Undirected graph contains a cycle
If a graph with $n$ vertices and $n$ edges there must a cycle?
ViewPager and fragments — what's the right way to store fragment's state? Im trying to understand the solution about this, and all the other solutions are.. a bit weird for me and dont answer all my questions!  Reference :   Im using the answer there as reference, but my understanding is a bit different!     Find the error in the following proof that all horses are the same   color. CLAIM: In any set of h horses, all horses are the same color.      PROOF: By induction on h.      Basis: For h = 1. In any set containing just one horse, all horses   clearly are the same color. Induction step: For k > I assume that the   claim is true for h = k and prove that it is true for h = k + 1. Take   any set H of k + 1 horses. We show that all the horses in this set are   the same color. Remove one horse from this set to obtain the set H1   with just k horses. By the induction hypothesis, all the horses in H,   are the same color. Now replace the removed horse and remove a   different one to obtain the set H2 . By the same argument, all the   horses in H2 are the same color. Therefore all the horses in H must be   the same color, and the proof is complete.   Here's what i have come up with :  Suppose a set of 1 horse. $H_a = { horse_{1} }$  Obviously, all the horses in set $H_a$ are the same, since theres just 1 horse.  So using that as the Basis of the induction we can proceed and try to prove its true for $N$ horses.  -- I dont fully understand the part below   As an example : set $H_0$ :   $H_0 = { horse_{1}, horse_{2}, horse_{3}... horse_{n} }$  In my understanding, its because, I can make a subset  formed by 1 horse :  $K_1  = {horse_{1}}$ $K_2  = {horse_{2}}$ ...  I don't get it. I clearly know that this proof is a lie.  And for me, its because you cant claim theres a relation on a set with size $&gt;= 2$ , since :  $H_1 = { horse_{1}, horse_{2}}$  Theres nothing that says $horse_{1}$ equals $horse_{2}$. The basis was for one horse. But that alone is not a proof.  So what should I do ? Where in Europe can I purchase a car without residency? I am not a resident anywhere and plan to travel for the foreseeable future, mostly in Europe. My country has good visa-free access to most of Europe, so it is conceivable I could remain without any residency for quite some time.  I enjoy driving, and maintaining cars, and I would like to purchase a car somewhere in Europe.  Where, if anywhere, in Europe (or outside of Europe, but with the ability to drive the car in Europe) can I legally purchase and insure a car as a non-resident? I am open to creative solutions like setting up a company to own the car, if the capital and ongoing paperwork requirements for the company are not too onerous.    I am aware that I could save money, get places quicker, etc, by taking trains or flying. I have a passion for cars and driving. Suggesting that I not buy a car is out of scope of this question. I believe that the set of all $\mathbb{R\to R}$ continuous functions is $\mathfrak c$, the . However, I read in the book &quot;Metric spaces&quot; by Ó Searcóid that set of all $[0, 1]\to\mathbb{R}$ continuous functions is greater than $\mathfrak c$:  &quot;It is demonstrated in many textbooks that $\mathbb{Q}$ is countable, that $\mathbb{R}$ is uncountable, that every non-degenerate interval is uncountable, that the collection of continuous functions deﬁned on  $[0,1]$ is of a greater cardinality than $\mathbb{R}$, and that there are sets of greater and greater cardinality.&quot;  I understand that (via composition with the continuous function $\tan$ or $\arctan$) these sets of continuous functions have the same cardinality. Therefore, which claim is correct, and how do I prove this? Find an element by id and replace its contents with php Fit a table into one page Do gravitational lenses work on gravitational waves? How does Wild Shape work with special ways to calculate your AC? Among the effects of Wild Shapes, these two:     Your game Statistics are replaced by the Statistics of the beast, but   you retain your alignment, personality, and Intelligence, Wisdom, and   Charisma scores. You also retain all of your skill and saving throw   proficiencies, in addition to gaining those of the creature. If the   creature has the same proficiency as you and the bonus in its stat   block is higher than yours, use the creature's bonus instead of yours.   If the creature has any legendary or lair actions, you can't use them.      You retain the benefit of any features from your class, race, or other   source and can use them if the new form is physically capable of doing   so. However, you can't use any of your Special senses, such as   Darkvision, unless your new form also has that sense (PHB p.67, "Wild Shape")   Seem to come into conflict with each other when things like Draconic Resilience:     When you aren't wearing armor, your AC equals 13 + your Dexterity   modifier. (PHB p.102, "Draconic Bloodline")   and Barbarians:     While you are not wearing any armor, your Armor Class equals 10 + your Dexterity modifier + your Constitution modifier. You can use a Shield and still gain this benefit.   and Monks:     Beginning at 1st level, while you are wearing no armor and not wielding a Shield, your AC equals 10 + your Dexterity modifier + your Wisdom modifier.   and Stone Sorcerers:     As an action, you can gain a base AC of 13 + your Constitution   modifier if you aren’t wearing armor, and your skin assumes a stony   appearance. This effect lasts until you end it as a bonus action, you   are incapacitated, or you don armor other than a shield.   How would these resolve? How to mount ufs file system under Debian testing? In order to mount my /dev/sda4  under /mnt/ufs_mount on Debian Testing , i have tried the following commands (it works fine under Debian Jessie):  modprobe ufs mount -r -t ufs -o ufstype=ufs2 /dev/sda4 /mnt/ufs_mount   and:  mount -t ufs -o ufstype=ufs2 /dev/sda4 /mnt/ufs_mount   But i get this error:  mount: /dev/sda4 is write-protected, mounting read-only mount: wrong fs type, bad option, bad superblock on /dev/sda4,    missing codepage or helper program, or other error     In some cases useful info is found in syslog - try    dmesg | tail or so.   The output of: dmesg | tail :  [ 1136.965142] ufs: ufs_fill_super(): bad magic number [ 1255.758946] ufs: ufs_fill_super(): bad magic number [ 2098.945757] ufs: ufs was compiled with read-only support, can't be mounted as read-write [ 2098.946045] ufs: You didn't specify the type of your ufs filesystem        mount -t ufs -o ufstype=sun|sunx86|44bsd|ufs2|5xbsd|old|hp|nextstep|nextstep-cd|openstep ...        &gt;&gt;&gt;WARNING&lt;&lt;&lt; Wrong ufstype may corrupt your filesystem, default is ufstype=old [ 2098.967212] ufs: ufs_fill_super(): bad magic number [ 2927.982112] perf: interrupt took too long (2504 &gt; 2500), lowering kernel.perf_event_max_sample_rate to 79750   Edit My sources.list:  deb http://httpredir.debian.org/debian/ stretch main deb-src http://httpredir.debian.org/debian/ stretch main  deb http://security.debian.org/debian-security stretch/updates main deb-src http://security.debian.org/debian-security stretch/updates main   And uname -a:  Linux debian 4.6.0-1-amd64 #1 SMP Debian 4.6.4-1 (2016-07-18) x86_64 GNU/Linux   How to mount ufs file system under Debian testing? I had the Facebook and Instagram apps installed on my iPhone (latest iOS). I deleted both apps and checked that there was no data from either app stored on iCloud (I went to Settings - iCloud - Manage storage and there was no mention of them storing anything on iCloud). Then I reinstalled Instagram, opened it, clicked on Login and the page still showed “Login as MYFACEBOOKNAME”.   How is this possible? How can I delete my FB login info once and for all on my iPhone? How to create a cross-browser sub document with iframe or shadow dom? How can I programmatically force-stop scrolling in a UIScrollView? How to make an ImageView with rounded corners? Using the expression "the same" for a previously mentioned item Detecting unused packages in LaTeX document I have quite a bad habit of copy-and-pasting my packages from one LaTeX document to another. Is there any way we can detect unused packages (i.e. \usepackage commands) so that they can be removed from the LaTeX document? Can we add a section in our vf page and add components to it Can we add a section in our vf page and add components to it. I mean split my vf page into two parts and then add components to it. How to dismiss the Windows 8.1 Upgrade banner? As of today, my Windows 8 system is displaying a large banner covering about a quarter of the screen, demanding that I upgrade to Windows 8.1 for free.  The banner helpfully informs me that that I can continue to use the computer while the upgrade is downloading.  There's a button to go to the store for the upgrade, but no other choices, such as No Thanks, Maybe Later, or Absolutely Not.    I can't find a way to dismiss the banner, and I can start programs but can't give them focus while the banner is present.  So I'm dead in the water.  How can I fix this?  Bringing up Task Manager doesn't give me any options.    My employer doesn't yet support Windows 8.1 (which is also an issue, but beyond my control) so I won't be able to connect to work if install the upgrade.  So there is no way on God's green earth that I'm going to install this upgrade now.    UPDATE: The banner went away after the second restart and I was able to run regedit as @magicandre suggested.  The banner hasn't reappeared.  So far, so good. Why cannot a set be its own element? Run multiple instance of a program in parallel How can you combine all lines that end with a backslash character? Using a common command line tool like sed or awk, is it possible to join all lines that end with a given character, like a backslash?  For example, given the file:  foo bar \ bash \ baz dude \ happy   I would like to get this output:  foo bar bash baz dude happy