MongoDB Array matching 6 from 10
Mongo find documents where array contains x values of given array
Is there any way to kill a Thread? Is it possible to terminate a running thread without setting/checking any flags/semaphores/etc.? Travelling with two different passports On iPhone it's very easy to clone an app (running multiple copies of the app on the same device); e.g.:    Is it possible to do this on Android? Does disabling root login enhance security? Internet Accounts spontaneously forgets passwords and I can't reauthorize them I've had this problem before, but I was able to resolve it that time. macOS 10.13.4, everything up to date. Yesterday, Internet Accounts forgot my LinkedIn password and would not reauthenticate. Today, it forgot my Facebook password.   I've tried:   App specific password. Removing the account and re-adding it. Turning off 2-factor authentication. Deleting associated entries from Keychain. Logging out of iCloud, deleting Keychain entries, removing and re-adding the account, then logging back in.   First of all, what the hell—why does this keep happening?! Secondly, what else can I do to fix this short of reinstalling macOS? How does Mage Armor interact with Unarmored Defense? What is a nicer, less immature saying?  I love using 'A Freudian slip', but that is only applicable in certain situations. How can I create a new extensible symbol? Given an arbitrary thin vertical symbol, can I create a new math delimiter from it (and if so, how)?   Edit: Ideally, I would like to be able to create an extensible symbol, either built out an existing normal symbol, or even made from scratch.  Edit 2: I am pretty sure that what I am after is an explanation of how to create a .tfm file for an extensible symbol (which Werner's answer describes the general structure of, but not how to make one).    For example, let's say I wanted to use the \dagger () or \wr () symbol as a delimiter, and write expressions like    \[\left\ldagger \sum_{n=1}^\infty a_n \right\rdagger\]   where the symbol scaled appropriately. Would there be any distinction between declaring a symmetric delimiter (like \vert), as opposed to one that has inherently different "left" and "right" versions like [ and ]?  Now, even being unaware of the inner workings of TeX, I imagine that there has to be lots of information in the fonts for how a delimiter is to be scaled in different situations, and if I try to make my own delimiter, this information won't be present. Would I need to get deep into FontForge or equivalent to achieve my custom delimiters, or is there an easier way? JMeter : How to record HTTPS traffic? How is money held in escrow for a home purchase from a IRA treated by the IRS? What are neutral density filters and how do I use them to create long exposures in daylight? I've seen a few related posts on the site but nothing specifically on this point:  I'd like to be able to take long exposure photos during daylight (like the examples in ). I'm told that you can do this by using a neutral density filter. So my question is: what is a neutral density filter exactly, and how is it used to achieve that smooth long exposure effect? phpmyadmin not working due to missing extensions I am about to use phpmyadmin on Ubuntu 16.04. But it shows that the page is not working. the error log says error 500   PHP Fatal error:  require_once(): Failed opening required '/usr/share/php/php-gettext/gettext.inc' (include_path='.:/usr/share/php') in /usr/share/phpmyadmin/libraries/select_lang.lib.php on line 477   At first it says that the mysqli|mysql extension is missing. but after sleepless hours I forget what I've been done to it. I am using php7. Proving the identity functions is continuous iff $\tau_1\supseteq\tau_2$ Exercise: Let $\tau_1$ and $\tau_2$ be two topological spaces on a  set $X$. Then $\tau_1$ is said to be a finer topology than $\tau_2$ if $\tau_1\supseteq\tau_2$. Prove the identity function $f:(X,\tau_1)\to (X,\tau_2)$ is continuous if and only if $\tau_1$ is a finer topology than $\tau_2$.  Attempted proof: Let $A\in\tau_1$. If $f$ is continuous then $\exists B\in\tau_2$ such $f(A)\subset B\implies A\subset B$ for any $A\in\tau_1$ such that all open sets of $\tau_1$ are open sets of $\tau_2$ hence $\tau_1\supseteq \tau_2$. If $\tau_1$ is a finer topology than $\tau_2$, for any $B\in\tau_2\implies B\in\tau_1$ which implies that $f^{-1}(B)=B$, then $B\in\tau_1$ which proves $f$ to be continuous. Questions: Is my proof right? If not. What is wrong? What are alternatives? Thanks in advance! Is there an IDictionary implementation that, on missing key, returns the default value instead of throwing? Prove that $a^3+b^3+c^3 \geq a^2b+b^2c+c^2a$ Let $a,b,c$ be positive real numbers. Prove that $a^3+b^3+c^3\geq a^2b+b^2c+c^2a$.   My (strange) proof:  $$ \begin{align*} a^3+b^3+c^3 &amp;\geq a^2b+b^2c+c^2a\\ \sum\limits_{a,b,c} a^3 &amp;\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a^2 &amp;\geq \sum\limits_{a,b,c} ab\\ a^2+b^2+c^2 &amp;\geq ab+bc+ca\\ 2a^2+2b^2+2c^2-2ab-2bc-2ca &amp;\geq 0\\ \left( a-b \right)^2 + \left( b-c \right)^2 + \left( c-a \right)^2 &amp;\geq 0 \end{align*} $$  Which is obviously true.    However, this is not a valid proof, is it? Because I could just as well have divided by $a^2$ rather than $a$:  $$ \begin{align*} \sum\limits_{a,b,c} a^3 &amp;\geq \sum\limits_{a,b,c} a^2b\\ \sum\limits_{a,b,c} a &amp;\geq \sum\limits_{a,b,c} b\\ a+b+c &amp;\geq a+b+c \end{align*} $$  Which is true, but it would imply that equality always holds, which is obviously false. So why can't I just divide in a cycling sum?  Edit: Please don't help me with the original inequality, I'll figure it out. Given a message and DES encrypted form of said message, is it possible to efficiently compute the key used to encrypt the data? Schwarz's Lemma, fixed points question This is from an old qualifying examination question.  If f is holomorphic in the unit disk $D$ and $|f(z)|&lt;1$ for all $z\in D$. Suppose also that $f$ has two distinct fixed points in $D$ then $f(z)=z$ for every $z\in D$  I know that I have to use the Schwarz's lemma and may be make use of Möbius transformations. I tried setting $g(z)=\phi_a\circ f\circ \phi_{-a}$. But that does not seem to work because I don't see why $|g(z)|=|z|$ for some non zero $z$.   Any helpful hints are greatly appreciated.  Edit: Of course the $a$ above is one of the fixed points. Why taking components of a component of a vector is invalid? Discrepancy with lsof command when trying to get the count of open files per process “He is small enough to walk.” But baby wasn't walking so why didn't they use “small” instead of “small enough”?