Advanced Algorithms
Piyush Kumar
(Lecture 4: MaxFlow MinCut Applications)
Welcome to COT5405
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• Homework 1 is due NOW.
• Preliminary Programming assignment 1 is attached
on the back of the homework solutions
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Today
• Complete Preflow Push
• Discuss Programming assignment 1
– Preliminary Deadline: 26th
• Bipartite matching/Hall’s theorem
• Application of MaxFlow MinCut
– Image segmentation
– Edge Disjoint paths
Applications
Max Flow Min Cut
Bipartite Matching
•
Matching.
– Input: undirected graph G = (V, E).
– M  E is a matching if each node appears in at most edge in M.
– Max matching: find a max cardinality matching.
Bipartite Matching
• Bipartite matching.
– Input: undirected, bipartite graph G = (L  R, E).
– M  E is a matching if each node appears in at most edge
in M.
– Max matching: find a max cardinality matching.
1
1'
2
2'
matching
1-2', 3-1', 4-5'
L
3
3'
4
4'
5
5'
R
Bipartite Matching
• Bipartite matching.
– Input: undirected, bipartite graph G = (L  R, E).
– M  E is a matching if each node appears in at most edge
in M.
– Max matching: find a max cardinality matching.
1
1'
2
2'
max matching
1-1', 2-2', 3-3' 4-4'
L
3
3'
4
4'
5
5'
R
Bipartite Matching
•
Max flow formulation.
– Create digraph G' = (L  R  {s, t}, E' ).
– Direct all edges from L to R, and assign infinite (or unit)
capacity.
– Add source s, and unit capacity edges from s to each node in L.
– Add sink t, and unit capacity edges from each node in R to t.
G'
1
1
s
L

1'
2
2'
3
3'
4
4'
5
5'
1
t
R
Bipartite Matching:
Proof of Correctness
• Theorem. Max cardinality matching in G = value of max flow
in G'.
• Pf. 
– Given max matching M of cardinality k.
– Consider flow f that sends 1 unit along each of k paths.
– f is a flow, and has cardinality k. ▪
G
1
1'
2
2'
3
3'
4
5
1
1

1'
2
2'
3
3'
4'
4
4'
5'
5
5'
s
1
t
G'
•
•
Bipartite Matching:
Proof of Correctness
Theorem. Max cardinality matching in G = value of max flow in G'.
Pf. 
– Let f be a max flow in G' of value k.
– Integrality theorem  k is integral and can assume f is 0-1.
– Consider M = set of edges from L to R with f(e) = 1.
• each node in L and R participates in at most one edge in M
• |M| = k: consider cut (L  s, R  t) ▪
1
1
s
G'

1'
1
1
1'
2
2'
3
3'
2
2'
3
3'
4
4'
4
4'
5
5'
5
5'
t
G
Perfect Matching
• Def. A matching M  E is perfect if each node appears in
exactly one edge in M.
• Q. When does a bipartite graph have a perfect matching?
• Structure of bipartite graphs with perfect matchings.
– Clearly we must have |L| = |R|.
– What other conditions are necessary?
– What conditions are sufficient?
Perfect Matching
• Notation. Let S be a subset of nodes, and let N(S) be the
set of nodes adjacent to nodes in S.
• Lemma. If a bipartite graph G = (L  R, E), has a perfect
matching, then |N(S)|  |S| for all subsets S  L.
• Pf. Each node in S has to be matched to a different node in N(S).
1
1'
2
2'
No perfect matching:
S = { 2, 4, 5 }
L
3
3'
4
4'
5
5'
N(S) = { 2', 5' }.
R
Marriage Theorem
• Marriage Theorem. [Frobenius 1917, Hall 1935] Let G = (L
 R, E) be a bipartite graph with |L| = |R|. Then, G has a
perfect matching or there is exists S  L such that
|N(S)| < |S|.
Pf: When there is no perfect matching, Use max-flow min-cut theorem
on a bipartite graph, to find the cut(A’,B’) < n.
Let A = L ∩ A’
Claim 1: N(A)  A’
Claim 2: |R ∩ A’| >= N(A)
Claim 3: c(A’,B’) = |L ∩ B’ | + |R ∩ A’| < n
Bipartite Matching:
Running Time
• Which max flow algorithm to use for bipartite matching?
– Generic augmenting path: O(m val(f*) ) = O(mn).
– Capacity scaling: O(m2 log C ) = O(m2).
– Shortest augmenting path: O(m n1/2).
• Non-bipartite matching.
– Structure of non-bipartite graphs is more complicated,
but
well-understood. [Tutte-Berge, Edmonds-Galai]
– Blossom algorithm: O(n4). [Edmonds 1965]
– Best known: O(m n1/2).
[Micali-Vazirani 1980]
7.10 Image Segmentation
15
Image segmentation
Image Segmentation
• Image segmentation.
– Central problem in image processing.
– Divide image into coherent regions.
• Ex: To recognize a hand (or
face/iris/fingerprint) from the
background. (And find out if the
person is known to the system)
Image Segmentation
•
Foreground / background segmentation.
– Label each pixel in picture as belonging to
foreground or background.
– V = set of pixels, E = pairs of neighboring pixels.
– ai  0 is likelihood pixel i in foreground.
– bi  0 is likelihood pixel i in background.
– pij  0 is separation penalty for labeling one of i
and j as foreground, and the other as background.
•
Goals.
– Accuracy: if ai > bi in isolation, prefer to label i in foreground.
– Smoothness: if many neighbors of i are labeled foreground, we
should be inclined to label i as foreground.
– Find partition (A, B) that maximizes:  a   b 
p
i A
foreground
background
i
jB
j
ij
(i, j)  E
A {i, j}  1
Image Segmentation
•
Formulate as min cut problem.
– Maximization.
– No source or sink.
– Undirected graph.
•
Turn into minimization problem.
– Maximizing  a i   b j 
i A
jB
 pij
(i, j)  E
A {i, j}  1
is equivalent to minimizing
 i  V a i   j  V b j  

– or alternatively
 a j   bi 
jB

i A
a constant
 pij
(i, j)  E
A {i, j}  1
 a i   bj 
i A
jB
 pij
(i, j)  E
A {i, j}  1
Image Segmentation
• Formulate as min cut problem.
– G' = (V', E').
– Add source to correspond to foreground;
add sink to correspond to background
– Use two anti-parallel edges instead of
undirected edge.
pij
pij
pij
aj
s
i
pij
bi
G'
j
t
Image Segmentation
• Consider min cut (A, B) in G'.
– A = foreground.
cap(A, B) 
 a j   bi 
jB
i A
 pij
if i and j on different sides,
pij counted exactly once
(i, j)  E
i A, j  B
– Precisely the quantity we want to minimize.

aj
i
s
A
G'
pij
bi
j
t
Edge Disjoint paths
Edge Disjoint Paths
• Disjoint path problem. Given a digraph G = (V, E) and two
nodes s and t, find the max number of edge-disjoint s-t
paths.
• Def. Two paths are edge-disjoint if they have no edge in
common.
• Ex: communication networks.
s
2
5
3
6
4
7
t
Edge Disjoint Paths
• Disjoint path problem. Given a digraph G = (V, E) and two
nodes s and t, find the max number of edge-disjoint s-t
paths.
• Def. Two paths are edge-disjoint if they have no edge in
common.
• Ex: communication networks.
s
2
5
3
6
4
7
t
Edge Disjoint Paths
•
Max flow formulation: assign unit capacity to every edge.
1
s
1
1
1
1
1
1
1
1
1
1
1
t
1
1
•
•
Theorem. Max number edge-disjoint s-t paths equals max flow
value.
Pf. 
– Suppose there are k edge-disjoint paths P1, . . . , Pk.
– Set f(e) = 1 if e participates in some path Pi ; else set f(e) = 0.
– Since paths are edge-disjoint, f is a flow of value k. ▪
Edge Disjoint Paths
•
Max flow formulation: assign unit capacity to every edge.
1
s
1
1
1
1
1
1
1
1
1
1
1
t
1
1
•
•
Theorem. Max number edge-disjoint s-t paths equals max flow
value.
Pf. 
– Suppose max flow value is k.
– Integrality theorem  there exists 0-1 flow f of value k.
– Consider edge (s, u) with f(s, u) = 1.
• by conservation, there exists an edge (u, v) with f(u, v) = 1
• continue until reach t, always choosing a new edge
– Produces k (not necessarily simple) edge-disjoint paths. ▪
can eliminate cycles to get simple paths if desired
Menger’s theorem
• In every directed graph with nodes s and
t, the maximum number of edge-disjoint st paths is equal to the minimum number of
edges whose removal separates s from t.
• Special case of MaxFlow MinCut theorem.
• We cud use this theorem instead of the
MaxFlow MinCut theorem in the proof of
Hall’s theorem.
References
• R.K. Ahuja, T.L. Magnanti, and J.B.
Orlin. Network Flows. Prentice Hall,
1993. (Reserved in Dirac)
• K. Mehlhorn and S. Naeher. The
LEDA Platform for Combinatorial and
Geometric Computing. Cambridge
University Press, 1999. 1018 pages