∑ ∑

the primal follows
min ∑ u e l e
e∈E
s.t. ∑ l e ≥ 1, ∀p ∈ P(s,t)
e:e∈p
l e ≥ 0, ∀e ∈ E
where l e can be interpreted as the length of the edge e in OPT D and u e l e therefore corresponds to the volume
of the edge. So the objective of the dual is to find the volume-minimizing assignment of lengths to the edges
so that that every s −t path has length at least 1.
Theorem 1. Max-flow = Min-cut.
Proof. First, we show that the dual above has 0−1 basic feasible solution. For an s−t cut ( S,S ) , let E ( S,S )
denote the edges crossing the cut and let u ( S,S ) denote the sum of capacities of edges crossing the cut. For
any s −t cut ( S,S ) , let l e = 1 for all e ∈ E ( S,S ) and l e = 0 for all others. Clearly, this is a feasible solution
for the dual. It follows from weak duality theorem, max-flow ≤ OPT D ≤ integral min-cut.
Next, we prove the other direction. Let l e be the length function and let d (v) be the shortest distance
from s to v for all v ∈ V according to the lengths l e . Note that d (s) = 0 and d (t) ≥ 1. Consider ρ ∈ U.A.R [0,1)
and let S ρ = {v ∈ V |d (v) ≤ ρ}, it follows
E [ u ( )]
S ρ ,S ρ = ∑ u e Pr ( e ∈ E ( ))
S ρ ,S ρ
e∈E
Note that Pr ( e ∈ E ( S ρ ,S ρ
))
≤ le , we obtain
E [ u ( )]
S ρ ,S ρ ≤ ∑ u e l e
e∈E
This implies that there exists a ρ with u ( )
S ρ ,S ρ ≤ ∑e∈E u e l e , i.e. min-cut ≤ OPT D , which proves that maxflow
= min-cut by strong duality theorem. Observe that for any u e ’s, there exists an integer s−t cut ( )
S ξ ,S ξ
with cost at most ∑ e∈E u e l e . This cut corresponds to
{1 e ∈ E ( )
S ξ ,S ξ
y e =
0 otherwise
Clearly, y is also feasible for the dual. Now we have ∑ e∈E u e y e ≤ OPT D , it follows that ∑ e∈E u e y e = OPT D
and l e = y e in OPT D . Thus, we have proved the integrality of l e ’s.
3.2 Global Min-Cut
Let us then consider the global min-cut problem. We show how to formulate a linear program and its dual
for the global min-cut problem. The definition of the global min-cut problem is given as follows.
Definition 2. (Global Min-Cut) Given in input an undirected graph G = (V,E) we want to find the subset
A ⊆ V such that A ≠ /0, A ≠ V , and the number of edges with one endpoint in A and one endpoint in V − A
is minimized.
9-2