Recently I was looking through developer options on my Moto X and I was interested by the option to do runtime in either dalvik or ART. After doing some research I decided to change my runtime to ART. I was wondering if this was a good choice or if I should continue running my phone on Dalvik. Let me know what you think.
ART has better performance and better battery life on paper at least.  Should I consider switching to it on my moto g? What are the results I can expect? What precautions should I take? Is it easy to turn back to dalvik?
More than one time dimension We know that space-time dimensions are 3+1 macroscopically, but what if 2+2?  Obviously it is tough to imagine two time dimensions, but mathematically we can always imagine as either having two parameters $t_1$ and $t_2$ or else in Lorentz matrix $$\eta_{00} = \eta_{11} = -1$$ and, $$\eta_{22} = \eta_{33} = 1.$$  Is there any physical reason for not taking this, like the norms become negative or something else? Proving $R-S$ contains a prime ideal when $S$ is a multiplicative set I'm mainly trying to prove that      If $0\not \in S\subseteq R$ is a multiplicative subset of a   commutative ring $R$ with identity. Then $R-S$ contains a prime ideal.   Now, by using Zorn's lemma, one can show that $R-S$ contains a maximal ideal I. In fact, I is a prime ideal as well, so we need to show this.To do that, I assumed the opposite, i.e $\exists a,b \in R$ s.t $ab\in I$, but $a,b \not \in I$. Then looked each case. However, I stuck when $a,b \in R-[S\cup I]$. The following is what I've done for this case;  Proof:  In this case, our assumptions are $a,b\in R-[S\cup I]$ and $I + (a) = J$, where $J$ is an ideal in $R$ and $J \cap S \not = \emptyset$.   These imply that; $S^{-1}(I + (a)) = S^{-1} J$, but since $J \cap S \not = \emptyset$, by a theorem that we proved in the class, $S^{-1} J = S^{-1}R$.Therefore, it must be true that $\exists (i\in I, r\in R, s_1,s \in S)$ s.t  $$\frac{i+ra}{s_1} =\frac{1}{s}$$ but this implies $\exists s_2 \in S$ s.t  $$ s_2(is + sra -1 s_1) = 0 \quad \Rightarrow \quad (s_1 s )i + (s_1 s r) a = (s_1 s )1,$$ ,so $$(b s_1 s )i + (s_1 s r) ab = (s_1 s )b.$$ Observe that by our assumption $ab\in I$, so $(s_1 s)b \in I$.  Now, at the point, I don't know how to proceed, so I would appreciate any help or hint because I stuck at this point for over a week now.  Edit  Thanks for the great answers; however, but I'm specifically interested in how to continue from the point where I stuck. Changing parameter values according to other parameter choice in Processing script using QGIS 3? Honestly, navigation through Area 51 is a bit confusing and seems to be much different from the other SE sites.    The feature that is very lacking is the footer with all the sites you are subscribed to so that a user can easily return to a different SE site. This one really bugs me.  Can this be easily fixed?  Also, unlike all the other sites there is no "meta" site to add comments specifically about that site.    Again Area 51 seems out of sync with the other SE sites. I have over 400 open-ended responses. The total has been coded by one rater, and a 10% identical sample of the total has been coded by 2 raters. There is a choice of 14 nominal codes, of which the rater can use any number per case, though most cases have between 1-4 codes. Which statistical test is the most appropriate to determine if I have inter-rater agreement on the 10% sample?   Also am I right in assuming that I will need to treat each code as a variable and look for the level of agreement for each variable?   I'm using SPSS 21.  Much thanks, Kellie Can the system leave a deletion reason when it deletes a question? they were question banned; it turned out it (in part) it was because of deleted questions.   was deleted by the Community user.  I'm aware it will delete old posts that have a negative score, but for a minute or two, I was really puzzled as to why it was unilaterally deleted without some notation left.  Can we have the Community user leave a notation in a post's history as to why the post was deleted?    Something like the following would suffice:     Post deleted automatically due to having negative votes and no activity for three months Inverse public-key encryption I'm in the need of an inverse public-key encryption method where you encrypt with the private key and decrypt with the public key. In other words an asymetric encryption where you can't generate the encryption key from the decryption key. Does such encryption exist? If yes which, and which would be the fastest?  If nothing like that exists I would be glad if you helped me out with my situation:    I need to authenticate users logging in from my Java application in the way that the response from the server cannot be falsified and thus giving the user full access to the application. Using public key encryption was an obvious method, I would decrypt data received from the client to confirm the credentials. This won't work because if someone manages to decompile and get the private key from the code, they would be able to generate the public key, which would make them able to crack my application again. So any ideas how I can get this secure? If $ a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}<1$ If $\displaystyle a,b,c\in \left(0,\frac{\pi}{2}\right)\;,$ Then prove that $\displaystyle \frac{\sin (a+b+c)}{\sin a+\sin b+\sin c}&lt;1$   $\bf{My\; Try::}$ Using $$\sin(a+\underbrace{b+c}) = \sin a\cdot \cos (b+c)+\cos a\cdot \sin (b+c)$$  $$ = \sin a\cdot (\cos b\cos c-\sin b\sin c)+\cos a(\sin b\cos c+\cos b\sin c)$$  $$ = \sin a\cos b\cos c-\sin a\sin b\sin c+\cos a \sin b\cos c+\cos a\cos b\sin c$$  Now how can i solve it after  that , Help required, Thanks Is a finite group uniquely determined by the orders of its elements? I put 4 different layers into my ArcMap document, but only one of them will show up on the map! I should be seeing the other 3 layers as well. When I do a Zoom to layer, I can see one layer, but I cannot figure out how to make the other 3 show up at once on top of each other.  Before I opened ArcMap, I used ArcCatalog where I right-clicked 3/4 of the shapefiles and  clicked Properties &rarr; Coordinate system &rarr; Import and then clicked on my Trees layer, which has the coordinate system I want the other 3 to have. But when I opened ArcMap the 4 layers did not overlap, nor could I move them above or below each other. What is the Taylor series of $$\ln \frac{1-x^2}{1+x^2}$$   ?   I started by evaluating the first derivatives but the more I go, the more complicated they are and I can't identify a pattern. Autosegmental Representation in tikz-qtree This is what I want to achieve    and this:    But all I've been able to make with tikz-qtree is the following:  \documentclass[12pt,a4]{article} \usepackage[utf8]{inputenc} \usepackage{tipa,tikz,tikz-qtree} \begin{document}  \begin{tikzpicture} [baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ p ] ] [.R [.$\mu$ a ] [.$\mu$ : ] ] ] \end{tikzpicture} \begin{tikzpicture} [baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ p ] ] [.R [.$\mu$ a ]  ] ] \end{tikzpicture}   And this:  \begin{tikzpicture}[baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ l ] ] [.R [.$\mu$ a ] [.$\mu$ l ] ] ] \end{tikzpicture} \begin{tikzpicture} [baseline] \tikzset{frontier/.style={distance from root=90pt}} \Tree [.$\sigma$ [.O [ : ] ] [.R [.$\mu$ a ] [ l ] ] ] \end{tikzpicture} \end{document}   Which renders:    and    The important part here is the linking from the leaves to more than one parent (it would be nice if I had the option of making the line dashed too). Now, pst-asr doesn't quite achieve what I want it to. It's important that I have the moras ('mu's). (Also it's a pstricks package, which means I must typeset it in DVIPSPDF.) I have tried a bit of the forest-package (especially forest-GP1), but I can't make that work as I want either. Lastly, some have recommended xyling, but I find that package very difficult to use.  Any recommendations?  Thanks. I recently got a Windows 10 laptop. Opposite to my Macbook, I must swipe right-to-left to go to the right virtual desktop. Is there a way to reverse this, so I swipe left-to-right to go to the right virtual desktop? The problem is, it seems when I enable a disabled module, the module's schema version in the system table is automatically updated to the latest version supplied by hook_update_N.  So my current module with the schema version -1 is modified to 7106 when I enable the module, and no hook_update_N functions are run.   So when I come to the next step of running database updates, no hook_update_N functions are run because the module thinks it's already the correct version.  The only method I can see is to manually set the module version in hook_enable, however I'll then need to update the module code to remove the hook_enable to make sure it doesn't run if the module is ever re-enabled in the future, so obviously this approach is a bit dirty.  I'm hoping there's a better solution? I am sitting inside a closed room with a cement ceiling just above my head. The pressure of the atmosphere is exerted on the ceiling from above(outside). But solid cement not being a fluid does not transport that pressure into my room. But i still feel the atm pressure which has no way to be transported into the room.I do not boil to death or anything. How does this happen? Today, when you accept an answer you did not provide, you get a +2 reputation, however, if you accept your own answer, you get +0.  I expect the reason for this is to stop people gaming the system by asking and answering their own questions repeatedly.  However, this has the downside of providing a disincentive of accepting your own answer when it seems the best and instead accepting an inferior one in order to get the reputation bonus.  Therefore, I propose that the accept your own answer is only worth zero when there are no answers provided by anyone other than yourself.  I feel this keeps the intent of preventing gaming the system for reputation, while removing disincentive from picking the best answer, even if it is your own. Do animagi get to pick their animal? Say that I want to become an Animagus, and am really hoping to be a certain animal (say, a stag). I study a whole bunch and manage to perfect the technique. Do I get to pick the animal I wind up as, so that I wind up being a stag like I want? Or is it something where I just find out what I turn into (and if I wind up as a dung beetle, tough luck)?    This is not a duplicate of  because that question is about which animals an Animagus can potentially transform into. Mine is about if I can decide which potential animal I wind up being. PHP: Return all dates between two dates in an array I can't seem to access instance members of the surrounding class from inside an enum, as I could from inside an inner class. Does that mean enums are static? Is there any access to the scope of the surrounding class's instance, or do I have to pass the instance into the enum's method where I need it?  public class Universe {     public final int theAnswer;      public enum Planet {         // ...         EARTH(...);         // ...          // ... constructor etc.          public int deepThought() {             // -&gt; "No enclosing instance of type 'Universe' is accessible in this scope"             return Universe.this.theAnswer;         }     }      public Universe(int locallyUniversalAnswer) {         this.theAnswer = locallyUniversalAnswer;     } } I am using Ubuntu 12.10.  Please outline the different ways I can upgrade to 13.04.