Re-ordering deprecated solutions to the bottom Since some software is constantly changing the solutions for the questions are also changing quite quickly.   E.g. , with different versions of the software, the API to the third-party tool changes. It ends up with legacy answers ranking highest because of the highest votes.   Is there a way to re-rank the deprecated solutions to the bottom?
Should outdated answers be voted down? On Stack Overflow, there are many questions that relate to currently-developing technologies such as HTML5 and CSS3. The accepted answers to these questions may become outdated, and better answers may appear. However, because the answers are accepted, it is likely that they will continue to be upvoted by users reading the answers, and the OP will likely not notice that a better, more up-to-date answer may exist.  Should these outdated answers be voted down? If not, what should we do with them?
I own a lot of games on Steam, and I'm overall a big fan of the service.  However, one thing about it really bothers me, and I'm wondering if anything can be done about it.  Let's say I have two different online, multiplayer games, both of which I legally own.   If I bought these games normally, I could be playing one of them online, while my fiance could play the other one online on another computer.     However, with Steam, I can only be signed in on one computer at a time.  So, despite the fact that I have bought about 40 games, if I'm playing one of them, someone else in my house is not able to play any of the others.  Is there a way to get around this?  I don't think this is illegal or immoral as I own both of the games in question, and only one person is playing the game at a given time (same IP address and the same household, so it should be easy for Steam to verify that I am not sharing my games unfairly).    If I had bought simple boxed copies, this wouldn't be an issue, but since I became such a fan of Steam, I can't play any of those games at the same time as my fiance.    Is there anything that can be done? Chrome 100% CPU usage, again - software_reporter_tool.exe software_reporter_tool.exe belonging to Chrome running at 100% cpu, fan full speed.  Deleting it doesn't fix anything, because it gets redownloaded again! Disabling "Protect me from bla bla" doesn't make it stop running.  Please anyone have any solutions? "Temporarily unreachable" when using fetch as Googlebot from Webmasters Tools The page I'm trying to fetch is public.  The server logs show no request.  Google reports that any page on my site is "Temporarily unreachable". It won't even fetch the verification file, a static HTML file that was successfully used to verify with the webmaster tool application itself.   My site is running on the .  What is causing this? Why apt can't find the source code of any package? What is the difference between "Redirection" and "Pipe"? This question may sound a bit stupid, but I can not really see the difference between redirection and pipes.  Redirection is used to redirect the stdout/stdin/stderr, e.g. ls &gt; log.txt.  Pipes are used to give the output of a command as input to another command, e.g. ls | grep file.txt.  But why are there two operators for the same thing?  Why not just write ls &gt; grep to pass the output through, isn't this just a kind of redirection also? What I am missing? Enumerate paragraphs and subparagraphs like sections but with letters Given two separate annulus with centers $[C_1,C_2]$ and their corresponding radii being $[R_1,r_1]$ and $[R_2,r_2]$ respectively, larger radius being $R$. There are methods to look at whether they are overlapping or not. If they overlap how to find the overlapping area. Is there any close form expression for it?  equation of first annulus   $r_1^2&lt;(x-x_{c1})^2 + (y-y_{c1})^2 &lt;R_1^2$   equation of second annulus  $r_2^2&lt;(x-x_{c2})^2 + (y-y_{c2})^2 &lt;R_2^2$  If there are three circles overlapping, we get overlapping area as solution to set of 4 in-equations by least squares approach(Though we don't end up getting total area but, area within tangents of circles ). Expand the username auto-completion to everyone who can be notified Making sense of principal component analysis, eigenvectors & eigenvalues Adding layers to GeoPackage using PyQGIS Solve $$u_{xx}-3u_{xt}-4u_{tt}=0$$ where $u(x,0)=x^2$ and $u_t(x,0)=e^x$.  My workings so far: I have factored the differential equation in the following way: $$(\delta_x-4\delta_t)(\delta_x+\delta_t)=0$$ where $\delta_x=\frac{\delta}{\delta x}$ etc. Now if we let $v$ be the solution to $(\delta_x+\delta_t)u$ then we have the following two equations: \begin{eqnarray*} (\delta_x+\delta_t)u=u_x+u_t=v\\ (\delta_x-4\delta_t)v=v_x-4v_t=0 \end{eqnarray*} Now for $v$ we simply find $$v=h(t+4x)$$ where $h$ is an arbitrary function of one variable. Now what remains is find $u$ such that $$u_x+u_t=h(t+4x)$$ I am stuck here, I thought about making a change of variables $\zeta=x+t$ and $\eta=x-t$ and thus using the product rule to show that $u_x=u_{\zeta}+u_{\eta}$ and $u_t=u_{\zeta}-u_{\eta}$ and thus $u_x+u_t=2u_{\zeta}$ and we need to solve $$u_{\zeta}=h(t+4x)$$ (I left out the factor 2 because $h$ is an arbitrary function). Do I simply integrate now and conclude $$u=f(\zeta)h(t+4x)+g(\eta)=f(x+t)h(t+4x)+g(x-t)$$ This seems wrong to me...  Some help would be greatly appreciated! How to remove redundant {} from a nested list of lists? Eclipse menus are cut off or don't show I'm using Ubuntu 13.10 (with JDK 1.7.0_45).  I've downloaded and uncompressed Eclipse Kepler for Java EE Developers.  The software menus are messed up. I didn't notice similar problems in other software.  When I click any menu in Eclipse, the drop-down seems to be "cut". Sometimes the drop-down shows, but the options are invisible.  I tried Eclipse Indigo, but the same problem occurs. $\lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite How do I prove that $ \displaystyle\lim_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite? How to log source file name and line number in Python I have a Queue&lt;T&gt; object that I have initialised to a capacity of 2, but obviously that is just the capacity and it keeps expanding as I add items.  Is there already an object that automatically dequeues an item when the limit is reached, or is the best solution to create my own inherited class? I'm trying to make a liquid simulation with Mantaflow where the flow of the liquid respects the flow object's local rotation, so it looks like water spraying in different directions. It looks like the options on the flow object only work for 'world' coordinates and doesn't take into account the coordinates of the object. How do I make the water 'fire' (like a hose) into a specific direction that uses local coordinates instead of world ones? The spectral radius of normal operator back-up: which patitions to back-up before upgrading/reinstalling I have Ubuntu13.04 on my laptop. I have to upgrade it to the next release or else I would install a fresh release after deleting my current system.  I have already installed and configured my current system a lot and I can't take the burden of doing it all over again if I install the next release. Now I would like to know that what all back-ups are required to be taken beforehand so that I would recover all my files/installations/configurations etc. In other words I should have everything on my new system (simply by restoring from back-ups) that I possess currently.  My current system has /, /boot, /home &amp; swap partitions which were created by me when I had freshly installed the O.S. Vieta Jumping: Related to IMO problem 6, 1988: If $ab + 1$ divides $a^2 + b^2$ then $ab + 1$ cannot be a perfect square. The famous IMO problem 6 states that if $a,b$ are positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $\frac{ a^2 + b^2}{ab + 1 }$ is a perfect square, namely, $gcd(a,b)^2$.  How about a modification of this problem:    If $a,b$ are (strictly) positive integers, such that $ab + 1$ divides $a^2 + b^2$, then $ab + 1$ cannot be a perfect square.    I am looking for a proof of the claim above, or a counter-example.  One possible approach towards a proof is the following:   Suppose there is such a pair $(a,b)$ as above, then by the famous IMO problem 6 from 1988, $\frac{ a^2 + b^2}{ab + 1 } = g^2$ where $g = gcd(a,b)$. Since $ab + 1$ is a perfect square, $a^2 + b^2 = c^2$ for some integer $c$. So that $(a,b,c)$ is a Pythagorean tripple, therefore there exists positive integers $n,m,l$ with $n$ coprime to $m$, such that $a = l(n^2 - m^2)$ and $b = 2lnm$ - by Euclids formula.  Then by plugging in $a$ and $b$ in terms of $n,m,l$ into the original equation, and solving for $l$, it is possible to obtain the following:  There exists positive coprime integers $n,m$ such that   $\frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)}$ is a perfect square.  If we put the quotient above into a program, as in this python code snipet:  N = 1000  for n in range(1,N):     for m in range(n+1, N):         A = (n*n + m*m + 1)*(n*n + m*m - 1)         B = 2*m*n*(n+m)*(m-n)         if A % B == 0:             print(A/B)   The quotient always is 2, regardless of whether or not $n$ and $m$ are co prime! So if the very strong implication   $2mn(n+m)(n-m) \mid (n^2 + m^2 + 1)(n^2 + m^2 - 1) \implies \frac{(n^2 + m^2 + 1)(n^2 + m^2 - 1)}{2mn(n+m)(n-m)} = 2$    holds, then the original problem will be solved.