Where can I find / download the source code that is used for the Markdown editor used by Stack Overflow? This is what it looks like:
Where is the open source project for the WMD editor used on this site?
Remove empty directory from CVS? I'm not quite sure how this happened, but somehow a completely empty hierarchy of directories has ended up in my repository:  com/ com/companyname/ com/companyname/blah/ com/sun/ com/sun/java/ com/sun/java/jax_rpc_ri/   I think what happened was that these directories did have files in them, but then a developer realized he/she shouldn't have checked them in in the first place since these are by-products of the build process, so he/she removed the files but somehow the empty directories are left in the repository as ancient relics.  How can I remove this from CVS? The only results I seem to be able to find on google say that there shouldn't be a need to remove empty directories as CVS won't keep them around in the first place, and that the -P (prune) options to cvs update should remove them from the working directory  - which is zero help if you actually have empty directories in your repository.  A cvs remove and cvs commit doesn't seem to take care of this situation:  $ cvs remove -Rf com cvs remove: Removing com cvs remove: Removing com/companyname cvs remove: Removing com/companyname/blah cvs remove: Removing com/sun cvs remove: Removing com/sun/java cvs remove: Removing com/sun/java/jax_rpc_ri $ cvs commit com cvs commit: Examining com cvs commit: Examining com/companyname cvs commit: Examining com/companyname/blah cvs commit: Examining com/sun cvs commit: Examining com/sun/java cvs commit: Examining com/sun/java/jax_rpc_ri $ ls -l com total 24 drwxrwxr-x  2 matt matt 4096 Oct 15 14:38 CVS drwxrwxr-x  9 matt matt 4096 Oct 15 14:38 companyname drwxrwxr-x  4 matt matt 4096 Oct 15 14:38 sun   It's still there!  Does SVN have this weird behavior too? After upgrading to Firefox 57 Quantum many website have rendering issues, such as empty frames, missing text or fonts, etc. I use Kubuntu 16.04 64 bit.  So far, I've tried:   Restarting Firefox in safe mode Disabling "hardware acceleration"   Wiping out my profile Purging Firefox and reinstalling Running Firefox from a clean user account   But nothing helped so far.  Any tips on how to solve this are much appreciated Constraints of relativistic point particle in Hamiltonian mechanics I try to understand constructing of Hamiltonian mechanics with constraints. I decided to start with the simple case: free relativistic particle. I've constructed hamiltonian with constraint:  $$S=-m\int d\tau \sqrt{\dot x_{\nu}\dot x^{\nu}}.$$  Here $\phi=p_{\mu}p^{\mu}-m^2=0$  $-$ .  Then $$H=H_{0}+\lambda \phi=\lambda \phi.$$   So, I want to show that I can obtain from this Hamiltonian the same equation of motion, as obtained from Lagrangian.  But the problem is that I'm not sure what to do with $\lambda=\lambda(q,p)$. I tried the following thing:  $$\dot x_{\mu}=\{x_{\mu},\lambda \phi\}=\{x_{\mu},\lambda p^2\}-m^2\{x_{\mu},\lambda\}=\lambda\{x_{\mu},p^2\}+p^2\{x_{\mu},\lambda\}-m^2\{x_{\mu},\lambda\}$$$$=2\lambda \eta_{\mu b} p^b+p^2\{x_{\mu},\lambda\}-m^2\{x_{\mu},\lambda\}=2\lambda \eta_{\mu b} p^b+p^2\frac{\partial \lambda}{\partial p^{\mu}}-m^2\frac{\partial \lambda}{\partial p^{\mu}},$$  $$\dot \lambda=\{\lambda, \lambda \phi \}=\{\lambda,\lambda p^2\}-m^2\{\lambda,\lambda\}=\lambda\{\lambda,p^2\}+p^2\{\lambda,p^2\}=2\lambda\eta_{ak}p^{a}\frac{\partial \lambda}{\partial x^{k}},$$  $$\dot p_{\mu}=\{p_{\mu},\lambda p^{2}-m^2\lambda \}=p^{2}\{p_{\mu},\lambda\}-m^2\{p_{\mu},\lambda\}=-p^{2}\frac{\partial \lambda}{\partial x^{\mu}}+m^2\frac{\partial \lambda}{\partial x^{\mu}}.$$  If we recall that $p^2-m^2=0$, then we get from the third equation: $\dot p=0$, and from the first: $$\dot x_{\mu}=2\lambda\eta_{ak}p^{a}.$$  So we have    $\dot x_{\mu}=2\lambda\eta_{\mu b}p^{b}.$ $\dot \lambda=2\lambda\eta_{ak}p^{a}\frac{\partial \lambda}{\partial x^{k}}.$ $\dot p=0.$   But I don't know what to do next. Can you help me? In a document I'm writing, I'm using the familiar  option for my floating tables, and I  them. I'd like to avoid repeating that every time I'm creating a table. I assume it's easy to add something in my preamble that makes sure that these two options are used every time I begin a new table environment?  \documentclass{article} \begin{document} \begin{table}[hbt] \centering \begin{tabular}{*3{l}} 1 &amp; 2 &amp; 3\\ a &amp; b &amp; c\\ \end{tabular} \end{table} \end{document} Why can't we see light travelling from point A to B? Let's say we have a cloud of dust which is a lightyear across and someone shoots a beam of light from point A to B , why it is not possible for an observer far far away to see the light while it travels through the cloud at the speed of light? In Star Wars: "The Force Awakens," how could this lightsaber have turned up where it did? In The Force Awakens:      Maz Kanata tells Rey that the lightsaber she just picked up and had visions from belonged to Anakin and Luke and now it's hers.   The lightsaber that Luke got from Obi-Wan Kenobi in A New Hope was lost/destroyed when Luke fought Darth Vader in Star Wars: The Empire Strikes Back.  In Return of the Jedi Darth Vader compliments Luke on building a new lightsaber and tells Luke that his skills are now complete.   How could that possibly be the lightsaber that Anakin used originally, and that was then passed on to Luke, when that lightsaber no longer exists?  Writers notes, director's notes, interviews, cast interviews, etc. are welcome. Thanks. Prove that $L(V,W)$ forms a vector space Let $V$ and $W$ be vector spaces over a field $F$. Let $L(V,W) = \{T:V\to W : T \text{ is linear} \}$, that is, $L(V,W)$ is the collection of all linear functions from $V$ to $W$. For $S,T \in L(V,W)$ and $a \in F$, define new functions $(S+T):V \to W$ and $ aT :V \to W$ by $(S+T)(x) = S(x)+T(x)$ ; $(aT)(x) = a(T(x))$.   Prove that $S+T$ belongs to $L(V,W)$ and $aT$ belongs to $L(V,W)$. Taking back turns, and forgotten privileges Understanding the differential $dx$ when doing $u$-substitution A real $2 \times 2 $ matrix $M$ such that $M^2 = \tiny \begin{pmatrix} -1&0 \\ 0&-1-\epsilon \\ \end{pmatrix}$ , then : A real $2 \times 2 $ matrix $M$ such that $$M^2 =  \begin{pmatrix} -1&amp;0 \\ 0&amp;-1-\epsilon  \\   \end{pmatrix}$$ (a) exists for all $\epsilon &gt; 0$. (b) does not exist for any $\epsilon &gt; 0$. (c) exists for some $\epsilon &gt; 0$. (d) None of the above is true.  Attempt: I couldn't think of any theory base to prove / disprove the existence of such a matrix. Could someone please give me a hint on how to go about this problem? Thank you very much for your help in this regard. Function overloading in C If two gravitational waves came in contact with each other what would happen? In another question entirely, what happens when a higher gravitational field interacts with a weaker one. The thing with these tags is, you can always (or almost always) use a specific product/technology/language tag which would render the company-name tag completely redundant (see Appendix C for sloppy non-proof). For example, what relevant information would  tag add to  already tagged with ? What would  tag add to  already tagged with ? Another way of looking at this: how much useful information do you lose if you don't tag a question with a company name? Note: I'm not talking about brand/company/organizations like , , ,  because those names also refer to websites/web services with which you can interact via API or from which you can extract data. Appendix C: The &quot;about&quot; test The relationship between a tag and a question can be expressed as such  This question is about [tag]  In case of ,  This question is about [microsoft]  Fair enough. But in reality people are using these company-name tags like this  This question is about something developed by or related to [microsoft]  But if that's the case, you might as well just go with  This question is about [something developed by or related to microsoft]  where  can be , , ... you get the idea. How to calculate temperature in Wire? Let's say I have 1m of 2,5mm² copper wire, insulated with H07V-K. The wire carries 40A and environment temperature is 25°C. The wire is suspended in free air. How hot will the wire get? And how can I calculate it for other values?  The use case is a 12V-230V AC converter I want to hook up to my car and determine the minimum wire thickness required. Because it's only used from time to time and not buried in a wall I suspect I can use thinner wires. This is a temporary thing and doesn't need to be up to code. Administrator URL '&' changed to '&amp;' The cartesian product $\mathbb{N} \times \mathbb{N}$ is countable I'm examining a proof I have read that claims to show that the Cartesian product $\mathbb{N} \times \mathbb{N}$ is countable, and as part of this proof, I am looking to show that the given map is surjective (indeed bijective), but I'm afraid that I can't see why this is the case. I wonder whether you might be able to point me in the right direction?  Indeed, the proof begins like this:  "For each $n \in \mathbb{N}$, let $k_n, l_n$ be such that $n = 2^{k_n - 1} \left(2l_n - 1 \right)$; that is, $k_n - 1$ is the power of $2$ in the prime factorisation of $n$, and $2 l_n - 1$ is the (necessarily odd) number $\frac{n}{2^{k_n - 1}}$."  It then states that $n \mapsto \left(k_n , l_n \right)$ is a surjection from $\mathbb{N}$ to $\mathbb{N} \times \mathbb{N}$, and so ends the proof.  I can intuitively see why this should be a bijection, I think, but I'm not sure how to make these feelings rigorous?   I suppose I'd say that the map is surjective since given any $\left(k_n , l_n \right) \in \mathbb{N} \times \mathbb{N}$ we can simply take $n$ indeed to be equal to $2^{k_n - 1} \left(2l_n - 1 \right)$ and note that $k_n - 1 \geq 0$ and thus $2^{k_n - 1}$  is both greater or equal to one so is a natural number (making the obvious inductive argument, noting that multiplication on $\mathbb{N}$ is closed), and similarly that $2 l_n - 1 \geq 2\cdot 1 - 1 = 1$ and is also a natural number, and thus the product of these two, $n$ must also be a natural number. Is it just as simple as this?  I suppose my gut feeling in the proving that the map is injective would be to assume that $2^{k_n - 1} \left(2 l_n - 1 \right) = 2^{k_m - 1} \left(2 l_m - 1 \right)$ and then use the Fundamental Theorem of Arithmetic to conclude that $n = m$. Is this going along the right lines? The 'implicit' definition of the mapping has me a little confused about the approach.    On a related, but separate note, I am indeed aware that if $K$ and $L$ are any countable sets, then so is $K \times L$, so trivially, taking the identity mapping we see trivially that this map is bijective and therefore that $\mathbb{N}$ is certainly countable (!), and thus so is $\mathbb{N} \times \mathbb{N}$. Hence, it's not really the statement that I'm interested in, but rather the exciting excursion into number theory that the above alternative proof provides. Let $f(z)$ be a holomorphic function on a disk of radius $&gt;1$ about the origin, and suppose that $|f(z)|\leq 1$ for $z$ with $|z|=1$ and $\text{Re}(z)&gt;0$ and $|f(z)|\leq 2$ for $z$ with $|z|=1$ and $\text{Re}(z)\leq 0$.  How does one show the bound $|f(0)|\leq \sqrt{2}$? A ring $R$ is a Boolean ring provided that $a^2=a$ for every $a \in R$. How can we show that every Boolean ring is commutative? Implied string comparison, 0=='', but 1=='1' I'tm trying to show in SQLcmd window the results of these queries:  SELECT SYSDATETIME() AS "date";  PRINT 'EXTRAÍDO POR:' SELECT SUSER_NAME() AS "Owner"  PRINT 'PASSWORD SETTINGS'; exec xp_loginconfig;  PRINT 'PASSWORD SETTINGS'; select * from sys.sql_logins;   And using this command in SQLcmd:  sqlcmd -S myServer -i inputpath -s, -o outputPath(like C:\log.xls) -e   So the sqlcmd output the file and does not show nothing in sqlcmd screen. How can i print these "Print" commands in the sqlcmd screen? (command promt).  When I run it on SSMS works well, like:  (1 row affected) OWNER:  (1 row affected) PASSWORD SETTINGS  (8 rows affected) PASSWORD SETTINGS  I would like to show in the prompt command this, like in the SSMS.  Thanks!