Juust to indicate that theethere is a solution to the problem: the trivial solution for the smallest graph with $k$ edge-disjoint Hamilton paths between a given pair of vertices is to take a graph $G'$ with $k$ edge-disjoint Hamilton cycles and split one of its vertices.
The smallest graph with two edge-disjoint Hamilton cycles is $K_5$ and so the smallest graph that solves the original problem is a vertex-split $K_5$ depicted below:
Juust to indicate that thee is a solution to the problem: the trivial solution for the smallest graph with $k$ edge-disjoint Hamilton paths between a given pair of vertices is to take a graph $G'$ with $k$ edge-disjoint Hamilton cycles and split one of its vertices.
The smallest graph with two edge-disjoint Hamilton cycles is $K_5$ and so the smallest graph that solves the original problem is a vertex-split $K_5$ depicted below:
Juust to indicate that there is a solution to the problem: the trivial solution for the smallest graph with $k$ edge-disjoint Hamilton paths between a given pair of vertices is to take a graph $G'$ with $k$ edge-disjoint Hamilton cycles and split one of its vertices.
The smallest graph with two edge-disjoint Hamilton cycles is $K_5$ and so the smallest graph that solves the original problem is a vertex-split $K_5$ depicted below:
Juust to indicate that thee is a solution to the problem: the trivial solution for the smallest graph with $k$ edge-disjoint Hamilton paths between a given pair of vertices is to take a graph $G'$ with $k$ edge-disjoint Hamilton cycles and split one of its vertices.
The smallest graph with two edge-disjoint Hamilton cycles is $K_5$ and so the smallest graph that solves the original problem is a vertex-split $K_5$ depicted below:
