Download Dual of Maximum Matching Problem

Polyhedral Optimization
Lecture 2 – Part 2
M. Pawan Kumar
[email protected]
Slides available online http://cvn.ecp.fr/personnel/pawan/
Outline
• Flows and Cuts
• Bipartite Matching and Vertex Cover
Functions on Arcs
D = (V, A)
s
1 1
v1
4
6
3 3
v3
10 8
v2
7
v4
3 3
t
Function f: A  Reals
Excess function Ef(v)
2
5
Arc capacities c(a)
Incoming value
Outgoing value
Functions on Arcs
D = (V, A)
s
1 1
v1
4
6
3 3
v3
10 8
v2
7
v4
3 3
t
Function f: A  Reals
Excess function Ef(v)
2
5
Arc capacities c(a)
Σain-arcs(v) f(a)
Outgoing value
Functions on Arcs
D = (V, A)
s
1 1
v1
4
6
3 3
v3
10 8
v2
7
v4
3 3
t
Function f: A  Reals
Excess function Ef(v)
2
5
Arc capacities c(a)
Σain-arcs(v) f(a)
Σaout-arcs(v) f(a)
Functions on Arcs
D = (V, A)
s
1 1
v1
4
6
3 3
v3
10 8
v2
7
v4
3 3
t
Function f: A  Reals
Excess function Ef(v)
2
5
Arc capacities c(a)
f(in-arcs(v))
f(out-arcs(v))
Ef(v1)
-6
Functions on Arcs
D = (V, A)
s
1 1
v1
4
6
3 3
v3
10 8
v2
7
v4
3 3
t
Function f: A  Reals
Excess function Ef(v)
2
5
Arc capacities c(a)
f(in-arcs(v))
f(out-arcs(v))
Ef(v2)
14
Excess Functions of Vertex Subsets
Excess function Ef(U)
s
1 1
v1
4
6
3 3
v3
10 8
v2
2
5
7
v4
3 3
t
Incoming Value
Outgoing Value
Excess Functions of Vertex Subsets
Excess function Ef(U)
s
1 1
v1
4
6
3 3
v3
10 8
v2
2
5
7
v4
3 3
t
Σain-arcs(U) f(a)
Outgoing Value
Excess Functions of Vertex Subsets
Excess function Ef(U)
s
1 1
v1
4
6
3 3
v3
10 8
v2
2
5
7
v4
3 3
t
Σain-arcs(U) f(a)
Σaout-arcs(U) f(a)
Excess Functions of Vertex Subsets
Excess function Ef(U)
s
1 1
v1
4
6
3 3
v3
10 8
v2
2
5
7
Ef({v1,v2})
v4
3 3
t
f(in-arcs(U))
f(out-arcs(U))
8
Excess Functions of Vertex Subsets
Excess function Ef(U)
s
1 1
v1
4
6
3 3
v3
10 8
v2
2
5
7
Ef({v1,v2})
-6 + 14
v4
3 3
t
f(in-arcs(U))
f(out-arcs(U))
Ef(U) = ΣvU Ef(v)
Outline
• Flows and Cuts
– s-t Flow
– s-t Cut
– Flows vs. Cuts
– Incidence Matrix
– Max-Flow Problem
– Dual of Max-Flow Problem
• Bipartite Matching and Vertex Cover
s-t Flow
Function flow: A  R
s
1
v1
8
6
3
v3
v2
2
5
7
Flow is non-negative
For all vertex expect s,t
v4
3
t
Flow of arc ≤ arc capacity
Incoming flow
= Outgoing flow
s-t Flow
Function flow: A  R
s
1
v1
8
6
3
v3
v2
2
5
7
Flow is non-negative
For all vertex expect s,t
v4
3
t
flow(a) ≤ c(a)
Incoming flow
= Outgoing flow
s-t Flow
Function flow: A  R
s
1
v1
8
6
3
v3
v2
2
5
7
flow(a) ≥ 0
For all vertex expect s,t
v4
3
t
flow(a) ≤ c(a)
Incoming flow
= Outgoing flow
s-t Flow
Function flow: A  R
s
1
v1
8
6
3
v3
v2
2
5
7
flow(a) ≥ 0
For all v  V \ {s,t}
v4
3
t
flow(a) ≤ c(a)
Incoming flow
= Outgoing flow
s-t Flow
Function flow: A  R
s
1
v1
8
6
3
v3
v2
2
5
7
flow(a) ≥ 0
For all v  V \ {s,t}
v4
3
t
flow(a) ≤ c(a)
Σ(u,v)A flow((u,v))
= Outgoing flow
s-t Flow
Function flow: A  R
s
1
v1
8
6
3
v3
v2
2
5
7
t
flow(a) ≤ c(a)
flow(a) ≥ 0
For all v  V \ {s,t}
v4
Σ(u,v)A flow((u,v))
3
= Σ(v,u)A flow((v,u))
s-t Flow
Function flow: A  R
s
1
v1
8
6
3
v3
v2
2
5
7
flow(a) ≥ 0
For all v  V \ {s,t}
v4
3
t
flow(a) ≤ c(a)
Eflow(v) = 0
s-t Flow
Function flow: A  R
s
1 1
v1
4
6
3 3
v3
10 8
v2
2
5
7
flow(a) ≥ 0
For all v  V \ {s,t}
v4
3 3
t
flow(a) ≤ c(a)
Eflow(v) = 0
✗
s-t Flow
Function flow: A  R
s
-1 1
v1
8
6
-1 3
v3
v2
2
5
7
flow(a) ≥ 0
For all v  V \ {s,t}
v4
-1 3
t
flow(a) ≤ c(a)
Eflow(v) = 0
✗
s-t Flow
Function flow: A  R
s
11
v1
8
6
13
v3
v2
2
5
7
flow(a) ≥ 0
For all v  V \ {s,t}
v4
13
t
flow(a) ≤ c(a)
Eflow(v) = 0
✓
Value of s-t Flow
s
1
v1
8
6
3
v3
v2
- Incoming flow of s
2
5
7
v4
3
t
Outgoing flow of s
Value of s-t Flow
s
1
v1
8
6
3
v3
v2
7
v4
3
t
Eflow(t)
Σ(s,v)A flow((s,v))
- Σ(u,s)A flow((u,s))
2
5
-Eflow(s)
Value of s-t Flow
s
11
v1
8
6
13
v3
v2
7
v4
13
t
Eflow(t)
Σ(s,v)A flow((s,v))
- Σ(u,s)A flow((u,s))
2
5
-Eflow(s)
Value = 1
Outline
• Flows and Cuts
– s-t Flow
– s-t Cut
– Flows vs. Cuts
– Incidence Matrix
– Max-Flow Problem
– Dual of Max-Flow Problem
• Bipartite Matching and Vertex Cover
Cut
D = (V, A)
Let U be a subset of V
10
v1
3
v3
v2
2
5
v4
C is a set of arcs such that
• (u,v)  A
• uU
• v  V\U
C is a cut in the digraph D
Cut
D = (V, A)
U
10
v1
3
v3
What is C?
v2
{(v1,v2),(v1,v4)} ?
2
5
V\U
{(v1,v4),(v3,v2)} ?
v4
✓
{(v1,v4)} ?
Cut
D = (V, A)
U
V\U
What is C?
10
v1
3
v3
v2
✓
2
5
{(v1,v2),(v1,v4),(v3,v2)} ?
v4
{(v4,v3)} ?
{(v1,v4),(v3,v2)} ?
Cut
D = (V, A)
V\U
U
What is C?
10
v1
3
v3
v2
2
5
✓
{(v1,v2),(v1,v4),(v3,v2)} ?
{(v3,v2)} ?
v4
{(v1,v4),(v3,v2)} ?
Cut
D = (V, A)
10
v1
3
v3
v2
2
5
C = out-arcs(U)
v4
Capacity of Cut
10
v1
3
v3
v2
Sum of capacity of all
arcs in C
2
5
v4
Capacity of Cut
10
v1
3
v3
v2
Σa  C c(a)
2
5
v4
Capacity of Cut
U
10
v1
3
v3
v2
2
5
V\U
v4
3
Capacity of Cut
V\U
U
10
v1
3
v3
v2
2
5
v4
15
s-t Cut
D = (V, A)
s
1
v1
8
6
3
v3
v2
A sink vertex “t”
v4
C is a cut such that
• sU
• t  V\U
2
5
7
3
t
A source vertex “s”
C is an s-t cut
Capacity of s-t Cut
s
1
v1
8
6
3
v3
v2
2
5
7
v4
3
t
Σa  C c(a)
Capacity of s-t Cut
s
1
v1
8
6
3
v3
v2
2
5
7
v4
3
t
5
Capacity of s-t Cut
s
1
v1
8
6
3
v3
v2
2
5
7
v4
3
t
17
Outline
• Flows and Cuts
– s-t Flow
– s-t Cut
– Flows vs. Cuts
– Incidence Matrix
– Max-Flow Problem
– Dual of Max-Flow Problem
• Bipartite Matching and Vertex Cover
Flows vs. Cuts
An s-t flow function flow: A  Reals
An s-t cut C such that s  U, t  V\U
Value of flow ≤ Capacity of C
Flows vs. Cuts
Value of flow = -Eflow(s)
= -Eflow(s) - ΣvU\{s} Eflow(v)
= -Eflow(U)
= flow(out-arcs(U))
- flow(in-arcs(U))
≤ Capacity of C
- flow(in-arcs(U))
Max-Flow Min-Cut Theorem
Value of max flow = Capacity of min cut
Ford and Fulkerson, 1958
“Combinatorial” proof
We want to use polyhedral techniques
Outline
• Flows and Cuts
– s-t Flow
– s-t Cut
– Flows vs. Cuts
– Incidence Matrix
– Max-Flow Problem
– Dual of Max-Flow Problem
• Bipartite Matching and Vertex Cover
Incidence Matrix
a1
s
2
4
3
v1
v2
a2
a3
a4
s
v1
v2
1
2
t
n vertices
m arcs
t
n x m matrix
a5
Incidence Matrix
a1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
s
-1
v1
1
v2
0
t
0
a2
a3
a4
n x m matrix
a5
Incidence Matrix
a1
a2
s
-1
-1
v1
1
0
v2
0
1
t
0
0
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
a3
a4
n x m matrix
a5
Incidence Matrix
a1
a2
a3
s
-1
-1
0
v1
1
0
-1
v2
0
1
1
t
0
0
0
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
a4
n x m matrix
a5
Incidence Matrix
a1
a2
a3
a4
s
-1
-1
0
0
v1
1
0
-1
-1
v2
0
1
1
0
t
0
0
0
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
n x m matrix
a5
Incidence Matrix
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
n x m matrix
Incidence Matrix
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
n x m matrix
Incidence Matrix
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
n x m matrix
?
Incidence Matrix
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
1
-1
0
v2
0
1
-1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
n x m matrix
Incidence Matrix
Each column contains exactly one +1
Each column contains exactly one -1
Rest of the column entries are 0
Linearly dependent rows, determinant = 0
Property
Digraph D = (V,A)
Incidence matrix M
M is a TUM
Proof?
Proof Sketch
Consider a submatrix S of M of size k x k
Mathematical induction: det(S) ∈ {0,+1,-1}
Trivial for k = 1
Assume it is true for k < m
Proof Sketch
Consider a submatrix S of M of size m x m
Case I: S contains a column of all zeros
det(S) = 0
Proof Sketch
Consider a submatrix S of M of size m x m
Case II: S contains a column with one non-zero
±1
sT
0
S’
Use induction assumption
Proof Sketch
Consider a submatrix S of M of size m x m
Case III: All columns contain two non-zeros
Sum of column = 0 (linearly dependent rows)
Hence proved
Outline
• Flows and Cuts
– s-t Flow
– s-t Cut
– Flows vs. Cuts
– Incidence Matrix
– Max-Flow Problem
– Dual of Max-Flow Problem
• Bipartite Matching and Vertex Cover
Max-Flow Problem
Variable x of size m x 1
s
2
4
3
v1
1
Flow in arc ‘ai’ is xi
v2
2
x≥0
t
n vertices
m arcs
Linear constraint
Max-Flow Problem
Capacity vector c
s
2
4
3
v1
2
4
v2
x
1
≤
2
1
t
2
n vertices
m arcs
3
Linear constraint
Max-Flow Problem
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
Incidence Matrix M
Max-Flow Problem
s
2
a2
a3
a4
a5
1
0
-1
-1
0 x
4
3
v1
a1
1
v2
2
t
n vertices
m arcs
v1
x1
Incoming
Flow
- x3
- x4
Outgoing
Flow
Max-Flow Problem
s
2
a2
a3
a4
a5
1
0
-1
-1
0 x
4
3
v1
a1
1
v2
2
t
n vertices
m arcs
v1
x1
- x3
Excess
Flow
- x4
Max-Flow Problem
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
Incidence Matrix M
Max-Flow Problem
s
2
a2
a3
a4
a5
0
1
1
0
-1 x
4
3
v1
a1
v2
v2
1
2
t
n vertices
m arcs
x2
+x3
Incoming
Flow
- x5
Outgoing
Flow
Max-Flow Problem
s
2
a2
a3
a4
a5
0
1
1
0
-1 x
4
3
v1
a1
v2
v2
1
2
t
n vertices
m arcs
x2
+x3
- x5
Excess
Flow
Max-Flow Problem
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
Incidence Matrix M
Max-Flow Problem
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
Incidence Submatrix M’
Is M’ a TUM?
YES
Max-Flow Problem
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
M’x = 0 Flow conversation
Linear constraint
Max-Flow Problem
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
Any other constraints?
Max-Flow Problem
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
w
Objective function?
Max-Flow Problem
a1
a2
a3
a4
a5
s
-1
-1
0
0
0
v1
1
0
-1
-1
0
v2
0
1
1
0
-1
t
0
0
0
1
1
s
2
4
3
v1
1
v2
2
t
n vertices
m arcs
w
maxx wTx
Max-Flow Problem
maxx
wTx
s.t.
0≤x≤c
M’x = 0
Integer polyhedron?
No, c may not be an integer vector
If c ∈ Zm, there exists an integer max flow
Outline
• Flows and Cuts
– s-t Flow
– s-t Cut
– Flows vs. Cuts
– Incidence Matrix
– Max-Flow Problem
– Dual of Max-Flow Problem
• Bipartite Matching and Vertex Cover
Dual of Max-Flow Problem
maxx
wTx
s.t.
0≤x≤c
M’x = 0
y≥0
z
Dual of Max-Flow Problem
min(y,z)
cTy
s.t.
y≥0
yT + zTM’ ≥ wT
Integer polyhedron?
YES
Optimal integer solution (y*,z*)
Dual of Max-Flow Problem
Define set U ⊆ V
Source s ∈ U
Sink t ∉ U
Cut C separates U and V\U
Capacity of C ≤ cTy*
Proof?
vi∈ U if z*i ≥ 0
Proof Sketch
Let ai = (vj,vk) ∈ out-arcs(C)
s
2
v1
4
3
1
v2
M’ =
1
0
-1
-1
0
0
1
1
0
-1
2
t
w=
y* ≥ 0
0
0
0
1
y*T + z*TM’ ≥ wT
1
Proof Sketch
Let ai = (vj,vk) ∈ out-arcs(C)
s
2
v1
4
3
1
y*1 ≥ 0
z*1 < 0
y*1 + z*1 ≥ 0
y*1 ≥ 1
v2
2
t
Capacity included in cTy*
y* ≥ 0
y*T + z*TM’ ≥ wT
Proof Sketch
Let ai = (vj,vk) ∈ out-arcs(C)
s
2
v1
4
3
y*3 ≥ 0
z*1 ≥ 0, z*2 < 0
v2
y*3 + z*2 - z*1 ≥ 0
1
y*3 ≥ 1
2
t
Capacity included in cTy*
y* ≥ 0
y*T + z*TM’ ≥ wT
Proof Sketch
Let ai = (vj,vk) ∈ out-arcs(C)
s
2
v1
4
3
1
y*5 ≥ 0
z*2 ≥ 0
y*5 - z*2 ≥ 1
y*5 ≥ 1
v2
2
t
Capacity included in cTy*
y* ≥ 0
y*T + z*TM’ ≥ wT
Proof Sketch
Let ai = (vj,vk) ∈ out-arcs(C)
z*j ≥ 0
z*k < 0
y*i + z*k – z*j ≥ 0
Therefore y*i ≥ 1
Capacity of C ≤ cTy*
Max-Flow Min-Cut Theorem
Capacity of any cut ≥ Value of any flow
C must be the minimum cut
Strong duality
Capacity of C ≤ cTy* = Value of max flow
Max-Flow Min-Cut Theorem
Capacity of any cut ≥ Value of any flow
C must be the minimum cut
Computing min cut is an “easy” problem
Strong duality
Capacity of C = cTy* = Value of max flow
Outline
• Flows and Cuts
• Bipartite Matching and Vertex Cover
Undirected Graph
G = (V, E)
v0
v1
v4
Simple graph
v2
v3
v5
v6
No parallel edges
No loops
Parallel edges
Loop
Walk
G = (V, E)
v0
v1
v4
v2
v3
V = {v1,…,vn}
v5
E = {e1,…,em}
v6
Sequence P = (v0,e1,v1,…,ek,vk), ei = (vi-1,vi)
v0, (v0,v4), v4, (v4,v2), v2, (v2,v5), v5, (v5,v4), v4
Path
G = (V, E)
v0
v1
v4
v2
v3
V = {v1,…,vn}
v5
E = {e1,…,em}
v6
Sequence P = (v0,e1,v1,…,ek,vk), ei = (vi-1,vi)
Vertices v0,v1,…,vk are distinct
Connected Graph
G = (V, E)
v0
v1
v4
v2
v3
V = {v1,…,vn}
v5
E = {e1,…,em}
v6
Connected?
There exists a walk from one vertex to another
Simple Connected Graphs
G = (V, E)
v0
v1
v4
v2
v3
v5
v6
We will assume the graph is simple
We will assume the graph is connected
Bipartite Graph
G = (V, E)
v0
v1
v4
v2
V = V1 ∪ V2
v5
V1 ∩ V2 = ϕ
V1 ≠ ϕ
v3
V2 ≠ ϕ
v6
No edge (u,v) ∈ E such that u ∈ V1 and v ∈ V1
No edge (u,v) ∈ E such that u ∈ V2 and v ∈ V2
Bipartite Graph
G = (V, E)
v4
v2
V = V1 ∪ V2
v5
Bipartite?
V1 ∩ V2 = ϕ
V1 ≠ ϕ
V2 ≠ ϕ
No edge (u,v) ∈ E such that u ∈ V1 and v ∈ V1
No edge (u,v) ∈ E such that u ∈ V2 and v ∈ V2
Bipartite Graph
G = (V, E)
v0
v1
v4
V = V1 ∪ V2
V1 ∩ V2 = ϕ
v2
Bipartite?
V1 ≠ ϕ
V2 ≠ ϕ
No edge (u,v) ∈ E such that u ∈ V1 and v ∈ V1
No edge (u,v) ∈ E such that u ∈ V2 and v ∈ V2
Bipartite Graph
G = (V, E)
v0
v1
v4
v2
v5
V1 ∩ V2 = ϕ
V1 ≠ ϕ
Bipartite?
v3
V = V1 ∪ V2
V2 ≠ ϕ
v6
No edge (u,v) ∈ E such that u ∈ V1 and v ∈ V1
No edge (u,v) ∈ E such that u ∈ V2 and v ∈ V2
Outline
• Flows and Cuts
• Bipartite Matching and Vertex Cover
– Bipartite Matching
– Vertex Cover
– Matching vs. Cover
– Incidence Matrix
– Maximum Matching Problem
– Dual of Maximum Matching Problem
– Generalization
Bipartite Matching
G = (V, E)
v0
v1
v4
v2
v3
G is bipartite
v5
v6
A matching is a set of disjoint edges
No two edges share a vertex
Bipartite Matching
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
Matching?
v3
v6
A matching is a set of disjoint edges
No two edges share a vertex
Bipartite Matching
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
Matching?
v3
v6
A matching is a set of disjoint edges
No two edges share a vertex
Bipartite Matching
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
Matching?
v3
v6
A matching is a set of disjoint edges
No two edges share a vertex
Size of Matching
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
Size?
v3
v6
Number of edges in the matching
Size of Matching
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
Size?
v3
v6
Number of edges in the matching
Bipartite Matching Problem
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
Optimal solution?
v3
v6
Find a matching with the maximum size
Outline
• Flows and Cuts
• Bipartite Matching and Vertex Cover
– Bipartite Matching
– Vertex Cover
– Matching vs. Cover
– Incidence Matrix
– Maximum Matching Problem
– Dual of Maximum Matching Problem
– Generalization
Vertex Cover
G = (V, E)
v0
v1
v4
v2
v3
G is bipartite
v5
C⊆V
v6
No edge (u,v) ∈ E such that u ∉ C and v ∉ C
C covers all the edges
Vertex Cover
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
C⊆V
Vertex cover?
v3
v6
No edge (u,v) ∈ E such that u ∉ C and v ∉ C
C covers all the edges
Vertex Cover
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
C⊆V
Vertex cover?
v3
v6
No edge (u,v) ∈ E such that u ∉ C and v ∉ C
C covers all the edges
Vertex Cover
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
C⊆V
Vertex cover?
v3
v6
No edge (u,v) ∈ E such that u ∉ C and v ∉ C
C covers all the edges
Size of Vertex Cover
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
Size?
v3
v6
Number of vertices in the vertex cover
Size of Vertex Cover
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
Size?
v3
v6
Number of vertices in the vertex cover
Vertex Cover Problem
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
Optimal solution?
v3
v6
Find a vertex cover with the minimum size
Outline
• Flows and Cuts
• Bipartite Matching and Vertex Cover
– Bipartite Matching
– Vertex Cover
– Matching vs. Cover
– Incidence Matrix
– Maximum Matching Problem
– Dual of Maximum Matching Problem
– Generalization
Question
Graph G = (V, E)
Not necessarily bipartite
Matching M ⊆ E
Vertex Cover C ⊆ V
|M| ≤ |C|?
|C| ≤ |M|?
Question
Graph G = (V, E)
Not necessarily bipartite
Matching M ⊆ E
Vertex Cover C ⊆ V
|M| ≤ |C|
Proof?
Kőnig’s Theorem
For bipartite graphs,
Size of max matching = size of min cover
Kőnig, 1931
“Combinatorial” proof
We want to use polyhedral techniques
Outline
• Flows and Cuts
• Bipartite Matching and Vertex Cover
– Bipartite Matching
– Vertex Cover
– Matching vs. Cover
– Incidence Matrix
– Maximum Matching Problem
– Dual of Maximum Matching Problem
– Generalization
Incidence Matrix
v0
v1
v2
v3
n vertices
m edges
Incidence Matrix
v0
v1
V1
v3
n vertices
m edges
v2
Incidence Matrix
v0
v1
V2
v3
n vertices
m edges
v2
Incidence Matrix
e1
v0
e2
e3
v0
v1
v2
v3
v1
v3
n vertices
m edges
v2
n x m matrix
e4
Incidence Matrix
e1
v0
v1
v2
v3
n vertices
m edges
v0
1
v3
0
v1
1
v2
0
e2
e3
n x m matrix
e4
Incidence Matrix
e1
e2
v0
1
1
v3
0
0
v1
1
0
v2
0
1
v0
v1
v2
v3
n vertices
m edges
e3
n x m matrix
e4
Incidence Matrix
e1
e2
e3
v0
1
1
0
v3
0
0
1
v1
1
0
1
v2
0
1
0
v0
v1
v2
v3
n vertices
m edges
n x m matrix
e4
Incidence Matrix
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
v0
v1
v2
v3
n vertices
m edges
n x m matrix
Incidence Matrix
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
v0
v1
v2
v3
n vertices
m edges
n x m matrix
Incidence Matrix
e1
e2
e3
e4
One 1 per column v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
V1
Incidence Matrix
One 1 per column
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
V2
Incidence Matrix
1x
-1 x
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
Add rows to get a vector of zeros
Linearly dependent, determinant = 0
1
V1
V2
Property
Bipartite graph G = (V,E)
Incidence matrix M
M is a TUM
Proof?
Proof Sketch
Consider a submatrix S of M of size k x k
Mathematical induction: det(S) ∈ {0,+1,-1}
Trivial for k = 1
Assume it is true for k < m
Proof Sketch
Consider a submatrix S of M of size m x m
Case I: S contains a column of all zeros
det(S) = 0
Proof Sketch
Consider a submatrix S of M of size m x m
Case II: S contains a column with one 1
1
sT
0
S’
Use induction assumption
Proof Sketch
Consider a submatrix S of M of size m x m
Case III: All columns contain two non-zeros
Incidence matrix of a bipartite graph, det(S) = 0
Hence proved
Outline
• Flows and Cuts
• Bipartite Matching and Vertex Cover
– Bipartite Matching
– Vertex Cover
– Matching vs. Cover
– Incidence Matrix
– Maximum Matching Problem
– Dual of Maximum Matching Problem
– Generalization
Maximum Matching Problem
Variable x of size m x 1
v0
Edge ‘ei’ in matching if xi = 1
v1
v2
Otherwise xi = 0
v3
n vertices
m edges
xi ∈ {0,1}
Integer constraint
Maximum Matching Problem
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
v0
v1
v2
v3
n vertices
m edges
Incidence Matrix A
Maximum Matching Problem
v0
v0
v1
1
1
0
0
x
v2
x 1 + x2
v3
n vertices
m edges
Number of edges incident on v0
Maximum Matching Problem
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
v0
v1
v2
v3
n vertices
m edges
Incidence Matrix A
Maximum Matching Problem
v0
v1
v2
v3
0
0
1
1
x
x 3 + x4
v3
n vertices
m edges
Number of edges incident on v3
Maximum Matching Problem
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
v0
v1
v2
v3
n vertices
m edges
Incidence Matrix A
x
Maximum Matching Problem
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
v0
v1
v2
v3
n vertices
m edges
Ax ≤ 1
Linear constraint
x
Maximum Matching Problem
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
v0
v1
v2
v3
n vertices
m edges
Any other constraints?
Maximum Matching Problem
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
v0
v1
v2
v3
n vertices
m edges
Objective function?
Maximum Matching Problem
e1
e2
e3
e4
v0
1
1
0
0
v3
0
0
1
1
v1
1
0
1
0
v2
0
1
0
1
v0
v1
v2
v3
n vertices
m edges
maxx 1Tx
Maximum Matching Problem
maxx
1Tx
s.t.
0≤x≤1
Ax ≤ 1
x ∈{0,1}m
Integer polyhedron?
YES
Maximum Matching Problem
maxx
1Tx
s.t.
0≤x≤1
Ax ≤ 1
Maximum bipartite matching is “easy”
Integer polyhedron?
YES
Maximum Matching Problem
maxx
1Tx
s.t.
0≤x
Why?
Ax ≤ 1
Maximum bipartite matching is “easy”
Integer polyhedron?
YES
Outline
• Flows and Cuts
• Bipartite Matching and Vertex Cover
– Bipartite Matching
– Vertex Cover
– Matching vs. Cover
– Incidence Matrix
– Maximum Matching Problem
– Dual of Maximum Matching Problem
– Generalization
Dual of Maximum Matching Problem
maxx
1Tx
s.t.
0≤x
y≥0
Ax ≤ 1
z≥0
Dual of Maximum Matching Problem
min(y,z)
1Tz
s.t.
0≤z
zTA - yT = 1T
0≤y
Dual of Maximum Matching Problem
minz
s.t.
1Tz
0≤z
zTA ≥ 1T
Integer polyhedron?
YES
Optimal integer solution z*
Dual of Maximum Matching Problem
Define set C ⊆ V
vi∈ C if z*i > 0
C is a vertex cover of the graph
Size of C ≤ 1Tz*
Proof?
For each vi ∈ C, z*i ≥ 1
Dual of Maximum Matching Problem
Size of any cover ≥ Size of any matching
C must be the minimum cover
Strong Duality
Size of C ≤ 1Tz*
= Size of maximum matching
Dual of Maximum Matching Problem
Size of any cover ≥ Size of any matching
C must be the minimum cover
Strong Duality
Size of C = 1Tz*
= Size of maximum matching
Minimum bipartite vertex cover is “easy”
Outline
• Flows and Cuts
• Bipartite Matching and Vertex Cover
– Bipartite Matching
– Vertex Cover
– Matching vs. Cover
– Incidence Matrix
– Maximum Matching Problem
– Dual of Maximum Matching Problem
– Generalization
b-Matching
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
n x 1 vector b
bv: +ve integer
v3
v6
b-Matching
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
n x 1 vector b
bv ∈ Z+
v3
v6
b-matching is a function x : E → Z+
∑e = (u,v) ∈ E x(e) ≤ bv
Matching?
b=1
b-Matching Problem
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
n x 1 vector b
bv ∈ Z+
v3
Given w : E → Z+
v6
max wTx
s.t. x is a b-matching
w-Vertex Cover
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
m x 1 vector w
w e ∈ Z+
v3
v6
w-vertex cover is a function z : V → Z+
For all e = (u,v), z(u) + z(v) ≥ we Vertex Cover?
w-Vertex Cover Problem
G = (V, E)
v0
v1
v4
v2
G is bipartite
v5
m x 1 vector w
w e ∈ Z+
v3
Given b : V → Z+
v6
min bTz
s.t. z is a w-vertex cover
b-Matching Problem
maxx
wTx
s.t.
0≤x
Ax ≤ b
x is integer
Integer polyhedron?
YES
b-Matching Problem
maxx
wTx
s.t.
0≤x
Ax ≤ b
b-matching is “easy”
Integer polyhedron?
YES
Dual of b-Matching Problem
minz
s.t.
bTz
0≤z
zAT ≥ w
w-vertex cover is “easy”
Integer polyhedron?
YES
Questions?