Let $G$ be a bipartite graph with a perfect matching. Prove that $G$ has a node $v$ for which every edge incident to $v$ belongs to a perfect matching.   I don't think that this question is true if all the edges incident to $v$ belongs to the same perfect matching.  My attempt: Suppose $v$ is not incident to any perfect matching of $G$. This implies that $G$ cannot have a perfect matching, which is a contradiction.  Am I on the right way, thanks for your comment.
A matching or independent edge set in a graph is a set of edges without common vertices.    A perfect matching (also called 1-factor) is a matching which matches all vertices of the graph.    a bipartite graph (or bigraph) is a graph whose vertices can be divided into two disjoint sets and (that is, and are each independent sets) such that every edge connects a vertex in to one in.    Now imagine a bipartite graph like G which has a perfect matching. Prove that there exists a vertex in G like V such that every edge connected to G is in a perfect matching.  Note : I tried to prove it but i have nothing to start with ! Thanks in advance.  Edit : The graph is a simple undirected graph and i don't know anything about "strongly" connected graphs.
In shell scripts used for unit testing with dynamic libraries in a directory other than the typical @rpath, I have previously been able to set DYLD_FALLBACK_LIBRARY_PATH to set the directory containing the libraries. Under 10.11.1, bash seems to ignore attempts to set this environment variable:  $ sh -x testscript.sh + DYLD_FALLBACK_LIBRARY_PATH=/Users/something/testinglibs + export DYLD_FALLBACK_LIBRARY_PATH + exec printenv   and DYLD_FALLBACK_LIBRARY_PATH is not present in printenv's output.  Is this a security-related hack in 10.11's shell? I haven't been able to find this change documented in man pages or online. OpenLayers low resolution and/or shifts in QGIS print composer? I'm having trouble with the OpenLayers Plugin of Qgis and the map composer:  If I create an OSM-background layer and if I want to export this, the OSM-Layer looks perfectly all right in the normal qgis program window. But in the map composer and after export the layer has shifted relatively to my other shape layers (EPSG:32633 - WGS 84 / UTM zone 33N).  The second thing is that the output resolution of the exported osm-layer is very, very poor.   A really bad way of getting around this whole trouble would be to increase the screen resolution and make a screenshot of the map composition window of qgis. But I don't think this would be very professional. It also would cause a lot of pain :)   I'm using Qgis 1.8.0-Lisboa under Linux. The openlayers plugin is version 0.92. Using \@-ified punctuation while using \frenchspacing After reading  about spacing and punctuation, I was left wondering about an issue that didn't seem to be addressed: using \frenchspacing together with the modified punctuation \@., \@?, etc.  As far as I can tell visually, using \frenchspacing together with these commands works correctly (i.e., it seems like nothing breaks, and the extra space is suppressed as normal), but are there problems that can arise from doing so?   Moreover, when using \frenchspacing, "should" I still use \@. and its friends where it would ordinarily be necessary, in case someone wants to remove \frenchspacing later? Would accommodating (hypothetical) future editors in this way be "standard", "above and beyond the call of duty", or "crazy"? What do Android application permissions mean? When installing an application, the application lists permissions that it needs to perform its  functions.    I am creating this list of the the system defined permissions and a description of what they mean. It is a community wiki so if new permissions are added in the future they can be added to this list. Importing a Single Symbol From a Different Font How can I prove that the additive group of rationals is not isomorphic to a direct product of two nontrivial groups? I am working through Paolo Aluffi's new GSM text on my own (self-study). On page 63, he asks the reader to     Prove that $\Bbb{Q}$ is not the direct product of two nontrivial groups.   For some context, this is an exercise following a section entitled "The category Grp".  I am assuming that he means "is not isomorphic to the direct product of two nontrivial groups", and I can see two possible ways to proceed with this proof, but have been unsuccessful with either approach.    Approach 1:  Show that the additive group of rationals has a property that is preserved by isomorphism that the direct product of two nontrivial groups does not have or vice-versa.  This seems challenging unless I can significantly narrow down the properties that a direct product of two nontrivial groups that was isomorphic to $\Bbb{Q}$ would necessarily have.  Approach 2:  Considering the section in which this question occurs, show that if $G$ and $H$ are nontrivial groups and $\Bbb{Q} \cong G \times H$, then there are homomorphisms $\varphi_{G}:\Bbb{Q} \rightarrow G$ and $\varphi_{H}:\Bbb{Q} \rightarrow H$ which do not factor or do not factor uniquely through the product $G \times H$.  This would be a contradiction, as $G \times H$ is a final object in the category Grp.  I would greatly appreciate suggestions on how to proceed further with either of these approaches or with alternate approaches. My printer does not print light magenta. (Canon MP220 - CL-38) I have a Canon MP 220, with new cartridges. Prints fine everything, but it won't print light magenta. I cleaned it and deep cleaned it many times. I have no idea what could be the problem. Cartridge models: CL-38 and PG-37.  Any ideas ?  LE: Now it won't print magenta at all! Isomorphism involving Eilenberg-Maclane space, Tors. Let $\pi$ be a group and let $K(\pi, 1)$ be a connected CW complex such that $\pi_1(K(\pi, 1)) = \pi$ and $\pi_q(K(\pi, 1)) = 0$ for $q \neq 1$. Does there exist an isomorphism between $H_*(K(\pi, 1); A)$ and $\text{Tor}_*^{\mathbb{Z}[\pi]}(A, \mathbb{Z})$? Why are there only derivatives to the first order in the Lagrangian? Why is the Lagrangian a function of the position and velocity (possibly also of time) and why are dependences on higher order derivatives (acceleration, jerk,...) excluded?  Is there a good reason for this or is it simply "because it works". What to write in visa application form and what documents to submit as an unregistered freelancer? There are many visa applicants who are working as an unregistered freelancer (photographer, construction contractor, fashion designer, journalist etc ). Some of these freelancers make really handsome monthly income and don't really need to work as a full time employee. Most of them receive variable amount of payments (cash, cheques, demand drafts etc) When it comes to filling a visa application form (USA, UK, Schengen etc), there is always an employment section in all visa form, eg for Schengen visa employment form :    This is a very important factor to make an application really strong.  Questions:   1 Unregistered freelancers can just simply write their profession? what documents to produce along with the application?  2 Bank statements with variable methods of payments could convince the visa officer that these are all legitimate payments? I just proved that any finite group of order $p^2$ for $p$ a prime is abelian. The author now asks to show that there are only two such groups up to isomorphism. The first group I can think of is $G=\Bbb Z/p\Bbb Z\oplus \Bbb Z/p\Bbb Z$. This is abelian and has order $p^2$. I think the other is $\Bbb Z/p^2 \Bbb Z$.   Now, it should follow from the fact that there is only one cyclic group of order $n$ up to isomorphism that these two are unique up to isomorphism. All I need to show is these two are in fact not isomorphic. It suffices to show that $G$ as before is not cyclic. But this is easy to see, since we cannot generate any $(x,y)$ with $x\neq y$ by repeated addition of some $(z,z)$.  Now, it suffices to show that any other group of order $p^2$ is isomorphic to either one of these two groups. If the group is cyclic, we're done, so assume it is not cyclic. One can see that $G=\langle (1,0) ,(0,1)\rangle$. How can I move on? launch sms application with an intent Negative pressure How is negative pressure created in a fluid system?Isn't it counter intuitive that we are reducing the pressure of a system containing no molecules(zero pressure) to a lower value? $\textit{Why}$ the cases in 4 $\nmid n^2 - 3?$ (bases cases in modular induction) This question has already been answered . My question is, when we suppose the hypothesis is not true (proof by contradiction), namely that $4 \mid (n^2 - 3)$, we get $n^2 - 3 = 4k$ for some integer $k$ and then we have two cases. But where does the two cases come from? In addition, why does putting them into two cases work to prove the statment? My thought process would be to try to prove this statement by contradiction and somehow figure out a way to produce a contradiction similar to how $\sqrt{2}$ is proved by contradiction. Any help is great! Let me know if I need to clarify more! There are four coins in the main hallway, and a number of them in the Boos in the yard. Is there anything to gain from them other than health? Do I get anything for collecting them, perhaps in a particular order? I have a set of aerial images which I am trying to clip by their overlapping areas, for use in a more involved program I am writing.  Is there a command line based tool out there with an easy method for extracting the overlapping areas from a pair of images?  The images themselves do not have georeferencing, but they DO come with a shapefile containing bounding boxes of the image collects, as well as IMU data from image collection.  I know this may be difficult since the images will have slight perspective differences, but I was hoping it would still be possible--especially since I am trying to maintain and utilize these perspective differences for something else. How to remove warnings like `Unknown media type ...` People like myself are Stack Exchange trusted users with thousands of reputation points and hundreds of badges.  In other words, we've proven ourselves as people who are assets to Stack Exchange and are a key component in making it so profitable.  As such, I was surprised when I posted  on the  site, and then shortly afterwards tried to  on that same meta site, but was blocked with a message in red stating I had to wait 40 minutes before posting again.  Why is this limit present for trusted Stack Exchange contributors, and can it be changed? I'm pretty new to this and this is my first question here, any help would be greatly appreciated.  So basically what I want to achieve is a floor material as seen in this photo...    ...the reflections are very crisp but dim.  With a glossy shader I get the following results:    First picture: reflections are blurry  Second picture: roughness of the material looks right but reflection is bright red  Third picture: If tinted in any other color than white, it seems to produce the result I want. Reflection is sharp and gets tinted blue. What I need is the reflection to be tinted white.  Thanks for any help. How to reduce the number of authors above >3 to 'First Author et al., (year)'?