Download - Polyhedral Optimization Lecture 4 – Part 3 M. Pawan Kumar [email protected] Slides available online
Polyhedral OptimizationLecture 4 – Part 3
M. Pawan Kumar
Slides available online http://cvn.ecp.fr/personnel/pawan/
• Matroid Mapping
• Matroid Union
Outline
Mapping
M’ = (S’, I’)
s’1
s’2
s’3
s’4
s’5
s1
s2
s3
s4
s5
f: S’ → S
f(X’) = {f(s), s X’} for any X’ S’∈ ⊆
I = {f(X’), X’ ∈ I’}
Bijective
M = (S, I)
Mapping
M’ = (S’, I’)
s’1
s’2
s’3
s’4
s’5
s1
s2
s3
s4
s5
f: S’ → S
Is M a matroid?
I = {f(X’), X’ ∈ I’}
Bijective
M = (S, I)
Mapping
M’ = (S’, I’)
s’1
s’2
s’3
s’4
s’5
s1
s2
s3
s4
s5
f: S’ → S
I = {f(X’), X’ ∈ I’}
One-to-One
M = (S, I)
Is M a matroid?
Mapping
M’ = (S’, I’)
s’1
s’2
s’3
s’4
s’5
s1
s2
s3
s4
s5
f: S’ → S
I = {f(X’), X’ ∈ I’}
M = (S, I)
Is M a matroid? YES Proof?
Proof Sketch
M = (S, I) is a subset system Proof is trivial
Let X ∈ I and Y ∈ I, with |X| < |Y|
There has to exist s Y\X, X {s} ∈ ∪ ∈ I
Let X’ ∈ I’ with f(X’) = X and |X’| = |X|
Let Y’ ∈ I’ with f(Y’) = Y and |Y’| = |Y|
X’ and Y’ are not necessarily unique
Proof Sketch
M = (S, I) is a subset system Proof is trivial
Let X ∈ I and Y ∈ I, with |X| < |Y|
There has to exist s Y\X, X {s} ∈ ∪ ∈ I
Let X’ ∈ I’ with f(X’) = X and |X’| = |X|
Let Y’ ∈ I’ with f(Y’) = Y and |Y’| = |Y|
Choose X’ and Y’ by maximizing |X’ ∩ Y’|
Proof Sketch
|X’| < |Y’|
There has to exist s’ Y’\X’, X’ {s’} ∈ ∪ ∈ I’
If f(s’) = s X, then X {s} ∉ ∪ ∈ I M is a matroid
Proof Sketch
|X’| < |Y’|
There has to exist s’ Y’\X’, X’ {s’} ∈ ∪ ∈ I’
Let f(s’) = s X∈
There exists s’’ X’ such that f(s’’) = s ∈
s’’ Y’∉ Why?
Otherwise, |Y’| > |Y| since f(s’) = f(s’’)
Proof Sketch
|X’| < |Y’|
There has to exist s’ Y’\X’, X’ {s’} ∈ ∪ ∈ I’
Let f(s’) = s X∈
There exists s’’ X’ such that f(s’’) = s ∈
s’’ Y’∉ X’’ = X’ – {s’’} + {s’}
|X’’ ∩ Y’| > |X’ ∩ Y’| Contradiction
• Matroid Mapping– Inverse Function– Rank Function
• Matroid Union
Outline
Inverse Function
M’ = (S’, I’)
s’1
s’2
s’3
s’4
s’5
s1
s2
s3
s4
s5
f: S’ → S
I = {f(X’), X’ ∈ I’}
M = (S, I)
f-1(s) = {s’ S’, f(s’) = s}∈ f-1(s1)? {s’1,s’2}
Inverse Function
M’ = (S’, I’)
s’1
s’2
s’3
s’4
s’5
s1
s2
s3
s4
s5
f: S’ → S
I = {f(X’), X’ ∈ I’}
M = (S, I)
f-1(s) = {s’ S’, f(s’) = s}∈ f-1(s3)? {s’4}
Inverse Function
M’ = (S’, I’)
s’1
s’2
s’3
s’4
s’5
s1
s2
s3
s4
s5
f: S’ → S
I = {f(X’), X’ ∈ I’}
M = (S, I)
f-1(X) = ∪s S∈ f-1(s) f-1({s1,s3})? {s’1,s’2,s’4}
• Matroid Mapping– Inverse Function– Rank Function
• Matroid Union
Outline
Mapping
M’ = (S’, I’)
s’1
s’2
s’3
s’4
s’5
s1
s2
s3
s4
s5
f: S’ → S
I = {f(X’), X’ ∈ I’}
M = (S, I)
Given U S, what is r(U)?⊆
Rank Function
Is r(U) = r’(f-1(U))? NO
|f-1(s)| can be greater than 1 for some s U∈
Let us construct sets X’ ⊆ f-1(U)
X’ contain at most 1 pre-image of each s U∈
How?
Rank Function
Is r(U) = r’(f-1(U))? NO
|f-1(s)| can be greater than 1 for some s U∈
Let us construct sets X’ ⊆ f-1(U)
X’ contain at most 1 pre-image of each s U∈
Partition Matroid PU
Parts = f-1(s), s U ∈ Limits = 1 for all parts
Rank Function
Take the intersection of M’ and PU
r(U) = max |X|, X is independent in M’ and PU
Why?
Matroid Intersection Theorem
r(U) = minT⊆U {|U\T| + r’(f-1(T))}
• Matroid Mapping
• Matroid Union
Outline
Matroid Union
M1 = (S1, I1) M2 = (S2, I2)
S1 and S2 are disjoint
M = (S1 S∪ 2, {X1 X∪ 2, X1 ∈I1 , X2 ∈I2})
Is M a matroid? YES
Proof is trivial
Matroid Union
M1 = (S1, I1) M2 = (S2, I2)
Is M a matroid? YES
Proof?
M = (S1 S∪ 2, {X1 X∪ 2, X1 ∈I1 , X2 ∈I2})
Matroid Union
M1 = (S1, I1) M2 = (S2, I2)
Make a copy of S’1 of S1
Make a copy of S’2 of S2
Matroid Union
M’1 = (S’1, I’1) M’2 = (S’2, I’2)
Make a copy of S’1 of S1
Make a copy of S’2 of S2
S’1 and S’2 are disjoint
M’ = (S’1 S’∪ 2, {X’1 X’∪ 2, X1 ∈I’1 , X2 ∈I’2})
Matroid
Matroid Union
M’1 = (S’1, I’1) M’2 = (S’2, I’2)
f: S’1 S’∪ 2 → S1 S∪ 2
M = (S1 S∪ 2, {X1 X∪ 2, X1 ∈I1 , X2 ∈I2})
rM(U) = minT U ⊆ {|U\T| + r1(T∩S1) + r2(T∩S2)}
Left as exercise !!
Matroid