What is a 3-regular graph?

A 3-regular graph is known as a cubic graph. A strongly regular graph is a regular graph where every adjacent pair of vertices has the same number l of neighbors in common, and every non-adjacent pair of vertices has the same number n of neighbors in common.

Is there a 3-regular graph on 9 vertices?

Question 38. We have shown the regular graphs of degree 2 on 8 vertices in Q21; there are no others. There are no graphs that are regular of degree 3 on 9 vertices. Why? (How many edges would such a graph have?)

How many edges will be there in a 3-regular graph of 6 vertices?

This graph being 3−regular on 6 vertices always contain exactly 9 edges. As this graph is not simple hence cannot be isomorphic to any graph you have given. A graph on 6 vertices is regular of degree 3 if and only if its complement is regular of degree 2.

How many simple graphs are there on a set of 8 vertices?

I think the total number of edges for a graph with 8 vertices would be: n(n−1)/2 which would yield 28. total number of set with 28 elements is 228.

Can a 3-regular graph have 5 vertices?

For a graph to be 3-regular on 5 vertices, the degree of each vertex must be 3. A graph cannot have a non-integer number of edges such as 7.5, so there is NO way for there to be a 3-regular graph on 5 vertices.

Can you draw a 3 normal graph with 7 vertices?

We know that the sum of the degrees in a graph must be even (because it equals to twice the number of its edges). Hence, there is no 3-regular graph on 7 vertices because its degree sum would be 7 · 3 = 21, which is not even.

How many graphs with 3 vertices are there?

There’s 3 edges, and each edge can be there or not. So 2^3=8 graphs. Unless you’re counting graphs up to isomorphism, in which case there’s only 4.

Does a 3-regular graph of 14 vertices exist?

If k 1 = 4 and k 2 = 4 , then G is isomorphic to Q 4 and hence, by Theorem 1.1, there is a 3-regular, 3-connected subgraph of G on 14 vertices.

How many graphs does 4 vertices have?

There are 11 simple graphs on 4 vertices (up to isomorphism). Any such graph has between 0 and 6 edges; this can be used to organise the hunt.

Are there any 3 regular graphs with odd number of vertices?

Closed 5 years ago. Do there exist any 3-regular graphs with an odd number of vertices? I’m starting a delve into graph theory and can prove the existence of a 3-regular graph for any even number of vertices 4 or greater, but can’t find any odd ones. Corrollary: The number of vertices of odd degree in a graph must be even.

Which is the degree of a regular graph?

In a complete graph of N vertices, each vertex is connected to all (N-1) remaining vertices. So, degree of each vertex is (N-1). So the graph is (N-1) Regular. For a K Regular graph, if K is odd, then the number of vertices of the graph must be even.

What makes the product of a regular graph odd?

For a K regular graph, each vertex is of degree K. Sum of degree of all the vertices = K * N, where K and N both are odd.So their product (sum of degree of all the vertices) must be odd. This makes L.H.S of the equation (1) is a odd number. So L.H.S not equals R.H.S.

How are graphs divided according to edge connectivity?

A classification according to edge connectivity is made as follows: the 1-connected and 2-connected graphs are defined as usual. This leaves the other graphs in the 3-connected class because each 3-regular graph can be split by cutting all edges adjacent to any of the vertices.