Polyhedral Optimization Lecture 4 – Part 3 M. Pawan Kumar Slides available online

Matroid Mapping Matroid Union Outline

Mapping M’ = (S’, I’ ) s’ 1 s’ 2 s’ 3 s’ 4 s’ 5 s1s1 s2s2 s3s3 s4s4 s5s5 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 s1s1 s2s2 s3s3 s4s4 s5s5 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 s1s1 s2s2 s3s3 s4s4 s5s5 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 s1s1 s2s2 s3s3 s4s4 s5s5 f: S’ → S I = {f(X’), X’ ∈ I’ } M = (S, I ) Is M a matroid? YESProof?

Proof Sketch M = (S, I ) is a subset systemProof 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 systemProof 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 s1s1 s2s2 s3s3 s4s4 s5s5 f: S’ → S I = {f(X’), X’ ∈ I’ } M = (S, I ) f -1 (s) = {s’ ∈ S’, f(s’) = s} f -1 (s 1 )?{s’ 1,s’ 2 }

Inverse Function M’ = (S’, I’ ) s’ 1 s’ 2 s’ 3 s’ 4 s’ 5 s1s1 s2s2 s3s3 s4s4 s5s5 f: S’ → S I = {f(X’), X’ ∈ I’ } M = (S, I ) f -1 (s) = {s’ ∈ S’, f(s’) = s} f -1 (s 3 )?{s’ 4 }

Inverse Function M’ = (S’, I’ ) s’ 1 s’ 2 s’ 3 s’ 4 s’ 5 s1s1 s2s2 s3s3 s4s4 s5s5 f: S’ → S I = {f(X’), X’ ∈ I’ } M = (S, I ) f -1 (X) = ∪ s ∈ S f -1 (s) f -1 ({s 1,s 3 })? {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 s1s1 s2s2 s3s3 s4s4 s5s5 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 P U Parts = f -1 (s), s ∈ U Limits = 1 for all parts

Rank Function Take the intersection of M’ and P U r(U) = max |X|, X is independent in M’ and P U Why? Matroid Intersection Theorem r(U) = min T ⊆ U {|U\T| + r’(f -1 (T))}

Matroid Mapping Matroid Union Outline

Matroid Union M 1 = (S 1, I 1 )M 2 = (S 2, I 2 ) S 1 and S 2 are disjoint M = (S 1 ∪ S 2, {X 1 ∪ X 2, X 1 ∈ I 1, X 2 ∈ I 2 }) Is M a matroid?YES Proof is trivial

Matroid Union M 1 = (S 1, I 1 )M 2 = (S 2, I 2 ) Is M a matroid?YES Proof? M = (S 1 ∪ S 2, {X 1 ∪ X 2, X 1 ∈ I 1, X 2 ∈ I 2 })

Matroid Union M 1 = (S 1, I 1 )M 2 = (S 2, I 2 ) Make a copy of S’ 1 of S 1 Make a copy of S’ 2 of S 2

Matroid Union M’ 1 = (S’ 1, I’ 1 )M’ 2 = (S’ 2, I’ 2 ) Make a copy of S’ 1 of S 1 Make a copy of S’ 2 of S 2 S’ 1 and S’ 2 are disjoint M’ = (S’ 1 ∪ S’ 2, {X’ 1 ∪ X’ 2, X 1 ∈ I’ 1, X 2 ∈ I’ 2 }) Matroid

Matroid Union M’ 1 = (S’ 1, I’ 1 )M’ 2 = (S’ 2, I’ 2 ) f: S’ 1 ∪ S’ 2 → S 1 ∪ S 2 M = (S 1 ∪ S 2, {X 1 ∪ X 2, X 1 ∈ I 1, X 2 ∈ I 2 }) r M (U) = min T ⊆ U {|U\T| + r 1 (T∩S 1 ) + r 2 (T∩S 2 )} Left as exercise !! Matroid