Proof that the existence of a Hamilton Path in a bipartite graph is NP-complete I tried to solve the above NP-completeness exercise by making a bipartite graph from a general one (undirected) by inserting a vertice in the middle of every edge of the first (general) graph. This generates problems as suggested here: Finding the flaw in a reduction from Hamiltonian cycle to Hamiltonian cycle on bipartitie graphs 
Anyone can give a hint on how to make a bipartite graph from a general one without using the above method and how the hamiltonian property can be passed to it?
Given $G = (V,E)$, define $\tilde{G} = (\tilde{V},\tilde{E})$ by adding vertices $i^+$ and $i^-$ to $\tilde{V}$ for each $i\in V$ and edges $i^-j^+$ and $i^+j^-$ for each edge $ij\in E$.

It's not too hard to check that if $|V|$ is odd, $G$ has a Hamiltonian cycle if and only if $\tilde{G}$ is.  If $|V|$ is even, just add one new vertex of degree 2 with neighbors that have an edge between them.

So far, we've seen that Hamiltonian Cycle in bipartite graphs is hard.  To reduce this to Hamiltonian Path, just continue in the way you normally would to reduce Hamiltonian Cycle to Hamiltonian Path.