Complexity of Ford-Fulkerson

• Let U = max {(i,j) in A} uij.• If S = {s} and T = N\{s}, then u[S,T] is at

most nU.• The maximum flow is at most nU.

– At most nU augmentations.

• Each iteration of the inner while loop is O(m):– Each arc is inspected at most once– Finding is O(n)– Updating the flow on P is O(n)

• Complexity is O(nmU).

Pathological Example

s 1

2

3

5

(0,106) (0,106)

(0,106)(0,106)

(0,1) t

An Augmenting Path

s 1

2

3

5

(1,106) (0,106)

(1,106)(0,106)

(1,1) t

v = 1

Residual Network

s 1

2

3

5

106-1 106

106-1106

0 t111

An Augmenting Path in the Residual Network

s 1

2

3

5

106-1 106

106-1106

0 t111

Updated Flow

s 1

2

3

5

(1,106) (1,106)

(1,106)(1,106)

(0,1) t

v = 2

Updated Residual Network

s 1

2

3

5

106-1 106 -1

106-1106 -1

1 t01 111

Next Augmenting Path in the Residual Network

s 1

2

3

5

106-1 106 -1

106-1106 -1

1 t01 111

This will take 2 million iterations to find the maximum flow!

Polynomial Max Flow Algorithms (Chapter 7)

• Always augment along the shortest augmenting path in the residual network.– O(n2m)

• Always augment along the maximum-capacity augmenting path in the residual network.– O(nm log U)

• Goldberg’s algorithm (preflow-push) with highest-label implementation.– O(n2m1/2)