How to pull and push commit in gitlab?
Git for beginners: The definitive practical guide
How do I tell which point-and-shoot cameras take good low light photos? I have a Canon Powershot A710 and I previously had a previous Powershot A-series model. I love them due to the high amount of control they offer. Sadly, they both suck in low light conditions, their flash is very unresponsive, and the camera takes several seconds preparing itself for the next shot after a flash shot. And even then, the pictures they take in low light/flash are as bad as good are those with good light.  I'd like a camera which gives a lot of control options, but what do I need to look for to find one that's great in low light, with or without flash? Do you use "a" or "an" before acronyms / initialisms? 99% of the time, I'm clear on when I should use "a" versus "an." There's one case, though, where people &amp; references I respect disagree.  Which of the following would you precede with "a" or "an," and why?   FAQ FUBAR SCUBA   [Note: I've read the questions  and , but the rules given there don't necessarily apply here.]    [Edited to add]  Here's a shorter (and hopefully clearer) version of the question…  In written English, which is correct (and why): "a FAQ" or "an FAQ"?  Some references with differing opinions:   an: the  and  a: the Microsoft Manual of Style for Tech Publications, 3e either: the  and Antisymmetric Relations Object not visible in render I have a model which you can see here (camera perspective).    In the second image, you can see how the render turned out.    I'm at my wit's end trying to figure our why it's not showing up in the render. If anyone needs more information, just ask and I'll get it to you. I've already checked the normals and they're not the cause. The error message is :the grub-efi-amd64-signed package failed to install into /target/. Without the GRUB bootloader the system will not boot  I'm trying to install on HP pavilion g6-2303 TX using USB.  How do I fix this problem? How to prove sum of powers property of roots of unity? What is the proper way to handle data between scenes? I am developing my first 2D game in Unity and I have come across what seems an important question.  How do I handle data between scenes?  There seems to be different answers to this:   Someone mention using , while other people told me this should be used to store other things like screen brightness and so on. Someone told me that the best way was to make sure to write everything into a savegame everytime that I changed scenes, and to make sure that when the new scene loads, get the info from the savegame again. This seemed to me wasteful in performance. Was I wrong? The other solution, which is the one I have implemented so far is to have a global game object that isn't destroyed between scenes, handling all the data between scenes. So when the game starts, I load a Start Scene where this object is loaded. After this ends, it loads the first real game scene, usually a main menu.   This is my implementation:  using UnityEngine; using UnityEngine.UI; using System.Collections;  public class GameController : MonoBehaviour {      // Make global     public static GameController Instance {         get;         set;     }      void Awake () {         DontDestroyOnLoad (transform.gameObject);         Instance = this;     }      void Start() {         //Load first game scene (probably main menu)         Application.LoadLevel(2);     }      // Data persisted between scenes     public int exp = 0;     public int armor = 0;     public int weapon = 0;     //... }   This object can be handled on my other classes like this:  private GameController gameController = GameController.Instance;   While this has worked so far, it presents me with one big problem: If I want to load directly a scene, let's say for instance the final level of the game, I can't load it directly, since that scene does not contain this global game object.  Am I handling this problem the wrong way? Are there better practices for this kind of challenge? I would love to hear your opinions, thoughts and suggestions on this issue.  Thanks How do I count the number of occurrences of a char in a String? I have the string   a.b.c.d   I want to count the occurrences of '.' in an idiomatic way, preferably a one-liner.  (Previously I had expressed this constraint as "without a loop", in case you're wondering why everyone's trying to answer without using a loop). How can I get a complete list of fields for a standard object? Changing CSV layer attribute value from text to number in QGIS? I read in a CSV file for linking to a .shp file.   All the variables that are just numbers came in as text.   How do I change the text variables to numeric in QGIS? What is the best format to capture addresses for a universal user-base? What is the best and most usable format to capture address information from a universal user base? I was thinking to have the system detect the country by IP address and have that selected by default, and then state/province droplist would be populated accordingly. But the issue is some countries have states and provinces and some do not. Some have postal codes and some do not. Is there a universal standard to capture complete address info to ensure efficiency for most users?  FYI: I did research on here and while there are some questions that are close, none are about this exact same question (unless I missed something) I read on  that Ubuntu will no longer use the familiar file size units we all know by now (kB, MB, GB, TB) and switch to a different IEC standard (KiB, MiB, GiB, TiB). If this is true, I would like to know what's the reasoning behind this change, and the impact (if any) this change has, especially with multiplatform applications or applications run with Wine. Show that if $a$ is an odd integer and $b$ is an even integer then $\gcd(a,b)=\gcd(a,b/2)$ Show that if $a$ is an odd en integer and $b$ is an even integer then $\gcd(a,b)=\gcd(a,b/2)$.  I understand that since $a$ is not divisible by $2$ but $b$ is, the gcd of $a$ and $b$ also can't be divisible by $2$ but I'm getting stuck using this to help me. Align equation left Here is my split equation:  \documentclass{article} \usepackage{amsmath} \begin{document} \begin{equation} \begin{split} a_{ijk} =&amp; \frac {Pr(M_{I} =2  \&amp;  M_J=1 \&amp; M_K =1 | I=i , J=j , K=k)}{Pr (M_I =1 \&amp; M_J =1 \&amp; M_K=1 | I=i , J=j)} \\             =&amp; \frac {\mu_{ijk211}}{\mu_{ijk111}}\\ \end{split} \end{equation} \end{document}   I want to align this equation to the left. Is this possible in an equation environment? How can we get the command line of a running application? Will everything eventually be a single black hole? Given the idea that gravity is always a factor (even though greater distances are excessively tiny), would everything eventually end up forming a single black hole?  My thinking is that, with infinite time, all mass will eventually be pulled together. How to concatenate characters in java? How do you concatenate characters in java? Concatenating strings would only require a + between the strings, but concatenating chars using + will change the value of the char into ascii and hence giving a numerical output. I want to do System.out.println(char1+char2+char3... and create a String word like this.  I could do   System.out.print(char1); System.out.print(char2); System.out.print(char3);   But, this will only get me the characters in 1 line. I need it as a string. Any help would be appreciated.  Thanks Explaining Horizontal Shifting and Scaling I always find myself wanting for a clear explanation (to a college algebra student) for the fact that horizontal transformations of graphs work in the opposite way that one might expect.  For example, $f(x+1)$ is a horizontal shift to the left (a shift toward the negative side of the $x$-axis), whereas a cursory glance would cause one to suspect that adding a positive amount should shift in the positive direction. Similarly, $f(2x)$ causes the graph to shrink horizontally, not expand.  I generally explain this by saying $x$ is getting a "head start". For example, suppose $f(x)$ has a root at $x = 5$. The graph of $f(x+1)$ is getting a unit for free, and so we only need $x = 4$ to get the same output before as before (i.e. a root). Thus, the root that used to be at $x=5$ is now at $x=4$, which is a shift to the left.  My explanation seems to help some students and mystify others. I was hoping someone else in the community had an enlightening way to explain these phenomena. Again, I emphasize that the purpose is to strengthen the student's intuition; a rigorous algebraic approach is not what I'm looking for. List of Figures + Beamer? I want to have a List of Figures in my Beamer presentation.  I spent some time searching the web for a solution and only found desperate people telling others that it's not possible, i.e.  or .  I also made some experiments on my own, i.e. "abuse" tocloft to do the job. However, even \usepackage{tocloft} causes pdflatex to fail with ! LaTeX Error: \l@section undefined. (and lying to tocloft by defining the command as empty beforehand doesn't to any good either).  I can't really see why it should be that difficult to make a List of Figures in beamer, since \tableofcontents works flawlessly.   I'm not afraid of doing some coding on my own, but I'm not deep enough into TeX to really understand the magic behind the "List of Figures", "Table of Contents" and alike.  Is anyone out there having suggestions or links that might help me figure this out?  \documentclass{beamer}  % uncomment the following line % to get the error message % `! LaTeX Error: \l@section undefined.`  %\usepackage{tocloft}   % uncomment the following lines INSTEAD % to lie to tocloft % and get the next error message  %\makeatletter %\newcommand\l@section{} %\makeatother %\usepackage{tocloft}  \begin{document}  \begin{frame}{A frame} \begin{figure} \mbox{A} \caption{An 'A'} \end{figure} \end{frame}  \begin{frame}{List of Figures} \listoffigures \end{frame}  \end{document}   EDIT: Solution  Starting from Marco Daniels answer to my question, I modified his code to reflect my needs. Here is what I came up with. Please feel free to comment, improve and reuse this.  \RequirePackage{ifthen}  \makeatletter  \AtEndDocument{%   \clearpage   \beamer@tempcount=\c@page\advance\beamer@tempcount by -1%   \if@filesw   \newwrite\tf@lof   \immediate\openout\tf@lof\jobname.lof\relax   \newwrite\tf@lot   \immediate\openout\tf@lot\jobname.lot\relax   \fi }  \long\def\beamer@makecaption#1#2#3#4{%   \def\insertcaptionname{\csname#1name\endcsname}%   \def\insertcaptionnumber{\csname the#1\endcsname}%   \edef\insertframenumber{\theframenumber}%   \ifthenelse{\equal{#3}{\empty}}{%       \def\insertlistcaption{#2}%   }{%     \def\insertlistcaption{#3}%   }   \def\insertsource{#4}%   \def\insertcaption{#2}%   \ifthenelse{\equal{#1}{figure}}{%       \addtocontents{lof}{\protect\listoffigureformat{\insertcaptionnumber}{\insertlistcaption}{\insertframenumber}{\insertsource}}{}{}%   }{}   \ifthenelse{\equal{#1}{table}}{%       \addtocontents{lot}{\protect\listoftableformat{\insertcaptionnumber}{\insertlistcaption}{\insertframenumber}}{}{}%   }{}   \nobreak\vskip\abovecaptionskip\nobreak   \sbox\@tempboxa{\usebeamertemplate**{caption}}%   \ifdim \wd\@tempboxa &gt;\hsize   \usebeamertemplate**{caption}\par   \else   \global \@minipagefalse   \hb@xt@\hsize{\hfil\box\@tempboxa\hfil}%   \fi   \nobreak\vskip\belowcaptionskip\nobreak% }  \def\listoffigureformat#1#2#3#4{%   \makebox[2ex][r]{#1}%   \hspace{1ex}%   {\usebeamercolor[fg]{bibliography entry author} #2}%   \ifthenelse{\equal{#4}{\empty}}{}{ -- #4}%   \dotfill%   \makebox[2ex][r]{#3}\par% } \def\listoffigures{%   \setlength{\leftskip}{3ex}   \setlength{\parindent}{-3ex}   \@starttoc{lof}% } \def\listoftableformat#1#2#3{\makebox[2ex][r]{#1}\hspace{1ex}#2\dotfill\makebox[2ex][r]{#3}\par} \def\listoftables{%   \setlength{\leftskip}{3ex}   \setlength{\parindent}{-3ex}   \@starttoc{lot}% }  \long\def\@caption#1[#2]#3{   \par\nobreak   \begingroup     \@parboxrestore     \if@minipage       \@setminipage     \fi     \beamer@makecaption{#1}{\ignorespaces #3}{\ignorespaces #3}{\ignorespaces #2}\par\nobreak     \endgroup}  \makeatother Let $p$ be an odd prime number. Prove that $$\left( \frac{1 \cdot 2}{p} \right) + \left( \frac{2 \cdot 3}{p} \right) + \left( \frac{3 \cdot 4}{p} \right) + \cdots + \left( \frac{(p-2)(p-1)}{p} \right) = -1$$ where $\left( \frac{a}{p}\right)$ is the Legendre symbol.  This seems to be a tricky one! I've tried using the property $\left( \frac{ab}{p} \right)=\left( \frac{a}{p}\right) \left( \frac{b}{p} \right)$ and prime factoring all the non-primes but to no avail. I had a quick thought of induction haha, but that was silly. I tried factoring the common Legendre symbols like $\left( \frac{3}{p}\right) \left[ \left( \frac{2}{p} \right) + \left( \frac{4}{p} \right) \right]$ but that didn't bring anything either. And I've been looking for pairwise cancellation with $1$ term leftover, but that does not seem to work.  Can you help?