How can I take two separate backups?   Contacts only Full backup   I do not want to install any 3rd party program. I do not want to store my information on cloud. I just want to store it directly from phone to PC.  I use Android on Samsung G5.
I know I'm asking for the close-to-impossible -- as a real full-backup certainly requires root permissions. But still there are many folks around not wanting to root their devices -- afraid of voiding their warranties or bricking their devices, or just not having their devices supported by any rooting method.  I also know there were a lot of questions asked already on this topic, and I checked all of them having the backup tag -- none covers the entire topic, all just have pieces, and especially people new to Android (but other non-power-users as well) would have a hard time figuring out the best way.  So I want a detailed canonical answer to the question:  Not involving root-solutions, how to achieve a backup as close as possible to a full backup?  Please do not give simple one-liner answers like "Use app xyz [full-stop]." If mentioning an app (and I doubt a single app would be the answer here), include what parts are covered by it -- as well as ideas on how to fill the gaps it leaves. (And to make it absolutely clear: I know the first thing coming to mind is Titanium Backup -- I use it myself, but this does not apply here: This question is strictly about methods for non-rooted devices!) Moreover, the answer should be as generic as possible in terms of applying to a wide range of devices (i.e. it should not be restricted to one device or manufacturer).  I already mentioned having investigated all questions here tagged backup. These will certainly be helpful providing details for answers here, so I will list up the topmost ones concerning my question:            All these provide some details for my question (so feel free to get inspired by them) -- but I feel there are still details missing. Also it would be helpful to have a summary -- say, an answer including the full description of "the maximum possible".  A last criterium: While I'd consider cloud-based solutions legit, I'd favour local solutions. If you can provide both: all the better. But some people value privacy quite high (I'm one of them), and thus would not trust their data to some cloud service.    Thanks to Ryan, we now have a solution for a really complete backup of all apps and their data -- which I didn't dare to hope for! The only disadvantage: His solution only works for less than 10% of all Android users; namely those whose devices are running at least with Android 4.x.  So again, I ask the close-to-impossible: Any solutions applicable at least on Gingerbread (2.3.x) or, better, even Froyo (2.2.x) -- so they cover the majority of current devices? These solutions may involve multiple tools (try to keep it as simple as possible though; think of your mother to follow your instructions ;)  And keep in mind: NO ROOT!    To further motivate high-quality answers (and hopefully avoid low-quality ones), I will be a harsh one here: As I rarely do, I will down-vote answers not fitting the mentioned criteria on one end -- but on the other end, I will setup a bounty for the best answer (which cannot be done immediately, but earliest after 2 days).
Multiplication in Permutation Groups Written in Cyclic Notation I didn't find any good explanation how to perform multiplication on permutation group written in cyclic notation. For example, if $$   a=(1\,3\,5\,2),\quad b=(2\,5\,6),\quad c=(1\,6\,3\,4), $$  then why does $ab=(1\,3\,5\,6)$ and $ac=(1\,6\,5\,2)(3\,4)$? The answer to tactical downvoting problem? As we know, the tactical downvoting (I'm talking especially about SO, as I don't know about other sites) is more and more becoming a serious problem, as I see, the community can't stop this behavior without some sacrificing.  As a suggestion, I think it makes sense to prevent an answerer from downvoting any answer on the same question, this -sure- has a harm, but IMO, it's less harmful than the tactical downvoting issue, and to make the things a bit more easier, we can prevent an answerer from downvoting a question without posting a comment to that answer before.  And vice versa, if a user downvoted an answer, he shouldn't be able to post another one unless he posted a comment first, to prevent working-around that system, so a user can't downvote other answers before posting his own to avoid posting comments.  EDIT, more clarification: I don't say wrong answers shouldn't be downvoted, BUT if you want to downvote an answer and post your own, you'll have to post a comment, so we make sure you're not just downvoting right answers tactically. Why does Javascript getYear() return 108? Working in a group blender project Geometric intuition of improper integrals I am aware that the area under the curve of $\frac{1}{x}$ is infinite yet the area under the curve of $\frac{1}{x^2}$ is finite.   Calculus and series wise, I understand what is going on, but I can't seem to get a good geometric intuition of the problem. Both curves can be shown to converge to $0$ (the curves themselves, not the area), and on the interval from $1$ to infinity, the two curves have nothing intrinsically different. Can someone please provide me with an good geometric intuition of what's going on? I can't find anything on the web, people seem to not want to explain it geometrically. How to solve this probability problem analytically (instead of using simulation)? Probability of Bill wins the game if he goes first Editing Gsettings unsuccesful when initiated from cron In the category $\mathbf{Set}$ is "the product of an empty set of sets a one-element set"? I was reading  on Category Theory and it said (paraphrased to add context):     Exercise 4: Explain why in $\textbf{Set}$ (the Category of Sets), the product of an empty set of sets is a one-element set.   which I think is incorrect. The product of two empty sets (or any number) is empty because we are considering:  $$ \emptyset \times \emptyset = \{ (a,b) : a \in \emptyset, b \in \emptyset \} = \emptyset$$  where $a \in \emptyset , b \in \emptyset$ are false, so the above is the $\emptyset$ which is NOT a one element set (its a zero element set).   This should be trivial so I am assuming I am somewhere mis reading the natural language of the exercise. Someone help me catch where is it and what the answer should be? i.e. whats being asked and the answer? Suppose that $ f $ is entire and that for each $ z $, either $ |f(z)| \leq 1 $ or $ |f^\prime (z) |\leq 1 $. Prove that $ f $ is a linear polynomial. My question is in the title. I'm a little lost in how to solve this problem. There is a hint associated with the problem that states the following:   Use a line integral to show that $ |f(z)| \leq A + |z| $ where A = $\max\{1, |f(0)|\}$  Thanks for the help. How old should you be when you begin Jedi training? Anakin Skywalker was about nine years old (see script link, below) during "The Phantom Menace". Yet according to various people, he was too old (or even "far too old") to begin training as a Jedi:   QUI-GON: "I'm afraid not. Had he been born in the Republic, we would have identified him early, and he would have become Jedi, no doubt...he has the way. But it's too late for him now, he's too old." OBI-WAN: "The boy will not pass the Council's tests, Master, and you know it. He is far too old." MACE WINDU: "He is too old. There is already too much anger in him." and even ANAKIN himself: "I'm with Qui-Gon...but...they're not going to let me be a Jedi. I'm too old."   All quotes above from .  However, during "The Empire Strikes Back" Yoda gives Luke's age (18 in chapter IV according to the ) as the reason to deny him training only after first saying that Luke is:   too impatient too angry not ready is unfocused ("Never his mind on where he was") and reckless   All from .  I would have thought that, if at 9 Anakin was too old, Yoda would have seen Luke at 18+ and started off by saying, "No chance, he has".  So, what is the age to begin Jedi training? Double orthogonal complement is equal to topological closure From :  Error propagation It was common to discuss the &quot;error propagation&quot; properties as a selection criterion for a mode of operation. It might be observed, for example, that a one-block error in the transmitted ciphertext would result in a one-block error in the reconstructed plaintext for ECB mode encryption, while in CBC mode such an error would affect two blocks.  I read  similar post before. And I can understand that in CBC decryption, only two blocks are affected due to error propagation (error in 1 block). But, for the same case in CBC encryption, why are still only two blocks affected? It looks like all the following blocks will be affected. Proving the universal mapping property for polynomial rings According to the Page 87 in Kruschke: Doing Bayesian Data Analysis, the author says that the mean of a distribution is a value that minimizes the variance of a probability distribution, for example, a normal distribution. The following is what is mentioned in the page:  It turns out that the value of $ M $ that minimizes $ ∫dx p(x)(x−M)^2$ is $E[X] $. In other words, the mean of the distribution is the value that minimizes the expected squared deviation. In this way, the mean is a central tendency of the distribution.  I have read the paragraph and kind of understood what the author is trying to say but I wonder how this can be written mathematically using the above equation or any other ways. execute command on a defined range of directories Is it possible to 'partition' a directory listing - say in to blocks of some number, to perform different actions on each 'range'?  For example, lets say I have the following directories inside another folder:  $ ls test abc/ def/ hij/ klm/ nop/ qrs/ tuv/ wxy/ zzz/   And I want to perform some action on the first 3 directories, another on the second 3, and so on.  My thought was that perhaps a loop across numbers would work, based on the output of something like ls | nl (but before anyone mentions it, I know parsing ls is a no-no!)  This obviously doesn't work, but illustrates my point I hope:  for dir in `ls | nl`;  do     do-something-with ${dir{1..3}} # Where $dir has taken on some numerical value linked to the folder)     do-something-with ${dir{4..6}}  # And so on...     do-something-with ${dir{n-3..n}}  done   The folders I intend to actually do this on, can be worked on in any order (i.e. the final splits can be totally arbitrary), but they have no logical naming consistency - by which I mean they can't be organised sensibly alphabetically or numerically based on any key within the directory name themselves. Is given metric space $(M, d)$. Let $A\cap B = \emptyset; \,\,\text{dist}(A,B):=\inf\{d(x,y):x\in A, y\in B\}$. $A, B$ are both closed sets. Is it possible that $\text{dist}(A,B)=0$?  The first thought comes into mind is that obviously $\text{dist}(A,B)&gt;0$, but possibly there are some tricky $d$ and $A, B$ so that it's untrue.  Thanks in advance! Count the number occurrences of a character in a string How should I setup separate MX records for a subdomain? Let's say I have a domain that I run a web application on, for example cranketywidgets.com, and I'm using  for handling email for people working on that domain, for example, support@ cranketywidgets.com, jane@cranketywidgets.com, joe@cranketywidgets.com and so on.  Google's own mail services aren't always the best for sending automated reminder emails, comment notifications and so on, so the current solution I plan to pursue is to create a separate subdomain called mailer.cranketywidgets.com, run a mail server off it, and create a few accounts specifically for sending these kinds of emails.  What should the MX records and A records look like here for this?  I'm somewhat confused by the fact that MX records can be names, but that they must eventually resolve to an A record. What should the records look like here?  cranketywidgets.com - A record to actual server like 10.24.233.214  cranketywidgets.com - MX records for Google's email applications  mailer.cranketywidgets.com - MX name pointing to server's IP address  I would greatly appeciate some help on this - the answer seems like it'll be obvious, but email spam is a difficult problem to solve. Let $q: \mathbb{N} \to \mathbb{Q}$ be a bijection and denote the image of $k \in \mathbb{N}$ by $q_k$.  Let $f: \mathbb{R} \to (0,1)$, $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} $$  Show that:  1) $f$ is strictly increasing  2) $f$ is continuous at $x$ iff $x \in \mathbb{R} \setminus \mathbb{Q}$.  My work:  Let $x, y \in \mathbb{R}$, $x &lt; y$. Then there exists $q_n \in \mathbb{Q}$ s.t. $x &lt; q_n &lt; y$. Now $$ f(x) = \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} &lt; \underset{q_k \leq x}{\sum_{k \in \mathbb{N}}} 2^{-k} + 2^{-n} \leq \underset{q_k \leq y}{\sum_{k \in \mathbb{N}}} 2^{-n} = f(y) $$ so $f$ is strictly increasing.  I think my proof of 1) is ok, but how can I prove 2)? I saw this question  but it doesn't have a full solution. What is the full solution to the 99 bottles of beer on the wall for Vim adventures?  Context: for the sake of being a good SO member I will give more details but a full solution will answer my question. It seems from reading the other post that my register is incorrect for some reason. I can't get the &lt;NL&gt;. I think that's the issue since pasting won't work...