What does **dict mean in python formatting expression?
What does ** (double star/asterisk) and * (star/asterisk) do for parameters?
Inequality vs. Inequation, Equation vs. Equality When talking about statements that two terms are mathematically (un-)equal (e.g., 1 = 1, 1 &lt; 2, or 1 != 2), what is the correct notion for such mathematical statements? Are there special cases w.r.t. "mixed" relations like x &lt;= y? Can some notions be used interchangeably?  It seems many online sources use "equation" and "inequality" but there is also a . Apart from personal posts in forums, I have not found any sources using "equality" in such a situation, yet. Systematically, I would think that "equation" and "inequation" would make more sense than "equation" and "inequality", but not everything (probably very few things) in natural languages have developed according to a purely logical system.  Note that this question is more general than  for that it particularly asks for the relation between the notion for equations and the one for inequalities. Equations are not part of the existing question at all. Why are there 2 packages for some applications (e.g. Chromium) in Ubuntu Software? Is "I'm being angry" ungrammatical? Which is better: side quests after killing the warrior or Playthrough 2? My question is, once I beat the warrior should I do all the sidequests in Normal before starting TVHM or should I go right to TVHM once it becomes available?    I just bought the game and have been playing with some friends who are higher level than me so that we could enjoy the Tiny Tina DLC together.  As a result, I leveled up pretty fast without doing much of the main story quest or side quests.  Now I'm a level 32 commando and haven't finished the main storyline in normal (Just finished "Where angels fear to treat Pt 2").  I get almost no XP from kills on missions.  I still have tons of sidequests.  I imagine some of them might give good loot, or be fun to play.  It it better to go to Playthrough 2 or should i do the sidequests in Normal first?  EDIT: changed question title and some wording for clarity Continuously copy all file changes from one folder to another on change (Windows) I'm working on a massive Java web application which is kept under centralized version control, but the source files are used to build and run the actual server, which then copies all the files I work on into a random temporary directory while it's running.  My problem is that I need to either rebuild this huge application to see any changes to my files, or keep track of all my changes and copy over the files I've changed back to their source location before committing them back to source control.  My question is: is there a command line script I can run that will monitor any changes in the source folder and automatically copy changed files across to temporary folder X?  The idea is to remove the human point of failure in having to manually mitigate the two-locations problem. We want to find the solution $x$ to the congruence system  $$\begin{align} x &amp;\equiv r_1 \!\!\!\pmod{\!m_1}\\ x &amp;\equiv r_2 \!\!\!\pmod{\!m_2}\end{align},\ \ {\rm where}\ \  \gcd(m_1, m_2) = 1$$  These can be rewritten as  $$\begin{align} x &amp;= r_1 + m_1 j\\ x &amp;= r_2 - m_2k\end{align}$$  for unknown integers $j, k$. So we set those two equations as equal and rearrange them to  $$m_1j + m_2k = r_2 - r_1$$  Now assume we perform $\text{egcd}(m_1, m_2)$, which gives us the solution to  $$m_1u + m_2v = \gcd(m_1, m_2)$$  Let the integer $\,h\,$ make $\,h \gcd(m_1, m_2) = r_2 - r_1$.  Then $\qquad hm_1u + hm_2v = h\gcd(m_1, m_2) =  r_2 - r_1 = m_1j + m_2k$  At this point I am lost because I don't think we can just assume  $hm_1u = m_1j$ and extract $j = hu$.   How do I get the values of $j$ or $k$ so I can get the value of $x$? How to Understand the Definition of Cardinal Exponentiation I'm having trouble understanding the definition of cardinal exponentiation.  Let's start with the definitions / claims I've been given:  For any finite sets $A,B$, such that if $|A|=a$ and $|B|=b\neq 0$ then the number of functions from $A$ to $B$ is $b^a$. ( I'm guessing this is only used when both sets are finite ).  For any sets $A,B$, $B^A$ is the set of functions from $A$ to $B$.  For every set $A$, $|P(A)|=|\left \{ 0,1 \right \}^A|$  Now for the definition and why I don't understand it:  Let $a,b$ be cardinal numbers.  Let $A,B$ be sets such that $|A|=a, |B|=b$.  Then $|B^A|=b^a$.  How analagous to normal exponentiation is this and where does in differ when we introduce cardinals like $\aleph_0$ or $C$?  Is $2^2=4$ and is $\frak c^2=c\cdot c=c\times c$? I am trying to install a Ralink wifi adapter in Ubuntu Desktop 14.04.  The system is not detecting the wifi.   lsusb lists the device as:  Bus 002 Device 004: ID 148f:7601 Ralink Technology, Corp.  I have the rt2800 USB driver listed in "/etc/modules"; Following some instructions on the web, I tried this command:  echo 148F 7601 | sudo tee /sys/bus/usb/drivers/rt2800usb/new_i  lsmod | grep rt2800 returns the following:  rt2800usb              26581  0  rt2x00usb              20041  1 rt2800usb rt2800lib              83150  1 rt2800usb rt2x00lib              48886  3 rt2x00usb,rt2800lib,rt2800usb mac80211              545990  3 rt2x00lib,rt2x00usb,rt2800lib crc_ccitt              12627  1 rt2800lib How to test multiple variables against a single value? single email limit exceeded salesforce very frequently..? Is changing your passport a good strategy for removing a negative immigration history in Schengen? When certain nationalities of people want to travel in the Schengen zone, they are required to apply for a Schengen visa. This is a straight-forward process for most, but when an applicant has had problems with removals or refusals they may be reluctant to apply because they want to avoid the complexities of serial visa refusals. This reluctance may lead some into wondering how their history can be expunged, and this line of thought inevitably leads to the idea of getting their passport replaced with a new one.  This sounds like a great formula for people who have had prior refusals or removals.  In essence they can become 'reincarnated' as a first-time traveller and hence avoid the unpleasantness of revealing their prior interactions with the Schengen authorities.  The relevant parts of the application form are shown here...    Nothing asks about a prior removal or refusal, and the applicant is left with simple denials for 3 questions in order to present themselves as a first-time traveller.  Assuming the person is happy to set aside the moral issues associated with denying their history, are there any holes to this strategy?  This question is scoped to Schengen visa applications so that it will not be unmanageably broad, but answers from other regimes (particularly the affluent Commonwealth and the USA) are acceptable as long as the main points are relatively consistent with Schengen. A malevolent computer with AI takes over a woman's house with the intent of hybrid computer/human procreation. The evil computer seals off the house with the woman in it and keeps her prisoner while it prepares. I read the tale sometime in the late 80's or the early 90's. It was one of many of my missing books that got yardsaled by my in-laws when I wasn't looking. It may also have been made into a movie, but I didn't see the film. Why is feed-forward mechanism used in hash functions? The compression function of SHA-1 when used in Davies-Meyer mode adds its input to the chaining values at the final step. For the first message block, the IV is used as the input and in the next step, the previous block's output (as specified by Merkle-Damgård) is used. These values are fed to the last round of the compression function.  Why would removing the feed-forward part (i.e.,the final step where the input value is added to the result to get the final output) make the hash function insecure? Apart from the fact that the compression function would not satisfy desired randomness properties (being a permutation for each fixed message). I'd never heard of anycast until a few seconds ago when I read "".  The wikipedia "" article on it is quite formal and doesn't really evoke a mental picture of how it would be used.  Can someone explain in a few informal sentences what "anycast" is, how you configure it (just in a general sense), and what its benefits are (what does it make easier)? Bibliographies from multiple .bib files I would like to use multiple .bib files for my article, such as file1.bib, file2.bib. However, I would like my references not to be separated (such as &quot;Primary sources&quot;, &quot;Secondary sources&quot;, etc.). These bibliographies just need to be merged into one. One way to do this is to copy-paste these .bib files into one .bib file, and this would solve the problem. However, I would like to keep it split into several .bib files, by topic (file1.bib contains articles on topic1, file2.bib contains articles on topic2, and so on). I tried to do \bibliographystyle{alpha}  \bibliography{file1} \bibliography{file2}  but this just gives me the same bibliography twice! Is there a way to make RC4 (ARCFOUR) secure, or is it completely broken? I need a method to authenticate a process with another in order to establish interprocess communication between them, to prevent malicious processes from trying to hook onto the system. Currently I just send a password in plaintext between the two processes and the host process just performs a string comparison with its copy of the password, and will disconnect the connecting process if they don't match. It was fine when I was limited to connecting each process on localhost, but obviously this is not secure if someone just happens to have access to a network between each process and could sniff the packets. Do note that I only care about encryption for the password exchange and I don't really need it to secure the communication.  So basically I need a encryption algorithm that doesn't need to protect its contents well but does need to protect its key, and needs to be as fast as possible with minimal RAM usage and I/O overhead. Yes, data exchange is more CPU-bound than I/O bound since the two processes will be doing loads of calculations and they most likely will be connecting through localhost. Now, I came across RC4, which has a throughput of over twice that of AES-128, but it is evident that it has some weaknesses and I have a lot of questions about how to implement it in a secure way.  Well first question - it's in the title. Is RC4 completely broken and should be avoided? I hear a lot of "it is secure if properly used" and then a lot of "use of RC4 is discouraged."  I was planning to keep the plaintext password on both processes and have them SHA-224 hashed, and concatenate the hash with a 32-bit IV as a 256-bit key for RC4. Are there any security risks with this approach? Perhaps I could then hash the concatenated IV and hashed password with SHA-256? I'm well aware that WEP was cracked because it used a 24-bit IV, but a 32-bit IV has over four billion more unique values. Should I avoid incrementing the IV after each packet like WEP?  And when it comes to RC4-drop[n], what is a reasonable value for n? 256, 512, 768 bytes? How would you implement drop n on a library whose only public operation is encrypting a byte array - pass a byte array of length n after the engine is initialized with the key? Would that mean you would have to basically process n empty bytes after every time you change the IV (which happens after every packet)?  Sources:   EDIT: after reading all these excellent responses, I realized how ignorant I sounded about the whole subject. I had no experience with cryptography before this day, and I only was able to read about RC4's risks from the limited Google results that I could barely follow. thank you to everyone who has been patient with me!  Correct me if I'm wrong, but I believe a better solution is to either use AES (which is not slower because re-keying RC4 after each packet is expensive) or an universal message authentication code. I can easily work with AES, but my understanding of MAC is a bit shaky. An IV is only secure if the update sequence is randomized and unpredictable, meaning I would have to send a 128 bit IV for every message that is sent, which can add a 400% overhead to the short, 4 byte packets that are commonly sent by my application. HMAC can add more than twice that amount if SHA-256 is used. I am going to have to investigate this matter further. Why does the following code does'nt work when i am using PDFLaTeX or just LaTeX ? I tried it in different editors as well.  This is the code   $\{f^'_n\}$   Surprisingly it works on Math.SE site Show that $u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$ is in $W^{1,n}(U)$, where $U=B(0,1)\subset\mathbb{R}^n$. The entire problem statement is:  Let $n&gt;1$ and let $U=B(0,1)\subset\mathbb{R}^n$. Show that $u:U\to\mathbb{R}$ given by  $$u(x)=\ln\left(\ln\left(1+\frac{1}{|x|}\right)\right)$$ is in $W^{1,n}(U).$  My attempt at the proof is as follows:  To show that $u\in W^{1,n}(U)$ it suffices to show that $|Du|\in L^{n}(U)$. Consider $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}|x|^{-2}\frac{x_i}{|x|},$$ which simplifies to, $$u_{x_i}=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{x_i}{|x|^3}.$$ Thus, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+\frac{1}{|x|}}\frac{1}{|x|^2}$$ which can be manipulated to, $$|Du|=\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{1+|x|}\frac{1}{|x|}.$$  Moreover, since $U=B(0,1)$ we can bound the second term from above which gives, $$|Du|\leq\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}$$  Thus, we can instead show that $$\int_U\left(\frac{1}{\ln\left(1+\frac{1}{|x|}\right)}\frac{1}{|x|}\right)^n&lt;\infty.$$  Converting this into polar coordinates we have with $r=|x|$, $$\int_0^R\int_{\partial B(0,r)}\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\,dSdr=\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^n\int_{\partial B(0,r)}dSdr$$ Note that $\int_{\partial B(0,r)}\,dS=n\alpha(n)r^{n-1}$, which is the surface area of $B(0,r)$. Thus, we have then $$n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\frac{1}{r}\right)^nr^{n-1}\,dr=n\alpha(n)\int_0^R\left(\frac{1}{\ln\left(1+\frac{1}{r}\right)}\right)^n\frac{1}{r}\,dr$$  And so this is where I get stuck in the problem, since I don't how I can evaluate that last integral to show it's finite.  Thank you for any help or feedback! What formats supporting animation are suitable for the web? Floating point precision while using Python's max()