Ragesh Jaiswal

CSE, IIT Delhi

CSL 356: Analysis and Design of

Algorithms

Network Flow: Recap.

Ford-Fulkerson algorithm:

Given network with integer capacities, find a source-to-sink

path and push as much flow along the path as possible.

Update the residual capacity of edges in the residual graph.

Repeat.

Proof of correctness:

The algorithm terminates.

Max-flow-min-cut theorem: In every network flow the

maximum value of an flow is equal to the minimum capacity of an cut.

Network Flow:

Ford-Fulkerson algorithm:

Given network with integer capacities, find a source-to-sink

path and push as much flow along the path as possible.

Update the residual capacity of edges in the residual graph.

Repeat.

Proof of correctness:

The algorithm terminates.

The capacities are integers.

What if the capacities are not integers? Does the algorithm terminate?

Max-flow-min-cut theorem: In every network flow the

maximum value of an flow is equal to the minimum capacity of an cut.

Network Flow: A simple example where the Ford-Fulkerson algorithm does

not terminate.

s

q r s p

t

100 100

100 100

100

100

1 1

1 = 2

Network Flow: A simple example where the Ford-Fulkerson algorithm does

not terminate.

s

q r s p

t

1 1

1 = 2

100 100

100 100

100

100

Network Flow: A simple example where the Ford-Fulkerson algorithm does

not terminate.

s

q r s p

t

1 0

1 = 2

100 100

100 100

99

99

Network Flow: A simple example where the Ford-Fulkerson algorithm does

not terminate.

Suppose inductively, the residual capacities of edges (, ), (, ), and (, ) are 1, 0, . Consider next four flows.

s

q r s p

t

1 0

1 = 2

Network Flow:

s

q r s p

t

1 0

1 = 2

A simple example where the Ford-Fulkerson algorithm does

not terminate.

Suppose inductively, the residual capacities of edges (, ), (, ), and (, ) are 1, 0, . Consider next four flows.

Network Flow:

s

q r s p

t

+1 0

1 = 2

A simple example where the Ford-Fulkerson algorithm does

not terminate.

Suppose inductively, the residual capacities of edges (, ), (, ), and (, ) are 1, 0, . Consider next four flows.

Network Flow:

s

q r s p

t

+1 0

1 = 2

A simple example where the Ford-Fulkerson algorithm does

not terminate.

Suppose inductively, the residual capacities of edges (, ), (, ), and (, ) are 1, 0, . Consider next four flows.

Network Flow:

s

q r s p

t

+1 0

1 = 2

A simple example where the Ford-Fulkerson algorithm does

not terminate.

Suppose inductively, the residual capacities of edges (, ), (, ), and (, ) are 1, 0, . Consider next four flows.

Network Flow:

s

q r s p

t

+1 0

1 = 2

A simple example where the Ford-Fulkerson algorithm does

not terminate.

Suppose inductively, the residual capacities of edges (, ), (, ), and (, ) are 1, 0, . Consider next four flows.

Network Flow:

s

q r s p

t

0 +1 +2

1 = 2

A simple example where the Ford-Fulkerson algorithm does

not terminate.

Suppose inductively, the residual capacities of edges (, ), (, ), and (, ) are 1, 0, . Consider next four flows.

Network Flow:

s

q r s p

t

0 +1 +2

1 = 2

A simple example where the Ford-Fulkerson algorithm does

not terminate.

Suppose inductively, the residual capacities of edges (, ), (, ), and (, ) are 1, 0, . Consider next four flows.

Network Flow:

s

q r s p

t

+1 0 +2

1 = 2

A simple example where the Ford-Fulkerson algorithm does

not terminate.

Suppose inductively, the residual capacities of edges (, ), (, ), and (, ) are 1, 0, . Consider next four flows.

Network Flow: A simple example where the Ford-Fulkerson algorithm does

not terminate.

The total value of the flow converges to (1 + 2) =

4 + 5.

The max flow is 201.

Network Flow

Strongly polynomial time algorithm for max-flow

Network Flow: Edmonds-Karp

Max-Flow //Edmonds-Karp algorithm

- Start with a flow such that () = 0

- while there is an path in

- Find an path with least hop-length - Execute the augmenting path algorithm to obtain

- Update to and to - return

s

u

v

t

1000000

1

1000000

1000000

1000000

Let (, ) denote the hop-length of the shortest path from to in .

Claim 1: For all , , (, ) either remains same or increases with each flow augmentation.

Network Flow: Edmonds-Karp

Let (, ) denote the hop-length of the shortest path from to in .

Claim 1: For all , , (, ) either remains same or increases with each flow augmentation.

Proof:

Let be the flow just before the first augmentation that decreases the shortest distance of some vertex. Let be the flow after this augmentation.

Let be the vertex with minimum value of (, ) whose shortest path length was reduced.

Let be the vertex just before v in the shortest path from s to in .

Network Flow: Edmonds-Karp

Claim 1: For all , , (, ) either remains same or increases with each flow augmentation.

Proof: Let be the flow just before the first augmentation that decreases the shortest

distance of some vertex. Let be the flow after this augmentation.

Let be the vertex with minimum value of (, ) whose shortest path length was reduced.

Let be the vertex just before in the shortest path from to in .

We have:

(, ) = (, ) 1

(, ) (, )

Claim: (, ) is not present in .

Proof: Since otherwise,

(, ) (, ) + 1 (, ) + 1 = (, ).

Network Flow: Edmonds-Karp

Claim 1: For all , , (, ) either remains same or increases with each flow augmentation.

Proof: Let be the flow just before the first augmentation that decreases the shortest

distance of some vertex. Let be the flow after this augmentation.

Let be the vertex with minimum value of (, ) whose shortest path length was reduced.

Let be the vertex just before in the shortest path from to in .

We have:

(, ) = (, ) 1

(, ) (, )

Claim: (, ) is not present in .

Proof: Since otherwise,

(, ) (, ) + 1 (, ) + 1 = (, ).

This means that (, ) was in the augmenting path. This means: (, ) = (, ) 1 (, ) 1 (, ) 2

Network Flow: Edmonds-Karp

Claim 2: The total number of flow augmentations in the

EdmondsKarp algorithm is ().

Proof: An edge is said to be critical while augmentation if it is the

bottleneck edge.

Claim: Any edge can become critical at most (/2) times.

Proof:

Consider any edge (, ). Let be the flow just before (, ) becomes critical. The we have

(, ) = (, ) + 1 (1)

After this the edge (, ) disappears. Let be the flow just before the augmentation that brings back edge (, ). Then we have (, ) = (, ) + 1 (2)

Using (1) and (2) we get:

Network Flow: Edmonds-Karp

Claim 2: The total number of flow augmentations in the

EdmondsKarp algorithm is ().

Proof: An edge is said to be critical while augmentation if it is the bottleneck edge.

Claim: Any edge can become critical at most (/2) times.

Proof:

Consider any edge (, ). Let be the flow just before (, ) becomes critical. The we have

(, ) = (, ) + 1 (1)

After this the edge (, ) disappears. Let be the flow just before the augmentation that brings back edge (, ). Then we have (, ) = (, ) + 1 (2)

Using (1) and (2) we get:

(, ) = (, ) + 1 (, ) + 1 = (, ) + 2

The shortest distance has increased by 2 between the instances when (, ) becomes critical.

Network Flow: Edmonds-Karp

Claim 2: The total number of flow augmentations in the

EdmondsKarp algorithm is ().

Theorem: The running time of Edmonds-Karp algorithm is

(2).

Network Flow: Edmonds-Karp

End