I am new to Android and have been an iOS user for a long time. I've seen a lot of tutorials to migrate my contacts from my iPhone to Android however all of them say I have to use gmail.  I take privacy very seriously and don't like the idea of have to upload all of my contact details into google's database especially after all of the leaks in the media as to how they are used.  I've never used Apple's cloud service for the same reason.  Question: How can I add my contacts from my iPhone as a flat file or perhaps via my Apple Address Book (on my mac) directly to my Android (Galaxy Nexus) WITHOUT having to give all of the data to Gmail (which I seldom use).  Thanks very much
I know someone who just got an Atrix 4G and previously had an iPhone 3GS. I was asked how he could move his contacts from the iPhone/iTunes onto the Android device. I have never owned an iPhone and have had limited interaction with them. Does anyone know the simplest method of getting iPhone contacts onto an Android device?
Implementing the decorator pattern in Python Isomorphism of a product $C_n \times C_m$ of cyclic groups with the cyclic group $C_{mn}$ Given that $C_n$ is a cyclic group of order $n$, what conditions must integers $n$ and $m$ satisfy such that the group $C_n \times C_m$ is isomorphic to C$_{mn}$?   So I attempted to investigate a few lower-order cyclic groups to find a pattern. In particular, I looked at $C_2 \times C_3$ and compared it to $C_6$. Here are a couple of diagrams:  $C_2 \times C_3$ -   $C_6$ -   They Cayley tables look pretty different (identity elements are all over the place for $C_2 \times C_3$) so I concluded that they are not isomorphic.  I did the same for $C_2 \times C_4$ and $C_8$:  $C_2 \times C_4$ -   $C_8$ -   Again, I arrived at the conclusion that the 2 groups are not isomorphic due to their Cayley table structure.  It seems I'm stuck now. Are my conclusions wrong for the examples I examined? If so, how do I find the conditions for the general case of comparing $C_m \times C_n$ with $C_{mn}$? I have an issue with my website: Google isn’t using my meta description, instead it’s using the content of the web page and I’m really confused.  Reason I’m confused is because I have checked and I don’t have duplicate page titles, description nor keywords, neither do I have any robots.txt for my website. My page title, description and keyword is unique, and page description is about 150 characters.  I have my site in Google Webmaster Tools and I checked for errors, but found none.  So I really don’t know why Google Search isn’t using my meta description. I'm well aware of the combinatorial variant of the proof, i.e. noting that each formula is a different representation for the number of subsets of a set of $n$ elements. I'm curious if there's a series of algebraic manipulations that can lead from $\sum\limits_{i=0}^n \binom{n}{i}$ to $2^n$. How to minimize $|z_1 - z_2|^2 + |z_1 - z_4|^2 + |z_2 - z_3|^2 + |z_3 - z_4|^2$? If $z_1,z_2,z_3,z_4 \in \mathbb{C}$ satisfy $z_1 + z_2 + z_3 + z_4  =  0$ and $|z_1|^2 + |z_2|^2 + |z_3|^2 + |z_4|^2 = 1$, then the least value of $|z_1 - z_2|^2 + |z_1 - z_4|^2 + |z_2 - z_3|^2 + |z_3 -  z_4|^2$ is $2$.   I solved this by taking two of the values to be $\frac{1}{2}$ and other two as $-\frac{1}{2}$. But I couldn't prove this in an efficient manner. Discriminant analysis vs logistic regression Array#each vs. Array#map ShareLaTeX compilation flags Is it possible and if so how can you pass flags to the ShareLaTeX compiler which one would have passed as parameters to the LaTeX compiler?  I am using the minted package and need to pass the -shell-escape flag. Restricting SSH sessions for accounts on ubuntu server May someone please show me how I would restrict accounts/people logging into an ubuntu 10.04 server by SSH to a single active session at a time for each account that logs into the server.  Regards I accidentally removed all the packages starting with the name jpeg with:  sudo apt-get remove jpeg*   This screwed up a lot of things, removed Unity, Firefox, Flash player etc.  I somehow managed to recover the desktop and unity but then a lot of things are still buggy. For example .rpm files and .tag.gz just won't open up at all.  What I want is a way to fix everything that's amiss. Something that is the equivalent of a 'repair installation' with a windows disc.  Any tips, anyone? How find this limit $\lim\limits_{x\to 0^{+}}\frac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$ I tried running Android 4.1.1 in VirtualBox, but it was too slow and sluggish to do anything. I came across Windroy which runs Android as a Windows app, but it won't run Google Play.  I want to run this on my laptop.  Any help would be greatly appreciated. Show that the set of polynomials with rational coefficients is countable. Triangle Inequality of the Cartesian product with Max function Let $(X,d)$ be a metric space.  Define $$d'((x,y),(z,w))=max\{d(x,z),d(y,w)\}.$$   I'm trying to prove the triangle inequality for this, but really don't have a clue how.  Any tips or suggestions would be greatly appreciated! Why don't we add a question with the same tag on multiple Stack Exchange sites? So this homework question is in the context of $\mathbb{R}$ only, and we are using Lebesgue measure.  The sum $A+B$ is defined to be $A+B=\{a+b|a\in A,b\in B\}$. The question is: If $m(A),m(B)&gt;0$, is it necessarily the case that $A+B$ contains an open interval?  This question is quite annoying, since it is quite difficult to calculate the sum even after you construct an example. However, it seems like no matter how I constructed a set of positive measure, there is a region that have a lot of point in it (yet this region is not necessarily dense). This suggest that when summed up, the holes should be covered up somehow and thus there would be an open interval, and so the answer is yes. Sorry for a rather vague idea, but I'm just not sure what to do with this problem. I have a feeling that somehow topology would have a way to capture precisely the idea above about a region with a lot of point.  (also, I think my middle school teacher mentioned this once, and I remembered the answer to be yes; but that is quite a while ago though)  Thank you. Any help will be appreciated.  EDIT2: So Camilo's answer is gone. I am looking at the convolution method right now. Because the convolution is continuous, if it is nonzero I am done. But then there is no guarantee that the convolution is nonzero even if $x\in A+B$. I still need help with this direction. Can someone give me some hints?  EDIT: Camilo's answer below essentially suggest the method of getting 2 open interval from regularity theorem, one cover a subset of A and one cover a subset of B such that the sum of the measure of these two subset is bigger than the measure of the larger interval. user99680's answer point to a wiki which show a method that essentially simply acquire an open set. After some thinking, I believe that these method simply can't be made to work. I constructed this example in the hope that it would convince you that it is the case:  Let $I_{0},I_{1},\ldots$ be disjoint open interval such that the distance between any 2 of them is nonzero and $m(I_{n})=9^{-n}$ and all of them are inside $[0,2]$. Let $J_{0},J_{1},\ldots$ be disjoint open interval such that the distance between any 2 of them is at least $3$ and $m(J_{3n})=m(J_{3n+1})=m(J_{3n+2})=\frac{m(I_{n})}{3}$. Let $O=\bigcup I_{n}$ and $H=\bigcup J_{n}$. Construct $A\subset O$ such that $m(A\bigcap I_{n})=\frac{3m(I_{n})}{4}$. Construct $B\subset H$ such that $m(B\bigcap J_{n})=\frac{3m(J_{n})}{4}$. Now if you apply Camilo's method, it is possible that your $I$ would be one of the $I_{n}$ above, and your $J$ is one of the $J_{m}$ above. No matter which $I_{n},J_{m}$ you have, the argument does not work. Rewrite $m=3q+r$ where $0\leq r\leq 2$. If $q&lt;n$ then $m(J_{m})&gt;m(I_{n})$ so the longer interval have length $\frac{m(I_{q})}{3}$; $m(A\bigcap I_{n})$ is at most $\frac{3m(I_{q+1})}{4}=\frac{m(I_{q})}{12}$ and $m(B\bigcap J_{m})=\frac{3m(J_{m})}{4}=\frac{m(I_{q})}{4}$. Thus $m(A\bigcap I_{n})+m(B\bigcap J_{m})$ is no larger than the larger interval. If $n\geq q$ then $m(I_{n})&gt;m(J_{m})$ so the larger interval is of length $m(I_{n})$; $m(A\bigcap I_{n})=\frac{3m(I_{n})}{4}$ and $m(B\bigcap J_{m})$ is at most $\frac{3m(J_{3n})}{4}=\frac{m(I_{n})}{4}$. Once again $m(A\bigcap I_{n})+m(B\bigcap J_{m})$ is no larger than the larger interval. Hence the Camilo's argument does not work. user99680's method does not work either, since you get $O$ and $H$ which is even bigger than any interval. Mac OS X 10.6 executable not found without full path I just installed Apache via MacPorts. It seems that my Mac was absolutely confused about which version of the Apache executables to run.  After moving the Apache executables that ship with the Mac to a directory that is not listed in the PATH variable, trying to run the httpd built by MacPorts fails even though the correct directory (/opt/local/apache2/bin) is listed in the PATH variable.  If I navigate to the directory /opt/local/apache2/bin and type the command httpd I still get the error message     -bash: httpd: command not found   If I type the command with the full path /opt/local/apache2/bin/httpd it works fine.  I've run the command alias to see if something was clashing but the only thing listed is:  alias wget='curl -O'   How do I find what is intercepting the command and preventing the executable being found in the directory, even when I'm inside the same directory?  By the way, the httpd file is executable:  -rwxr-xr-x  1 root      admin  442496  9 May  2012 httpd Why is cd not a program? Calculate E[X] from incomplete data? The exercise I'm doing describes the random variable $X$ as the following  | Number of cars |  0  |  1 |  2 |  3 |  4+ |  | % of families  |  15 | 45 | 25 | 13 |  2  |   Then it asks me to evaluate $E[X]$. But if there could be cases with 4+ cars, wouldn't that make it impossible to calculate the expectancy? Improve the editing flow with predefined options for "Edit Summary"