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## Natural Information Without Truth

In Information Without Truth, the Veridicality Thesis for both natural (environmental) and non-natural (semantic) information is rejected.

The Veridicality Thesis for Natural Information (VTN) can be stated:

(VTN) If a signal s being F carries natural information about an object o being G, then o is G.

This is like Dretske’s definition of information flow, and it entails that if an agent A has natural information about an object o being G when they receive a signal of s being F, then o is G.

Contrary to this, in Information Without Truth a Probability Raising Thesis for Natural Information (PRTN) is given, where the transmission of natural information involves nothing more than the truth of the following probabilistic claim:

(PRTN) If a signal s being F carries natural information about an object o being G, then P(o is G | s is F) > P(o is G | $\neg$ (s is F))

It should be noted that they footnote this definition with the following:

One of us has argued that signal s being F can also carry (negative) natural information about o being G by lowering the probability that o is G.

So what would a general definition about carrying natural information look like? It seems to me that something like this would do the job:

If a signal s being F carries natural information about an object o being G, then P(o is G) $\neq$ P(o is G | s is F)

## New Book on The Philosophy of Information

http://ukcatalogue.oup.com/product/9780199232383.do

Luciano Floridi presents a book that will set the agenda for the philosophy of information. PI is the philosophical field concerned with (1) the critical investigation of the conceptual nature and basic principles of information, including its dynamics, utilisation, and sciences, and (2) the elaboration and application of information-theoretic and computational methodologies to philosophical problems. This book lays down, for the first time, the conceptual foundations for this new area of research. It does so systematically, by pursuing three goals. Its metatheoretical goal is to describe what the philosophy of information is, its problems, approaches, and methods. Its introductory goal is to help the reader to gain a better grasp of the complex and multifarious nature of the various concepts and phenomena related to information. Its analytic goal is to answer several key theoretical questions of great philosophical interest, arising from the investigation of semantic information.


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Nginx as Reverse Proxy and Cache for Thin/Sinatra

posted 2012-Jan-26
— updated 2012-Jan-27

Phrogz.net runs a variety of different web applications, such as the page you’re reading now and ObjJob. Most of the web applications are written in Ruby, using the Sinatra as the web framework and Thin as the server. All of these apps, however, sit behind a single Nginx process that acts as the web server for all of Phrogz.net.

As shown above, a single nginx.conf file controls multiple websites. You use a server{ … } block for each logical server, providing the host name used to match the rules therein.

Most of the magic takes place in the try_files command:

• The text in blue—$uri—says, “If you can find a file in the root directory matching the request, just send that first.” • Requests for http://phrogz.net/favicon.png and http://phrogz.net/images/foo.gif will have Nginx serve the file straight from the hard drive. • The text in green—/cache$uri/index.html /cache\$uri.html—says, “Next, look in the cache folder and see if you can find a file matching the request, but ending in .html. Oh, and if the request ends in a slash, look for a file named index.html inside that folder.”
• A request for http://phrogz.net/foo would serve up /var/www/phrogz.net/public/cache/foo.html if that file exists.
• A request for http://phrogz.net/foo/bar/ would serve up /var/www/phrogz.net/public/cache/foo/bar/index.html if that path and file exists.
• Finally, the text in beige—@proxy—says, “If you couldn’t find any files, then just serve up the @proxy.” Which passes off control to the sinatra.phrogz upstream. Which, as shown, uses a round-robin between two different web servers (Thin processes) running on different ports.

A future article will cover using sinatra/cache to automatically populate the cache directory based on controller requests. With just this, however, you have the ability to crawl your website and save out static copies and have them served up in place of the dynamic ones. For example, the ObjJob list of Properties takes almost 10 seconds for Haml to create the 450k of code, but it is served immediately from a disk-based static copy on the server.

format: plain


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Find the area of the region bounded by the curves $y=6-x^{2}$ and $y=x+4$.
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$\textbf{Answer:}$

$A= - \frac{9}{2}$

$\textbf{Explanation:}$

Let $y_1(x) = 6-x^2$ and $y_2(x) = x+4$

Find the end points of the area, for the end points, we have that;

$y_1(x) =y_2(x)$

$6-x^2 = x+4$

$x^2 +x-2=0$

$(x-1)(x+2)=0$

$\implies x=1$ and $x=-2$

$\implies a=-2$ and $b=1$

To see which of the functions is greater, we need to evaluate them at a point between $-2$ and $1$, lets choose $0$

$y_1(0) = 6-(0)^2 = 6$

$y_2(0) = 0+4=4$

$y_1(x)$ is the greater one, hence we have

$A = \int^b_a (y_1(x) -y_2(x)) dx$

$A = \int^1_{-2} ((6-x^2)-(x+4))dx$

$A = \int^1_{-2} (x^2+x-2 )dx$

$A = \frac{1}{3}x^3|^1_{-2}+\frac{1}{2}x^2|^1_{-2}-2x|^1_{-2}$

$A = 3- \frac{3}{2} +6$
by Bronze Status (6,040 points)

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# Thread: General Forms of Recursive Series

1. ## General Forms of Recursive Series

Hi,

I have a set of expressions which are related. The relationship is apparent in the multipliers in each line.

$\displaystyle 3rd: {\it t_1}+{\it t_3}$
$\displaystyle 4th: {\it t_1}+{\it t_2}+{\it t_3}+{\it t_4}$
$\displaystyle 5th: {\it t_1}+2\,{\it t_2}+2\,{\it t_3}+2\,{\it t_4}+{\it t_5}$
$\displaystyle 6th: {\it t_1}+3\,{\it t_2}+4\,{\it t_3}+4\,{\it t_4}+3\,{\it t_5}+{\it t_6}$
$\displaystyle 7th: {\it t_1}+4\,{\it t_2}+7\,{\it t_3}+8\,{\it t_4}+7\,{\it t_5}+4\,{\it t_6}+{\it t_7}$
$\displaystyle 8th: {\it t_1}+5\,{\it t_2}+11\,{\it t_3}+15\,{\it t_4}+15\,{\it t_5}+11\,{\it t_6}+5\,{\it t_7}+{\it t_8}$
$\displaystyle 9th: {\it t_1}+6\,{\it t_2}+16\,{\it t_3}+26\,{\it t_4}+30\,{\it t_5}+26\,{\it t_6}+16\,{\it t_7}+6\,{\it t_8}+{\it t_9}$
$\displaystyle 10th: {\it t_1}+7\,{\it t_2}+22\,{\it t_3}+42\,{\it t_4}+56\,{\it t_5}+56\,{\it t_6}+42\,{\it t_7}+22\,{\it t_8}+7\,{\it t_9}+{\it t_{10}}$

eg. t1 is always multiplied by 1

the multiplication of t2 in each case increments by 1. [0 1 2 3 4 5 6 ....]

the multiplication of t3 in each case increments using the previous increment intervals [0 1 2 3 4 5 6 ....] to give [1 1 2 4 7 11 ...].

the multiplication of t4 in each case increments using the previous increment intervals [1 1 2 4 7 11 ...] to give [0 1 2 4 8 15 26 ...]

My problem is this...

I would like to give a general solution for the Nth order case using some sort of series notation but although there is a pattern clear in the multipliers I am unsure how to go about this.

Does anyone have any experience with this type of problem or how to go about formulating a general solution.

Thanks
Alex

2. Hello, asouthern!

I have a set of expressions which are related.
The relationship is apparent in the multipliers in each line.

. . $\displaystyle \begin{array}{ccccc}\text{3rd:} &\qquad\qquad\qquad\qquad\qquad t_1+ t_3 \\ \text{4th:}& \qquad\qquad\qquad\qquad\qquad t_1 + t_2 + t_3 + t_4 \\ \end{array}$
. . $\displaystyle \begin{array}{cc} \text{5th:} & t_1 + 2t_2 + 2t_3 + 2t_4 + t_5 \\ \text{6th:} & t_1 + 3t_2 + 4t_3 + 4t_4 + 3t_5 + t_6 \\ \text{7th:} & t_1 + 4t_2 + 7t_3 + 8t_4 + 7t_5 + 4t_6 + t_7\\ \text{8th:} & t_1 + 5t_2 + 11t_3 + 15t_4 + 15t_5 + 11t_6 + 5t_7 + t_8 \\ \text{9th:} & t_1 + 6t_2 + 16t_3 + 26t_4 + 30t_5 + 26t_6 + 16t_7 + 6t_8 + t_9 \\ \text{10th:} &t_1+7t_2+22t_3+42t_4+56t_5+56t_6+42t_7+22t_8+7t_9 +t_{10} \end{array}$
I see an even stronger pattern, but . . .

You have a variation of Pascal's Triangle . . .

. . $\displaystyle \begin{array}{cccccccccccc} 3 &&1\;\;0\;\;1\\ 4 && 1\;\;1\;\;1\;\;1\\ 5 && 1\;\;2\;\;2\;\;2\;\;1\\ 6 && 1\;\;3\;\;4\;\;4\;\;3\;\;1\\ 7 && 1\;\;4\;\;7\;\;8\;\;7\;\;4\;\;1\\ 8 && 1\;5\;11\;15\;15\;11\;5\;1\\ 9 && 1\;\;6\;16\;26\;30\;26\;16\;6\;1 \\ 10 && 1\;7\;22\;42\;56\;56\;42\;22\;7\;1\end{array}$

Each number is the sum of the two numbers directly above it.

In row 3, we have: $\displaystyle [1\:0\:1]$ which represents the polynomial: $\displaystyle x^2 + 1$

Multiply by $\displaystyle (x+1)\!:\;\;(x+1)(x^2+1) \:=\:x^3 + x^2 + x + 1$
. . and its coefficients are in row 4.

Multiply by $\displaystyle (x+1)\!:\;\;(x+1)(x^3+x^2+x+1) \:=\:x^4 + 2x^3 + 2x^2 + 2x + 1$
. . Its coefficients are in row 5.

Multiply by $\displaystyle (x+1)\!:\;\;(x+1)(x^4 + 2x^3 + 2x^2 + 2x + 1) \:=\:x^5 + 3x^4 + 4x^3 + 4x^2 + 3x + 1$
. . Ites coefficients are in row 6.

So I see the powerful pattern embedded in the triangle,
. . but so far, I haven't been able to express this in a closed form.

3. Originally Posted by asouthern
Hi,
I have a set of expressions which are related. The relationship is apparent in the multipliers in each line.

$\displaystyle 3rd: {\it t_1}+{\it t_3}$
$\displaystyle 4th: {\it t_1}+{\it t_2}+{\it t_3}+{\it t_4}$
$\displaystyle 5th: {\it t_1}+2\,{\it t_2}+2\,{\it t_3}+2\,{\it t_4}+{\it t_5}$
$\displaystyle 6th: {\it t_1}+3\,{\it t_2}+4\,{\it t_3}+4\,{\it t_4}+3\,{\it t_5}+{\it t_6}$
$\displaystyle 7th: {\it t_1}+4\,{\it t_2}+7\,{\it t_3}+8\,{\it t_4}+7\,{\it t_5}+4\,{\it t_6}+{\it t_7}$
$\displaystyle 8th: {\it t_1}+5\,{\it t_2}+11\,{\it t_3}+15\,{\it t_4}+15\,{\it t_5}+11\,{\it t_6}+5\,{\it t_7}+{\it t_8}$
$\displaystyle 9th: {\it t_1}+6\,{\it t_2}+16\,{\it t_3}+26\,{\it t_4}+30\,{\it t_5}+26\,{\it t_6}+16\,{\it t_7}+6\,{\it t_8}+{\it t_9}$
$\displaystyle 10th: {\it t_1}+7\,{\it t_2}+22\,{\it t_3}+42\,{\it t_4}+56\,{\it t_5}+56\,{\it t_6}+42\,{\it t_7}+22\,{\it t_8}+7\,{\it t_9}+{\it t_{10}}$

...

Alex
From where are you getting the $\displaystyle t_i$ values?
Could you list ALL of the t values at the start?
for example:

$\displaystyle t_0 = -1$
$\displaystyle t_1 = 1$
$\displaystyle t_2 = 3$
$\displaystyle t_3 = 5$
$\displaystyle t_4 = 7$
$\displaystyle t_5 = 9$

which would give:
$\displaystyle S_0 = ?$
$\displaystyle S_1 = ?$
$\displaystyle S_2 = ?$
$\displaystyle S_3 = t_1 + t_3$ = 6
$\displaystyle S_4 = t_1 + t_2 + t_3 +t_4$ = 16
$\displaystyle S_5 = t_1 + 2 t_2 +3 t_3 + 2 t_4 + t_5$ = 45

Soroban shows a modified Pascal Triangle for the coefficients,
and I'm sure he can modify the binomial expansion to give coefficients for any n.

Do you have method that generates the t values or are they already permanently defined?

Is it possible that the $\displaystyle t$ values and the coefficients are related?

4. ## The relation between the t terms

Hi,

Thanks for your replies, I agree the pascal triangles do seem the strongest relationship. As for the definition of the t values. Well,

$\displaystyle t_i$ is shorthand for $\displaystyle r\cos{\theta_i}$

r is constant, the angle changes with every order N, so that,

$\displaystyle \theta_i = \frac{\pi\cdot(1+2(i-1))}{2\cdot{N}}$

i = 1,2...,N

I didn't include the terms for when N =1 and N =2 as they are too simple.

Thanks
Alex

5. Also,

In my mind the general representation could be,

$\displaystyle =\sum_{i = 1}^N(a_i\cdot{t_i})$

e.g.
for N = 4,

$\displaystyle a_i =[1,1,1,1]$

e.g.
for N = 5,

$\displaystyle a_i =[1,2,2,2,1]$

So some way to form the series $\displaystyle a_i$ for any order N, would be sufficient.

Any advice on this would be greatly appreciated.

Thanks Again
Alex


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Why is the Fourier (or cosine) transform decorrelating?

The discrete Fourier transform (DFT) and the discrete cosine transform (DCT) both decompose a signal into its frequency-domain spectrum. One property that I have seen praised across various domains such as image processing, audio/speech processing and more, is that it tends to produce decorrelated coefficients. For example, subsequent coefficients of a signal might tend to be statistically correlated, but its Fourier- or cosine-transformed counterpart does not exhibit such behaviour. As a result, the statistical properties of the latter (e.g. covariance matrix) can be modeled much more efficiently using diagonal matrices instead of dense matrices, among other benefits.

However, I do not immediately see why this decorrelation should hold true in general. What are the reasons behind this fact? Where is the connection between the DFT or DCT's frequency-decomposing property and the described decorrelating property?

• Can you please provide a few examples or references to support your claim? I've never heard of DFT/DCT being describe as "decorrelators" and I've done a ton of audio processing. – Hilmar Jun 11 at 12:56

1 Answer

Whether that decorrelation happens depends on the signals you put in – it's not a general property of the DFT!

Especially, when you model your signal as sum of narrowband signals, then you'll find that due to the DFT of a narrowband signal being concentrated on one or very few bins, and practically 0 elsewhere, the decorrelation simply stems from the fact that it's easy to represent the same signal in frequency domain with fewer non-zero coefficients than in time domain.

The fact that these non-zero coefficients are then uncorrelated amongst each other depends again on the signal model, and isn't widely true – for example, if you have a line spectrum DFT, for example from observing a rectangular wave digital signal, the lines aren't uncorrelated at all.


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# Start a new discussion

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## Site Tag Cloud

Vanilla 1.1.10 is a product of Lussumo. More Information: Documentation, Community Support.

1. Grothendieck Construction

Ammar Husain

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeJul 19th 2019
• (edited Jul 19th 2019)

Thanks for the addition. I have edited a little:

Have added hyperlinks (just enclose technical keywords in double square brackets!).

Have changed the notation for the delooping groupoid of $G$ from $G$ to $\mathbb{B}G$.

Adjusted wording a little.

Now it reads like so, but be invited to edit further:

Writing $\mathbb{B} G$ for the category with a single object $\ast$ and the group $G$ as its hom set (i.e. the delooping groupoid of $G$), define a functor $F \colon \mathbb{B}G \to$ Cat to send that single object to the delooping groupoid of $\Gamma$, i.e. $* \mapsto \mathbb{B}\Gamma$ and to send the morphisms $G \to Aut(\Gamma)$ according to the given action of $G$ on $\Gamma$.

Then the delooping of the semidirect product group $\Gamma \rtimes G$ arises as the Grothendieck construction of this functor:

$\mathbb{B}( \Gamma \rtimes G) \simeq \int_{\mathbb{B}G}F$
2. This is taken from here: https://mathoverflow.net/a/96256 (with a little bit more detail)

Anonymous


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# Understanding Cryptography by Christof Paar and Jan Pelzl - Chapter 4 Solutions - Ex4.10

- 20 mins

## Exercise 4.10

In the following, we check the diffusion properties of AES after a single round. Let

be the input in 32-bit chunks to a 128-bit AES. The subkeys for the computation of the result of the first round of AES are $W_0,...,W_7$ with 32 bits each are given by

Use this book to figure out how the input is processed in the first round (e.g., SBoxes). For the solution, you might also want to write a short computer program or use an existing one. In any case, indicate all intermediate steps for the computation of ShiftRows, SubBytes and MixColumns!

1. Compute the output of the first round of AES to the input W and the subkeys $W_0,...,W_7$.
2. Compute the output of the first round of AES for the case that all input bits are zero.
3. How many output bits have changed? Remark that we only consider a single round — after every further round, more output bits will be affected (avalanche effect).

### Solution

Feedback kindly provided by Lisa Roy in the comment section has allowed me to correct some errors in earlier versions of this answer. Given that we have now converged upon a common answer, I can be reasonably sure of correctness.

Note: From now on the state is represented as a 4x4 grid of bytes expressed in hexadecimal. This makes other transformations easier to represent also. The ordering of the bytes within the grid is as follows:

$\begin{array}{|c|c|} \hline A_0 & A_4 & A_8 & A_{12} \\ \hline A_1 & A_5 & A_9 & A_{13} \\ \hline A_2 & A_6 & A_{10} & A_{14} \\ \hline A_3 & A_7 & A_{11} & A_{15} \\ \hline \end{array}$

1. This question is slightly confusingly expressed and it took me a while to figure out what is meant. The keys $k_0$ and $k_1$ are given as a single array ($W_0,...,W_7$). In fact, $k_0 = W_0,...,W_3$ and $k_1 = W_4,...,W_7$. As such the first step is the initial $k_0$ key adition which occurs before we get into applying the rounds. The first word of the input is the only one that isn’t all-zeroes. As such, the state after $k_0$ addition is:

Note: From now on the state is represented as a 4x4 grid of bytes expressed in hexadecimal. This makes other transformations easier to represent also.

$\begin{array}{|c|c|} \hline 2A & 28 & AB & 09 \\ \hline 7E & AE & F7 & CF \\ \hline 15 & D2 & 15 & 4F \\ \hline 16 & A6 & 88 & 3C \\ \hline \end{array}$

It’s worth noting that the four columns correspond exactly to the $k_0 = (W_0,...,W_3)$ subkey bytes with the exception of the first cell, which was slightly altered by the XOR operation.

After applying the ByteSubstitution (S-box) layer, the state is as follows:

$\begin{array}{|c|c|} \hline E5 & 34 & 62 & 01 \\ \hline F3 & E4 & 68 & 8A \\ \hline 59 & B5 & 59 & 84 \\ \hline 47 & 24 & C4 & EB \\ \hline \end{array}$

The ShiftRows layer is a very simple transformation. The first row remains unchainged. The other rows are rotated right by a number of positions; the second by 3, the third by 2 and the fourth by 1. This gives the following:

$\begin{array}{|c|c|} \hline E5 & 34 & 62 & 01 \\ \hline E4 & 68 & 8A & F3 \\ \hline 59 & 84 & 59 & B5 \\ \hline EB & 47 & 24 & C4 \\ \hline \end{array}$

The final transformation (other than the $k_1$ addition) is the MixColumn layer. This involves a Galois Extension Field matrix multiplication with the following description:

$\left( \begin{array}{c} C_0 \\ C_1 \\ C_2 \\ C_3 \\ \end{array} \right) = \left( \begin{array}{c} 02 & 03 & 01 & 01 \\ 01 & 02 & 03 & 01 \\ 01 & 01 & 02 & 03 \\ 03 & 01 & 01 & 02 \\ \end{array} \right) \left( \begin{array}{c} B_0 \\ B_5 \\ B_{10} \\ B_{15} \\ \end{array} \right)$

The C values refer to the outputted column. The B values refer to the input columns which were the output of the ShiftRows layer. The indexes here are preserved from prior to the shift. Each new column can be calculated left-to-right using this procedure.

it should also be noted that $03$ here would refer to $x + 1$ in $GF(2^8)$. The same logic is used to turn a hex byte into a Galois Extension Field polynomial. The reduction polynomial is the AES primitive polynomial:

As such the calculation to be performed is as follows for each of the columns:

$\left( \begin{array}{c} 02 \times E5 + 03 \times E4 + 01 \times 59 + 01 \times EB \\ 01 \times E5 + 02 \times E4 + 03 \times 59 + 01 \times EB \\ 01 \times E5 + 01 \times E4 + 02 \times 59 + 03 \times EB \\ 03 \times E5 + 01 \times E4 + 01 \times 59 + 02 \times EB \\ \end{array} \right)$
$\left( \begin{array}{c} 02 \times 34 + 03 \times 68 + 01 \times 84 + 01 \times 47 \\ 01 \times 34 + 02 \times 68 + 03 \times 84 + 01 \times 47 \\ 01 \times 34 + 01 \times 68 + 02 \times 84 + 03 \times 47 \\ 03 \times 34 + 01 \times 68 + 01 \times 84 + 02 \times 47 \\ \end{array} \right)$
$\left( \begin{array}{c} 02 \times 62 + 03 \times 8A + 01 \times 59 + 01 \times 24 \\ 01 \times 62 + 02 \times 8A + 03 \times 59 + 01 \times 24 \\ 01 \times 62 + 01 \times 8A + 02 \times 59 + 03 \times 24 \\ 03 \times 62 + 01 \times 8A + 01 \times 59 + 02 \times 24 \\ \end{array} \right)$
$\left( \begin{array}{c} 02 \times 01 + 03 \times F3 + 01 \times B5 + 01 \times C4 \\ 01 \times 01 + 02 \times F3 + 03 \times B5 + 01 \times C4 \\ 01 \times 01 + 01 \times F3 + 02 \times B5 + 03 \times C4 \\ 03 \times 01 + 01 \times F3 + 01 \times B5 + 02 \times C4 \\ \end{array} \right)$

This produces the following state:

$\begin{array}{|c|c|} \hline 54 & 13 & 3C & 7D \\ \hline 36 & 34 & A2 & FC \\ \hline 95 & 86 & 36 & D4 \\ \hline 44 & 3E & 3D & D6 \\ \hline \end{array}$

All that’s left to do after this is the KeyAddition layer for $k_1 = W_4,...,W_7$:

$\begin{array}{|c|c|} \hline F4 & 9B & 1F & 57 \\ \hline CC & 60 & 01 & 90 \\ \hline 6B & AA & 0F & A2 \\ \hline 53 & 8F & 04 & D3 \\ \hline \end{array}$

Therefore, the output of the first round of encryption is as follows (remembering to read it out column-wise):

2. For the case that the input is all-zeroes, the state after the $k_0$ key addition will be as follows (only the first byte is different from part 1):

$\begin{array}{|c|c|} \hline 2B & 28 & AB & 09 \\ \hline 7E & AE & F7 & CF \\ \hline 15 & D2 & 15 & 4F \\ \hline 16 & A6 & 88 & 3C \\ \hline \end{array}$

After applying the ByteSubstitution (S-box) layer, the state is as follows:

$\begin{array}{|c|c|} \hline F1 & 34 & 62 & 01 \\ \hline F3 & E4 & 68 & 8A \\ \hline 59 & B5 & 59 & 84 \\ \hline 47 & 24 & C4 & EB \\ \hline \end{array}$

After applying the ShiftRows layer, the state is as follows:

$\begin{array}{|c|c|} \hline F1 & 34 & 62 & 01 \\ \hline E4 & 68 & 8A & F3 \\ \hline 59 & 84 & 59 & B5 \\ \hline EB & 47 & 24 & C4 \\ \hline \end{array}$

Up until this point, only one byte (in fact only one bit) has been altered in compared to part 1. The MixColumn introduces some diffusion, and after this layer is applied, the state is as follows:

$\begin{array}{|c|c|} \hline 7C & 13 & 3C & 7D \\ \hline 22 & 34 & A2 & FC \\ \hline 81 & 86 & 36 & D4 \\ \hline 78 & 3E & 3D & D6 \\ \hline \end{array}$

The first column of the state now has a different set of values than in part 1, the others are unaltered. All that’s left to do after this is the KeyAddition layer for $k_1 = W_4,...,W_7$:

$\begin{array}{|c|c|} \hline F4 & 9B & 1F & 57 \\ \hline CC & 60 & 01 & 90 \\ \hline 6B & AA & 0F & A2 \\ \hline 53 & 8F & 04 & D3 \\ \hline \end{array}$

Therefore, the output of the first round of encryption is as follows (remembering to read it out column-wise):

3. We can see how many output bits have been altered by XORing the two output values together. This produces:

In this form, we can clearly see that only the first column is altered after the first round.

The 1s in the binary above correspond to output bits which have changed. There are ten of them, so therefore the number of output bits which have changed due to a 1 bit change in input, in this instance, is 10 after the first round. The ShiftRows operations in further rounds will diffuse this effect to other columns.

I wrote a python script which can perform the operations involved in computing an AES round:


text:
Enter the specific heat, change in temperature, and mass into the calculator to determine the total amount of heat absorbed.

## Heat Absorbtion Formula

The following formula is used to calculate the total amount of heat energy absorbed by a mass.

Q = m*C*ΔT

• Where Q is the total amount of heat energy absorbed by a mass
• C is the specific heat of the material
• ΔT is the change in temperature

This formula is also used for heat loss calculations, but the difference, in this case, is that the final temperature will be higher than the initial temperature.

Therefore, the formula for the change in temperature would be T2-T1 instead of T1-T2 in the case of cooling.

Heat absorption is the process of a material increasing in temperature due to it taking energy for a higher temperature material or solution.

The methods of heat transfer include conduction, convection, and radiation.

## Example Problem

How to calculate heat absorption?

First, determine the total mass of the object or solution that is absorbing the heat.

For this example problem, the total mass of the solution is found to be 50kg.

Next, determine the specific heat of the material.

In this case, the specific heat is found to be 400 J/kg-C.

Next, determine the total change in temperature. The initial temperature is 20 C and the final temperature is 60C. Therefore the change in temperature is calculated to be 60C-20C = 40C.

Finally, calculate the total amount of heat absorbed using the formula above:

Q = m*C*ΔT

Q = 50*400*40

Q = 800,000 J.


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# Power of the infinitesimal generator

Let $$A$$ be the infinitesimal generator of a $$C_0$$ semigroup of linear operators in a Banach space. Let $$n$$ be a positive integer $$n \geq 2$$? Is the power operator $$A^n$$ closed?

Here (setting $$A^1$$ $$:=$$ $$A$$, and denoting the domain of $$A$$ by $$\cal{D}(A)$$), the operator $$A^n$$ has been defined inductively for $$n=2,3...,$$, by $${\cal{D}}(A^n):=\{f: f\in {\cal{D}}(A^{n-1})\; and \; A^{n-1}f \in {\cal{D}}(A) \},$$ $$A^{n}f:=A (A^{n-1} f).$$


text:
# Limit at infinity of a uniformly continuous integrable function [duplicate]

This is an exercise from Berkeley preliminary exams, Fall 1983

Let $f : [0, \infty ) \rightarrow \mathbb{R} \$ be a uniformly continuous function with the property that

$\lim_{b \to \infty}\int_{0}^{b} f(x)dx \$

exists (as a finite limit). Show that

$\lim_{x \to \infty}f(x) = 0$

Obviously if the limit exists, it must be $0 \$; so the problem is to prove that the limit exists. Any hint ?

-

## marked as duplicate by Rudy the Reindeer, Benjamin Lim, Ilya, J. M., Brian M. ScottMay 3 '12 at 8:14

Are you sure that the last limit is for $x \to 0$? I guess it should be $x \to +\infty$. – Giuseppe Negro Mar 25 '12 at 10:18
This is Barbalat's Lemma. A solution can be found here: mathproblems123.wordpress.com/2009/10/01/barbalats-lemma And the last limit should be with $x \to \infty$, because changing the function near $0$ such that it still remains uniformly continuous does not affect the converge of the integral limit. – Beni Bogosel Mar 25 '12 at 10:24
@BeniBogosel: You required $f$ to be positive ($f \colon [0, \infty) \to [0, \infty)$) but this hypothesis is absent here. While your argument pushes through if $f \in L^1([0, \infty))$, I'm afraid it doesn't work if $f$ is allowed to change sign and $\int \lvert f(x)\rvert\, dx=+\infty$. – Giuseppe Negro Mar 25 '12 at 10:35
@Giuseppe Negro: I have edited – WLOG Mar 25 '12 at 11:10
@GiuseppeNegro: I know that the hypothesis is $f$ positive in my proof, but the proof does not make use of that hypothesis anywhere but in the place I say that $f(x_n) \to \ell > 0$, which can be assumed WLOG here (the idea is that we assume that there exists a sequence $x_n \to \infty$ such that $f(x_n)$ does not converge to zero). The proof works just fine if you assume that $f(x_n) \to \ell<0$. – Beni Bogosel Mar 25 '12 at 12:04

Suppose that $\lim_{x\to \infty} f(x)$ doesn't exist. Then there is an $\epsilon > 0$ and a sequence $x_n \to \infty$ such that $|f(x_n)| > \epsilon$ for all $n$ (because the limit, if existing, has to be 0). By uniform continuity there is a $\delta > 0$ such that $|f(x) - f(y)| < \frac{\epsilon}2$ if $|x-y| < \delta$. It follows that $|f(x)| > \frac{\epsilon}2$ if $|x-x_n| < \delta$ for some $n$. But now $|\int_{x_n-\delta}^{x_n+\delta} f(x)\, dx| > \epsilon\delta$ for all $n$ contradicting $\int_a^b f(x)\,dx \to 0$ for $a,b \to \infty$.
I have the same objection I had in comments to the main post: this argument certainly works if $f$ is non-negative, or if it is $L^1$, but I'm afraid it doesn't if $f$ is allowed to change sign and $\int \lvert f \rvert\, dx=+\infty$. Do you agree? – Giuseppe Negro Mar 25 '12 at 10:38
@GiuseppeNegro Why? If $\int_0^\infty f(x) \,dx$ exists as in inproper integral, we need to have $\int_a^b f(x) \, dx \to 0$ for $a, b\to \infty$, which yields that there is an $M$ such that $|\int_a^b f(x)\, dx| < \epsilon\delta$ for all $a,b \ge M$. But now I can choose $x_n$ with $x_n - \delta > M$. I don't see where I could have used $f \in L^1$. – martini Mar 25 '12 at 10:44
Oh yes, yes, you're right. The fact that $\int_a^bf(x)\, dx \to 0$ is certainly true and you argument contradicts it. I had mistakenly seen a contradiction with $\int\lvert f(x)\rvert \, dx <+\infty$. Thank you for clarifying! – Giuseppe Negro Mar 25 '12 at 15:09
@martini Did you mean "Suppose that $$\lim_{x \to \infty} f(x) \neq 0$$" in your first sentence? – Rudy the Reindeer May 2 '12 at 20:56
@MattN. As I wrote in the paratheses: If the limit exists, it has to be 0, for if it is $c > 0$ (wlog) say, then we would have $f(x) > \frac c2$ for $x \ge N$ ($N$ choosen properly) and $\int_0^\infty f(x)\, dx$ would diverge. – martini May 2 '12 at 20:58


text:
# [NTG-context] kerning with \quotation

Hans Hagen pragma at wxs.nl
Sun Mar 6 15:56:44 CET 2016

```On 3/6/2016 2:19 PM, Schmitz Thomas A. wrote:
> Hi all,
>
> short example in extra big size to make the effect visible:
>
> \setupbodyfont [termes,48pt]
>
> \mainlanguage [de]
>
> \starttext
>
> „Vater“
>
> \quotation{Vater}
>
> \stoptext
>
> In the first line, the character „ has a bit of kerning with the following V (could be even more, to my taste). With \quotation, the gap between them is very large. Could the kerning be applied in this case too? Or is it better to use the raw „? (After I have taught emacs to auto-insert \quotation{ and } when I type a quotation mark…)

\setupdelimitedtext
[method=font]

-----------------------------------------------------------------


text:
# Another oldie

• December 13th 2008, 09:23 AM
Soroban
Another oldie

A surveyor starts at the center of a circular park with radius 7 miles.

He walks 3 miles directly east and drives a stake into the ground at point $A.$

He walk directly north until he reaches the perimeter of the park.
Then he walks directly west until he is directly north of the center,
. . and drives a stake into the ground at point $B.$

Find the distance $\overline{AB}.$

.
• December 13th 2008, 10:10 AM
skeeter
here's the hint:

diagonals of a rectangle are ... ?
• December 13th 2008, 01:06 PM
CaptainBlack
Quote:

Originally Posted by Soroban
A surveyor starts at the center of a circular park with radius 7 miles.

He walks 3 miles directly east and drives a stake into the ground at point $A.$

He walk directly north until he reaches the perimeter of the park.
Then he walks directly west until he is directly north of the center,
. . and drives a stake into the ground at point $B.$

Find the distance $\overline{AB}.$

.

You have a rectangle one diagonal of which is a radius, AB is the other diagonal.

CB
• December 14th 2008, 07:40 AM
ice_syncer
Use pythagoras theorem, the shortest possible method, no need of trigonometry (Thinking)

CLUE: IT IS A SURD!!!!
• December 14th 2008, 11:26 AM
skeeter
Quote:

Originally Posted by ice_syncer
Use pythagoras theorem, the shortest possible method, no need of trigonometry (Thinking)

CLUE: IT IS A SURD!!!!

don't think so ...
• December 15th 2008, 02:49 AM
ice_syncer
Quote:

Originally Posted by skeeter
don't think so ...

no, use pythagoras theorem.
• December 15th 2008, 03:00 AM
samer_guirguis_2000
let me guess! 7 miles may be?
• December 15th 2008, 03:13 AM
Greengoblin
^^The diagonals of a rectangle are equal, so it equals the radius, 7 miles.
• December 20th 2008, 09:10 PM
Aryth
I know this is a while afterwards but, the solution is quite simple.

One diagonal is a radius, so the Northern leg of the trip must be:

$3^2 + b^2 = 7^2$

$b = \sqrt{40}$

So, the parallel side is also $\sqrt{40}$, so therefore the diagonal AB must be:

$3^2 + (\sqrt{40})^2 = c^2$

$7 = c$

And there you go.


text:
# Simultaneous Equations with Roots

## -

### 28/09/2021Solve Simultaneously

$x+y=11$ $x+y=7$

-/|\-

### Solution:

The equations imply that $x,y>0$.
Both equations have integers of the RHS
implying $x,y,\sqrt{x},\sqrt{y}$ are elements of the integers
moreover $x,y$ are perfect squares
so $0 and $0.
So $x\in \left\{1,4\right\}$ and $y\in \left\{1,4,9\right\}$.
Therefore $x=4$ and $y=9$
Are there any other solutions?

-oOo-

^
Contact Michael Andrew Smith on 0 77 86 69 56 06 or via mike.smith@physics.org

Expires 1 Year After Update

Site by M.A. Smith.


text:
# Tag Info

I think 't Hooft and Kugo are solving different problems. 't Hooft addresses the issue that the anomaly involves a topological term. As a result, in perturbation theory there is no theta dependence and the anomaly equation by itself does not solve the $U(1)$ problem. He shows that topological objects (semi-classically, instantons) generate theta ...


text:
MathOverflow will be down for maintenance for approximately 3 hours, starting Monday evening (06/24/2013) at approximately 9:00 PM Eastern time (UTC-4).

2 deleted 32 characters in body; added 26 characters in body; added 2 characters in body

I have formulated a linear program with binary indicator variables $z_i(a)$ which is equal to $1$ if the $i^{th}$ document is of rank $a$ and $0$ otherwise.

The other variables in the linear program, $z^1_{ij}(a), z^2_{ij}(a)$ are defined as follows:

\begin{eqnarray} z^1_{ij}(a) \equiv z_i(a) \sum_{b < a} z_j(b), \ z^2_{ij}(a) \equiv z_i(a) \sum_{b\geq a} z_j(b). \end{eqnarray}

I am trying to convert the above non-linear constraint to the following set of equivalent linear constraints:

$$z^1_{ij}(a) + z^2_{ij}(a) = z_i(a), \forall i, j, a$$

The problem I am facing is that, the above set of linear constraints are clearly not equivalent to the definition of $z^1_{ij}(a), z^2_{ij}(a)$. Any idea ideas if it is possible to represent convert such non-linear ranking type constraints as equivalent to linear constraints?

1

Is it possible to represent non-linear ranking type constraints as equivalent linear constraints?

I have formulated a linear program with binary indicator variables $z_i(a)$ which is equal to $1$ if the $i^{th}$ document is of rank $a$ and $0$ otherwise.

The other variables in the linear program, $z^1_{ij}(a), z^2_{ij}(a)$ are defined as follows:

\begin{eqnarray} z^1_{ij}(a) \equiv z_i(a) \sum_{b

I am trying to convert the above non-linear constraint to the following set of equivalent linear constraints:

$$z^1_{ij}(a) + z^2_{ij}(a) = z_i(a), \forall i, j, a$$

The problem I am facing is that, the above set of linear constraints are clearly not equivalent to the definition of $z^1_{ij}(a), z^2_{ij}(a)$. Any idea if it is possible to represent non-linear ranking type constraints as equivalent linear constraints?


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# get stuck in the middle of GMAT Prep

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get stuck in the middle of GMAT Prep [#permalink]

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25 Mar 2008, 18:31
I took 2 GMAT Prep tests..scored 42 in the quant first time and 44 in the quant second time...on both exams i got 2 of the first 12 wrong..then i would get 6-7 in a row wrong....after that it was about 60-40 right to wrong ratio...it seems the middle section is killing me...

why am i getting those middle ones wrong? are these 700-800 level questions or 600-700? not 1 in a row..not 2 in a row..but 5 or 6...is that middle killing me more than the end...or is the end killing me more than the middle or both equally....

my scores are ending up 630-650...i feel the quant is keeping me from the upper 600's

any advice is appreciated..i have noticed that the majority of the middle questions are data suff...
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Re: get stuck in the middle of GMAT Prep [#permalink]

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26 Mar 2008, 02:10
well i gave one of the gprep and scored 45 in quant.well for me i noticed only 2 out of 1st ten questions were wrong,even after that i never got at a stretch of more than5 or so questions wrong.it was like getting 2 wrong...then 3 correct etc....with the last two wrong bcos of short of time...i have to give other to analyse deeply.i guess concentrating on 1st ten questions r fine but provided u dont follow a slump of wrong questions...just try this in ur next exam...one favour...i scored a miserably low 25 in verbal.I noticed out of 1st ten questio i got many wrong..may be 6...and most of my sc were correct....only 2-3 wrong...but very bad cr.....rc was ok....was dat the reason for sucha low score since i ended up in 580...plz advice...how to improve verbal scores....giving it next friday...
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26 Mar 2008, 06:13
what helped me in CR was to read the question and narrow the answers down to 2..on my scratch paper i write A,B,C,D,E and scratch them off so that I'm left with two, this also helps me focus only on those remaining (i tried this technique in my head and I would spend more time trying to figure out which ones I scratched off)..i do this for every CR question...if i end up with only 1 letter I go back until I get two...If I end up with 3 I go back until I can get only two..then I focus on the two...you should easily be able to scratch off two of the letters because usually 2 are way off. Then focus on 2 answers...the rest is up to you.

1st goal...make sure the correct answer is one of the two choices..if not...keep practicing..

pretty much stuff verbal guides will teach
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26 Mar 2008, 09:22
thanks lagomez...ya i can easily eliminate two...but then i will try to formulate the ans without looking the ans choices....thanks once again....happy gmatting...
Re: get stuck in the middle of GMAT Prep   [#permalink] 26 Mar 2008, 09:22
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# Parabola

## Setup

The first thing we'll do is create a file containing the system describing the curve we want to decompose.

Since tropical geometric objects intersect the hyperplane at $$\infty$$, we need to be able to compute points in this hyperplane. That's right, we'll compute points at $$\infty$$! The tool we'll use is projectivization, a method of compactification which identifies all points on the same line as being the same point. This is accomplished by introducing a new variable to the system, called the homogenizing variable. We typically call it $$h$$.

For our parabola, the starting system looks like
$$y = x^2-1,$$
but we phrase problems in terms of a system vanishing, so we rewrite it as
$$-y+ x^2-1=0.$$
Now we homogenize. Every term (summand) in the system must have the same degree. The term $$y$$ has degree 1, $$x^2$$ has degree 2, and $$1$$ has degree 0. The max is 2. Hence, each term needs to have degree 2. We increase degree by multiplying by the new homogenizing variable $$h$$. Hence, the system becomes
$$h y + x^2 - h^2$$
But now there are three variables and a single constraint, so it looks like a dimension-two object, whereas the parabola we started with was of dimension one.

The second part of projectivization is to introduce an affine patch. It picks out a single point on each line which is identified in projective space. We'll choose a random patch,
$$p_h h + p_x x + p_y y -1 = 0.$$

Finally we are ready to write the Bertini input file to be used to decompose the curve.
Note the top portion, bracketed by CONFIG and END;. Bertini 1 passes its configuration via a config section at the top of the input file. We will be working with positive-dimensional varieties, so we indicate to Bertini to generate a set of witness sets by passing tracktype: 1;.

## Running

Now that we have our system set up in a Bertini input file, we will go through the steps to do the tropical decomposition.

### NID

The input for the tropical decomposition includes the witness set for the component to be decomposed. Hence, we need to generate that witness set. The Numerical Irreducible Decomposition (NID) contains that witness set, and the algorithm for computing it is accessed by using tracktype:1;.

All we need to do is call Bertini, assuming the input file is simply called 'input'. Hence, move to the folder containing the input file (which you should name something corresponding to the system, such as 'parabola'). This can be done either in a terminal, or in Matlab.

Then call Bertini. Let it run, for this system, it should complete very quickly, like less than one second. My output looks like this:
The most important part is the last bit, where it tells us we have a single dimension 1 classified component, of degree 2. Briefly, this means that a random complex intersection with this component will yield two points. Indeed, the witness set contains two points.

The output from Bertini is the 'witness data' file, contained in the working folder.

### Tropical Decomposition - $$\mathbb{C}$$

Now that we have the NID in 'witness_data', we can compute the tropical decomposition of the component.

To do so, we simply create a 'tropical_curve' object in Matlab. The constructor for this class, provided in this code, performs the decomposition. There are options you can pass in, such as whether you want the complex or real curve, and several numerical tolerances.

For now, let's stick to default mode, which does the decomposition of the complex curve. Here's what I get on my machine:
There's a lot going on in that output. Beyond that, a folder has been made in the current directory, containing the data for the run, and the tropical curve you made was automatically saved to a time-stamped .mat file. I just got:
• a folder called parabola_complex_tropical_decomposition
• a file called tropdecomp_parabola_complex_1601261210.mat.
The parabola part of these names is determined by the name of the working folder. The time stamp is evident.

The rays and multiplicities are the most important part of the tropical curve. If you look at the output, you can see that the object 'T' that we just made
• is not real
• is homogenized
• has some rays
• has some multiplicities
• and has a bunch of other data, including 'aux_data'.
To look at the rays, just call the name of the field in Matlab: 'T.rays'.
One thing to notice is that there are too many coordinates in some sense. Well, we homogenized, first thing. We introduced a new variable, $$h$$, and it appears at the front of the variable ordering. Let's de-homogenize. I built the call into Matlab:
Since tropical_curves are implemented as handle classes in Matlab, the call T.dehomogenize modifies T. It does not produce a new copy of T which is dehomogenized, but rather T itself changes. This is ok, the tropical curve automatically saved, so you can just re-load it from disk if you feel like you screwed something up.

Now let's look at the rays, and their multiplicities, without the 'extra' homogenizing variable mucking things up.
Sure enough, the rays are as expected. Columns in the matrix represent the rays. Ok, so we have rays and multiplicities, but do they balance? Ray-multiplicity structure balances for tropical curves, in the sense that they sum to 0, if you multiply each ray by its multiplicity, and add them all up for all rays, you should get 0. If not, then an error certainly occurred. Let's check.
They balanced! The curve was decomposed correctly.

### Tropical Decomposition - $$\mathbb{R}$$

We have successfully decomposed the parabola's complex curve. What about what we usually think of as a parabola, the real curve? The code I provide can compute real tropicalizations, too! It starts from the point of a complete NID, just as the $$\mathbb{C}$$ does. So we just call T = tropical_curve('real') and let the algorithm go.
The data automatically saves, and the bread crumbs from the Bertini runs are preserved in their own folder, similarly to the complex case.

Real tropical curves can be arranged according to how the curve crosses the hyperplane, something which can't be done for complex curves. The command 'arrange' for tropical real curves in Matlab will render the rays in space for me. First, let's dehomogenize, then call arrange
Ok, the code is letting me know that the command 'labelText' will allow for re-sizing of the fonts in the image. Whatever. For now let's just save it to disk, and put it here on this website.