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launch sms application with an intent What are the different endings in Dishonored? Why does extended piston become stuck if air above it is powered? To reproduce this problem make layout like this and then power the piston. Once the piston is extended, removing the power source will not retract it:    Since piston head block is transparent, I see no explanation for this. Also, even if it wasn't transparent, it still wouldn't power a piston. For example this doesn't power the piston:    But if I place redstone and THEN place the piston, it's powered:    It also remains powered no matter what, as long as there isn't a block update event:    I suppose this behavior can be used to make a block update detector, but in my case this sucks. I wanted such setup for automatic smooth stone generator (without any clocks):    But due to this bug, it won't work - the piston will be always extended. An infinite subset of a countable set is countable In my book, it proves that an infinite subset of a coutnable set is countable. But not all the details are filled in, and I've tried to fill in all the details below. Could someone tell me if what I wrote below is valid?  Let $S$ be an infinite subset of a countable set $T$. Since $T$ is countable, there exists a bijection $f: \mathbb{N} \rightarrow T$. And $S \subseteq T = \{ f(n) \ | \ n \in \mathbb{N} \}$. Let $n_{1}$ be the smallest positive integer such that $f(n_{1}) \in S$. And continue where $n_{k}$ is the smallest positive integer greater than $n_{k-1}$ such that $f(n_{k}) \in S$. And because $S$ is infinite, we continue forever. Now consider the function $\beta : \mathbb{N} \rightarrow S$ which sends $k \rightarrow f(n_{k})$.   So in order for $S$ to be countable, $\beta$ would have to be a bijection. $\beta$ is injective since if $f(n_{k}) = f(n_{j})$, then $n_{k} = n_{j}$ because $f$ is injective. We also can conclude $k = j$ because the various $n_{i}$ chosen were strictly increasing.   Edit: $\beta$ also needs to be surjective. So for every $r \in S$ there exists a $q \in \mathbb{N}$ such that $\beta(q) = f(n_{q}) = r$. We know that $f^{-1}(r) \in \{n_{1}, n_{2}, ..., n_{k} ... \}$, so we can let $f^{-1}(r) = n_{d}$. Thus $\beta(d) = f(n_{d}) = r$. Solve $f(x)f(2x^2) = f(2x^3+x)$ How to normalize data to 0-1 range? I am lost in normalizing, could anyone guide me please.  I have a minimum and maximum values, say -23.89 and 7.54990767, respectively.  If I get a value of 5.6878 how can I scale this value on a scale of 0 to 1. How to text-to-speech output using command-line? How to get speech output from entered text by using command-line?  Also facility to change speech rate, pitch, volume etc using simple command. How do I mount a hibernated NTFS partition? User systemd PulseAudio service Not much else to this question: I would like to know what I need to create a PulseAudio systemd service to be run with systemd --user.  I've tried a few different things, including various user services from Google, but I can't get it set up just right between dbus and everything else required for it.  Update: It seems to work ok with graphical sessions, but I'm still having some inconsistancy with dbus when logging in through console.  But PulseAudio is starting rather reliably, so I'm not overly concerned. I have a table with a varchar(max) column that stores text ~4-10KB in size. Data is just being added - never modified, it is a log. My problem is that a table grows too large in size - I would like to keep about 1000000 rows in this table.   I am limited to SQL Server Express and hit the limit on database size. Compression is not available in SQL Server Express and I afraid that Filestream may not be appropriate and the number of files would be too big.   What is the best solution in this case? Prove that "injective function $f:X\to Y$ exists" and "surjective function $g:Y\to X$ exists" is logically equivalent. Prove that   injective function $f:X\to Y$ exists surjective function $g:Y\to X$ exists   is equivalent.$(x,y\neq \emptyset)$   My approach  First,prove left to right.  If injective function $f:X\to Y$ exists, a set A can be defined like this. $$A=\{f(x):x\in X\}$$ then it is sure that $A\subset Y$, and $|A|=|X|$  And, if we define function $g$ like $$g(y)=\begin{cases}x &amp; \text{if $y\in A$ and } f(x)=y,\\x_0&amp; \text{if }y\notin A.\end{cases}$$ ($x_0$ is a fixed element of $X$)  Then, $g$ is a surjective function, so surjective function $g:Y\to X$ exists.  Now,prove right to left.  If surjective function $g:Y\to X$ exists, $\forall x \in X$, we can select a element $y\in Y$ that $g(y)=x$, which is clear that we can pick a different $y\in Y$ for $\forall x \in X$  if we define function $f$ as $f(x)=\text{one element in Y that g(that element)=$x$}$, f is a injective function, so injective function $f:X\to Y$ exists.    But, I am not sure that my approach is right, especially on proving right to left. Please check whether my approach is right, ways to improve it,  and some other ways of proving this question. Boot failure : No DEFAULT or UI configuration directive found! If there is a wizard that, by some means, has acquired proficiency in the use of shields, can said wizard use his shield in one hand and his arcane focus quarterstaff in the other and still cast somatic component spells?  Since the quarterstaff has been crafted to also be an arcane focus, and the Sage Advice Compendium states that you can provide the somatic components of spells with the same hand wielding the arcane focus, in this case an arcane focus quarterstaff, could that wizard still be able to perform somatic component spells while using a shield in the other?  I understand War Caster is the feat that allows you to cast somatic component spells even while using a shield and weapon or two weapons in each hand, but is this a simple way around those rules seeing as this "weapon" is your arcane focus? I would like to believe it is not, but I'm not fully fluent in the ways of spellcasting. How to solve ~(P → Q) : P & ~Q by natural deduction? I'm trying to solve the following by natural deduction:  ~(P → Q) : P &amp; ~Q  It's a trivial problem if identities are used, as can be seen by the following:   {1}      1.   ~(P → Q)                    Prem. {1}      2.   ~(~P ∨ Q)                   1 Mat. Impl. {1}      3.   ~~P & ~Q                    2 De Morgan's {1}      4.   P & ~Q                      3 DNE   Starting from step 2 of the previous proof, it's also fairly straightforward to reach a conclusion using natural deduction; i.e. starting from ~(~P ∨ Q), which is an identity of ~(P → Q):  ~(~P ∨ Q) :  P &amp; ~Q   {1}      1.   ~(~P ∨ Q)                   Prem. {2}      2.   ~P                          Assum. {2}      3.   ~P ∨ Q                      2 ∨I {1,2}    4.   ~(~P ∨ Q) & (~P ∨ Q)        1,3 &I {1}      5.   P                           2,4 RAA {6}      6.   Q                           Assum. {6}      7.   ~P ∨ Q                      6 ∨I {1,6}    8.   ~(~P ∨ Q) & (~P ∨ Q)        1,7 &I {1}      9.   ~Q                          6,8 RAA {1}      10.  P & ~Q                      5,9 &I   I also found that it's fairly easy to conclude ~Q:   {1}      1.   ~(P → Q)                    Prem. {2}      2.   Q                           Assum. {3}      3.   P                           Assum. {2,3}    4.   P & Q                       2,3 &I {2,3}    5.   Q                           4 &E {2}      6.   P → Q                       3,5 CP {1,2}    7.   ~(P → Q) & (P → Q)          1,6 &I {1}      8.   ~Q                          2,7 RAA ... {1,2}    12.  P & ~Q   In spite of how easy it looks, I can't figure out how to do it. I believe it can be done in 12 steps. Thanks! Is the density operator a mathematical convenience or a 'fundamental' aspect of quantum mechanics? In quantum mechanics, one makes the distinction between mixed states and pure states.  A classic example of a mixed state is a beam of photons in which 50% have spin in the positive $z$-direction and 50% have spin in the positive $x$-direction.  Note that this is not the same as a beam of photons, 100% of which are in the state $$ \frac{1}{\sqrt{2}}[\lvert z,+\rangle + \lvert x,+\rangle]. $$ It seems, however, that at least in principle, we could describe this beam of particles as a single pure state in a 'very large' Hilbert space, namely the Hilbert space that is the tensor product of all the $\sim 10^{23}$ particles (I think that's the proper order of magnitude at least).  So then, is the density operator a mathematical convenience, or are there other aspects of quantum mechanics that truly require the density operator to be a 'fundamental' object of the theory?  (If what I mean by this is at all unclear, please let me know in the comments, and I will do my best to clarify.) When do exactly we choose to use have or take? They look sort of having similar context     I have a bath      I take a bath   And by the way, I saw that      I take a chance   Can't be transformed to     I have a chance   Can it?   *Moreover, If there's a verb used like the bath, it means the bath is no more a verb but a noun right?  Also very important is that , can they be applied randomly to *a verb or a noun?  Like :  *  - I take a choose   Or    I take an option ( this one makes sense but I mean in the context of delexical verb)   ?  *EDIT: My mistake , bath is actually not "a verb transformed to a noun, but its absolutely a noun."  So my concerns would be if they can be used interchangeably? And be applied to random noun? Can you use a readied action to teleport out of the way of a weapon attack? Let's say that a player uses the ready action and says "If a character aims at me with a weapon, I'll use Teleport to move behind cover."  What happens? Does this work? If not, please cite the rules to explain why not. Display sheet name in Google Spreadsheets Is there any way to display a sheet name as the result of a function in a Google Spreadsheets (exactly like , except in Google Sheets)? Is there an option to create and Opt-in a user in the same API call? Coming from SketchUp one of the tools I miss most is the circle draw tool which lets you draw a circle on any face and then you could extrude it in/out. The only tool that I know of that comes close to this is the knife tool but this would be very impractical to cut out with the knife  Example:    Is there any tool in blender that can match this function? also how could you do this with squares?