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A complete Bipartite graph $K_{m,n}$ has a Hamiltonian cycle if and only if $m=n$.

I want to know if the following proof technique is correct. My proof will consider using proof by contradiction.

Assume that $m\not = n$. Let $H$ be the (Hamilton) cycle that goes through every vertex in $K_{m,n}$. $ H = v_0e_0v_1e_1...v_ie_i$. Since $K$ is bipartite, the cycle must alternate between the vertices on each side. Since $m\not = n$ there exists a $v_a = v_b, a < b$ inside cycle H. This leads to a contradiction since a cycle cannot have repeating vertices. Hence, a complete Bipartite graph $K_{m,n}$ has a Hamilton cycle if and only if $m= n$.

Is this correct?

Learner
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Zee
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  • It could be phrased a bit more clearly, but yes, you’ve proved that if there is a Hamilton cycle, then $m=n$. You’ve not proved that if $m=n$, there’s a Hamilton cycle. Admittedly, this is pretty trivial, but it does have to be addressed. – Brian M. Scott Dec 12 '14 at 00:27
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    Your proof looks good. I would justify the existence of a repeated vertex using the pigeonhole principle. You also say here: "This leads to a contradiction since a cycle cannot have repeating vertices." I would note more strongly that a Hamiltonian cycle visits each vertex exactly once. It clarifies for the reader. – ml0105 Dec 12 '14 at 00:28
  • @BrianM.Scott How would one prove the inverse? Contradiction again? Would a direct proof be more appropriate? – Zee Dec 12 '14 at 00:32
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    Direct: just label the vertices $u_1,\ldots,u_n$ and $v_1,\ldots,v_n$ in the two parts and write down a Hamilton cycle. – Brian M. Scott Dec 12 '14 at 00:33
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    And don't forget there is no Hamilton cycle if $m=n=1$. – Especially Lime Mar 22 '18 at 13:42

2 Answers2

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A complete bipartite graph $K_{m,n}$ is Hamiltonian if and only if $m = n$ , for all $m, n \geq 2$.

Proof: Suppose that a complete bipartite graph $K_{m, n}$ is Hamiltonian. Then, it must have a Hamiltonian cycle which visits the two partite sets alternately. Therefore, there can be no such cycle unless the two partite sets have the same number of vertices. If $m = n = 1$, it is clear that $K_{m, n}$ contains no Hamiltonian cycle.

Thus, we get $m = n$, for all $m, n \geq 2$.

Conversely, it is easy to see a Hamiltonian cycle $x_1, y_1, x_2, y_2, x_3, y_3, \dots, x_n, y_n, x_1$ for such graphs.

Elen Khachatryan
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Divyansh Hardia
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Other direction can be prove in following way. As noted any cycle in bipartite graph will be of even length, and will alternate between the vertices of partite-sets. That means any Hamiltonian cycle in $K_{n,n}$ will have equal number of elements from both the partite-sets and as it covers entire vertex set, together we get $m=n$.

user3138972
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