Does there exist a simple graph with five vertices of the following degrees?

A simple graph has no parallel edges nor any loops. There are only 5 vertices, so each vertex can only be joined to at most four other vertices, so the maximum degree of any vertex would be 4. Hence, you can’t have a vertex of degree 5.

Can a simple graph exist with 15 vertices each of degree 5?

Therefore by Handshaking Theorem a simple graph with 15 vertices each of degree five cannot exist.

How many edges does a complete graph of 5 vertices have?

It has ten edges which form five crossings if drawn as sides and diagonals of a convex pentagon.

Is it possible to construct a graph with 15 vertices such that 5 of the vertices have degree 2 and remaining vertices have degree 3?

Solution: This is not possible by the handshaking theorem, because the sum of the degrees of the vertices 3 ⋅ 5 = 15 is odd. Theorem: An undirected graph has an even number of vertices of odd degree.

How many edges are there in a graph with 4 vertices each of degree 5?

This is a repeat of Q. 20. For 3 vertices the maximum number of edges is 3; for 4 it is 6; for 5 it is 10 and for 6 it is 15. For n,N=n(n−1)/2.

How do you know if a simple graph exists?

Note that a simple graph is a graph with no self-loops and parallel edges….Stopping conditions:

1. All the elements remaining are equal to 0 (Simple graph exists).
2. Negative number encounter after subtraction (No simple graph exists).
3. Not enough elements remaining for the subtraction step (No simple graph exists).

Is there a 3 regular graph on 9 vertices?

A 3–regular graph is one where all the vertices have the same degree equal to 3. If we try to draw the same with 9 vertices, we are unable to do so. This is as the sum of the degrees of the vertices has to be even and for the given graph the sum is , which is odd.