My fiancee and I are trying to find a game that we can play together, and neither one of us is a strategy fan. The whole sandboxy aspect of Civ seemed like it might be a nice way to get our feet wet. The one issue is we tend to have conflicting schedules.  Is it possible to start a two player game, make a turn, then log out. Then she can log in, make her turn, and log out. Sort of like one would do with games like "Chess with friends"?   Or does it require that we both sit down and play a game from beginning to end? Can't seem to find an answer to this. Thanks!
The Civilization series of games is famous for having a way to play by e-mail. This allows people to take a long time on their turn without holding up the other players, as you do in a real-time game. Yet, I can not find any method to play by e-mail in the game menus. Does it exist? If so, how do you do it?
How do you file a bug report for Windows? I have come across an issue that I feel is a  in Windows 7. What is the official process for submitting an issue to Microsoft from a consumers perspective as an end user of Windows? I was chatting with a friend after I saw his photos in Russia (Petersburg). He is a white British dude who has been to many, many, many countries. On the other hand, I am a pure Arab who looks like a typical Middle Easterner or North African, who has also been to many countries, but never to Russia or its neighbors.  Anyway, it seemed like he really liked Russia, so I told him it will be in my list. He told me "you do not belong".  At first, I wasn't sure what he did mean. He started explaining and told me stuff like "you will be hassled for no reason in some parts of the city" and "the police will stop you every few hundred meters asking for money". He made sure to make me believe that's all because I do not belong.  Can someone clarify this? Of course, some of this is hyperbole to help convey the point, but to what extent?  Note: The friend is . If you think the question is wrong, feel free to downvote him ;) I use the babel package in order to write in greek with font "kerkis".  I want to add code in english ("courier new" font) to my latex document but I get some difficulties.  Can anybody help me, please. Thank you in advance.  \documentclass[12pt]{article} \usepackage[english,greek]{babel} \usepackage[utf8x]{inputenc} \usepackage{graphicx} \usepackage{fancybox} \usepackage{verbatim} \usepackage{courier} \usepackage{kerkis} \usepackage{float} \usepackage{geometry} \usepackage{fancyhdr} \usepackage{titlesec} \usepackage{tocloft} \usepackage{hyperref} \usepackage{afterpage} \usepackage{color}    \usepackage{listings}  \renewcommand{\lstlistingname}{Κώδικας}  \lstset{  %matlab   backgroundcolor=\color{white},     basicstyle=\normalsize,    % COURIER NEW?        commentstyle=\color{green},    % comment style   escapeinside={\%*}{*)},           extendedchars=true,               frame=single,                              keywordstyle=\color{blue},       % keyword style   numbers=left,                     numbersep=5pt,                     numberstyle=\normalsize\color{black},    rulecolor=\color{black},           stepnumber=1,                      stringstyle=\color{magenta},     % string style }   \begin{document}  \selectlanguage{greek} Ελληνικά: \textlatin{this word means greek}  \selectlanguage{english} Now I write in english  \selectlanguage{greek} Ελληνικά  %\selectlanguage{english}???  \begin{lstlisting}[caption = \textlatin{example1.m}] { % Entropy E = 0; for i = 1:J     if p(i) ~= 0         E = E + p(i)*log2(1/p(i));     end end } \end{lstlisting}  \end{document} What's the difference when using GET or POST method? Which one is more secure? What are (dis)advantages of each of them?  () How do I watch a file for changes? How should I parse user input in a text adventure game? Parsing user commands in a text adventure is a spectrum from 's simple "go north" to some mind-bogglingly clever ones in .  I seem to remember reading nice how-tos in computer magazines back in the 80s, but now I find almost nothing on the 'net except a .  How would you do it?    Update: I went with the simplest approach possible in . I am trying to set up a wordpress multisite with MAMP Pro that will allow me to have two different domains (one domain for each language): dev.domainenglish.com dev.domainfrench.com I have already set up the primary english virtual host and secondary virtual french host. I have installed the wordpress multi site on the primary English virtual host. Next I created a &quot;sub site&quot; within the primary english domain multi site that has the location  dev.domainenglish.com/french/ Then I tried to use the MU Domain mapping plugin to make it so that when you visit dev.domainfrench.com the theme and content posted on dev.domainenglish.com/french/ will be visible. As of now when you visit dev.domainfrench.com the page loads the standard MAMP Pro index.html instead of my wordpress subsite:  the virtual host was set up successfully. If you can see this page, your new virtual host was set up successfully. Now, &gt;web contents can be added and this placeholder page1 should be replaced or &gt;deleted.&quot;  I understand that this plugin was built for live sites. But is there anyway I can achieve my goal of having two domains under one Wordpress multisite on my localhost? Thank you for your time! How to execute a Matlab script from C++ Recall the definition of the Weyl Chambers: A Weyl Chamber is a region of $V \setminus \bigcup_{\alpha \in \Phi} H_{\alpha}$, where $V$ is underlying Euclidean space, and $H_\alpha$ the hyperplane that is perpendicular to $\alpha$ the elements of  $\Phi$ in $V$.   I want to show that:   The Weyl chambers are open, convex and connected.   My attempts:   A single hyperplane $H_\alpha$ separates the space $V$ into two half-spaces: positive half-space $H_\alpha^+=\{x \in V \mid (\alpha,x)&gt;0\}$ and negative half-space $H_\alpha^-=\{x \in V \mid (\alpha,x)&lt;0\}$. It is clear that $V \setminus H_\alpha = H_\alpha^+\cup H^-_\alpha.$ As the result, the Weyl chambers can be regarded as components of  $$V \setminus \bigcup_{\alpha \in \Phi} H_{\alpha}= \bigcap_{\alpha \in \Phi} \left( V \setminus H_{\alpha} \right)= \bigcap_{\alpha \in \Phi} \left( H_\alpha^+\cup H_\alpha^- \right).$$ By the discussion above, a Weyl chamber is simply a $\textbf{intersection}$ of finite half-spaces. It is the case since one may write $$\left( H_\alpha^+\cup H_\alpha^- \right) \cap \left( H_\beta^+\cup H_\beta^- \right) =\left( H_\alpha^+\cap H_\beta^+ \right)  \cup\left( H_\alpha^+\cap H_\beta^- \right)  \cup \left( H_\alpha^-\cap H_\beta^+ \right)  \cup\left( H_\alpha^-\cap H_\beta^- \right).$$ Therefore openness and convexity of the Weyl chambers are straightforward since half-spaces are open and convex, so are their intersections. Moreover, we would replace $V$ by $\Bbb R^{\dim V}$ since they are isomorphic. Now we can prove that the Weyl chamber is connected by showing that every convex set in $\Bbb R^{\dim V}$ is connected. And I'm sure that proof of this statement existed. So am I done? Nested roots sequence, how to prove it's monotone and bounded? Let $a\ge1$ and define the sequence $(x_n)$ recursively by: $$x_1 = \sqrt{a}$$ $$x_{n+1}= \sqrt{a+x_n}$$  Here's what I did:  Plugging in some values makes it seem as if the sequence is increasing. I first solve the equation $t=\sqrt{a+t}$, equivalently $t^2 -t -a =0$, to get an idea what's going on. The roots are $$t_{1,2}= \frac{1 \pm \sqrt{1+4a} }{2}.$$ Therefore, between the roots, that is for $$t\in \left( \frac{1 - \sqrt{1+4a} }{2}, \frac{1 + \sqrt{1+4a} }{2}\right)$$  we have $t^2-t-a \le 0 \iff t \le \sqrt{a+t}$. Here's my attempt at showing inductively that $$\forall n\in \mathbb{N}, \ x_n \le \frac{1 + \sqrt{1+4a} }{2}$$  For $n=1$, the statement is true (simple arithmetic). Suppose it's true for $n$. Consider  $$x_{n+1}= \sqrt{a+x_n}\le \sqrt{a+\frac{1 + \sqrt{1+4a} }{2}}=\frac{1 + \sqrt{1+4a} }{2},$$  because $\frac{1 + \sqrt{1+4a} }{2}$ satisfies $x=\sqrt{x+a}$.  So this establishes the sequence $(x_n)$ is bounded. Now, by induction again, I try to show it's monotonically increasing. The base case $x_1\le x_2$ is obvious. Assume (strong induction) $x_{n-1}\le x_n$ for all $n&gt;1$. But since we have $$x_1\le x_n \le \frac{1 + \sqrt{1+4a} }{2}, $$   then $x_n$ lies in the appropriate interval, so $x_n \le \sqrt{a+x_n}=x_{n+1}.$  Is this proof correct? Can the condition on $a$ be relaxed to $a&gt;0$? Maybe even small negative $a$? And could someone provide me with another proof of convergence of $x_n$? I absolutely hate this one! Does $E = mc^2$ apply to photons? Who provides the Internet service to Internet Service Providers (ISPs)? I've been wondering recently about how the infrastructure of the Internet really works.    I know that I have an Internet Service Provider (ISP) that supplies my connection to the Internet.   But what I don't know is: Who provides the Internet to the ISP?  And who supplies it to them? Is there a never-ending loop that eventually connects us all together? It isn't any of the words on this  and I believe it started with the letter 'e' (though please offer your word even if it doesn't).  Example sentence: Because he researched what he had heard from his friends, he was able to refute many of their _________________.  Factoid seems to be the closest one, but the word I am looking for only refers to a belief that is false. How can one make Firefox ignore my GTK theme entirely? I am trying to remove all files and subdirectories in a directory. I used rm -r to remove all files, but I want to remove all files and subdirectories, excluding the top directory itself.  For example, I have a top directory like images. It contains the files header.png, footer.png and a subdirectory.  Now I want to delete header.png, footer.png and the subdirectory, but not images.  How can I do this in linux? How to handle a deliberate duplicate question I got  question in review. First sentence is:     I know there is a  question, but no answers.   And indeed the two questions are virtually identical. I started to flag as duplicate - but the original question had no answer at the time, so it wouldn't be correct and apparently . (I didn't go so far as to try to submit the flag).  What is the proper action here? How do we want such a question to be handled? Should both (or more) questions stay open indefinitely until either is answered and somebody gets the good idea to do the duplicate linking?   As the new question is essentially a bump to the front page (perhaps with re-tagging), especially in the case where the OP knows that a previous question exist, perhaps something should be done to enable bumping re-emphasizing without writing the new question.  As I wrote this,  actually got an answer. So the old one should now be closed as a duplicate of the new - but my question remains for the case where all known/found duplicate questions have no answer?  Ironically, just before asking this, the "Questions that may already have your answer" gave me ... (Although the suggestion has no answer). So it has been asked before - the comment there even suggest the ability to re-emphasize. String comparison in Python: is vs. == How to prevent water from freezing Windows 10 upgrade assistant stuck at 99% Installed Windows 7 Ultimate (64-bit) on a fresh PC and after installing service pack 1 started the upgrade to Windows 10. The assistant app downloads and verifies but gets stuck at 99% in the upgrading window phase. Even after hours it remains stuck there.  I've tried to reboot the machine and that doesn't help either. I've googled and found out that others have faced the exact problem but so far there isn't any solution that works.  Has anyone seen that situation and is there a fix for it? A question about the arctangent addition formula. In the arctangent formula, we have that:  $$\arctan{u}+\arctan{v}=\arctan\left(\frac{u+v}{1-uv}\right)$$  however, only for $uv&lt;1$. My question is: where does this condition come from? The situation is obvious for $uv=1$, but why the inequality?  One of the possibilities I considered was as following: the arctangent addition formula is derived from the formula:  $$\tan\left(\alpha+\beta\right)=\frac{\tan{\alpha}+\tan{\beta}}{1-\tan{\alpha}\tan{\beta}}.$$  Hence, if we put $u=\tan{\alpha}$ and $v=\tan{\beta}$ (which we do in order to obtain the arctangent addition formula from the one above), the condition that $uv&lt;1$ would mean $\tan\alpha\tan\beta&lt;1$, which, in turn, would imply (thought I am NOT sure about this), that $-\pi/2&lt;\alpha+\beta&lt;\pi/2$, i.e. that we have to stay in the same period of tangent.  However, even if the above were true, I still do not see why we have to stay in the same period of tangent for the formula for $\tan(\alpha+\beta)$ to hold. I would be thankful for a thorough explanation.