Whose full bipartite and 3-partite graphs are eulerian/hamiltonian graphs? > Which full bipartite ($K_{r,s}$) and 3-partite ($K_{r,s,t}$) graphs are eulerian/hamiltonian graphs? What I have found so far: 1. bipartite/eulerian $\rightarrow 2|r \land 2|s$ 2. $3$-partite/eulerian $\rightarrow 2|r \land 2|s \land 2|t$ 3. bipartite/hamiltonian $\rightarrow r=s$ 4. $3$-partite/hamiltonian $\rightarrow s=r+t$ (and symmetric cases) or $ s=r+t + 1$ Shortly speaking: $1.$ and $2. $ is a result of the fact that graph has eulerian cycle $\leftrightarrow$ $2|deg(v)$ $ v\in V(G)$ $3.$ and $4.$ we have $k$-partite graphs so when we go to one part, we must be able to return. Do you agree with that?

A graph is eulerian iff all its vertex degrees are even. For a complete 3-partite graph $K_{r,s,t}$, this is true if $r + s$, $r+t$, and $s+t$ are all even, which happens iff $r$, $s$, and $t$ are all the same parity.

Ore's theorem implies that a complete 3-partite graph is always hamiltonian unless one part has more than half the vertices. The converse also holds in this case.