1. Polyhedral Optimization Lecture 4 – Part 2 M. Pawan Kumar pawan.kumar@ecp.fr Slides available online http://cvn.ecp.fr/personnel/pawan/

2. Uncrossing Operation Span of Maximal Chain Maximal Chains for Matroid Pair Matroid Intersection Polytope Outline

3. Tight Set M = (S, I ) is a matroid Independent set polytope P(M) x s ≥ 0, for all s ∈ S ∑ s ∈ U x s ≤ r M (U), for all U ⊆ S Consider x ∈ P(M) A ⊆ S is tight if x(A) = ∑ s ∈ A x s = r M (A)

4. Uncrossing Operation Given x ∈ P(M), let A and B be tight. A ∪ B is tight Proof?A ∩ B is tight

5. Proof Sketch Given x ∈ P(M), let A and B be tight. r M (A ∪ B) + r M (A ∩ B)≥ x(A ∪ B) + x(A ∩ B) Why? = x(A) + x(B)Why? = r(A) + r(B)Why? ≥ r(A ∪ B) + r(A ∩ B) Why? All inequalities are equalities !!

6. Proof Sketch Given x ∈ P(M), let A and B be tight. r M (A ∪ B) + r M (A ∩ B)= x(A ∪ B) + x(A ∩ B) r M (A ∪ B) ≥ x(A ∪ B) r M (A ∩ B) ≥ x(A ∩ B)

7. Proof Sketch Given x ∈ P(M), let A and B be tight. A ∪ B is tight A ∩ B is tight r M (A ∪ B) + r M (A ∩ B)= x(A ∪ B) + x(A ∩ B) r M (A ∪ B) = x(A ∪ B) r M (A ∩ B) = x(A ∩ B) Hence proved !!

8. Uncrossing Operation Span of Maximal Chain Maximal Chains for Matroid Pair Matroid Intersection Polytope Outline

9. Chain of Tight Sets Consider x ∈ P(M) Chain C = {A 1, A 2, A 3, …, A k } A i is tight with respect to xr(A i ) = x(A i ) A i is a proper subset of A i+1 A i ⊂ A i+1 By convention, A 0 = null set

10. Maximal Chain of Tight Sets Consider x ∈ P(M) Chain C = {A 1, A 2, A 3, …, A k } Number of sets k is maximum possible for x

11. Span of Maximal Chain Consider x ∈ P(M) Maximal chain C = {A 1, A 2, A 3, …, A k } Incidence vector v i for A i U is tight with respect to x v U ∈ span({v i, i=1,…,k}) Proof?

12. Proof Sketch Case I: U is not a subset of A k Define A = A k ∪ U A is tight with respect to x Add A to chain C to get a bigger chain Contradiction Why?

13. Proof Sketch Case II: U is a subset of A k and there exists j s.t. U ∩ (A j \A j-1 ) ⊂ A j \A j-1 U cuts A j \A j-1 into two non-empty parts Define A = (U ∩ A j ) ∪ A j-1 Contradiction U ∩ (A j \A j-1 ) ≠ null A is tight A can be inserted between A j and A j-1 in chain C

14. Proof Sketch Case III: U is a subset of A k and for all j U ∩ (A j \A j-1 ) = A j \A j-1 U does not cut A j \A j-1 into two non-empty parts There must exist a J ⊆ {1,2,…,k} Hence proved U ∩ (A j \A j-1 ) = null v U = ∑ j ∈ J v j -v j-1 OR

15. Uncrossing Operation Span of Maximal Chain Maximal Chains for Matroid Pair Matroid Intersection Polytope Outline

16. Maximal Chains of Matroid Pair M 1 = (S, I 1 ) M 2 = (S, I 2 ) Consider vertex x ∈ P(M 1 ) ∩ P(M 2 ) Maximal chain C = {A 1,…,A k } wrt x for M 1 Maximal chain D = {B 1,…,B l } wrt x for M 2 {v T |T ∈ C ∪ D} has m ≥ |supp(x)| l.i. vectors Proof?

17. Proof Sketch x s ≥ 0, for all s ∈ S ∑ s ∈ U x s ≤ r 1 (U), for all U ⊆ S ∑ s ∈ U x s ≤ r 2 (U), for all U ⊆ S P(M 1 ) ∩ P(M 2 )x is a vertex |S| tight linearly independent constraints |S|-supp(x) constraints of form x s ≥ 0

18. Proof Sketch ∑ s ∈ U x s ≤ r 1 (U), for all U ⊆ S ∑ s ∈ U x s ≤ r 2 (U), for all U ⊆ S x is a vertex supp(x) linearly independent constraints of form ∑ s ∈ U x s = r 1 (U) or ∑ s ∈ U x s = r 2 (U) U ∈ span{v T |T ∈ C ∪ D} Hence proved !!

19. Uncrossing Operation Span of Maximal Chain Chains for Matroid Pair Matroid Intersection Polytope Outline

20. Matroid Intersection Polytope M 1 = (S, I 1 ) M 2 = (S, I 2 ) P(M 1 ∩ M 2 ) ⊆ P(M 1 ) ∩ P(M 2 ) Proof?

21. Matroid Intersection Polytope M 1 = (S, I 1 ) M 2 = (S, I 2 ) P(M 1 ∩ M 2 ) = P(M 1 ) ∩ P(M 2 ) Proof? Induction on |S|Trivial for |S| = 0 Let x be a vertex of P(M 1 ) ∩ P(M 2 ) x ∉ P(M 1 ∩ M 2 )

22. Proof Sketch Case I: x s = 0 for some s ∈ S Remove x s from x to obtain x’ Delete s to obtain M 1 \s and M 2 \s x’ ∈ P(M 1 \s) and x’ ∈ P(M 2 \s) Why? x’ ∈ P(M 1 \s ∩ M 2 \s) Why? x ∈ P(M 1 ∩ M 2 ) Why?

23. Proof Sketch Case II: x s = 1 for some s ∈ S Remove x s from x to obtain x’ Contract s to obtain M 1 /s and M 2 /s x’ ∈ P(M 1 /s) and x’ ∈ P(M 2 /s) Why? x’ ∈ P(M 1 /s ∩ M 2 /s) Why? x ∈ P(M 1 ∩ M 2 ) Why?

24. Proof Sketch Case III: 0 < x s <1 for all s ∈ S |supp(x)| = |S| {v T |T ∈ C ∪ D} has m ≥ |S| l.i. vectors Every A j \A j-1 has at least two elementsWhy? |C| ≤ |S|/2|D| ≤ |S|/2

25. Proof Sketch Case III: 0 < x s <1 for all s ∈ S |supp(x)| = |S| {v T |T ∈ C ∪ D} has m ≥ |S| l.i. vectors Every A j \A j-1 has at least two elements |C| = |S|/2|D| = |S|/2

26. Proof Sketch Case III: 0 < x s <1 for all s ∈ S |supp(x)| = |S| {v T |T ∈ C ∪ D} has m ≥ |S| l.i. vectors Every A j \A j-1 has at least two elements |C| = |S|/2|D| = |S|/2 A k = S B l = S |C ∪ D| ≤ |S| -1 Contradiction

27. Matroid Intersection Polytope M 1 = (S, I 1 ) M 2 = (S, I 2 ) P(M 1 ∩ M 2 ) = P(M 1 ) ∩ P(M 2 ) Not true for 3 matroids

28. Matroid Intersection Polytope M 1 = (S, I 1 ) M 2 = (S, I 2 ) P(M 1 ∩ M 2 ∩ M 3 ) ≠ P(M 1 ) ∩ P(M 2 ) ∩ P(M 3 ) NP-hard problem max |X| such that X ∈ I 1 ∩ I 2 ∩ I 3 M 3 = (S, I 3 )