Maximum Bipartite Matching Algorithm

1
Maximum Bipartite Matching Algorithm

• Explanation and Proof
• by Sean Rose

2
Goals of Presentation

• Show that the algorithm performs as claimed
• At termination show
• C found by algorithm is a node cover
• M found by algorithm is a matching
• M C
• Discuss the complexity of the algorithm

3
Finding a Node Cover
X
Y

• A node cover C ? V is a set of vertices such that
  every edge in G is incident with a node in C
• Konigs Theorem In a bipartite graph, a matching
  M is maximum, and a node cover C is minimum IFF
  M C
• If algorithm ends at step 1 then
• M X Y
• let C X
• then M C
• so C is a minimum node cover

C
M
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Finding a Node Cover
G
X
Y
Td
a
a
f
f
b
b
c
g
g
d
d
h
Even Odd
e
i
M
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Finding a Node Cover
X
Y

• Let P be the nodes not belonging to an
  alternating tree
• P c,e,h,i
• Note that half of the nodes in P belong to X and
  the other half to Y
• Let CP P ? X
• CP c,e
• Let No be all odd nodes
• No f,g
• Claim C CP ? No is a node cover
• C c,e,f,g

a
f
M
b
P
No
Ne
c
g
d
h
e
i
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Finding a Node Cover
X
Y

• Assume C isnt a node cover
• then there exists an edge not incident with any
  node in No or any node in Cp
• 3 possibilities for e uv
• u?Ne, v?No
• u?P, v?No
• u?Ne, v?P
• such an edge cannot exist, so C is a node cover

a
f
M
P
No
b
e?
Ne
e?
c
g
e?
d
h
e
i
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Prove M is a matching

• A matching is a set of edges without common
  vertices
• Initially
• M Ø, so M is trivially a matching
• At each augmentation step
• M M ? P, for an M-augmenting path P
• Recall the definition of an M-augmenting path
• a path joining two nodes u,v such that the edges
  are alternatively in, and not in M, and neither
  u, nor v are saturated by M

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Prove M is a matching
G
Matching M
Matching M M ? P
M-augmenting path P
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Prove M is a matching

• Assume M M ? P isnt a matching?
• There exists two edges in M that share a vertex
• 4 possibilities
• e1,e2 ? M
• e1,e2 ? P
• e1 ? M, e2 ? P
• e1 ? P, e2 ? M
• M M ? P never contains two such edges, thus M
  is a matching

y
x
z
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Prove M C

• Let U be nodes not saturated by M
• If U Ø then
• terminate at step 1
• M X Y V/2
• C X
• C X M

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Prove M C
X
Y

• If U ? Ø then
• M 1/2 V\U
• Each alternating tree has
• one more even node than odd nodes
• one node in U (the root)
• Ne No U
• Three types of disjoint vertices
• V Ne ? No ? P
• V Ne No P
• V 2No P U
• V\U 2No P
• 1/2 V\U No 1/2 P
• Recall
• half of the nodes in P are in X
• CP P?X
• CP 1/2 P
• C No ? CP
• C No CP
• C No 1/2 P

a
f
U d
M
b
P
No
Ne
c
g
d
h
e
i
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Algorithm Structure

• Step 0 Initialization
• Step 1 Termination Test
• Step 2 Beginning Forest Growth
• Step 3 Edge Selection
• Step 4 Growth
• Step 5 Augmentation
• Step 6 Termination

S0
S1
S2
S3
S5
S4
S6
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Algorithm Complexity
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Algorithm Complexity

• Total Complexity max1, n, n2,mn
  O(mn)
• Hopcroft-Karp algorithm has complexity O(mn1/2)
• Method based upon fast matrix multiplication has
  complexity O(n2.376)
• better in theory
• slower in practice

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Questions
?