Let $G(V, E)$ be an undirected final and simple graph. $\bar{G}(V, E')$ a simple graph on the same vertices, while $e \in E'$ iff $e \notin E$ Prove that if G is not a connected graph, then $\bar{G}$ is a connected graph.  This is my idea: Let $v_1, v_2$ $(v_1 \neq v_2)$ be vertices in $V$, assume there isn't such a route $v_1 \overset{e_1}{\rightarrow} ... \overset{e_m}{\rightarrow} v_2$, while $e_i \in E'$. therefore in particular $(v_1, v_2) \notin E'$, therefore $(v_1, v_2) \in E$. this is true for every $(v_1, v_2)$ in contradiction to G not being connected. Is this ok? I feel like there is something wrong here... *I translated it so if anything is not clear please tell me.
I was tasked to prove that when given 2 graphs $G$ and $\bar{G}$ (complement), at least one of them is a always a connected graph.  Well, I always post my attempt at solution, but here I'm totally stuck. I tried to do raw algebraic manipulations with # of components, circuit ranks, etc, but to no avail. So I really hope someone could give me a hint on how to approach this problem.
How extended must my extended paper be? The best stack-up possible with a four-layer PCB? I'm designing a 4 layer PCB and I know that the standard stack-up is  Signals GND VCC Singals  (GND and VCC can be switched depending on the layer with more signals) The problem is, I don't really want to connect all ground pins through vias, there are just too many of them ! maybe because I'm not used to 4 layer PCBs, anyway, I've read a  by Henry W. Ott about a different stack-up  GND Signals Signals GND  (Where the power is being routed with wide traces on the signal planes) According to him, this is the best stack-up possible with a four-layer PCB, for the following reasons: 1.Signal layers are adjacent to ground planes. 2.Signal layers are tightly coupled (close) to their adjacent planes. 3.The ground planes can act as shields for the inner signal layers. (I think this requires stitching ??) 4.Multiple ground planes lower the ground (reference plane) impedance of the board and reduce the common-mode radiation. (don't really understand this one) One problem is cross-talk, but I really don't have any signals in the third layer, so I don't think that corss-talk will be an issue with this stack-up,am I correct in my assumption ? Note: The highest frequency is 48MHz, there's a wifi module on the board too. Smooth option sometimes produces incorrect arrow tips in PGFplots It seems that the smooth option applied to lines sometimes effects the arrow tips and produces incorrect results.  No problem with vertical lines, but any other seems to have a problem. Is this a known bug, or is there a reason why the smooth should not be applied to straight lines?  Updated: to show problem does NOT occur with TikZ, and problem also occurs with curves if smooth option is applied.  \documentclass{article} \usepackage{pgfplots} \begin{document}  \tikzstyle{MyStyle}      =[-&gt;, black, ultra thick,       ] \tikzstyle{MyStyleSmooth}=[-&gt;, black, ultra thick, smooth]  \begin{tikzpicture} \begin{axis}     % 'smooth' option seems to have no effect on lines '1' and '2'     \addplot [MyStyle      , green ] coordinates{(0,0) (0,1)} node [above] {1};     \addplot [MyStyleSmooth, blue  ] coordinates{(1,0) (1,1)} node [below] {2};     %     % But, can't seem to have the 'smooth' on lines, '4', '5' and '6'     \addplot [MyStyle      , orange] coordinates{(2,0)   (3,0)   } node [right] {3};     \addplot [MyStyleSmooth, red   ] coordinates{(4,0)   (2,0.5) } node [below] {4};     \addplot [MyStyleSmooth, red   ] coordinates{(2,1)   (3,1)   } node [left ] {5};     \addplot [MyStyleSmooth, red   ] coordinates{(3,0.8) (2,0.8) } node [right] {6};      % Similar problem with curves: "non-smooth" is ok, "smooth" is not     \addplot[MyStyle,       mark=none, domain=0:1,samples=50, blue] (x,x*x+1);     \addplot[MyStyleSmooth, mark=none, domain=0:1,samples=50, red ] (x,x*x+2);  \end{axis} \end{tikzpicture}  \begin{tikzpicture} % These are all fine     \draw [MyStyle      , green ] (0,0)   -- (0,1)    node [above] {1};     \draw [MyStyleSmooth, blue  ] (1,0)   -- (1,1)    node [above] {2};     %     \draw [MyStyle      , orange] (2,0)   -- (3,0)    node [right] {3};     \draw [MyStyleSmooth, red   ] (4,0)   -- (2,0.5)  node [below] {4};     \draw [MyStyleSmooth, red   ] (2,1)   -- (3,1)    node [right] {5};     \draw [MyStyleSmooth, red   ] (3,0.8) -- (2,0.8)  node [left ] {6}; \end{tikzpicture} \end{document} Is there a single word that is a more vicious connotation of "competition"? I would like to show that  $$\forall n\in\mathbb{N}^*, \quad \sqrt{\frac{n}{n+1}}\notin  \mathbb{Q}$$  I'm interested in more ways of proofing this.  My method :  suppose that $\sqrt{\frac{n}{n+1}}\in  \mathbb{Q}$ then there exist $(p,q)\in\mathbb{Z}\times \mathbb{N}^*$ such that $\sqrt{\frac{n}{n+1}}=\frac{p}{q}$ thus   $$\dfrac{n}{n+1}=\dfrac{p^2}{q^2} \implies nq^2=(n+1)p^2 \implies n(q^2-p^2)=p^2$$ since $p\neq q\implies p^2\neq q^2$ then  $$n=\dfrac{p^2}{(q^2-p^2)}$$  since $n\in \mathbb{N}^*$ then $n\in  \mathbb{Q}$   I'm stuck here and I would like to see different ways to prove $\sqrt{\frac{n}{n+1}}\notin  \mathbb{Q}$ Trying to dual boot Ubuntu 14.10 with Windows 10, last one not detected My keyboard doesn't have an Fn key so I can't get the coordinates to show up. I don't know how to get into the multiplayer settings and Shift + F3 doesn't work either.  How can I fix this? Grammar-checking tool for use with LaTeX? I am looking for a grammar-checking tool, i.e., something that checks for mistakes such as "two dog" or "a books".  It has to work on Linux (Ubuntu).  Also, since I use macros to generate some of the prose, it has to somehow work with LaTeX.  I don't think that such checkers are LaTeX-aware, so it has to work with some compilation product of LaTeX: DVI or PDF.    If you suggest a grammar checker that checks plain text, please explain how to you extract the text.  catdvi does a nice job, but it leaves some garbage like page numbers. With ubuntu 19.10 I have been attempting to pair my Airpods with my laptop. There seems to be a problem with pairing and system is unable to add them as Bluetooth device. This happened after an upgrade from 19.04. Changing ControllerMode in /etc/bluetooth/main.conf seems to have no affect. Any help appreciated. I am trying to proof the following algorithm to see if a there exists a path from u to v in a graph G = (V,E).   Reach := {u} LastCardinality := 0 while LastCardinality &lt; |Reach| do   LastCardinality := |Reach|   Next := null   foreach i &lt;- Reach do     foreach j &lt;- V do       if(i,j) exists in E then         Next := Next + {j}   Reach := Reach + Next return Reach   So far I have got that the output needs to be Reach = Goal where Goal = {r | r exists V, there is a path from u to r}. I also have three invariants that I am pretty sure will help prove this.   Reach is always a subset of Goal and Next is always a subset of Goal. After an arbitrary K times through the while loop, for every t that exsits in V, if t has a path of length k+1 from u, then t exists in Reach If Reach is a subset of Goal before the while loop then LastCardinality &lt; |Reach|   I know that to finish up the proof, I need to prove termination, the invariants, and correctness but I have no idea how. I think I need to use induction on the while loop but I am not exactly sure how.   How do I proof those three characteristics about an algorithm? What metadata does a LibreOffice file contain and how to remove it? How do I change player name in Minecraft smp in offline mode on a Mac? find a $\int_{x}^{g(x)} f(t)\,dt = 1$ Let $f : [0, 2] → \mathbb{R}$ be a continuous, positive function and  $$\int_{0}^1 f(x)\,dx = \int_{1}^2 f(x)\,dx= 1$$ For each $x \in [0, 1]$, prove that there exists a unique function $g(x) \in [1, 2]$ such that $$\int_{x}^{g(x)} f(t)\,dt = 1$$ and $g \in C^1$. Well, I think in $g(x) = x +1$, but I'm not sure. Any hint is appreciated.  Thank you. Demonstrate that if $f$ is surjective then $X = f(f^{-1}(X))$ I haven't been able to do this exercise:     Let $f: A \rightarrow B$ be any function. $f^{-1}(X)$ is the inverse   image of $X$. Demonstrate that if $f$ is surjective then $X = f(f^{-1}(X))$ where $X \subseteq B$.   Since $X \subseteq B$, all the elements in $X$ belong to the codomain of $f$.  Since $f$ is surjective, it means that all elements in the codomain $B$ have some preimage in $A$. Given that $X \subseteq B$, all elements in $X$ must also have a preimage in $A$.  Have $\triangle = f^{-1}(X)$, $\triangle$ is now a set containing the preimages of the elements in $X$. Because of this, $\triangle \subseteq A$.  If we evaluate $f(\triangle)$, we...... nope, I don't know what I'm doing now.    What do you think? Random weighted choice Consider the class below that represents a Broker:  public class Broker {     public string Name = string.Empty;     public int Weight = 0;      public Broker(string n, int w)     {         this.Name = n;         this.Weight = w;     } }   I'd like to randomly select a Broker from an array, taking into account their weights.  What do you think of the code below?  class Program     {         private static Random _rnd = new Random();          public static Broker GetBroker(List&lt;Broker&gt; brokers, int totalWeight)         {             // totalWeight is the sum of all brokers' weight              int randomNumber = _rnd.Next(0, totalWeight);              Broker selectedBroker = null;             foreach (Broker broker in brokers)             {                 if (randomNumber &lt;= broker.Weight)                 {                     selectedBroker = broker;                     break;                 }                  randomNumber = randomNumber - broker.Weight;             }              return selectedBroker;         }           static void Main(string[] args)         {             List&lt;Broker&gt; brokers = new List&lt;Broker&gt;();             brokers.Add(new Broker("A", 10));             brokers.Add(new Broker("B", 20));             brokers.Add(new Broker("C", 20));             brokers.Add(new Broker("D", 10));              // total the weigth             int totalWeight = 0;             foreach (Broker broker in brokers)             {                 totalWeight += broker.Weight;             }              while (true)             {                 Dictionary&lt;string, int&gt; result = new Dictionary&lt;string, int&gt;();                  Broker selectedBroker = null;                  for (int i = 0; i &lt; 1000; i++)                 {                     selectedBroker = GetBroker(brokers, totalWeight);                     if (selectedBroker != null)                     {                         if (result.ContainsKey(selectedBroker.Name))                         {                             result[selectedBroker.Name] = result[selectedBroker.Name] + 1;                         }                         else                         {                             result.Add(selectedBroker.Name, 1);                         }                     }                 }                   Console.WriteLine("A\t\t" + result["A"]);                 Console.WriteLine("B\t\t" + result["B"]);                 Console.WriteLine("C\t\t" + result["C"]);                 Console.WriteLine("D\t\t" + result["D"]);                  result.Clear();                 Console.WriteLine();                 Console.ReadLine();             }         }     }   I'm not so confident. When I run this, Broker A always gets more hits than Broker D, and they have the same weight.  Is there a more accurate algorithm?  Thanks! Good color palette, usable for the color blind. I'm designing a chart right now, that will slowly have more items added to it. I'd like to create an list/order of colors for the items, that will be most usable for the color blind to differentiate between each item. Any ideas how/where to start? How can I take a photo with everything in focus with my DSLR? I am quite new to DSLRs and one of the first things I noticed was the incredible focus points and appearance of depth that you can achieve. That is great in a lot of scenarios, but not in all. When filming a landscape as a whole, I do not want to have the focus on that single tree, but on the whole skyline.  Is there a way by which you can sort-of "disable" the focus, so that the raw image is recorded without any added blur? I suppose something as simple as this is only a small function you can turn off, but I can't seem to find it in the manual of my camera. Java Command line arguments I am trying to detect whether the 'a' was entered as the first string argument. New database server hardware and software inserts slower than old setup Does any particle ever reach any singularity inside the black hole?