Min-cut with maximal number of edges

Question

I’ve searched for a solution for this problem for some time now, it is out of an algorithm question sheet.

We know that in order to find the minimal amount of edges in a flow graph’s min-cut we need to find the maximal flow, $c_o$,set the capacity of the edges of the min-cut to $1$ and for for the rest $e\in E$ set $c(e)=c(e)\cdot (|E|+1)+1$ and find the max flow $c_n$ in the new graph such that $c_n=c_o\cdot (|E|+1)+k$, in which $k$ is the minimal number of edges in the graph’s min-cut.

Our problem is about finding the maximal number of edges in a graph’s min-cut.

Deducting one instead of adding would have the same effect, hence I’ve thought that perhaps if we add some number $\epsilon \in \mathbb{R} $ until the max flow increases, that could tell us something.

However I got stuck in that thought, unable to find ways on how to progress.

Answer 1

Here is a simple algorithm that is based on your idea to reduce the capacity function.

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Suppose the given flow network is directed graph $G=(V,E)$ with source $s$, sink $t$ and capacity function $c$.

Let $\epsilon>0$ be a constant that is small enough. Consider a new capacity function $c^-$ that is $\epsilon $ less than $c$, i.e., $c^-(e)=c(e)-\epsilon$ for each edge $e$.

Let $\mathcal C_1$ and $\mathcal C_2$ be two $s$-$t$ cuts of $G$. Abusing the notations $c$ and $c^-$, let

• $c(\mathcal C_1)$ and $c^-(\mathcal C_1)$ be the capacity of $\mathcal C_1$ wrt to $c$ and $c^-$ respectively.
• $c(\mathcal C_2)$ and $c^-(\mathcal C_2)$ be the capacity of $\mathcal C_2$ wrt to $c$ and $c^-$ respectively.

Let $\|\mathcal C_1\|$ be the number of edges in $\mathcal C_1$ i.e. in the cut-set of the $s$-$t$ cut $\mathcal C_1$ (which are the edges that start with nodes in the part of $\|\mathcal C_1$ that contains $s$ and end with nodes in the other part of $\|\mathcal C_2\|$ that contains $t$). We have $\|\mathcal C_2\|$ similarly.

Claim: $c^-(\mathcal C_1)<c^-(\mathcal C_2)$ $\iff$ either $c(\mathcal C_1)<c(\mathcal C_2)$ or $c(\mathcal C_1)=c(\mathcal C_2)$ and $\|\mathcal C_1\|=\|\mathcal C_2\|$.
Proof: $c^-(\mathcal C_1)=c(\mathcal C_1)-\|\mathcal C_1\|\epsilon$.
$c^-(\mathcal C_2)=c(\mathcal C_2)-\|\mathcal C_2\|\epsilon$.
$$c^-(\mathcal C_1)-c^-(\mathcal C_2)= c(\mathcal C_1)-c(\mathcal C_2)-(\|\mathcal C_1\|-\|\mathcal C_2\|)\epsilon.$$ Since $\epsilon>0$ is small enough and $-|E|<\|\mathcal C_1\|-\|\mathcal C_2\|<|E|$, the claim is true.

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The claim means any min-$s$-$t$-cut wrt to $c^-$ must be a min-$s$-$t$-cut wrt to $c$ with the maximum number of edges. In other words, the maximum number of edges in any min-$s$-$t$-cut of the flow network $(G,c)$ is the number of edges in a min-$s$-$t$-cut of the flow network $(G, c^-)$.

So we have the following algorithm.

1. Construct $c^-$ such that $c^-(e)=c(e)-\epsilon$ for a small-enough constant $\epsilon>0$.
2. Compute a min-$s$-$t$-cut of $G$ wrt to $c^-$. Return the number of edges in it.

There are two ways to implement step 1.

• If $c$ is integer valued, we can choose $\epsilon=\frac1{|E|}$.
• In all cases, we can represent the scalar capacity $c(e)$ as vector $(c(e),0)$. Let $\epsilon=(0,1)$. $c^-(e)=(c(e),-1)$. Use the natural arithmetic and comparison order for the vectors. In this way, $\epsilon$ is realized as an "infinitesimal positive" capacity.

The time-complexity of the algorithm is about the same as the time-complexity to compute a max-flow of a flow network, such as $O(VE^2)$ for Dinic's algorithm.