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1. If a graph with n vertices has n ~ 1 edges must it be a tree? Why or why not?...

Question

1. If a graph with n vertices has n ~ 1 edges must it be a tree? Why or why not?

1. If a graph with n vertices has n ~ 1 edges must it be a tree? Why or why not?



Answers

Show that every connected graph with $n$ vertices has at least $n-1$ edges.

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

Okay, So for this problem, it's saying that G is a simple graph with inverted sees. So Part A is asking is ST if saying that G is a tree if, and only if it is connected and has an minus one edges. So because it's an if and only if there's going to be two parts to this proof, there's gonna be a going from the left to right statement. And if then statement part, the second half of this is gonna be an if then going from right to left. So I want to just kind of talk about a few things. So for growing left or right, I can simply state retreat definition but saying that it's yours connection in the end, no circuits. So I can use both of these to my advantage. So going from left to right, brats, half of the proof right there is just stating that definition and then doing for the left side. If I have inverted sees within minus one edges, then I can use induction proof to prove that that is actually the case now going from the rights again, the connection being a tree. You know that kind of fits in together, but I kind of want to bring in the whole entire circuit bring So I'm going to kind of do a negation and kind of supposed that there's from the since I'm gonna suppose there are circuits and what I'm gonna do is I'm gonna remove those until gone. So what I want to show is that I can Onley grifter are inverted, sees so inverted sees um, then removing. I will be removing in minus one. Vergis ease. Until I get to the point, I can't be that anymore. So once you remove the circuits and there's connections there, then there's gonna have to be in minus one edges. Okay, so I am going to go ahead and start on the proof. So going for the left direction or get on estate. My tree definitions of truth g is a tree. There is one Vertex. Then there is, um, it. But it is connected and has no simple circuits. Well, then I want to use, um he performed, so I'm gonna talk about how it's gonna be induction proof. So when an equals one that there is one vertex and no edges, so in minus one, Maybe one minus one, Which is gonna be zero. We want to assume earthy and equals. K statement holds four and minus one equals K minus one edges. Suppose he has K plus one vergis ease. Wanna let Vertex be? I want to show that if I remove a vertex, it's gonna become Cavour to seize. Which would that be? K minus one edges similar. Vertex be, um be a leaf. Four parents for text A. So, if be were removed from G Renji prime would he produced having hey, Vergis ease. G prime would still hold connection with no circuits. Brin g Prime would have k minus one bridges. Yes, he is a tree if it is connected with an minus one urges. And the first part of that is completely gone. So the for the right direction Russell let g b a connected simple graph within minus one. Edges someone a Go ahead. Just say that that's the case. So it's connected. It's not in minus one edges. So I want to go ahead and negate this and show that there's okay. So suppose G is not a tree than the removal oven edge would keep G prime connected as there is a circuit present, G but taking it out. Um, now suppose G prime is not a tree in the removal of an edge would keep g double prime connected as there is a circuit present in G prime. So the process. Okay, so this process would continue until a connected graph is attained with no circuits. Hey, this would require at most in minus one steps, as there are in minus one edges to remove us. Inverted sees exists in G and he is a tree. And of course, that would be the end of the proof. Now, for part B, this one's going to be a little bit different. So first off the the left right is going to be the same as part is a He's left basically, but we could just refer back to that one and call it a day. Now, as faras the as faras the rights going to the right direction case of for the right direction, they actually get the problem actually gives us a hint in it. So it tells us that we need to show that, um, that G is connected. If it has no simple circuit and it's in minus one edges. So we need to show that it cannot have more than one connected component for each Vertex. Okay, so in order to do that, what we need to do is this would be a good opportunity. Whenever you have something that's gonna it's not you're gonna have no more than a certain number. You always want to do a negation type of Chris. So suppose it does have its not supposed to happen or what you're trying to prove that it does. And that way, if you can prove a contradiction and you can, you know, continue warrant. So for this one, this is thesafeside proof, or this has already been proven in part A of this problem with the direction proof. So now we can just jump into the right direction. So what I want to do is I won't identify. So let g a simple graph with no simple circuits and has in minus one edges with man verte sees. I want to suppose this is where my doing what I'm not supposed to be going has more then one connected component such that any to Vergis ease such a C and B would have two possible adds. So if they have two possible paths, that means that Brin G would contain a circuit. Rich is a contradiction. US. G is a tree if, and only if it has no several circuits and has an minus one edges not saying to the proof.


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