Can anyone help me figure out what this "find" command with curly braces is saying? I found some old notes where someone suggest I run the following command to set some file permissions:  find /perl/lib/ -type f -exec chmod o+r \{\} \;   Would you happen to know what the \{\} part of the statement is doing?  Is it even necessary?  I tested running it a few times on some test files and it seems to do the same thing as removing the backslashes:  find /u01/app/oracle/product/12102/perl/lib/ -type f -exec chmod o+r {} \;   Any help you can offer would be appreciated.
GNU find and masking the {} for some shells - which? The man page for GNU find states:    -exec command ;     [...] The  string `{}'  is  replaced  by the current      file name being processed everywhere it occurs in the      arguments to the command, not just in arguments where      it is alone, as in some  versions  of  find.     Both  of  these  constructions might need to be escaped      (with a `\') or quoted to protect them from expansion      by the shell.     That's from the man to find (GNU findutils) 4.4.2.   Now I tested this with bash and dash, and both don't need to have the {} being masked. Here is a simple test:  find /etc -name "hosts" -exec md5sum {} \;    Is there a shell, for which I really need to mask the braces? Note, that it doesn't depend upon whether the file found contains a blank (invoked from bash):   find ~ -maxdepth 1 -type d -name "U*" -exec ls -d {} \;  /home/stefan/Ubuntu One   This changes if the found file is passed to a subshell:  find ~ -maxdepth 3 -type d -name "U*" -exec bash -c 'ls -d {}' \;  ls: cannot access /home/stefan/Ubuntu: No such file or directory ls: cannot access One: No such file or directory   which can be solved by:  find ~ -maxdepth 3 -type d -name "U*" -exec bash -c 'ls -d "$0"' {} \;   in contrast to:   find ~ -maxdepth 3 -type d -name "U*" -exec bash -c 'ls -d "{}"' \;  /home/stefan/Ubuntu One   but that's not what the man page is talking about, is it? So which shell treats {} in a different way?
Why isn't reflexivity redundant in the definition of equivalence relation? An equivalence relation is defined by three properties: reflexivity, symmetry and transitivity.  Doesn't symmetry and transitivity implies reflexivity? Consider the following argument.  For any $a$ and $b$, $a R b$ implies $b R a$ by symmetry. Using transitivity, we have $a R a$.  Source: Exercise 8.46, P195 of Mathematical Proofs, 2nd (not 3rd) ed. by Chartrand et al Instant e-mail notifications of answers to questions I'd like to have Instant notifications by email of answers to my questions.  Can this be implemented? How to solve "Extra junk at end of file" when using apt? I'm trying to configure some proxies for a guest virtual machine and now I'm getting:  sudo apt-get update E: Syntax error /etc/apt/apt.conf.d/95proxy:4: Extra junk at end of file   The content of the file is:  cat /etc/apt/apt.conf.d/95proxy   Acquire::http::proxy  "http://10.0.0.60:3128/" Acquire::ftp::proxy "http://10.0.0.60:3128/" Acquire::https::proxy "http://10.0.0.60:3128/" Whats the simplest way to edit and add files to "/var/www"? Having installed the web server is there a simple way to set a user able to use the graphic interface to copy files and directories to the local web server /var/www  I gave myself administrative privileges in Ubuntu but it still doesn't allow copies. 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: I'm trying some things with my shell. This includes the following command:  kill $$  My expectation was that the shell would log me out and terminate itself. But instead the bash persists without change.   Any idea why?  I ran this command on different environments (archlinux, debian). The behavior was always the same. Navier-Stokes system I'm using Ubuntu and I'm trying to connect to another machine in a different network using remote desktop.   In Windows7 I have made the following in order to activate remote desktop:   I've gone to computer -> properties -> remote settings I've selected the option: "Allow connections from computers running any version of Remote Desktop I've opened "Windows Firewall with Advanced Security" In inbound rules I've enabled the rules for remote desktop (public and domain)   I have also installed Remmina in the Ubuntu machine.  For configuring it I did the following steps:   Selected the RDP protocol In the server input I have written the Windows machine public IP. In username / password I have typed my login credentials (the same as my Windows admin account)   But when I try to connect I get this error message:  "Unable to connect to RDP server 89.130.251.160"  If I ping my Windows7 machine, I have a correct response.  Any suggestions? Split file into multiple small files, separate by the newline symbol Is there a utility that split file by newline symbol? e.g if a file contains the following lines,  aa bbb cccc   If I want to split it to 3 files, the desired output would be:  aa, bbb And cccc (in 3 different files)  I already checked the split command, it only cut file by file sizes, not what I want.  If I don't wrote a utility myself, is there any standard tool to use? Do I need a Schengen visa if I have UK permanent residency? How to get the pid of the last executed command in shell script? I want to have a shell script like this:  my-app &amp; echo $my-app-pid   But I do not know how the get the pid of the just executed command.  I know I can just use the jobs -p my-app command to grep the pid. But if I want to execute the shell multiple times, this method will not work. Because the jobspec is ambiguous. Double-column listing package Not too well up on LaTeX, but I want to use the listings package for listings spanning a two-column format.  Unfortunately, asterisks cannot save me (like for figures/tables); this doesn't work:  \begin{lstlisting*} ... \end{lstlisting*}   I've Googled about, but got no satisfactory answer, and getting a bit desperate. Should be simple... hope I'm not missing something stupid. Perhaps it's possible to embed this into a separate spanning float?  Help?  (P.S., I've posted this on the main stackoverflow site as well...  if this is considered a duplicate, please comment and I'll remove the other. Getting a bit desperate fighting a deadline.) A story about a guy who helped humanity by being held in a prison I remember a story I have read buy could not remember the author the name or anything apart from the basic plot. The story is about some future where humanity has to fight with some aliens (not 100% sure). The story tells about a guy who was some kind of activist and who was in a prison two or more times and somehow helped humanity by being held in a prison (can't remember details). I think the story starts by saying that there is a statue in his honor that is not very well known but he did more to help humanity than famous politicians.   The story is most likely written by some famous English-speaking author, though I read it in a translation. Automorphisms of the complex plane How can it be shown that $$Aut(\mathbb{C})=\{f|f(z)=az+b,a\neq 0\},$$ where an automorphism of $\mathbb{C}$ is defined as a bijective entire function with entire inverse?  If $f$ is of the form $f(z)=az+b$, with $a\neq 0$, then obviously $f$ is in $Aut(\mathbb{C})$. How can I prove the converse? If $f(x)=x-\int^{\frac{\pi}{2}}_0f(x)\sin(x)dx$ find the explicit function for $f(x)$ in its simplest form Inverse of $f(x)=\sin(x)+x$ What is the inverse of   $$f(x)=\sin(x)+x.$$  I thought about it for a while but I couldn't figure it out and I couldn't find the answer on the internet.  What about  $$f(x)=\sin(a \cdot x)+x$$  where $a$ is a known real constant.  Thank you for taking the time to read this question! Sorry if this has been asked before... If there are injective homomorphisms between two groups in both directions, are they isomorphic? If I have two groups, $G$ and $H$ and two injective homomorphisms $\phi:G \to H$ and $\psi: H \to G$, then by the first isomorphism theorem applied to $\phi$, we have that $G \cong \mathrm{Im} (\phi)$, a subgroup of $H$. Similarly, $H$ is isomorphic to a subgroup of $G$. For finite groups, this guarantees that $G \cong H$ but does this hold for infinite groups?  Weird things can happen for infinite groups, e.g $n\mathbb Z \subsetneq \mathbb Z$ but $n \mathbb Z \cong \mathbb Z$. I'm wondering if this kind of thing stops an isomorphism from occurring.  I think this question generalises to rings, modules, fields and so on. Can this be answered for all of these structures? if $a,b$ are nilpotent elements of a commutative ring $R$, show that $a+b$ is also nilpotent Iterate over all pairs of consecutive items in a list What is the norm of this bounded linear functional? Let $a$, $b$ be two arbitrary but fixed real numbers such that $a &lt; b$, let $C[a,b]$ denote the normed space of all continuous real (or complex) valued functions defined on $[a,b]$ with the maximum norm, and let $f$ be the functional defined on $C[a,b]$ as $$f(x):= \int_{a}^{b} x(t) x_0(t) dt $$ for all $x$ in $C[a,b]$, where $x_0$ is a fixed element of $C[a,b]$.   Now $f$ is evidently linear and bounded. How to compute the norm of $f$? What is this norm?   And what is the norm of the bounded linear functional $g$ defined as follows: $$g(x):= \int_{a}^{(a+b)/2} x(t) \cdot dt - \int_{(a+b)/2}^{b} x(t) \cdot dt$$ for all $x$ in $C[a,b]$?