I'm using Stack Overflow. I got 35 points there yesterday. Today 2/1 points were subtracted from the total (i.e. 33/34 points), but downvotes / negative points are not visible in my reputation history. What's happening here?
Yesterday I had 3,606 reputation but today I have 3,592.  My profile shows that I lost 6 reputation when a user was removed, but that would put me on 3,600 - 8 reputation has magically disappeared.  Adding up all of my reputation does work out at 3,591 (I'm assuming by default new users have 1 rep, so that would make sense).  Somewhere a reputation change since yesterday isn't being displayed. I don't know what this would be for, but -8 would lead me to believe that someone removed their up-vote on an answer I'd posted which already had a down-vote (could it be from that deleted user?).  I don't know if this is a bug or if there is simply a delay between reputation updating, but I figured it was worth reporting.    Sadly I cannot prove that I had 3,606 reputation yesterday.
In many C/C++ macros I'm seeing the code of the macro wrapped in what seems like a meaningless do while loop.  Here are examples.  #define FOO(X) do { f(X); g(X); } while (0) #define FOO(X) if (1) { f(X); g(X); } else   I can't see what the do while is doing.  Why not just write this without it?  #define FOO(X) f(X); g(X) Persist sort/filter selections Can I change a user's keyboard input? "User accounts" or "users account" The question is as below:      Show that each vertex $v$ of a tree $T$ appears $deg(v)-1$ times in the Prüfer encoding of $T$, where $deg(v)$ is the degree of $v$ in $T$.   For this, I think there are 3 scenarios a leaf should be considered for: (1) when it is a leaf as part of a tree; (2) a leaf that has been discarded; (3) the remaining edge. I feel like this will result in a short proof which is only two (or so) statements long, but at the same time, feel it is too simple.   EDIT: I've had a look at this again and narrowed it down to 2 cases:   Case 1: The vertex $v$ in question has $deg(v)\ge2$. $v$ then loses its neighbours as the algorithm is applied repeatedly until $v$ becomes a leaf itseelf. At this point, $v$ is attached to another neighbour, leaving an edge. This means the algorithm can no longer be applied and we have obtained the final Prüfer Encoding for the tree $T$.  Case 2: The vertex $v$ has $deg(v)\ge 2$. It loses its neighbours as the algorithm is applied repeatedly but is deleted itself at some point in the process.   Have I gone about this wrong?  Thank you. While I am familiar with Descartes and the oft-quoted idea, I think therefore I am, I am not familiar with any philosophy that actively sought to rationalize that anybody else exists.  Are there philosophers who analyzed this question? Or is it taken for granted that it is (im)possible for anybody to validate their own existence to somebody else? How did the phenomenon of doubling words come about? I am referring to phrases such as: "Do you like her, or do you like like her." Can someone provide an explanation of this? There are many more examples but none come to mind at the moment. What is the autocorrelation for a random walk? Seems like it is really high, but this is counterintuitive to me. Can somebody please explain? I am very confused by this issue and would appreciate a detailed, insightful explanation. Thanks a lot in advance! Although I have installed the OpenLayers plugin in QGIS 2.4, I cannot see it in the list when I click on Plugins.    Does anyone else have this issue? "Basis extension theorem" for local smooth vector fields How to stop HTTPS requests for non-ssl-enabled virtual hosts from going to the first ssl-enabled virtualhost (Apache-SNI) I hope that title is clear.  How do I prevent HTTPS requests for non-ssl-enabled virtual hosts from going to the first ssl-enabled virtualhost (setup is Apache-SNI).  For example, using my abbreviated config below, requests for https://example.com (a non-ssl vhost) are being served by Apache at the ssl-enabled vhost https://example.org. I'd like to disable that behavior and possibly reply with the appropriate HTTP response code (unsure of what that is).  It may not even be possible, but I thought I'd ask.  # I actually have a SNI setup, but it's not demonstrated here. # I don't think it's relevant in this situation.  NameVirtualHost *:80 NameVirtualHost *:443  &lt;VirtualHost *:80&gt;     ServerName example.org &lt;/VirtualHost&gt;  &lt;VirtualHost *:443&gt;     ServerName example.org &lt;/VirtualHost&gt;  &lt;VirtualHost *:80&gt;     ServerName example.com &lt;/VirtualHost&gt;   EDIT: Maybe a mod_rewrite rule in the first ssl-vhost? Calculating the average area of the states listed in states.txt Need someone to calculate the average area, in square miles, of the states listed in states.txt. States.txt (file): State           Population   SqMi      Region  ------------------------------------------ California           39.37   163.7       West Texas                29.36   268.6      South Florida              21.73    65.7      South New_York             19.33    54.7  NorthEast Pennsylvania         12.78    46.1  NorthEast Illinois             12.59    57.9    Midwest Ohio                 11.69    44.8    Midwest Georgia              10.71    59.4      South North_Carolina       10.60    53.8      South Michigan             9.97    96.7    Midwest  Output should look like this: Average Area: 82.85 square miles  Note my awk program will be ran like this: awk –f &lt;file.awk&gt; states.txt At some point, the filesystem on my digital audio player has become read-only. I cannot copy files into it or remove files on it.  Are there some possible reasons for the player's file system to change permissions in this way?  I tried using chmod:  $ sudo chmod a+rw SGTL\ MSCN/  chmod: changing permissions of `SGTL MSCN/': Read-only file system   SGTL MSCN is the mount point of the digital audio player.  What else can I try to make it writeable? Ways of getting three of a kind in a 52 card deck This question has probably been asked before, but just to be clear here, I am NOT asking for the answer, I know the answer.  What i want to know is why my solution is not equivalent to the actual solution.  Three of a kind in a deck means you draw 3 cards of the same rank, and 2 other cards each of different ranks.  So. The solution in my book is the following: (13 C 1)(4 C 3) * (12 C 2)(4 C 1)(4 C 1)  The logic is the following:  1) Choose 1 out of the 13 ranks, then for the 4 cards in the chosen rank, pick 3.  2) Choose a combination of 2 ranks from the remaining 12, then for the 4 cards in each rank, choose 1.  Total combinations: 54912  My solution is the following:  ((13 C 1)(4 C 3)) * ((12 C 1)(4 C 1)) * ((11 C 1)(4 C 1))  1) My logic for the 3 of a kind is the same as the solution.   2) Choose 1 of the remaining 12 ranks, and 1 card from the 4 cards in the rank.  3) Choose 1 of the remaining 11 ranks, and 1 card from the 4 cards in the rank.  Total combinations: 109824  My solution says that the combinations are exactly double of the book's solution.  I'm just very confused as to why this is. If anyone could explain this to me I would be very appreciative, thank you. Select groups based on number of unique / distinct values Compute $\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx$ I'm having trouble computing the integral: $$\int \frac{\sin(x)}{\sin(x)+\cos(x)}\mathrm dx.$$ I hope that it can be expressed in terms of elementary functions. I've tried simple substitutions such as $u=\sin(x)$ and $u=\cos(x)$, but it was not very effective.  Any suggestions are welcome. Thanks. I have a simple question.     over centuries / for centuries   what is the difference?  Thank you. Colorize symbols with lstdefinelanguage in latex I want to put some symbols in green color with lstdefinelanguage in latex, but it seems that symbols are not recognized.  The symbols I need to be colored are =, > and &lt;. Is it possible? How can I do this?  The main file:  \documentclass{article}   \usepackage{graphicx,color} \usepackage{booktabs} \usepackage{listings}   \lstdefinelanguage{NeoIDL}{   sensitive = true,   keywords = [1]{module, resource, enum, annotation, for, import, entity, path,   @get, @post, @put, @delete, require, ensure, otherwise, call},   %ndkeywords={int, \char{}  },   %ndkeywordstyle=\color{blue}\bfseries,   morekeywords=[2]{&gt;, &lt;, ==},   keywordstyle=[2]\color{green},   numbers=left,   numberstyle=\scriptsize,   stepnumber=1,   numbersep=8pt,   showstringspaces=false,   breaklines=true,   frame=top,   comment=[l]{//},   morecomment=[s]{/*}{*/},   commentstyle=\color{purple}\ttfamily,   stringstyle=\color{red}\ttfamily,   morestring=[b]',   morestring=[b]"   }  \begin{document}  \begin{figure}[htb] \begin{small} \lstinputlisting[language=NeoIDL,firstnumber=1]{store_pos_servico.neo} \end{small} \caption{Service 1} \end{figure}   \end{document}   The file to be formatted:  module store {    (...)     resource order  {      path = "/order/{id}";       @post   int postOrder (Order order)     require (call store.getOrder(id) == "NotFound"),     otherwise "InvalidPrecondition"    }; } Scattering, Perturbation and asymptotic states in LSZ reduction formula I was following Schwarz's book on quantum field theory. There he defines the asymptotic momentum eigenstates $|i\rangle\equiv |k_1 k_2\rangle$ and $|f\rangle\equiv |k_3 k_4\rangle$ in the S-matrix element $\langle f|S|i\rangle$ as the eigenstates of the full Hamiltonian i.e., $H=H_0+H_{int}$. Therefore, the states $|i\rangle=|k_1 k_2\rangle$ is defined as $$|k_1 k_2\rangle=a_{k_1}^{\dagger}(-\infty) a_{k_2}^{\dagger}(-\infty)|\Omega\rangle$$ where $|\Omega\rangle$ is the vacuum of the full interacting theory. Then the LSZ reduction formula connects the S-matrix element $\langle f|S|i\rangle$ to the Green' functions of the interaction theory defined as $$G^{(n)}(x_1,x_2,...x_n)=\langle \Omega|T[\phi(x_1)\phi(x_2)...\phi(x_n)]|\Omega\rangle.$$ Here are a few doubts. Doubt 1 When the particles are far away, the interaction can be considered to be adiabatically switched off. Therefore, at $t=\pm\infty$ the states are really free particle states and should have been written as $$|k_1 k_2\rangle=a_{k_1}^{\dagger}(-\infty) a_{k_2}^{\dagger}(-\infty)|0\rangle$$ and $$|k_3 k_4\rangle=a_{k_3}^{\dagger}(+\infty) a_{k_4}^{\dagger}(+\infty)|0\rangle$$ where $|0\rangle$ is the vacuum of the free theory.  I do not understand why these the states $|i\rangle$ and $|f\rangle$ are derived from $|\Omega\rangle$ instead of $|0\rangle$. Doubt 2 The initial and final states were derived from the vacuum of the interacting theory $|\Omega\rangle$. According to my understanding, this suggests that the states $|i\rangle\equiv |k_1 k_2\rangle$ and $|f\rangle\equiv |k_3 k_4\rangle$  are eigenstates of the full Hamiltonian $H$. Since then there is no perturbation, there should not be any scattering or transition at all.  More references Even Peskin &amp; Schroeder, Bjorken &amp; Drell, Srednicki take the same approach as Schwartz; they too define the external momentum eigenstates to the eigenstate of the full Hamiltonian $H$. However, if the system was initially in a stationary state why should it undergo a transition in absence of any perturbation? I'm working on a Ubuntu system, and my client has completely forgotten his administrative password. He doesn't even remember entering one; however it is there.  I've tried the suggestions on the website, and I have been unsuccessful in deleting the password so that I can download applets required for running some files. Is there a solution?