I have the following problem: I am writing a couple of algorithms using algorithm2e. I have a bunch of longer if statements and longer if conditions. The lines are broken, but the new lines are not indented properly:  \begin{document}    \begin{algorithm}[H]     \SetAlgoLined     \KwInput{Parameters...}      \KwOutput{$\emptyset$}      \If{the lines through $x_{1}, x_{2}$ and $y_{1}, y_{2}$ intersect in a single point $z$         which lies on the circle $C$        }     {       Do something \;     }     \caption{Algoritm test...}   \end{algorithm}  \end{document}   I could force line breaks at certain positions using \, but that does not fix the indentation problems either. I saw the solution provided here:   But I don't understand what is happening in that code and I was not able to reproduce the result with my problem. How can I indent the code correctly (is it possible to use something like an mbox?)?
I am trying to (partially) reproduce the algorithm displayed in Figure 4 from   My problem is with the multi-line conditions present in the "IFs". I want them to be indented, but \Indp does not seem to take effect if a new line \\ is not present in the same line.  This is what I tried:  \begin{algorithm}[htb]     \SetAlgoNoLine%     \SetNoFillComment     \SetKwIF{If}{ElseIf}{Else}{if}{:}{elif}{else:}{}%      \If{(SYNs from both hosts and \\             \Indp SYNs in earliest NFs of both hosts and \\             \Indp hosts' earliest packets differ)}{         Host with earliest NF is originator      }     \If{(SYN only from one host and \\         \Indp SYN is in connection's earliest NF)}{     Host is originator     }     \If{one host uses port 20 (ftp-data)}{     Host is originator     }     \If{only one host uses a well-known port}{     Host is responder     }     \If{start of hosts' first packets differ}{     Host with earliest NF is originator     }     Arbitrarily choose originator     \caption{Pseudo code for determining connection server side}     \label{algo:serverside2} \end{algorithm}   For instance, the line "SYNs in earliest NFs of both hosts and" is indented correctly, but "hosts' earliest packets differ" is not.
There is currently no way to delete questions or answers in the iOS app, even with the required Trusted User privilege. We can edit and vote, but not delete.  Can this option please be added to the app? I'm setting up an Nginx web server and and want to perform load balancing using the round robin method.  This is what I have found thus far:  upstream backend  {    server backend1.example.com;    server backend2.example.com;    server backend3.example.com; }   According to the page they says i need to add the above lines to my nginx.conf file and upload it to server to load balance.  I do not understand what it means by server backend3.example.com;.  How do I setup a backend instance? Flowchart using Tikz I'm trying to create a fine flowchart that is a bit complicated. Therefore I need a block for a predefined process and a connector block as well as a data block. Can someone help me? In comment is what I'd like to see in my code to keep my code nicely arranged. The rest of the code is a simple example for the Tikz site.  \documentclass{article}  \usepackage[latin1]{inputenc} \usepackage{tikz} \usetikzlibrary{shapes,arrows} \begin{document} \pagestyle{empty}   % Define block styles \tikzstyle{decision} = [diamond, draw, fill=blue!20,      text width=4.5em, text badly centered, node distance=3cm, inner sep=0pt] \tikzstyle{block} = [rectangle, draw, fill=blue!20,      text width=5em, text centered, rounded corners, minimum height=4em] \tikzstyle{line} = [draw, -latex'] \tikzstyle{cloud} = [draw, ellipse,fill=red!20, node distance=3cm,     minimum height=2em]  %\tikzstyle{subroutine} = [draw,rectangle split, rectangle split         horizontal,rectangle split parts=3,minimum height=1cm]  %\tikzstyle{connector} = [draw,circle,...] %\tikzstyle{data} = [draw, trapezium, ...]  \begin{tikzpicture}[node distance = 2cm, auto]     % Place nodes     \node [block] (init) {initialize model};     \node [cloud, left of=init] (expert) {expert};     \node [cloud, right of=init] (system) {system};     \node [block, below of=init] (identify) {identify candidate models};     \node [block, below of=identify] (evaluate) {evaluate candidate models};     \node [block, left of=evaluate, node distance=3cm] (update) {update     model};     \node [decision, below of=evaluate] (decide) {is best candidate     better?};     \node [block, below of=decide, node distance=3cm] (stop) {stop};     % Draw edges     \path [line] (init) -- (identify);     \path [line] (identify) -- (evaluate);     \path [line] (evaluate) -- (decide);     \path [line] (decide) -| node [near start] {yes} (update);     \path [line] (update) |- (identify);     \path [line] (decide) -- node {no}(stop);     \path [line,dashed] (expert) -- (init);     \path [line,dashed] (system) -- (init);     \path [line,dashed] (system) |- (evaluate); \end{tikzpicture}   \end{document} I think the questions says it all! :)  By TeX, I mean any derivative of TeX as long as the code is clearly a derivative of TeX, i.e. that it could belong on TeX.SX (and not pure Lua for example).  I'll give two answers myself so you see what I mean. Changing website favicon dynamically I have a web application that's branded according to the user that's currently logged in. I'd like to change the favicon of the page to be the logo of the private label, but I'm unable to find any code or any examples of how to do this. Has anybody successfully done this before?  I'm picturing having a dozen icons in a folder, and the reference to which favicon.ico file to use is just generated dynamically along with the HTML page. Thoughts? How to track the number of tag wikis I edited? I want to see the number of tag wiki that I edited till now. Also I want to see which are accepted and which are rejected with the detail of all those tags. How can I know that? Also I want to know the progress for Research Assisstant badge. I am getting the following error in the top menu bar:     An error occurred.Please run the Package Manger form the right-click menu or apt-get in a terminal to see what is wrong.The error message was:'Error:BrokenCount>0'.This usually means that your installed packages have unmet dependencies. I've read that abelian categories can naturally be enriched in $\mathbf{Ab}.$ How does this actually work? Wikipedia defines the notion of an abelian category as follows ().     A category is abelian iff         it has a zero object,   it has all binary products and binary coproducts, and   it has all kernels and cokernels.   all monomorphisms and epimorphisms are normal.      It later explains that an abelian category can naturally be enriched in $\mathbf{Ab},$ as a result of the first three axioms above.  How does this actually work? Show that the set of all finite subsets of $\mathbb{N}$ is countable. What makes the number of neutrons the number of proton similar? In basic chemistry, we are taught that an atom has roughly the same number of neutrons and number of protons, this doesn't seems to hold for larger atoms, but it is always roughly proportional (i.e. you seldom find an atom with 100 protons but 1 neutron, that just does not happen). Why is that? Quantifiers "most" vs. "most of" Exercise 12 from SEC. 82 of Finite-Dimensional Vector Spaces - 2nd Edition by Paul R. Halmos.  If $A$ and $B$ are linear transformations on a finite-dimensional inner product space, and if $\textbf{0} \le A \leq B$, then $\det A \leq \det B$.  Notation: $0 \le A$ means that $A$ is positive.  If $\det B = 0$, choose $x \in \ker B$, then $0 \le ((B-A)x,x) = (Bx, x) - (Ax, x) = -(Ax, x) \le 0$. Therefore $\ker B \subseteq \ker A$ and $0 = det A \le det B = 0$. How to prove for the case when $B$ is invertible? Intuitively, I feel that since both $A$ and $B$ are positive their proper values are also positive and $B$'s proper values should be greater the $A$'s and same for determinants. Complex analysis : If $z =re^{i\theta}$, then prove that $|e^{iz}| =e^{-r\sin\theta}$ Problem :   If $z =re^{i\theta}$, then prove that $|e^{iz} | =e^{-r\sin\theta}$   My working :   $z = re^{i\theta} = r(\cos\theta + i\sin\theta)$   $\Rightarrow iz = ir(\cos\theta +\sin\theta) $ = $-r\sin\theta +ir\cos\theta $  $\Rightarrow e^{iz} =e^{(-r\sin\theta +ir\cos\theta)} = e^{-\sin\theta} e^{ir\cos\theta} $   $\Rightarrow |e^{iz}| = |e^{-r\sin\theta}||e^{ri\cos\theta}|$   Now please guide how to proceed further to get the result... thanks. Why doesn't pressing enter return '\n' to getch()? #include &lt;stdio.h&gt; #include &lt;conio.h&gt; main() {     char ch,name[20];     int i=0;     clrscr();     printf("Enter a string:");     while((ch=getch())!='\n')     {         name[i]=ch;         i++;     }     name[i] = '\0';     printf("%s",name); }   When I give "abc" as input and if I press enter it's not working. Can anyone let me know why the condition ch=getch() != '\n' is not becoming false when I press enter? I have also observed that ch is taking \r instead of \n. Kindly let me know. Thanks Prove that $\lim_{x \to \infty}\big(\frac{x}{x-1}\big)^x$ is also $e$. Trying to make sense out of the idea that $100\%$ continuous decay is $\frac{1}{e}$, I thought about this:  You can express $1+\frac{1}{x}$ as $\frac{x+1}{x}$, such that $\big(1+\frac{1}{x}\big)^x = \big(\frac{x+1}{x}\big)^x$  And you can express $1-\frac{1}{x}$ as $\frac{x-1}{x}$, such that $\big(1-\frac{1}{x}\big)^x = \big(\frac{x-1}{x}\big)^x =\frac{1}{\big(\frac{x}{x-1}\big)^x}$  Now $\big(\frac{x}{x-1}\big)^x$ and $\big(\frac{x+1}{x}\big)^x$ look very similar, and I can imagine (and see on Mathematica) that $\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x$ also approaches $e$.  However I'm clueless about limit proofs, how do you prove that?  EDIT: Just had a last second insight.  Can you say that $$\lim_{x \to \infty} \big(\frac{x}{x-1}\big)^x = \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot \lim_{x \to \infty} \big(\frac{x}{x-1}\big)$$ $$= \lim_{x \to \infty} \big(\frac{x}{x-1}\big)^{x-1} \cdot 1$$ $$=\lim_{x \to \infty} \big(\frac{x+1}{x}\big)^x = e ?$$   (not sure about that last limit transition...I mean I'm pretty sure it's true, just not sure how to write it; I would appreciate feedback on that, thank you) If $\lambda_{max} = n$, then $n\times n$ positive, reciprocal matrix A is consistent At the end of . the author states that  Suppose we have $n \times n$ matrix $A$ having only positive elements and satisfying the property $a_{ij}=1/a_{ji}$ (a matrix satisfying this property is called a reciprocal matrix).  If its largest eigenvalue $\lambda_{max}$ is equal to $n$, then the matrix $A$ satisfies the property (consistency property) $a_{ij} a_{jk} = a_{ik}$ where $i,j,k=1,2,...,n$.  I'm not convinced this theorem is true. Could anyone help? $$ \sum_{n=1}^{\infty} \frac{(\sin n+2)^n}{n3^n}$$  Does it converge or diverge?   Can we have a rigorous proof that is not probabilistic?   For reference, this question is supposedly a mix of real analysis and calculus. Do there exist general conditions under which we can conclude that continuity on a topological space is detected by $\mathbb{R}$? Whenever $X$ is a topological space, let us say that continuity on $X$ is detected by $\mathbb{R}$ iff for all functions $f : X \rightarrow Y$ where $Y$ is another topological space, we have that if every continuous function $c : \mathbb{R} \rightarrow X$ satisfies "$f \circ c$ is continuous", then $f$ is itself continuous.   Clearly, continuity on $\mathbb{R}$ is detected by $\mathbb{R}$ It is shown  that continuity on $\mathbb{R}^2$ is detected by $\mathbb{R}$. Clearly, it is not the case that continuity on $\mathbb{Q}$ is detected by $\mathbb{R}$, since every continuous mapping $\mathbb{R} \rightarrow \mathbb{Q}$ is constant.      Question. Do there exist general conditions underwhich we can conclude that continuity on a topological space is detected by $\mathbb{R}$? Is it appropriate to use the salutation "Dear All" in a work email? I have observed that in my work place, whenever a mail is sent to more than one person( like an information, meeting request or a notice etc.), the mail starts with the salutation "Dear All". This, somehow, doesn't seem correct to me. For me, a salutation should address the reader individually, at least in its singular form, as in "Dear Colleagues".  Am I correct in thinking that? I have never found experimental evidence that measuring one entangled particle causes the state of the other entangled particle to change, rather than just being revealed.  Using the spin up spin down example we know that one of the particles will be spin up and the other spin down, so when we measure one and find it is spin up we know the other is spin down. Is there any situation after creation that the particle with spin up will change to spin down?