Suppose that G is a complete bipartite graph with at least 4

Suppose that G is a complete bipartite graph with at least 4 vertices. Give a necessary and sufficient condition for G to have a Hamilton cycle. Briefly justify your answer.

Solution

We proceed by induction on n. For n = 1, the only graph with 1 vertex and 0 edges is K1, which is a tree. Suppose that every connected graph with 1 vertices and 2 edges is a tree. Let G be a connected graph with 4 and edges. First we claim that G has a vertex v of degree 1 Indeed,If every vertex had degree at least 2, then the right-hand side of the above equation would be at least 2n and that is not possible. Since G is connected, every vertex has degree at least 1 and, therefore, there must be a vertex v of degree 1.

A Hamilton path is a path in a graph G that passes through every vertex exactly once ,the Hamilton path must pass through each vertex exactly once and we do not worry about the edges, while an Euler path must pass through every edge exactly once and we do not worry about the vertices. that is the difference in between them ,

Let G be a connected, simple graph with N vertices, where N 3. If the degree of each vertex is at least n/2, then G has a Hamilton circuit.sufficient conditions to determine if a graph has a Hamilton circuit and path.