Applications of the Max-Flow Min-Cut Theorem. S-T Cuts SF D H C A NY 5 6 2 4 5 4 7 S = {SF, D, H},...

Applications of the Max-Flow Min-Cut

Theorem

S-T Cuts

SF

D

H

C

A

NY

5

6

2

4

5

4

7

S = {SF, D, H}, T={C,A,NY}[S,T] = {(D,A),(D,C),(H,A)}, Cap [S,T] = 2+4+5 = 11

S = {SF, D, A,C}, T = {H, NY}[S,T] = {(SF,H),(C,NY)}, Cap[S,T] =6 + 4 = 10

Partition the Nodes into SetsS and T.

[S,T] = Arcs from Nodes in Sto Nodes in T.

Maximum-Flow Minimum-Cut Theorem

Removing arcs (D,C) and (A,NY) cuts off SF from NY.

The set of arcs{(D,C), (A,NY)} is an s-t cut with capacity 2+7=9.

The value of a maximum s-t flow = the capacity of a minimum s-t cut.

SF

D

H

C

A

NY

5

6

2

4

5

4

7

SF

D

H

C

A

NY

5

64

5

4

Network Connectivity

An s,t vertex separator of a graph G=(V,E) is a set of vertices whose removal disconnects vertices s and t.

The s,t-connectivity of a graph G is the minimum size of an s,t vertex separator.

The vertex-connectivity of G is the min{s,t-connectivity of G: (s,t) in V}.

Example

1 8

2

3

4

5

6

7

Some Vertex Separators For s=2, t=7: {3, 4, 6}, {1, 4, 8}2,7-connectivity = 3

Some Vertex Separators For s=1, t=8: {4, 5, 6}, {2,3}1,8-connectivity = 2

Vertex-Connectivity = 2

Menger’s Theorem

Given and undirected graph G and two nonadjacent vertices s and t, the maximum number of vertex-disjoint (aside from sharing s and t) paths from s to t is equal to the s,t-connectivity of G.

Maximum Flow Formulation

For G=(V,E), construct the network G’=(N,A) as follows– For each vertex v in V/{s,t}, add nodes v and v’– Add arc (v,v’) with capacity 1– For each edge (u,v) in E, add arcs (u’,v) and (v’,u) with

infinite capacity– For each edge (s,v) in E, add arc (s,v) with infinite

capacity– For each edge (v,t) in E, add arc (v’,t) with infinite

capacity

Proof Lemma 1

– Each set of k vertex-disjoint s,t paths in G, corresponds to exactly one integral flow of value k in G’.

Lemma 2– Each s,t cut of finite capacity c corresponds to an s,t

vertex-separator of size c in G Result follows from Max-Flow Min-Cut

Theorem

Finding the Vertex-Connectivity of G=(N,A)

Let Node 1 be the Source Node s Let c = |N| For i = 2 … |N|

– Let t = i– If s-t Connectivity < c Then

Let c = s-t Connectivity

Return c

Find Vertex-Connectivity of G with |N|-1 Maximum Flow Computations

Mining for Gold Divide a cross-section of earth into a grid of equally-sized

blocks. Given the cost of excavating a full block and the market

value of a full block of gold, determine an optimal shape for the mine.

Constraints on the shape of your mine require that a block of earth cannot be excavated unless the three blocks above it (directly, and diagonally left and right) have also been excavated.

Solution To get the gold in block (2,4), we need to

excavate blocks (1,3), (1,4), (1,5) and (2,4). Revenue = $500, Cost = $400, Profit = $100. Take block (2,4).

To get the gold in block (3,1), we need to excavate blocks (3,1), (2,1), (2,2), (1,1), (1,2) and (1,3). Revenue = $500, Cost = $500*, Profit = $0. Don’t take block (3,1).

* Assuming (1,3) Already Excavated

Maximum Flow Formulation Each Block is a Node Add Infinite Capacity Arc from Node (i,j) to Node (i-1,j-1),

(i-1,j),(i-1,j+1) Add Source Node s and Sink Node t Wij = Vij - C Where

Vij = Value of Gold in Block (i,j)

C = Cost of Excavating a Block If Wij < 0, Add an Arc from Node (i,j) to t with Capacity -

Wij If Wij > 0 Add an Arc from s to Node (i,j) with Capacity Wij

Maximum Flow Network

1,1 1,2 1,3 1,4 1,5

2,1 2,2 2,3 2,4 2,5

3,1 3,2 3,3 3,4 3,5

4,1 4,2 4,3 4,4 4,5

s

t

400

400

100 100100 100

100

Minimum Cut

1,1 1,2 1,3 1,4 1,5

2,1 2,2 2,3 2,4 2,5

3,1 3,2 3,3 3,4 3,5

4,1 4,2 4,3 4,4 4,5

s

t

400

400

100 100100 100

100

Cut:S={s,(2,4)(1,3), (1,4), (1,5)}

T: All Other Nodes

Meaning of the Minimum Cut Excavate all Blocks in S

– Only Finite-Capacity Arcs in Minimum Cut– Blocks in S Can be Excavated Without Excavating Blocks in T

Arcs in [S,T]– Arcs from s to Nodes in T

• Capacity = Value of Gold in T

– Arcs from Nodes in S to t • Capacity = Cost of Excavating S

Cap[S,T] =Value of Gold in T + Cost of Excavating S

Maximizing Profit Finding a Maximum Flow Minimizes:

Value of Gold in T + Cost of Excavating S Equivalent to Maximizing:

F = - (Value of Gold in T + Cost of Excavating S ) Equivalent to Maximizing:

Value of All Gold + F =

Value of Gold in S - Cost of Excavating S =

Profit for Excavating S