GPG error: The following signatures were invalid: ########
How do I fix the GPG error "NO_PUBKEY"?
How do I block unwanted callers without jailbreaking? Similes and Metaphors - are similes a subset of metaphors? Today in the feed on the Android Stack Exchange app there is a Stack Overflow blog post. This post had the site icon next to it for  rather than the proper site logo (Meta).        Stack Exchange Android App Version 1.0.73 and 1.0.74 Android 5.1 on a Moto X (2013) running stock firmware.  Same behavior on a Nexus 7 (flo) running Android 6.0 (stock).  This issue also happens on the iOS app.    The most recent featured SO blog post (as of 17 Nov) is showing up with the "edX" site logo.  The logo is seemingly randomly incorrect. Trying to wrap my head around the ways to properly handle collision in a game designed around components.  I see many examples have some sort of PhysicsComponent that gets added to the entity's list of components but the actual implementation is confusing me.  For this to work, the PhysicsComponent would need access to the world around it. This doesn't make intuitive sense to me. Shouldn't a component be unaware of not only its container (the entity), but its container's container (the world)?  To me, it sounds like the level or scene should maintain a list of these entities and every game update, loop through the entities to determine which collide.  My question is firstly, whether or not this is good design, and secondly, how to determine which entities can collide. I suppose solid entities could implement an empty IRigidBody interface so the level could determine which entities in the list support collision. But is this breaking the component design?  Instead, should they contain an empty RigidBody component? This may actually be better because it may not always be empty and this approach is more future-proof. The only problem with this is the complexity. The scene would have to loop through not only every entity, but also every entity's components to determine if it had this RigidBody component.  Thirdly, when they do collide both entities should be informed somehow and I am unsure on how to accomplish this either.  Let's say that both entities contained a HealthComponent and when they collided both their healths would be decreased by some arbitrary value, 5. I suppose it would be the scene's responsibility to handle this when it detects a collision between two entities?  But then is the scene responsible for too much? I could see this possibly getting out of hand and becoming unwieldy when the scene is responsible for many things entities should not (?) have access to.  Edit: Question updated with more details. Why can in-class initializers only use = or {}? Is it possible to make the default scene fill two monitors at startup? I know how to set up the default scene by pressing Ctrl+U, but whenever I startup Blender it always fills only one of my screens, even if I set it to fill both before setting the default scene. I end up having to resize it every time I start it up. Is there a way to set this to happen by default? I am asked to fill the genotypes in the spaces provided but looks like I am aving a little bit of trouble. can anyone help? thanks in advance! I need to write a Java Comparator class that compares Strings, however with one twist.  If the two strings it is comparing are the same at the beginning and end of the string are the same, and the middle part that differs is an integer, then compare based on the numeric values of those integers.  For example, I want the following strings to end up in order they're shown:   aaa bbb 3 ccc bbb 12 ccc ccc 11 ddd eee 3 ddd jpeg2000 eee eee 12 ddd jpeg2000 eee   As you can see, there might be other integers in the string, so I can't just use regular expressions to break out any integer.  I'm thinking of just walking the strings from the beginning until I find a bit that doesn't match, then walking in from the end until I find a bit that doesn't match, and then comparing the bit in the middle to the regular expression "[0-9]+", and if it compares, then doing a numeric comparison, otherwise doing a lexical comparison.  Is there a better way?  Update I don't think I can guarantee that the other numbers in the string, the ones that may match, don't have spaces around them, or that the ones that differ do have spaces. Handling registration ID changes in Google Cloud Messaging on Android Automatic line breaking of long lines of text? One thing that seems to be missing in the Top&nbsp;Bar is a link to the Stack&nbsp;Exchange portal. It only displays under the "Current&nbsp;Community" section when you are in the community, but it disappears as soon as you go to another one.  Would it be possible to add a link that would always be there no matter which community you are currently on? This is my first question.   Let $a_1, a_2,\ldots, a_k$ be natural numbers $\leq n$ with $m$ prime factors.  Let $p_1, p_2, \ldots, p_r$ be the prime numbers $\leq n$.   Let $$C_{m,n} = \frac{\sum_{i=1}^{k}a_i}{\sum_{i=1}^{r}p_i}$$ Does the following limit exist when $m\geq 2$? Obviously, when $m=1$,  $C_{1,n}=1$.  $$\lim_{n\to \infty} C_{m,n}$$  I have some guesses but not defendable enough.  My best,  The FM. A graph $G$ is bipartite if and only if $G$ can be coloured with 2 colours Evaluating $\lim\limits_{n\to\infty} e^{-n} \sum\limits_{k=0}^{n} \frac{n^k}{k!}$ I'm supposed to calculate: $$\lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}$$ By using WolframAlpha, I might guess that the limit is $\frac{1}{2}$, which is a pretty interesting and nice result. I wonder in which ways we may approach it. Is there a way to forget node names between TikZ pictures? I have a document with several TikZ pictures in it.  I generally use simple names for the nodes such as:  \node (A) at .....  \node (B) at .....   and so forth.  Because of this I have name clashes between the various pictures.  That's generally not a problem.  However I find myself in a position where I may or may not have defined a certain node within a given picture.  So I use the following snippet to test whether the node has been defined:  \makeatletter \long\def\ifnodedefined#1#2#3{%%   \@ifundefined{pgf@sh@ns@#1}{#3}{#2}} \makeatother   Unfortunately this fails if I've already created a node by the same name in another picture.  Is there a way to scope the definitions of nodes or is there a way to undefine nodes once you no longer need them?  P.S.  I'm aware that I could embed each picture in a standalone file and import it using \includegraphics.  Unfortunately, this approach will not work for the document I'm creating.  MWE  \documentclass{article} \usepackage{tikz} \usetikzlibrary{calc}  \makeatletter \long\def\ifnodedefined#1#2#3{%%   \@ifundefined{pgf@sh@ns@#1}{#3}{#2}} \makeatother  \setlength{\parindent}{0pt} \pagestyle{empty} \begin{document}   There's no node \textbf{D} here, so I draw a triangle:\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1); %  \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}     There is a node \textbf{D} here, so I draw a square:\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1);   \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}      There isn't supposed to be a node \textbf{D} here.  I'm expecting a triangle, but get a square.\par \begin{tikzpicture}    \coordinate (A) at (0,0);   \coordinate (B) at (1,0);   \coordinate (C) at (1,1); %  \coordinate (D) at (0,1);    \ifnodedefined{D}     {  \draw (A) foreach \myn in {B,C,D} {  -- (\myn) } --cycle;  }     {  \draw (A) foreach \myn in {B,C}   {  -- (\myn) } --cycle;  }  \end{tikzpicture}  \end{document}   produces: How do I solve the following second order partial differential equation?  $4u_{xx}+5u_{xy}+u_{yy}+u_x+u_y=2$  I have classified the equation to be hyperbolic and changed variables to obtain the canonical form as $u_{\epsilon\nu}={1\over{3u_\nu}}-{8\over9}$ which I believe is correct but I am struggling to find the general solution? A step by step solution would be appreciated. Custom macro for multicolumn with & without xparse The following example has two normal LaTeX \newcommand macros and two xparse \NewDocumentCommand for specifying multicolumn text. The only difference between each pair of commands is one has the alignment character as part of the macro and one does not.  However, the xparse version that requires the alignment character to be outside of the macro does NOT work and results in Misplaced \omit. error.  \documentclass{article} \usepackage{xparse}  \newcommand{\MulticolumnA}[3]{&amp;\multicolumn{#1}{#2}{#3}} \newcommand{\MulticolumnB}[3]{ \multicolumn{#1}{#2}{#3}}  \NewDocumentCommand{\MulticolumnX}{m m m}{&amp;\multicolumn{#1}{#2}{#3}} \NewDocumentCommand{\MulticolumnY}{m m m}{ \multicolumn{#1}{#2}{#3}}  \begin{document} \[ \begin{tabular}{lllllll}     {a} \MulticolumnA{4}{l}{A-abcdefghijkli}\\     {a}&amp;\MulticolumnB{4}{l}{B-abcdefghijkli}\\     {a} \MulticolumnX{4}{l}{X-abcdefghijkli}\\     %{a}&amp;\MulticolumnY{4}{l}{Y-abcdefghijkli}\\     {col1}&amp;col2&amp;{col3}&amp;{col4}&amp;{col5}\\ \end{tabular} \] \end{document}   Above compiles as is, but does not compile if the one line is uncommented. Solving $ax \equiv c \pmod b$ efficiently when $a,b$ are not coprime I know how to compute modular multiplicative inverses for co-prime variables $a$ and $b$, but is there an efficient method for computing variable $x$ where $x &lt; b$ and $a$ and $b$ are not co-prime, given variables $a$, $b$ and $c$, as described by the equation below?   $ a x \equiv c \mod b $   For example, given  $ 154x \equiv 14 \mod 182 $, is there an efficient method for computing all the possibilities of $x$, without pure bruteforce?  Please note that I'm not necessarily asking for a direct solution, just a more optimized one.    I do not believe that the Extended Euclidean Algorithm will work here, because $a$ and $b$ are not co-prime.   Edit:  Follow up question, since the first one had a shortcut:  Could the be computed efficiently as well?  $12260x \equiv 24560 \mod 24755$.  $107$ needs to be one of the computed answers. Where is the "Add vector layer"-button in GRASS 6.4.4. plug-in QGIS 2.10.1? I've just downloaded and installed the new version of QGIS 2.10.1 with GRASS 6.4.4.  When opening a new GRASS mapset and importing layers, I missed the button "Add a new vector layer". I've updated the GRASS plugin, but again the button wasn't there (all other GRASS symbols except "add raster layer" were visuable).  It's maybe a stupid question of a GIS beginner, but what can I Do? Is there an option to add (not import) a vector layer via the GRASS toolbox? What effect does the Barr body have, in relation to female Turner syndrome? Why do persons with  have  developmental abnormalities, when normal XX-females do not, even though they only have 1 active X chromosome? From what I know, one X-chromosome is inactivated into a Barr body in normal XX-females, but these obviously don't show the symptoms of Turner syndrome. What function does the inactive chromosome have?