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Applications of the Max-Flow Min-Cut Theorem

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Given the cost of excavating a full block and the market value of a full block ... C = Cost of Excavating a Block. If Wij 0, Add an Arc from Node (i,j) to t ... – PowerPoint PPT presentation

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Title: Applications of the Max-Flow Min-Cut Theorem


1
Applications of the Max-Flow Min-Cut Theorem
2
S-T Cuts
Partition the Nodes into Sets S and T. S,T
Arcs from Nodes in S to Nodes in T.
S SF, D, H, TC,A,NY S,T
(D,A),(D,C),(H,A), Cap S,T 245 11 S
SF, D, A,C, T H, NY S,T
(SF,H),(C,NY), CapS,T 6 4 10
3
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 279.
  • The value of a maximum s-t flow the capacity of
    a minimum s-t cut.

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
    mins,t-connectivity of G (s,t) in V.

5
Example
6
2
1
8
4
3
7
5
Some Vertex Separators For s2, t7 3, 4, 6,
1, 4, 8 2,7-connectivity 3 Some Vertex
Separators For s1, t8 4, 5, 6,
2,3 1,8-connectivity 2 Vertex-Connectivity
2
6
Mengers 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.

7
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

8
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

9
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 lt c Then
  • Let c s-t Connectivity
  • Return c
  • Find Vertex-Connectivity of G with N-1 Maximum
    Flow Computations

10
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.

11
(No Transcript)
12
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. Dont take block (3,1).
  • Assuming (1,3) Already Excavated

13
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,j1)
  • 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 lt 0, Add an Arc from Node (i,j) to t with
    Capacity -Wij
  • If Wij gt 0 Add an Arc from s to Node (i,j) with
    Capacity Wij

14
Maximum Flow Network
100
100
100
100
100
400
400
15
Minimum Cut
Cut Ss,(2,4)(1,3), (1,4), (1,5) T All Other
Nodes
100
100
100
100
100
400
400
16
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
  • CapS,T
  • Value of Gold in T Cost of Excavating S

17
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
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