Sample input file: abcd &lt;Space&gt; 1234 &lt;Space&gt;  Desired output: abcd 1234
I want to remove all empty lines from a file. Even if the line contains spaces or tabs it should also be removed.
How does vampirism work? A PC in our group was off on his own following a recurring villain of our groups. He was paralyzed where he could move, and the vampire bit his neck. The vampire said "now you get to remain 19 (years old) forever" before he left the cave. This was all done in RP.  We found him later, and healed up his bite marks. Our characters know of vampires, specifically only the one that bit him (recurring villain), but don't really know how vampirism works, because it isn't a super common thing in this world yet.   My question as the player is:  How does RAW vampirism work in 5e?  Is it a disease that develops over time? Is it something where they have to die and be brought back as a vampire? machine wakes only on local magic packet I've got a Debian server, which I would like to wake up remotely using a tool like wakeonlan.  I have assigned the server a static internal IP 192.168.0.14 and set up the appropriate port forwarding in the router (which otherwise has a very limited admin interface).  When I am logged in to the same network, I can wake the server using the subnet IP, but I cannot wake the server using the public IP of the router.  What I do not understand: When the server is running and I listen on the server with netcat nc -ulv -p 9, I can see the packet coming in in both cases (both when sending to the subnet IP and when sending to the public IP of the router).  If the packet arrives in both cases, why does the machine wake up only in one of them? Avoid large spaces between text and equations I use the options  \renewcommand{\baselinestretch}{1.25} \setlength{\parskip}{0pt} \setlength{\parindent}{0pt}    in the preamble of my document and most of the time the document looks like intended. Once in a while though, especially when I use multiline-equations, I get huge spaces right before and after the equation and the surrounding text, for example:    Is there any way I can avoid that without removing the \parskip option? What does an icon on the tag mean, for example the "rg" before the sql-server tag as pictured below? I have a table that looks like this:  +----------------------------------------+ |Name                | kode      | jum   | +----------------------------------------+ | aman               |kode1      | 2     | | aman               |kode2      | 1     | | jhon               |kode1      | 4     | | amir               |kode2      | 4     | +--------------------+-----------+-------+   How can I make a view like this with MySQL?                      kode1    kode2     count aman                  2         1        3   jhon                  0         4        4 amir                  0         4        4 What is wrong in this proof: That $\mathbb{R}$ has measure zero Consider $\mathbb{Q}$ which is countable, we may enumerate $\mathbb{Q}=\{q_1, q_2, \dots\}$. For each rational number $q_k$, cover it by an open interval $I_k$ centered at $q_k$ with radius $\epsilon/2^k$.  The total length of the intervals is a geometric progression that sums up to $\epsilon$.  Each real number is arbitrarily close to a rational number since $\mathbb{Q}$ is dense in $\mathbb{R}$. Thus, each real number is in one of the open intervals.  Thus the entire real line is covered by the union of the $I_k$, thus $\mathbb{R}$ is a null set with measure zero.  Clearly there is something wrong in the above proof, however I am not sure where is it?  Thanks for any help. Why does the multi-paragraph quotation rule exist? I have 12.10 ubuntu and when I try to remove / install programs I get notification :      Software can't be installed or removed because the authentication service is not available.    (org.freedesktop.PolicyKit.Error.Failed: ('system-bus-name', {'name':  ':1.104'}): org.debian.apt.install-or-remove-packages   When I log in as root user (instead of user account that I regulary use) i have no problems. "Open With" dialog ignores my selection I am running Windows 7 x64. I have a file type that I want to associate with Microsoft Access 2003.  I right-click on the file and select Open With->Choose Default program. In the "Open With" dialog, I select the browse button. I choose Program Files (x86)\Microsoft Office\Office11\msaccess.exe and press "Open".  I end up back at the "Open With" dialog, but nothing has changed. Access is not listed as a program to use to open the file with. The original program (Adobe Acrobat) that is associated with this program is still selected.  Any tips on getting this association to work? How to use wc and piping to find how many files and directories are in a certain directory? How can I use word counter (wc) and piping to count how many files or directories are in the /usr/bin directory? Looking for title of a fantasy fiction with huge war, invader on airship with warlock, island fortress with sorcerer, mnemonic storyteller and dragon This was a thick book that I read over fifteen years ago that I never finished reading because someone borrowed it and didn't return it to the school library. I want to find its title.  The story takes place in a kingdom that is preparing for the a cycle of war, where every time the winds changes (50 or 100 years), invaders will try to cross the sea from another continent using airships with air balloons, filled with soldiers and warlocks. The kingdom has made several man-made island fortress along their coastline, filled with soldiers with war machines, sorcerers, and enchanters who act as the main guard/defence for the kingdom, calling lightning and fireballs, and using machines to stop and destroy and bring down as many of the invader airships as possible while countering the warlocks on board.  The book involved a man and a girl whose journey brought them apart in preparing for the imminent war.   The women found herself adept in magic and went to the academy in the capital to learn to be a sorceress.   The man found a different calling. With a good memory, he was sent to learn to become a mnemonic storyteller (to train with battle staff and fighting skill and mnēmonikos skill) who carried the history, literature of the entire culture in their heads. They used their knowledge to convey the kingdom's literature, poetry, history and stories to the people from generation to generation.  Both eventually "graduate". Sadly, and with great emotion, both have to part ways. She was sent to a fortress island and began her role as a sorcerer who will stand guard there. He, as a storyteller, began his own journey without her, going from town to town, village and soldier barrack, across the kingdom, telling stories of great men and their deeds, and courageous heroes who fought the war, to uplift the spirit of the people and bring valor to the soldiers stationed at the border and people in the coming war.   In brief, he meets a dragon (who used to be involved in the war in ages past), long thought to be extinct. He finds himself able to communicate with it and has a "born" friendship with it. The dragon teaches him to become a "dragon rider", a long-lost ability of his people (due to the "extinction" of dragons.)   The wind changes with the dread of war hanging on the kingdom.  Airships start to blacken the sky again with soldiers and war machines battling each other. Is it possible to find power armor in the wasteland? If you send somebody out in the wasteland for long enough with good enough luck, is it possible to find power armor or any other legendary gear? I'm working on a 2nd order passive low pass filter, consisting of two passive low pass filters chained together.         &ndash; Schematic created using   Let \$ H(s) = H_1(s)H_2(s) \$ where \$ H_1(s) \$ and \$ H_2(s) \$ are the transfer functions for each separate filter stage.  Then \$ |H(s)| = |H_1(s)||H_2(s)| \$  Knowing the magnitude of a passive low pass filter,   $$|H(s)| = \dfrac1{\sqrt{ (\omega R_1C_1)^2 + 1} } \times \dfrac1{\sqrt{ (\omega R_2C_2)^2 + 1} } = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}} $$  Then trying to find the cutoff frequency:  $$ \left(\dfrac1{\sqrt{2}}\right)^2  = \dfrac1{\sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)}} $$ $$ 2  = \sqrt{((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1)} $$ $$ 4  = ((\omega R_1C_1)^2 + 1)((\omega R_2C_2)^2 + 1) $$ $$ 4  = (\omega R_1C_1)^2(\omega R_2C_2)^2 + (\omega R_1C_1)^2 + (\omega R_2C_2)^2 + 1 $$  And I'm stuck. Research on the web tells me \$ \omega_c = \dfrac1{\sqrt{R_1C_1R_2C_2}} \$, but I can't find why? Can anybody show my the derivation to find this? Is the proof of these identities possible, only using elementary differential and integral calculus? If it is, can anyone direct me to the proofs? ( or give a hint for the solution )  1)$$\int_0^\infty { e^{-x^2} \ln x }\,dx = -\tfrac14(\gamma+2 \ln 2) \sqrt{\pi} $$  2)$$\int_0^\infty { e^{-x} \ln^2 x }\,dx = \gamma^2 + \frac{\pi^2}{6} $$  3) $$\gamma = \int_0^1 \frac{1}{1+x} \sum_{n=1}^\infty x^{2^n-1} \, dx$$  and lastly,  4) $$\zeta(s) = \frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)} \prod_\rho \left(1 - \frac{s}{\rho} \right) e^{s/\rho}\!$$  I personally think the last is obtained from a simple use of the Weierstrass factorization theorem. I'm unsure as to what substitution is used.  $\gamma$ is the Euler-Mascheroni constant and $\zeta(s)$ is the Riemannian Zeta function.  Thanks in advance. How to use \def to redefine \end{split}? When I use \def to redefine \end{split}, there is an error.  \documentclass{article} \usepackage{amsmath}  \begin{document} \def\be{\begin{equation}} \def\ee{\end{equation}} \def\bs{\begin{split}} \def\es{\end{split}}  \be     \bs         a\\         b     \es \ee  \end{document}   The error is:      Runaway argument?      a\ b \es \ee       ! Paragraph ended before \split was complete.   However, if I don't use \def\es{\end{split}} and just modify the code to be  \documentclass{article} \usepackage{amsmath}  \begin{document} \def\be{\begin{equation}} \def\ee{\end{equation}} \def\bs{\begin{split}} \def\es{\end{split}}  \be     \bs         a\\         b     \end{split} \ee  \end{document}   it's OK. What's the problem? Thank you! I have a model similar to below with a continuous outcome, continuous predictor and a categorical variable. the goal of this analysis is to make inferences on the group variable. The issue is that y is bounded by x in the form of y >= -x. When the model makes predictions you can see in the plot below that the predicted value is outside the possible range. See the last plot.   How can this be remedied? A custom loss function to penalize going beyond the bound and use gradient descent regression? Or is there some other way? Thank you.   x =   rep(seq(-10,10,1),each=5)  y =  rep(0,length(x) ) weights = sample( seq(1,1000,1) ,length(x)) groups = rep( letters[1:5], times =length(x)/5  )   ## the outcome y has the relationship y &gt;= -x for(i in 1:length(x)){   y[i] = sample(  seq(-x[i],-x[i]+ ifelse(x[i]&lt;0 ,                                            sample(seq(1,10,.1),1),                                           sample(seq(5,40,.1),1) ),1)                        ,1) }   dat =data.frame(x = x, y = y, weights = weights, group = groups) model = lm(dat = dat, y ~ x + group + group * x, weights = weights) dat$FIT= predict(model, dat, allow.new.levels = TRUE)  library(ggplot2) ggplot(data = dat, aes(x = x, y = y, color = group))+geom_point( aes(size = weights))+ geom_line(data = dat, aes(x = x, y = FIT, color = group))    #data for new predictions newdata = data.frame( x = seq(-100,100,1) ,                        group = rep("c", length( seq(-100,100,1) )),                       bound = -seq(-100,100,1)                       )  ## when pridicting new data the outcome y has the relationship y &gt;= -x ## but the predicted vales are below the bound. See how red line is below blue to the left of zero new_preds  = predict(model, newdata, allow.new.levels = TRUE) ggplot(data = newdata, aes(x = x, y = new_preds, color = group))+geom_line()  +   geom_line(data = newdata, aes(x = x, y = bound, color = "green")) solution for apex cpu time limit exceeded exception I am trying to parse url-encoded strings that are made up of key=value pairs separated by either &amp; or &amp;amp;.   The following will only match the first occurrence, breaking apart the keys and values into separate result elements:  var result = mystring.match(/(?:&amp;|&amp;amp;)?([^=]+)=([^&amp;]+)/)   The results for the string '1111342=Adam%20Franco&amp;348572=Bob%20Jones' would be:  ['1111342', 'Adam%20Franco']   Using the global flag, 'g', will match all occurrences, but only return the fully matched sub-strings, not the separated keys and values:  var result = mystring.match(/(?:&amp;|&amp;amp;)?([^=]+)=([^&amp;]+)/g)   The results for the string '1111342=Adam%20Franco&amp;348572=Bob%20Jones' would be:  ['1111342=Adam%20Franco', '&amp;348572=Bob%20Jones']   While I could split the string on &amp; and break apart each key/value pair individually, is there any way using JavaScript's regular expression support to match multiple occurrences of the pattern /(?:&amp;|&amp;amp;)?([^=]+)=([^&amp;]+)/ similar to PHP's preg_match_all() function?  I'm aiming for some way to get results with the sub-matches separated like:  [['1111342', '348572'], ['Adam%20Franco', 'Bob%20Jones']]   or   [['1111342', 'Adam%20Franco'], ['348572', 'Bob%20Jones']] Proof that disjoint cycles commute I'm trying to prove that in a symmetric group two disjoint cycles commute. But I suspect that something is not right about my proof (a sense of vagueness). Some hints would be appreciated.  Here's my proof:  Let $\sigma=(s_1 s_2 ... s_n)$, $\tau=(t_1 t_2 ... t_m)$ for some integers $m, n$. That is, $s_i \neq t_j$ for any $i \in [1, n]$, $j \in [1, m]$. Now, $\sigma\tau=(s_1 ... s_n)(t_1 ... t_m)$ is a cycle of disjoint permutations, which cannot be represented by any other disjoint permutations. Thus $\sigma\tau=\tau\sigma$. Why don't self-closing script elements work?