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Show that if G is a 3-regular simple connected graph with facesof degree 4 and 6 (squares and hexagons), then it must containexactly 6 squares....

Question

Show that if G is a 3-regular simple connected graph with facesof degree 4 and 6 (squares and hexagons), then it must containexactly 6 squares.

show that if G is a 3-regular simple connected graph with faces of degree 4 and 6 (squares and hexagons), then it must contain exactly 6 squares.



Answers

Show that if $G$ is a bipartite simple graph with $v$ vertices and $e$ edges, then $e \leq v^{2} / 4 .$

So this question We're trying to show that a simple graph with an inverted sees must be connected if it has greater than for it. But sorry, greater than and minus one times online, it's to over to edges. So 11 way to do this is to show that if this simple draft so our logic is, if the graph is not connected, then C max number of edges is n minus one times and minus two over two. That's one way to solve this problem, right? Because if the mass number of edges in a non connected graft is this number, then that means that if you have more than this number of edges than you must be connected. So that's the essentially the overview of how we're going to solve this problem. So without loss of generality, supposed that the graph is not connected, then let's say one component has Cavor two seats where K is just a number from like 12 and minus one. Okay, Verte season. One component That means that there are n minus Cavor theses split among at least one other component, right? We don't know if they're all in a separate component where if they're like, all separate at the one other important, right? So if we show that this graph has can't have more than this, number of edges were good. So we want to maximize maximize the number of edges in this graph in this graph in the graph above right. And the way to do that is by assuming right. As we've said before, the maximum number of edges in a simple graph occurs when the graph is complete. And so let's assume that this is a complete component. So if we maximize that, we get that this we have cages to Verte season this complete component and then to maximize the number of edges. And this we assume that they're old in one component and that if there's a complete component as well, so the number of met. So we have maximized number of edges and this graph by assuming that both of these components are complete and then that gives us ah que squared minus NK plus n squared minus two end over to that's This is the maximum number edges in the end, this in this craft, and so we want to maximize this expression right. So we realized if we look at its Ah, this is a problem shaping upward, right? If we find its Vertex, its vertex occurs by calculus at an over to we just take the derivative. So we way we get to K minus and is equal to zero. So and is able to care word to okay is able to end over to Sorry and sorry. But Texans here, that means that it's Massouma will be at the bounds, right? It will be greatest farthest away from an over to So we know that bound the domain, right? For for a k iss 12 and minus one. Luckily, that's symmetric about an over two. So basically, this, uh, this expression, right. Um So in other words, this expression in quotation marks is maximized that que goes one and cables and minus one. They will both give us the same answer. So let's see what the expression What? This Ah, this evaluates to ATC equals one. So he was one. This expression becomes one minus and plus n square blind stand over to is able to end squared minus three. Annual plus 2/2 is equal the end minus one times n minus 2/2. Well, uh, so in other words, in other words, the mass me number edges in a graph that has Cavor season one component and M minus capers use. Put him on. At least one other component is this number. So we found that the max number of edges in a disconnected graft on Enver disease is an minus. One times n minus 2/2. And we're done, right? So in other words, if you have more than that, if you have more than minus one times that minutes to over two edges, you must be a connected graph. And that's what we're looking for. So to review, Um, a simple graph is connected, if that's more than this number edges. So we flipped the from law on its side and we were like, Okay, how can we look at this from another lens? Ah, the other perspective is that if the graph is not connected, then it can't have more than this number of edges, right? And so that's what we were seeking the show. So essentially we assumed that the graph was not connected. In other words, that had Cavour to season One component for que is between one and M minus one and the other end minus Cavour to Cesaire, split among at least whether one other component. So if we show that the maximum number of edges in this graph is this number, then we're done. We have to do is maximize number of edges and dis craft, which we have done by assuming so. The way to do that would be to assume that this is one component it is complete that will give you this many agents. That's an accident number of edges and this component. And then to maximize this part, we assume that these end my escape overseas are all in one component and they're all complete. They're all connected with each other. So the maximum number of edges we get this expression and then we need to maximize this expression for for K, for the domain for the domain in which four K is between one and M minus one. And so we realized that the domain that the problem is maximized at cables one and chemicals and minus one. And so the maximum number of edges is is once we work it out and minus one times in lines to over to, and we're done

Question on the one show that if he is connected, then it is possible to revert seems to disconnect you. It's an only if he is not. For one, reading a person's problem is to show that if G is complete, then we cannot renews ver disease to disconnect G. And alternatively, if he is not complete, then we can. We're news, Berta. Seems you disconnect, Cheat, right? Doing so essentially showing that these two parts are both true will give us the information needed to draw that conclusion of from 51. But first off, let's assume that G is a connected graph on N bird of seed straight. So if you remove one vertex, then we'll get connected. Graph on n minus one Bird sees right because if we remove one for attacks, then the degree at all the remaining burgesses goes down by exactly one. So all and minus one Burgess sees will have degree and minus two because they the edges connecting. They're all connecting all pairs. All possible pairs of these end minus one remaining burr disease will still be intact. There still be edges. So basically, if we remove moving J verjus, he's done right Gives us a complete graph on n minus J on all and minus J remaining emergencies. So this goes on forever. And so you cannot in conclusion, you cannot for a new cannot disconnect g by removing bird disease. Because no matter how many read, remove, we'll just get a complete draft on the remaining Vergis ease in the graph. So the first part of the statement is complete, right? We've shown that if G is complete and we no matter what we worked burgesses removed g will always remain connected. Now, if G is not complete, this is the second part of the problem. If he is not complete, then there exists UV, a pair of urgencies union G such that you ve is not an edge. Right? Because if there was an edge connecting every pair of distinct for two season G, then G would be complete. He used a is not complete. We confined such a pair. You be such a UV UV Vergis is such that they're not connected by edge. No, the way we're maneuver theses in order to create a disconnected graph, it's just removed. All Vergis is But you envy, right? Is If we do so, we will be left with two overseas, no edges. And this is not a complete craft. And so this will results in disconnect in this connected G. And for that reason, if she is not complete, we can remover disease to disconnect. Right? We've proven both these two statements because that's all we need. We're done. The problem is done. Inclusion. We just split the problem into two parts. One was approving the statement regarding a complete craft almost proving statement regarding a non complete breath that's improved. Both of those are, you know.

One show that if if G is a graph with an verses, then no more than an over two edges can be colored the same in an edge coloring of Jesus. So, um well, we know we know that two edges, uh, that have, um They have the same colors, have no common points, so we know that two edges that's have the same colors. Look, I'm in weeks, okay? No. Okay. Now, in case we call her end over two edges with the same color, then the graph can have more than well end times and over two, which is equal to end mercy's. So, um, we're state this as, um Well, in a case where we color more Dan and over too, eh? Jews with the same color. Um, then, um, the graph can have more. There are no. Well, then two times and over two, which is equal to n first. Okay, so, while hence in any graph g with an adversities. So heads and any graph any graph g with en versus, um, no more then and over two inches. Candy colored hopes can be would be cholera rules colored, um, with the same color in an edge. Coloring of G. Okay. Color with the same. Call him and okay, that's calling Orgy. Yeah,

You're even so grab show that is congruent to zero or one bod. Four. I suppose that G is a self complimentary simple graph for Texas V an Edge, said he. So since she's self complementary, it follows that G is Isom or thick, too. It's compliment, and we have that the compliment is going to be the and the complement of E. And we have that since the rights of more FIC, the number edges in G is equal to the number of edges in the complement of gene. Now suppose that member of courtesies is end. We're not in, but instead what's USVI? Then we have that the previous problem. The number of edges in the graph g must be less than or equal to the number of urgencies times the number of courtesies minus one over to. And since there could be at most this many edges and since GMG compliment have the same number of edges you have, that smacks the number of edges going to equal to number edges G plus number of edges in the complement of G just going to be two times number of edges. G since those are equal, so it follows that the number of edges is equal to Beatem Zeon minus one before, but the number of edges has to be an integer. So it follows that three times V minus one must be congruent to zero marred. For now we have that. If V is even follows, that B minus one is odd and so every times three minus one is even, this is odd. Then B minus one is even. And so it follows that we times remind us worn. It's still even so, we'll be times V minus one is also congruent to zero mod to And so it follows that in the case that V is even, we have that for v times. We might just want to be confirmed as your Ahmad. For it follows that he must be congruent to zero mod for the B minus. One is bod Likewise, it is odd. It follows that b minus one b can grow into twos your ahmad for so we have that he is congruent to zero for one model For this is what we wanted to prove


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