Linear Programming Optimization

1
3.3 Applications of Maximum Flow and Minimum Cut
Bipartite Matchings and Covers

• A cover of an undirected graph G is a set C of
  nodes such that every edges of G has at least one
  end in C.
• For any matching M and any cover C, each edge in
  M has an end in C and the corresponding nodes in
  C are all distinct.
• gt M ? C (primal-dual relation)
• Thm 3.14 (Königs Theorem, 1931)
• For a bipartite graph G,
• max M M a matching min C C a
  cover

2
(Transform to max flow problem)
P
Q
a
a
A
A
b
b
B
B
c
c
C
C
r
s
d
d
D
D
f
F
f
F
h
h
H
H
urp1
uqs1
I
I
i
i
upq?
3
(Maximum matching and corresponding flow)
a
a
A
A
b
b
B
B
c
c
C
C
r
s
d
d
D
D
f
F
f
F
h
h
H
H
urp1
uqs1
I
I
i
i
upq?
node cover
4
(Identifying x-augmenting path)
a
a
A
A
b
b
B
B
c
c
C
C
r
r
s
s
d
d
D
D
f
F
f
F
h
h
H
H
urp1
uqs1
urp1
uqs1
I
I
i
i
upq?
upq?
node reachable from r
5
a
A
a
A
b
b
B
B
c
c
C
C
r
r
s
s
d
D
d
D
f
F
f
F
h
h
H
H
urp1
uqs1
urp1
uqs1
I
I
i
i
upq?
upq?
Cover C (pink) (P\A) ? (Q?A) Capacity of cut
P\AQ?AC
Let A blue nodes\r
6
Q\A
P\A
s
r
P?A
Q?A

• no edge from P?A to Q\A
• Cover C (P\A) ? (Q?A)
• Capacity of cut P\AQ?AC
• Running time O(mn)

7
(Halls Theorem)

• System of distinct representatives
• Given a set Q and a family (S1, S2, , Sk) of
  subsets of Q, a system of distinct representative
  (SDR) is a set q1, , qk of distinct elements
  of Q such that qi ?Si for 1 ? i ? k.
• Halls Theorem
• The family (S1, S2, , Sk) has an SDR if and
  only if for every subset I of 1, , k we have
  ?i ? I Si ? I .
• (The family (S1, S2, , Sk) does not have an SDR
  if and only if there is a set I such that ?i ?
  I Si lt I . )

8
q1
q1
S1
S1
q2
q2
S2
S2
q3
q3
S3
S3
q4
q4
S4
S4
q5
q5
S5
S5
q6
q6
The family has a SDR iff the corresponding
bipartite graph has a matching of size k.
9
q1
q1
S1
S1
q2
q2
S2
S2
q3
q3
S3
S3
q4
q4
S4
S4
q5
q5
S5
S5
q6
q6
Circled nodes have ?i ? I Si lt I , hence
no SDR.
10
Dilworths Theorem

• Chains and Antichains in Partially Ordered Sets
• A list of pairs x ? y satisfying the following
  conditions is called a partial order
• (i) no x ? x
• (ii) x ? y and y ? z together imply x ? z
• A set whose elements appear in a partial order
  is called a partially ordered set
• A subset C of a partially ordered set is called
  a chain if
• x ? y or y ? x whenever x, y ? C and x ?
  y
• (The elements can be enumerated as z1, z2, , zr
  in such a way that z1 ? z2, z2 ? z3, , zr-1 ?
  zr )
• A subset A of a partially ordered set is called
  an antichain if
• x ? y for no x, y ?A

11

• Dilworths Theorem
• In every partially ordered set P, the smallest
  number of chains whose union is P equals the
  largest size of an antichain
• application guided tour, airplane assignment
• pf) text exercise 3.37. We consider a different
  approach here.
• ex) Set X a, b, c, d, e, f, g, tours
• partial order a ? c, a ? d, a ? f, a
  ? g, b ? c, b ? g, d ? g, e ? f, e ? g
• ( need 3 guides, a ? d ? g, b ? c, e ? f
  )
• Construct a bipartite graph by doubling
  each node v as v and v.
• There exists an edge xy iff x ? y.

12
Matching M defines k n - M chains (guides)
ex) (i) a ? d ? g (ii) b ? c
(iii) e ? f In the matching, each tour has at
most one successor and at most one predecessor.
Otherwise, M would not be a matching. By
concatenating successive tours, we obtain a set
of chains (A chain may consists of a single
tour) Let k be the number of chains and ri be the
length (number of nodes) of i-th chain. Then
a
b
a
c
d
b
c
e
f
d
g
e
f
Hence min k ? n - M for max M Converse can
be shown too, i.e. k chains define n k sized
matching M.
g
13
Cover C

• Different reasoning using Königs Theorem
• Let M be a largest matching and C be min cover.
  Define a set A of trips as follows
• z ?A iff z?C and z? C.
• No x ? y with x, y ?A is possible
• ? no guide can take care of two different trips
  in A.
• need at least A guides.
• But A has at least n - C elements (each node
  in C makes at most one tour ineligible for
  membership in A)
• Hence every schedule must use at least
• A ? n - C n - M different guides.
• ? The schedule defined by M, using only n -
  M guides, is optimal.

a
A
b
a
c
d
b
c
e
f
d
g
e
f
g
14

• Proof of Dilworths Theorem
• The partially ordered set P may be thought of as
  a set of tours. Chains are those sets of tours
  that can be taken care of by one guide. The
  described procedure gives the smallest k of
  chains. Along with the chains, the procedure
  also finds an antichain A of size at least k.
• Since every antichain A shares at most one
  element with each of the chains C1, C2, , Ck ,
  it must satisfy A ? k ? A. So A is a
  largest antichain and its size is k (plug in A
  for A). ?
• Reference Linear Programming, Vasek Chvatal,
  1983, Freeman.

15
Optimal Closure in a Digraph

• Situation If we want to choose project v, then
  we must choose project w also. We want to choose
  the set of projects which gives maximum profit
• Model as a problem on a digraph. Use directed
  arc vw if project w precedes project v. Find
  the maximum benefit set C ? V such that ?(C) ?.
  ( closure of G)
• Example open pit mining problem.
• Partition the volume to be considered for
  excavation into small 3-dimensional blocks.
• Block v produces profit of bv (profit of ore in
  v cost of excavating v )
• If we need to excavate block w to excavate block
  v , we include arc vw in G.
• Want to find a closed set in G which maximizes
  profit.

16

• Convert to min-cut problem (max-flow problem)
• Divide the nodes into two sets A and B.
• v ?A if bv ? 0 and v ? B if bv lt 0.
• Add a source r and a sink s to the graph and
  add arcs rv for each v ?A and arcs vs for each
  v ?B.
• Capacity on the arcs are urv bv for each v
  ?A and uvs - bv for each v ?B.
• The upper bounds on the original arcs are
  infinite.
• A set C of nodes is closed if and only if the cut
  C ? r has finite capacity.
• If the capacity of C ? r is finite, then it
  equals

Since ?v ?A bv is a constant, minimizing the
capacity of C ? r amounts to maximizing
?v ?C bv .
17
(Example)
2
-7
?
2
?
C
7
4
r
4
1
?
?
-1
3
?
3
2
3
s
?
-3
?
1
2
?
-1
18
Mengers Theorems

• Ref Graph Theory with Applications, J. A. Bondy,
  U. S. R. Murty
• Lemma Let G be a digraph in which each arc has
  unit capacity. Then
• (a) The maximum (r, s) flow is equal to the
  maximum number m of arc disjoint directed (r,
  s)-paths in G and
• (b) The capacity of a minimum (r, s)-cut is
  equal to the minimum number n of arcs whose
  deletion destroys all directed (r, s)-paths in G.
• Thm Let r, s be two nodes of a digraph G. Then
  the maximum number of arc-disjoint directed (r,
  s)-paths in G is equal to the minimum number of
  arcs whose deletion destroys all directed (r,
  s)-paths in G.
• (pf) From above Lemma and max-flow and min-cut
  theorem.

19

• Thm Let r, s be two nodes of a graph G. Then
  the maximum number of edge-disjoint (r, s)-paths
  in G is equal to the minimum number of edges
  whose deletion destroys all (r, s)-paths in G.
• (pf) Replace each edge of G by two oppositely
  oriented arcs and apply the previous results. ?
• Lemma A graph G is k-edge-connected iff any two
  distinct nodes of G are connected by at least
  k-edge-disjoint paths.

20

• (node connectivity)
• Thm Let r, s be two nodes of a digraph G, such
  that r is not joined to s. Then the maximum
  number of node-disjoint directed (r, s)-paths in
  G is equal to the minimum number of nodes whose
  deletion destroys all directed (r, s)-paths in G.
• (pf) Construct G as follows ( except r and s)

21

• Thm Let r and s be two nonadjacent nodes of a
  graph G. Then the maximum number of
  node-disjoint (r, s)-paths in G is equal to the
  minimum number of nodes whose deletion destroys
  all (r, s)-paths.
• Cor A graph G with n ? k1 is k-node connected
  iff any two distinct nodes of G are connected by
  at least k node-disjoint paths.