The Ford–Fulkerson Algorithm, Exams of Art

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The Ford–Fulkerson Algorithm

Math 482, Lecture 26

Misha Lavrov

April 6, 2020

A summary of the last lecture

In the previous lecture, we found a high-value flow in a network by starting with the zero flow and repeating the following procedure: (^1) Find an augmenting path. (^2) Use it to augment the flow as much as possible.

s t

a (^) b

c (^) d

0 / 10

0 / 10

0 / 12

0 / 10 0 / 4

0 / (^40) / 8

0 / 4

A summary of the last lecture

In the previous lecture, we found a high-value flow in a network by starting with the zero flow and repeating the following procedure: (^1) Find an augmenting path. (^2) Use it to augment the flow as much as possible.

s t

a (^) b

c (^) d

10 / 10

10 / 10

10 / 12

4 / 10 4 / 4

0 / (^44) / 8

0 / 4

A summary of the last lecture

In the previous lecture, we found a high-value flow in a network by starting with the zero flow and repeating the following procedure: (^1) Find an augmenting path. (^2) Use it to augment the flow as much as possible.

s t

a (^) b

c (^) d

10 / 10

10 / 10

12 / 12

6 / 10 4 / 4

2 / (^44) / 8

0 / 4

A summary of the last lecture

In the previous lecture, we found a high-value flow in a network by starting with the zero flow and repeating the following procedure: (^1) Find an augmenting path. (^2) Use it to augment the flow as much as possible.

s t

a (^) b

c (^) d

10 / 10

8 / 10

12 / 12

8 / 10 4 / 4

4 / (^46) / 8

2 / 4

Eventually, there are no more augmenting paths.

The final residual graph

We can see this in the residual graph for the final flow obtained:

s (^) t

a (^) b

c (^) d

10

2 8

2

8

2

2

4

4

12

2 6

The residual graph theorem

Theorem Suppose that we have a network (N, A) and a feasible flow x such that there is no s, t-path in the residual graph. Then:

The residual graph theorem

Theorem Suppose that we have a network (N, A) and a feasible flow x such that there is no s, t-path in the residual graph. Then:

Let S be the set of all nodes reachable from s in the residual graph. Let T be the set of all other nodes. The cut (S, T ) has the same capacity as the value of x.

The residual graph theorem

Theorem Suppose that we have a network (N, A) and a feasible flow x such that there is no s, t-path in the residual graph. Then:

Let S be the set of all nodes reachable from s in the residual graph. Let T be the set of all other nodes. The cut (S, T ) has the same capacity as the value of x.

In particular, x is a maximum flow and (S, T ) is a minimum cut.

In our example, we take S = {s, c} and T = {a, b, d, t}. The capacity of this cut is csa + ccb + ccd = 10 + 4 + 4 = 18, same as the value of x.

Applying the definition

In the cut (S, T ) defined in the residual graph theorem, the residual graph has no arcs from S to T.

Applying the definition

In the cut (S, T ) defined in the residual graph theorem, the residual graph has no arcs from S to T. What does that mean? Recall: Whenever xij < cij for an arc (i, j) ∈ A, the residual graph has an arc i → j. Whenever xij > 0 for an arc (i, j) ∈ A, the residual graph has an arc j → i.

Applying the definition

In the cut (S, T ) defined in the residual graph theorem, the residual graph has no arcs from S to T. What does that mean? Recall: Whenever xij < cij for an arc (i, j) ∈ A, the residual graph has an arc i → j. Whenever xij > 0 for an arc (i, j) ∈ A, the residual graph has an arc j → i. Therefore: For every arc (i, j) with i ∈ S and j ∈ T , xij = cij.

Another equation for the value

Lemma For any cut (S, T ), v (x) =

i∈S

j∈T

xij −

i∈T

j∈S

xij.

(We proved this at the end of Lecture 23.)

Another equation for the value

Lemma For any cut (S, T ), v (x) =

i∈S

j∈T

xij −

i∈T

j∈S

xij.

(We proved this at the end of Lecture 23.)

Example: S = {s, a, b} and T = {c, d, t}.

s (^) t

a (^) b

c (^) d

10 / 10

8 / 10 12 / 12

8 / 10 4 / 4

4 / (^46) / 8

2 / 4