So there's  that seems to be a bot. It keeps getting closed and deleted, but the OP reposts it pretty quickly.  seems to be a bot with some built-in search words, which it looks for in comments. If none of them are found, it defaults to "troll". It then inserts the found word in a template response.   How should we go about dealing with this user and such posting behavior? One diagnostic red flag that comes to mind is to pay special attention to posts that tag many programming languages as this one did. If too many programming languages are tagged in a question, it may be indicative of the OP not knowing what they are asking.
has now probably made around 15 accounts. He consistently uses the , , ,  and  tags, and does not ever do anything constructive.  was his latest.  Unfortunately I haven't been bookmarking his other questions, but here are some of his previous accounts:          He has also gone by many other names (Iraq War Veteran is the one that comes to mind), but those accounts have been deleted.    Is there nothing we can do about this? Deleting and suspending his accounts just seems to give him more reason to create other accounts. Is there a way to block him entirely?    Edit:  seems to have been deleted by Community ♦
Words to describe people with good/poor vocabularies Words to describe people with good/poor vocabularies. I am looking for words which I can use to describe characters and I would also like to describe their vocabularies, and explain how their vocabularies stem from their upbringing If parsing the output of ls is dangerous because it can break on some funky characters (spaces, \n, ... ), what's the best way to know the number of files in a directory?  I usualy rely on find to avoid this parsing, but similarly, find mydir | wc -l will break for the same reasons.  I'm working on Solaris right now, but I'm looking for a answer as portable across different unices and different shells as possible. Splitting a string into words and punctuation I'm trying to split a string up into words and punctuation, adding the punctuation to the list produced by the split.  For instance:  &gt;&gt;&gt; c = "help, me" &gt;&gt;&gt; print c.split() ['help,', 'me']   What I really want the list to look like is:  ['help', ',', 'me']   So, I want the string split at whitespace with the punctuation split from the words.  I've tried to parse the string first and then run the split:  &gt;&gt;&gt; for character in c: ...     if character in ".,;!?": ...             outputCharacter = " %s" % character ...     else: ...             outputCharacter = character ...     separatedPunctuation += outputCharacter &gt;&gt;&gt; print separatedPunctuation help , me &gt;&gt;&gt; print separatedPunctuation.split() ['help', ',', 'me']   This produces the result I want, but is painfully slow on large files.  Is there a way to do this more efficiently? iPhone Apps in Waiting mode for the last 4 days I tried to download the "Dictionary Free" app from the App Store four days ago, but it's been stuck in "Waiting" ever since. I've tried everything: rebooting the phone, restoring from an iTunes backup, side-loading the app from iTunes, but nothing has worked.  When I turn Airplane Mode on and then off again, I get an alert popup that says "Unable to download application. It will be available for download when you log in to the iTunes Store on your computer".  I've tried what it recommends, but nothing seems to work. We know that if you (using Polyjuice), take the form of someone who had been injured, you assume that injured form (e.g. Barty/Moody). If you are injured, presumably you would be uninjured while under Polyjuice. My question is, is this in real-time? That is, if you take a Polyjuice, and while you're still under its effects, the person being transformed into gets injured (or even dies), will that injury be transferred onto you? And conversely, if you get injured while in someone else's body, will that transfer over to your real body after the Polyjuice wears off, will it disappear, or will it go to the person you turned into? What is an intuitive explanation of the Expectation Maximization technique? Differences between LuaTeX, ConTeXt and XeTeX Getting video dimension / resolution / width x height from ffmpeg Lille Metro schedule... is it 24/7? The M1 and M2 lines in Lille, France have a confusing . It's 7 days per week, but it looks like it's "running" basically 24h hours per day (minus some ten-minute delay). I never took the metro later than 11 PM when I was in Lille last year, but people told me that it stops just before midnight. But... it looks like it starts right back up again immediately, after about 10 minutes.  What are the actual hours of the Lille M1 and M2 lines? I have a hard time believing that they're 24/7... Using MySQL with Entity Framework I've used git clone to make a local copy of a GitHub repository containing a  I'd like to use. The local copy contains all of the source for the program; but as a newbie to Ubuntu, I've no idea of what tools or steps are involved in compiling and installing the driver.   I'd really appreciate if anyone could provide a step-by-step guide? I have posted a proof below, and would appreciate it if someone could review it for accuracy. Thanks!   Problem:   Let n $\in$ $\mathbb{Z}$ with $n$ $\ge$ 3. Prove the following:   (a) Z(D$_{2n}$) = 1 if $n$ is odd.   (b) Z(D$_{2n}$) = {1, r$^k$} if $n$ = $2k$.   Note that $r$ and $s$ generate D$_{2n}$ with the group presentation { $r$,$s$ | $r$$^n$ = $s$$^2$ = 1, $rs$ = $sr$$^{-1}$ }   Proof:  part (a)   Let $n$ $\ge$ 3 where $n$ $\in$ $\mathbb{Z}$ and $n$ is odd.   For any x $\in$ Z(D$_{2n}$), x must commute with both $s$ and $r$, since both are in $D_{2n}$. Note that if $x$ commutes with both $s$ an $r$, then $x$ commutes with any element of $D_{2n}$ since $r$ and $s$ generate D$_{2n}$.    Then for any x$\in$ Z(D$_{2n}$), we have $xr$ = $rx$, where x is of the form $x$ = s$^j$r$^w$ (any such element can be arranged into this form via the relation $rs$ = $sr$$^{-1}$ and the fact that $r$ and $s$ generate D$_{2n}$). Note that since r and s have finite order, the exponents are modulo n and modulo 2 for $r$ and $s$ respectively.   Then we have  s$^j$r$^w$$r$ = $r$s$^j$r$^w$.   Which implies s$^j$r$^{w+1}$ = s$^j$r$^{w +- 1}$  In the case where the power of r on the RHS is $w$ - 1, it is clear we cannot have equality unless |r| = 1, which it is not. Note that if $j$ is even then we have only the $w$ + 1 case. Hence $x$ may be of the form $1$$r$$^w$ = $r$$^w$.   Also we have that x must commute with $s$, hence $s$$x$ = $x$$s$.   Then $s$$r$$^w$ = $r$$^w$$s$.  Which implies $s$$r$$^w$ = $s$$r$$^{-w}$.   Applying s$^{-1}$ to both sides we arrive at the task of finding when r$^w$ = r$^{-w}$ i.e when $w$ = $-w$. But since r has finite order we have:   $$ \bar{w} = (n-1)*w $$   where $\bar{w}$ is the residue class of w modulo n. Hence for some a $\in$ $\mathbb{Z}$$_+$   $$ w + an = wn - w $$  $$ 2w = wn - an $$ $$ 2w = n(w-a) $$   Now since 0 $\lt$ w $\le$ n we have that the LHS is greater than 0. Also since n $\ge$ w then (w-a) $\le$ 2, since otherwise the RHS would be greater than the LHS. So since n is odd we must have that (w-a) is even and hence (w-a) = 2, implying that 2w = 2n and hence n = w.   Hence x = $r$$^w$ = $r$$^n$ = 1. So Z(D$_{2n}$) = 1.   part (b)   From the argument above any x $\in$ Z(D$_{2n}$) must be of the form $s$$^j$$r$$^w$ for j even and j $\in$ $\mathbb{Z}$. Then x is of the form $r$$^w$ for $0$ $\lt$ w $\le$ $n$. Since $n$ is even we have the equation below is satisfied when $2w$ = $n$ or $w$ = $n$, since (w-a) $\le$ 2 but also (w-a) $\ge$ 0 for (w-a) $\in$ $\mathbb{Z}$$_+$.   $$ 2w = n(w - a) $$   Hence $r$$^w$ $\in$ Z(D$_{2n}$) when $2w$ = $n$, as desired. I'm having trouble solving a Fitch Style Proof and I was hoping someone would be able to help me.  Premises: p ∧ q  Goal: q ∨ r  Would be an assumption (p ∧ q => q ∨ r) in first step a correct one?   Then, q ∨ r Implication Elimination: 2, 1  Why is it not correct?  Thank you very much for any help! Two  have been announced in LoL.  They have an ability that makes them be able to recall together.                                  What are the limits to this ability? Can the 'non-recaller' jump in at the last second and get some kind of instant recall?  Can somebody with the setup to get on PBE test it and tell us what are the limitations of this ability? How to prove the midpoint of a chord is also the midpoint of the line segment defined by the points of intersection of other two chords with it? Integral $\int^{ \pi /2}_{0} \ln (\sin x)\ dx$ $$\int^{ \pi /2}_{0} \ln (\sin x)\  dx$$  The answer is $- \frac{\pi}{2} \ln 2$.  I have changed it into $$\frac{1}{2} \int^{1}_{0} t d \ln t^{2}$$  But I didn't get the answer with it. What can cause "Discovered - currently not indexed" in the new GWT Explicit solution for linear regression with two predictors I have some samples of data of the form $x,y$ and $z=f(x,y)$. I wish to fit a plane $$z = Ax + By + C$$ to the data with the smallest mean square errors.  I have found an "answer" in section 3 of  document, but it is left in the form of some equations left to be solved. I just about have the ability to solve these equations, but the process gets so messy that the likelihood of me making a mistake is quite high. Surely someone has written out the full solution longhand somewhere (it might be called "closed form") anyway, in the form $$A=\ldots,$$ $$B=\ldots,$$ $$C=\ldots.$$  EDIT: Maybe "closed form" is the wrong phrase. So let me be clear. I want an explicit solution for $A$, $B$, and $C$, and not a solution that ends with "So now if you can solve these equations then you can discover the values of $A$, $B$ and $C$". Game states & passing data between scenes If GCD $(a_1,\ldots, a_n)=1$ then there's a matrix in $SL_n(\mathbb{Z})$ with first row $(a_1,\ldots, a_n)$ Let $n \geq 2$. Let $a_1, a_2, \ldots, a_n$ be $n$ integers such that $\gcd\left(a_1, a_2, \ldots, a_n\right) = 1$. Prove that there exists a matrix in $\operatorname{SL}_n\left(\mathbb{Z}\right)$ whose first row is $\left(a_1, a_2, \ldots, a_n\right)$.   Since the gcd of the integers $a_1,\ldots, a_n$ is $1$, there exists weights $x_i \in \mathbb{Z}$ such that $a_1x_1+\cdots+ a_nx_n=1$. My two ideas are (a) to brute force construct an $n\times n$ matrix with first row $a_1,\ldots ,a_n$ and to construct the remaining rows such that the determinant is $\sum a_ix_i=1$ or (b) to use induction.  (a) (Constructive) This is tedious since once I find a way to construct the remaining $n-1$ rows to ensure that $a_1x_1$ appears in the determinant, I am not sure how to modify these $n-1$ rows to ensure  that only the terms $a_ix_i$ appear in the cofactor expansion. If such a matrix exists, I'd like to see it.  (ii) (Non-constructive) If I proceed by induction then the base case $n=2$ is settled since I can choose the 2nd row to be $-x_2, x_1$ so that the determinant is $a_1x_1-a_2(-x_2)=1$. However, I'm not sure how to use the inductive hypothesis to show that if I can construct such an $n\times n$-matrix then I can construct an $\left(n+1\right) \times \left(n+1\right)$-matrix with the desired property. In particular, if the gcd $(a_1,\ldots ,a_{n+1})$ is $1$, it is not necesarry that the gcd of any $n$ of these terms is $1$, so induction may not even apply here.  How can I construct such a matrix or prove that one exists (without necessarily constructing it)?