Lecture Notes Discrete Optimization - Applied Mathematics

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Figure 7: A bad instance for the Ford-Fulkerson algorithm (left). Suppose that B is a
large integer. The algorithm alternately augments one unit of flow along the two paths
〈s,u, p,t〉 and 〈s, p,u,t〉. The flow after two augmentations is shown on the right. The
algorithm needs 2B augmentations to find a maximum flow.
Proof.
f(X,V \ X)= f(X,V)− f(X,X)= f(X,V)= f(s,V)+ f(X− s,V)= f(s,V)=| f|.
Intuitively, it is clear that if we consider an arbitrary cut (X, ¯X) of G, then the total flow
f(X, ¯X) that leaves X is at most c(X, ¯X). The next lemma shows this formally.
Lemma 5.4. The flow value of any flow f in G is at most the capacity of any cut (X, ¯X)
of G, i.e., f(X, ¯X)≤c(X, ¯X).
Proof. By Lemma 5.3, we have
| f|= f(X, ¯X)= ∑
∑
u∈X v∈V\X
f(u,v) ≤ ∑
∑
u∈X v∈V\X
c(u,v)=c(X, ¯X).
A fundamental result for flows is that the value of a maximum flow is equal to the minimum
capacity of a cut.
Theorem 5.3 (Max-Flow Min-Cut Theorem). Let f be a flow in G. Then the following
conditions are equivalent:
1. f is a maximum flow of G.
2. The residual graph G f contains no augmenting path.
3. | f|=c(X, ¯X) for some cut (X, ¯X) of G.
Proof. (1)⇒(2): Suppose for the sake of contradiction that f is a maximum flow of G
and there is an augmenting path P in G f . By Lemma 5.2, we can augment f along P and
obtain a flow of value strictly larger than| f|, which is a contradiction.
(2)⇒(3): Suppose that G f contains no augmenting path. Let X be the set of nodes that
are reachable from s in G f . Note that t /∈ X because there is no path from s to t in G f .
That is, (X, ¯X) is a cut of G. By Lemma 5.3, | f| = f(X, ¯X). Moreover, for every u∈X
and v∈ ¯X, we must have f(u,v) = c(u,v) because otherwise (u,v)∈E f and v would be
part of X. We conclude| f|= f(X, ¯X)=c(X, ¯X).
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