Maximum flow = minimum cut

• See Jamboard. More than one cut: minimum capacity = 14.
• Note: Across the cut, in the max flow, all edges are saturated (to capacity) and all returning edges are avoided (zero).

Hostile Dogs

• Take the directed graph representing their neighborhood (with each street being an edge and each intersection being a vertex) and assigning a capacity of 1 to every street.
  • Ensures that only one dog can use that street.
  • Direct the edges toward the vet.
• You would need a flow of at least 2.
  • 2 dogs would leave s and 2 dogs would enter t.

• Note: a minimum cut of 1 would be a street both dogs were forced to walk on.

Maxflow-Mincut Theorem

• Lemma. The value of any feasible \((s,t)\)-flow is less than or equal to the capacity of any \((S,T)\)-cut.
• Corollary. If you can find a cut and a flow such that the capacity of the cut equals the value of the flow, you have found a maximum flow.
  • In this situation, all the edges from \(S\) to \(T\) will be saturated, and all the edges from \(T\) to \(S\) will be avoided.
• Theorem. You can always find such a cut and flow.

Maxflow-Mincut Theorem. In every flow network with source \(s\) and target \(t\), the value of the maximum \((s,t)\)-flow is equal to the capacity of the minimum \((s,t)\)-cut.

Proof. Ford-Fulkerson Algorithm

Residual Graph

Given a flow network \(G\) and a flow \(f\), the residual graph \(G_f\) is the graph with the same vertex set as \(G\), but whose edges indicate how much \(f\) could be increased/decreased on each edge of \(G\).

• If \(u\rightarrow v\) in \(G\) has spare capacity \(x\), then \(u\rightarrow v\) has capacity \(x\) in \(G_f\).
• If \(u \rightarrow v\) in \(G\) has flow value \(y\), then the reverse edge \(v \rightarrow u\) in \(G_f\) has capacity \(y\).

See Figure 10.5.

Group exercise

Draw the residual graph for the given graph. (See Jamboard)

Augmenting paths

Now suppose \(G_f\) has a path from \(s\) to \(t\).

• Let \(F\) be the capacity of the smallest edge in this path.
  • So this path can handle a flow of \(F\).
• Now augment the flow \(f\) in \(G\):
  • Add \(F\) to edge \(u \rightarrow v\) if \(u\rightarrow v\) is in the augmenting path.
  • Subtract \(F\) from edge \(u \rightarrow v\) if the reverse \(v \rightarrow u\) is in the augmenting path.

See Figure 10.5 again.

Flow stays feasible when augmented

• Since \(F\) is the smallest common increment/decrement, the capacity limits are still obeyed.
• Since the augmenting path is itself a flow, the conservation property is still satisfied at each vertex.
  • Details in book.

No augmenting path? That’s a min cut!

Suppose there is no path from \(s\) to \(t\) in \(G_f\).

• Let \(S\) be the vertices reachable from \(s\) in \(G_f\), and let \(T\) be the complement.
• There is no edge from \(S\) to \(T\) in \(G_f\).
  • So every edge from \(S\) to \(T\) in \(G\) is saturated.
  • And every edge from \(T\) to \(S\) in \(G\) is avoided.

That means we’ve found a cut with the same capacity as the value of the flow.

• This proves the theorem! (You can always find a matching cut/flow).
• This gives us the algorithm too.

Group exercise

• Find an augmenting path on your residual graph.
• Augment the flow.
• Update the residual graph.

Repeat until no path from \(s\) to \(t\) exists.

Ford-Fulkerson Method/Algorithm

Start with a zero flow and construct the residual graph \(G_f\). Repeat the following until there’s no path from \(s\) to \(t\) in \(G_f\).

• Find an augmenting path in the residual graph.
• Augment the flow.
• Update the residual graph.

Ford-Fulkerson Time?

Each iteration is linear time: \(O(E)\) since the graph is connected.

• Assume capacities are integers.
  • So \(f\) and \(F\) will always be integers.
• Let \(|f^*|\) be the value of the maximum flow.
  • Each iteration increases the flow by \(F\), which is at least 1.
  • So there could only be \(|f^*|\) iterations, worst case.

Total Time: \(O(E|f^*|)\).

There are faster max flow algorithms

• Maximum flow algorithms are still an active area of research.
• Current best algorithm: Orlin (2012)
  • Time: \(O(VE)\)

So when estimating running times of algorithms that use maximum flows, you can assume the maximum flow part has running time \(O(VE)\).