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Time course experiment with limma voom
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Entering edit mode
@mahesmuniandy-7955
Last seen 1 day ago
Helsinki

I am working on a medication switch-control comparison (RNA sequence data, human tissue) and the results have confused me somewhat. I wonder if anyone can help me make sense of my findings.

I have data for 30 people (half of them have been on the same medication they have taken for years, half of them have had their medication switched to a new one). For each one of them, I have RNA seq data at baseline and 6 months after medication switch/continued medication. I have performed analysis with limma-voom and find that there is a 30% overlap (significantly different genes) between people who have stayed on the same medication and people who have switched medication. Top 20 genes (sorted by p-value) have 80% overlap – fold changes (LFC more than 1) that are quite similar, in the same direction. The clinician has advised me that the results are not possible to explain in any biological sense - nor with any time effect. We started thinking that there could be plate effect (technical effect) in the sequencing platform and found that the genes that are differentially expressed are also correlating with the plate position (correlations as high as 0.8). I can’t adjust for plate position in my model because my phenotype is also correlating with plate position; both sample type (medication switch, control) and visit (baseline, 6 months). My question is on how to remove this plate effect when I have such a strong correlation to phenotype. I fear I now have data I can’t work with - any advice is appreciated.

Mahes Muniandy (University of Helsinki)

design <- model.matrix(~ 0 + TS +  Age + Sex + Smoking  + BMI )
v <- voom(dge_AT, design)
corfit <- duplicateCorrelation(v,design,block=ID)
v <- voom(dge_AT,design,block= ID,correlation=corfit$consensus) fit <- lmFit(v,design,block=ID,correlation=corfit$consensus)
cm <- makeContrasts(
ctrl = TScontrol_V2-TScontrol_V1,
treat = TStreatment_change _V2- TStreatment_change _V1,
difference = ((TStreatment_change _V2- TStreatment_change _V1)-( TScontrol_V2-TScontrol_V1
, levels=design)

limma • 158 views
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Entering edit mode

I've never known plate position to be an important factor in an RNA-seq experiment. What sequencing technology are you using and how does plate position enter into it?

How have you correlated genes with plate position? It isn't obvious to me what you mean by that.

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Illumina Stranded total RNA prep (with ribo depletion) was the platform used. The top genes that were different between time points in both treatment and control were mostly mt-DNA genes. Same fold changes in both conditions. Many other genes also behaved similarly in both conditions. Because we couldn’t find a biological explanation, we wondered if the platform caused a huge technical effect. Initially, we thought that maybe the ribo depletion step did not act on all samples in the same manner. We plotted the mt-genes against the rRNA counts and didn’t see any patterns. Then we plotted the mt-DNA genes against the plate column position (10 columns) and saw a linear relationship – the mt-DNA gene counts increased steadily as the column number progressed from left to right. We then did Spearman correlation to correlate the column number with all genes. Out of 20,000 genes, 4,000 correlated significantly with column position.

Samples on the plate Columns 1-4 : baseline (both people who stayed on the same medication and people who switched medication) Columns 5-8 : 6 months (both people who stayed on the same medication and people who switched medication) Columns 9-10 : people without the disease.

For the top genes, there was steady increase in gene counts from baseline to 6 months to people without the disease. I didn’t plot all the genes to check, just the top 10.

Any ideas or thoughts on how I might approach this is hugely appreciated. Thank you so much for your time.

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I asked you for information about the sequencing technology. You've told me the library preparation protocol but not the sequencing technology. I still don't know how you have sequenced the samples or for what purpose you use a plate with 10 columns. I do not know of any sequencing technology with 10 plate columns.

Regarding the correlations, you would need to adjust read counts for sequencing depth before correlating with plate position. Perhaps you did that, but you don't mention doing so.

Problems ribo depletion would decrease counts sizes and statistical power but would not bias the DE analysis for most genes. Even if ribo depletion was inconsistent between samples, it would not seriously affect a limma analysis because TMM normalization would correct for unequal ribo depletion.

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Entering edit mode
@gordon-smyth
Last seen 1 hour ago
WEHI, Melbourne, Australia

Your question relates to a delicate research question that is far beyond what could be addressed in an online forum like this. If you're not a bioinformatics specialist yourself, then I recommend that you find a good statistical bioinformatician to collaborate with to explore whether your analysis questions can be solved. If the analysis can be improved then the bioinformatician would be a co-author of the final paper. If not, then they might be able to recommend how to redesign your experiment in the future.

I am still not clear for what purpose you are using a plate with 10 columns. However, if you have confounded the treatment groups in your experiment with plate position, and plate position has a strong technical effect, then it may not be possible to rescue your experiment.

Having said that, I would personally not analyse a balanced RNA-seq experiment like this using duplicateCorrelation. I would prefer to include patient ID in the linear model and estimate separate treatment effects for cases and active treatments. Then it would not be necessary to adjust for any patient characteristics like age, sex, smoking or BMI. Such an approach would not eliminate a plate effect, if one exists, but it would be more robust to baseline differences between the subjects.


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# self doubt

87 views

if the flip flops are going in these states that is starts from 0 to 5 then 4 and then again 5 .Then what is the mod of the circuit??

edited
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i am getting 3
0

complete question is this

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its mode 2

counter itself mean like 3bit up counter 0-1-2-3-4-5-6-7-0 ....
0

it is initially in  0 state then it goes to 5 and then  to 4 and then to 5

so i thought it would be 3

are u sure for 2 @minal

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i think mod will be 2 .. lets see someone else verify

modulus of the  counter is decided by number of state found in cycle of transition state diagram.here mod will be 2 .

## Related questions

1
361 views
Was solving the GO 2019 pdf when I encountered this question asked in TIFR 2017. Although I have understood this question, I have one doubt- It is mentioned in the question that for a 3-bit number, the ordering (000, 100, 101, 111, 110, 010, 011, 001) is one of the possible Gray codes. Then, how many such orders are there for an n-bit number?
How addition of $+3$ and $-3$ in sign magnitude form is $-5?$

format: plain


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# laughinghan/mathquill forked from mathquill/mathquill

### Subversion checkout URL

You can clone with
or
.
Magic textboxes where you can type math as easily as writing!
JavaScript CSS
This branch is 10 commits behind mathquill:master.
Fixing improper css that causes ugly font. (No italic where it should be in math environment.)
latest commit f73e1d57c5
authored
 Failed to load latest commit information. script Dec 2, 2013 src Feb 26, 2014 test Nov 18, 2013 .gitattributes Jun 12, 2012 .gitignore Dec 9, 2012 BUILDING Dec 19, 2012 CHANGELOG.md Nov 10, 2013 Makefile Dec 6, 2013 README.md Nov 9, 2013 package.json Nov 24, 2013

# MathQuill

by Han and Jay. Current development is proudly supported by Desmos, whose awesome graphing calculator makes extensive use of Mathquill.

Please note that this is a beta version, so bugs and unimplemented features are all over the place.

## Usage

(Note: Requires jQuery 1.4.3+. Google CDN-hosted copy recommended.)

• the stylesheet
• the fonts in the font/ directory relative to mathquill.css (or change your copy of mathquill.css to include from the right directory)
• the script (unminified)

then on your webpages include the stylesheet

<link rel="stylesheet" type="text/css" href="/path/to/mathquill.css">

and after jQuery, the script

Then wherever you'd like to embed LaTeX math to be rendered in HTML:

\frac{d}{dx}\sqrt{x}

or have an editable math textbox:

f(x)=?

Note that for dynamically created elements that weren't in the HTML DOM on document ready, you will need to call our jQuery plugin after inserting into the visible HTML DOM:

$('x^2').appendTo('body').mathquill() or .mathquill('editable') MathQuill has to perform calculations based on computed CSS values. If you mathquill-ify an element before inserting into the visible HTML DOM, then once it is visible MathQuill will need to recalculate: $('\sqrt{2}').mathquill().appendTo('body').mathquill('redraw')

Any element that has been MathQuill-ified can be reverted:

$('.mathquill-embedded-latex').mathquill('revert');  Manipulating the HTML DOM inside editable math textboxes can break MathQuill. Currently, MathQuill only supports a limited scripting API: • To access the LaTeX contents of a mathquill-ified element: $('x^{-1}').mathquill().mathquill('latex') === 'x^{-1}'

• To render some LaTeX in a mathquill-ified element:

\$('').mathquill().appendTo('body').mathquill('latex','a_n x^n')

• To write some LaTeX at the current cursor position:

someMathQuillifiedElement.mathquill('write','\\frac{d}{dx}')

• To insert a LaTeX command at the current cursor position or with the current selection:

someMathQuillifiedElement.mathquill('cmd','\\sqrt')

A Note On Changing Colors:

To change the foreground color, don't just set the color, also set the border-color, because the cursor, fraction bar, and square root overline are all borders, not text. (Example below.)

Due to technical limitations of IE8, if you support it, and want to give a MathQuill editable a background color other than white, and support square roots, parentheses, square brackets, or curly braces, you will need to, in addition to of course setting the background color on the editable itself, set it on elements with class matrixed, and then set a Chroma filter on elements with class matrixed-container.

For example, to style as white-on-black instead of black-on-white:

#my-math-input {
color: white;
border-color: white;
background: black;
}
#my-math-input .matrixed {
background: black;
}
#my-math-input .matrixed-container {
filter: progid:DXImageTransform.Microsoft.Chroma(color='black');
}

(This is because almost all math rendered by MathQuill has a transparent background, so for them it's sufficient to set the background color on the editable itself. The exception is, IE8 doesn't support CSS transforms, so MathQuill uses a matrix filter to stretch parens etc, which anti-aliases wrongly without an opaque background, so MathQuill defaults to white.)

## Building and Testing

To hack on MathQuill, you're gonna want to build and test the source files you edit. In addition to make, MathQuill uses some build tools written on [Node][http://nodejs.org/#download], so you will need to install that before running make. (Once it's installed, make automatically does npm install, installing the necessary build tools.)

• make builds build/mathquill.{css,js,min.js}
• make dev won't try to minify MathQuill (which can be take nonzero time)
• make test also doesn't minify MathQuill, but it also builds mathquill.test.js, which is used in test/unit.html

## Understanding The Source Code

All the CSS is in mathquill.css. Most of it's pretty straightforward, the choice of font isn't settled, and fractions are somewhat arcane, see the Wiki pages "Fonts" and "Fractions".

All the JavaScript that you actually want to read is in src/, build/ is created when you run make just to contain a cat'ed and minified version of all that.

### Overview of how things fit together:

(Just skim the logic, but do read the starred comments, definitions and method signatures.)

In comments and internal documentation, :: means .prototype..

intro.js defines some simple sugar for the idiomatic JS classes used throughout MathQuill, plus some globals and opening boilerplate.

• Classes are defined using Pjs, and the variable _ is used by convention as the prototype.

tree.js defines the abstract classes for the JS objects that make up the edit tree.

• A Node is a node in the tree.
• A Fragment is a range of siblings in the tree. This is used, for example, for selections.

• The edit tree has two kinds of nodes: commands and blocks

• blocks, like the root block, can contain any number of commands
• commands, like x, 1, +, \frac, \sqrt (clearly siblings in the tree) contain a fixed number of blocks
• symbols like x, y, 1, 2, +, - are commands with 0 blocks
• All edit tree nodes are instances of MathElement
• blocks are instances of MathBlock
• commands are instances of MathCommand
• symbols are instances of Symbol

cursor.js defines the "singleton" classes for the visible blinking cursor and highlighted selection.

• The methods on Cursor pretty much do what they say on the tin. They're how the tree is supposed to traversed and modified.

rootelements.js defines the edit tree root elements, and a function createRoot() that attaches event handlers to the jQuery-wrapped HTML elements:

• Some root elements can actually be in others, so rather than attaching handlers in the constructor, createRoot() is called on the actual root element. Except \editable{}s need text input event handlers that aren't attached to the static math containing them...it's a little messy.
• Event delegation is used in 2 ways:
• in the HTML DOM, the root span element of each MathQuill thing is delegated all the events in it's own MathQuill thing
• keyboard events usually end up triggering their analogue in the edit tree on the cursor, which then bubble upwards
• in the edit tree, the root MathElement is delegated most of these virtual keyboard events
• for example, RootMathBlock::keydown()
• some special commands do intercept these events, though

textarea.js handles all the HTML DOM events necessary to emulate a textarea, using a hidden textarea.

symbols.js defines classes for all the symbols like & and \partial, and adds the constructors to CharCmds or LatexCmds as used by Cursor::write().

commands.js defines classes for all the commands like \frac and /, and adds the constructors to CharCmds or LatexCmds.

publicapi.js defines the public jQuery::mathquill() method and on document ready, finds and mathquill-ifies .mathquill-editable and so on elements.

outro.js is just closing boilerplate to match that in intro.js`.


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# How does bending in space-time caused by mass energy translate into acceleration of object? [duplicate]

Let say we have 2 similar apples separated by a distance apart, just their mass energy alone is sufficient to cause bending in the space-time. I think it is this bending in the space-time that cause both the apples to accelerate but there shouldn't be any external forces to induce the acceleration, I've skimmed at the math it just says no bending no acceleration and that's not helping. I like to understand how bending in space-time cause the apple to accelerate?

okay I know the math shows that time is a vector(or scalar or even tensor*) too but that makes the matter even worst for me!

*just want to insured myself!


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# How to calculate log (1÷0.8500)?

###### Question:

How to calculate log (1÷0.8500)?

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# Elementary formula for permutations?

Suppose I fix $n$ and let $\sigma_k$ represent the $k$th permutation of $S_n$ with respect to some ordering (whatever ordering might serve my purpose). Is there an elementary formula for $\sigma_k(i)$ which requires only $i, k,$ and $n$?

Is one known for small $n$, perhaps even as small as 4?

-
I'm not sure I understand your question. Do you mean, is there some ordering of $S_n$ for which there is an "elementary formula" for $f(n,k,i)$, where this represents "the image of $i$ under the $k$th permutation of $S_n$"? (and what does "elementary formula" mean?) –  Arturo Magidin Nov 21 '11 at 3:23
I mean elementary formula in that the image of $i$ under the permutation $\sigma_k$ can be computed without means of ridiculous things like the gamma function, i.e., with elementary functions. And to clarify, I would be fine picking any ordering for which such a formula exists, but I really want to know what the formula is; the grand prize is to have such a formula for all $n$, which is based only on $n, i, k$. –  JeremyKun Nov 21 '11 at 3:36
Of course, I could imagine there are multiple formulas based on different orderings (if one exists, why not many?), which is why I'm saying pick any ordering you like. –  JeremyKun Nov 21 '11 at 3:42

It depends upon what you mean by elementary. If the ordering is lexicographic, the first element of the $k^{\text(th)}$ permutation of $S_n$ is $\lfloor \frac{k}{(n-1)!}\rfloor$ (assuming the first element of the set is $0$).This leads to an easy recursive function to find the whole $k^{\text(th)}$ permutation.
@Bean: No, I was referring to a function $k \to S_n$, viewing the elements of $S_n$ as strings of $n$ naturals in the range $[1,n]$. This is how to get the first entry in the string. Then for the second you use the same function from $k-\lfloor \frac{k}{(n-1)!}\rfloor(n-1)!\to S_{n-1}$ regarding the set as $\{1,2,3,\ldots n\}$ missing the one you already chose. –  Ross Millikan Nov 21 '11 at 3:44
Should the "floor" be a "1 + floor"? For any $k < (n-1)!$ the first element of the first permutation will be 0... –  JeremyKun Nov 21 '11 at 3:55
@Bean: right you are if indexed from $1$. I will clarify. –  Ross Millikan Nov 21 '11 at 4:55
Sorry but, how does this answer the question? Where is $i$? (Or, have I misunderstood the question?) Moreover, it seems to me that unranking a permutation requires a loop. –  James Nov 21 '11 at 22:32


text: #every PyTorch Module object has a self.training boolean which can be used. The aim of this post is to implement a variational autoencoder (VAE) that trains on words and then generates new words. denoising, 3.) Imports. download the GitHub extension for Visual Studio. In general, I would use a minimum of 32 filters for most real world problems. The Autoencoders, a variant of the artificial neural networks, are applied very successfully in the image process especially to reconstruct the images. , $W_{out}$ = $$\frac{W_{in} + 2 × padding[1] - dilation[1] × (kernel_size[1] - 1) - 1}{stride[1]} + 1$$, $H_{out}$ = ($H_{in}$ - 1) × stride[0] - 2 ×padding[0] + dilation[0] × (kernel_size[0] - 1) + output_padding[0] + 1, $W_{out}$ = ($W_{in}$ - 1) × stride}[1] - 2 ×padding[1] + dilation[1] × (kernel_size[1] - 1) + output_padding[1] + 1, Convolutional Denoising Auto Encoder with Maxpool2d and ConvTranspose2d. If nothing happens, download the GitHub extension for Visual Studio and try again. model -- the PyTorch model / "Module" to train, loss_func -- the loss function that takes in batch in two arguments, the model outputs and the labels, and returns a score. While training my model gives identical loss results. The input is binarized and Binary Cross Entropy has been used as the loss function. anomaly detection, 4.) Now let jump to our layer1 which consists of two conv2d layers followed by ReLU activation function and BatchNormalization.self.layer1 takes 3 channels as an input and gives out 32 channels as output.. In this article we will be implementing an autoencoder and using PyTorch and then applying the autoencoder to an image from the MNIST Dataset. This way we can create a Denoising Autoencoder! I'm looking for the kind of stuff you have in this HW, detailed results showing what you did/tried, progress, and what you understood / learned. Denoising Auto Encoders (DAE) In a denoising auto encoder the goal is to create a more robust model to noise. This … Note: This tutorial uses PyTorch. The framework can be copied and run in a Jupyter Notebook with ease. To train your denoising autoencoder, make sure you use the “Downloads” section of this tutorial to download the source code. Aquí nos gustaría mostrarte una descripción, pero el sitio web que estás mirando no lo permite. Variational Autoencoder Code and Experiments 17 minute read This is the fourth and final post in my series: From KL Divergence to Variational Autoencoder in PyTorch.The previous post in the series is Variational Autoencoder Theory. I hope that you will learn a lot, and I will love to know your thoughts in the comment section. # PyTorch stores gradients in a mutable data structure. From the reconstructed image it is evident that denoising CNN Auto Encoders are the more accurate and robust models. 2) Compare the Denoising CNN and the large Denoising Auto Encoder from the lecture numerically and qualitatively. #Now we just need to update all the parameters! In this post, we will be denoising text image documents using deep learning autoencoder neural network. They have some nice examples in their repo as well. 21: Output of denoising autoencoder For example, an autoencoder trained on numbers does not work on alphabets. One application of convolutional autoencoders is denoising. We use this to help determine the size of subsequent layers, dnauto_encode_decode_conv_convtranspose_big, dnauto_encode_decode_conv_convtranspose_big2, # 8 * 28 *28 to 8 * 14 *14 #2 stride 2 kernel size make the C*W*H//4 or (C,W//2,H//2) shaped. Despite its sig-ni cant successes, supervised learning today is still severely limited. Quoting Wikipedia “An autoencoder is a type of artificial neural network used to learn efficient data codings in an unsupervised manner. Because the autoencoder is trained as a whole (we say it’s trained “end-to-end”), we simultaneosly optimize the encoder and the decoder. Supra-ventricular Premature or Ectopic Beat (SP or EB) 5. So it will be easier for you to grasp the coding concepts if you are familiar with PyTorch. In this paper, we propose the "adversarial autoencoder" (AAE), which is a probabilistic autoencoder that uses the recently proposed generative adversarial networks (GAN) to perform variational inference by matching the aggregated posterior of the hidden code vector of the autoencoder … The autoencoder is a neural network that learns to encode and decode automatically (hence, the name). The aim of … #Lets find out validation performance as we go! Show transcript Advance your knowledge in tech . The motivation is that the hidden layer should be able to capture high level representations and be robust to small changes in the input. The denoising autoencoder network will also try to reconstruct the images. Wow, above an beyond on this homework, very good job! This means that we can only replicate the output images to input images. Sharing the transposed weights allows you to reduce the number of parameters by 1/2 (training each decoder/ encoder one layer at a time). Denoising CNN Auto Encoder's with ConvTranspose2d. A Pytorch Implementation of a denoising autoencoder. In other words, we would like the network to somehow learn the identity function f (x) = x f (x) = x. pos_edge_index (LongTensor): The positive edges to train against. please tell me what I am doing wrong. Convtranspose layers have the capability to upsample the feature maps and recover the image details. I did the dimensionality reduction example today. Application to image denoising. Detecting Medical Fraud (Part 2) — Building an Autoencoder in PyTorch Published on February 5, 2020 February 5, 2020 • 28 Likes • 1 Comments Sparse autoencoder 1 Introduction Supervised learning is one of the most powerful tools of AI, and has led to automatic zip code recognition, speech recognition, self-driving cars, and a continually improving understanding of the human genome. The denoising CNN Auto Encoder models are clearly the best at creating reconstructions than the large Denoising Auto Encoder from the lecture. The Overflow Blog Podcast 287: How do you make software reliable enough for space travel? So, an autoencoder can compress and decompress information. However, if there are errors from random insertion or deletion of the characters (= bases) in DNA sequences, then the problem is getting more complicated (for example, see the supplemental materials of the HGAP paper ). I have tried different layerd Denoising CNN Auto Encoders and most of networks have able to capture even minute details from the original input. I start off explaining what an autoencoder is and how it works. Denoising CNN Auto Encoder is better than the large Denoising Auto Encoder from the lecture. 2) Create noise mask: do(torch.ones(img.shape)). 2 shows the reconstructions at 1st, 100th and 200th epochs: Fig. The dataset is available on my Google Drive. By generating 100.000 pure and noisy samples, we found that it’s possible to create a trained noise removal algorithm that … Variational AEs for creating synthetic faces: with a convolutional VAEs, we can make fake faces. In future articles, we will implement many different types of autoencoders using PyTorch. #Now we are just grabbing some information we would like to have, #moving labels & predictions back to CPU for computing / storing predictions, #We have a classification problem, convert to labels. Normal (N) 2. Get all the quality content you’ll ever need to stay ahead with a Packt subscription – access over 7,500 online books and videos on everything in tech. I am training an autoencoder for a multiclass classification problem where I transmit 16 equiprobable messages and send them through a denoising autoencoder … Remember that a good project dosn't necessarily have to be working/complete. The get_dataset method will download and transform our data for our model.It takes one argument train is set to true it will give us a training dataset and if it is false it will give us a testing dataset. dimensionality reduction, 2.) GitHub Gist: instantly share code, notes, and snippets. It's simple: we will train the autoencoder to map noisy digits images to clean digits images. It's simple: we will train the autoencoder to map noisy digits images to clean digits images. In the context of computer vision, denoising autoencoders can be seen as very powerful filters that can be used for automatic pre-processing. Denoising Text Image Documents using Autoencoders. In this model, we assume we are injecting the same noisy distribution we are going to observe in reality, so that we can learn how to robustly recover from it. Denoising CNN Auto Encoder's with noise added to the input of several layers. We apply it to the MNIST dataset. Each part consists of 3 Linear layers with ReLU activations. A denoising autoencoder tries to learn a representation (latent-space or bottleneck) that is robust to noise. Premature Ventricular Contraction (PVC) 4. Denoising Autoencoder. Q&A for Work. In this blog post, we created a denoising / noise removal autoencoder with Keras, specifically focused on signal processing. introducing noise) that the autoencoder must then reconstruct, or denoise. This method returns a DataLoader object which is used in training. Denoising autoencoders are an extension of the basic autoencoders architecture. This is a follow up to the question I asked previously a week ago. The Linear autoencoder consists of only linear layers. #Otherwise, it will have old information from a previous iteration. A Short Recap of Standard (Classical) Autoencoders. Which one is better? However, there still seems to be a few issues. Building Denoising Autoencoder Using PyTorch . Detecting Medical Fraud (Part 2) — Building an Autoencoder in PyTorch Published on February 5, 2020 February 5, 2020 • 28 Likes • 1 Comments An LSTM Autoencoder is an implementation of an autoencoder for sequence data using an Encoder-Decoder LSTM architecture. PyTorch Implementation. converting categorical data to numeric data. Example convolutional autoencoder implementation using PyTorch - example_autoencoder.py Single heartbeat from a previous iteration, or the CIFAR10 dataset common in real-world scenarios scaled down MNIST dataset. 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text:
# Do we only need Choice for higher order-statements (Zorn’s lemma as example)?

In this earlier post, I asked why the axiom of choice, which is an axiom in set theory, is used in areas that are not set theory, such as group theory. The answer was that, whenever choice is used, it is used to prove something about the set of models of group theory, and hence is a statement about sets rather than about groups.

Now I’m trying to understand this for Zorn’s lemma specifically. Zorn’s lemma says:

Zorn’s Lemma — Suppose a partially ordered set $$P$$ has the property that every chain in $$P$$ has an upper bound in $$P$$. Then the set $$P$$ contains at least one maximal element.

The axiom of choice is used in the proof of Zorn’s lemma. This at first sight confused me, since Zorn’s lemma seems like a theorem about partial orders, not about sets in general. But then I realized that it is a second-order statement, rather than first order, since it is about subsets of $$P$$. I don’t fully understand the proof of Zorn’s lemma, since I’m somewhat confused still by the axiom of choice, so let me ask a question to clarify:

Question 1. Is it true that the proof of Zorn’s lemma uses the axiom of choice on totally ordered subsets of $$P$$, rather than on $$P$$ itself?

This gives me the more general conjecture:

Question 2. Is it true that whenever we have a first-order statement about a structure (such as posets), we never need the axiom of choice to prove it, but if the statement is higher-order, we may need it?

Note that (in the context of partial orders) I’m talking about first-order statements within the theory of partial orders, not e.g. first-order statements about the set of models of the theory of partial orders, or e.g. about the category of posets.

• What does the question even mean? Higher order statements? Extend your language to be multisorted and everything becomes first-order. Jun 2, 2019 at 7:55
• The existence of a maximal element is a first-order statement. Yes, the condition on upper bounds is not first-order in the language of orders, but it is not what we want to prove anyway. Jun 2, 2019 at 7:57

For example, if ZFC proves that every group satisfies some first-order sentence $$\varphi$$, then so does ZF - the point being that the language $$\langle e,*,{}^{-1}\rangle$$ of groups is well-orderable (indeed, finite).
• @user56834 Re: your first question, remember that a first-order language is just a set of symbols of appropriate type. If choice fails, there are some sets which are non-well-orderable; well, this leads to sets of symbols which are non-well-orderable. For example, consider the language consisting of an $n$-ary function symbol for each $n$-ary function on $\mathbb{R}$ (this language shows up in nonstandard analysis, for example). This set of symbols is incredibly huge; it has size $2^{2^{\aleph_0}}$. So if the powerset of the reals isn't well-orderable, neither is this language. (cont'd) Jun 3, 2019 at 13:10
• @user56834 Re: your first question, remember that a first-order language is just a set of symbols of appropriate type. If choice fails, there are some sets which are non-well-orderable; well, this leads to sets of symbols which are non-well-orderable. For example, consider the language consisting of an $n$-ary function symbol for each $n$-ary function on $\mathbb{R}$ (this language shows up in nonstandard analysis, for example). This set of symbols is incredibly huge; it has size $2^{2^{\aleph_0}}$. So if the powerset of the reals isn't well-orderable, neither is this language. (cont'd) Jun 3, 2019 at 13:10
• Of course such languages actually occur only very rarely in practice; basically every first-order language you actually use is even finite! But they do crop up from time to time. Re: your second question, model existence is "dual" to entailment: "$T$ entails $\varphi$" is equivalent to "There is no model of $T\cup\{\neg\varphi\}$." Showing that choice plays no role in model existence arguments of a certain kind also shows that it plays no role in semantic entailment arguments of that same kind. Jun 3, 2019 at 13:12


text:
# A question about the possibilities of GAP

Let $$R=\mathbb{Z}/1024\mathbb{Z}$$ and $$G=GL(3,R)$$. Let $$H$$ be the subgroup of $$G$$ consisting of all matrices with determinant $$1$$ which are congruent to the identity matrix modulo the ideal $$4R$$. Let

$$A=\begin{pmatrix} 1 &0&512\\ 0& 1&0\\0&0 &1\end{pmatrix}$$ and $$B=\begin{pmatrix} 1 &4&2\\ 8& 1&4\\16 &8 &1\end{pmatrix}$$.

Can one use GAP (or other software) to find out if $$A\in B^H$$? (Here $$B^H$$ is the subgroup of $$G$$ generated by the set $$\{ B^h\mid h\in H\}$$.)

• Yes, in principle, and this is quite easy to do for smaller rings -- but $\operatorname{GL}(3, R)$ has order $2^{84} \cdot 3 \cdot 7$, so this might take quite some time in your case (and this is not a limitation unique to GAP!). Jun 15, 2021 at 14:51
• About what amount of time are we talking here? Jun 18, 2021 at 7:48
• Yes, if we wanted to write down all elements this would be hopeless. But grouop theoretic algoriths try to describe the large group using a much smaller data structure, which makes it feasible to do this kind of calculation. Jul 11, 2021 at 16:21

It is contained. Here is the way I did the calculation with GAP. It ought to work nicer, but there is a stupid technical issue in the way that makes it hard to implement a membership test in a subgroup. (The issue is basically that we cannot guarantee that elements will always lie in one big parent group.)

Thus, one needs to do things in a somewhat pedestrian way. You will need my matgrp package for some calculations in the large groups over residue class rings to work:

gap> LoadPackage("matgrp");
true
gap> A:=[[1,0,512],[0,1,0],[0,0,1]];
[ [ 1, 0, 512 ], [ 0, 1, 0 ], [ 0, 0, 1 ] ]
gap> B:=[[1,4,2],[8,1,4],[16,8,1]];
[ [ 1, 4, 2 ], [ 8, 1, 4 ], [ 16, 8, 1 ] ]
gap> r:=Integers mod 1024;
(Integers mod 1024)
gap> Ar:=A*One(r);
[ [ ZmodnZObj( 1, 1024 ), ZmodnZObj( 0, 1024 ), ZmodnZObj( 512, 1024 ) ],
[ ZmodnZObj( 0, 1024 ), ZmodnZObj( 1, 1024 ), ZmodnZObj( 0, 1024 ) ],
[ ZmodnZObj( 0, 1024 ), ZmodnZObj( 0, 1024 ), ZmodnZObj( 1, 1024 ) ] ]
gap> Br:=B*One(r);
[ [ ZmodnZObj( 1, 1024 ), ZmodnZObj( 4, 1024 ), ZmodnZObj( 2, 1024 ) ],
[ ZmodnZObj( 8, 1024 ), ZmodnZObj( 1, 1024 ), ZmodnZObj( 4, 1024 ) ],
[ ZmodnZObj( 16, 1024 ), ZmodnZObj( 8, 1024 ), ZmodnZObj( 1, 1024 ) ] ]
gap> G:=GL(3,r);
GL(3,Z/1024Z)
gap> FittingFreeLiftSetup(G);; # build a data structure for group order
gap> Size(G);
406199075390515402701275136

Next we look at the image modulo 4. We construct it also as a permutation group (so we can use permutation group functionality for homomorphisms:

gap> rf:=Integers mod 4;
(Integers mod 4)
gap> gens:=List(GeneratorsOfGroup(G),m->List(m,r->List(r,x->Int(x)*One(rf))));;
gap> q:=Group(gens);
<matrix group with 5 generators>
gap> Size(q);
86016
gap> Size(GL(3,rf));
86016
gap> isop:=IsomorphismPermGroup(q);;
gap> p:=Image(isop);

We now construct a map from this permutation group to the matrix group and collect generators for its co-kernel -- these are evaluated relators for the factor that together will generate the kernel of the reduction map on $$G$$:

gap> reverse:=GroupGeneralMappingByImagesNC(p,G,GeneratorsOfGroup(p),
> GeneratorsOfGroup(G));;
gap> it:=CoKernelGensIterator(reverse);
<iterator>
gap> hg:=[];;
gap> for i in it do
>   if not IsOne(i) then
>   fi;
> od;
gap> Length(hg);
448

This number of generators is a bit too large. We just pick 20 random ones and verify they still generate the kernel. (Would iterate/try more generators if not):

gap> preH:=Group(List([1..20],x->Random(hg)));
<matrix group with 20 generators>
gap> FittingFreeLiftSetup(preH);;
gap> Size(G)/Size(preH);
86016

Now for determinant 1. We use the same idea for the homomorphism onto the determinant, and get $$H$$ as kernel

gap> dets:=List(GeneratorsOfGroup(preH),DeterminantMat);;
gap> d:=Group(det); # GAP will issue a harmless warning that it assumes the elements indeed are invertible
<group with 20 generators>
gap> Size(d);
256
gap> isop:=IsomorphismPermGroup(d);;
gap> p:=Image(isop);;
gap> reverse:=GroupGeneralMappingByImagesNC(p,preH,GeneratorsOfGroup(p),
> GeneratorsOfGroup(preH));;
gap> it:=CoKernelGensIterator(reverse);;
gap> hg:=[];;
gap> for i in it do
>   if not IsOne(i) then
>   fi;
> od;
gap> Length(hg);
4873
gap> H:=Group(List([1..20],x->Random(hg)));;FittingFreeLiftSetup(H);;
gap> Size(preH)/Size(H);
256

Now we are ready to calculate $$B^H$$. We use a standard closure algorithm, starting with $$B$$ and then take conjugates of generators with generators of $$H$$ until no new conjugates arise, that is the group is normal.

To test membership $$x\in S$$, we check (this is the kludge I mentioned) whether $$|S|=|\langle S,x\rangle|$$. (Indeed this ought to be better, but it still beats hand-calculations.)

gap> bco:=[Br];;
gap> sub:=Group(bco);;
gap> FittingFreeLiftSetup(sub);;
gap> Size(sub);
512
gap> for i in bco do
>   for j in GeneratorsOfGroup(H) do
>     x:=i^j;
>     t:=Group(Concatenation(bco,[x]));
>     FittingFreeLiftSetup(t);
>     if Size(t)<>Size(sub) then
>       sub:=t;
>     fi;
>   od;
> od;

We now use the same kludgy element test to see whether $$A\in B^H$$:

gap> Size(sub);
4503599627370496
gap> t:=Group(Concatenation(bco,[Ar]));
<matrix group with 11 generators>
gap> FittingFreeLiftSetup(t);;
gap> Size(t);
4503599627370496

The order stays the same after adding $$A$$, thus $$A\in B^H$$.

• Thank you very much! In my question I demanded that the elements of $H$ have determinant $1$. Was this taken into account in your solution? Jul 12, 2021 at 13:05
• Sorry, I had overlooked this extra condition. Is included now. Since $B^H$ stays the same, the result is the same. Jul 12, 2021 at 16:08
• Thanks again! I have a question, regarding We now construct a map from this permutation group to the matrix group and collect generators for its co-kernel -- these are evaluated relators for the factor that together will generate the kernel of the reduction map on G. Could you explain why these elements generate the kernel of the reduction map? Jul 14, 2021 at 9:14
• As far as this question is concerned, we verified that they generate a subgroup of correct index (and could verify that they reduce to zero mod 4). More generally, this is the specification of the function (which is used to calculate kernels of homomorphisms) -- see e.g. Holt, Handbook of CGT. Jul 14, 2021 at 14:19
• Is it possible to get a (short) expression of $A$ as a product of generators $B^h$ of $B^H$? Jul 23, 2021 at 7:55


text:
$(A \cong B \; \; \; \wedge \;\;\; \gamma:A \rightarrowtail B) \implies \gamma:A \twoheadrightarrow B \;\;$?

If objects $A$ and $B$ (in some category $\mathcal{C}$) are isomorphic, and if some morphism $\gamma:A \rightarrow B$ is monic, does it follow that $\gamma$ is also epic?

Thanks!

-

This isn't even true in Set. For example, $f:\mathbb{N} \rightarrow \mathbb{N}$, $f(n)=n+1$ is monic but not epic.


text:
mersenneforum.org SIQS diagnostics
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2019-05-16, 20:55   #12
henryzz
Just call me Henry

"David"
Sep 2007
Cambridge (GMT/BST)

2·32·7·47 Posts

Quote:
 Originally Posted by bsquared Is that time spent just building new A polys or does it also include setup time for the new poly (e.g., computing A^-1 mod p for all factor base p)? 10% seems high either way but if that's just for building new A polys then it seems *really* high. In contrast I usually spend < 3% total for building new A's and all of the setup for new A's. Seems reasonable given your other parameters. I typically see anywhere from 50 to 150 A's at 50 digits, using 6 factors per A, but I use a far smaller factor base and sieve interval. It is all a tradeoff.
That is including all the set up. The two bits that take the time are calculating A^-1 mod p and then multiplying that with all the generators for each factor of A mod p. I am using the built in BigInteger in c# which is quite slow. I might experiment with a couple of alternatives libraries. In the past I have seen BigInteger have a terrible runtime order but have less overhead for smaller numbers. Looking at my modular inversion code I can see improvements.

I think you can see why I have a larger sieve interval.

I have just realized that I was still skipping half the Bs so calculating As is taking half the time it was. It is still very slow for smaller numbers especially.

Last fiddled with by henryzz on 2019-05-16 at 21:06

2019-05-17, 02:32   #13
tabidots

May 2019

110002 Posts

Quote:
 Originally Posted by bsquared I would ask, how are you determining smoothness exactly? In other words, the sieve value above (or below) which you try to find a relation. If your polynomials and roots are good then maybe it is as simple as tweaking your smoothness bound so as to generate more potential relation reports. The value I use is obtained from a bunch of empirically determined magic numbers, loosely related to Contini's suggestion along the lines of: "compute the number of bits in M/2*sqrt(N/2), then subtract some slack". M is the size of your sieve interval.
Thanks for this suggestion. I haven't gotten around to properly debugging the polynomials as jasonp suggested, but I sure hope the fix is as simple as this!

I realized that my threshold was quite far off. For SIQS, I did not change the procedure I found in the Python implementation for MPQS, which is full of that person's magic numbers:

1. Calculate the maximum possible g(x) value: (M*sqrt(2n)) / 2

2. Define the base threshold: 0.735 * log10(max_gx)

3. Determine the minimum "significant" prime: floor(3 * base_threshold) (p below this won't be used in the sieve)

4. Calculate the fudge factor: 0.25 * sum(log10(p) for all p below min_sig_prime)

5. Calculate the final threshold: base_threshold - fudge_factor

One thing I noticed (perhaps because in MPQS, I have factored one "a" out of the polynomial) is that my max_gx value for SIQS is completely off.

When you say

Quote:
 "compute the number of bits in M/2*sqrt(N/2), then subtract some slack"
the number whose bits you're counting is the max_gx value, right? Is that

M / (2 * sqrt(N/2)) or
(M/2) * sqrt(N/2) ?

I can't seem to get my max_gx value to approach anywhere near the maximum values actually returned by the polynomials.

I decided to try using Msieve's parameters for F & M for SIQS. So for the 50-digit number above (55720182748350551450373114705225729163236062899069), M is 65536.

One polynomial I just generated is
g(x) = 2723740665885560669400118386023188081 x^2 +
930016696893326215892951094451829278 x +
5687450887644869965258458870102862443574037010

g(65536) = 6010987144693123589465802296837959857061524974

which is not even on the same planet as:

current max_gx = (M*sqrt(2N)) / 2 = 3494811551456090780950364160
M / (2 * sqrt(N/2)) = 6.144776667872878E-19
(M/2) * sqrt(N/2) = 1747405775728045390475198464

I have probably made some stupid mistake somewhere, right?

Last fiddled with by tabidots on 2019-05-17 at 02:39

2019-05-17, 13:25   #14
bsquared

"Ben"
Feb 2007

2×1,789 Posts

Quote:
 Originally Posted by tabidots Is that M / (2 * sqrt(N/2)) or (M/2) * sqrt(N/2) ?
Contini suggests taking "a" approximately equal to sqrt(2N)/M, so that gmax is about M*sqrt(N/2). In my code I treat M as the entire sieve interval (+ and - sides) which is why I divide M by 2. Sorry for the confusion.

With $A \approx \sqrt{2N}/M$, you should be generating values with size of about 70 bits. Here are a few that my code came up with for the 50 digit example:

Code:
A = 978059782600599905911
B = 1406266881656945627063
C = -2107894421575738271129468804544
A = 975874113426336640897
B = 1085138790265600306933
C = -2112615482004038794930218933912
A = 982775394573367733491
B = 1201046251131435906861
C = -2097780196401273381630906077752
So something seems to be amiss with how you generate polynomials as the coefficients (and the resulting values of g(x)) are about 50 bits too large. That would definitely explain the difficulty in finding smooths!

2019-05-17, 13:52   #15
tabidots

May 2019

23·3 Posts

Quote:
 Originally Posted by bsquared So something seems to be amiss with how you generate polynomials as the coefficients (and the resulting values of g(x)) are about 50 bits too large. That would definitely explain the difficulty in finding smooths!
Ah! I seem to be misunderstanding something very fundamental here.

In MPQS, one side of the congruence was obtained from values of ax + b, the "unsquared" polynomial, and the other side was taken from (ax + b)^2, or a(ax + 2b + c).

The coefficients I posted were not the raw "a, b, c" but rather a^2, 2ab, b^2 - N, since g_{a,b}(x) = (ax + b)^2 - N = (a^2)(x^2) + 2ab + b^2 - N. Since a is not a square in SIQS, I cannot factor out the a, and the polynomial must remain in its full form.

But I am still looking for a congruence between ax + b and (a^2)(x^2) + 2ab + b^2 - N, right? Or am I looking for a congruence between something else?

Last fiddled with by tabidots on 2019-05-17 at 13:53

2019-05-17, 14:37   #16
bsquared

"Ben"
Feb 2007

1101111110102 Posts

Quote:
 Originally Posted by tabidots Ah! I seem to be misunderstanding something very fundamental here. In MPQS, one side of the congruence was obtained from values of ax + b, the "unsquared" polynomial, and the other side was taken from (ax + b)^2, or a(ax + 2b + c). The coefficients I posted were not the raw "a, b, c" but rather a^2, 2ab, b^2 - N, since g_{a,b}(x) = (ax + b)^2 - N = (a^2)(x^2) + 2ab + b^2 - N. Since a is not a square in SIQS, I cannot factor out the a, and the polynomial must remain in its full form. But I am still looking for a congruence between ax + b and (a^2)(x^2) + 2ab + b^2 - N, right? Or am I looking for a congruence between something else?
You can construct b such that b^2 - N is divisible by a. Then A factors out and you have $g_{a,b}(x) = a(ax^2+2bx+c)$, i.e., all g(x) are divisible by A. For each smooth g(x)/a you find, add the factors of A to the relation. In mpqs that wasn't needed since A was chosen to be a square. Now they participate in the matrix step more directly.

Contini's thesis starting on page 9 describes how to choose B so that B^2-N is divisible by A, and so that enumerating all such possible B's can be done quickly using a Gray code.

So, looking at the poly you posted again, I get that:
Code:
A=sqrt(2723740665885560669400118386023188081)=1650375916537066009
B=930016696893326215892951094451829278/(2*a) = 281759048824692071
But (B^2 - N) is not divisible by A, so maybe look at how you are generating B (Contini, bottom of page 9).

 2019-05-17, 15:58 #17 tabidots   May 2019 23·3 Posts Weird... I don't think I changed anything, but all my polynomials are testing fine now for $b^2 \equiv N (\textrm{mod } a)$, (though I don't have the original data that generated the polynomial I wrote earlier, so I can't be sure). A and B are around ~70 bits and C is around ~100 bits. I changed the RHS polynomial to $ax + 2b + c$, where $c = \frac{(b^2 - N)}{a}$. LHS is still $ax + b$. (RHS are the values that get converted to a matrix. LHS are the values that just get multiplied together.) All my $soln_1, soln_2$ solutions for the polynomial are checking out, as well. But it still takes a long time to get smooths. I played around with the threshold, and setting it very low made it grab almost 100 smooths per A. But the sieving took an eternity. It might have worked out to the same amount of time in total. Felt like at least 10 minutes, and then it seemed to hang when trying to find a factor (not sure what happened there). My sieving code can't be the roadblock either, because my MPQS just factored this number in ~83s. What else can I diagnose?
 2019-05-17, 16:49 #18 tabidots   May 2019 23·3 Posts Here is some sample test output (just a tiny bit, obviously, because the data generated by this program is oceanic). Don't know how helpful it will actually be, but just in case: Code: N = 55720182748350551450373114705225729163236062899069 M = 65536 Max g(x): 345916436897199940228470734848 Current smooths: 0/1200 (0 primes used) ------- New A: 161193073467281037817 B: 251205167917836684051 C: -345673554618068592277051911604 Max g(x) from this poly: 346678350229398117877754253900 B: 237418390873394944367 C: -345673554659860363590871822140 B: 116211583075761509923 C: -345673554925767324491631150420 B: 129998360120203249607 C: -345673554904709089668306466860 B: 47900535511707982579 C: -345673554995315416711249773684 ..... I do notice that the max-gx value is still a bit screwy, in that for the very very first polynomial (first b of first a) where I printed the value of g(M), g(M) is actually higher than max-gx ($M \sqrt{\frac{N}{2}}$). But it's not that far off, so I imagine that's not the reason this is performing so poorly. Last fiddled with by tabidots on 2019-05-17 at 16:50
2019-05-17, 22:12   #19
Till

"Tilman Neumann"
Jan 2016
Germany

1D916 Posts

Quote:
 Originally Posted by jasonp This process is what led me to believe that Contini's thesis has a typo in the arithmetic for computing the next B given the previous one; instead of adding the next root you have to subtract it.
I'ld confirm that.

@tabidots:

The size of your A- and B-parameters looks good.

If I run your last example, then my program uses a bigger prime base (size=1732, pmax=32621) and a smaller sieve array (size=11520). The Knuth-Schroeppel multiplier used is 37. My A-parameters have 8 factors instead of yours having 6 of them. These differences must not mean much, but adjustment is a big issue in SIQS.

The biggest performance bug I had was not permitting primes p > M in the prime base.

2019-05-18, 00:33   #20
tabidots

May 2019

23·3 Posts

Quote:
 Originally Posted by Till I'ld confirm that.
I found the typo that was being referred to in your superbly-documented code. I actually had been referring to your code previously as well (to the extent that I can read Java), but for some reason I did not think to hunt for the sign-switch typo in it.

So, here's something strange—in my code I have z, which is what you call -1^CeilTerm.

Previously, when I simply changed the line

Code:
b_new = b + 2 * z * Bs[nu - 1]
(paraphrasing from Clojure) to

Code:
b_new = b - 2 * z * Bs[nu - 1]
it didn't do anything, which is why I wasn't getting anywhere before.

After looking at your code, I tried changing z to be the negative of itself before setting b_new, and now it picks up smooths at what I would consider to be a more normal pace. Wow!

Strange, because it should be the same result, but whatever, that performance hurdle has been cleared. *shrug*

Other gotchas: I was recalculating c upon every new b, and the new c values were not divisible by a. So you have to re-use the same c (as well as the same a); only b changes.

Now I am actually dealing with the opposite problem—my program is too eager to look for a factor, and finds relations way too fast, haha (in that there is no way 40 or 50 relations will produce a congruence for a number fo this size). So I will have to edit that bit and see what happens when I gather more relations than the factor base, not just the used primes.

Quote:
 Originally Posted by Till The Knuth-Schroeppel multiplier used is 37... The biggest performance bug I had was not permitting primes p > M in the prime base.
I tried to translate the K-S multiplier algorithm from your code when I was working on SQUFOF, but I couldn't seem to get it working properly. Not only that, but I found that square-free multipliers were often not the fastest; in some cases, they didn't even find a factor, and actually the best multipliers were perfect squares! So I abandoned K-S. Maybe it's different for SIQS?

Last fiddled with by tabidots on 2019-05-18 at 00:45 Reason: clarity; emojis don't work

 2019-05-18, 01:00 #21 tabidots   May 2019 1816 Posts Okay, so my SIQS is still not finding a factor even after gathering a number of relations greater than the size of the factor base. Speed aside, that means sieving is okay; as I said, my MPQS works well, so linear algebra is okay as well. So, I must be mistaken in how I'm looking for congruences. Just want to clarify the following: The RHS (main) polynomial values are $ax + 2b + c$. c does not change when b changes, only when a changes. From the main polynomial, one a has already been factored out and should be multiplied back in when finding factors (in my code I do this after adding the exponent vectors). The LHS (smaller) values, which just get multiplied together in the end, are $ax + b$ whose b changes with every iteration, just like the main polynomial. EDIT: I realized why changing the sign of z made an impact—it affected how the new set of solutions to the polynomial are calculated! So now my sieve picks up lots of smooths per polynomial, but every new "b" after the first (for each "a") seems to be wrong somehow. b^2 - N is not divisible by a, and the solutions calculated with the Bainv2 values are also not producing zeros. Hmm... Last fiddled with by tabidots on 2019-05-18 at 01:51
 2019-05-18, 02:40 #22 tabidots   May 2019 23×3 Posts Ah, scratch everything I just said. I realized that instead of changing the sign of z, I was subtracting it from 1. Man, complicated algebra is really confusing in Lisps. So no wonder nothing was working despite picking up so many smooths—they must have all been completely wrong values! Anyway, I thought I was getting somewhere, but I am back where I started (or not even there—since I changed the polynomial, now I can't find factors at all). I clearly don't know what I'm doing and I think I'm just going to give up for now. Thanks all for your help and patience.

 Similar Threads Thread Thread Starter Forum Replies Last Post cgy606 Factoring 1 2016-10-21 13:37 henryzz Factoring 2 2013-08-26 23:02 patrickkonsor Factoring 3 2008-10-20 12:03 lythanhnguyen Hardware 3 2007-09-11 06:19 R.D. Silverman Hardware 3 2006-11-17 16:06

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Consider a 5—stage pipeline processor used to execute 200 number of instructions and among those 100 instructions cause 3 stall cycles each. What is the total cycles required for these operation if CPI is not equal to one.

### 1 comment

$(200+5-1) + (3*100) = 504$.

You can also try this by table filling

 1 2 3 4 5 6 7 …. 100 101 102 103 104 105 106 107 108 109 110 111 112 ….. 500 501 502 503 504 I1 F D E M W I2 F D E M W I3 F D E M W . . I100 F D E M W I101 $\phi$ $\phi$ $\phi$ F D E M W I102 $\phi$ $\phi$ $\phi$ F D E M W . $\phi$ $\phi$ $\phi$ I200 F D E M W

100 instructions with no stalls :

$5 + (100-1)* 1 = 104$ cycles

Next 100 with 3 stall cycles :

$108 + (100-1)*4 = 504$ cycles


text:
Professor John Horton Conway: sphere packing problem

Thursday 7th May 2020

The famous British mathematician and Princeton Professor John Horton Conway passed away on Saturday 11 April 2020 from the Coronavirus. One of his many contributions to maths was in the study of sphere packing, where mathematicians study the different ways to arrange non-overlapping spheres in a confined space. A typical sphere packing problem is to find an arrangement in which the spheres fill as much of the space as possible such as packing equally sized oranges into a box with a defined volume.

In two dimensions this is called circle packing, where the aim is to fit as many equal sized circles into a confined space as possible without the circles overlapping. This is particularly relevant today since the social distancing recommendations say that people must stay a distance of at least two metres apart from each other. One way to visualise this application of circle packing is to consider each person to be at the centre of a circle of radius one metre, where the circles are not allowed to overlap. In this scenario, four circles (or people) can be ‘packed’ into a square of side length four metres.

But what about five circles? What is the smallest square which can hold five circles of radius one metre if the circles are not allowed to overlap? Hint: the arrangement must be as shown.

$$2+2\sqrt{2}$$


text:
# 6.16 Free and Local Variables¶

When you want to refer to a variable that is not local to your function, use a free declaration. free Variables declared to be free free variable are assumed to be defined globally variable:free in the variable:global workspace. global variable

This is a global workspace variable.

counter := 0

 0

Type: NonNegativeInteger

This function refers to the global counter.

f() ==
free counter
counter := counter + 1

Type: Void

The global counter is incremented by 1.

f()

   Compiling function f with type () -> NonNegativeInteger
+++ |*0;f;1;G82322| redefined

 1

Type: PositiveInteger

counter

 1

Type: NonNegativeInteger

Usually FriCAS can tell that you mean to refer to a global variable and so free isn’t always necessary. However, for clarity and the sake of self-documentation, we encourage you to use it.

Declare a variable to be local when you do not want to refer to variable:local a global variable by the same name. local variable

This function uses counter as a local variable.

g() ==
local counter
counter := 7

Type: Void

Apply the function.

g()

 7

Type: PositiveInteger

Check that the global value of counter is unchanged.

counter

 1

Type: NonNegativeInteger

Parameters to a function are local variables in the function. Even if you issue a free declaration for a parameter, it is still local.

What happens if you do not declare that a variable x in the body of your function is local or free? Well, FriCAS decides on this basis:

1. FriCAS scans your function line-by-line, from top-to-bottom. The right-hand side of an assignment is looked at before the left-hand side.
2. If x is referenced before it is assigned a value, it is a free (global) variable.
3. If x is assigned a value before it is referenced, it is a local variable.

Set two global variables to 1.

a := b := 1

 1

Type: PositiveInteger

Refer to a before it is assigned a value, but assign a value to b before it is referenced.

h() ==
b := a + 1
a := b + a

Type: Void

Can you predict this result?

h()

   Compiling function h with type () -> PositiveInteger
+++ |*0;h;1;G82322| redefined

 3

Type: PositiveInteger

[a, b]

 [3,1]

Type: List PositiveInteger

What happened? In the first line of the function body for h, a is referenced on the right-hand side of the assignment. Thus a is a free variable. The variable b is not referenced in that line, but it is assigned a value. Thus b is a local variable and is given the value a+1=2. In the second line, the free variable a is assigned the value b+a which equals 2+1=3. This is the value returned by the function. Since a was free in h, the global variable a has value 3. Since b was local in h, the global variable b is unchanged—it still has the value 1.

It is good programming practice always to declare global variables. However, by far the most common situation is to have local variables in your functions. No declaration is needed for this situation, but be sure to initialize their values.

Be careful if you use free variables and you cache the value of your function (see ugUserCache ). Caching only checks if the values of the function arguments are the same as in a function call previously seen. It does not check if any of the free variables on which the function depends have changed between function calls.

Turn on caching for p.

)set fun cache all p

function p will cache all values.

Define p to depend on the free variable N.

p(i,x) == ( free N; reduce( + , [ (x-i)^n for n in 1..N ] ) )

Type: Void

Set the value of N.

N := 1

 1

Type: PositiveInteger

Evaluate p the first time.

p(0, x)

 x

Type: Polynomial Integer

Change the value of N.

N := 2

 2

Type: PositiveInteger

Evaluate p the second time.

p(0, x)

 x

Type: Polynomial Integer

If caching had been turned off, the second evaluation would have reflected the changed value of N.

Turn off caching for p.

)set fun cache 0 p

Caching for function p is turned off

FriCAS does not allow fluid variables, that is, variables variable:fluid bound by a function f that can be referenced by functions called by f. fluid variable

Values are passed to functions by reference: a pointer to the value is passed rather than a copy of the value or a pointer to a copy.

This is a global variable that is bound to a record object.

r : Record(i : Integer) := [1]

 [i=1]

Type: Record(i: Integer)

This function first modifies the one component of its record argument and then rebinds the parameter to another record.

resetRecord rr ==
rr.i := 2
rr := [10]

Type: Void

Pass r as an argument to resetRecord.

resetRecord r

 [i=10]

Type: Record(i: Integer)

The value of r was changed by the expression rr.i:=2 but not by rr:=[10].

r

 [i=2]

Type: Record(i: Integer)

To conclude this section, we give an iterative definition of Fibonacci numbers a function that computes Fibonacci numbers. This definition approximates the definition into which FriCAS transforms the recurrence relation definition of fib in ugUserRecur .

Global variables past and present are used to hold the last computed Fibonacci numbers.

past := present := 1

 1

Type: PositiveInteger

Global variable index gives the current index of present.

index := 2

 2

Type: PositiveInteger

Here is a recurrence relation defined in terms of these three global variables.

fib(n) ==
free past, present, index
n < 3 => 1
n = index - 1 => past
if n < index-1 then
(past,present) := (1,1)
index := 2
while (index < n) repeat
(past,present) := (present, past+present)
index := index + 1
present

Type: Void

Compute the infinite stream of Fibonacci numbers.

fibs := [fib(n) for n in 1..]

 [1,1,2,3,5,8,13,21,34,55,…]

Type: Stream PositiveInteger

What is the 1000th Fibonacci number?

fibs 1000

434665576869374564356885276750406258025646605173717804024_
8172908953655541794905189040387984007925516929592259308_
0322634775209689623239873322471161642996440906533187938_
298969649928516003704476137795166849228875

Type: PositiveInteger

As an exercise, we suggest you write a function in an iterative style that computes the value of the recurrence relation having the initial values . How would you write the function using an element OneDimensionalArray or Vector to hold the previously computed values?


text:
# Find the probability of having 3 cards of the same suit and 2 cards of the same suit in a 5 card hand from a standard 52 card deck? (Method)

I'm trying to understand why my method doesn't work to answer the question: What is the probability of having 3 cards of the same suit and $2$ cards of the same suit (but a different suit than the first three) in a $5$ card hand from a standard $52$ card deck?

The method that seems to work uses combinations. You do: $${4 \choose 1}\cdot{13 \choose 3}\cdot{3 \choose 1}\cdot{13 \choose 2} = 267,696$$ To find the number of successful possibilities and then $${52 \choose 5} = 2,598,960$$ To find the number of total possibilities. So the answer is: $$\frac{267,696}{2,598,960}$$

I'm wondering why figuring out the individual card chances and then multiplying them doesn't work. My method goes like this: $$\frac{12}{51}\cdot\frac{11}{50}\cdot\frac{39}{49}\cdot\frac{12}{48}$$

After you've picked your first card you have a $\frac{12}{51}$ chance of getting a matching second card, then you've a $\frac{11}{50}$ chance of getting your third card to match.

For the other two cards you've a $\frac{39}{49}$ chance, because you still have three suits left. You need the next one to match, so now you have a $\frac{12}{48}$ chance.

Shouldn't multiplying all of those individual probabilities yield the same answer as the first method? I realize that there are different ways to order the individual probabilities, but when they're all multiplying, shouldn't it all work out to be the same?

I've noticed that multiplying the answer to my method by $10$ makes it equal to the first method's answer. Is there a reason that works?

-

Your calculation is the chance that you will draw three matching cards first, followed by a pair. But when you are looking for a full house, the second card doesn't have to match the first, so you have missed some possibilities. The factor $10$ is because there are ${5 \choose 3}=10$ different ways to choose the positions of the five draws that will be the three of a kind.

-
Thanks for the answer! I think I'm starting to understand why it would work like that. But why is it 5C3 instead of the 5P5? I'd always thought that you multiplied by the permutations of all the individual ways to choose. –  user74928 Apr 29 '13 at 4:10
@user74928: because the three of a kind are interchangeable, so you are choosing which three positions in the draw order they occupy. Your calculation already allows for changing the order of the three matching cards and of the two cards in the pair. –  Ross Millikan Apr 29 '13 at 13:34

(P(13,2) * P(13,3) * C(4,2) * [5!/2!3!]) / P(52,5) = 429/8330

To find the number of successful possibilities:
P(13,2) --> different ways to draw 2 cards from 13 cards of the same suit
P(13,3) --> different ways to draw 3 cards from 13 cards of the same suit
C(4,2) --> different ways to draw 2 separate suits from 4 suits
(5!/2!3!) --> different ways to draw 2 of one card type and 3 of another

To find the total number of possibilities:
P(52,5) --> ways to draw 2 cards from a 52 card deck

-


text:
A result of too much time on my hands and too many hours into Breath of the Wild.

View the code on GitHub and check out a live demo of the project here!

## Background

During spring break 2018, I planned on going on my first ever ski trip with my friend (and current roommate). I had gotten everything packed to hit the road, but the day before the trip I became so sick that I could not in good conscience go with his family to Colorado. Instead, I spent the entire week bedridden on the couch, playing the new Legend of Zelda game that my brother had recently gotten for his Switch. This video game brought me sheer wonder and joy as I explored the amazing open world, and completed just a fraction of all the adventures that the game had to offer. I tore through the game over the next 8 days and eventually was able to defeat the final boss before my semester resumed on Monday.

Fast forward a year and a half, and I was again on break from college (this time in the winter), without much to do. Again, I was playing copious amounts of Zelda, and pointedly ignoring the goal that I had set for myself of updating my personal website. With these two topics in my mind, I thought about how nice the game loading screen would look on a website, and thus this project was born.

## The web part

Because this was just a simple, meaningless project that I was trying to finish ASAP, the tech involved is not anything to write home about. I have an index.html file with a bunch of divs (maybe not the most semantic of HTML), a styles.sass for the styles and looping animations, and main.js that handles all of the timed animations, as well as the code to display the game tips. SASS and Jquery were used for no other reason than to allow me to write code faster, even with the extra overhead.

### Getting started

What seemed like a relatively straightforward animation actually had a lot of moving parts once I broke it down. There were so many subtle transitions, interactions, and details that I only noticed once I started playing the loading screen frame by frame. Some of the most interesting to me were:

• The subtle glow of the triforce logo in the bottom left once the animation is complete.
• The bounce of the Vah Naboris icon (the camel-shaped one, second from the left).
• The wipe-in/out animation of the top part of the screen.

I spent a large amount of time taking screenshots, timing animations on my phone and ensuring everything was as close to the original as possible. This include matching colors, sizing, and icons seen in the original game. Because many of the icons here are exclusively shown in the loading screen, it was hard to find an existing repository of assets that included the icons I needed, but I eventually found a zip dump of the game files that included the original images (and even found a similar typeface!!).

### So many animations

The divine beast animations in particular were very time-consuming to create. There are 4 of them, each with different animations running on different timers, conveying different emotions. However, the process of creatin and slowly tweaking them to match the game was actually very fun, and I think well worth the effort. For example, here is the code for one of the divine beasts (compiled from SASS -> CSS)

@keyframes vah-naboris {
0% {
bottom: 0;
right: 0;
}

15% {
bottom: 4px;
right: 3px;
}

30% {
right: 6px;
bottom: 0;
}

35% {
right: 6px;
bottom: 2px;
}

40% {
right: 6px;
bottom: 0;
}

43% {
right: 6px;
bottom: 1px;
}

46% {
right: 6px;
bottom: 0;
}

60% {
right: 6px;
}

100% {
bottom: 0;
right: 0;
}
}

As you can see, the animation has 9 steps, and each step of the animation usually only shifts the image by just a few pixels. But altogether, it leads to this smooth, continuous animation:

Animation of the divine beast Vah Naboris

### Getting the details right

This project really was a labor of love, and I tried to ensure that I got every detail right to the best of my ability. This included setting unique game tips that could be swapped by the user (either by pressing ‘a’, ‘enter’, or clicking on the screen), making sure the animations and styles still appeared natural on mobile devices, and even creating a master mode version to mirror the loading screen of DLC.


text:
## Converting series of real numbers to some formula [closed]

I have a finite series of real numbers without any pattern in them. Can we represent such a series with a formula?

Thanks

-
1, 3, 5, 7, ... it's already a formula. This isn't a real question. – Wadim Zudilin Jan 8 2011 at 12:27
Yes, and in infinitely many ways! – Harry Gindi Jan 8 2011 at 12:28
That was not my question @wadim, i've edited it, take a look again. – Adam Jan 8 2011 at 12:33
I counted 7 Adams and 2 Evas (without lastnames) on MO. I don't see a pattern but it's a reasonable formula. – Wadim Zudilin Jan 8 2011 at 12:35
^ arrow for truth. – Harry Gindi Jan 8 2011 at 12:39


text:
# GATE Questions & Answers of Steel Structures Civil Engineering

#### Steel Structures 42 Question(s) | Weightage 05 (Marks)

A steel column of ISHB 350 @72.4 kg/m is subjected to a factored axial compressive load of 2000 kN. The load is transferred to a concrete pedestal of grade M20 through a square base plate. Consider bearing strength of concrete as $0.45f_{ck}$, where $f_{ck}$ is the characteristic strength of concrete. Using limit state method and neglecting the self weight of base plate and steel column, the length of a side of the base plate to be provided is

In a fillet weld, the direct shear stress and bending tensile stress are 50 MPa and 150 MPa, respectively. As per Is 800: 2007, the equivalent stress (in MPa, up to two decimal places) will be ______

The dimensions of a symmetrical welded I-section are shown in the figure.

The plastic section modulus about the weaker axis (in cm3, up to one decimal place) is ______

A fillet weld is simultaneously subjected to factored normal and shear stresses of 120 MPa and 50 MPa, respectively. As per IS 800 : 2007, the equivalent stress (in MPa, up to two decimal places) is ______

Four bolts P, Q, R and S of equal diameter are used for a bracket subjected to a load of 130 kN as shown in the figure.

The force in bolt P is

A prismatic propped cantilever beam of span $L$ and plastic moment capacity $M_p$ is subjected to a concentrated load at its mid-span. If the collapse load of the beam is $\alpha\frac{M_p}L$ , the value of $\alpha$ is ______

A column is subjected to a load through a bracket as shown in the figure.

The resultant force (in kN, up to one decimal place) in the bolt 1 is ___________

Two plates of 8 mm thickness each are connected by a fillet weld of 6 mm thickness as shown in the figure.

The permissible stresses in the plate and the weld are 150 MPa and 110 MPa, respectively. Assuming the length of the weld shown in the figure to be the effective length, the permissible load P (in kN) is  ____________

The semi-compact section of a laterally unsupported steel beam has an elastic section modulus, plastic section modulus and design bending compressive stress of $500\mathrm{cm}^3$ ,$650\mathrm{cm}^3$ and, $200\;\mathrm{MPa}$ respectively. The design flexural capacity (expressed in kNm) of the section is _________

A propped cantilever of span L carries a vertical concentrated load at the mid-span. If the plastic moment capacity of the section is $M_p$, the magnitude of the collapse load is

Two plates are connected by fillet welds of size 10 mm and subjected to tension, as shown in the figure.  The thickness of each plate is 12 mm.  The yield stress and the ultimate tensile stress of steel are 250 MPa and 410 MPa, respectively.  The welding is done in the workshop $\left(\gamma_{mw}=1.25\right)$.

As per the Limit State Method of IS 800: 2007, the minimum length (rounded off to the nearest higher multiple of 5 mm) of each weld to transmit a force P equal to 270 kN (factored) is

Two bolted plates under tension with alternative arrangement of bolt holes are shown in figures 1 and 2. The hole diameter, pitch, and gauge length are d, p and g, respectively.

Which one of the following conditions must be ensured to have higher net tensile capacity of configuration shown in Figure 2 than that shown in Figure 1 ?

A fixed-end beam is subjected to a concentrated load $\left(P\right)$ as shown in the figure. The beam has two different segments having different plastic moment capacities $(M_p,\;2M_p)$ as shown.

The minimum value of load $\left(P\right)$ at which the beam would  collapse (ultimate load) is

A bracket plate connected to a column flange transmits a load of 100 kN as shown in the following figure. The maximum force for which the bolts should be designed is _____ kN.

For formation of collapse mechanism in the following figure, the minimum value of Pu is cMp/L. Mp and 3Mp denote the plastic moment capacities of beam sections as shown in this figure. The value of c is________.

Prying forces are :

A steel member ‘M’ has reversal of stress due to live loads, whereas another member ‘N’ has reversal of stress due to wind load. As per IS 800:2007, the maximum slenderness ratio permitted is:

The ultimate collapse load (P) in terms of plastic moment Mp by kinematic approach for a propped cantilever of length L with P acting at its mid-span as shown in the figure, would be

A steel section is subjected to a combination of shear and bending actions. The applied shear force is V and the shear capacity of the section is Vs. For such a section, high shear force (as per IS:800-2007) is defined as

The tension and shear force (both in kN) in each bolt of the joint, as shown below, respectively are

A prismatic beam (as shown below) has plastic moment capacity of Mp, then the collapse load P of the beam is

As per IS 800:2007, the cross-section in which the extreme fiber can reach the yield stress, but cannot develop the plastic moment of resistance due to failure by local buckling is classified as

Two steel columns P (length L and yield strength fy=250 MPa) and Q (length 2L and yield strength fy=500 MPa) have the same cross-sections and end-conditions. The ratio of buckling load of column P to that of column Q is:

A propped cantilever made of a prismatic steel beam is subjected to a concentrated load P at mid span as shown.

If load P=80 kN,find the reaction R(in kN) (correct to 1-decimal place)using elastic analysis. __________

A propped cantilever made of a prismatic steel beam is subjected to a concentrated load P at mid span as shown.

If the magnitude of load P is increased till collapse and the plastic moment carrying capacity of steel beam section is 90 kNm, determine reaction R(in kN)(correct to 1-decimal place) using plastic analysis. __________

In a steel plate with bolted connections, the rupture of the net section is a mode of failure under

Two plates are connected by fillet welds of size 10 mm and subjected to tension, as shown in the sketch. The thickness of each plate is 12 mm. The yield stress and the ultimate tensile stress of steel are 250 MPa and 410 MPa, respectively. The welding is done in the workshop (${\gamma }_{mw}$ = 1.25). As per the Limit State Method of IS 800:2007, the minimum length (rounded off to the nearest higher multiple of 5 mm) of each weld to transmit a force P equal to 270 kN is

For the fillet weld of size "s" shown in the adjoining figure the effective throat thickness is

The value of W that results in the collapse of the beam shown in the adjoining figure and having a plastic moment capacity of Mp is

The adjoining figure shows a schematic representation of a steel plate grider to be used as a simply supported beam with a concentrated load. For stiffeners, PQ(running along the beam axis) and RS(running between the top and bottom flanges) which of the following pairs of statements will be TRUE?

A double cover butt riveted joint is used to connect two flat plates of 200mm width and 14mm thickness as show in the figure. There are twelve power driven rivets of 20mm diameter at a pitch of 50mm in both directions on either side of the plate. Two cover plates of 10mm thickness are used. The capacity of the joint in tension considering bearing and shear ONLY, with permissible bearing and shear stresses as 300 MPa respectively is

Two plates, subjected to direct tension, each of 10 mm thickness and having widths of 100 mm and 175 mm, respectively are to be fillet welded with an overlap of 200 mm. Given that the permissible weld stress is 110 MPa and the permissible stress in steel is 150 MPa, then length of the weld required using the maximum permissible weld size as per IS:800-1984 is

In the theory of plastic bending of beams, the ratio of plastic moment to yield moment is called

The square root of the ratio of moment of inertia of the cross section to its cross sectional area is called

A 12 mm thick plate is connected to two 8 mm plates, on either side through a 16 mm diameter power driven field rivet as shown in the figure below. Assuming permissible shear stress as 90 MPa and permissible bearing stress as 270 MPa in the rivet, the rivet value of the joint is

Consider the following statements for a compression member:

I. The elastic critical stress in compression increases with decrease in slenderness ratio.
II. The effective length depends on the boundary conditions at its ends.
III. The elastic critical stress in compression is independent of the slenderness ratio.
IV. The ratio of the effective length to its radius of gyration is called as slenderness ratio.

The TRUE statements are

The shape of the cross-section, which has a largest shape factor, is

Rivets and bolts subjected to both shear stress $\left({\tau }_{\nu f,cal}\right)$ and axial tensile stress $\left({\sigma }_{tf},cal\right)$ shall be so proportioned that the stresses do not exceed the respective allowable stresses , and the value of $\left(\frac{{\tau }_{vf,cal}}{{\tau }_{vf}}+\frac{{\sigma }_{if,cal}}{{\sigma }_{if}}\right)$ does not exceed

A continuous beam is loaded as shown in the figure below. Assuming a plastic moment capacity equal to MP, the minimum load at which the beam would collapse is

A steel flat of rectangular section of size 70 × 6 mm is connected to a gusset plate by three bolts each having a shear capacity of 15 kN in holes having diameter 11.5 mm. If the allowable tensile stress in the flat is 150 MPa, the maximum tension that can be applied to the flat is


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# Addendum of a gear is equal to-

This question was previously asked in
NPCIL ST ME (07/11/2019, Shift-2)
View all NPCIL Scientific Assistant Papers >
1. Circular Pitch
2. 0.3 CP
3. 0.3183 CP
4. 0.3683 CP

Option 3 : 0.3183 CP

## Detailed Solution

Concept:

• It is the radial distance between pitch circle and the addendum circle.
• The addendum is the height by which a tooth of a gear projects beyond (outside for external, or inside for internal) the standard pitch circle or pitch line.
• Also, the radial distance between the pitch diameter and the outside diameter
• In the gear drive, Standard Addendum = 1 module
• But we know that, addendum (a) $$a=\frac{1}{DP}=\frac{CP}{\pi }=0.318~CP$$

• Circular pitch: It is the distance measured on the circumference of the pitch circle from a point of one tooth to the corresponding point on the next tooth. It is usually denoted by CP. Mathematically,

• Circular pitch, $$CP = \frac{{\pi d}}{T}$$

• Diametral Pitch: It is the ratio of the number of teeth to the pitch circle diameter in millimetres. It is denoted by DP. Mathematically,

• $$DP = \frac{T}{d}= \frac{\pi}{CP}$$

• Dedendum: It is the radial distance between the pitch circle and the root circle.
• Working depth: This is the distance of engagement of two mating teeth and is equal to the sum of the addendums of the mating teeth of two gears.
• Whole depth: This is the height of a tooth and is equal to the addendum plus dedendum.
• Face width: This is the width of the gear and it is the distance from one end of a tooth to the other end.
• Face of a tooth: This is the surface of the tooth between the pitch circle and the outside circle.
• Flank of tooth: This is the surface of the tooth between the pitch circle and root circle.
• Clearance: This is the radial distance between the top land of a tooth and the bottom land of the mating tooth.
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text:
## Arithmetic Reasoning Flash Card Set 914300

 Cards 10 Topics Commutative Property, Distributive Property - Multiplication, Exponent to a Power, Factorials, Greatest Common Factor, Negative Exponent, Percentages, Rates, Rational Numbers

#### Study Guide

###### Commutative Property

The commutative property states that, when adding or multiplying numbers, the order in which they're added or multiplied does not matter. For example, 3 + 4 and 4 + 3 give the same result, as do 3 x 4 and 4 x 3.

###### Distributive Property - Multiplication

The distributive property for multiplication helps in solving expressions like a(b + c). It specifies that the result of multiplying one number by the sum or difference of two numbers can be obtained by multiplying each number individually and then totaling the results: a(b + c) = ab + ac. For example, 4(10-5) = (4 x 10) - (4 x 5) = 40 - 20 = 20.

###### Exponent to a Power

To raise a term with an exponent to another exponent, retain the base and multiply the exponents: (x2)3 = x(2x3) = x6

###### Factorials

A factorial has the form n! and is the product of the integer (n) and all the positive integers below it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

###### Greatest Common Factor

The greatest common factor (GCF) is the greatest factor that divides two integers.

###### Negative Exponent

A negative exponent indicates the number of times that the base is divided by itself. To convert a negative exponent to a positive exponent, calculate the positive exponent then take the reciprocal: $$b^{-e} = { 1 \over b^e }$$. For example, $$3^{-2} = {1 \over 3^2} = {1 \over 9}$$

###### Percentages

Percentages are ratios of an amount compared to 100. The percent change of an old to new value is equal to 100% x $${ new - old \over old }$$.

###### Rates

A rate is a ratio that compares two related quantities. Common rates are speed = $${distance \over time}$$, flow = $${amount \over time}$$, and defect = $${errors \over units}$$.

###### Rational Numbers

A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form $${a \over b}$$ where a is the numerator and b is the denominator. An improper fraction ($${5 \over 3}$$) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number ($$1 {2 \over 3}$$) which has a whole number part and a fractional part.