00:01
We are given special graphs and we are asked to find an adjacency matrix for each of these graphs.
00:07
Part a, special graph is kn, the complete graph with n vertices.
00:16
So first notice that this graph has n vertices, which i'll call.
00:27
And we also have that this implies that the adjacency matrix is going to be an n -by -n matrix.
00:41
And moreover, we have that every vertex is...
00:53
Is connected to every, i guess i should say adjacent to every other vertex.
01:11
So it follows that the degree of a vertex v is going to be n minus 1.
01:21
And recall from a previous problem we have that the sum of the rows.
01:31
This is going to be the degree, i guess the sum of row i.
01:38
Is the degree of vi minus the number of loops at vi.
01:48
Now, there are no loops at vi in this graph.
01:54
This is a simple graph, and so this is simply going to be equal to the degree of vi, which is n -1.
02:02
So we have that the sum of each row is going to be n -1, and we have that vi is not adjacent.
02:17
To vi, so the main diagonal is just zeros, and every other entry is going to be ones.
02:30
So we obtain a matrix that looks like this, where we have all zeros down the diagonal and ones everywhere else.
03:35
In part b, we are given the graph cn.
03:41
This is the cycle on n vertices.
03:43
So number of vertices is n, and we have that degree.
03:49
Of each vertex is going to be two.
03:59
And finally we have that no loops are allowed.
04:05
And so by a previous problem we have that the sum of the rows, well instead of thinking of it that way, we have that vi is adjacent to vj if and only if i is equal to j plus or minus 1 mod and so i draw the adjacency matrix, have say v1, v2 all the way down to vn, and v2 all the way over to vn.
05:09
And we have that for one.
05:11
For each vi, vi is not adjacent to itself.
05:14
So we're going to have zeros down the main diagonal, and we have that v1's adjacent to v2.
05:31
Likewise, this is asymmetric matrix, since it's undirected.
05:34
V2 is also adjacent to v1.
05:40
We have that v1 is not adjacent to v3.
05:53
We do have that v2 is adjacent to v3.
05:58
And so we get sort of this checkerboard pattern of zeros and ones.
06:06
And so the form of the matrix is actually going to depend on the value of n.
06:14
Well, no, it won't.
06:18
So we know that v1 is going to be adjacent to vn.
06:21
So we'll have ones in these corners.
06:27
And again, there's zeros down the main diagonal...