I made the experience that flags I place for marking a question (or answer) as low quality gets disputed very often. So I think I must have some sort of a misunderstanding on when to use them.  For example, I marked the question  as low-quality, which then got disputed.  The question later got closed due to "put on hold as unclear what you're asking", so there was definitely something wrong with this question. However for me it was clear what he was asking; it was just hard to understand what he was doing.  So when should I use them then? If I find a question to be really low quality, but everything seems fine to me, why not use this flag?
The "Very Low Quality" (VLQ) flag reason is really annoying me.   It is the analog to the mommy mommy flag (which my 1.5-year-old child raises quite frequently). It provides no real clue as to what the flagger really wants us to do with the post.   Does she want us to close the question?  Does she want us to edit it into shape?  Does she want us to delete the question?  Does she want us to ban the user?  Does she want us to jump around and have a party?    Should we get rid of this flag reason?   Is it just me? Should and can we replace it with something better?     EDIT I accepted Jeff's answer. The tougher wording has heavily decreased the amount of very low quality flags. We now see only about 25% of the amount we used to and they are usually valid. Additionally, we now only allow very low quality on posts with a score of 0 or lower.
I'm working on moving my current server setup to newer hardware, and migrating from ubuntu karmic koala to lucid lynx. Currently i'm using gw6c (compiled from the gogo6 website, as opposed to the version from the repositories) to get ipv6 access for my systems. On the karmic koala system, i used simple init.d script to get the ipv6 client started   #! /bin/sh /usr/local/gw6c/bin/gw6c -f /usr/local/gw6c/bin/gw6c.conf   I figured since this runs at any runlevel, it should translate to  respawn console none  start on startup  stop on shutdown  script exec /usr/local/gw6c/bin/gw6c -f /usr/local/gw6c/bin/gw6c.conf emit free6_ipv6_started end script   this works fine started from initctrl, but it apparently fails to start when it boots. - its status being stop/waiting. It works fine (and respawns) when started otherwise.Any ideas on where i'm going wrong, and what would be the appropriate 'start on' arguement?  EDIT: the exact error is 'init: gw6c main process (xxx) ended with status 8' followed by the process respawning , with xxx being a PID i suspect. I'm also half suspecting this is cause gw6c starts before networking does, and i need my eth0 up before gw6c is I tried to prove this question but without a success: Let $K_1$ and $K_2$ be definable sets, prove that $K_1\cup K_2$ is definable.  What I tried to do is to assume: $K_1=Ass(X)=\{ v|v \vDash X \}$ and $K_2=Ass(Y)=\{ v|v \vDash Y \}$ and to say that $K_1\cup K_2=Ass(X \cap Y)$. but I couldn't prove that: $K_1\cup K_2 \supseteq Ass(X \cap Y)$ .   I would like to get help with this proof How to get reflections from 3D to show up on real images Andrew Price released a nice tutorial on how to nicely put 3D objects on top of a real image:   I have seen images, such as this:  which have 3D elements over images and have actual reflections from the object showing up on the wet asphalt.  How can this effect be achieved using render passes (or any other way)?  Here is another example (from ): Sum of two squares modulo p I have heard somewhere that for all primes $p$, for all $k$, there exist $x, y$ s.t. $x^2 + y^2\equiv k \pmod{p}$? I recall that the proof is very elementary, but I can't remember such a proof. How would one prove this? One way is to use Cauchy-Davenport, but I don't think that this is the simplest solution. How to make unit selection circles merge? I would like to know how to make this effect of merged circle selection. Here are images to illustrate:     Basically I'm looking for this effect:    How the merge effect of the circles can be achieved ? I didn't found any explanation concerning this effect. I know that to project those texture I can develop a decal system but I don't know how to create the merging effect.  If possible, I'm looking for purely shaders solution. I was trying to fix my audio problems using suggestions from . One of the comments suggested I execute this and only this:     killall pulseaudio   So I did, and I restarted my computer just in case, but all it did was remove my soundcard and I still couldn't hear anything. Now I don't know how to get my soundcard back.  Please note that I am extremely inexperienced when it comes to Ubuntu. How to obtain $f(x)$, if it is known that $f(f(x))=x^2+x$? The replier's From email address is modified when Marketing Cloud RMM forwards an email to the routing address We've set up RMM to forward replies to routing address and SFMC adds our subdomain to the replier's From email address (i.e. replier@gmail.com becomes replier.gmail.com@oursubdomain.com). However, their reply-to address remains the same.  It seems like DMARC Forward Reply Rewrite is enabled, but we were advised not to touch it as that may affect email deliverability.  Is there any workaround to this? And what will happen if we disable this feature? What does the caret (‘^’) mean in C++/CLI? I just came across this code and a few Google searches turn up no explanation of this mysterious (to me) syntax.  Hashtable^ tempHash = gcnew Hashtable(iterators_);  IDictionaryEnumerator^ enumerator = tempHash-&gt;GetEnumerator();   What the heck does the caret mean? (The gcnew is also new to me, and I asked about that .) Select node without namespace with XPath Number of ways in which $5$ boys and $5$ girls are ordered in such a way that exactly $4$ girls stand consecutively in the queue Let $n$ be the number of ways in which $5$ boys and $5$ girls can stand in a queue in such a way that all the girls stand consecutively in the queue. Let $m$ be the number of ways in which $5$ boys and $5$ girls in such a way that exactly $4$ girls stand consecutively in the queue. Then find $\frac{m}{n}$.    I could find $n$ correctly as $6!\times 5!$, but I am finding $m$ as $7!\times 4!\times \binom{5}{1}$, but the book says $m$ is $6!\times 4!\times 5\times \binom{5}{4}$. Where have I gone wrong? What is the proper logic to find $m$? How to solve 5x5 grid with 16 diagonals? How to write the word 'smart phone' in a research article? Smart phone Smartphone Smart-phone   I am really looking for some help regarding hyphen (-) issue in text found in research article. Is there any standard method how to use it? Currently the migration path options from SO are:   Meta.SO Serverfault SuperUser Webmasters Programmers   If we are going to be limited to 5 options, shouldn't we periodically re-evaluate which sites are most appropriate?   (10k only) indicate that the top sites for the last 90 days are:   Superuser - 564 - 5% rejected ServerFault - 278 - 10% rejected Tex - 157 - 13% rejected DBA - 144 - 2% rejected Sharepoint - 127 - 1% rejected Programmers - 126 - 26% rejected CodeReview - 100 - 3% rejected Meta.SO - 81 - 18% rejected Wordpress - 75 - 14% rejected Webmasters - 58 - 8% rejected   All these migrations for sites out of the list required a flag and a moderator to process it.  This is just extra work for the mod team!  If we had the top 5 destinations by volume as the 5 options, we would have had a net savings of 160 flags to be processed.  I suspect we will actually see these numbers change quite a bit if it's easier for users to put questions in the appropriate spot without moderator intervention.  Can we have a review of the migration path list based on migration volume?  Think of the flags! Why is it dangerous to initialize weights with zeros? Is there any simple example that demonstrates it? FFmpeg - Apply blur over face Winsock2's select() on fd 0 (stdin) fails Bijection between an open and a closed interval Recently, I answered to this problem:     Given $a&lt;b\in \mathbb{R}$, find explicitly a bijection $f(x)$ from   $]a,b[$ to $[a,b]$.   using an "iterative construction" (see below the rule).  My question is: is it possible to solve the problem finding a less exotic function?  I mean: I know such a bijection cannot be monotone, nor globally continuous; but my $f(x)$ has a lot of jumps... Hence, can one do without so many discontinuities?    W.l.o.g. assume $a=-1$ and $b=1$ (the general case can be handled by translation and rescaling). Let:  (1) $X_0:=]-1,-\frac{1}{2}] \cup [\frac{1}{2} ,1[$, and  (2) $f_0(x):=\begin{cases}     -x-\frac{3}{2} &amp;\text{, if } -1&lt;x\leq -\frac{1}{2} \\ -x+\frac{3}{2} &amp;\text{,     if } \frac{1}{2}\leq     x&lt;1\\ 0 &amp;\text{, otherwise} \end{cases}$,  so that the graph of $f_0(x)$ is made of two segments (parallel to the line $y=x$) and one segment laying on the $x$ axis; then define by induction:  (3) $X_{n+1}:=\frac{1}{2} X_n$, and  (4) $f_{n+1}(x):= \frac{1}{2} f_n(2 x)$  for $n\in \mathbb{N}$ (hence $X_n=\frac{1}{2^n} X_0$ and $f_n=\frac{1}{2^n} f_0(2^n x)$).  Then the function $f:]-1,1[\to \mathbb{R}$:  (5) $f(x):=\sum_{n=0}^{+\infty} f_n(x)$  is a bijection from $]-1,1[$ to $[-1,1]$.  Proof: i. First of all, note that $\{ X_n\}_{n\in \mathbb{N}}$ is a pairwise disjoint covering of $]-1,1[\setminus \{ 0\}$. Moreover the range of each $f_n(x)$ is $f_n(]-1,1[)=[-\frac{1}{2^n}, -\frac{1}{2^{n+1}}[\cup \{ 0\} \cup ]\frac{1}{2^{n+1}}, \frac{1}{2^n}]$.  ii. Let $x\in ]-1,1[$. If $x=0$, then $f(x)=0$ by (5). If $x\neq 0$, then there exists only one $\nu\in \mathbb{N}$ s.t. $x\in X_\nu$, hence $f(x)=f_\nu (x)$. Therefore $f(x)$ is well defined.  iii. By i and ii, $f(x)\lesseqgtr 0$ for $x\lesseqgtr 0$ and the range of $f(x)$ is:  $f(]-1,1[)=\bigcup_{n\in \mathbb{N}} f(]-1,1[) =[-1,1]$,  therefore $f(x)$ is surjective.  iv. On the other hand, if $x\neq y \in ]-1,1[$, then: if there exists $\nu \in \mathbb{N}$ s.t. $x,y\in X_\nu$, then $f(x)=f_\nu (x)\neq f_\nu (y)=f(y)$ (for $f_\nu (x)$ restrited to $X_\nu$ is injective); if $x\in X_\nu$ and $y\in X_\mu$, then $f(x)=f_\nu (x)\neq f_\mu(y)=f(y)$ (for the restriction of $f_\nu (x)$ to $X_\nu$ and of $f_\mu(x)$ to $X_\mu$ have disjoint ranges); finally if $x=0\neq y$, then $f(x)=0\neq f(y)$ (because of ii). Therefore $f(x)$ is injective, hence a bijection between $]-1,1[$ and $[-1,1]$. $\square$ How do we know what type of gauge field to add to a theory? I've been watching  and in one lecture, he discusses a very simple gauge theory.  We have a complex scalar field $\phi(x)$ with Lagrangian  $$\mathscr{L} = \partial_\mu \phi^* \partial^\mu \phi - V(\phi^* \phi)$$  and we want to do the gauge transformation $\phi(x) \to e^{i\theta(x)} \phi(x)$.  But the derivative term in the Lagrangian is not invariant if $\theta$ varies from place to place, so we add a new vector field $A_\mu$, and define it to transform like $$A_\mu \to A_\mu + \partial_\mu \theta$$ under the gauge transformation.  Then we change the Lagrangian to use covariant derivatives $D_\mu = \partial_\mu + iA_\mu$ instead of ordinary ones; now when the gauge transform is done all the derivatives of $\theta$ cancel out, and the result is invariant.  My question is: was there any freedom in choosing to add a vector field, with that specific gauge transformation law?  Are there any other ways we could get gauge invariance here—by using a different transformation law, or with a different type of field (say, a scalar, tensor, or spinor) with some other transformation law?  More generally, how can we tell what type of field is needed and what its gauge transformation law should be, to get some particular Lagrangian to be invariant under some particular gauge transformation? How do you iterate backwards through an STL list?