Polyhedral OptimizationLecture 4 – Part 3

M. Pawan Kumar

pawan.kumar@ecp.fr

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