Hamilton cycles for bipartite graphs I know that a hamilton cycle exists for a bipartite graph $K_{m,n}$ if and only if $m=n$
But my question is why is it not possible to have a bipartite graph if $m=n=1$
I mean we would go from $x$ to $y$ and back to $x$
Why this is not allowed, We don't have a restriction on the edges, I mean we are allowed to use an edge more than once in a Hamilton cycles unlike Euler cycles.
So I am just surprised why is it not possible, does someone have an explanation for that ?
Because according to the definition of a Hamilton cycle, It's basically a cycle that begins and ends on the same vertex and we don't have restrictions as I said on the usage of the edges more than once.
 A: I think this is really a conventions question, similar to whether we should say that $0$ divides $0$, or if we should call $1$ prime or composite or neither. 
For a graph with at least $3$ nodes, repeated edge visitations in a Hamilton cycle are prohibited. This happens as a consequence, though, rather than by definition: If we visit an edge multiple times, we must be visiting both of that edge's vertices multiple times, provided there are at least $3$ vertices. This of course violates the conditions for a Hamilton cycle, as we can only visit our start/end vertex twice, to "close up" the cycle (visiting all other vertices exactly once).
In the case of a path/bipartite/whatever-you-want-to-call-it graph with two nodes, I could see convention/definition going either way, whether we should say a Hamilton cycle exists. I don't think it would be very consequential whatever convention we choose (unlike deciding that $1$ shouldn't be prime), but I don't do any serious graph theory so perhaps there are good reasons.
