I tend to answer my own questions if they remain unanswered for some time, and I wondered this:  If I edit my question within 12 hours before or after answering it, like I do with others' questions, will that count as progress toward the , , and  answer ?
The difference between a poor or meh question and a stellar question can often simply be someone understanding it and providing it a great answer. I can't begin to count the number of times I've justified re-opening a question as a moderator by saying:     Look at the answer it got, though. This isn't something we want deleted, this is something we want fixed, because it's obviously valuable. Someone with domain knowledge can easily edit that question based on the answer it received.   This doesn't only apply to borderline-poor or poor questions – sometimes a good edit means the difference between 100 and 10,000 people finding something through typing things into Google. Good titles are hard.   We think we've come up with a way to provide incentive for good, opportunistic edits to happen more often, ideally without questions needing to go through closure or collecting a centimeter of dust before they do. What if we simply rewarded folks that answered and edited questions with a special, relatively difficult to earn badge?  Let's say that you answer something, and:   You edit the question 12 hours before or after answering it. This allows you to edit now, answer later &mdash; or answer now and edit later, when you have the time. Your edit isn't rolled back, or outright rejected if it was a suggested edit The question is not closed for any reason, even simply being a duplicate Your answer has a score of 1 or higher   &hellip; then you've done something that we probably want to recognize. You understood something, you provided the knowledge that you have, and then you provided an edit to make sure that more people in need of this knowledge can find it, while raising the overall quality of the site.  Here's the proposed tiers for the badge (within the above stated criteria):   Explainer (Bronze): Answered &amp; edited 10  questions Refiner (Silver): Answered &amp; edited 50 questions Illuminator (Gold): Answered &amp; edited 500 questions   We think this is going to be pretty hard to game. You can't go leaving streaks of terse snark all over the place on questions you know are going to be closed anyway, and junk edits aren't going to fly.  We're talking about rewarding folks that take a tiny bit more time out of their day just to write a more descriptive title so that folks can find the awesome answer that they wrote. Just doing that alone can make a big difference. Sure, it's a tad hard to get at the gold level, but it's supposed to be. If you routinely edit questions that you answer and have done so for some time, you'd probably get it the day we roll it out.  While this is technically something that we came up with on the quality initiative ( | ) &mdash; this badge could be earned on all of our sites.  What do you think? What did we miss? What could be better about the idea?
Is "remove" a reserved keyword in Google Chrome? We have played a 7-team competition, all seven teams met, we eliminated the team with the lowest points. After that the remaining six teams played knock out games and the three winning teams will go direct to semi-finals.  We decided that 4 teams should be picked to go for semi-final and that the best loser will join the other three to make it four teams to battle at the semi-final and final.  During the quarter final games, one team is beaten 2-0, another is beaten 3-0 and the third team is beaten in a penalty shoot-out by 4-0.  Now should you want a best loser to join the other three teams making four teams to play in the semi-final, who among these three teams would be in the best losers position. Pygame - blitting from x and y cordinates of image Can "Mr", "Mrs", etc. be used with a first name? Is it correct to use Mr/Mrs with a first name? The only solution that I have been able to come up with is $d=1$, but because $d&gt;2$, I can't seem to figure out how to find a solution. Capture characters from standard input without waiting for enter to be pressed I am buing assamble self PC for playing games.  I am wondering, all the drivers, which are most likely for Windows, can i just use them on Ubuntu?  So if i understand you right, it will automaticaly detect the drivers, no need for installs ? or it find custom drivers for ubuntu.  This is the setup, will there be any problem with it? Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$. Why do we get the same differential equations from both principles? Surely there is a fundamental connection between them? When written out, the two seem to have nothing in common.   $$\sum _i ( \mathbf F _i - \dot{\mathbf p}_i) \cdot \delta \mathbf r _i = 0$$  $$S[q(t)] = \int ^{t_2} _{t_1} \mathcal L (q,\dot{q},t)dt$$  After playing with d'Alembert's principle  we find that we can rewrite the whole thing as $$\sum _i \left[ \frac{d}{dt} \left( \frac{\partial T}{\partial \dot{q}_i} \right) - \frac{\partial T}{\partial q_i}-Q_i\right] \delta q_i$$  This can further be rewritten under certain conditions so we get the exact form of the E-L equation.   It seems to me that both ways of arriving at the result are fundamentally different.   A function must obey the E-L equations in order to minimize the action over a path, but when we look at the virtual work, it appears that they come from the fact that (quoting Goldstein) "particles in the system will be in equilibrium under a force equal to the actual force plus a 'reversed effective force'."  I think I understand the principle of stationary action, I can see how it leads to the E-L equations, but d'Alembert's Principle seems so arbitrary, I can't see any motivation for it. sed fails with "unknown option to `s'" error I have been reading through the  and was curious about the  declaration. I personally have never used it, however I am interested in exploring the language.   What is a good example of using friend?    Reading the FAQ a bit longer I like the idea of the &lt;&lt; &gt;&gt; operator overloading and adding as a friend of those classes. However I am not sure how this doesn't break encapsulation. When can these exceptions stay within the strictness that is OOP? Let $R$ a ring, and let $$\displaystyle R[[x]]=\left\{\sum_{k=0}^{\infty}a_k x^k\;\middle\vert\; a_k\in R\right\}$$ with addition and multiplication as defined for polynomials.  We have that $R[[x]]$ is a ring containing $R[x]$ as a subring.  How to prove that if $a_0\in R$ is a unit, then  $\displaystyle\sum_{k=0}^{\infty}a_k x^k$ is a unit in $R[[x]]$? How to show $\lim_{n \to \infty} a_n = \frac{ [x] + [2x] + [3x] + \dotsb + [nx] }{n^2} = x/2$? What happens if both players have the ability to pump their creature an unlimited number of times? Say that both players each had 2 , 2 , and a . (Or any other way to get unlimited mana, and a Nantuko Shade).  Player 1 attacks with his , and player 2 blocks with his . Before damage is assigned, each has chances to pump up their  as much as they want. They can't just declare them to be "infinite," but each can always one-up the other player. What happens?  It seems there would be a loop of events, but would one player eventually be forced to break the loop as per rule 714.3? Is it considered the same game state after each player pumps their creature a bunch as it was before?  This was discussed at length on the Draw3Cards Magic Q&amp;A site (which is no longer available), but no conclusion was reached. Some felt that the game would simply be a draw, others felt that the active player would be forced to make a different decision to break the loop. Others felt that you would just have to call a judge and see what he decides. Solutions to $y^2 = x^3 + k$? The equation $y^2 = x^3 + k$ for $k = (4n-1)^3 - 4m^2$, with $m, n \in \mathbb{N}$ and no prime number that p is congruent to 1 modulo 4 divids m, doesn't have any answer and its proof can be obtained by using quadratic reciprocity law.  Do you know answers of this equation for two or three different values of $k$? In addition, do you know any reference about that? How can I tell which cards or sets are currently legal in various formats? So, for example, when I type man ls I see LS(1). But if I type man apachectl I see APACHECTL(8) and if I type man cd I end up with cd(n).  I'm wondering what the significance of the numbers in the parentheses are, if they have any. Dark matter and dark energy references I've been looking for questions about dark matter, and I've read some very interesting answers. However, I desire too look into it deeply.   This is not actually a question. I'm asking the community to recommend interesting references to understanding dark matter and dark energy.   I accept all sort of references: notes, books, scientific papers etc.  Let us assume some background on classical physics, thermodynamics and basics about quantum theory. How to add a script to Startup Applications from the command line? I have a shell script. I want to configure my script to run automatically during startup. I know how to do this from GUI. But I wanted to do this from terminal. How can I do this?  On a primary research I found that the file needs to be moved to /etc/int.d/ directory. But this operation needs sudo permission. I wanted to do this without super user permissions.   I also found that there are files under ~/.config/autostart/ which are having some settings regarding the startup applications. But I don't know how to edit them to achieve this.   Can someone tell me the exact procedure to achieve this? The boundary of union is the union of boundaries when the sets have disjoint closures Assume $\bar A\cap\bar B=\emptyset$. Is $\partial (A \cup B)=\partial A\cup\partial B$, where $\partial A$ and $\bar A$ mean the boundary set and closure of set $A$?   I can prove that $\partial (A \cup B)\subset \partial A\cup\partial B$ but for proving $\partial A\cup\partial B\subset \partial (A \cup B)$ it seems not trivial. I tried to show that for $x\in \partial A\cup\partial B$ WLOG, $x\in \partial A$ so $B(x)\cap A$ and $B(x)\cap A^c$ not equal to $\emptyset$ but it seems not enough to show the result.