How to get a reflex edge on a non-convex mesh
Narrow-phase collision detection algorithms
Convert DOS line endings to Linux line endings in Vim If I open files I created in Windows, the lines all end with ^M. How do I delete these characters all at once? Is there a simple way to flatten a list of iterables with a list comprehension, or failing that, what would you all consider to be the best way to flatten a shallow list like this, balancing performance and readability?  I tried to flatten such a list with a nested list comprehension, like this:  [image for image in menuitem for menuitem in list_of_menuitems]   But I get in trouble of the NameError variety there, because the name 'menuitem' is not defined.  After googling and looking around on Stack Overflow, I got the desired results with a reduce statement:  reduce(list.__add__, map(lambda x: list(x), list_of_menuitems))   But this method is fairly unreadable because I need that list(x) call there because x is a Django QuerySet object.  Conclusion:   Thanks to everyone who contributed to this question.  Here is a summary of what I learned.  I'm also making this a community wiki in case others want to add to or correct these observations.  My original reduce statement is redundant and is better written this way:  &gt;&gt;&gt; reduce(list.__add__, (list(mi) for mi in list_of_menuitems))   This is the correct syntax for a nested list comprehension (Brilliant summary !):  &gt;&gt;&gt; [image for mi in list_of_menuitems for image in mi]   But neither of these methods are as efficient as using itertools.chain:  &gt;&gt;&gt; from itertools import chain &gt;&gt;&gt; list(chain(*list_of_menuitems))   And as @cdleary notes, it's probably better style to avoid * operator magic by using chain.from_iterable like so:  &gt;&gt;&gt; chain = itertools.chain.from_iterable([[1,2],[3],[5,89],[],[6]]) &gt;&gt;&gt; print(list(chain)) &gt;&gt;&gt; [1, 2, 3, 5, 89, 6] What are the subcategories of ordered sets / groups? There are several distinct usages:   kernel density estimation   kernel trick   kernel smoothing   Please explain what the "kernel" in them means, in plain English, in your own words. How can I do development on my Android device? I have a pretty long commute to work and own an android phone. I figure it would be nice to do some simple programming on the way. I'm not picky on which language to code on. I usually have an internet connection on the train.  I've been looking through the android market but haven't found any apps that offer a "development environment" or an editor with a compiler. Is any development ON the Android possible? Does an app for this exist? Bijection between an open and a closed interval Recently, I answered to this problem:     Given $a&lt;b\in \mathbb{R}$, find explicitly a bijection $f(x)$ from   $]a,b[$ to $[a,b]$.   using an "iterative construction" (see below the rule).  My question is: is it possible to solve the problem finding a less exotic function?  I mean: I know such a bijection cannot be monotone, nor globally continuous; but my $f(x)$ has a lot of jumps... Hence, can one do without so many discontinuities?    W.l.o.g. assume $a=-1$ and $b=1$ (the general case can be handled by translation and rescaling). Let:  (1) $X_0:=]-1,-\frac{1}{2}] \cup [\frac{1}{2} ,1[$, and  (2) $f_0(x):=\begin{cases}     -x-\frac{3}{2} &amp;\text{, if } -1&lt;x\leq -\frac{1}{2} \\ -x+\frac{3}{2} &amp;\text{,     if } \frac{1}{2}\leq     x&lt;1\\ 0 &amp;\text{, otherwise} \end{cases}$,  so that the graph of $f_0(x)$ is made of two segments (parallel to the line $y=x$) and one segment laying on the $x$ axis; then define by induction:  (3) $X_{n+1}:=\frac{1}{2} X_n$, and  (4) $f_{n+1}(x):= \frac{1}{2} f_n(2 x)$  for $n\in \mathbb{N}$ (hence $X_n=\frac{1}{2^n} X_0$ and $f_n=\frac{1}{2^n} f_0(2^n x)$).  Then the function $f:]-1,1[\to \mathbb{R}$:  (5) $f(x):=\sum_{n=0}^{+\infty} f_n(x)$  is a bijection from $]-1,1[$ to $[-1,1]$.  Proof: i. First of all, note that $\{ X_n\}_{n\in \mathbb{N}}$ is a pairwise disjoint covering of $]-1,1[\setminus \{ 0\}$. Moreover the range of each $f_n(x)$ is $f_n(]-1,1[)=[-\frac{1}{2^n}, -\frac{1}{2^{n+1}}[\cup \{ 0\} \cup ]\frac{1}{2^{n+1}}, \frac{1}{2^n}]$.  ii. Let $x\in ]-1,1[$. If $x=0$, then $f(x)=0$ by (5). If $x\neq 0$, then there exists only one $\nu\in \mathbb{N}$ s.t. $x\in X_\nu$, hence $f(x)=f_\nu (x)$. Therefore $f(x)$ is well defined.  iii. By i and ii, $f(x)\lesseqgtr 0$ for $x\lesseqgtr 0$ and the range of $f(x)$ is:  $f(]-1,1[)=\bigcup_{n\in \mathbb{N}} f(]-1,1[) =[-1,1]$,  therefore $f(x)$ is surjective.  iv. On the other hand, if $x\neq y \in ]-1,1[$, then: if there exists $\nu \in \mathbb{N}$ s.t. $x,y\in X_\nu$, then $f(x)=f_\nu (x)\neq f_\nu (y)=f(y)$ (for $f_\nu (x)$ restrited to $X_\nu$ is injective); if $x\in X_\nu$ and $y\in X_\mu$, then $f(x)=f_\nu (x)\neq f_\mu(y)=f(y)$ (for the restriction of $f_\nu (x)$ to $X_\nu$ and of $f_\mu(x)$ to $X_\mu$ have disjoint ranges); finally if $x=0\neq y$, then $f(x)=0\neq f(y)$ (because of ii). Therefore $f(x)$ is injective, hence a bijection between $]-1,1[$ and $[-1,1]$. $\square$ Shell scripting: Way to parse parameters from command line myscript [-a a-arg] [-c c-arg] [-b] [-e] somedirectory   Given that I want my shell script be invoked at the command line using the above parameters - where [these brackets] denote that they are optional - what is the best method to parse them? Rank correlation statistics comparison I am trying to understand the relative behavior of the following rank correlation statistics:   Spearman coefficient Kendall Tau / Concordance percentage Normalized Gini coefficient (area under curve of percentage captured versus percentage observations) Normalized Area under ROC curve (for binary classifiers)   I don't believe any of these are functionally related to the others.  The accepted answer  references  and Spearman and Kendall are highly correlated (as one would expect).  Are there good intuitions behind/discussions of relative (across datasets) or absolute (for a given dataset) differences for (any pair of) these measures? Where to insert the exclamation mark Where would I insert an exclamation mark and a comma in the below sentence?     "Look at that boy" he exclaimed. Aligning three figures next to each other based on text width Problem:  I am trying to align three figures next to each other with text centered to each figure. This is going well until the box for the third box does not adjust after the text width.  Minimal Working Example (MWE):  \documentclass{article} \usepackage{graphicx} \usepackage{caption}  \begin{document}  \begin{figure}[!tbp]   \centering   \begin{minipage}[b]{0.2\textwidth}     \includegraphics[width=\textwidth]{icon-password.eps}     \caption*{Login system}   \end{minipage}   \hfill   \begin{minipage}[b]{0.2\textwidth}     \includegraphics[width=\textwidth]{icon-shopping.eps}     \caption*{Shopping cart}   \end{minipage}   \hfill   \begin{minipage}[b]{0.2\textwidth}     \includegraphics[width=\textwidth]{icon-clock.eps}     \caption*{Temporary information}   \end{minipage} \end{figure}  \end{document}   Outputs:   Desired outcome:  To get the width to adjust after the text so that "Temporary information" can be written out in one line instead of two. Need help identifying a comic book about a wizard I need help with identifying the comicbook, what follows is my best effort to gather whatever I remember from reading it years ago (circa year 2000, but I'm positive it was older than that):   The main character was a guy with white hair, who wanted to become powerful.  Through some training and a deal of some kind he was able to become a mage of sorts. The training was set on some other world, or possibly even parts were happening in different dimension. His powers might have had something with travelling between dimensions, but I'm not  100% sure. Also, I seem to recall there was a hard limit of how many times he could use them. Then followed a part where he was famous, but someone framed him into a scandalous plot (it was a sexual scandal, I think) using his clone or a doppelganger. At the end of the issue     the main character is seen trapped in some strange state, unable to move or escape using his powers (this might have to do something with the deal he made earlier). I remember the scene at the end as this: he is suspended in mid-air, his friends look at him in disbelief while some malevolent figure explains to them that he is now trapped forever.  His name was the title of the comic. The second and third parts were set on futuristic-looking planet.   Rings any bells? I'd like to read it again only I can't google it out. How can I combine hashes in Perl? What is the best way to combine both hashes into %hash1?  I always know that %hash2 and %hash1 always have unique keys.  I would also prefer a single line of code if possible.    $hash1{'1'} = 'red'; $hash1{'2'} = 'blue'; $hash2{'3'} = 'green'; $hash2{'4'} = 'yellow'; I have two internet connection from two different ISP. ADSL on eth0 and 4G on wlan0. Is there any way that if one ISP fails, the other take over automatically without a physical router and by Host OS (Ubuntu 18.04)? I'm new to the world of Apple, I bought a new Macbook Pro, mid-2020, Intel i7, 16GB RAM. I'm a developer, I run apps like VS Code, Sublime Text, Slack, Chrome (Multiple Windows), I play videos/songs in the background. I have been keeping a track of the energy consumption on my Mac, and it only runs for 4 hours. I have reset my SMC after I upgraded to Big Sur. Is 4 hours normal? or should I go to the service center? Update on 3 Feb 2021: I submitted my Mac to the Apple service center they said they conducted a few tests and the conclusion was that the battery is fine (I don't know what type of test they did). So I guess I'll be using it on AC Update on 12 Feb 2021: Took it to the service center, they didn't find any issues so I started using Safari for work and my battery has improved by 1 hour on Average. $2$ out of $3$ property of the unitary group Compute $e^{e^z}$, where z is a complex number I am self-studying the MAT 67 linear algebra course and I got stuck at a problem in Chapter 2:     Compute the real and imaginary parts of $e^{e^z}$ for $z$ $\in$ $\Bbb C$.   I tried to expand it using Euler's formula but it seemed endless. Any help? Is there a standard for public key encrypting and signing short messages? Some systems, that operate over flaky bearers, need to send signed, encrypted messages that are entirely self contained - i.e. send a single 300 byte UDP rather than have a 10Kb, two-way, certificate conversation over TCP.  Assume the destination has a private key and already knows the source's public key, and the source has a private key and already knows the destination's public key.  Given this assumption, is there a standard for encrypting/signing a small, but arbitrary length, message?  (I'm well aware of the dangers of rolling my own encryption scheme.)  Would there be a minimum ciphertext size that is related to the public key size?  (I'm thinking plain PKI encryption for plaintexts shorter than the key, and a PKI encrypted AES key for longer plaintexts.) If a,b,c,d are positive integers with a sum of 63, what is the maximum value of ab + bc + cd ? (By using calculus to solve it)) If $a,b,c,d$ are positive integers with    $$a+b+c+d=63,$$   what is the maximum value of $$ab+bc+cd?$$   I treated this as an optimization problem and tried to solve it but I did not make any progress. I am using the algorithmic package to typeset an algorithm. I would like to have connecting lines between the beginning and end of a loop or conditional like in the algorithm2e package. Is there an easy way to accomplish this, or am I better off switching to algorithm2e? We need a profanity filter I searched SE.Meta and was unable to find such question.  Someone just swore at us in a  as an  (warning: profanity, may be NSFW), sure, Arqade (profanity in phrases in games) or some SE (ELU, ELL or Anime.SE maybe?) can sometimes contain profanity as a small parts of question.  I think that a part should be added to the review system that will automatically hide questions/answers with profanity, just like the close vote system, the question/answer will be on hold and will be voted to be shown or be deleted.  Tip, if this gets implemented, u can be typed as \/ or v, and most other can be written as leet (l = 1 etc). Maybe it can be called "suspicious activity", and can be triggered when there is too many number or Unicode characters. (I can write hi as |-| |, |-|I, |-|l, l-li, hî, 1-1i, H1 and many more)