Bipartite Graph with partition $V$ = $X $$\quad\cup Y $.
Suppose graph $G$ is not Hamiltonian but has a Hamiltonian path.
Is it possible for $|X|$ = $|Y|$?
From what I understand this shouldn't be possible if $G$ is not hamiltonian.
I am confused on this idea because my book does not go over Hamiltonian paths at all and mainly focuses on circuits/cycles.
Any help will be appreciated!
On
Let G be bipartite and have a hamiltonian Path but G is not hamiltonian. If $|X|$=$|Y|$ then set $X$ and set $Y$ have the same number of vertices. We can show a hamiltonian Path by starting at a vertex $u$ in $X$ and end at a vertex $w$ in $Y$. Such that u1,w1,...,wn. Thus a hamiltonian Path as long as all vertices have been transversed in alternating fashion.
I believe this shows it's possible.
All paths are bipartite. In particular, a path has a Hamiltonian path; namely, itself. Consider the path on 5 vertices. It has parts of size 2 and 3 respectively.
Edit: In light of your clarification, take a perfect matching on $2n$ vertices. The parts have equal size but the graph is not connected. So therefore, not Hamiltonian. Nor does it have a Hamiltonian path.