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How many edges has a tree with 63 vertices?...

Question

How many edges has a tree with 63 vertices?

how many edges has a tree with 63 vertices?



Answers

How many edges does a tree with $10,000$ vertices have?

Okay, So this problem ask tells us that as how many edges are there in a forest of tea trees containing a total of in Verdecia. So first off, we have in minus one edges, and we're gonna say that t I is going to contain in I vergis is so if that's the case that we have in wine plus into plus all the way to end, I is going to equal in That's all of the verdicts. Ease Internet shell for all of the different trees and all the different forests. So I can stay as far as my edges I'm gonna have in one minus one plus into a minus one plus in I minus one is going to equal in one plus into plus and I minus one plus one and all the way to that ice one now, So we already know that this is going to be in, and we know that this is going to be tee times, so it's gonna be in mind

Hi there. We're gonna look at the question concerning the number of Vergis is that a full five very tree with one 100 internal vergis is has. Now, that's actually a pretty straightforward formula to solve this. Ah, but let's derive that form. You know, I'm going to do that by, ah, starting with just the simplest form of the problem. We can, um, instead of worrying about a full five very tree, let's just worry about ah, really simple binary tree. Um, and that could look something like this. We just have one, ah, parent for attacks with two Children coming off it. Now that's a full binary tree, since each for Texas, either terminating or it has two Children, these two on the bottom are terminating, and the one on the top is the parent. And ah, it also only has one internal vertex because an internal vertex is defined as anything that has ah, Children or other Vergis ease directionally coming off of it Now. Since this is written with the direction with the direction of flow going down, then this top Vertex is the only one that has things coming off of it, and this is the only internal vertex. Ah, hence we're looking at Ah ah, simple tree with just one internal vertex. Ah, but then it obviously has three total vergis, ese. So the question we're trying to answer is, how do we get from this number to this number? Well, if you think about it, if there's one internal vertex in a binary tree, that means there's one vertex that has Children and it's a full binary tree. Then it should have two Children coming off of it. So this one top vertex, um, leads to two and a There you go. Ah, one parent plus two Children equals the three Vergis ese that Ah, we want Ah, so that's a pretty simple solution. Doesn't really tell us. Exactly. The formula is still, um, but let's look at that Something slightly more complicated. Let's take a similar binary tree, but then add on extra layer to it. Now, this is still a full binary tree. Each parent has two Children. Um, but ah, now we're looking at three levels. It has a height of three. Ah, So, ah, in this case, there are now, um, three internal Verdecia because there's three of our Texas here that have ah Children coming off of them. And, uh, when you look at their seven total vergis ease, now we're looking at three and seven. Now, let's keep the same logic we had for the last tree and say, Okay. Ah, if there's three total, um, parent Verdecia is that means each parent, vergis e. Has to Children coming off of them. Three intern overseas times two, that equals six total Vergis is, um so this means we know their six Children. Ah, no. What of these? Seven Vergis ese? Which six of those counting? Well, it's actually counting, and I'll mark this in green Mark, these six and green here, it's actually counting all the Children. Um, so if you could define any of these these Children than, ah, they're being counted in the six. So that's this one, This one and these four down here, it's not counting this top one. Ah, because the top one here is ah, parent. And it has no ah has no overseas above it that are leading to it hence were able to count all six. But then we need to add one at one. Here, get back to our seven total vergis, ese. Ah. So the form that becomes a little more apparent because you see that if we know that there are em branches off of every ah, every parent. So, uh, listen a new tab, Uh, if there's, um, on m number of branches. So in this case, it was Ah, it was a binary tree. So that was to, um Then you could multiply that by the number of internal Vergis. Is that our That's the number of total parents. If you're thinking in that terms, and will this label that as capital I for internal Ah, and the times I gives you the total number of Children? But that leaves off that top one. Because no matter how many, no matter what m r I is, the tree will always look something like this where it's gonna have a bunch of stuff below it. But there's be one parent of the top that's not counted. Ah, in the number of Children. Remember this the number of Children. Um, so since that was not counted in this am I, we just add one to that and that gives us our total number of vergis ese so going back to the original problem, um m was equal to five. I was equal to ah ah, 100. Um and thus the formula gives us that very Vergis ease equals five times 100 plus one, which equals 501. There's your answer.

Okay, So you're asked how maney edges does. A full binary tree with 1000 internal vergis is half. So recall that a full Emily tree with I internal Vergis is has well and is equal to m I plus one. Vergis is, and a tree with advertises has n minus one edges. So then it follows that a full emery tree with I intern advertises has e being equal thio m I edges. Okay, so suppose that t is a full binary tree with 1000 intern overseas. Well, here we have that m being equal to two. And we have that I is equal to 1000. So then, um therefore, e equals m I. That's two times 1000 which is 2000. So therefore, a full binary tree with 1000 intern over disease has 2000 edges. All right, all right. 2000 inches

We're told that a regular graph we of Degree four has 10 edges and were asked to find number of vergis ease. This graph has so since this is a regular graph with 10 edges and each Vertex has degree, for you have by the hand shaking your um, that's to times the number of edges 10 is equal to. It's some overall Vergis ease of the degree of the Vertex V. This is equal to the sum over all the vergis ease be of or and then if the number of Vergis ease is, say N, this is equal to four times en. So we had that four ends equal to two times 10 March 20 or that's N is equal to fine, so the graph has five Burgess sees.


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