Question

Let G be a graph.Prove that if G is k-regular and has girth at least four, then G has at least 2k vertices. Determine all such graphs with exactly 2k vertices.

Answer 1

To prove that a k-regular graph G with girth at least four has at least 2k vertices, assume G has fewer than 2k vertices and derive a contradiction.

Step-by-step explanation:

To prove that a k-regular graph G with girth at least four has at least 2k vertices, we can use the following approach:

1. Assume G has fewer than 2k vertices and derive a contradiction.
2. Since G is k-regular, each vertex of G is incident to exactly k edges.
3. Since the girth of G is at least four, each vertex's k edges must be incident to k different vertices.
4. Therefore, if G has fewer than 2k vertices, there must exist at least two vertices that share a common neighbor, resulting in a cycle of length less than four, which contradicts the assumption that the girth is at least four.

Therefore, G must have at least 2k vertices to satisfy the given conditions. To determine all such graphs with exactly 2k vertices, further analysis and enumeration are necessary.