Find the number of paths of length n between any two nonadjacent vertices in K3,3 for the values

step 1 So for this question, we're given K3, 3 of bipartite graph. And we're trying to find the number

Find the number of paths of length n between any two...

Key Concepts

Adjacency Matrix and Matrix Exponentiation

The adjacency matrix of a graph is a square matrix used to represent its structure, with matrix entries indicating whether pairs of vertices are adjacent. Raising this matrix to a power n provides a method to count the number of paths of length n between vertices, as the (i, j) entry of the resulting matrix equals the number of distinct paths of that length from vertex i to vertex j. This tool is especially useful in solving combinatorial path-counting problems in graph theory.

Parity Considerations in Bipartite Graphs

In bipartite graphs, the vertices are divided into two independent sets such that no two vertices within the same set are adjacent. Consequently, any path connecting vertices from the same set must have an even number of edges, while paths connecting vertices from different sets must have an odd number. This parity constraint is fundamental when analyzing the existence and counting of paths between vertices in bipartite graphs.

Complete Bipartite Graphs

A complete bipartite graph is a graph whose vertices can be divided into two disjoint sets such that every vertex in one set is connected to every vertex in the other set, and there are no edges among vertices within the same set. This concept is fundamental in graph theory, as it models relationships between two different classes of objects and serves as a key example in studying graph properties, connectivity, and bipartiteness.

Counting Paths in Graphs

Counting paths in graphs involves determining the number of ways to travel from one vertex to another using a sequence of edges, where the path length is defined by the number of edges. This concept is pivotal in combinatorial analysis within graph theory and is applied in fields such as network analysis and computer science, often using methods like recurrence relations or matrix exponentiation to account for various restrictions like designated starting and ending vertices.

Transcript

00:01 So for this question, we're given k3, 3 of bipartite graph.

00:05 And we're trying to find the number of pads of length, 2, 3, 4, and 5 between non -adjacent vertices.

00:27 And so one way to do this, and this might be what you first think of, is to create the six by six matrix, uh, adjacency matrix of k33.

00:42 And then the way to do that and then just, you know, square it to find your answer for two, cube it for three, et cetera.

00:52 But the way i'm going to solve this problem is a little different because, uh, i think squaring six by six matrices aren't isn't, you know, six by six matrix multiplication isn't exactly the most elegant or, you know, it's a little tedious.

01:08 So, because, so we, first off, all we have to do is, uh, if we think about it mathematically, right, um, each edge, so we can split, uh, the graph into halves, right? like the top half and the bottom half, because each, um, edge connects, goes from the top half, one, two, three to a vertex in the bottom half.

01:37 That's why it's k33, a bar.

01:39 Bipartite graph.

01:40 So essentially, we're trying to find the number of paths from that, like, go to the same half, right? so non -adjacent vertices are always in the same half, right? so basically, non -adjacent.

01:54 One is non -adjacent with two or three.

01:56 So we're trying to find the number of paths of length two from one to two, one to three, et cetera.

02:03 First off, i want to note that we can solve three and five easily.

02:10 We note that essentially dividing the graph into two halves, the number of paths between non -adjacent vertices means the number of paths in which you stay in the same half, right? your end destiny, your start point and your end vertex are in the same half, either the top half or the bottom half.

02:32 Additionally, each edge goes from each edge, you have to switch halves, right? so if you start in the top half and you take an edge, no matter which edge you take, right, for a vertex one, right? no matter which edge you take, 1 ,4, 1 .5, or 1 6, you'll go to the bottom half.

02:47 So, likewise, if you take any vertex from a bottom, if you start at a bottom half vertex and you take any edge, you'll arrive at the top half.

02:57 So for that reason, right, we want to, for three and five, you want to stay in the same half or stay in the same half, which you can't do, right? so basically you have to end in the same half in the same half as you started.

03:24 And that's a path that you want.

03:27 But each edge makes you you switch halves right as we've just discussed so if you take three edges if you start in the top you're going to go to the bottom so three edges will go top bottom top bottom right after three any path of length three in other words will make you switch halves because if you take an even number of edges you will stay in the same half.

04:00 If you take an odd number of edges, you will switch which half you end in.

04:07 So we want to stay in the same half.

04:09 We want to end in the same half, but that's impossible with three, with paths of length three or five.

04:16 So for these paths, you will never end on a non -adjacent vertex.

04:27 Because that landing in a non -indjacent, ending in a non -adjacent vertex means that you stayed in the same half or you ended in the same half as you started but as you said already odd number of edges means that you will be guaranteed to end in a different end in the other half so for that reason there are no pads between non -adjacent vertices of length three or five so for three and five right part b and part d the answer is zero there are no pads now for part a right pads of length two.

05:07 We just want to, for example, any path from one to two, right? all we have to do is choose that there are three paths from one to two of length two, right? we can go one, four, two, one five, two, one six, two for three as well, one four, three, one five, three, one six three.

05:28 And so we see that there are three paths of length two between any two non -adjacent vertices.

05:37 If you're not convinced, we can, one, we can say that, right, if we count the number of paths from vertex 1 to vertex 2 that are length 2, we can say that 1 and 2 are, we are like they're generalized vertices.

05:55 This is chosen without loss of generality.

05:58 These are arbitrary.

05:59 If you switch them, you'll have three paths of length 2 no matter what.

06:03 Another way to think of this is that you can choose which vertex.

06:07 To go to, right? so you know that you're going to choose four, five, or six as your first edge connecting one with something...

Please give Ace some feedback