Recently I have been writing a chatbot, and in part of the process it expands contractions. While doing this, I have found odd behavior for the contraction "isn't". There are two different way to expand this contraction:     isn't this great &rightarrow; is this not great?   or:     what isn't great &rightarrow; what is not great   Essentially, in some cases the "not" goes after the following word (the word following the contraction); in others it doesn't. I can't seem to nail down in which cases this happens and in which cases it doesn't. Is there some rule to specify how it expands?
I refer to the usage of "don't" as an imperative to tell someone what not to do. As in,     Hey! Don't you dare touch that button!   When it is used in the interrogative or as part of a statement, "don't" can be expanded like most other contractions. It either just breaks up right in the sentence, or surrounds the subject.     Don't you love me, Juanromeo? -> Do you not love me, Juanromeo?      I don't love you, Marijuliet. -> I do not love you, Marijuliet.   However, I have no idea how one breaks up the imperative. "Do you not dare[...]" sounds like it's a lead up to a question, "You do not dare[...]" comes across as a statement moreso than a command, and "Do not you dare[...]" just sounds silly.  "Don't" is a bit unique among "not" contractions (at least that I know of) because it is the only one that can be used in an imperative sense. Couldn't, shouldn't, wouldn't, haven't, isn't, doesn't, aren't, can't... none of these can be used in an imperative sense because you cannot command someone to could, should, etc.  Is it possible to expand the don't contraction when it is used as part of a command? Or is its unique case such that it exists as its own linguistic construct?
Is a sub-shell the same thing as a child-shell There are this two names: a subshell and a child-shell.  Yes, a child process will be started by any of this:  sh -c 'echo "Hello"' ( echo "hello" ) echo "$(echo "hello") echo "hello" | cat   Are all equivalent and share the same name? Do all share the same properties?    POSIX :     A shell execution environment consists of ....   But the last paragraph of above link has this:     A subshell environment shall be created as a duplicate of the shell environment, except that signal traps that are not being ignored shall be set to the default action.    And specially:     Command substitution, commands that are grouped with parentheses, and asynchronous lists shall be executed in a subshell environment. Additionally, each command of a multi-command pipeline is in a subshell environment; ....   The sh -c 'echo "Hello"' is not included there, should that be called a subshell also? Options to hide reputation information on Stack Overflow? Show that the operator norm is submultiplicative We had in our lecture on numerical analysis the following: Let $\mathrm{Lin}(X,Y)$ be the set of all linear maps $X\rightarrow Y$. Let $A\in\mathrm{Lin}(\mathbb R^l,\mathbb R^n)$ and $B\in\mathrm{Lin}(\mathbb R^n,\mathbb R^m)$ and $\|C\|_{op}:=\max_{\|x\|\leq1}\|C(x)\|$. Then our lecturer followed $\|A\circ B\|_{op}\leq\|A\|_{op}\cdot\|B\|_{op}$. So he didn't prove it and so I've tried it by my own. My attempt: $$\begin{aligned} \|A\circ B\|_{\mathrm{op}} &amp;= \max_{\|x\| &lt; 1}\|(A \circ B)(x)\| \\ &amp; \leq \max_{\|x\| &lt; 1} \|A\|_{\mathrm{op}} \|B(x)\| \\ &amp;= \|A\|_{\mathrm{op}} \max_{\|x\|\leq1}\|B(x)\|\\ &amp;= \|A\|_{\mathrm{op}}\|B\|_{\mathrm{op}} \end{aligned}$$ But this seems too easy. I am really interested in a nice proof so anybody could help? Thanks a lot! How can I use the "signature" function in Preview on Mac Mini with no camera? I have a Mac Mini with no attached camera. I would like to use the feature of Preview that allows you to insert a signature. This relies on taking an image of your signature using an isight/facetime/webcam. I do not have on.  I do have a MacBook pro on which I use the signature function. Is there a way to copy a file from the the MacBook Pro to the Mac Mini to enable this function on the Mini? How can you use optional parameters in C#? Note: This question was asked at a time when C# did not yet support optional parameters (i.e. before C# 4).  We're building a web API that's programmatically generated from a C# class. The class has method GetFooBar(int a, int b) and the API has a method GetFooBar taking query params like &amp;a=foo &amp;b=bar.   The classes needs to support optional parameters, which isn't supported in C# the language. What's the best approach? Unable to uninstall Wine apps Building a wormhole We regularly get questions about wormholes on this site. See for example  and . Various wormhole solutions are known, of which my favourite is  because it's closest to what every schoolboy (including myself many decades ago) thinks of as the archetypal wormhole.  The trouble is that Visser has pulled the same trick as  of starting with the required (local) geometry and working out what stress-energy tensor is required to create it. So Visser can tell us that if we arrange exotic string along the edges of a cube the spacetime geometry will locally look like a wormhole, but we know nothing about what two regions of spacetime are connected.  My question is this: suppose I construct a Visser wormhole by starting in Minkowksi spacetime with arbitrarily low densities of exotic matter and gradually assembling them into the edges of a cube, how would the spacetime curvature evolve as I did so?  I'm guessing that I would end up with something like Wheeler's bag of gold spacetime. So even though I would locally have something that looked like a wormhole it wouldn't lead anywhere interesting - just to the inside of the bag. I'm also guessing that my question has no answer because it's too hard to do any remotely rigorous calculation. Still, if anyone does know of such calculations or can point me to references I would be most interested. Example where schema are better than multiple databases In several database systems (e.g., PostgreSQL and MS SQL Server) there is a concept of schema. There are many questions out there asking what schema are and how they are used; here is a . Usually the answer boils down to "namespace" and "permissions". However, what remains unclear to me is why a schema has any advantage over just adding a new database. A different database allows one to have both separate namespaces and permission assignments.  Can someone provide an example in PostgreSQL where multiple schema in a single database provide an advantage over multiple databases without schema? From a purely organizational perspective they appear equivalent. Show question ID When I want to close a question as duplicate using the app, I have to enter the question ID as usual. Problem is, as far as I can tell, the question IDs are shown nowhere! Please add the ID somewhere. Google Chrome error: Error initializing NSS without a persistent database It was working properly before, suddenly it stopped working. When I tried to open by clicking on its icon, but nothing happens.  As already here are many questions related to this, but none of them are working for me. I already checked  and .  I uninstalled Chrome by following the instructions from .  sudo apt-get purge google-chrome-stable rm ~/.config/google-chrome/ -rf   Again I installed Chrome by:  sudo apt-get install google-chrome-stable   But the problem is still thesame.  When I enter google-chrome in terminal I get following error.  [12289:12323:0723/014456:ERROR:nss_util.cc(744)] Error initializing NSS with a persistent database (sql:/home/maddy/.pki/nssdb): libsoftokn3.so: cannot open shared object file: No such file or directory [12289:12323:0723/014456:ERROR:nss_util.cc(212)] Error initializing NSS without a persistent database: NSS error code: -5925 [12289:12323:0723/014456:FATAL:nss_util.cc(214)] nss_error=-5925, os_error=0 Immigration check between Northern Ireland/Republic of Ireland and Great Britain Do immigration checks exist between Northern Ireland and England/Scotland? (not just for flights but also ferry/bus/train) Do immigration checks exist between the Republic of Ireland and England/Scotland? Newton's 3rd law... hitting drywall (which I break) vs hitting a brick (which breaks me)? According to the Third Newton's law of motion:        For every action there is an equal and opposite reaction.   So, I understand that if I hit a brick wall with $50\, \mathrm{lbs}$ of force, the brick wall also hits me with $50\, \mathrm{lbs}$ of force (usually painfully). In this instance, Newton's third law makes sense. What I'm confused about is, if I hit a patch of drywall with $50\, \mathrm{lbs}$ of force, it's probably going to break, and due to the lack of pain in my hand, I can tell it did not hit me back with $50\, \mathrm{lbs}$ of force.   How does Newton's third law apply to situations when one object or the other is destroyed? It certainly seems like at that point it is incapable of delivering the full force of my blow back to me. What happens with the energy? How to do binary calculation？ I'm reading Baby Rudin and exercise 28 of chapter 2 reads "Prove that every closed set in a separable metric space is the union of a (possibly empty) perfect set and a set which is at most countable." I don't understand why the set must be closed in order for this to be true.  My proof is as follows, and I don't make any reference to the closedness of the set:  Let $E$ be a set in a separable metric space.  Let $B$ be a countable base of that metric space.  Let $P$ be the set of condensation points of $E$.  I claim that $P$ is perfect and $E-P$ is at most countable.  If $p$ is a limit point of $P$, then every neighborhood of $p$ includes at least one point $q$ of P.  For a neighborhood $N_{p}$ of $p$, choose a neighborhood $N_{q} \subset N_{p}$ of $q$.  Since $q$ is a condensation point of $E$, uncountably many points of $E$ are in $N_{q}$, and thus in $N_{p}$.  So $p\in P$ and $P$ is closed.  If $p \in P$, then $p$ is a condensation point of $E$, so every neighborhood $N_{p}$ of $p$ must include uncountably many points of $E$.  If $p$ is not a limit point of $P$, then there exists a neighborhood $N_{p}$ of $p$ such that $N_p$ includes no condensation points of $E$ except for $p$.  This means that every point $q\in N_{p}-\{p\}$ must have a neighborhood $N_{q}$ that does not have uncountably many points of $E$.  For each $b_{k}\in B \cap \mathscr{P}(N_{p})$, choose such a neighborhood $N_{q_{k}} \supset b_{k}$ of some point $q\in N_{p}-\{p\}$.  Now $\{N_{q_{k}}\}$ covers $N_{p}$, and $\bigcup_{k=1}^{\infty}N_{q_{k}}\supset N_{p}$ has at most countably many points of $E$, for it is a union of countably many sets that have at most countably many points of $E$.  This contradicts the assumption that $p\in P$.  So $P$ is perfect.  Let $S=E-P$. There are no condensation points in $S$.  So there exists an open cover $\{N_s\}$, where each $N_s$ is a neighborhood of $s \in S$ with at most countably many elements of $S$. For each $b_{k}\in B \cap \mathscr{P}(S)$, choose such a neighborhood $N_{s_{k}} \supset b_{k}$ of some point $s\in S$.  Then we have a countable subcover $\{N_{p_{k}}\}$ of $S$. As $N_{p_{k}} \cap S$ is at most countable for all $k$, $S$ must be at most countable.  So what am I doing wrong?  Or is Rudin's assumption of closedness extraneous? This is something I've been curious about since starting this whole 'meta' thing a couple years back.     Let's say a user posts a feature request.   Under what circumstances is the feature-request read by staff? If staff decline to do anything (I'd guess it's some sort of triage), under what circumstances is the feature-request reevaluated by staff?     Let's say a user posts a bug.    Under what circumstances is the bug tested? What generally determines whether a verified bug is actionable? (And/or what are the criterion for deciding what to do?)     For both of the above, would it be possible to see a general-but-relatively-detailed flow for what happens to these questions? This story had a researcher landing on an uncharted planet to see if it was any good for mining some type of mineral/metal.  Whilst there he encounters this type of ball that chases him all over the planet. He ends up finding this ball’s collection of other things it has caught including animals all the way back to the dinosaurs.  On the ball it has the writing '+/-7.  In the end it chases him for so long that he weighs too little for the balls 'collection'.  Any ideas greatly appreciated. Listings - Only highlight between delimiters I want to implement a function name color highlighting and I use the listings package. My 'regex' for this is def %s( where def (including space) is the first delimiter %s is the function name and ( is the second delimtier. Now my aim is to color only %s in purple. So basically I want to only touch the part between the delimiters and the delimiters should stay untouched.  Here is what I tried so far:  \documentclass{article} \usepackage{xcolor,listings} \usepackage[ngerman]{babel} \usepackage{fontspec} \setmonofont{Courier New}  \begin{document}  \lstset{language=Python,     escapebegin=\color{orange},     keywordstyle=\color{blue}\bfseries,     commentstyle=\color{orange!80!red},     stringstyle=\ttfamily\color{gray},     %moredelim = [is][\color{blue}\bfseries def \mdseries\color{purple}]{def\ }{(},     moredelim = [s][\color{purple}]{def\ }{(},     showstringspaces=false}‎ \lstset{literate=%    *{0}{{{\color{red}0}}}1     {1}{{{\color{red}1}}}1     {2}{{{\color{red}2}}}1     {3}{{{\color{red}3}}}1     {4}{{{\color{red}4}}}1     {5}{{{\color{red}5}}}1     {6}{{{\color{red}6}}}1     {7}{{{\color{red}7}}}1     {8}{{{\color{red}8}}}1     {9}{{{\color{red}9}}}1 }  ‎‎\begin{lstlisting}[basicstyle=\ttfamily,mathescape] def fancy_function_name(page_url = 'tex.stackexchange.com'):     return page_url.split('.')[0] \end{lstlisting}  \end{document}   This code will generate:   You can see that it matches the right place but is too greedy and colors the delimiters too. So I thought about the uncommented line: %moredelim = [is][\color{blue}\bfseries def \mdseries\color{purple}]{def\ }{(}, If I use this one and uncomment the line below I get this result:    It's nearly what I wan't but I cannot add the last, now missing ( in black color.  Does anyone has a solution for this without pretouching the code and add all the @@ symbols? Does there exists an Isomorphism between $\mathbb{C}^* $ under multiplication with $\mathbb{C}$ under addition.  Attempt In my opinion It does not exists.  If there exists such an isomorphism say $\phi :\mathbb{C}^* \to \mathbb{C}$ then, from the fact that Isomorphism takes identity to identity we have $$\phi \left( 1 \right)=0$$  From the fact      Given $\phi:G \to H$ be an isomorphism, then For a fixed integer $k$ and a fixed group element $b$ in $G$, the   equation $x^k=b$ has the same number of solutions in $G$ as does   the equation $x^k=\phi\left(b\right)$ in $H$.   We have $x^4=1$ and $x^4=0$ must have same number of solutions, but the solutions to these equations are 4 and 1 only. How can I get the source code of a Python function? Suppose I have a Python function as defined below:  def foo(arg1,arg2):     #do something with args     a = arg1 + arg2     return a   I can get the name of the function using foo.func_name. How can I programmatically get its source code, as I typed above? Script Tag Error - Aura Lightning Component