I wonder if its possible to use Java as a scripting language in unity to create a basic 2d game?
I am fairly new to Unity3D experience , but I have a very good knowledge of Java and Android development. I am really confused that if Java is at all needed for developing android applications? I read somewhere in Unity documentation that adding behaviour to objects in Unity requires use of scripts, and that unity only supports C# , .Net , and Boo scripts. Is there no use of Java at all?
Non-Exchange address in Exchange GAL We just made the switch to Google Apps for part of our domain. We disabled the accounts that are now on Google Apps and mail routes correctly. Our problem is that now we have no way of seeing those users in the Global Address List.  So we have these users in our Active Directory, they have no mailbox on Exchange, and we want to be able to see/search for their addresses in the GAL. Is there any way to do this?  This is Exchange 2007, if that matters. What should I do when I have lenses with different filter thread sizes? Proof that the real numbers are countable: Help with why this is wrong I was just thinking about this recently, and I thought of a possible bijection between the natural numbers and the real numbers.  First, take the numbers between zero and one, exclusive.  The following sequence of real numbers is suggested so that we have bijection.    0.1, 0.2, 0.3, 0.4, 0.5, 0.6, 0.7, 0.8, 0.9, 0.01, 0.02, ... , 0.09, 0.10, 0.11, ... , 0.99, 0.001, 0.002, ... , 0.999, 0.0001, etc.  Obviously, this includes repeats, but this set is countable.  Therefore, the set of all numbers between zero and one is a subset of the above countable set, and is thus countable.  Then we simply extend this to all real numbers and all the whole numbers themselves, and since the real numbers, as demonstrated above, between any two whole numbers is countable, the real numbers are the union of countably many countable sets, and thus the real numbers are countable.    Please help me with this.  I understand the diagonalization argument by Cantor, but I am curious specifically about this proof which I thought of and its strengths and flaws.  Thanks. Could Dark Matter form black hole? Some people speculate that the mysterious dark matter in the universe could be tiny black holes. But on the other side, could dark matter particles attract each other by gravity and finally form a black hole? Since dark matter is even more abundant than normal matter, the dark matter black hole should not be rare. I am travelling from Panama to Los Angeles and around the States for approx 80 days then flying back to Panama and making my way up Central America for approx 3 months before flying back to LA to fly home to Australia.    Will this be OK with an ESTA?    Does it count as spending too long in the States?    I have been told that Panama is not far enough away. What is the difference between the "...", '...', $'...', and $"..." quotes in the shell? Sometimes I see shell scripts use all of these different ways of quoting some text: &quot;...&quot;, '...', $'...', and $&quot;...&quot;. Why are there so many different kinds of quote being used? Do they behave differently or affect what I can do inside them? Wrong place for hyperref link in scaled tikz picture I have a TikZ picture with a node containing a minipage with an itemize environment; each item is a hyperref to other sections of the document.   When scaling the image (for example with \begin{tikzpicture}[scale=.7, transform shape] I've a strange behavior: links in itemize are not in the right place: they present a somewhat vertical offset, as if they don't scale as the rest of image.   Here the MWE:  \documentclass[a4paper]{article} \usepackage{tikz} \usepackage[pdftex]{hyperref} \begin{document}  Here something interesting\label{text:interesting-one} \newpage  Here something interesting, too\label{text:interesting-two} \newpage  and here there are very very interesting stuff\label{text:interesting-three} \newpage  \newpage Now the problem: links rectangles have some offset, making links unusable in case of long itemize  \begin{figure}[htb] \begin{tikzpicture}[scale=.7, transform shape] \node at (0,0) (mynode) { \begin{minipage}{4cm} \begin{itemize} \item \hyperref[text:interesting-one]{First link} \item \hyperref[text:interesting-two]{Second link} \item \hyperref[text:interesting-three]{Third link} \item \hyperref[text:interesting-one]{First link again} \item \hyperref[text:interesting-two]{Second link again} \item \hyperref[text:interesting-three]{Third link again} \end{itemize} \end{minipage} }; \end{tikzpicture} \end{figure} \end{document}   And here a snapshot of the problem    Note: This is evident using the default settings of hyperref package, which draws the red rectangle around links, but it took me some time to found it when only the text is coloured, i.e. when using \usepackage[linkcolor=blue,colorlinks=true]{hyperref} How to make a half cylinder? I am having a tough time building a half cylinder and have not found any tutorial showing me how to do that. How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort? I will keep it short and take only an extract (most important part) of   the old task.      $$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$   What I have done is a lot work and time consuming, I have "simply" solved it. But I think with a lot less work, there would be an easier and faster way. It's just I cannot see it : /  If anyone wants see, here is my long solution which I'm not happy with:  $$\frac{n(n+1)(2n+1)+6(n+1)^2}{6}=\frac{(n^2+2n+n+2)(2n+3)}{6} \Leftrightarrow$$  $$\Leftrightarrow \frac{(2n^3+n^2+2n^2+n)+6n^2+12n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} \Leftrightarrow$$  $$\Leftrightarrow \frac{2n^3+3n^2+n+6n^2+12n+6}{6}=\frac{2n^3+3n^2+6n^2+9n+4n+6}{6} \Leftrightarrow$$  $$\Leftrightarrow \frac{2n^3+9n^2+13n+6}{6}=\frac{2n^3+9n^2+13n+6}{6}$$ Understanding of the theorem that all norms are equivalent in finite dimensional vector spaces The following is a well-known result in functional analysis:     If the vector space $X$ is finite dimensional, all norms are equivalent.     Here is the standard proof in one textbook. First, pick a norm for $X$, say  $$\|x\|_1=\sum_{i=1}^n|\alpha_i|$$ where $x=\sum_{i=1}^n\alpha_ix_i$, and $(x_i)_{i=1}^n$ is a basis for $X$. Then show that every norm for $X$ is equivalent to $\|\cdot\|_1$, i.e.,  $$c\|x\|\leq\|x\|_1\leq C\|x\|.$$ For the first inequality, one can easily get $c$ by triangle inequality for the norm. For the second inequality, instead of constructing $C$, the  is applied to construct a contradiction.   The strategies for proving these two inequalities are so different. Here is my question,      Can one prove this theorem without Bolzano-Weierstrass theorem?   UPDATE:     Is the converse of the theorem true? In other words, if all norms for a vector space $X$ are equivalent, then can one conclude that $X$ is of finite dimension? A real $2 \times 2 $ matrix $M$ such that $$M^2 =  \begin{pmatrix} -1&amp;0 \\ 0&amp;-1-\epsilon  \\   \end{pmatrix}$$ (a) exists for all $\epsilon &gt; 0$. (b) does not exist for any $\epsilon &gt; 0$. (c) exists for some $\epsilon &gt; 0$. (d) None of the above is true.  Attempt: I couldn't think of any theory base to prove / disprove the existence of such a matrix. Could someone please give me a hint on how to go about this problem? Thank you very much for your help in this regard. Python idiom to return first item or None I'm sure there's a simpler way of doing this that's just not occurring to me.  I'm calling a bunch of methods that return a list.  The list may be empty.  If the list is non-empty, I want to return the first item; otherwise, I want to return None.  This code works:  my_list = get_list() if len(my_list) &gt; 0: return my_list[0] return None   It seems to me that there should be a simple one-line idiom for doing this, but for the life of me I can't think of it.  Is there?  Edit:  The reason that I'm looking for a one-line expression here is not that I like incredibly terse code, but because I'm having to write a lot of code like this:  x = get_first_list() if x:     # do something with x[0]     # inevitably forget the [0] part, and have a bug to fix y = get_second_list() if y:     # do something with y[0]     # inevitably forget the [0] part AGAIN, and have another bug to fix   What I'd like to be doing can certainly be accomplished with a function (and probably will be):  def first_item(list_or_none):     if list_or_none: return list_or_none[0]  x = first_item(get_first_list()) if x:     # do something with x y = first_item(get_second_list()) if y:     # do something with y   I posted the question because I'm frequently surprised by what simple expressions in Python can do, and I thought that writing a function was a silly thing to do if there was a simple expression could do the trick.  But seeing these answers, it seems like a function is the simple solution. How to allocate aligned memory only using the standard library? Is this algebraic identity obvious? $\sum_{i=1}^n \prod_{j\neq i} {\lambda_j\over \lambda_j-\lambda_i}=1$ What is the official name for the "special" ASP.NET tags like this:  &lt;%# %&gt; &lt;%= %&gt; &lt;%@ %&gt; &lt;%$ %&gt;   I can't seem to figure out the conceptual or well known name for these, so I'm having trouble searching for more info.  As a bonus, can anyone give me a quick rundown of all of the possible "special tags" and what each one of them does (or point me to a resource)? I was working on a worksheet for class and came across this problem:      Let $\varphi:R\to S$ be an onto ring homomorphism with the property that every ideal of $R$ is principal. Show that the same must be true for $S$.    I honestly do not know how to do this one! Do all droids have a four alphanumeric character name and does this have any significance? Showing that Y has a uniform distribution if Y=F(X) where F is the cdf of continuous X Let $X$ be a random variable with a continuous and strictly increasing c.d.f. function $F$ (so that the quantile function $F^{−1}$ is well-deﬁned). Deﬁne a new random variable $Y$ by $Y = F(X)$. Show that $Y$ has a uniform distribution on the interval $[0, 1]$.  My initial thought is that $Y$ is distributed on the interval $[0,1]$ because this is the range of $F$.  But how do you show that it is uniform? Storing two kinds of point geometry in shapefile or creating multipoint shapefiles? Following , can I have a mixture of two kinds of geometry in a shapefile of type point. Just like I have about shapefiles of type polygon or I should always create a shapefile of type multipoint in order to support multipoints in my layer?    My problem is as you see at , I can't use ogrinfo in order to understand that the geometry type of a multipoint shapefile will be known as wkbMultipoint or wkbPointwhen I use this line of code:    OGRwkbGeometryType GeometryType = poLayer -&gt;GetGeomType();     for a layer of type multipoint before programming.     Will OGR know a layer of type multipoint as wkbpoint or wkbmultipoint?   When we create a shapefile in ArcCatalog, we have point and multipoint, but not multipolyline or multipolygon for polyline and polygon.Why? I have encountered the definition of Cantor set and its property. Cantor set is constructed by removing the middle third open set of each interval .So each time we get some union of closed sets. As $F_0=[0,1]$ $F_1=[0,1/3]\cup [2/3,1]$ $F_2=[0,1/9]\cup [2/9,1/3]\cup [2/3,7/9]\cup [8/9,1]$ So on , That for $F_n$ we get the Union of $2^n$ closed interval . I know that finite union of closed interval is closed, but this is not true for arbitarly union. As I had a counterexample:  $\cup$ {{1/n}|for $n$ $\in N$} as this is not closed as $0$ is not in  that union. So what is the argument here to say that $F_n$ is closed for any $ n$ . As we are using this fact to prove the Cantor set to be closed as $F=\cap_{n\to \infty} F_n$ i.e arbitarly intersection of closed set is closed . Where am I misunderstanding? Any help will be appreciated.