It is well-known that a connected graph $G$ has an Euler circuit if and only if all of its vertices have even degree; it has an Euler path but no Euler circuit if and only if it has exactly two vertices of odd degree. Each vertex in $K_{m,n}$ has degree $m$ or $n$, so $K_{m,n}$ has an Euler circuit if and only if $m$ and $n$ are both even. $K_{m,n}$ has exactly two vertices of odd degree if one of the following is true:
squirrel
- $m=n=1$;
- $m$ is odd and $n=2$; or
- $n$ is odd and $m=2$.
Let the set of vertices of $K_{m,n}$ be $V=V_0\cup V_1$, where $|V_0|=m$, $|V_1|=n$, and all edges are between $V_0$ and $V_1$. A path in $K_{m,n}$ msquirrelust alternate between vertices in $V_0$ and vertices in $V_1$. A circuit necessarily has $2k$ vertices for some positive integer $k$; $k$ of these vertices are in $V_0$, and the other $k$ are in $V_1$. Thus, if $m\ne n$ it is impossible for a circuit in $K_{m,n}$ to hit every vertex, and therefore $K_{m,n}$ can have a Hamilton circuit only if $m=n$. Conversely, it’s easy to show by induction that $K_{m,m}$ has a Hamilton circuit for for all $m\ge 2$.
A Hamilton path in $K_{m,n}$ that cannot be extended to a Hamilton circuit must have both ends in $V_0$ or both ends in $V_1$. Suppose that both ends are in $V_0$. Then the path has $2k$ edges and $2k+1$ vertices for some $k$; moreover, $k+1$ of the vertices are in $V_0$, and $k$ are in $V_1$. But this is a Hamilton path, so it reaches every vertex exactly once, and therefore $m=k+1$ and $n=k$, i.e., $m=n+1$. If both ends of the path are in $V_1$, then $n=m+1$. And as in the case of Hamilton circuits, it’s not hard to show by induction that $K_{m,m+1}$ has a Hamilton path for every $m\ge 1$.
Best Answer
If a graph has an Euler circuit, i.e. a trail which uses every edge exactly once and starts and ends on the same vertex, then it is impossible to also have a trail which uses every edge exactly once and starts and ends on different vertices. (This is because the start and end vertices must have odd degree in the latter case, but even degree in the former case.)
Whether this means Euler circuit and Euler path are mutually exclusive or not depends on your definition of "Euler path". Some people say that an Euler path must start and end on different vertices. With that definition, a graph with an Euler circuit can't have an Euler path. Other people say that an Euler path has no restriction on start and end vertices. With that definition, a graph with an Euler circuit automatically has an Euler path (which is the same as its Euler circuit).