This graphic was adapted from Hansen, McDonald, and Newey (2010), I am just trying to reproduce something similar in LaTeX.
I need to draw a tree, and constantly using dotty and including the PNGs is starting to be a pain. Is there a way to get LaTeX to draw it's own trees that doesn't involve learning an entirely new language like TikZ?
Remap a key combination to another combination, e. g. Super+Ctrl+Shift+J -> Ctrl+Shift+Left Libvirt: how to pass qemu command line args? I have the current libvirt 1.2.20 i want to pass to my machine create with virt-manager some qemu flags. I edited the file and add    &lt;/devices&gt;   &lt;qemu:commandline&gt;           &lt;qemu:arg value='-L pc-bios'/&gt;   &lt;/qemu:commandline&gt; &lt;/domain&gt;   But doesn't work i also tried to edit from virsh  virsh edit domain   and put   &lt;qemu:commandline&gt;           &lt;qemu:arg value='-L pc-bios'/&gt;   &lt;/qemu:commandline&gt;   But give me this error  errore: XML document failed to validate against schema: Unable to validate doc against /usr/share/libvirt/schemas/domain.rng Element domain has extra content: qemu:commandline  Failed. Try again? [y,n,i,f,?]:    How to pass correctly qemu command line? Thanks I want to open a file for reading, the C++ way. I need to be able to do it for:   text files, which would involve some sort of read line function. binary files, which would provide a way to read raw data into a char* buffer. Why can't root on one machine change nfs mounted content from another machine? On my NFS server, I have the following export defined:  #NFS exports Database /shared -alldirs -network=192.168.1 -mask=255.255.255.0   On my NFS client:  192.168.1.7:/shared /shared nfs rw 0 0   Obviously, as root on the server, I can do whatever I want.  On the client however, my regular user 'gabe' can make changes to the nfs mount (assuming I have permissions to), but root cannot.    As my regular user:   gabe@client$ cd /shared gabe@client$ ls -l total 8 drwxrwxrwx  4 gabe  wheel   512 Mar 20 19:20 tmp gabe@client$ cd tmp gabe@client$ touch test.txt gabe@client$ rm test.txt   As root:  # cd /shared/tmp # touch test.txt touch: test.txt: Permission denied   Again, this is all on the NFS client side of things, and I suspect perhaps it has something to do with the -maproot option.  This is the first time I'm setting up NFS and I just noticed this peculiarity.  I'm going to do some reading now, to see if I can figure this out, but if anyone has any insight, I would appreciate it. I downloaded the July 30th update for Pokemon Go and when I open the app I'm prompted to login, But once I login, it redirects me to this page on Cookies::     If I click "Done" in the top left, it prompts me to login again! I've tried reinstalling the game, restarting my phone, etc. I can't much enable Cookies on the inline Safari browser. It doesn't make sense, and cookies are on in Safari (because I don't use Safari and they're enabled by default).  How can I fix this? I have this input:  sdkxyosl 1 safkls 2 asdf--asdfasxy_asd 5 dkd8k  jasd 29 sdi44sw 43 asasd afsdfs 10 rklyasd 4   I need this output:  sdi44sw 43 dkd8k  jasd 29 asasd afsdfs 10 asdf--asdfasxy_asd 5 rklyasd 4 safkls 2 sdkxyosl 1   So i need to sort the lines by the last column.  I don't know how many columns are in one line.  I just can't figure it out, how to do it. I don't have "perl powers". I just have ~average scripting powers with sed, awk, cut, etc..   Does somebody know how to do it? Is there any truth to the idea that you shouldn't multiply seasonings when multiplying a recipe? I often hear cooks discuss that idea that if you are multiplying, say, a recipe for 4 to be a for 32, that you shouldn't multiply some herbs, spices, or salt. I've never understood any reason why this should be true. Maybe something to do with surface area to volume ratios, or cooking times? Does anyone have a real explanation, or is this nonsense? Find the least value of $ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$ Find the least value of $$ \sec^6 x +\csc^6 x + \sec^6 x\csc^6 x$$ I tried AM greater than equal to GM But that's for finding maximum value.  This can probably be solved with calculus but I don't know for some reason I can't find the answer . Which is $80$. I have a problem with the interpretation of a test result in which the p-value is 8.6e-28. How should it be interpreted? What is the e for?  Specifically, I used White's Test for the homoscedasticity assumption of a linear regression:  White's general test statistic :  150.0535  Chi-sq( 9)  P-value =  8.6e-28 I have a problem. I am going to make a minecraft animation in blender. But when I put a minecraft texture on a 3d cube that is 16x16x16 it is very low resolution is there a way to get the texture in a better resolution or is there a way to make blender make the texture better?  I hope you have an answer. Thanks in advance I want to add a feature class within a Geodatabase to a map in ArcGIS Pro (1.4.1) using arcpy. I am converting some python code from ArcGIS Desktop to ArcGIS Pro.   This issue is very similar to:    and      However, instead of adding a feature class (or layer) to a map using the Python Window I am doing this from a  Script within a Toolbox. I have included a code snippet below.  lyrTest = r"C:\data\test.gdb\Layer1"  aprx = arcpy.mp.ArcGISProject("CURRENT") aprxMap = aprx.listMaps("MainMap")[0]  lyrFile = arcpy.mp.LayerFile(lyrTest) aprxMap.addLayer(lyrFile)   I get the following error message:  Traceback (most recent call last): File "C:\data\AddLayers.py", line 5, in &lt;module&gt;     aprxMap.addLayer(lyrTest) File "c:\users\user\appdata\local\programs\arcgis\pro\Resources\arcpy\arcpy\utils.py", line 191, in fn_ return fn(*args, **kw) File "c:\users\user\appdata\local\programs\arcgis\pro\Resources\arcpy\arcpy\_mp.py", line 1048, in addLayer return convertArcObjectToPythonObject(self._arc_object.addLayer(*gp_fixargs((add_layer_or_layerfile, add_position), True))) ValueError: C:\data\test.gdb\Layer1 Failed to execute (AddLayers). Drawing simple edge, no arrow I have  \begin{tikzpicture}[-&gt;,auto,node distance=3cm,   thick,main node/.style={draw,font=\bfseries}]   \node[main node] (111) at (0,1) {$X_{1,1}^{(1)}$};   \node[main node] (121) at (0,2) {$X_{1,2}^{(1)}$};   \node[main node] (1n1) at (0,4) {$X_{1,n}^{(1)}$};   \node[main node] (211) at (0,6) {$X_{2,1}^{(1)}$};   \node[main node] (221) at (0,7) {$X_{2,2}^{(1)}$};   \node[main node] (2n1) at (0,9) {$X_{2,n}^{(1)}$};   \node[main node] (n11) at (0,11) {$X_{n,1}^{(1)}$};   \node[main node] (n21) at (0,12) {$X_{n,2}^{(1)}$};   \node[main node] (nn1) at (0,14) {$X_{n,n}^{(1)}$};   \node[main node] (112) at (2.5,1) {$X_{1,1}^{(2)}$};   \node[main node] (122) at (2.5,2) {$X_{1,2}^{(2)}$};   \node[main node] (1n2) at (2.5,4) {$X_{1,n}^{(2)}$};   \node[main node] (212) at (2.5,6) {$X_{2,1}^{(2)}$};   \node[main node] (222) at (2.5,7) {$X_{2,2}^{(2)}$};   \node[main node] (2n2) at (2.5,9) {$X_{2,n}^{(2)}$};   \node[main node] (n12) at (2.5,11) {$X_{n,1}^{(2)}$};   \node[main node] (n22) at (2.5,12) {$X_{n,2}^{(2)}$};   \node[main node] (nn2) at (2.5,14) {$X_{n,n}^{(2)}$};   \node[main node] (11k) at (6,1) {$X_{1,1}^{(k)}$};   \node[main node] (12k) at (6,2) {$X_{1,2}^{(k)}$};   \node[main node] (1nk) at (6,4) {$X_{1,n}^{(k)}$};   \node[main node] (21k) at (6,6) {$X_{2,1}^{(k)}$};   \node[main node] (22k) at (6,7) {$X_{2,2}^{(k)}$};   \node[main node] (2nk) at (6,9) {$X_{2,n}^{(k)}$};   \node[main node] (n1k) at (6,11) {$X_{n,1}^{(k)}$};   \node[main node] (n2k) at (6,12) {$X_{n,2}^{(k)}$};   \node[main node] (nnk) at (6,14) {$X_{n,n}^{(k)}$};   \node[main node] (11n2) at (9.5,1) {$X_{1,1}^{(n-2)}$};   \node[main node] (12n2) at (9.5,2) {$X_{1,2}^{(n-2)}$};   \node[main node] (1nn2) at (9.5,4) {$X_{1,n}^{(n-2)}$};   \node[main node] (21n2) at (9.5,6) {$X_{2,1}^{(n-2)}$};   \node[main node] (22n2) at (9.5,7) {$X_{2,2}^{(n-2)}$};   \node[main node] (2nn2) at (9.5,9) {$X_{2,n}^{(n-2)}$};   \node[main node] (n1n2) at (9.5,11) {$X_{n,1}^{(n-2)}$};   \node[main node] (n2n2) at (9.5,12) {$X_{n,2}^{(n-2)}$};   \node[main node] (nnn2) at (9.5,14) {$X_{n,n}^{(n-2)}$};   \node[main node] (11n1) at (12,1) {$X_{1,1}^{(n-1)}$};   \node[main node] (12n1) at (12,2) {$X_{1,2}^{(n-1)}$};   \node[main node] (1nn1) at (12,4) {$X_{1,n}^{(n-1)}$};   \node[main node] (21n1) at (12,6) {$X_{2,1}^{(n-1)}$};   \node[main node] (22n1) at (12,7) {$X_{2,2}^{(n-1)}$};   \node[main node] (2nn1) at (12,9) {$X_{2,n}^{(n-1)}$};   \node[main node] (n1n1) at (12,11) {$X_{n,1}^{(n-1)}$};   \node[main node] (n2n1) at (12,12) {$X_{n,2}^{(n-1)}$};   \node[main node] (nnn1) at (12,14) {$X_{n,n}^{(n-1)}$};   \path     (111) edge (112)         edge (122)         edge (1n2)     (121) edge (112)         edge (122)         edge (1n2)     (1n1) edge (112)         edge (122)         edge (1n2)     (121) edge[--,dashed] (1n1); \end{tikzpicture}   I just want simple lines, no arrows. How to manage with edge to achieve that? Thanks Yesterday, I tried to dualboot Ubuntu 16.04 onto my hard drive, creating a new partition for Ubuntu alongside my existing Windows 8. First, I disabled the fast-boot option in Windows 8, and still GRUB did not load, and it went directly to Windows boot-up. Then I did some research and found this article:     Basically, I issued the BCDEdit command in Windows, and entered in the following code:  bcdedit /set {bootmgr} path \EFI\ubuntu\grubx64.efi   When I reboot, I get a basic GRUB2 terminal, and after I type in the exit command, then a second Shell prompt appears, and after typing the exit command again, the GRUB2 menu to select the OS to boot finally appears... with only Ubuntu and Ubuntu memory test as options, not Windows 8.  After booting into Ubuntu, I tried to run the Boot-repair Utility, but that shows this following error:   How do I approach this problem? The Pastebin log is here:   Thanks for the help! Are security.ubuntu.com updates eventually merged into normal updates? Upon my observation, seems new packages will first appear in security.ubuntu.com, but if you upgrade later, they will appear in XX.archive.ubuntu.com (depends on your config),  Was that true ? What is happening here, when I use squared loss in logistic regression setting? I am trying to use squared loss to do binary classification on a toy data set. I am using mtcars data set, use mile per gallon and weight to predict transmission type. The plot below shows the two types of transmission type data in different colors, and decision boundary generated by different loss function. The squared loss is $\sum_i (y_i-p_i)^2$ where $y_i$ is the ground truth label (0 or 1) and $p_i$ is the predicted probability $p_i=\text{Logit}^{-1}(\beta^Tx_i)$. In other words, I am replace logistic loss with squared loss in classification setting, other parts are the same. For a toy example with mtcars data, in many cases, I got a model &quot;similar&quot; to logistic regression (see following figure, with random seed 0).  But in somethings (if we do set.seed(1)), squared loss seems not working well.  What is happening here? The optimization does not converge? Logistic loss is easier to optimize comparing to squared loss? Any help would be appreciated.  Code d=mtcars[,c(&quot;am&quot;,&quot;mpg&quot;,&quot;wt&quot;)] plot(d$mpg,d$wt,col=factor(d$am)) lg_fit=glm(am~.,d, family = binomial()) abline(-lg_fit$coefficients[1]/lg_fit$coefficients[3],        -lg_fit$coefficients[2]/lg_fit$coefficients[3]) grid()  # sq loss lossSqOnBinary&lt;-function(x,y,w){   p=plogis(x %*% w)   return(sum((y-p)^2)) }  # this random seed is important for reproducing the problem set.seed(0)  x0=runif(3) x=as.matrix(cbind(1,d[,2:3])) y=d$am opt=optim(x0, lossSqOnBinary, method=&quot;BFGS&quot;, x=x,y=y)  abline(-opt$par[1]/opt$par[3],        -opt$par[2]/opt$par[3], lty=2) legend(25,5,c(&quot;logisitc loss&quot;,&quot;squared loss&quot;), lty=c(1,2)) What happened to this user's rep, badges, and name? This user is only displaying 56 rep.   But surely (s)he has much more than that.        The user has also never given out any bounties.  Also, badges are missing: "Authoring" versus "Writing" Pressure at base of 3 different dam designs The question asks, For which design will the water pressure at the base of the dam wall be the greatest?  The answer is all are equal.    This doesn't make sense to me.  Using the formula: P (pressure) = p (atmospheric pressure) + rho (density) * g (gravity) * y (depth below water surface)    What is confusing me is the y value, the depth below the surface. Yes both red points are the same depth below the water surface. But doesn't the red point in dam III only "see" a smaller column of water above it and so shouldn't it have less pressure on it than the red point in dam II?    Also, I know the formula would give the points at the base of dam I and dam II equal. But why doesn't the extra water in dam II (shaded in black) cause it to have a higher pressure at the base? So assume $Y$ and $X$ are exponentially distributed with parameters $y_1$, and $x_1$ respecitively.  assume c is a constant.  I am having huge trouble to understand the integration of the following expression.  $P(Y&lt;c/u(X))$  $=\int_{t}^{\infty}f_X(x)\int_{0}^{c/u(x)}f_Y(y)dydx +\int_{0}^{t}f_X(x)\int_{c/u(x)}^{\infty}f_Y(y)dydx$  where t is the cross-point that $u(x)$ change sign    here $c/u(x)$ is given by the plot below, t is the point crossing the zero:   Confusion:   I don't understand the second integration of the second term "$\int_{c/u(x)}^{\infty}f_Y(y)dydx$", this isn't right because Y an X only defined for y>0, and x>0. So it's the first quadrant in this plot. How is this command legal? "> file1 < file2 cat" Assuming file2 already exists, the command   &gt; file1 &lt; file2 cat   appears to be copying the content of file2 to file1.  But I cannot understand this structure.  I understand that "Nothing" is being directed to file1 (creating or erasing its content). Then the content of file2 is being directed to file1.  Why is cat after file2? How does it know to cat file2 if the operands are not in the correct order?