Is there a mathematical method to figure out what room the Mysterious Stranger will go into next after we previously got him?
When ever I hear the music to indicate the Mysterious Stranger has arrived, I can never find him. It is almost like he is not in the vault.  What is a good way to catch the Mysterious Stranger? Where does he usually show up?
Appositives with possessiveness? I would like to say something like     The boy, Adam's, favorite toy was a bike.   What is the proper way to say this? how to glob every hidden file except current and parent directory I want to glob every hidden file and directory, but not the current (.) and parent directory (..).  I am using bash.  Observe current behaviour:  $ ls -a .  ..  ...a  ...aa  ..a  ..aa  .a  .aa  .aaa  a $ echo * a $ echo .* . .. ...a ...aa ..a ..aa .a .aa .aaa   I would like .* to behave like this  $ echo .* ...a ...aa ..a ..aa .a .aa .aaa   There is the shell option dotglob  $ shopt -s dotglob   that works in a way; now I can use * to glob everything (hidden or not) but not . and ..  $ echo * ...a ...aa ..a ..aa .a .aa .aaa a   but now I can't differentiate between hidden or not. Also, .* still globs . and ..  $ echo .* . .. ...a ...aa ..a ..aa .a .aa .aaa   Is there a way to make .* not expand to . and ..? Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python? If $\Gamma$ is satisfiable then $\Gamma$ is consistent We say that $\Gamma$ is consistent if for no $\phi$, then both $\Gamma\vdash \neg\phi$ and $\Gamma\vdash\phi$. We say that $\Gamma$ is satisfiable if there is a structure $M$ s.t. $\models_M\gamma[s]$ for alle $\gamma\in\Gamma$.  I want to show that if $\Gamma$ is satisfiable, then $\Gamma$ is consistent.  I'm thinking we assume that $\Gamma$ is not consistent, and then show that $\Gamma$ is not satisfiable. So if $\Gamma$ is not consistent then both $\Gamma\vdash \neg\phi$ and $\Gamma\vdash\phi$ for some $\phi$. But by soundness theorem then $\Gamma\models\neg\phi$ and $\Gamma\models\phi$, meaning for structure $M$, we have $\models_M\neg\phi[s]$ (whereas $\not\models_M\phi[s]$) and $\models_M\phi[s]$. So would this mean that $\Gamma$ is not satisfiable? I'm a little confused, because wouldn't this instead be a contradiction? What tips and advice do you have for photographing the Aurora Borealis? A friend has been kind enough to lend me a DSLR for a few months.  It's my first foray into DSLR photography, or indeed any kind of photography beyond simply aiming my Canon IXUS at something and clicking the button.    Over the last few weeks, I've been busy studying, practising and trying to learn as much as possible ahead of a planned trip to Norway where I hope to see, and get some photos of, the Aurora Borealis.  I currently have:   Canon Rebel Xsi (450D) Canon 50mm f/1.4 L  Canon 18-55m kit lens Aluminium full height tripod with ball head Cable release Spare battery &amp; memory cards   I'm prepared to spend the night wrapped up outdoors, in the hope that they will make an appearance. I'd like some advice on how to get the most out of this kit for the trip:   How should I go about setting up to get pics of the Aurora?  How might I achieve optimum focus? From what I've read, I'll need to use the tripod and cable release, with IS turned off, on continuous shooting mode with long exposure times. But I'd like some more specific tips in terms of which aperture and ISO settings might work best? Should I use one of the lenses above, or would it be worthwhile renting a better, wide angle lens; if so, which would you recommend? Take the manifold: The graph of $|x|$ on $(-1,1)$, with the induced topology from $\mathbb{R}^2$. This is a topological manifold, which is homeomorphic to $(-1,1)$ by projection. Is it a differentiable manifold? I believe it is, because we can take an atlas with only the projection, and then we will have only one transition map which is the identity, and therefore differentiable. But I'm not sure I get the definition of a differentiable manifold right...  Thanks! I'm trouble with exercise 1.8 of Carroll's Space-Time and Geometry:     If $\partial_\nu T^{\mu \nu} = Q^\mu$, what physically does the spatial vector $Q^i$ represent? Use the dust energy momentum tensor to make your case.    First of all, since the momentum energy tensor of dust $T^{\mu \nu}=\rho U^\mu U^\nu$ is symmetric, one has the energy-momentum conservation law $$ \partial_\nu T^{\mu \nu} = \partial_\nu T^{\nu \mu}  =0.$$ A vector $Q$ is constant in any frame related by a Lorentz transformation, while this isn't the case for the components $Q^\mu$. But here $Q=0$ and hence $Q^\mu=0$ also for every component $\mu =0,\cdots,4$?   Even if a straightforwardly apply the equation: $$ Q^i = \partial_\nu T^{i \nu} = (\partial_\nu \rho) U^i U^\nu + \rho(\partial_\nu U^i)  U^\nu + \rho U^i(\partial_\nu  U^\nu ),$$ doesn't seems to yield something...   Is my first reasoning right ? Otherwise: can someone give me a hint ?  Thanks I'm trying modeling things like this using oracle:  An account can have 1~5 members  So I'll have a account table and a member table with a FK of account. But is that possible to check if the account have less than 5 members?  I've tried constraint and trigger but oracle doesn't allow sub queries in constraint or condition of trigger like  CHECK( SELECT ...) or IF count(*) &gt;= 5  which make things getting hard. Can CBC ciphertext be decrypted if the key is known, but the IV not? Similar to Lifepod 1993, but not it This film is very similar to Lifepod 1993, but it takes place in a large ship, not a pod.  I can remember three characters... a technical engineer, being the one sabotaging the ship. And also this military cut character (who looks a lot like Kane from Mortal Kombat) and another girl (I think it was long, wavy black hair).  The "plot" twist that sends this one apart, at least back in the 1990´s, is that the ending comes to the fact there is only enough food and oxygen for one survivor and no escape pods or cryotubes, only a device that "could" work as a cryotube. For some reason that I cant remember the "Kane" dude chooses to die instead, by taking poison (which had been previously used on other surviving members). But the girl "tricks" him and drugs him instead, then places him in the "cryotube" and she stays awake to await for the rescue team.  Of course, they get salvaged, and he is still okay, surprised that she tricked him and saved him. Rings a bell for anyone? Is definitely a B-movie. Preventing LibreOffice from using smart quotes (“ ”) instead of (dumb) straight quotes? I use LibreOffice Calc to make lists of programs with different parameters to run on an hpc cluster. For each batch of commands I must do a find and replace on the smart quotes, as they cause errors in command line arguments. It would be nice to cut this step out of my work flow. Does anyone know how I might change the default punctuation used?  Edit: Apparently the title font transforms dumb quotes into smart quotes.  Smart quotes have a different symbol depending on if they are at the start or end (“ ”)  Dumb quotes use the same symbol for the start and the stop ( " ) Can I publish parts of the Ph.D thesis as a paper in a journal? Training on the full dataset after cross-validation? General Information about Eigenvalues for an 3x3 symmetric matrix Prove $(1-a)(1-b)(1-c)(1-d)\geq abcd$ if $a^2+b^2+c^2+d^2=1$ Let $a,b,c,d\geq0$, $a^2+b^2+c^2+d^2=1$ Prove $\displaystyle (1-a)(1-b)(1-c)(1-d)\geq abcd$ I mutiplied both with $\displaystyle (1+a)(1+b)(1+c)(1+d)$ to use $1-a^2=b^2+c^2+d^2$ and try using the cauchy-schwarz and holder but it is doesn't work. At work we have a Domain Controller running Server 2008 R2. Our desktop support group has the ability to join computers to the domain using their network credentials. We are now running into a problem where if the name of the computer that is being joined is the same as an existing computer on the domain, the existing computer will get a "The trust relationship has failed..." error and will not authenticate at all unless someone logs in as a local admin, changes the computer name and rejoins it to the domain.  I want to make it so that Desktop Support can only join computers with unique names to the domain. If the name already exists on the domain the operation should fail. Looking through TechNet it looks like this functionality exists but I can't seem to find out how to enable it.  Any ideas? Is there a more efficient way to clamp real numbers than using if statements or ternary operators? I want to do this both for doubles and for a 32-bit fixpoint implementation (16.16). I'm not asking for code that can handle both cases; they will be handled in separate functions.  Obviously, I can do something like:  double clampedA; double a = calculate(); clampedA = a &gt; MY_MAX ? MY_MAX : a; clampedA = a &lt; MY_MIN ? MY_MIN : a;   or  double a = calculate(); double clampedA = a; if(clampedA &gt; MY_MAX)     clampedA = MY_MAX; else if(clampedA &lt; MY_MIN)     clampedA = MY_MIN;   The fixpoint version would use functions/macros for comparisons.  This is done in a performance-critical part of the code, so I'm looking for an as efficient way to do it as possible (which I suspect would involve bit-manipulation)   EDIT: It has to be standard/portable C, platform-specific functionality is not of any interest here. Also, MY_MIN and MY_MAX are the same type as the value I want clamped (doubles in the examples above). Is there a way to get the raw SQL string executed when calling PDOStatement::execute() on a prepared statement? For debugging purposes this would be extremely useful. How do we know that certain quantum effects are random? Is it really possible to break the speed of light by flicking your wrist with a laser pointer?