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How many subgraphs do the following graphs have?The graph with degree sequence 3,2,2,1_A graph with degree sum of 20.The complete graph on n vertices:...

Question

How many subgraphs do the following graphs have?The graph with degree sequence 3,2,2,1_A graph with degree sum of 20.The complete graph on n vertices:

How many subgraphs do the following graphs have? The graph with degree sequence 3,2,2,1_ A graph with degree sum of 20. The complete graph on n vertices:



Answers

How many edges does a graph have if its degree sequence is 5, 2, 2, 2, 2, 1? Draw such a graph.

This question. We are us. We are given us degree sequins off a symbol graph and we ask, How many ages are there in this graph and asked to draw it. So we're just gonna draw it and cow the it just later. Right The way to right, Tao, a graft from degree sequence like this. You can do it in many ways. But what I usually do is that we start with the largest number, right. It will represent politics with the most degree and we draw it. Just all of them took equal to that degree. And now this First what is this is done. They there will be nothing else attached to it. We can we can call it a or something. This is the first day Now the seconds that we look at the next number. Right now this work is is how total up tree degree. But now it has one degree from the first politics. Let's let's call this be you need to moss. So I would just added to two others. What is in doesn't matter which as long as it doesn't want a lift there their degree numbers. All right, so now be has three degree already. We still left with another one with three degrees. Right. And you call this see? So this need one one edges. But where would this go? Can it go to this? What is we can because it already has two degree, right? And so the last one that we can map to is the one with only one degree. So it has to go there. All right, on this is I was in Bo graph. It has a total of seven edges. Okay. And that is it. Thanks.

We're given the graph in each part and were asked to find how maneuver theses and how many edges these graphs have. In part. They were given the complete graph with end nodes K. And so now, in the complete graph, we know that each Vertex shares an edge with the end minus one other virtus ease. And so it follows that for over dis ease. The degree of the vergis e is an minus one. Also, for I get too far. There are Enver disease, by definition, since the degree over Texas and minus one. We know by a theorem from the book called The Handshaking Lemma Sometimes but the handshaking, the're, um, as they call it, tells us that's the son of the degrees of the Vergis ease for overseas in the Vertex set is going to be equal to to em For some integer em, however we have that this is the same as the sum for over theses and V of and minus one is equal to two m, and this tells us that we have end times and minus one equals two m so that M, which is the number of edges, in fact, is equal to end times and minus one over to. And so we have. This complete graph is going to head end times and minus 1/2. Edges in Part B were given the cycle with inverted sees. CN, by definition, has enver theses, and we have that. It also has sort of easy to see. Intuitively, you can prove it rigorously you want and edges. This is because each Vertex has in edge coming off of it. And so there's a 1 to 1 correspondence between the Vergis ease and edges in Part C. We're given the wheel with an overseas, so this is going to have, well, not Enver Toussie's, but, um, the wheel. And this is going to be the number of Vergis ease in the cycle. CNN, which is inverted sees plus one for the extra Vertex in the center, says it's going to be n plus one Burgess ease, and we have it. The number of edges is going to be well. We have the number of edges from the cycle, which was and edges plus and then have that the central Vertex is going to have and more edges. We're not to each of the in outer Vergis ease. And so this craft has to end edges in Part d. We're given the complete bi partite graph K m n for some integers, m and n. And by definition, since we have that V one has inverted seas and has be Tuas Enver to seize the total number of urgencies is the verses in V one m plus the very season be too. And which is M plus and Morrissey's and now thinking about, say, Vertex. The one from Vertex said V one. You know that the degree of this Vertex is going to be well, there's Enver Theses in V two, which view one has to have an edge exactly one edge with each of them. So the degree is going to be n and we have. That's for all v twos. And yet ah Burgess E set V two. It follows that the degree of V two is going to be well for each vertex in one. There has to be an edge to be to, and so this is going to be M. And so it follows that by the handshaking, the're, um, two times the number of edges E. I'll call it is the sun over all verte season The Vertex set v off the degree of the your texts And this is the same since V one and V tour this joint as some of all the ones in a V one of the degree of the plus the sum for all these in B two of the degree of V And we have from earlier that this is going to be the some for all these in the one, uh and plus the sum over all Visa and be too Uh, I m and we have it. There are end Verdecia is in V one and vs and the ones this is going to be equal to m times and plus and there are inverted season V to this is going to be end times m. And so this is going to be to M and and therefore we have a number of edges. Is m times and edges. Oh, in part e were given que of then this is the end cube and we have that This is the graph which represents the, uh to to the end, ways to write a binary string of length and And so, if all of theirs to to the end vergis ease and to find the number of edges. Well, think about how que en is going to be constructed. So starting with Q one, this is simply we have possibilities of the binary could be zero or one. And so we just have scrapped with two verte season an edge. So there's one edge and consider que two. And now we have the possibility of 00 and 01 That's from the old graph butts. To get to this new graph, we also have the possibilities of 10 and 11 So we're going to add two more vergis ease and an edge between them, and we want to connect any two representations which differ in exactly one digit or one bit. I should say binary. So we have that there's at between 00 and 10 and judge between 1101 and so we end up with four edges. So notice that this is equal to two squared, which is the same as two times two to the first noticed that one is the same as 2 to 0, just the same as okay, the same as one times 2 to 0 and all. Draw. So now consider I'm going to say that in general Que en is going to have end times two to the n minus one edges. So in order to prove this, you could use induction. So you've already shown this is true for the case and equals wanted to So suppose that's que en has end times to the n minus one edges for some end greater than or equal to one. It follows that Hugh N Plus one is constructed and this was in the book by making two copies of que en clodagh top and bottom. And this is going to contain two times end times to to the n minus one edges already and then ad ones and front of top bits. I guess I should say the top strings and zeros in front of the bottom strings and then connect a top and bottom verdicts or vergis ease if and only if their strings it for by exactly one it. And so the question is, how many edges does this at? Well, Notice said In the case of Q two, this added to edges So we had it one edge with 00110 and we had one edge with 01 and 11 and everything about Q three. We have essentially two Q twos, so like squares. And to connect these. There's really only going to be one other Vertex to connect to across the space. So this is going to ad another the number of Verdecia uses and edges. And to find the other Vertex it differs by exactly one bit we have to do is find the one that we added a one in front of instead of zero. So we have that the total number of edges is going to be equal to and plus two times and minus two to the end, minus one just equal to end plus one times two to the end. Just what we wanted to show. And so it follows that que en has end times two to the n minus one edges for all natural numbers

Were given a number of Vergis ease. We were asked to find how many Nano some ice Isom or fix simple graphs there are with this number of emergencies party. We're told the number of urgencies end is to Yeah, I noticed that we have a simple graph with two Vergis is se A and B. You owe me graph. We could draw these flu to possible grass so either have A and B with no edge between them or the the courtesies A and B with exactly one edge between them. So it follows that there are two Nanos, more thick grafts, and the reason they're nice amorphous is because the verge issues where Jason didn't want and not a Jason on the other in Part B were given any close three. So this is going to be a simple graph with three very sees a be I see. And so to find a number of minus more photographs First, you could have that all of the vergis ease are isolated ver disease. That's one possibility. It could also have that one of the Vergis season isolated Vertex and the other two are pendant vergis ease. We could also have that two of the Vergis ease our pendant courtesies, and one of the Vergis ease has degree to. So, in essence, this is the graph. No, this is a just a path, I think to. And finally, we could also have a graph where each Vertex has degree to says the complete graph on three Vergis ease. So it follows that for more analytical standpoint, the graph can have anywhere from zero of through three edges, and we have that graphs with the same number of edges are Isom or FIC, and we can see it's just a renaming points to give one of the graphs we've shown above. So it follows that the number of 90 smart forfeit graphs is for 90 days. Some more Think grass. Finally, in part C. We're given that n equals four, so we have four Burgess ease, ABC and D. Now we have that the maximum number edges is going to be four times 3/2, which is six. So the graph can have 01 all the way up to six edges total. And of course, if the graph has zero edges, there's only one not Isom or thick graph And this is simply the graph where all the Hendon courtesies? Uh huh Oliver theses are isolated. Vergis ease now consider if there is one edge, but we have it all graphs with one edge our ice and more thick. And to see why we construct such a graph say, there's an edge between A and B. We contain any other graph with exactly one edge by simply re labeling, so we have one, not a some more thick graph. Next, let's consider there are two edges in the graph, and this is actually going to have two different cases. So suppose that the edges are adjacent, that is, they both contain a common vertex since, for example, we'd have sand between and be an edge between being see Well, it follows that by re labeling vergis ease. We has that. All other graphs with two edges which are adjacent are the same upto Isom more physicians. So we have 19 ice amorphous graph here, and in the second case, suppose that the edges are not adjacent, that is, they don't have any end points in common. So, for example, we have graph with an edge connecting a and be photographed with an edge connecting C and D. Well, the only their graphs of this kind they're all the same as other. Twice a morph is, um this graph. So it follows that this case provides one more non isom or FIC graph. So it follows that if we have two edges, you are too non I so more thick grafts. Now consider the case where we have three edges me. A few different cases we have the edges could form a triangle. And clearly in this case, we have that all graphs of this type are the same upto isom dwarfism. There's one non isom or thick graph here. I got this call now Also, consider that we have three edges. We could also form a path so it would look something like this is likely to be B is connected to C and C expected to D. And we have that Any graph of this type is the same up to Isom Morph ism says provides one more non I so more thick graph you can obtain the other graphs, but simply re labeling Vergis ease and finally consider case where if we have three edges, the edges form a star in particular. This is the star like this. So I put a B and C around the outside and D in the middle. And so we have an edge from a two D image for me to be attribute to see And we have that Yes, we have another graph which forms a star. These two graphs rice, um or fix and we buy relabel, inverted sees. And so it follows that provides one more non I so more thick graph. And so it follows that case of three edges provides us with three non a Samore FIC grafts. And next, let's consider the case of four use. Now consider we have one possible graph is simply a square, and we have that. If this is graph G, it follows that you complement of this graph shown here G bar and another possibility. So in this graph we made it so that every point had a degree of two. But suppose that instead one of the Virgin sees has a degree of one. So have a k'nex to be connects to see next to D. But then we'll have that a connects to see again so call this H. Well, clearly we have that G and H seems like they certainly aren't Isom or FIC. But to see why. Consider also that the complement of this graph each bar looks like this, and this is another graph two edges. You recall that we have a graph is going to be nice. Amore faked another graph if and only if it's complimentary. Graph is nice, amorphous to the other grafts. Complementary graph. So it follows that since there are two kinds of or to nine ice amorphous graphs with two edges, there are two non I so more thick grafts on four edges running out of space here. But now let's consider five edges. Well, if we were tohave five edges well, I'm married this and blues with easier distinguish. This means that this is graph G than the complement of G has. Well, exactly one edge. There's only one non I so more thick graph with one edge, and therefore there's only 19 some orphan graph with five edges. And finally consider the case where we have six edges. Well, in this case, the graph is simply going to be K four. Alternatively, we have that complement of this graph has zero edges. And of course you shouldn't. There's only one nice amorphous graph zero edges in there for one night dysmorphic craft with six edges, he said. The total number of nice, amorphous graphs is going to be one plus one plus two plus three plus two plus one plus one. This is equal to 11. And so you're sure that there are 11 nice, amorphous graphs, if any equals four.

This question. We are asked what? The gray sequins off a complete by party Graff came and would be so first we look at a complete by a tiger. Have you called the two part Hi, we and you. Since it is a complete by part I ish were takes in one pot. I rue how edges too. Off the opposing party Where takes cyst. And so his degree on on the Wheaties I degree off each vertex ruby in always right because it touched everything on the use I Conversely, the degree of what takes on you had high view equal to him because it touched everything on this side on the res I and so that's pretty much we have all the degrees off every word. Texas. We just have to arrange them, Eden. Even a man knew that. Give the greater number first. So we have two kisses right when m is granted, and then the work takes that has degree am gonna come first, and there are in off them. So we're gonna have these sequins and conversely as well similarly, b, you ve have Sorry, this doesn't make sense. You ve how in greater than him. Now we're gonna have sequins that in Come first, followed by him and their numbers gonna be reverse as well. But this makes sense. It's nothing complicated. So here you have it for complete by Patrick am in, you're gonna have a sequence of degree as just him and end in some order. Thank you.


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