The proof theoretic strength of the Steinitz exchange theorem - EACA 2006

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The proof theoretic strength of the Steinitz exchange theorem - EACA 2006

  1. 1. The proof theoretic strength of theSteinitz Exchange TheoremMichael SoltysMcMaster University, CanadaFull version of this paper in Discrete Applied Mathematics.1
  2. 2. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysA propositional proof system (PPS) is just a PolyTimefunction f : Σ∗−→ Σ∗, where range(f) = TAUTOLOGIES.x ∈ Σ∗is a proof of τ iff f(x) = τ.A PPS f is poly bounded iff there exists a polynomial p suchthat for all τ, ∃x ∈ Σ∗, such that |x| ≤ p(|τ|) and f(x) = τ.A poly bounded PPS exists ⇐⇒ NP = co-NP.An automatizable, poly bounded, PPS exists ⇐⇒ P = NP.Context: complexity theory, automated reasoning, “reversemathematics”, . . .2/25
  3. 3. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael Soltys• Open Question: is there a poly bounded PPS?• Program: show that stronger and stronger PPS are not polybounded.• Next Step in the Program: show a separation between Fregeand Extended Frege.Frege: lines are boolean formulas.Extended Frege: lines are boolean circuits.(Allow abbreviations in proofs: a ≡ φ)3/25
  4. 4. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysCandidates for the separation:universal principles of matrix algebra,for example:• AB = I → BA = I• pA(A) = 0, pA is the char poly of A, Cayley-Hamilton Thm• det(AB) = det(A) det(B)4/25
  5. 5. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysStart with the following aim:Show “NC2concepts” prove the Cayley-Hamilton Thmnegandor5/25
  6. 6. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysWe encode a set of vectors {v1, v2, . . . , vn} as a matrixT = [v1v2 . . . vn].Steinitz Exchange Theorem (SET): if T is total, and E islinearly independent, then there exists F ⊆ T, such that |F| = |E|,and (T − F) ∪ E is total.∃X, TX = I ∧ (∀Y = 0, EY = 0) → ∃F ⊆ T, |F| = |E| ∧ ∃X, (T − F ∪ E)X = ISo SET is a ΠB2 formulas of QLA.6/25
  7. 7. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysGiven T, E, the matrix F can be computed in NC2.Suppose E = [e1e2] and T = [t1t2t3t4], then consider[e1e2] [e1e2t1] [e1e2t1t2] [e1e2t1t2t3] [e1e2t1t2t3t4]Independently for every i = 0, 1, 2, 3, ifRank([e1e2t1 . . . ti]) = Rank([e1e2t1 . . . ti+1])then put ti+1 in F.Rank can be computed in NC2with Mulmuley’s algorithm.7/25
  8. 8. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysMulmuley’s AlgorithmaM is n × n matrix, pM (x) its char. poly. (Berkowitz’s alg.)Geometric Rank : usual rank.Algebraic Rank : n − (highest power of x that divides pM (x)).Claim: RankG(M) = RankG(M2) ⇒ RankG(M) = RankA(M).Given M, let M be 0 MtM 0Clearly, RankG(M) = 12 RankG(M ).M = M · diag(1, y, y2, . . . , y2n−1); RankG(M ) = RankG((M )2).Here y is an indeterminate, and M is a matrix over the field F(y).aParallel Linear Algebra, by Joachim von zur Gathen, chapter in Synthesisof Parallel Algorithms.8/25
  9. 9. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysMulmuley’s AlgorithmaM is n × n matrix, pM (x) its char. poly. (Berkowitz’s alg.)Geometric Rank : usual rank.Algebraic Rank : n − (highest power of x that divides pM (x)).Claim: RankG(M) = RankG(M2) ⇒ RankG(M) = RankA(M).Given M, let M be 0 MtM 0Clearly, RankG(M) = 12 RankG(M ).M = M · diag(1, y, y2, . . . , y2n−1); RankG(M ) = RankG((M )2).Here y is an indeterminate, and M is a matrix over the field F(y).aParallel Linear Algebra, by Joachim von zur Gathen, chapter in Synthesisof Parallel Algorithms.9/25
  10. 10. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysMulmuley’s AlgorithmaM is n × n matrix, pM (x) its char. poly. (Berkowitz’s alg.)Geometric Rank : usual rank.Algebraic Rank : n − (highest power of x that divides pM (x)).Claim: RankG(M) = RankG(M2) ⇒ RankG(M) = RankA(M).Given M, let M be 0 MtM 0Clearly, RankG(M) = 12 RankG(M ).M = M · diag(1, y, y2, . . . , y2n−1); RankG(M ) = RankG((M )2).Here y is an indeterminate, and M is a matrix over the field F(y).aParallel Linear Algebra, by Joachim von zur Gathen, chapter in Synthesisof Parallel Algorithms.10/25
  11. 11. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysMulmuley’s AlgorithmaM is n × n matrix, pM (x) its char. poly. (Berkowitz’s alg.)Geometric Rank : usual rank.Algebraic Rank : n − (highest power of x that divides pM (x)).Claim: RankG(M) = RankG(M2) ⇒ RankG(M) = RankA(M).Given M, let M be 0 MtM 0Clearly, RankG(M) = 12 RankG(M ).M = M · diag(1, y, y2, . . . , y2n−1); RankG(M ) = RankG((M )2).y is an indeterminate, and M is a matrix over the field F(y).aParallel Linear Algebra, by Joachim von zur Gathen, chapter in Synthesisof Parallel Algorithms.11/25
  12. 12. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysSET can be shown with PolyTime concepts.Can it be shown with NC2concepts?12/25
  13. 13. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysSET proves in QLA the following principles1. (∃B = 0)[AB = I ∨ AB = 0],2. The columns of an n × (n + 1) matrix are linearly dependent,3. Every matrix has an annihilating polynomial,4. AB = I ⊃ BA = I,5. Existence of An, and the Cayley-Hamilton Thm.13/25
  14. 14. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysQLA can prove the existence of powers of a matrix from SET.Let POW(A, n) be the formula:∃ X0X1 . . . Xn (∀i ≤ n)[X0 = I ∧ (i < n ⊃ Xi+1 = Xi ∗ A)]Show QLA (∃B = 0)[AB = I ∨ AB = 0] ⊃ POW(A, n).14/25
  15. 15. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysLet N be the n2× n2matrix consisting of n × n blocks which are allzero except for (n − 1) copies of A above the diagonal zero blocksa.Then Nn= 0, and (I − N)−1= I + N + N2+ . . . + Nn−1=I A A2. . . An−10 I A . . . An−2.........0 0 0 . . . I.Set C = I − N.Show that if CB = 0, then B = 0, using induction on the rows ofB, starting with the bottom row.Using (∃B = 0)[CB = I ∨ CB = 0], conclude that there is a B suchthat CB = I.aA taxonomy of problems with fast parallel algorithms, by Stephen Cook.15/25
  16. 16. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysLet N be the n2× n2matrix consisting of n × n blocks which are allzero except for (n − 1) copies of A above the diagonal zero blocksa.Then Nn= 0, and (I − N)−1= I + N + N2+ . . . + Nn−1=I A A2. . . An−10 I A . . . An−2.........0 0 0 . . . I.Set C = I − N.Show that if CB = 0, then B = 0, using induction on the rows ofB, starting with the bottom row.Using (∃B = 0)[CB = I ∨ CB = 0], conclude that there is a B suchthat CB = I.aA taxonomy of problems with fast parallel algorithms, by Stephen Cook.16/25
  17. 17. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysLet N be the n2× n2matrix consisting of n × n blocks which are allzero except for (n − 1) copies of A above the diagonal zero blocksa.Then Nn= 0, and (I − N)−1= I + N + N2+ . . . + Nn−1=I A A2. . . An−10 I A . . . An−2.........0 0 0 . . . I.Set C = I − N.Show that if CB = 0, then B = 0, using induction on the rows ofB, starting with the bottom row.Using (∃B = 0)[CB = I ∨ CB = 0], conclude that there is a B suchthat CB = I.aA taxonomy of problems with fast parallel algorithms, by Stephen Cook.17/25
  18. 18. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysLet N be the n2× n2matrix consisting of n × n blocks which are allzero except for (n − 1) copies of A above the diagonal zero blocksa.Then Nn= 0, and (I − N)−1= I + N + N2+ . . . + Nn−1=I A A2. . . An−10 I A . . . An−2.........0 0 0 . . . I.Set C = I − N.Show that if CB = 0, then B = 0, using induction on the rows ofB, starting with the bottom row.Using (∃B = 0)[CB = I ∨ CB = 0], conclude that there is a B suchthat CB = I. Finally, show that B = I + N + N2+ · · · + Nn−1.aA taxonomy of problems with fast parallel algorithms, by Stephen Cook.18/25
  19. 19. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysStrong Linear Independence (SLI)if {v1, . . . , vm} are n × 1, non-zero, linearly dependent vectors, thenthere exists a 1 ≤ k < m such that{lin. indep.v1, . . . , vk, vk+1lin. dep., vk+2, . . . , vm}19/25
  20. 20. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysCsanky’s algorithm (NC2) for computing the characteristic poly. ofa matrix uses Newton’s symmetric polynomials:s0 = 1,sk =1kki=1(−1)i−1sk−itr(Ai)pA(x) := s0xn− s1xn−1+ s2xn−2− · · · ± snx0.20/25
  21. 21. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysTheorem: QLA proves the Cayley-Hamilton Thm. from SETand SLI.The 12 steps proof :(1) pA is the characteristic polynomial of the matrix A ascomputed by Csanky’s algorithm.(2) Let W = {ei, Aei, . . . , Anei}.(3) By SET, W must be linearly dependent.(4) By SLI there exists a k ≤ n such thatW0 = {ei, Aei, . . . , Ak−1ei} is linearly independent and k is thelargest such index.21/25
  22. 22. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael Soltys(5) Akei can be written as a linear combination of the vectors inW0.Let c1, . . . , ck be the coefficients of this linear combination, so thatif g(x) = xk+ c1xk−1+ · · · + ck, then g(A)ei = 0.(6) Let Ag be the k × k companion matrix of g,0 0 0 . . . 0 −ck1 0 0 . . . 0 −ck−10 1 0 . . . 0 −ck−2.........0 0 0 . . . 1 −c122/25
  23. 23. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael Soltys(7) LAP proves pAg = g, and so LAP proves (pAg (A))ei = 0.(8) Extend W0 to B = W0 ∪ {ej1 , . . . , ejn−k}.Existence of B follows from SET:let T = B0 = the standard basis,let E = W0, which is linearly independent,let F = B0 − {ej1 , . . . , ejn−k},so B = (T − F) ∪ E.A ∼Ag E10 E223/25
  24. 24. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael Soltys(9) LAP proves that if C1 ∼ C2 then pC1 (x) = pC2 (x)(tr(A) = tr(PAP−1), since tr(AB) = tr(BA)).(10) LAP proves that ifC =C1 ∗0 C2then pC(x) = pC1 (x) · pC2 (x).(11) ∴ pA(A)ei = (pAg (A) · pE(A))ei = pE(A) · (pAg (A)ei) = 0.(12) This is true for all ei in the standard basis, and so pA(A) = 0.24/25
  25. 25. EACA06 The proof theoretic strength of the Steinitz Exchange TheoremMichael SoltysQuestion:Can algorithms be proven correct within the complexity classesthat they run in ?25/25

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