. Inverse element For each a in G, there exists an element b in G such that a ⋅ b = e and b ⋅ a = e, where e is the identity element. This is also true in matrices. In mathematics, an identity element, or neutral element, is a special type of element of a set with respect to a binary operation on that set, which leaves any element of the set unchanged when combined with it.
Compare Ninja Foodi Grill Models, Working Out On 5 Hours Of Sleep, 4 Week Glute Workout Plan Pdf, Rim Rock Trail - Colorado, Hemingway Gritti Palace, Bread And Pastry Questions And Answers, Bayer Tapeworm Dewormer For Dogs Side Effects, Equity Release Calculator Nationwide, Low Calorie Bread, Lemon Pistachio Shortbread Cookies, Utmb Pt Bridge Program, Njit Library Resources, Kim Man Bong Price, | 1,007 | 4,056 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4 | 4 | CC-MAIN-2021-49 | latest | en | 0.934717 |
http://mathhelpforum.com/algebra/209756-how-factored.html | 1,529,580,009,000,000,000 | text/html | crawl-data/CC-MAIN-2018-26/segments/1529267864139.22/warc/CC-MAIN-20180621094633-20180621114633-00282.warc.gz | 210,150,258 | 9,910 | I can't figure out how this is true. how was this factored?
Originally Posted by kingsolomonsgrave
I can't figure out how this is true. how was this factored?
$\displaystyle x^2-x-6=(x+2)(x-3)$.
what happened to the x in the numerator?
In order to subtract the two functions they must first have the same denominator.
If you multiply 4/(x-3) by x+2 you should get (4x+8) / (x^2-x-6)'
Then multiply 1/(x+2) by x-3 and your answer should be (x-3) / (x^2-6).
You then combine the numerator by like terms all over the denominator. It should look like (4x-x+8-(-3) / (x^2-x-6)
4x - x = 3x and 8 - (-3) = 11 (Using skip change change) | 208 | 634 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.890625 | 4 | CC-MAIN-2018-26 | latest | en | 0.946203 |
https://www.diychatroom.com/threads/generator-output-and-transfer-switch.110486/#post-684647 | 1,656,927,392,000,000,000 | text/html | crawl-data/CC-MAIN-2022-27/segments/1656104364750.74/warc/CC-MAIN-20220704080332-20220704110332-00288.warc.gz | 763,902,244 | 19,707 | 1 - 4 of 4 Posts
#### aquasport17
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Hello,
I have a 120/240V generator which is rated for 7200W continuous, and has one L14-30 for 120/240V, one L5-30 for 120V, and one 20A duplex. The L14-30 connection has two 30A breakers on the generator panel - I assume one for each pole. So assuming this generator can push 60A @ 120V or 30A at 240V (or some combination), what is the best way to maximize the connection to the transfer switch? In other words, is the L14-30 a 30A feed or a 60A feed?
My issue seems to be that it seems silly to have a 30A transfer switch fed by a 30A generator feed when my generator is capable of [email protected]
It seems that I would use a switch like the GenTran 3026 or 3028, but how do I get the full 60A if needed?
Thanks,
Jeff
#### AandPDan
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You're confusing watts and amps.
30 amp X 240 volts = 7200 watts.
60 amp X 120 volts = 7200 watts.
Between each hot leg and neutral (120 volts) on your generator you can draw 30 amps which is 3600 watts. Between both hot legs (240 volts) you can draw 30 amps, 7200 watts.
What you do is use the proper L14-30 connector to your generator panel. The panel distributes the power onto two buses. You can then put any combination of 120 volt and 240 volt devices, up to 7200 watts load. It is important that you try to balance the load equally on the panel. You can't for example, draw 4000 watts on one leg and 3200 on the other. You'll trip the generator breaker.
If you look at the Gentran 3026 for example, the watt meters are to help you balance the load. Note: they only read up to 3750 watts per leg, they don't read amps.
#### aquasport17
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Okay I sort of figured that - but didn't realize I needed to balance the load. Balancing the load seems tricky to me since the appliances in use, and 240V well pump cycling could vary at any time. But I get the concept now. how close does it need to be?
So assuming I wire as you stated, are the remaining receptacles on the generator available for use so long as I don't exceed rated wattage? I would assume they are.
Thanks!
#### AandPDan
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The closer you can balance the better for the generator.
The well pump, being 240 volt is already going to be balanced. | 644 | 2,389 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.59375 | 3 | CC-MAIN-2022-27 | latest | en | 0.911059 |
https://www.physicsforums.com/threads/heat-engines-and-efficiency.148044/ | 1,603,583,716,000,000,000 | text/html | crawl-data/CC-MAIN-2020-45/segments/1603107885059.50/warc/CC-MAIN-20201024223210-20201025013210-00494.warc.gz | 861,831,577 | 17,249 | # Heat Engines and efficiency
## Homework Statement
Suppose that two heat engines are connected in a series, such that the heat exhaust of the first engine is used as the heat input of the second (attached diagram below). The efficiencies of the engines are e1 and e2, respectively. Show that the net efficiency of the combination is given by:
e(net)= e1 + (1-e1)e2
## Homework Equations
e(max)= 1 - Tc/Th
e= w/Qh = Qh-Qc/Qh = 1 - Qc/Qh
Qc/Qh = Tc/Th
## The Attempt at a Solution
I broke up the diagram into two free-body diagrams allowing me to for solve e1 and e2:
e1 = 1-Th/Tm (Am I allowed to apply Qc/Qh = Tc/Th into 1 - Qc/Qh ?[/B])
and
e2= 1-Tm/Tc
Applying e1 and e2 in the given equation:
e(net)= e1 + (1-e1)e2
= (1-Th/Tm) + [1-(1-Th/Tm)](1-Tm/Tc)
which leaves
=1-Tm/Tc (The total efficiency of the engine equals to only the second engine because all the heat input eventually ends up there? )
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OlderDan
Homework Helper
I think you want the starting point for e(net) to be the ratio of total work (w_t) to Qh. Instead of starting with the relationship you are trying to prove, start with the fundamental definition of efficiency
e(net) = w_t/Qh = (w1 + w2)/Qh
some of your assumptions appear really bad to me. You are not working with Carnot engines are you?
andrevdh
Homework Helper
The thermal efficiency of a heat engine is a measure of converting the "input heat" to the engine to useful mechanical work
$$\varepsilon = \frac{W}{Q_H}$$
if the engines are in series the $$Q_{C1}$$ will become the $$Q_{H2}$$.
Note that the thermal efficiency of a heat engine would be 100% if $$Q_C = 0$$.
#### Attachments
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Last edited:
The thermal efficiency of a heat engine is a measure of converting the "input heat" to the engine to useful mechanical work
$$\varepsilon = \frac{W}{Q_H}$$
if the engines are in series the $$Q_{C1}$$ will become the $$Q_{H2}$$.
Note that the thermal efficiency of a heat engine would be 100% if $$Q_C = 0$$.
And figure out work by substituting W= Qh-Qc?
Does Qm play a role in this or does it not effect the work of the heat engine?
some of your assumptions appear really bad to me. You are not working with Carnot engines are you?
No, I guess I'm working with heat pumps. I see now that I should've not used e(max) since that would be used when figuring out the effieciency of a Carnot engine...
OlderDan
Homework Helper
And figure out work by substituting W= Qh-Qc?
Does Qm play a role in this or does it not effect the work of the heat engine?
Qm does play a role. Qm is the heat output of the first engine and the heat input of the second engine. Write the two work contributions in terms of the three heat values, and write the individual efficiencies in terms of the respective work to heat-input ratios.
andrevdh
Homework Helper
And figure out work by substituting W= Qh-Qc?
Does Qm play a role in this or does it not effect the work of the heat engine?
Yes you can calculate the work done by a heat engine this way. It will apply for both engines. What it means is that the engine converts the input heat (energy) to usefull work and some left-over heat. A heat engine will always need to shed some left-over heat since it can only function between a hot and cold reservior. That is it can only function if heat can flow between a hot and cold region.
In the case of this problem the heat output of engine one becomes the heat input of engine 2:
$$Q_m = Q_{C1} = Q_{H2}$$
We are therefore assuming that no heat escapes from the mid reservior. It just acts to transfer the heat from the first to the second engine.
Last edited:
I think you should use
enet=(WA+WB)QH1 | 1,018 | 3,751 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.859375 | 4 | CC-MAIN-2020-45 | longest | en | 0.897963 |
https://onlinejudge.org/board/viewtopic.php?f=21&t=7554&p=101005 | 1,597,078,032,000,000,000 | text/html | crawl-data/CC-MAIN-2020-34/segments/1596439736057.87/warc/CC-MAIN-20200810145103-20200810175103-00505.warc.gz | 356,704,489 | 8,881 | ## 802 - Lead or Gold
Moderator: Board moderators
mf
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### Re: 802 - Lead or Gold
Oh, now I see. I knew that trick, I just forgot its name.
But I think its use is totally unnecessary here. In fact, you don't even have to add any artificial variables. Here's LP formulation that I've used in my simplex solution:
Code: Select all
minimize (A+B+C) - \sum_{i=1}^n (a[i] + b[i] + c[i]) x[i]
subject to
\sum_{i=1}^n a[i] x[i] <= A
\sum_{i=1}^n b[i] x[i] <= B
\sum_{i=1}^n c[i] x[i] <= C
x[i] >= 0
That is, I just turn strict equalities into inequalities and try to minimize the sum of residuals, starting from zero basic feasible solution. The original problem is feasible if that sum reaches zero.
serur
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### Re: 802 - Lead or Gold
Also I noticed that in CLR -- without Stein -- they give no description of simplex algo.
I guess you have an English version of CLRS that you brought from one of your innumerable
trips around the world
If there is ever a war between men and machines, it is easy to guess who will start it (c) Arthur Clarke
mf
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### Re: 802 - Lead or Gold
Yes, they've added a chapter on linear programming only in 2nd edition of the book.
By the way, the book's third edition is coming this fall!
serur
A great helper
Posts: 251
Joined: Thu Feb 23, 2006 11:30 pm
### Re: 802 - Lead or Gold
I got AC with your LP formulation, as well. Of course, your approach is neat:
your just sum the residuals and change the order of summation and get the objective function (just dropping
a constant summand).
If there is ever a war between men and machines, it is easy to guess who will start it (c) Arthur Clarke
tryit1
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Posts: 119
Joined: Sun Aug 24, 2008 9:09 pm
### Re: 802 - Lead or Gold
i got AC on 10089 but this problem is giving me trouble when i've vectors like (0,0,q) or (0,q,r).
If all were positive i could convert it to 10089 by multiplying the lcm of (p,q,r) to make the ratio equations equal. But this zero is messing the whole thing. Any tricks , can we add newEPS=1e-3 in place of 0 and actual EPS=1e-9. but adding newEPS to each quantity of that will change the ratio and possibly solution.
mf
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Joined: Mon Feb 28, 2005 4:51 am
Location: Zürich, Switzerland
Contact:
### Re: 802 - Lead or Gold
i got AC on 10089 but this problem is giving me trouble when i've vectors like (0,0,q) or (0,q,r).
You're talking about weights in the desired (target) mixture? Well, if some component is zero, it's just the same problem with one less dimension. If target mixture is (a, b, 0), then the only mixtures you may take are also of form (x, y, 0). So it doesn't hurt to remove all mixtures which are not of this form, drop this 0 coordinate from all vectors, and you get a 2D problem.
(Note: I haven't actually tried to solve this problem by reducing to 10089, so maybe I'm saying here something wrong. I've got accepted with simplex method, and a method based on projection and convex hull.) | 928 | 3,192 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.171875 | 3 | CC-MAIN-2020-34 | latest | en | 0.9144 |
http://www.evi.com/q/convert_60_kilos_to_pounds | 1,418,881,187,000,000,000 | text/html | crawl-data/CC-MAIN-2014-52/segments/1418802765616.69/warc/CC-MAIN-20141217075245-00127-ip-10-231-17-201.ec2.internal.warc.gz | 641,071,915 | 9,008 | # convert 60 kilos to pounds
• 60 kilograms is 132.28 pounds.
• tk10publ tk10ncanl
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• 60 kg + how many pounds | 5,752 | 17,398 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.65625 | 3 | CC-MAIN-2014-52 | latest | en | 0.904898 |
http://dict.cnki.net/h_54471593000.html | 1,582,031,799,000,000,000 | text/html | crawl-data/CC-MAIN-2020-10/segments/1581875143695.67/warc/CC-MAIN-20200218120100-20200218150100-00080.warc.gz | 45,497,736 | 9,450 | 全文文献 工具书 数字 学术定义 翻译助手 学术趋势 更多
standard-form 的翻译结果: 查询用时：0.054秒
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standard-form 标准型(2)
标准型
INVERSELY SOLVING THE PARAMETER AND INNER WEIGHT OF A STANDARD-FORM RATIONAL QUADRATIC BEZIER CURVE 反求标准型有理二次Bezier曲线的参数与内权因子 短句来源 Given three non-collinear control vertices and a point on a conic inside the convex hull of the vertices, a standard-form rational quadratic Bezier curve can be determined. 给定不共线的三个控制顶点和位于这些顶点的凸包内在二次曲线上的一点，可以决定一条标准型有理二次Bezier曲线。 短句来源
“standard-form”译为未确定词的双语例句
By discussing the method of eliminating and simplifying the restriction of force and moment in this model, the paper develops the standard-form equation set suitable for programming in a computer, and programs it. 分析了此种末敏弹系统的受力情况 ,建立了系统的二体运动模型 ,并详细讨论了此模型中约束力及约束力矩的消元化简方法 ,推导出了适于上机编程的标准形式的方程组 ,并用其编制了诸元计算程序。 短句来源
相似匹配句对
Which is the standard? 采用什么标准评价? 短句来源 Standard 标准 短句来源 The standard waveform was given. 本工作给出标准波形,以利于缺陷的定位。 短句来源 the standard and interface of information; 信息标准及接口； 短句来源 T-DMB Standard and Its Application 地面数字多媒体广播(T-DMB)标准与应用 短句来源
我想查看译文中含有：的双语例句
standard-form
Additionally, the FAM was successful in predicting several gross indices of psychopathology as well as high-point elevations on the standard-form MMPI. The purpose of this paper is to analyze Tikhonov regularization in general form by means of generalized SVD (GSVD) in the same spirit as SVD is used to analyze standard-form regularization. This investigation suggests the truncated SVD as a favorable alternative to standard-form regularization in cases of ill-conditioned matrices with well-determined numerical rank. Der Taucher erreicht in der ?Standard"-Form eine Me?genauigkeit von 10-3μl, als Kapillartaucher eine Me?genauigkeit von 10-5μl. The primal projective standard-form variant of Karmarkar's algorithm for linear programming is applied to the duals of a sequence of linear programming relaxations of the combinatorial optimization problem. 更多 | 648 | 1,962 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.53125 | 3 | CC-MAIN-2020-10 | latest | en | 0.42754 |
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The divisors of 10 illustrated with Cuisenaire rods: 1, 2, 5, and 10 (Wikipedia).
Euler Problem 12 takes us to the realm of triangular numbers and proper divisors.
The image on the left shows a hands-on method to visualise the number of divisors of an integer. Cuisenaire rods are learning aids that provide a hands-on way to explore mathematics.
## Euler Problem 12 Definition
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be $1 + 2 + 3 + 4 + 5 + 6 + 7 = 28$. The first ten terms would be: $1, 3, 6, 10, 15, 21, 28, 36, 45, 55, \ldots$ Let us list the factors of the first seven triangle numbers:
1: 1
3: 1, 3
6: 1, 2, 3, 6
10: 1, 2, 5, 10
15: 1, 3, 5, 15
21: 1, 3, 7 ,21
28: 1, 2, 4, 7, 14, 28
We can see that 28 is the first triangle number to have over five divisors. What is the value of the first triangle number to have over five hundred divisors?
## Solution
Vishal Kataria explains a simple method to determine the number of divisors using prime factorization as explained by in his video below. The prime factorization of $n$ is given by:
$n = p^{\alpha_1}_1 \times p^{\alpha_2}_2 \times p^{\alpha_k}_k$
The number of proper divisors is:
$d = (\alpha_1 + 1) (\alpha_2 + 1) \ldots (\alpha_k + 1)$
The code reuses the prime factorisation function developed for Euler Problem 3. This function results in a vector of all prime factors, e.g. the prime factors of 28 are 2, 2 and 7.
The code to solve this problem determines the values for alpha using the run length function. This function counts the number of times each element in a sequence is repeated. The outcome of this function is a vector of the values and the number of times each is repeated. The prime factors of 28 are 2 and 7 and their run lengths are 2 and 1. The number of divisors can now be determined.
$28 = 2^2 \times 7^1$
$d = (2+1)(1+1) = 6$
The code to solve Euler Problem 12 is shown below. The loop continues until it finds a triangular number with 500 divisors. The first two lines increment the index and create the next triangular number. The third line in the loop determines the number of times each factor is repeated (the run lengths). The last line calculates the number of divisors using the above-mentioned formula.
i <- 0
divisors <- 0
while (divisors < 500) {
i <- i + 1
triangle <- (i * (i+1)) / 2
pf <- prime.factors(triangle)
alpha <- rle(pf)
divisors <- prod(alpha\$lengths+1)
} | 796 | 2,718 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 14, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 4.78125 | 5 | CC-MAIN-2020-50 | latest | en | 0.868097 |
https://ask.sagemath.org/users/156/pang/?sort=recent | 1,547,686,754,000,000,000 | text/html | crawl-data/CC-MAIN-2019-04/segments/1547583658662.31/warc/CC-MAIN-20190117000104-20190117022104-00028.warc.gz | 423,608,850 | 14,053 | 2018-11-08 12:48:48 -0600 received badge ● Famous Question (source) 2018-07-03 09:57:27 -0600 answered a question Getting convex hull of a set of points and plotting the polygon Also P1 = Polyhedron(vertices = [[-5,2], [4,4], [3,0], [1,0], [2,-4], [-3,-1], [-5,-3]]) P1.plot() 2018-01-31 10:00:25 -0600 commented answer Can I generate 3D plots in isometric projection? Very promising! I do see the menu when I right click on the plot, but the menu does not work, and the submenu does not display... 2018-01-29 08:18:14 -0600 commented question Is there CUDA support in Sage? You can probably "sage -pip" install pytorch, theano, caffe, tensorflow, etc, within the sage environment. pytorch for instance installs very smoothly: http://pytorch.org/ ... but that wouldn't be very integrated. I don't think you should expect symbolic integration or group theory in the GPU for some time. 2018-01-29 08:08:22 -0600 commented question How can you change the viewpoint of perspective (to infinity) in a 3d plot? That's related to the question: https://ask.sagemath.org/question/384... I don't know any simple way... 2018-01-28 16:35:35 -0600 commented answer Math document containing some Sage calculations Is TeXmacs still alive? I loved that thing, but it is no longer in the ubuntu repos and I cannot install it from source. It looks abandoned. 2018-01-28 16:26:53 -0600 commented answer Math document containing some Sage calculations Note that you must install rst2ipynb: sage -i rst2pynb Then you can use it sage -rst2ipynb source.rst out.ipynb 2017-08-08 13:58:36 -0600 asked a question Can I generate 3D plots in isometric projection? I'd like to show some plots in isometric projection. For certain plots, it's not hard to project the lines manually and do a 2d plot, but ideally, I'd like to rotate the plot with the mouse, and be able to show any 3d plot in isometric projection: do you know a way? 2017-04-28 10:38:04 -0600 answered a question Animation doesn't work This code will display the animation embeded in a jupyter notebook my_path = 'my.gif' a.gif(savefile=my_path) html(''%my_path) Warning: this saves the animation in the same folder where the notebook is. 2017-03-13 11:02:13 -0600 received badge ● Nice Question (source) 2017-01-26 12:21:42 -0600 answered a question How does one use the qhull optional package? I'll expand on tmonteil's last comment: there is an interface to qhull included on scipy! For example, this code computes the hull of a random set of 2D points: import numpy as np from scipy.spatial import ConvexHull points = np.random.rand(30, 2) # 30 random points in 2-D hull = ConvexHull(points) and this code plots it: my_plot = point2d(points) for simplex in hull.simplices: my_plot += line2d([points[v,:] for v in simplex]) my_plot While this code computes the hull of a random set of 3D points: import numpy as np from scipy.spatial import ConvexHull points = np.random.rand(30, 3) # 30 random points in 3-D hull = ConvexHull(points) and this code plots it: my_plot = point3d(points) for simplex in hull.simplices: my_plot += line3d([points[simplex[-1],:]]+[points[v,:] for v in simplex]) my_plot 2017-01-26 11:51:50 -0600 received badge ● Popular Question (source) 2016-12-13 04:01:43 -0600 received badge ● Notable Question (source) 2016-11-21 04:50:34 -0600 received badge ● Necromancer (source) 2016-11-02 02:30:34 -0600 received badge ● Necromancer (source) 2016-10-26 12:15:41 -0600 commented question Using sage for slideshows/presentations This thread is a bit old. These days the jupyter notebook has some reveal.js integration available when we use sage --notebook=jupyter. However it is not trivial to use. Should I ask for a sage tutorial for this feature here, or in a new question? 2016-10-10 11:15:39 -0600 received badge ● Popular Question (source) 2016-07-13 14:47:15 -0600 received badge ● Famous Question (source) 2016-05-12 08:22:08 -0600 commented question qepcad failing to replicate examples Can we mark this question as outdated? qepcad installs just fine, and I can think of no reason why anyone would want to read it now. 2016-02-21 18:14:59 -0600 received badge ● Famous Question (source) 2016-02-21 18:14:59 -0600 received badge ● Popular Question (source) 2016-02-21 18:14:59 -0600 received badge ● Notable Question (source) 2015-12-01 05:30:35 -0600 commented question elementary matrices for elementary row operations I don't understand the question. But this might be related: http://wiki.sagemath.org/interact/linear_algebra#The_Gauss-Jordan_method_for_inverting_a_matrix 2015-12-01 05:16:28 -0600 commented answer Reformulation of a semialgebraic set, without quantifiers I had problems installing sage from source, but I finally did it, and qepcap installed without problems. Thanks again, Thierry! 2015-11-24 04:20:21 -0600 asked a question Convert sagews into sws A colleague asked if it is possible to convert from sagews into sws, so that he can use the notebooks that his students create on Sage Cloud. I have found a sage-cloud thread which mentions this quick script. My motivations for writing this questions are: I thought ask.sagemath.org should have an answer to that question There will be better ways in the future There may be better ways now! So feel free to contribute and keep this question up to date! PD: For the record, sws files are compressed tar files that you can decompress with tar jxf %F. They contain an html file with code and comments, a pickle file with some python objects, a data folder with the attachments, and a cells folder with the images contained in the output of the cells. On the other hand, sagews files are plain text files with no attachments. 2015-11-15 03:33:19 -0600 received badge ● Popular Question (source) 2015-10-28 09:33:01 -0600 commented answer Reformulation of a semialgebraic set, without quantifiers I do have a trac account. I'll try first with a fresh sage install. My current sage install is old, and has been updated several times. I'll need a few days: if it works I'll comment here, and if it doesn't I'll fill a bug. Regards! 2015-10-28 06:27:21 -0600 commented question Reformulation of a semialgebraic set, without quantifiers For the record: I couldn't get qepcad to make a projection of codimension 2 using even 4 points, but I did get some significant results by restricting to a three dimensional space, so that qepcad only projects along lines. 2015-10-28 05:06:16 -0600 commented answer substitute expression instead of formal function symbol f.operator() is actually foo. 2015-10-28 04:57:48 -0600 answered a question Difference of performance bewteen a core i5 and core i7? If you plan to take exams and/or work with novices, I'd say it's worth it. It's not easy to jail each process and give it limited resources. If X students write an infinite loop, you server will collapse. The more threads and RAM you have, the greater X can get. If you divide the available RAM into the total number of students, you're wasting resources, it's better to estimate the number of students that can write an infinite loop that appends to a list, and not hit interrupt and/or restart on time. In my experience, the probability of a student blowing the server is higher on a exam than on a regular class, and it's higher in combinatorics or number theory, average on linear algebra, lower on calculus (unless the exercise involves symbolic integration of non-trivial functions). 2015-10-28 04:45:07 -0600 answered a question How can i set up a server that uses several computers? Old question, but still: have you looked at the code from the Sage cloud: https://github.com/sagemathinc/smc The Sage cloud seems to be actually easier to install in several machines than in only one >:O ! 2015-10-28 04:36:28 -0600 commented answer Reformulation of a semialgebraic set, without quantifiers I have bug traces etc: should I open a ticket? 2015-10-28 04:26:32 -0600 marked best answer Reformulation of a semialgebraic set, without quantifiers According to wikipedia: The Tarski–Seidenberg theorem states that a set in (n + 1)-dimensional space defined by polynomial equations and inequalities can be projected down onto n-dimensional space, and the resulting set is still definable in terms of polynomial identities and inequalities. That's great, and my question is if it can be done in practice, in a particular case. I have read that the original Tarski–Seidenberg algorithm is of little practical use, but maybe my problem is tractable with another algorithm: We have a finite set of points in C^2 (z_i,w_i), and I want to project onto the first C coordinate the intersection of the cones {(z,w): |z-z_i|>|w-w_i|}. It would be awesome, for example for plotting the set, if I could actually get the polynomial inequalities that define the projection. Getting only the higher dimensional strata is fine (an open set in C). Any idea? Thanks in advance. 2015-10-28 04:26:30 -0600 commented answer Reformulation of a semialgebraic set, without quantifiers I get a "RuntimeError: unable to start QEPCAD". I upgraded to sage 6.9 with ./sage -upgrade and I installed qepcad with ./sage -i qepcad The first upgrade reported errors when building the html docs, but nothing else, and everything seems to work fine. The installation of qepcad produced no errors (I think), and there are qepcad and saclib folders in sage/pkgs. 2015-10-27 10:31:54 -0600 received badge ● Associate Editor (source) 2015-10-27 10:31:11 -0600 commented answer Reformulation of a semialgebraic set, without quantifiers It just finished installing. Thanks for taking care of this packege, Thierry! 2015-10-27 10:19:49 -0600 commented answer Reformulation of a semialgebraic set, without quantifiers I downloaded the program and I'm calling it from the command line. I can't get it to work on any non-trivial example (it does work with all the examples from its documentation). The output is: Failure occurred in: GCSI (final check) Reason for the failure: Too few cells reclaimed. 2015-10-27 08:54:15 -0600 received badge ● Nice Answer (source) 2015-10-27 05:35:00 -0600 edited answer Reformulation of a semialgebraic set, without quantifiers I found the optional package qepcad: http://flask.sagenb.org/src/interface... mentioned in this site: http://mathoverflow.net/questions/159... 2015-10-27 04:31:05 -0600 commented question Similar command to Mathematica's Reduce[]? I just found about the optional package qepcad: maybe it works for you! | 2,805 | 10,454 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 2.578125 | 3 | CC-MAIN-2019-04 | longest | en | 0.896501 |
https://socratic.org/questions/jesus-purchased-cookies-for-a-party-he-bought-11-cookies-for-51-each-and-12-cook | 1,638,278,075,000,000,000 | text/html | crawl-data/CC-MAIN-2021-49/segments/1637964358973.70/warc/CC-MAIN-20211130110936-20211130140936-00431.warc.gz | 590,609,669 | 6,154 | # Jesus purchased cookies for a party. He bought 11 cookies for $.51 each and 12 cookies for$.62. How much money did he spend on cookies?
$13.05$ spent altogether
$11$ cookies @ $0.51 means: 11 xx$0.51 = $5.61 $12$cookies@ $0.62 means: 12 xx $0.62 =$7.44
In total he spent: $5.61+$7.44 = \$13.05 | 107 | 296 | {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 8, "mathjax_inline_tex": 1, "mathjax_display_tex": 0, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.9375 | 4 | CC-MAIN-2021-49 | latest | en | 0.949847 |
https://mytabsapp.com/methodology/best-answer-what-is-velocity-and-capacity-in-scrum.html | 1,632,626,029,000,000,000 | text/html | crawl-data/CC-MAIN-2021-39/segments/1631780057796.87/warc/CC-MAIN-20210926022920-20210926052920-00633.warc.gz | 454,662,022 | 19,752 | # Best answer: What is velocity and capacity in Scrum?
Contents
Velocity is based on actual points completed, which is typically an average of all previous sprints. Velocity is used to plan how many product backlog items the team should bring into the next sprint. Capacity is how much availability the team has for the sprint.
## How do you find velocity and capacity in Scrum?
Take the amount of story points your team completed in three past sprints, add them together, and divide by three (the amount of sprints). That average is your basic velocity. The more sprints you add to your velocity measurement, the more accurate your average.
## What does velocity mean in Scrum?
Velocity is an indication of the average amount of Product Backlog turned into an Increment of product during a Sprint by a Scrum Team, tracked by the Development Team for use within the Scrum Team. There is no such thing as a Good Velocity or a Bad Velocity.
## How is velocity used in Scrum?
Velocity is a measure of the amount of work a Team can tackle during a single Sprint and is the key metric in Scrum. Velocity is calculated at the end of the Sprint by totaling the Points for all fully completed User Stories.
## Why is velocity important in Scrum?
In Scrum, velocity help you to understand how long it will take your team to complete the product backlog. … It will be more accurately forecasting of how many stories a Team can do in a Sprint. For forecasting purposes the average of the last three or four Sprint’s Velocity should be used.
## How do we measure velocity?
Velocity (v) is a vector quantity that measures displacement (or change in position, Δs) over the change in time (Δt), represented by the equation v = Δs/Δt. Speed (or rate, r) is a scalar quantity that measures the distance traveled (d) over the change in time (Δt), represented by the equation r = d/Δt.
## What is velocity in agile with example?
Simply, velocity is the amount of work your team gets through in a set amount of time. As with the burndown charts that I talked about in my last post, velocity can be measured in person-hours, number of tasks, story points, or whatever other unit of measurement you use for estimating work.
## How do you increase velocity in Scrum?
5 Ways to Improve Sprint Velocity
1. Use Metrics Responsibly. You should not try to compare velocities across teams. …
2. Focus on Increasing Quality. Higher quality work can reduce the need to revise or fix work later, increasing productivity. …
4. Promote Focus and Consistency. …
5. Embrace Cross-Training.
## How is velocity calculated in Jira?
Velocity is calculated by taking the average of the total completed estimates over the last several sprints. So in the chart above, the team’s velocity is (17.5 + 13.5 + 38.5 + 18 + 33 + 28) / 6 = 24.75 (we’ve ignored the zero story point sprint).
## What is the difference between speed and velocity?
The reason is simple. Speed is the time rate at which an object is moving along a path, while velocity is the rate and direction of an object’s movement. Put another way, speed is a scalar value, while velocity is a vector.
## Who creates user stories in agile?
Anyone can write user stories. It’s the product owner’s responsibility to make sure a product backlog of agile user stories exists, but that doesn’t mean that the product owner is the one who writes them. Over the course of a good agile project, you should expect to have user story examples written by each team member.
## What is Sprint Backlog?
A sprint backlog is the set of items that a cross-functional product team selects from its product backlog to work on during the upcoming sprint. Typically the team will agree on these items during its sprint planning session. In fact, the sprint backlog represents the primary output of sprint planning.
## What are Scrum metrics?
The following metrics can help measure the work done by scrum teams and value delivered to customers:
• Sprint Goal Success. A Sprint Goal is an optional part of the scrum framework. …
• Escaped Defects and Defect Density. …
• Team Velocity. …
• Sprint Burndown. …
• Time to Market. …
• ROI. …
• Capital Redeployment. …
• Customer Satisfaction.
## How do you calculate velocity in agile?
How do I calculate the velocity for my agile team? Divide the number of backlog items or user story points that’s been delivered during the course of several sprints by the total number of days in those sprints.
IT IS INTERESTING: Is backlog grooming a scrum ceremony?
## What is a velocity in agile?
Velocity in Agile is a simple calculation measuring units of work completed in a given timeframe. Units of work can be measured in several ways, including engineer hours, user stories, or story points. … For example, to track Agile velocity, most Scrum teams measure the number of user points in a given sprint.
## What is definition of velocity?
The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. … Velocity is a physical vector quantity; both magnitude and direction are needed to define it. | 1,107 | 5,133 | {"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0} | 3.25 | 3 | CC-MAIN-2021-39 | latest | en | 0.946144 |
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