make different content between domain and IP Server
Setting up virtual host
Sometimes, I find that it is useful to do a Clay Render for my model, where all of the materials render with a single common "clay"-like material (see example image below). In the past, I have done this by:   Selecting all of my objects by using the A hotkey Manually deselecting the World object, camera, and lights in the Outliner Linking materials using Ctrl+L Manually setting to the same material in the Properties>Materials menu Render like normal   When I was done with my Clay render, I would revert to the old version by pressing Ctrl+Z a few time. However, this is a rather clunky solution, since I can't make any edits in the clay version without losing my old materials/textures data.  Is there a better way to do Clay renders, either with an addon or through some other technique such that I can make edits to my model and not lose my materials? If at all possible, I would like a single setting that allows me to quickly toggle clay-mode on and off in the same file.  Example "Clay Render": Animagi (Sirus specifically) and wands? How to open port via SSH tunnel? There is a system B with the following open servers: I have a web server listening on port 80 and a ssh server listening on 22. However, only port 22 is publicly available. Now, I would like to create some kind of tunnel, so that I can access B:80 from A. However, my client computer A that would like to connect to the system B is not publicly open, either.  So, all I have is a client computer A from where I would like to access the server B and, there, an open port 22. On A, no port is open or can be opened.  What (I think) I would need is to open locally (on A) some port that connects in some way through port 22 of B to port 80 on B.  Is this possibly without using any man-in-the-middle open servers with multiple ports? Related to : The question specifies morphs, but I wondered if this was true for Manifest as well? As Manifest is not an ability on the card, does  prevent it from being turned face up for its mana cost? Windows says RAM ran out while there is still 4 GB of physical memory available If I live in the US, for example, where current interest rates are practically nil, how can I take advantage of the fact that ? How did Harry remove the Philosopher's Stone from the Mirror of Erised? ‘How did I get the Stone out of the Mirror?’   ‘Ah, now, I’m glad you asked me that. It was one of my more brilliant ideas, and between you and me, that’s saying something. You see, only one who wanted to find the Stone – find it, but not use it – would be able to get it, otherwise they’d just see themselves making gold or drinking Elixir of Life.’   Philosopher's Stone - page 217 - Bloomsbury - chapter seventeen, The Man With Two Faces - Harry talking with Dumbledore   Dumbledore doesn't directly answer Harry's question, but rather describes under what circumstances a person could collect the Philosopher's Stone from the mirror. Did Harry have to do anything to retrieve the stone, such as reach into the mirror and take it? If not, what kind of magic could have removed the stone from the mirror and put it into Harry's possession?  A canon-based answer is preferred, such as the books, interviews with J.K. Rowling, or info from Pottermore, but subjective answers within the spirit of canon are welcome. Filtering questions by multiple tags I know you can select a tag and then click on unanswered to view the unanswered questions for that tag.  But is there a way to view questions only for a certain set of tags > 1 tag?  Mainly I'd like the union of tag1 with tag2 with tag3 with tag4. But also the intersection of tags would be nice as well. How can I set a dark theme in TeXstudio? After perusing , I'm considering switching from TeXmaker (in which I use the preset "dark theme") to its fork, TeXstudio.  What I mean by "dark theme" is a dark background with suitable text colours, which, in my experience, is easier on the eyes than a bright background is. Here is an example:    TeXstudio doesn't seem to offer any preset dark theme, though, which means I have to customise all the colours/syntax-highlighting settings from scratch to get a dark theme even remotely usable. This can prove very time-consuming.   I'm aware this might sound a bit cheeky but,   if you are a long-time TeXstudio user, and you have fine-tuned the editor colours to get a visually pleasant dark theme,   please consider posting the relevant lines from your settings (.txsprofile or .tmxprofile) file as an answer below.  TeXstudio users who would like to use a dark theme could then build on your settings to fine-tune their own. How to Update Multiple Array Elements in mongodb The meaning of various equality symbols Combining php arrays I have the following arrays in PHP (okay they are a bit bigger but the idea is what counts).  $array1 = array(1 =&gt; 'a', 2 =&gt; 'b'); $array2 = array(3 =&gt; 'c', 4 =&gt; 'd');   Essentially I want to combine the two arrays as if it were something like this  $array3 = array(1 =&gt; 'a', 2 =&gt; 'b', 3 =&gt; 'c', 4 =&gt; 'd');   Thanks Number of normals from a point to an ellipse From a typical point $P$ inside an ellipse, how many points $Q_i$ on the ellipse have $PQ_i$ normal to the ellipse?  Someone asked me at school many years ago but I don't think I worked it out. Necessary and Sufficient conditions for $(i \, j)$ and $(1 \, 2 \, \dotsc \, n)$ generate $S_n$. I have a homework question that asks     Find necessary and sufficient conditions on $1 \leq i &lt; j \leq n$ so that $(i \, j)$ and $(1 \, 2 \, \dotsc \, n)$ generate $S_n$.   Here is what I have done so far.  Call $c = (1 \, 2 \, \dotsc \, n)$.  I made the observation that $$ \overbrace{c c \dotsb c}^{n - j + 1} (i \, j) \overbrace{c^{-1} c^{-1} \dotsb c^{-1}}^{n - j + 1} = (1 + i - j + n \, 1). $$ Thus, $\langle (i \, j), c \rangle = \langle (1 \, i - j + n + 1), c \rangle$, so it suffices to look at $\langle (1 \, k), c \rangle$ for positive integers $2 \leq k \leq n$.  I've checked that, for the value $k=2$, $\langle (1 \, 2), c \rangle = S_n$.  This immediately implies $\langle (1 \, n), c \rangle = S_n$.  I suspect that no other values of $k$ will work, but am not sure how to prove it. $\mathbb Q$ is not the direct product of two non-trivial groups. I know this has been asked before, in a bunch of places, so I expect it to be closed as a duplicate at some point, but I'd like to know if my own proof works. If the proof is incorrect, please limit answers to pointing out the errors. If the proof is correct, then thoughts on improving it would be appreciated.  Prove that $\Bbb Q$ is not the direct product of two non-trivial groups.  —Aluffi exercise II.3.5. My proof Suppose that $\Bbb Q = G \times H$, neither $G$ nor $H$ trivial. First note that $0_G$ is the only element of $G$ with finite order: If $m\ne 0$ and $m\cdot g=0_G$, then $m\cdot(g,0_H)=0_{G\times H}$, contradicting the fact that $\Bbb Q$ has no non-zero elements of finite order. Suppose that for some $m\ne 0$ and some $n\ne 0$, $\pi_G(m/n)=0_G$. Then $0_G=n\cdot\pi_G\left(\frac m n\right)=\pi_G\left(n\cdot\frac m n\right)=\pi_G(m)=m\cdot\pi_G(1)$. Since $\pi_G(1)$ has finite order, we conclude that $\pi_G(1)=0$. Since $\Bbb Z=\langle 1\rangle$, $\pi_G$ maps every integer to $0_G$. Now let $a$ and $b$ be non-zero integers. Then $b\cdot\pi_G\left(\frac a b\right)=\pi_G\left(b\cdot \frac a b\right)=\pi_G(a)=0_G$. As above, this shows that $\pi_G\left(\frac a b\right)=0_G$. Thus if $\pi_G$ is not an injective function, then it must map everything to $0_G$, so $G$ will be a trivial group. The same result holds for $\pi_H$. Since by assumption, $G$ and $H$ are non-trivial, $\pi_G$ and $\pi_H$ are both injective functions, which is of course impossible. Why did Dumbledore drink the potion in the attempt to get a Horcrux? When Albus Dumbledore  finally reaches the island with Harry Potter and finds out that you can't get near the potion, he made a cup and drank it. Why couldn't he just have poured the potion on the "floor"? Why couldn't he have made a big cup to hold almost everything inside the basin? I have two laptops: Dell xps M1210 and Dell vostro 1400. On MS windows the output volume is so loud that I just need to set volume to around 20% to listen music and watch videos. I use Bose TriPort In-Ear Headphones most of the times. On ubuntu, even if I set 100% volume on the multimedia player, on the desktop volume manager as well, and also on the ncurses GUI of alsa mixer, I get very low volume output. Sometimes while playing some vidoes the volume is so low that I barely hear any words. I have struggled for two years since I started to use ubuntu to solve this mystery. The same sound card which gives excellent sound on MS windows fails miserably on GNU/Linux. Few weeks back I bought an Asus 100PX EEE PC. To my surprise even it also has the same issue. I have tried many versions of ubuntu and also I have done manual compiles of alsa modules without any success. I have asked and read about it in every possible forums and howto's. Nothing has helped me to address the issue. I'm trying to compute this integral, $$\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx \hbox{ where } 0&lt;a&lt;1$$ I drew a typical Pacman contour with branch cut at positive real axis. Then, we have $$\int_{\Gamma}\frac{\ln z}{z^{a}(z+1)}dz=\left(\int_{L_{1}}+\int_{L_{2}}+\int_{C_{R}}+\int_{C_{\epsilon}}\right)\frac{\ln z}{z^{a}(z+1)}dz$$ The function has a pole at $z=-1$. Then, by residue theorem, the whole integral becomes  \begin{align*} \int_{\Gamma}\frac{\ln z}{z^{a}(z+1)}dz&amp;=2\pi i\text{res}(\frac{\ln z}{z^{a}(z+1)};-1)\\&amp;=2\pi i\lim_{z\rightarrow-1}(z+1)\frac{\ln z}{z^{a}(z+1)}\\&amp;=2\pi i\lim_{z\rightarrow-1}\frac{\ln z}{z^{a}}\\&amp;=2\pi i\frac{\ln e^{i\pi}}{e^{i\pi a}}\\&amp;=2\pi i\frac{\ln1+i\pi}{e^{i\pi a}}\\&amp;=-\frac{2\pi^{2}}{(e^{i\pi})^{a}} \end{align*} On $L_1$, $z=xe^{i\epsilon}$ where $\epsilon\leq x\leq R$  On $L_{2}$, $z=xe^{i(2\pi-\epsilon)}$ where $\epsilon\leq x\leq R$  On $C_{R}$, $z=Re^{i\theta}$ where $\epsilon\leq\theta\leq2\pi-\epsilon$  Lastly on On $C_{\epsilon}$, $z=\epsilon e^{i\theta}$ where $\epsilon\leq\theta\leq2\pi-\epsilon$  So I tried to compute them separately and I anticipate the curves will both go to $0$ by ML inequality, and two linear integrals will consist of the term $\int_{0}^{\infty}\frac{\ln x}{x^{a}(x+1)}dx$ and hence simply solve for it. But I'm having trouble physically solving it out. I think the log is throwing me off. Can anybody please help me? How to block specific websites without using Parental Controls? Is there any efficient algorithm for finding subgroups of a given finite group? I am implementing an algorithm which finds every subgroup of given group.  Here's my algorithm.  Let $G$ be a group of order $n$ with elements $g_1,\cdots,g_n$.  Then I consider each $\langle g_i\rangle $ which are all cyclic subgroups of $G$.  Of course, if $\langle g_i\rangle =\langle g_j\rangle $, we append only $\langle  g_i\rangle $ into our result set.  We call all distinct subgroups of the form $\langle g_i\rangle $ to be the first-generation. (Note that the number of the first-generation subgroups is at most $n$.)  To get $(i+1)$-th generation from $i$-th generation inductively, we construct the subgroup $S$ from each two elements of $i$-th generation $S_1, S_2$ to be  $$S=\langle S_1,S_2\rangle .$$  Obviously, if generated subgroup is already in previous generations, we do not call it $(i+1)$-th generation, hence every subgroups of $G$ has the unique generation number.  We do this construction until the number of the typical generation subgroups to be $0$, since we get all subgroups of $G$.  This works perfectly to any groups, but unfortunately, it is not fast although I implemented additional steps to avoid repeated computation.  So I searched some other algorithms but they are all vague to understand.  One of the algorithm is called the cyclic extension algorithm, but I can not get it easily unless I have studied some backgrounds for this method.  If you have enough knowledge about the cyclic extension algorithm or some other algorithms, please give me an easy explanation. I am in undergrad. course of math dept. Thanks.