00:01
We were asked to find the number of non -isomorphic simple graphs with six vertices, where the degree of each vertex is three.
00:22
So i'll call these vertices a, b, c, t, e, and f.
00:38
And because these are simple graphs, we have by the handshaking theorem that two times the number of edges is going to be equal to the number of vertices six times the degree of each vertex, which is three.
00:54
So we have the number of edges is going to be nine.
01:02
So this means that if we have nine edges, it's consider the possibilities.
01:36
So we could have that a connects to b and b connects to c, and c connects to d, d connects to e, e connects to f, f connects to a.
01:51
So for all the vertices have a degree of two.
01:55
However, you've only used one, two, three, four, five, six edges.
02:01
So we have three edges left.
02:02
You want to make it so that each vertex has a degree of three.
02:09
So to do this, simply connect across the graph.
02:17
So have an edge with a and d as well as an edge with f and c and an edge with e and v.
02:28
This is a clearly valid construction here.
02:33
Another possibility.
02:50
So we'll have the same outer edges, and a is still connected to d.
02:57
However, instead of connecting f to c, let's connect f to b and connect e to c...