Presently I am teaching the Course "Graph Theory and Applications" to engineering students. I would like to teach some applications of graph theory in the class, like a stable marriage problem (matching), Scheduling problem (graph coloring), etc.  It will be helpful if some one provide some sources where I can find some interesting applications (based on planar graphs, connectivity, Ramsey theory etc)
What are the areas where graph theory can be applied?  Cause I wonder what applications this have on the real world.  Does it solve certain problems and stuff?  Areas such as communication networks and coding in IT, genomics, computation, and scheduling in operations research
How do I install applications in Ubuntu? Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic I'm reading a book about abstract algebra, but I'm having trouble solving this excercise: "Show that $(\mathbb{Q}^*,\cdot)$ and $(\mathbb{R}^*,\cdot)$ aren't cyclic"  Where $(\mathbb{Q}^*,\cdot)$ is the group of nonzero rational numbers under multiplication and $(\mathbb{R}^*,\cdot)$ is the group of nonzero real numbers under multiplication.    Here is my attempt for the first.  Suppose $(\mathbb{Q}^*,\cdot)$ is cyclic, then $\mathbb{Q}^*=\langle\frac{p}{q}\rangle=\{(\frac{p}{q})^n,n\in\mathbb{Z}\}$, where $p$ and $q$ are coprime.  $\frac{2p}{q}$ is also in $\mathbb{Q}^*$ so it must be equal to $(\frac{p}{q})^n$ for some $n\in\mathbb{Z}$.  To solve $\frac{2p}{q}=(\frac{p}{q})^n$, I take a logarithm of both sides and end up with $1+\log_\frac{p}{q}(2)=n$, since $n$ is an integer $\log_\frac{p}{q}(2)$ must be an integer too, but it is possible only when $\frac{p}{q}=2^{\frac{1}{k}}, k\in\mathbb{N}$, (i.e. $\frac{p}{q}$ is a k-th root of $2$), but $k$ must be $1$ for $2^\frac{1}{k}$ to be rational so $\frac{p}{q}=2$ contradicting the hypothesis of $p$ and $q$ being coprime.    However I don't know whether this is a proper proof and the same reasoning cannot be applied to $\mathbb{R}^*$, I'd like you to just give me an hint towards a proof, without telling me the whole proof, if possible. I deleted an answer which had no upvotes or downvotes, but after some time I saw that my reputation was -2 and the reason is linked to my search engine.    I want to know why this happened and what "linked to my serach engine" means? Create WiFi hotspot on ubuntu Arbitrary intersection of closed, connected subsets of a compact space connected? Let $(B_i)_{i\in I}$ be an indexed family of closed, connected sets in a compact space X. Suppose $I$ is ordered, sucht that $i &lt; j \implies B_i \supset B_j$.  Is $B = \bigcap_i B_i$ necessarily connected?  I can prove it, if I assume $X$ to be Hausdorff as well: If $B$ is not connected, then there are two disjoint, closed, nonempty sets $C$, $D$ in $B$, such that $C \cup D = B$. Now these sets are also closed in $X$, hence by normality there exist open disjoint neighborhoods $U$, $V$ of $C$ and $D$, respectively.  Then for all $i$: $B_i \cap U^c \cap V^c \ne \emptyset$, since $B$ is contained in $B_i$ and $B_i$ is connected. Thus we must also have  $$ B \cap U^c \cap V^c = \bigcap_i B_i \cap U^c \cap V^c \ne \emptyset $$  by compactness and the fact that the $B_i$ satisfy the finite intersection property. This is a contradiction to the choice of $U$ and $V$.  I can neither see a counterexample for the general case, nor a proof. Any hints would be greatly appreciated!  Thanks,  S. L. "Back" and "Install Now" buttons grayed out in 14.04 installer The "Back" and "Install Now" buttons are grayed out in the 14.04 installer. How do I restore them? Is there a unique solution for this quadratic matrix equation? Quadrilateral problem with midpoints of diagonals For an arbitrary convex quadrialteral $ABCD$:   $$AF=FC, BE=ED, GI=IH$$  Prove that points $F$, $E$ and $I$ are collinear.  I was able to solve the problem by using vectors and I think that it can be also solved analytically but I wanted some more elegant proof, more in the spirit of Euclid. What character / attack / stage combination can get me a combo of 100? For the first two challenge panels, I just used my golden hammers on the "get a combo of x," but now I'm seeing on challenge panel 3, the game expects you to get a combo of 100! I can barely get past a combo of 9 without the counter resetting. What character / stage combination can get you a combo of 100? Are there multiple, so I can more easily get the "get a total combo of 400 across all characters" challenge?  Additionally, what exactly constitutes a combo? Sometimes my combos are cut short and sometimes they're longer, so I must not understand how they're tracked. How do I correct an HSA distribution when I was refunded by a health provider? Does "FN" stand for or mean anything? Prove if $\operatorname{ord}(a)=m$, $\operatorname{ord}(b)=n$, and $\operatorname{gcd}(m,n)=1$, then $\operatorname{ord}(ab)=mn$.   I was reading this and was thinking how this proof would look like. I tried to do it and am not sure if this is correct. Here is what I did:  If $a$ and $b$ commute then $(ab)^{mn} = a^{mn} * b^{mn} = (a^m)^n * (b^n)^m = 1 * 1 = 1.$ So $ord(ab) | mn$.  Now, take $k = \operatorname{ord}(ab) = m^\prime * n^\prime$ where m' is relatively prime with $n$ and $n'$ is relatively prime with $m$. By the result above $m' | m$ and $n' | n$. Now we have $((ab)^k)^{m/m'} = 1$ since $(ab)^k=1$. But on the other hand, by the commutativity:  $$((ab)^k)^(m/m') =$$  $$((ab)^(m' * n'))^(m/m') =$$  $$ a^{n'm} * b^{n'm} = $$  $$(a^m)^{n'} * b^{n'm'}=$$  $$b^{n'm'} = 1$$  This implies that $n' * m'$ is divisible by $n$. But $m'$ is relatively prime with $n$, so we must have $n' = n.$ By symmetry, $m' = m$. So $ord(ab) = mn$.  Just to say this again, I want to prove if $a$ and $b$ commute and $m$ and $n$ are relatively prime then   $$ord(ab)=mn.$$   The comments are vague. I guess this must be the answer then and there probably is not another way to do this. Java - How to reduce the size of third-party jars to reduce the size of your application Using "${a:-b}" for variable assignment in scripts Recently I've noticed more and more sign-in / sign-up workflows defaulting to the sign-up option first and providing a minimal "Sign-In / Login" action somewhere else on the page. What's the reason behind this practice?  Examples where I've seen this recently: Number of bins when computing mutual information I want to quantify the relationship between two variables, A and B, using mutual information. The way to compute it is by binning the observations (see example Python code below). However, what factors determines what number of bins is reasonable? I need the computation to be fast so I cannot simply use a lot of bins to be on the safe side.  from sklearn.metrics import mutual_info_score  def calc_MI(x, y, bins):     c_xy = np.histogram2d(x, y, bins)[0]     mi = mutual_info_score(None, None, contingency=c_xy)     return mi Searching for questions that have ONLY a given tag (i.e. add a range operator for tags) $f$ is surjective iff it has a right inverse: using the axiom of choice and errors in ProofWiki plist or sqlite Change line spacing inside a node in TikZ