1. Query Lower Bounds for Matroids via Group Representations Nick Harvey

2. Matroids Definition A matroid is a pair (S, B ) where B ⊂ 2 S s.t. Example: B = { spanning trees of graph G } Sets in B are called bases Rank of matroid is |A| for any A ∈ B Exchange Property Let A and B ∈ B  a ∈ A\B,  b ∈ B\A s.t. B+a-b ∈ B

3. Matroids Definition A matroid is a pair (S, B ) where B ⊂ 2 S s.t. Example: B = { spanning trees of graph G } Applications –Generalize Graph Problems –Approximation Algorithms –Network Coding Exchange Property Let A and B ∈ B  a ∈ A\B,  b ∈ B\A s.t. B+a-b ∈ B b1⊕b2b1⊕b2 b1⊕b2b1⊕b2 b1⊕b2b1⊕b2 s t b1b1 b2b2

4. Multicast Network Coding Goal: Multicast source  sinks at maximum rate Algorithm: Can construct optimal solution in P Flow: [Jaggi et al. ’05], Matroids: [H., Karger, Murota ’05] Source Sinks a b a b ab a+b

5. Reversibility of Network Codes Sources Sinks a b ab a+b sasa sbsb tbtb tata G

6. Reversibility of Network Codes Flow: G feasible  G rev feasible Coding:  feasible G s.t. G rev not feasible [Dougherty, Zeger ’06] Sources Sinks a b ab a+b sasa sbsb tbtb tata G rev

7. Constructed from Fano and non-Fano matroids Reversibility of Network Codes [Dougherty, Zeger ’06]

8. Discrete Optimization Problems Matroid Intersection Bipartite Matching Non-Bip. Matching Network Flow Submodular Function Minimization Submodular Flow Matroid Matching Minimum Spanning Tree Matroid Greedy Algorithm Spanning Tree Packing Min-cost Arboresence Matroid Intersection Matroid Intersection Given matroids M 1 =(S, B 1 ) and M 2 =(S, B 2 ), is B 1 ⋂ B 2 = ∅ ?

9. n = # elements r = rank Unweighted Edmonds/Lawler ’68- ’75 O(nr 2 ) oracle Cunningham ’86O(nr 1.5 ) oracle Gabow-Xu ’89-’96O(nr 1.62 ) Harvey ’06O(nr 1.38 ) Matroid Intersection Algorithms Linear Matroids W = max weight Weighted Lawler/Edmonds ~’75O(nr 2 ) oracle Shigeno-Iwata ’95 O ( nr 1.5 log(rW) ) oracle Gabow-Xu ’89-’96 O ( nr 1.77 log(rW) ) Harvey ’07O(nr 1.38 W) Linear Matroids

10. n = # elements r = rank Unweighted Edmonds/Lawler ’68- ’75 O(nr 2 ) oracle Cunningham ’86O(nr 1.5 ) oracle Gabow-Xu ’89-’96O(nr 1.62 ) Harvey ’06O(nr 1.38 ) Linear Matroids Weighted Are these algorithms optimal? W = max weight Lawler/Edmonds ~’75O(nr 2 ) oracle Shigeno-Iwata ’95 O ( nr 1.5 log(rW) ) oracle Gabow-Xu ’89-’96 O ( nr 1.77 log(rW) ) Harvey ’07O(nr 1.38 W) Linear Matroids

11. Algorithm Computational Lower Bounds Strong lower bounds in unrestricted computational models are beyond our reach –5n - o(n) is best-known lower bound on circuit size for an explicit boolean function. [Iwama et al. ’05] –We believe 3SAT requires 2  (n) time, but best-known result is  (n). A super-linear lower bound for any natural problem in P is hopeless. Data

12. Query Lower Bounds Strong lower bounds can be proven in concrete computational models –Sorting in comparison model –Monotone graph properties [Rivest-Vuillemin ’76] –Volume of convex body Deterministic [Elekes ’86] Randomized [Rademacher-Vempala ’06] Our work –Matroid intersection, Submodular Function Minimization B Out In Data Black Box Algorithm Queries

13. Query Model for Matroids (Independence Oracle) Out In Matroid (S, B ) Algorithm “Yes” if  B ∈ B s.t. T ⊆ B “No” otherwise T⊆ST⊆S Example: if B = { spanning trees of graph G }, then query asks if T is an acyclic subgraph of G

14. Algorithms O(nr 2 ) queries [Lawler ’75] Matroid Intersection Complexity Are  (nr 2 ) queries necessary and sufficient to solve matroid intersection? D. J. A. Welsh, “Matroid Theory”, 1976.

15. Algorithms O(nr 2 ) queries [Lawler ’75] Matroid Intersection Complexity O(nr 1.5 ) queries [Cunningham ’86] Can one prove any non-trivial lower bound on # queries to solve matroid intersection? Are  (nr 2 ) queries necessary and sufficient to solve matroid intersection? D. J. A. Welsh, “Matroid Theory”, 1976.

16. Algorithms O(nr 2 ) queries [Lawler ’75] # queries Rank r 0nn/2 O(nr 1.5 ) queries [Cunningham ’86] Trivial LB Cunningham UB Matroid Intersection Complexity n 2n 0

17. Algorithms O(nr 1.5 ) queries [Cunningham ’86] # queries Rank r 0nn/2 Trivial LB Cunningham UB Via Dual Matroids Matroid Intersection Complexity n 2n 0

18. Algorithms O(nr 1.5 ) queries [Cunningham ’86] # queries Rank r 0nn/2 Trivial LB Cunningham UB Via Dual Matroids Optimal UB? Matroid Intersection Complexity n 2n 0

19. Matroid Intersection Complexity Algorithms O(nr 1.5 ) queries [Cunningham ’86] # queries Rank r 0n n 2n 0 n/2 Trivial LB Cunningham UB Via Dual Matroids Optimal UB? 1.58n New LB [Harvey ’08]

20. A family M of matroids, each of rank n/2 # oracle queries for any deterministic algorithm on inputs from M is:  (log 2 3) n - o(n) > 1.58n Lower Bound [Harvey ’08] Hard Instances Communication Complexity Rank Computation AliceBob M =M = 0101 1010 0101 1011

21. Hard Instances Bipartite Matching in Almost-2-Regular Graphs 1 3 2 4 Is there a perfect matching? Four vertices have degree 1

22. Hard Instances Bipartite Matching in Almost-2-Regular Graphs 1 3 2 4 Is there a perfect matching? No: if path from 1 to 2 Yes: otherwise Four vertices have degree 1

23. Alice given  ∈ S n and Bob given  ∈ S n Elements 1 and 2 are not in the same cycle Permutation  Permutation  -1 1 2 3 4 5 6 1’ 2’ 3’ 4’ 5’ 6’ In-Same-Cycle Problem: Are elements 1 and 2 in the same cycle of composition  -1 º  ? Permutation Formulation

24. LB from Rank Computation Let C be a matrix with rows and columns indexed by permutations in S n C ,  = where G = {  : 1 & 2 are in the same cycle of  } C is adjacency matrix of Cayley graph for S n with generators G 1 0 if  -1 º  ∈ G otherwise Main Result: rank C = (Moreover, it’s diagonalizable, all eigenvalues are integers, and they can be explicitly computed.) Corollary: # queries  log rank C = (log 2 3) n - o(n).

25. Main Proof A Tour of Algebraic Combinatorics Step 0: Young Tableaux Step 1: Decomposing G Step 2: Decomposing C Step 3: Block diagonalizing R Step 4: Diagonalizing X i Wrap-up 1 3 5 6 2 4 7 1 1 1 1 1 1 G = { (1,2) } × X 3 × X 4 … × X n

26. Young Diagrams Young diagram of shape =( 1, 2,..., k ) Standard Young Tableau of shape Main Result: rank C = # of SYT with n boxes such that 3  1 Row i has i boxes 1  2  … k >0 # boxes = n = ∑ i i Place numbers {1,..,n} in boxes Rows increase → Columns increase ↓ 1 1 2 2 6 6 8 8 3 3 5 5 9 9 4 4 11 7 7 10 (and some other minor conditions)

27. Main Proof Step 0: Young Tableaux Step 1: Decomposing G Step 2: Decomposing C Step 3: Block diagonalizing R Step 4: Diagonalizing X i Wrap-up 1 3 5 6 2 4 7 1 1 1 1 1 1 G = { (1,2) } × X 3 × X 4 … × X n

28. Decomposing S n Claim:  ∈ S n   = ◦ ( n,  -1 ( n ) ), where ∈ S n-1 Example: Let ∈ S 6 be Then ◦ ( 7, 3 ) ∈ S 7 is  ˜ 1 3 5 6 2 4 2 4 7 1 3 5 6  ˜  ˜  ˜

29. Decomposing S n Restatement: Let X i = { (j,i) : 1  j  i }. Then S n = X 2 × … × X n. 1 3 5 6 2 4 7 (2,2)  = = ◦ (1,3)◦ (2,4)◦ (3,5)◦ (5,6)◦ (3,7) Claim:  ∈ S n   = ◦ ( n,  -1 ( n ) ), where ∈ S n-1  ˜  ˜

30. Decomposing G Let G = {  : 1 & 2 are in the same cycle } Claim: Let X i = { (j,i) : 1  j  i }. Then G = { (1,2) } × X 3 × X 4 … × X n. 1 & 2 remain in the same cycle 1 3 5 6 2 4 7 (1,2)  = = ◦ (1,3)◦ (4,4)◦ (2,5)◦ (5,6)◦ (3,7)

31. Main Proof Step 0: Young Tableaux Step 1: Decomposing G Step 2: Decomposing C Step 3: Block diagonalizing R Step 4: Diagonalizing X i Wrap-up 1 3 5 6 2 4 7 1 1 1 1 1 1 G = { (1,2) } × X 3 × X 4 … × X n

32. Decomposing C Regular Representation Recall: C is defined C ,  = Definition: R(  ) is defined R(  ) ,  = Thus: C = ∑  ∈ G R(  ) 1 0 if  -1 º  ∈ G otherwise 1 0 if  -1 º  =  otherwise 111 111 111 111 111 111 1 1 1 1 1 1

33. Main Proof Step 0: Young Tableaux Step 1: Decomposing G Step 2: Decomposing C Step 3: Block diagonalizing R Step 4: Diagonalizing X i Wrap-up 1 3 5 6 2 4 7 1 1 1 1 1 1 G = { (1,2) } × X 3 × X 4 … × X n C = ∑  ∈ G R(  )

34. Decomposing R “Fourier Transform” 1 1 1 1 1 1 1 1 1 R()R()BR(  )B -1 Irreducible Representations Young Tableaux  change-of-basis matrix B block-diagonalizing R(  ) ′ ′′

35. Main Proof Step 0: Young Tableaux Step 1: Decomposing G Step 2: Decomposing C Step 3: Block diagonalizing R Step 4: Diagonalizing X i Wrap-up 1 3 5 6 2 4 7 1 1 1 1 1 1 G = { (1,2) } × X 3 × X 4 … × X n C = ∑  ∈ G R(  )

36. Let X i = { (j,i) : 1  j  i } Let Y (X i ) = ∑  ∈ X i BR(  )B -1, restricted to irreducible block 1 1 1 1 1 1 R()R() BR(  )B -1 Young Tableaux ′ ′′ Diagonalizing X i Jucys-Murphy Elements

37. Let X i = { (j,i) : 1  j  i } Let Y (X i ) = ∑  ∈ X i BR(  )B -1, restricted to irreducible block 1 11 1 11 11 1 11 1 1 1 1 1 1 1 ∑∈Xi R()∑∈Xi R() ∑  ∈ X i BR(  )B -1 Young Tableaux ′ ′′ Y (X i ) Diagonalizing X i Jucys-Murphy Elements

38. Let X i = { (j,i) : 1  j  i } Let Y (X i ) = ∑  ∈ X i BR(  )B -1, restricted to irreducible block Fact: Y (X i ) is diagonal (and entries known) 1 11 1 11 11 1 11 1 1 1 1 1 1 1 ∑∈Xi R()∑∈Xi R() ∑  ∈ X i BR(  )B -1 Young Tableaux ′ ′′ Y (X i )

39. Diagonalizing X i Jucys-Murphy Elements Let X i = { (j,i) : 1  j  i } Let Y (X i ) = ∑  ∈ X i BR(  )B -1, restricted to irreducible block Fact: Y (X i ) is diagonal (and entries known) Y (X 4 ) = Young Tableau SYT t1t1 t2t2 t3t3 t2t2 t3t3 t1t1 t2t2 t3t3 t1t1 Let X i = { (j,i) : 1  j  i } Let Y (X i ) = ∑  ∈ X i BR(  )B -1, restricted to irreducible block Fact: Y (X i ) is diagonal, and entry Y (X i ) t j,t j is c-r+1, where i is in row r and col c of t j. Content Value

40. Diagonalizing X i Jucys-Murphy Elements Let X i = { (j,i) : 1  j  i } Let Y (X i ) = ∑  ∈ X i BR(  )B, restricted to irreducible block Fact: Y (X i ) is diagonal (and entries known) 0 3 3 Y (X 4 ) = Young Tableau SYT t1t1 t2t2 t3t3 t2t2 t3t3 t1t1 t2t2 t3t3 t1t1 Let X i = { (j,i) : 1  j  i } Let Y (X i ) = ∑  ∈ X i BR(  )B -1, restricted to irreducible block Fact: Y (X i ) is diagonal, and entry Y (X i ) t j,t j is c-r+1, where i is in row r and col c of t j. Content Value

41. Main Proof Step 0: Young Tableaux Step 1: Decomposing G Step 2: Decomposing C Step 3: Block diagonalizing R Step 4: Diagonalizing X i Wrap-up 1 3 5 6 2 4 7 1 1 1 1 1 1 G = { (1,2) } × X 3 × X 4 … × X n C = ∑  ∈ G R(  )

42. Wrap-up Diagonalizing C Y ({ (1,2) }) ∙ Y (X 3 ) … Y (X n ) is diagonal  Y ({ (1,2) } × X 3 × … × X n ) is diagonal  Y ( G ) is diagonal  ∑  ∈ G BR(  )B -1 is diagonal  BCB -1 is diagonal (homomorphism) (Step 1) (Step 3) (Step 2) (Step 4)

43. Y ( G ) t j,t j  0  Y (X i ) t j,t j  0  i  3  If i  3 is in row c and col r of t j, then c-r+1  0  rank C = # SYT with 3  1 (and 2 below 1,...) Wrap-up What are eigenvalues of C? 1234 012 -2 01 c-r+1 10 0 0 0 Content Value Content Values 1 0 0 0 2 SYT t j No i  3 can go here  3  1 Y ( G )=Y ({ (1,2) } × X 3 × … × X n )

44. Main Proof Step 0: Young Tableaux Step 1: Decomposing G Step 2: Decomposing C Step 3: Block diagonalizing R Step 4: Diagonalizing X i Wrap-up 1 3 5 6 2 4 7 1 1 1 1 1 1 G = { (1,2) } × X 3 × X 4 … × X n C = ∑  ∈ G R(  ) QED