Is there any reason for doing something like:  a="abc"   or  a='abc'   The quotes seem completely superfluous, but then there might be some hideous corner case... I'm interested in Bash and Sh in particular, but answer about the other ones as well if you can and will.
The old advice used to be to double-quote any expression involving a $VARIABLE, at least if one wanted it to be interpreted by the shell as one single item, otherwise, any spaces in the content of $VARIABLE would throw off the shell.  I understand, however, that in more recent versions of shells, double-quoting is no longer always needed (at least for the purpose described above).  For instance, in bash:  % FOO='bar baz' % [ $FOO = 'bar baz' ] &amp;&amp; echo OK bash: [: too many arguments % [[ $FOO = 'bar baz' ]] &amp;&amp; echo OK OK % touch 'bar baz' % ls $FOO ls: cannot access bar: No such file or directory ls: cannot access baz: No such file or directory   In zsh, on the other hand, the same three commands succeed.  Therefore, based on this experiment, it seems that, in bash, one can omit the double quotes inside [[ ... ]], but not inside [ ... ] nor in command-line arguments, whereas, in zsh, the double quotes may be omitted in all these cases.  But inferring general rules from anecdotal examples like the above is a chancy proposition.  It would be nice to see a summary of when double-quoting is necessary.  I'm primarily interested in zsh, bash, and /bin/sh.
I'm trying to draw a rounded rectangle with round "holes" using  but I need the rectangle borders to be thick and "holes" borders to be thin. Is there a better way to do so than:  \begin{tikzpicture}[x=1mm, y=1mm]     % Background         \fill[fill=black!10] (1.5, 1.5) rectangle (15, -15);      % Fill     \fill[fill=black!2, even odd rule, rounded corners=0.5mm]          (0, 0) rectangle (10, -10)          (2, -2) circle[radius=1.25mm]         (2, -5) circle[radius=1.25mm];     % Rectangle border     \draw[thick, rounded corners=0.5mm] (0, 0) rectangle (10, -10);     % Holes borders     \draw[thin] (2, -2) circle (1.25mm);     \draw[thin] (2, -5) circle (1.25mm); \end{tikzpicture}   The result image should look like this: Can you create a strong blockcipher with small blocksize, given a strong blockcipher of conventional blocksize? Suppose I want a strong 20-bit blockcipher.  In other words, I want a function that takes a key (suppose the key is 128 bits), and implements a permutation from 20 bits to 20 bits.  The set of permutations should be close to a randomly-chosen subset of size $2^{128}$ of all $2^{20}!$ permutations on 20 bits.  I don't want you to build a new blockcipher from scratch.  Instead, assume you have a "strong" blockcipher like AES (128-bit blocksize, 128-bit keysize) and use this to build your 20-bit cipher.   Of course, your construction should be practical (i.e., you should be able to run it in both directions with reasonable amounts of time and memory). Trying to install ubuntu on a laptop. The "bios/uefi" has no uefi boot options &amp; the windows boot manager just kicks in. The only "option" is "boot from hard disk". So we tried to use efibootmgr to put grub to boot first.  When we run an efibootmgr -o xxxx,xxxx,xxxx command it seems like the order gets fixed but on reboot the windows boot manager still has control, then when we re-opened the live usb &amp; checked the boot order it remains unaffected.  What are other ways to change the efi boot order? Maybe by mounting the efi partition (in mnt) and doing what? The average of inverse square radius $r^{-2}$ for any state quantum $|n\ell\rangle$ using the radial Schrödinger equation Is circular reasoning always a fallacy? Intuitive explanation for formula of maximum length of a pipe moving around a corner? For one of my homework problems, we had to try and find the maximum possible length $L$ of a pipe (indicated in red) such that it can be moved around a corner with corridor lengths $A$ and $B$ (assuming everything is 2d, not 3d):    My professor walked us through how to derive a formula for the maximum possible length of the pipe, ultimately arriving at the equation $L = (A^{2/3} + B^{2/3})^{3/2}$.  The issue I have is understanding intuitively why this formula works, and exactly what it's doing. I understand the steps taken to get to this point, but there's an odd symmetry to the end result -- for example, is the fact that $\frac{2}{3}$ and its inverse are the only constants used just a coincidence, or indicative of some deeper relationship?  I also don't quite understand how the formula relates, geometrically, to the diagram. If I hadn't traced the steps myself, I would have never guessed that the formula was in any way related to the original problem.  If possible, can somebody give an intuitive explanation as to why this formula works, and how to interpret it geometrically?    Here's how he found the formula, if it's useful:  The formula is found by finding the maximum possible length of the pipe by expressing the length in terms of the angle $\theta$ formed between the pipe and the wall, and by taking the derivative to find when $\frac{dL}{d\theta} = 0$, which is the minimum of $\frac{dL}{d\theta}$ and is therefore when $L$ is the smallest:  $$ L = \min_{0 \leq \theta \leq \frac{\pi}{2}} \frac{A}{\cos{\theta}} + \frac{B}{\sin{\theta}} \\ 0 = \frac{dL}{d\theta} = \frac{A\sin{\theta}}{\cos^2{\theta}} - \frac{B\cos{\theta}}{\sin^2{\theta}} \\ 0 = \frac{A\sin^3{\theta} - B\cos^3{\theta}}{\sin^2{\theta}\cos^2{\theta}} \\ 0 = A\sin^3{\theta} - B\cos^3{\theta} \\ \frac{B}{A} = \tan^3{\theta} \\ \theta = \arctan{\left( \frac{B}{A} \right)^{\frac{1}{3}}} \\ $$  At this point, we can substitute $\theta$ back into the original equation for $L$ by interpreting $A^{1/3}$ and $B^{1/3}$ as sides of a triangle with angle $\theta$ and hypotenuse $\sqrt{A^{2/3} + B^{2/3} }$:  $$ \cos{\theta} = \frac{A^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \sin{\theta} = \frac{B^{1/3}}{ \sqrt{A^{2/3} + B^{2/3} }} \\ \therefore L = A^{2/3} \sqrt{A^{2/3} + B^{2/3} } + B^{2/3} \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3}) \sqrt{A^{2/3} + B^{2/3} } \\ L = (A^{2/3} + B^{2/3})^{3/2} \\ $$  The equation for the formula for the maximum length of the pipe is therefore $L = (A^{2/3} + B^{2/3})^{3/2}$. I came across this problem and I believe Lagrange's theorem is the key to its solution.  The question is:     Let $p$ be an odd prime. Prove that there is some integer $x$ such that $x^2 \equiv −1 \pmod p$ if and only if $p \equiv 1 \pmod 4$.   I appreciate any help.  Thanks. Set myself as owner of /etc with chown command now getting all kinds of errors Question of an isomorphism of $\epsilon_ 0$ and a subset of the rationals. Python super() raises TypeError In Python 2.5, the following code raises a TypeError:  &gt;&gt;&gt; class X:       def a(self):         print "a"  &gt;&gt;&gt; class Y(X):       def a(self):         super(Y,self).a()         print "b"  &gt;&gt;&gt; c = Y() &gt;&gt;&gt; c.a() Traceback (most recent call last):   File "&lt;stdin&gt;", line 1, in &lt;module&gt;   File "&lt;stdin&gt;", line 3, in a TypeError: super() argument 1 must be type, not classobj   If I replace the class X with class X(object), it will work. What's the explanation for this? I have two layers, one (A) has point features and the other (B) has polygons.  How do I produce a new table with some (the ID) or all the fields from the A table and some (the area code) or all from the B table where the B polygon contains the A point?  I am using QGIS. New passport and entrance Schengen I am from Brazil and travel a lot in Schengen countries. Under the rules of Schengen visa (3 months stay in a 6 months period). In my passport are the entrance and exit stamps, also to chec the days I stay in Schengen.  I need a new passport and will get this in Brazil. So in this new passport are NOT the stamps. Does this mean I start new with the Schengen visa rule? I seems like. The new passport is empty. Or do I miss something? It would be great because I am almost on my 90 days stay in the period of 180 days. What can I do if I forgot my Windows password? I got a brand new Windows 7 machine, installed the operating system, created one account and forgot its password. What can I do?  There is no external CD, the operating system is loaded from somewhere inside the machine.  I already tried to remember passwords and tried all candidates with all possible combinations of caps lock, num lock etc. Where to find my Favorites in Google Chrome? Where can I find my Favorites in Google Chrome ?  Where to find the folder that contains them ? How do you select the fastest mirror from the command line? I want to update my sources.list file with the fastest server from the command line in a fresh Ubuntu Server install. I know this is trivially easy with the GUI, but there doesn't seem to be a simple way to do it from from the command line? Problem with packages subfiles and import.sty after using TexLive Utility for updating Everything worked fine before using TexLive for updating my packages three days ago. Now it is not possible anymore to use the package subfiles   Here are the MWE  The minimum Preamble.tex file  \documentclass{amsart} \usepackage{subfiles}   The MWE.tex file  \documentclass[MWE.tex]{subfiles} \begin{document} MWE \end{document}   When I compile the MWE file I get on the console  This is pdfTeX, Version 3.14159265-2.6-1.40.20 (TeX Live 2019) (preloaded &gt;format=pdflatex)  restricted \write18 enabled. entering extended mode  (./MWE.tex  LaTeX2e &lt;2020-02-02&gt; patch level 5  L3 programming layer &lt;2020-02-25&gt; (/usr/local/texlive/2019/texmf-dist/tex/latex/subfiles/subfiles.cls  Document Class: subfiles 2020/02/14 v16 Multi-file projects (class)  Preamble taken from file `MWE.tex'  (/usr/local/texlive/2019/texmf-dist/tex/latex/tools/verbatim.sty)  (/usr/local/texlive/2019/texmf-dist/tex/latex/import/import.sty)  ! LaTeX Error: File `import.sty' not found.  Type X to quit or &lt;RETURN&gt; to proceed, or enter new name. (Default extension: sty)  Enter file name:   I checked and the import.sty file is in the right directory, and everything worked fine before. Do you have any idea about what happened (apart from the fact that I should not upgrade stuff before the official releases of TexLive I suppose  ...)? There aren't any new or popular questions. There are tons of questions in both of those tabs but the question count is 0 for both of those when I try to view them in the expanded view. A customer of mine wants to pay me from the usa to my account with N26 bank.N26 Bank is a German online bank with no branches but available to customers in the euro-zone.I opened my account online in Portugal.When my customer made the transfer and sent me the slip,i discovered the beneficiary name was completely wrong but the Iban and Swift was correct.Will the wire transfer go through with correct Iban and Swift but incorrect beneficiary name? Zero divisor in $R[x]$ Let $R$ be commutative ring with no (nonzero) nilpotents. If $f(x) = a_0+a_1x+\cdots+a_nx^n$ is a zero divisor in $R[x]$, how do I show there's an element $b \ne 0$ in $R$ such that $ba_0=ba_1=\cdots=ba_n=0$? Primary ideals in Noetherian rings For an $R$-module $M$ I have the following definition for a submodule $N\subset M$ to be $\mathfrak{p}$-primary: this is the case when $\text{Ass}(M/N) = \{\mathfrak{p}\}$, that is, $M/N$ is coprimary or $(0)$ is a primary submodule.  Now for an ideal $\mathfrak{q} \subset R$ this definition gives that $\mathfrak{q}$ is $\mathfrak{p}$-primary iff $\text{Ass}(R/\mathfrak{q}) = \{\mathfrak{p}\}$. It can be shown that this is equivalent with $\mathfrak{p}^n \subset \mathfrak{q}$ for some $n\in \mathbb{N}$ and the condition \begin{align*} \forall r,s \in R: rs \in \mathfrak{q}, r\not\in \mathfrak{p}\Rightarrow s \in \mathfrak{q}. \end{align*}     Now I suspect that if $R$ is Noetherian, and if $\mathfrak{q}$ is $\mathfrak{p}$-primary, then we have $\sqrt{\mathfrak{q}} = \mathfrak{p}$. I have shown this for PID's, but does this generaly hold?    And if so, is every ideal $\mathfrak{q}\subset R$ then $\sqrt{\mathfrak{q}}$-primary?   Many thanks.