Confusion about what password one should use for the `sudo` command Yester I asked  question, and tried to follow cl-netbox's advice:     Highlight the Ubuntu entry in the GRUB boot menu and press the E key.   Add nouveau.modeset=0 to the end of the linux line - press F10 to boot.   On login screen press Ctrl+Alt+F1 ... enter user name and password ...   Now execute the commands ...   After pressing Ctrl+Alt+F1 at the login screen and writing sudo apt-get update, the laptop asks me for my password (as per usual). However, the password that I normally always use does not work. I've no idea what password I should use. Anybody can help me?
How can I determine the sudo password? What is the default password of the sudo? I visit some forums, it tells that the default password is password, but it doesn't work, how can I determine the password?
If $[G' : G'']\leq p^2$, then $G'$ is abelian. Problem : Let $G$ be a p-group and $G'$ denote the commutator subgroup of $G$. If $[G' : G'']\leq p^2$, then $G'$ is abelian.  It is easy to prove it for the case of $[G' : G'']=1$ since G is solvable. But it is too hard to prove the other cases. Please give me any advice or solutions. Thanks in advance. Stack Overflow technology makes me write bad answers Today I had some time and felt like writing a relatively good answer on Stack Overflow; explaining both the underlying mechanisms and good practices, unlike usual declarative answers in a "do that" style.  It took me like an hour of time, $50 if I were free lance. It'll bring me nothing but 3-5 occasional views, and - if I get lucky - some ignorant comments.  You know, it's quite discouraging.  Instead of pride for the job that was done, I feel frustrated. I know, this answer may offer help to the OP, if he ever notices it (being already satisfied with other answers).  So the only fate the answer will meet is to sink forever, below thousands other questions, falling like snow.  So, being a programmer, I am thinking of efficiency. It's a waste of time to spend so much effort to answer one question when there is no way to [actually] reuse it. So, I wouldn't do it again but just turn back into short answers mode without explanations.       Seeing other answers, I don't think I am only one with such feeling.  The current policy is driving knowledgeable and educational users away, yet it attracts inexperienced ones in numbers. Ones who will tell you to use [some technology] which they never have laid their hands upon. Ones who write answers just because they have learned some "good practice" from other questions, and this "knowledge" becomes an endless source of rep points.  The idea of Stack Overflow is rotten. It was excellent when trees were high and traffic was low.  It turned bad now; it has become a honeypot for all the "enthusiast programmers" of the world, eager to share their 2 cents faster than you can say the word "close".  You're killing a great resource. Despite all the nice words and proper declarations, it discourages the reuse of the knowledge. And encourages fast on-site answers.  Of course, the rules will tell you contrary. It's all nice and proper (if you don't mind it's strict mutual exclusivity). But mechanically, technologically it works in the way I described.  I tried to write a tag wiki. You know, it was biggest waste of time in my life. As a matter of fact, it is not a wiki at all; you can't even link to a certain section. Not to mention that nobody ever has the idea to read tag wikis.  Reference questions? Don't make me laugh. There is no good mechanism to store and categorize links. Favorites, although intended for the different purpose, are a chunk of crap when you want to use it as a reference. So, you can't compete with 30-seconds answers: a question will get 5 answers before you can even find an appropriate link.  Okay, I can organize links myself but, there are some problems:   I am just a human. I cannot remember everything I am just a human and I cannot answer questions at rate of 1 qpm. One person cannot fight the Tide. It's helpless. When technology makes people writing fast [incomplete] answers -  no good will of proper declarations can do a thing.  Even if a question gets finally closed, it won't hurt anyone - the hydra grew two more already.   You'll call me a brute for insulting honest users and innocent volunteers, but that's fact: There are much more inexperienced users than experienced ones. And inexperienced users have much more spare time to hang over than experts who have their job to do. So, from that fact we can conclude that most of the answers on Stack Overflow have a low quality. Nobody cares. But I do.   There should be a technical way to limit the activity of not-so-knowledgeable-yet-eager-to-answer users. There should be a technical preference for the closing a question over answering it (have a silly answer for the silly question? welcome to comments). There should be a technical solution to accumulate and structure knowledge, a system that lets you choose a good answer instead of writing it by hand.   That's the only way. Otherwise Stack Overflow will remain an automated dispenser of bad practices.  Well, I just felt like letting it out again. Thank you for reading. 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 The question is $$\text{If }\, p, \ q\in \mathbb{N}, \;1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}=\frac{p}{q}.\qquad \text{Prove that } 1979\mid p.$$ So my solution went like this:  $$S=1-\frac12+\frac13-\frac14-\dotsb-\frac{1}{1318}+\frac{1}{1319}\Rightarrow S=\sum_{k=1}^{1319}\frac{1}{k}-2\sum_{k=1}^{659}\frac{1}{2k}=\sum_{k=660}^{1319}\frac{1}{k}$$ Now this is equivalent to finding the sum of the roots of the equation $$\left(x-\frac{1}{660}\right)\left(x-\frac{1}{661}\right)\dots \left(x-\frac{1}{1319}\right)=0$$. Rewriting this equation, we get $(660x-1)(661x-1)\dotsc(1319x-1)=0$ (we can discard the $\displaystyle \frac{1319!}{659!}$ since $1979$ is a prime). Calculating the sum of the roots, we get the required value to be equal to $(1320+659)\times 330$ which is obviously divisible by $1979$ (again discarding the coeffecient of $x^{660}$).  Now my question is: Using the same method of converting the problem into finding the sum of roots of a polynomial, we get from the original equation something like this: $$(x-1)\left(x+\frac{1}{2}\right)\dots \left(x-\frac{1}{1319}\right)=0 \Rightarrow (x-1)(2x+1)\dotsc(1319x-1)=0$$ So, the sum of roots would be $-1+2-3+4-\dotsc+1318-1319=-660$ But this is not divisible by $1979$. Why does the former solution work while the latter doesn't? Is there a flaw in my reasoning? Note: I know that the question can be solved by some other methods. Please only explain my given doubts and don't instead give another solution  I realised my mistake after some hours of pondering over it. (In the 1st equation) To get the required sum I would need to take calculate $\displaystyle -\frac{\text{coeffecient of }x^{659}\,}{\text{coeffecient of }x^{660}}$. However what I've done is calculate this:  $\displaystyle -\frac{\text{ coeffecient of }x}{\text{coeffecient of }x^{660}}$ which is actually equal to $\text{sum of roots taken }659 \text{ at a time} $. It was just a coincidence that the value I calculated was divisible by $1979$. The correct way, indeed, would be through the link supplied Which meta am I supposed to post on? I'm mostly active on  and we have a separate . Most of my questions on IT Security meta have answers with links here. So my question is, where am I supposed to post what i.e. feature request, bug report, behavior questions? How to round a number to n decimal places in Java Is there an easy way to convert a number to a word in PHP? SQL Server 2014 - Strange SP behavior during compilation How do electromagnetic waves travel in a vacuum? This is perhaps a total newbie question, and I will try to formulate it the best I can, so here it goes. How does an electromagnetic wave travel through for example, the vacuum of space?   I usually see that waves are explained using analogies with water, pieces of rope, the strings of a guitar, etc, but it seems to me that all those waves need a medium to propagate. In fact, from my point of view, in those examples the wave as a "thing" does not exist, it's just the medium that moves (involuntary reference to The Matrix, sorry).   But in space there is no medium, so how does a wave travel? Are there free particles of some sort in this "vacuum" or something? I believe , so supposedly there isn't a medium for the wave to travel through.   Moreover, I've seen other answers that describe light as a perturbation of the electromagnetic field, but isn't the existence of the field, potential until disturbed? How can it travel through something it does not exist until it's disturbed by the traveling light in the first place? (this last sentence is probably a big misconception by me). Is it possible to add new alphabets to unicode-math? In ordinary LaTeX, one can easily add new math alphabets by using \DeclareMathAlphabet. One is then able to use, say, cmsy, eus, and rsfs letters side-by-side. Unfortunately, there does not seem to be a similar mechanism in unicode-math, or at the very least \DeclareMathAlphabet doesn't do what it's supposed to do.  Question. How can I set up my own math alphabets in unicode-math?   Attempted workarounds  Since unicode-math has lots of math alphabets, one workaround might be to replace an unused alphabet with the desired one using \setmathfont[range=...-&gt;...]. This doesn't quite work right in my example:  \documentclass[12pt]{article} \usepackage{fontspec,xltxtra,xunicode} \defaultfontfeatures{Mapping=tex-text}  \usepackage{unicode-math}  \scrollmode  \setmathfont{xits-math.otf} \setmathfont[range={\mathscr,\mathbfscr},StylisticSet=0]{xits-math.otf} \setmathfont[range={\mathcal,\mathbfcal},StylisticSet=1]{xits-math.otf} \setmathfont[range={\mathtt-&gt;\mathcal}]{latinmodern-math.otf}  \begin{document} \[ \begin{array}{c} \mathcal{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \mathscr{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \mathtt{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \end{array} \]   I suspect the problem is that the codepoints of the calligraphic letters in Unicode are not contiguous.   Using \mathversion seems to work better:  \documentclass[12pt]{article} \usepackage{fontspec,xltxtra,xunicode} \defaultfontfeatures{Mapping=tex-text}  \usepackage{unicode-math}  \setmathfont{xits-math.otf}  \setmathfont[version=eucal]{latinmodern-math.otf} \newcommand{\eucal}[1]{\mathchoice %   {\mbox{\mathversion{eucal}$\displaystyle\mathscr{#1}$}} %   {\mbox{\mathversion{eucal}$\mathscr{#1}$}} %   {\mbox{\mathversion{eucal}$\scriptstyle\mathscr{#1}$}} %   {\mbox{\mathversion{eucal}$\scriptscriptstyle\mathscr{#1}$}}}  \setmathfont[range={\mathcal,\mathbfcal},StylisticSet=1]{xits-math.otf}  \begin{document} \[ \begin{array}{c} \mathcal{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \mathscr{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \\ \eucal{A B C D E F G H I J K L M N O P Q R S T U V W X Y Z} \end{array} \] \end{document}   But this feels very inefficient. How to minimize $|z_1 - z_2|^2 + |z_1 - z_4|^2 + |z_2 - z_3|^2 + |z_3 - z_4|^2$? If $z_1,z_2,z_3,z_4 \in \mathbb{C}$ satisfy $z_1 + z_2 + z_3 + z_4  =  0$ and $|z_1|^2 + |z_2|^2 + |z_3|^2 + |z_4|^2 = 1$, then the least value of $|z_1 - z_2|^2 + |z_1 - z_4|^2 + |z_2 - z_3|^2 + |z_3 -  z_4|^2$ is $2$.   I solved this by taking two of the values to be $\frac{1}{2}$ and other two as $-\frac{1}{2}$. But I couldn't prove this in an efficient manner. Can a plan "age" reach zero? Is it possible to delete multiple elements from a list at the same time? If I want to delete elements at index 0 and 2, and try something like del somelist[0], followed by del somelist[2], the second statement will actually delete somelist[3].  I suppose I could always delete the higher numbered elements first but I'm hoping there is a better way. Should I remove non-significant variables from my regression model I have run a multiple linear regression using stepwise regression to select the best model, however the best model returned has a non-significant variable. When I remove this the AIC value goes up indicating the model without the significant variable is a worse fit. Should I remove the non-significant predictor or should I leave it in as it is a better model? The new ask page is now live on the network! Possible Duplicate:             Each post has an indicator which summarises who the post is shared.    What is the difference between these two terms? Should I write summarises or summarizes? Search and replace a line in a file in Python I want to loop over the contents of a text file and do a search and replace on some lines and write the result back to the file. I could first load the whole file in memory and then write it back, but that probably is not the best way to do it.  What is the best way to do this, within the following code?  f = open(file) for line in f:     if line.contains('foo'):         newline = line.replace('foo', 'bar')         # how to write this newline back to the file Recently I switched from Google to DuckDuckGo and so far I am very pleased with the experience. "The Internet" tells me that one of DuckDuckGo's best features is the possibility to directly search on specific pages using bangs (defined prefixes). E.g. search for "!so [duck-duck-go]" on DuckDuckGo to get duck-duck-go tagged questions on Stackoverflow.  I was searching for a comprehensive list of all available bangs, but the only thing I found was the official  wich offers a search box and categorized lists, but no complete list of all 7,447* bangs.  Is there a possibility to find all bangs without clicking through all the categories on the bang-page?    (*) by time of writing, 2015-12-15 Calculating the total number of surjective functions It is quite easy to calculate the total number of functions from a set $X$ with $m$ elements to a set $Y$ with $n$ elements ($n^{m}$), and also the total number of injective functions ($n^{\underline{m}}$, denoting the falling factorial). But I am thinking about how to calculate the total number of surjective functions $f\colon X \twoheadrightarrow Y $.  The way I thought of doing this is as follows: firstly, since all $n$ elements of the codomain $Y$ need to be mapped to, you choose any $n$ elements from the $m$ elements of the set $X$ to be mapped one-to-one with the $n$ elements of $Y$. This results in $n!$ possible pairings. But the number of ways of choosing $n$ elements from $m$ elements is $\frac{m!}{(m-n)!\,n!}$, so the total number of ways of matching $n$ elements in $X$ to be one-to-one with the $n$ elements of $Y$ is $\frac{m!}{(m-n)!\,n!} \times n! = \frac{m!}{(m-n)!}$.  Now we have 'covered' the codomain $Y$ with $n$ elements from $X$, the remaining unpaired $m-n$ elements from $X$ can be mapped to any of the elements of $Y$, so there are $n^{m-n}$ ways of doing this. Therefore I think that the total number of surjective functions should be $\frac{m!}{(m-n)!} \, n^{m-n}$.  Is this anything like correct or have I made a major mistake here? Multiple disks vs single disk on virtualized environment We're planning to install a SQL Server on this physical architecture:  A hyper-v cluster based on windows server 2019 A cluster shared volume where all VMs live A full-flash SAN that contains the CSV  I've always read that is a common best practice to have multiple disks for data files, log files, temp db files ect... But that best practice is referred to a bare-bone installation. So, what I'm asking is: what are the actual benefits of doing so on the above described architecture (if any)?