Condition for Complete Bipartite Graph to be Semi-Hamiltonian

Theorem
Let $$K_{m, n}$$ be a complete bipartite graph.

Then $$K_{m, n}$$ is semi-Hamiltonian iff either:


 * $$m = n = 1$$;


 * $$m = n + 1$$ (or $$n = m + 1$$).

Proof
Let $$K_{m, n}$$ be a complete bipartite graph.


 * If $$m = n = 1$$ then $$K_{m, n} = K_{1, 1}$$ is trivially semi-Hamiltonian:


 * K1-1.png


 * If $$m = n + 1$$, we can construct a Hamiltonian path as follows.

Let $$K_{m, n} = \left({A|B, E}\right)$$ such that $$|A| = m, A = \left\{{u_1, u_2, \ldots, u_m}\right\}, |B| = n, B = \left\{{v_1, v_2, \ldots, v_m}\right\}$$.

We start at vertex $$u_1 \in A$$.

As $$K_{m, n}$$ is complete, there is an edge connecting $$u_1$$ to $$v_1$$.

From $$v_1$$, there is an edge connecting $$v_1$$ to $$u_2$$.

And so on.

There is therefore a Hamiltonian path $$\left({u_1, v_1, u_2, v_2, \ldots, u_{n-1}, v_{n-1}, u_n, v_n, u_m}\right)$$ (where of course $$u_m = u_{n+1}$$).

So when $$m = n + 1$$, $$K_{m, n}$$ has a Hamiltonian path.

However, from Condition for Bipartite Graph to be Hamiltonian, $$K_{m, n}$$ is only fully Hamiltonian if $$m = n$$.

Hence when $$m = n + 1$$, $$K_{m, n}$$ is semi-Hamiltonian.

Similarly, if $$n = m + 1$$ the same argument applies, but this time the Hamiltonian path is $$\left({v_1, u_1, v_2, y_2, \ldots, v_{n-1}, y_{n-1}, v_m, u_m, v_n}\right)$$ (where of course $$v_n = u_{m+1}$$).


 * Now, suppose $$K_{m, n}$$ is such that neither $$m = n + 1$$ nor $$m = n + 1$$, and it is not the case that $$m = n = 1$$.

If $$m = n$$, then from Condition for Bipartite Graph to be Hamiltonian $$K_{m, n}$$ is Hamiltonian, and therefore not semi-Hamiltonian.

If $$m > n+1$$, then we can once more trace a path through $$K_{m, n} = \left({A|B, E}\right)$$ as before, but at the end of the path there is a spare vertex in $$A$$ which has not been traversed which you can't get to without going abck to one of the vertices in $$B$$ that you have already visited.

Similarly if $$n > n+1$$.

Hence the result.