0

This was claimed in Reinhard Diestel's Graph Theory textbook. This doesn't seem true at all to me. By cut I mean that if in a graph $G=(V,E)$, we have a vertex partition $\{V_1, V_2\}$, then the set of all edges crossing this partition is called a cut. By varying the (2 element) partitions, we get all the cuts of $G$. One direction is obvious: If in a graph $G = (V,E)$ we have a vertex partition $\{V_1, V_2\}$ and $V_1$ is disconnected, take one connected component of $V_1$ and remove it from $V_1$, adding it to $V_2$ and we get a vertex partition with a smaller cut. However, for the other direction, we have the following counterexample: Consider $G = K^n$, then if we have the partition $\{\{1,2\},\{3,\cdots,n\}\}$, we have $2n-4$ edges in our cut. If we have the partition $\{\{1,2,3\},\{4,5,\cdots,n\}\}$, then we have $3n-9$ edges in our cut, yet both partitions have connected subgraphs so how could they be minimal? Am I misunderstanding the definitions or is the statement just plain wrong?

Alex Ravsky
  • 106,166

0 Answers0