The Ford-Fulkerson Algorithm

aka The Labeling Algorithm

Ford-Fulkerson Algorithmbegin x := 0; label node t; while t is labeled do begin unlabel all nodes; pred(j) := 0 for all j in N; label s; LIST := {s}; while LIST is not empty and t is not labeled do begin remove a node i from LIST; for all {j in N: (i,j) in A and rij > 0} do if j is unlabeled then pred(j) := i, label j, add j to LIST; end; if t is labeled then augment flow on path from s to t end;end;

Labeling Algorithm Example

s t1

2 4

3

(0,2)

(0,6)

5(0,5)

6

(0,5) (0,4)

(0,4)

(0,7)

The Residual Network G(x)

s t1

2 4

3 5

6

5

0

0

6

2

0

5

0

40

4

0

07

Iteration 1: LIST = {1}, Labeled = {1}

s t1

2 4

3 5

6

5

0

0

6

2

0

5

0

40

4

0

70

i = 1

Iteration 1: LIST = {1}, Labeled = {1}

• i = 1• LIST = {}• Arc (1,2)

– pred(2) =1– label 2– LIST = {2}

• Arc (1,3) – pred (3) = 1– label 3– LIST = {2, 3}

Iteration 1: LIST = {2,3}, Labeled= {1,2,3}

s t1

2 4

3 5

6

5

0

0

6

2

0

5

0

40

4

0

07

pred(2) = 1

pred(3) = 1

Iteration 1: LIST = {2,3}, Labeled = {1,2,3}

• i = 2• LIST = {3}• Arc (2,4)

– pred(4) =2– label 4– LIST = {3,4}

• Arc (2,5) – pred (5) = 2– label 5– LIST = {3,4,5}

• Arc (2,1)– residual capacity

of (2,1) = 0

Iteration 1: LIST = {3,4,5}, Labeled= {1,2,3,4,5}

s t1

2 4

3 5

6

5

0

0

6

2

0

5

0

40

4

0

07

pred(2) = 1

pred(3) = 1 pred(5) = 2

pred(4) = 2

Iteration 1: LIST = {3,4,5}, Labeled = {1,2,3,4,5}

• i = 3• LIST = {4,5}• Arc (3,5)

– 5 is already labeled

• Arc (3,1) – residual capacity of (3,1) = 0

Iteration 1: LIST = {4,5}, Labeled = {1,2,3,4,5}

• i = 4• LIST = {5}• Arc (4,2)

– residual capacity of (4,2) = 0• Arc (4,6)

– pred(6) =4– label 6– LIST = {5,6}

Iteration 1: LIST = {5,6}, Labeled= {1,2,3,4,5,6}

s t1

2 4

3 5

6

5

0

0

6

2

0

5

0

40

4

0

07

pred(2) = 1

pred(3) = 1 pred(5) = 2

pred(4) = 2

pred(6) = 4

Iteration 1: The sink is labeled

• Use pred labels to trace back from the sink to the source to find path P– P = 1 -> 2 -> 4 -> 6

= min {rij: (i,j) in P) = 2• Send 2 units of flow from to s to t

along path P

Flow x After Iteration 1

s t1

2 4

3

(2,2)

(0,6)

5(0,5)

6

(2,5) (2,4)

(0,4)

(0,7)

v = 2

The Residual Network G(x)

s t1

2 4

3 5

6

3

2

0

6

0

2

5

0

40

2

2

07

Iteration 2: LIST = {1}, Labeled = {1}• i = 1• LIST = {}• Arc (1,2)

– pred(2) =1– label 2– LIST = {2}

• Arc (1,3) – pred (3) = 1– label 3– LIST = {2, 3}

Iteration 2: LIST = {2,3}, Labeled={1,2,3}

s t1

2 4

3 5

6

3

2

0

6

0

2

5

0

40

2

2

07

p=1

p=1

Iteration 2: LIST = {2,3}, Labeled = {1,2,3}

• i = 2• LIST = {3}• Arc (2,4)

– residual cap (2,4) = 0

• Arc (2,5) – pred (5) = 2– label 5– LIST = {3,5}

• Arc (2,1)– residual capacity

of (2,1) = 0

Iteration 2: LIST = {3,5}, Labeled={1,2,3,5}

s t1

2 4

3 5

6

3

2

0

6

0

2

5

0

40

2

2

07

p=1

p=1 p=2

Iteration 2: LIST = {3,5}, Labeled = {1,2,3,5}

• i = 3• LIST = {5}• Arc (3,5)

– 5 is already labeled

• Arc (3,1) – residual

capacity of (3,1) = 0

• i = 5• LIST = {}• Arc (5,2)

– residual cap = 0

• Arc (5,3)– residual cap = 0

• Arc (5,6) – pred(6) = 5– label 6– LIST = {6}

Iteration 2: LIST = {6}, Labeled={1,2,3,5,6}

s t1

2 4

3 5

6

3

2

0

6

0

2

5

0

40

2

2

07

p=1

p=1 p=2

p=5

Iteration 2: The sink is labeled

• Use pred labels to trace back from the sink to the source to find path P– P = 1 -> 2 -> 5 -> 6

= min {rij: (i,j) in P) = 3• Send 3 units of flow from to s to t

along path P

Flow x After Iteration 2

s t1

2 4

3

(2,2)

(0,6)

5(0,5)

6

(5,5) (2,4)

(3,4)

(3,7)

v = 5

The Residual Network G(x)

s t1

2 4

3 5

6

0

5

0

6

0

2

5

0

13

2

2

04

Iteration 3: LIST = {1}, Labeled = {1}• i = 1• LIST = {}• Arc (1,2)

– residual capacity = 0• Arc (1,3)

– pred (3) = 1– label 3– LIST = {3}

Iteration 3: List = {3}, Labeled = {1,3}

s t1

2 4

3 5

6

0

5

0

6

0

2

5

0

13

2

2

04

p=1

Iteration 3: LIST = {3}, Labeled =

{1,3}• i = 3• LIST = {}• Arc (3,1)

– residual capacity = 0• Arc (3,5)

– pred (5) = 3– label 5– LIST = {5}

Iteration 3: List = {5}, Labeled = {1,3,5}

s t1

2 4

3 5

6

0

5

0

6

0

2

5

0

13

2

2

04

p=1 p=2

Iteration 3: LIST = {5}, Labeled =

{1,3,5}• i = 5, LIST = {}• Arc (5,2)

– pred(2) = 5– label 2– LIST = {2}

• Arc (5,3): residual capacity = 0• Arc (5,6)

– pred (6) = 5– label 6– LIST = {2,6}

Iteration 3: List = {2,6}, Labeled = {1,2,3,5,6}

s t1

2 4

3 5

6

0

5

0

6

0

2

5

0

13

2

2

04

p=1 p=2

p=5

p=5

Iteration 3: The sink is labeled

• Use pred labels to trace back from the sink to the source to find path P– P = 1 -> 3 -> 5 -> 6

= min {rij: (i,j) in P) = 4• Send 4 units of flow from to s to t

along path P

Flow x After Iteration 3

s t1

2 4

3

(2,2)

(4,6)

5(4,5)

6

(5,5) (2,4)

(3,4)

(7,7)

v = 9

The Residual Network G(x)

s t1

2 4

3 5

6

0

5

4

2

0

2

1

4

13

2

2

70

Iteration 4: LIST = {1}, Labeled = {1}• i = 1• LIST = {}• Arc (1,2)

– residual capacity = 0• Arc (1,3)

– pred (3) = 1– label 3– LIST = {3}

Iteration 4: List = {3}, Labeled = {1,3}

s t1

2 4

3 5

6

0

5

4

2

0

2

1

4

13

2

2

70

p=1

Iteration 4: LIST = {3}, Labeled =

{1,3}• i = 3• LIST = {}• Arc (3,1)

– 1 is labeled• Arc (3,5)

– pred (5) = 3– label 5– LIST = {5}

Iteration 4: List = {5}, Labeled = {1,3,5}

s t1

2 4

3 5

6

0

5

4

2

0

2

1

4

13

2

2

70

p=1 p=3

Iteration 4: LIST = {5}, Labeled =

{1,3,5}• i = 5• LIST = {}• Arc (5,2)

– pred(2) = 5– label 2– LIST = {2}

• Arc (5,6) – residual capacity = 0

Iteration 4: List = {2}, Labeled = {1,2,3,5}

s t1

2 4

3 5

6

0

5

4

2

0

2

1

4

13

2

2

70

p=1 p=3

p=5

Iteration 4: LIST = {2}, Labeled =

{1,2,3,5}• i = 2 LIST = {}• Arc (2,1)

– 1 is already labeled

• Arc(2,4)– residual capacity = 0

• Arc (2,5) – 5 is already labeled

Iteration 4: List = {}

• The sink is not labeled• Algorithm ends with optimal flow x

Correctness

• At the end of each iteration, the algorithm either augments the flow or terminates because it can’t label the sink.

• Let S be the set of labeled nodes when the algorithm terminates. Let T = N \ S.

• We need to show that when the algorithm terminates v = u[S,T] which implies x is a maximum flow.

Correctness: arcs in (S,T)

• rij = 0• rij = uij - xij + xji• 0 = uij – xij + xji• xij = uij + xji

Since xji must be at least zero and 0 xij uij,it follows that xij = uij.

i js

Correctness: arcs in (T,S)

• Suppose xij > 0• rji = uji – xji + xij• Implies rji > 0 (uji xji)

i j s

• Implies s can reach i in residual network which means i should have been labeled.

• Thus, xij = 0

Correctness

].,[

thisshowedjust We0

flow feasible a is x since

),(),(

),(),(),(),(

TSu

v

TSjiij

STjiij

TSjiij

u

xx

This means x is a maximum flow by Property 6.1.

Thus, the flows on the forward arcs in [S,T] are at the upper bound and there is no flow on any of the backwards arcs.

Complexity

• 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.• 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(nU x m) = O(nmU).