Notice that .
So (i) is impossible.
For (ii) let s be the number of degree six vertices and t be the number of degree three.
What is wrong with that?
I have an exercise about graphs which I cannot solve. Could you please help me?
The exercise is as follows:
Thank you very muchA simple graph, also called a strict graph, is an unweighted, undirected graph containing no graph loops or multiple edges. A well-known theorem states that the sum of the degrees of the vertices of a simple graph equals twice the number of edges of the graph.
Prove the following:
There is no simple graph with 12 vertices and 28 edges so that
(i) all vertices have degree 3 or 4
(ii) all vertices have degree 3 or 6