Today I've earned the "Yearling" badge. When I'm checking the other users who are all received the same badge in the , there I can see a strange behaviour.  For the user , the Yearling badge was awarderd two times in two days. The first one was awarded on 2019-12-19 16:50:12Z and the second one was awarded on 2019-12-20 10:20:09Z. When clicking on the both links, it is redirecting to the same user profile.  What is the reason behind that?  Screenshot for reference:
Formerly List of all badges with full descriptions.  What are badge name's requirements? Why didn't I get badge name ? Which badges can I earn multiple times?    Jump to:                    Note: Some badges are awarded based on score. The term score means the total number of upvotes minus the total number of downvotes.  Any badge with a *** next to it is one of 20 badges that count towards the  displayed on a moderator candidate in an election. The candidate score is a total of 40 points: the first 20 are awarded based on the user's rep divided by 1000 and rounded down; the second 20 represents the total number of unique badges earned of the 20 that count. Note that if the same badge is earned multiple times, it will only count once towards the candidate score.  Any badge with a &quot;(retired)&quot; next to it is no longer awarded, but is retained by users who previously earned them. See  for more info on what it means for a badge to be retired.       Visit the  of the  on any site to see a complete list of badges that you can filter by earned, unearned, or type.
I am using the 2D Animation preset and so far every new scene I made had the same bright white background. Now my two latest scenes (created with 'copy settings') have a light grey background even though they are set to white. There's also no difference in surface, nodes or strength.  There are apparently exist different background shades with an off-white color but selecting the white one doesn't help either I am a new QGIS user and i have a challenge to convert legend from right to left reading in QGIS like ArcGIS. Can I tap a creature with summoning sickness to pay "tap an untapped creature you control" costs? I have a  on the battlefield and I played a vampire earlier this turn. Can I tap both to transform Chosen of Markov even though I just played the vampire?  Extra Example: I play  to put two 1/1 white bird tokens into play. Can I now tap a both of them and another creature I control to pay Battle Screech's flashback cost of tapping three white creatures?   Obviously these creatures couldn't use any tap abilities until my next turn, but since the tapping is coming from a different source, I figured the situation might be different. Left-pad printf with spaces How can I pad a string with spaces on the left when using printf?  For example, I want to print "Hello" with 40 spaces preceding it.  Also, the string I want to print consists of multiple lines.  Do I need to print each line separately?  EDIT: Just to be clear, I want exactly 40 spaces printed before every line. Why is the Yang–Mills existence and mass gap problem so fundamental? Ref:   Can anyone please explain why the Yang–Mills Existence and Mass Gap problem is so important / fundamental to contemporary mathematics (and, presumably, theoretical / mathematical physics)?  If possible, please include the following in your answer / response:  (1) Why the problem is so fundamental / important - for both mathematics and theoretical / mathematical physics; (2) How the solution to the problem would impact research in the above two fields; (3) (If applicable) What progress has been made so far in resolving it; (references to papers, books, other resources would be helpful).  I'm an upper-level undergraduate in theoretical physics, and so am seeking an answer / response at such level. Homebrew putty installation does not make the putty command recognisable I would like to instal  with use of Homebrew. I'm performing the following command:  brew install putty   which gives the seemingly correct output:  ==&gt; Downloading https://homebrew.bintray.com/bottles/putty-0.67.el_capitan.bottle.tar.gz Already downloaded: /Users/thisuser/Library/Caches/Homebrew/putty-0.67.el_capitan.bottle.tar.gz ==&gt; Pouring putty-0.67.el_capitan.bottle.tar.gz 🍺  /usr/local/Cellar/putty/0.67: 11 files, 1.5M   However, the putty command is not recognised:  thisuser-541/41:~$ putty -bash: putty: command not found Where to put my code: plugin or functions.php? Which statistical test to analyse? I am trying to find out how inflammatory markers change in response to a mechanical device change. I have daily readings for the 5 days before the event and for the 5 days after the event. I am expecting a rise in inflammatory markers in the pre readings followed by a fall after the device change. I also need to find out how the change differs in two subgroups (sample sizes of the subgroups are n=21 and n=88). The aim is to find out if the change is significant, to consider these markers as predictors of the device failing and hence needing to be changed.    I would like to know what test would be appropriate to analyse this. I need to keep things simple as I and the people reading the results are non-statisticians. Math people:  The title is the question. I am convinced that uncountable sets exist, thanks to Cantor's diagonal proof.  It is not intuitively clear to me that uncountable ordinals or cardinals should exist.  What axioms of set theory are needed to accept their existence?  I think I have read here that the Axiom of Choice is not necessary.  I am reluctant to accept as "true" any theorems that rely on the Axiom of Choice.  EDIT: This question may be a duplicate: I just found  .  I apologize if this is so. Apt-get update with an @ in password error I want to update ubuntu 11.04 throught apt-get. So I added the following lines in apt.conf  export http_proxy=http://username:password@proxyserver.net:port/ export ftp_proxy=http://username:password@proxyserver.netport/  export http_proxy=http://deepak:Deepak@123@12.1.1.1:3128   My question is: how do I insert special characters to a username or password?  For instance: my password for proxy is Deepak@123 and it is getting an error. I have posted a proof below, and would appreciate it if someone could review it for accuracy. Thanks!   Problem:   Let n $\in$ $\mathbb{Z}$ with $n$ $\ge$ 3. Prove the following:   (a) Z(D$_{2n}$) = 1 if $n$ is odd.   (b) Z(D$_{2n}$) = {1, r$^k$} if $n$ = $2k$.   Note that $r$ and $s$ generate D$_{2n}$ with the group presentation { $r$,$s$ | $r$$^n$ = $s$$^2$ = 1, $rs$ = $sr$$^{-1}$ }   Proof:  part (a)   Let $n$ $\ge$ 3 where $n$ $\in$ $\mathbb{Z}$ and $n$ is odd.   For any x $\in$ Z(D$_{2n}$), x must commute with both $s$ and $r$, since both are in $D_{2n}$. Note that if $x$ commutes with both $s$ an $r$, then $x$ commutes with any element of $D_{2n}$ since $r$ and $s$ generate D$_{2n}$.    Then for any x$\in$ Z(D$_{2n}$), we have $xr$ = $rx$, where x is of the form $x$ = s$^j$r$^w$ (any such element can be arranged into this form via the relation $rs$ = $sr$$^{-1}$ and the fact that $r$ and $s$ generate D$_{2n}$). Note that since r and s have finite order, the exponents are modulo n and modulo 2 for $r$ and $s$ respectively.   Then we have  s$^j$r$^w$$r$ = $r$s$^j$r$^w$.   Which implies s$^j$r$^{w+1}$ = s$^j$r$^{w +- 1}$  In the case where the power of r on the RHS is $w$ - 1, it is clear we cannot have equality unless |r| = 1, which it is not. Note that if $j$ is even then we have only the $w$ + 1 case. Hence $x$ may be of the form $1$$r$$^w$ = $r$$^w$.   Also we have that x must commute with $s$, hence $s$$x$ = $x$$s$.   Then $s$$r$$^w$ = $r$$^w$$s$.  Which implies $s$$r$$^w$ = $s$$r$$^{-w}$.   Applying s$^{-1}$ to both sides we arrive at the task of finding when r$^w$ = r$^{-w}$ i.e when $w$ = $-w$. But since r has finite order we have:   $$ \bar{w} = (n-1)*w $$   where $\bar{w}$ is the residue class of w modulo n. Hence for some a $\in$ $\mathbb{Z}$$_+$   $$ w + an = wn - w $$  $$ 2w = wn - an $$ $$ 2w = n(w-a) $$   Now since 0 $\lt$ w $\le$ n we have that the LHS is greater than 0. Also since n $\ge$ w then (w-a) $\le$ 2, since otherwise the RHS would be greater than the LHS. So since n is odd we must have that (w-a) is even and hence (w-a) = 2, implying that 2w = 2n and hence n = w.   Hence x = $r$$^w$ = $r$$^n$ = 1. So Z(D$_{2n}$) = 1.   part (b)   From the argument above any x $\in$ Z(D$_{2n}$) must be of the form $s$$^j$$r$$^w$ for j even and j $\in$ $\mathbb{Z}$. Then x is of the form $r$$^w$ for $0$ $\lt$ w $\le$ $n$. Since $n$ is even we have the equation below is satisfied when $2w$ = $n$ or $w$ = $n$, since (w-a) $\le$ 2 but also (w-a) $\ge$ 0 for (w-a) $\in$ $\mathbb{Z}$$_+$.   $$ 2w = n(w - a) $$   Hence $r$$^w$ $\in$ Z(D$_{2n}$) when $2w$ = $n$, as desired. There are 100 ropes in a bag. In each step, two rope ends are picked at random, tied together and put back into a bag. The process is repeated until there are no free ends. What is the expected number of loops at the end of the process? Using Ubuntu 12.04. is there any way to hide a running process ?  That process shouldn't' display in system monitor,top,htop . There is no problem to remember the process id (PID) so observation of the process can be done easily .but i want to hide the process completely . is there any way ? How can I convert a string to boolean in JavaScript? Can I convert a string representing a boolean value (e.g., 'true', 'false') into a intrinsic type in JavaScript?  I have a hidden form in HTML that is updated based upon a user's selection within a list. This form contains some fields which represent boolean values and are dynamically populated with an intrinsic boolean value. However, once this value is placed into the hidden input field it becomes a string.  The only way I could find to determine the field's boolean value, once it was converted into a string, was to depend upon the literal value of its string representation.  var myValue = document.myForm.IS_TRUE.value; var isTrueSet = myValue == 'true';   Is there a better way to accomplish this? Viruses in wine? I am working on the following problem from group theory:     If $G$ is a group of order $2n$, show that the number of elements of $G$ of order $2$ is odd.   That is, for some integer $k$, there are $2k+1$ elements $a$ such that $a \in G,\;\; a*a = e$, where $e$ is the identity element of $G$. I am recording videos on iPhone and uploading them to my Linux server. Users can see these videos on website. If users are seeing videos on mac or iDevices the video is looking fine but when users are seeing that video on windows systems these videos are getting rotated.   I am assuming this is because of some apple's flag in the recorded video which Windows systems are not able to read. Can I record videos in some way so they don't get rotated on Windows systems? Or is there any way I can re-encode videos with proper interface orientation using FFmpeg?  I am open to both FFmpeg or iOS way. "'void' type not allowed here" error (Java) I have a project, and it has an object that has always rendered correctly, but one day when I open the project and make a test render, the object has disappeared. All the settings are normal, the object, and all of its children render display in the object outliner set to true. I really don't know what happened to it. Can someone tell me what I should look through to diagnose it? Sorry I cannot provide much detail. What I've explained is all I know. Edit: I am uploading the .blend file at  This is a Minecraft scene. It is the wolf that doesn't render. At one point it did, but now it just won't. The wolf has particle properties. I don't know if that matters, but it could. Look through all the wolf's properties to try and figure it out. Does $k+\aleph_0=\mathfrak{c}$ imply $k=\mathfrak{c}$ without the Axiom of Choice? I'm currently reading a little deeper into the Axiom of Choice, and I'm pleasantly surprised to find it makes the arithmetic of infinite cardinals seem easy. With AC follows the Absorption Law of Cardinal Arithmetic, which states that for $\kappa$ and $\lambda$ cardinal numbers, the larger infinite and the smaller nonzero, then $\kappa+\lambda=\kappa\cdot\lambda=\max(\kappa,\lambda)$.  I was playing around with the equation $k+\aleph_0=\mathfrak{c}$ for some cardinal $k$. From the above, it follows that $\mathfrak{c}=k+\aleph_0=\max(k,\aleph_0)$, which implies $k=\mathfrak{c}$.  I'm curious, can we still show $k=\mathfrak{c}$ without the Axiom of Choice? Is it maybe possible to bound $\mathfrak{c}-\aleph_0$ above and below by $\mathfrak{c}$? But then I'm not quite sure such algebraic manipulations even mean anything, or work like that here. Certainly normal arithmetic does not! Thanks.