Condition for Bipartite Graph to be Hamiltonian

Theorem
Let $G = \left({A \mid B, E}\right)$ be a bipartite graph.

Let $G$ be Hamiltonian.

Then $\left|{A}\right| = \left|{B}\right|$.

That is, there is the same number of vertices in $A$ as there are in $B$.

Proof
Let $G = \left({A \mid B, E}\right)$ be a bipartite graph.

To be Hamiltonian, a graph $G$ needs to have a Hamilton cycle: that is, one which goes through all the vertices of $G$.

As each edge in $G$ connects a vertex in $A$ with a vertex in $B$, any cycle alternately passes through a vertex in $A$ then a vertex in $B$.

Suppose WLOG that $\left|{A}\right| > \left|{B}\right|$, that is, that there are more vertices in $A$ than in $B$.

Let $\left|{A}\right| = m, \left|{B}\right| = n$.

Suppose $G$ has a Hamilton cycle $C$.

Let that cycle start at $u \in B$.

After $2n$ edges have been traversed, we will have arrived back at $u$ again, and all the vertices of $B$ will have been visited.

But there will still be $m - n$ vertices in $A$ which have not been visited.

Hence $C$ can not be a Hamilton cycle.

Note
The implication does not go both ways.

This graph:


 * NonHamiltonianBipartite.png

clearly fulfils the conditions (i.e. bipartite graph such that $|A| = |B|$) and is equally clearly not Hamiltonian.