How to send Alt+Ctrl+Del key sequence in a VirtualBox VM guest I have a   VM guest, which I run in vboxHeadless mode. The host is Linux 64-bit. The VirtualBox version is 4.2.12r84980.  The VM loads fine, and I am presented with the "windows server" greeting screen. That is, I have to press Alt + Ctrl + Del to log on.  I have tried numerous ways to send Alt + Ctrl + Del:   Remote desktop to the VM from the port that is assigned by VirtualBox headless. That is, &lt;VirtualBox host IP address>: VMguestPort. A-C-D doesn't work as the key sequence gets intercepted by the client PC at which I'm sitting and trying to log on to the VM guest. Remote desktop directly to the VM &lt;Windows Server 2008 VM guest IP address>:3389 doesn't work for the same reason.  gives an a button to send A-C-D to a VM guest. This doesn't work either. I read that from a command line I could use the command  VBoxManage controlvm win2k8_64 keyboardputscancode 1d 38 53  VBoxManage controlvm win2k8_64 keyboardputscancode 9D B8 D3    for key press and then key release sequence. This doesn't work either.  Why do these work? Especially with a Windows Server 2008 as a VM guest?    The suggested previous answer does not work (pressing Alt + Ctrl + End does not work either) and also, as I am running a headless server, there isn't any change for me to see the VirtualBox GUI. The best I can get is the phpVirtualBox web interface and in VirtualBox there isn't any "right Ctrl" capture key.
How do I press Ctrl+Alt+Delete in VirtualBox? I'm running VirtualBox with Windows Vista as host OS and Windows Server 2008 as guest OS. To login to Windows Server I have to press Ctrl+Alt+Delete, but my host OS reacts on that so I can not send it to VirtualBox. How can I do it?
I can't play Pokemon GO! What's wrong? Lets say I have a loop in Bash:  for foo in `some-command` do    do-something $foo done   do-something is cpu bound and I have a nice shiny 4 core processor. I'd like to be able to run up to 4 do-something's at once.  The naive approach seems to be:  for foo in `some-command` do    do-something $foo &amp; done   This will run all do-somethings at once, but there are a couple downsides, mainly that do-something may also have some significant I/O which performing all at once might slow down a bit. The other problem is that this code block returns immediately, so no way to do other work when all the do-somethings are finished.  How would you write this loop so there are always X do-somethings running at once? 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! What happens if you plug more than one charger in the new MacBook Pro (2016)? Apple's  decision to convert all the ports on their 2016 MacBook Pro lineup to USB-C ports contributes to blur the difference between ports that supply power (such as a power adapters) and ports that can power external devices (such as USB keys, mice or other gadgets).    This made me think about the following scenario: you sit at a table where your laptop is plugged in on the right, and don't notice that it is plugged in, so you decide to charge it with a second nearby charger that is laying around (i.e. belonging to someone who lives with you and who recently sat at the same spot).  What would happen then?   Would the computer charge twice as fast, would one of the chargers be dismissed by the system (which one?), or would it generate a surcharge that would destroy the computer?  I couldn't find anything about this online, but since I am always trying to force a USB in the MagSafe charger of my MacBook Air, I think that I could easily end up making this kind of mistake.  Note: My question asks what happens when two chargers are inserted into a MacBook Pro, but I am also wondering what would happend if 3, or 4 chargers were inserted. Although this scenario is unlikely to happen by mistake, it would constitute an obvious vulnerability to an exposed device, i.e. in a public space. What can you say about $f$ and $g$ in the case that $fg$ is 1) Injective, 2) Surjective - Cohn - Classic Algebra Page 15 Question: Are my proofs below valid?   In both cases we are using: $f:A\to B, g: B\to C$  Notation of your type converted: $(g\circ f)(x)=g(f(x))=xfg$    If $fg$ is injective what can be said about $f$ and $g$?  If $fg$ is injective then $f$ is injective:  Proof:  $(x_1fg = x_2fg \implies x_1=x_2)$  $x_1fg=x_2fg$ is clearly true when $x_1f=x_2f$  Which requires $x_1=x_2$, hence $f$ is injective.  If $f$ were not injective, then $x_1fg\ne x_2fg$  $\blacksquare$    If $fg$ is surjective what can be said about $f$ and $g$?  If $fg$ is surjective then $g$ is surjective:  Proof:  $(\forall c\in C)(\exists a\in A) |( a(fg))=c$  This means for $g$ every element must have had been mapped to. How can this be proved Mathematically?  well $(af)g=c$ is surjective, hence $(af)=(b\in B)$ and $bg=c$  $(\forall c\in g)(\exists b\in B)|(bg=c)$  Hence $g$ is surjective.  $\blacksquare$ Dealing with suggested edits on question to be closed as duplicate What are the current state-of-the-art convolutional neural networks? I'm interested in understanding which neural network architecture is currently the state of the art (sometimes abbreviated "SOTA") with respect to standard image classification tasks such as MNIST, STLN-10 and CIFAR. This is challenging because new results are published frequently, and it can be hard to keep up. Is there a resource or website which tracks the best results for these tasks? How to create an annuity timeline on LaTeX I was wondering, how can I create a timeline exactly like this one:    on LaTeX?  I am a Finance student, and write all of my notes using LaTeX. These timelines, are giving me a headache. Since we use a lot of different timelines because of annuities. I want to learn how to create one basic timeline, so that I can then change anything I want. But basically, I want something like this, with the arrows and everything.  Thank you so much guys! What does "yield break;" do in C#? I have seen this syntax in MSDN: , but I don't know what it does.  Does anyone know? Long ago someone demonstrated the following card trick with a standard 52-card deck:  (1) A volunteer selects 5 cards from a shuffled deck, which the performer does not see.  (2) The assistant puts exactly 4 of those in a pile and passes it to the performer.  (3) The performer can then identify the 5th card.  There are a few ways to execute this trick, one of which goes as follows.  (1) The assistant finds a suit common to 2 of the 5 cards, with values $x$ and $y$ modulo $13$.  (2) He then removes one of them, say $x$, such that $x-y \in [1..6]$ modulo $13$.  (3) He then puts the other one at the top of the pile.  (4) And he puts the 3 other cards after that in a permutation corresponding to $x-y$ modulo $13$.  Since (1) works by pigeonhole, (2) works because $\frac{13-1}{2} = 6$, and (4) works because there are 6 possible permutations of 3 cards, the trick succeeds. My question is whether the same trick can work with more than 52 cards. It is not obvious to me how to do so, nor compute the maximum number of (pairwise distinct) cards possible. Also, it seems to me that even if the trick can be made to work for $n$ cards, it may not work for $n-1$ cards! Any ideas? What techniques have been used to repeat geometry or other detail to the infinite horizon? When I composed my     I would have liked to have the camera pitch up enough to show the horizon, except I didn't have enough geometry to extend the cobblestone street and the sidewalk all the way to infinity (mostly because that would require infinite stuff).  Even in that shot it is clear that I need to widen the "sidewalk" to cover up the "sky"box.    Contrast with     where the ray tracer can use the procedurally replicated geometry.   is a failed attempt at faking out the horizon, even with low detail.  It fails because lighting's interaction with the geometry gives it away.  Some techniques that prevent you from having to deal with infinity include   put the scene in a box (usually a room) put the scene in a canyon (which is just a box without a top so you can see the relatively featureless sky) route the road into a tunnel, or around a corner and hide the horizon with other geometry (which is really just a variant on the canyon). have no detail on the ground at all (flat color or gradient)   Those are all dodging the question of interest to me:  What techniques have people used in blender to fake various kinds of detail repeating into the distance (instead of just obstructing the view of infinity)? In a bash shell script, writing a for loop that iterates over string values In bash, I know that it is possible to write a for loop in which some loop control variable i iterates over specified integers.  For example, I can write a bash shell script that prints the integers between 1 and 10: #!/bin/bash  for i in {1..10} do  echo $i done  Is it possible to instead iterate over a loop control variable that is a string, if I provide a list of strings?  For example, suppose that I have a string fname that represents a file name.  I want to call a set of commands for each file name.  For example, I might want to print the contents of fname using a command like this: #!/bin/bash  for fname in {&quot;a.txt&quot;, &quot;b.txt&quot;, &quot;c.txt&quot;} do  echo $fname done  In other words, on the first iteration, fname should have the value fname=&quot;a.txt&quot;, while on the second iteration, fname should have the value fname=&quot;b.txt&quot;, and so on.  Unfortunately, it seems that the above syntax is not quite correct.  I would like to obtain the output:  a.txt b.txt c.txt  but when I try the above code, I obtain this output:  {a.txt, b.txt, c.txt}  Can you please help me determine the correct syntax, so that I can iteratively change the value/contents of the variable fname?  Thank you for your time. What is a "Swap Area"? So I noticed this question has not been asked directly;  What is this "Swap Area" people keep referring to when it comes to installing Ubuntu? Careers - filter by location & tag On Careers I thought it wold be nice to be able to filter by , possibly multiple, in additional to location.  Nothing fancy, but something like ....        The image of tags is stolen from the SO side to convey the idea. I suspect functionality on the Careers side would need to be a bit different.  Recent tags would likely make sense.  Favorite tags, particularly the more specific ones might not (e.g., a job posting for , ).  Anyway, the image is just to give a sense of the idea. Showing that $Q_n=D_n+D_{n-1}$ Let $T_n$ be the set of permutations of $\{1,2,\ldots,n\}$ which do not have $i$ immediately followed by $i+1$ for $1\le i\le n-1$; in other words, let \begin{align} T_n=\{\sigma \in S_n: \sigma(i)+1\ne\sigma(i+1) \text{ for all } 1\le i\le n-1\} . \end{align} Let $Q_n$ be the number of elements of $T_n$.    Let $D_n$ be the number of  of $\{1,2,\ldots,n\}$.  It is not hard to show algebraically that $Q_n=D_n+D_{n-1}$, but I am having difficulty coming up with a combinatorial argument to show why this is true.  (I believe there are $D_{n-1}$ elements of $T_n$ that leave $n$ fixed, and $D_n$ elements of $T_n$ that move $n$, but I don't know how to justify this combinatorially.) What original Dr Who episodes are left when not watching unfinished stories? Sounds of the letter a How can I know, precisely, when to differentiate the sounds of the letter a, like in: apple and vault? Proof of continuity $f(x+y) = f(x) + f(y)$ Lost my Downloads folder both from Dock and Finder I did something stupid apparently and lost my Downloads folder both from Dock and then when I opened Finder to drag it down to the Dock it somehow got deleted in the desktop.  Now although I can see the folder exists (e.g. by searching in Spotlight) I do not know how to restore it in the Dock and in Finder as well. Any help please? moderncv: How to control the length of cvitems and cvlines?