Csr2011 june14 14_00_agrawal

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Csr2011 june14 14_00_agrawal

  1. 1. The Arithmetic Complexity of Euler Function Manindra Agrawal IIT Kanpur Computer Science Symposium in Russia, June 2011Manindra Agrawal (IITK) CSR 2011 1 / 33
  2. 2. Outline1 Euler Function and Permanent Polynomial2 Computing Euler Function3 Proof of First Theorem4 Proof of Second Theorem5 Black-box Derandomization of Identity Testing6 Open Questions and a Conjecture Manindra Agrawal (IITK) CSR 2011 2 / 33
  3. 3. Euler Function E (x) = (1 − x k ) k>0Defined by Leonhard Euler. Manindra Agrawal (IITK) CSR 2011 3 / 33
  4. 4. Relation to Partition NumbersLet pm be the number of partitions of m. Then 1 = pm x m . E (x) m≥0Proof. Note that 1 1 = k = ( x kt ). E (x) k>0 (1 − x ) k>0 t≥0 Manindra Agrawal (IITK) CSR 2011 4 / 33
  5. 5. Euler Identity ∞ 2 −m)/2 E (x) = (−1)m x (3m . m=−∞Proof. Set up an involution between terms of same degree and oppositesigns. Only a few survive. Manindra Agrawal (IITK) CSR 2011 5 / 33
  6. 6. Shape Over Complex Plane Undefined outside unit disk. Zero at unit circle. Bounded inside the unit disk: Red represents value 4 Black represents value 0Image created by Linas Vepstas (linas@linas.org) and released under theGnu Free Documentation License (GFDL). Borrowed fromhttp://en.wikipedia.org/wiki/File:Q-euler.jpeg. Manindra Agrawal (IITK) CSR 2011 6 / 33
  7. 7. Capturing SymmetriesDedekind Eta Function πiz η(z) = e 12 E (e 2πiz ).η(z) is defined on the upper half of the complex plane and satisfies manyinteresting properties: πi η(z + 1) = e 12 η(z). 1 √ η(− z ) = −izη(z).Proof. First part is trivial. Second part requires non-trivial complexanalysis. Manindra Agrawal (IITK) CSR 2011 7 / 33
  8. 8. Permanent Polynomial For any n > 0, let X = [xi,j ] be a n × n matrix with variable elements. Then permanent polynomial of degree n is the permanent of X : n per n (¯) = x xi,σ(i) . σ∈Sn i=1 It is believed to be hard to compute. Manindra Agrawal (IITK) CSR 2011 8 / 33
  9. 9. Permanent Polynomial For any n > 0, let X = [xi,j ] be a n × n matrix with variable elements. Then permanent polynomial of degree n is the permanent of X : n per n (¯) = x xi,σ(i) . σ∈Sn i=1 It is believed to be hard to compute. Manindra Agrawal (IITK) CSR 2011 8 / 33
  10. 10. Permanent Polynomial For any n > 0, let X = [xi,j ] be a n × n matrix with variable elements. Then permanent polynomial of degree n is the permanent of X : n per n (¯) = x xi,σ(i) . σ∈Sn i=1 It is believed to be hard to compute. Manindra Agrawal (IITK) CSR 2011 8 / 33
  11. 11. Outline1 Euler Function and Permanent Polynomial2 Computing Euler Function3 Proof of First Theorem4 Proof of Second Theorem5 Black-box Derandomization of Identity Testing6 Open Questions and a Conjecture Manindra Agrawal (IITK) CSR 2011 9 / 33
  12. 12. Two Families of Polynomials Let n 2 −k)/2 EΣ,n (x) = (−1)k x (3k k=−n and n EΠ,n = (1 − x k ). k=1 We have: E (x) = limn→∞ EΣ,n (x) = limn→∞ EΠ,n (x). EΣ,n (x) is a polynomial of degree 1 (3n2 + n) and EΠ,n (x) is a 2 polynomial of degree 1 (n2 + n). 2 A circuit family computing EΣ,n (x) or EΠ,n (x) can be viewed as computing E (x). We will consider arithmetic circuit families for computing EΣ,n (x) and EΠ,n (x). Manindra Agrawal (IITK) CSR 2011 10 / 33
  13. 13. Two Families of Polynomials Let n 2 −k)/2 EΣ,n (x) = (−1)k x (3k k=−n and n EΠ,n = (1 − x k ). k=1 We have: E (x) = limn→∞ EΣ,n (x) = limn→∞ EΠ,n (x). EΣ,n (x) is a polynomial of degree 1 (3n2 + n) and EΠ,n (x) is a 2 polynomial of degree 1 (n2 + n). 2 A circuit family computing EΣ,n (x) or EΠ,n (x) can be viewed as computing E (x). We will consider arithmetic circuit families for computing EΣ,n (x) and EΠ,n (x). Manindra Agrawal (IITK) CSR 2011 10 / 33
  14. 14. Two Families of Polynomials Let n 2 −k)/2 EΣ,n (x) = (−1)k x (3k k=−n and n EΠ,n = (1 − x k ). k=1 We have: E (x) = limn→∞ EΣ,n (x) = limn→∞ EΠ,n (x). EΣ,n (x) is a polynomial of degree 1 (3n2 + n) and EΠ,n (x) is a 2 polynomial of degree 1 (n2 + n). 2 A circuit family computing EΣ,n (x) or EΠ,n (x) can be viewed as computing E (x). We will consider arithmetic circuit families for computing EΣ,n (x) and EΠ,n (x). Manindra Agrawal (IITK) CSR 2011 10 / 33
  15. 15. Arithmetic Circuits for UnivariatePolynomials A circuit computing polynomial P(x) over field F takes as input x and −1; and outputs P(x). It is allowed to use addition and multiplication gates of arbitrary fanin over F . Size of a circuit is the number of wires in it. A depth two circuit family of size O(n2 ) can compute both EΣ,n (x) and EΠ,n (x) over any field as they are polynomials of degree O(n2 ). A depth three circuit family of size O(n) can compute EΠ,n (x) over any field: follows from definition. Manindra Agrawal (IITK) CSR 2011 11 / 33
  16. 16. Arithmetic Circuits for UnivariatePolynomials A circuit computing polynomial P(x) over field F takes as input x and −1; and outputs P(x). It is allowed to use addition and multiplication gates of arbitrary fanin over F . Size of a circuit is the number of wires in it. A depth two circuit family of size O(n2 ) can compute both EΣ,n (x) and EΠ,n (x) over any field as they are polynomials of degree O(n2 ). A depth three circuit family of size O(n) can compute EΠ,n (x) over any field: follows from definition. Manindra Agrawal (IITK) CSR 2011 11 / 33
  17. 17. Arithmetic Circuits for UnivariatePolynomials A circuit computing polynomial P(x) over field F takes as input x and −1; and outputs P(x). It is allowed to use addition and multiplication gates of arbitrary fanin over F . Size of a circuit is the number of wires in it. A depth two circuit family of size O(n2 ) can compute both EΣ,n (x) and EΠ,n (x) over any field as they are polynomials of degree O(n2 ). A depth three circuit family of size O(n) can compute EΠ,n (x) over any field: follows from definition. Manindra Agrawal (IITK) CSR 2011 11 / 33
  18. 18. Circuits for E (x) Can a higher depth circuit do significantly better? For some other polynomials, we can do substantially better. For example, log n−1 n−1 j (1 + x 2 ) = xi j=0 i=0 can be computed by a depth three circuit of size O(log n). However, it is not clear how to compute E (x) with no(1) sized circuits. Manindra Agrawal (IITK) CSR 2011 12 / 33
  19. 19. Circuits for E (x) Can a higher depth circuit do significantly better? For some other polynomials, we can do substantially better. For example, log n−1 n−1 j (1 + x 2 ) = xi j=0 i=0 can be computed by a depth three circuit of size O(log n). However, it is not clear how to compute E (x) with no(1) sized circuits. Manindra Agrawal (IITK) CSR 2011 12 / 33
  20. 20. Circuits for E (x) Can a higher depth circuit do significantly better? For some other polynomials, we can do substantially better. For example, log n−1 n−1 j (1 + x 2 ) = xi j=0 i=0 can be computed by a depth three circuit of size O(log n). However, it is not clear how to compute E (x) with no(1) sized circuits. Manindra Agrawal (IITK) CSR 2011 12 / 33
  21. 21. The Main TheoremsTheorem (First Theorem)Suppose every circuit family computing EΣ,n (x) over F , char(F ) > 2, hassize s(nΩ(1) ) for some s(m) ≥ (log m)2 . Then permanent polynomialfamily requires arithmetic circuits of size s(2Ω(n) ) over F .Theorem (Second Theorem)Suppose every circuit family computing EΠ,n (x) over F , char(F ) > 2, has O(1)size s(2Ω(s(n )) ) for some s(m) ≥ (log m)2 . Then permanent polynomialfamily requires arithmetic circuits of size s(2Ω(n) ) over Z.A weaker version of second theorem was recently shown by Pascal Koiran. Manindra Agrawal (IITK) CSR 2011 13 / 33
  22. 22. The Main TheoremsTheorem (First Theorem)Suppose every circuit family computing EΣ,n (x) over F , char(F ) > 2, hassize s(nΩ(1) ) for some s(m) ≥ (log m)2 . Then permanent polynomialfamily requires arithmetic circuits of size s(2Ω(n) ) over F .Theorem (Second Theorem)Suppose every circuit family computing EΠ,n (x) over F , char(F ) > 2, has O(1)size s(2Ω(s(n )) ) for some s(m) ≥ (log m)2 . Then permanent polynomialfamily requires arithmetic circuits of size s(2Ω(n) ) over Z.A weaker version of second theorem was recently shown by Pascal Koiran. Manindra Agrawal (IITK) CSR 2011 13 / 33
  23. 23. Outline1 Euler Function and Permanent Polynomial2 Computing Euler Function3 Proof of First Theorem4 Proof of Second Theorem5 Black-box Derandomization of Identity Testing6 Open Questions and a Conjecture Manindra Agrawal (IITK) CSR 2011 14 / 33
  24. 24. Multilinear Version of EΣ,n (x) (3n2 +n)/2 Let EΣ,n (x) = t=0 ct x t . Define (3n2 +n)/2 u t[j] Mn (z1 , z2 , . . . , zu ) = ct zj , t=0 j=1 where u = log(3n2 + n) − 1, t[j] is jth bit of t, and ct ∈ {−1, 0, 1} such that 2 u−1 EΣ,n (x) = Mn (x, x 2 , x 2 , . . . , x 2 ). The coefficient ct is computable in polynomial time given t: check if t = 1 (3m2 ± m) for some m; if it is, then ct = ±1, else ct = 0. 2 Using Valiant’s result on hardness of permanent, we get that 2c log n Mn (z1 , z2 , . . . , zu ) can be expressed as permanent of a matrix of size O(log n) for a suitable choice of constant c > 0. Manindra Agrawal (IITK) CSR 2011 15 / 33
  25. 25. Multilinear Version of EΣ,n (x) (3n2 +n)/2 Let EΣ,n (x) = t=0 ct x t . Define (3n2 +n)/2 u t[j] Mn (z1 , z2 , . . . , zu ) = ct zj , t=0 j=1 where u = log(3n2 + n) − 1, t[j] is jth bit of t, and ct ∈ {−1, 0, 1} such that 2 u−1 EΣ,n (x) = Mn (x, x 2 , x 2 , . . . , x 2 ). The coefficient ct is computable in polynomial time given t: check if t = 1 (3m2 ± m) for some m; if it is, then ct = ±1, else ct = 0. 2 Using Valiant’s result on hardness of permanent, we get that 2c log n Mn (z1 , z2 , . . . , zu ) can be expressed as permanent of a matrix of size O(log n) for a suitable choice of constant c > 0. Manindra Agrawal (IITK) CSR 2011 15 / 33
  26. 26. Multilinear Version of EΣ,n (x) (3n2 +n)/2 Let EΣ,n (x) = t=0 ct x t . Define (3n2 +n)/2 u t[j] Mn (z1 , z2 , . . . , zu ) = ct zj , t=0 j=1 where u = log(3n2 + n) − 1, t[j] is jth bit of t, and ct ∈ {−1, 0, 1} such that 2 u−1 EΣ,n (x) = Mn (x, x 2 , x 2 , . . . , x 2 ). The coefficient ct is computable in polynomial time given t: check if t = 1 (3m2 ± m) for some m; if it is, then ct = ±1, else ct = 0. 2 Using Valiant’s result on hardness of permanent, we get that 2c log n Mn (z1 , z2 , . . . , zu ) can be expressed as permanent of a matrix of size O(log n) for a suitable choice of constant c > 0. Manindra Agrawal (IITK) CSR 2011 15 / 33
  27. 27. Computing EΣ,n (x) Suppose permanent family can be computed by a circuit family of size s(2o(n) ) over F . Then, the polynomial family 2c log n Mn can be computed by a circuit family of size s(no(1) ). Let circuit C compute 2c log n Mn . j Modify C by replacing its input zj by x 2 . This adds O(log n) multiplication gates to C . Multiply the resulting circuit by 2−c log n in F (since char(F ) > 2, it always exists). The final circuit computes EΣ,n (x) and has size s(no(1) ), a contradiction. Manindra Agrawal (IITK) CSR 2011 16 / 33
  28. 28. Computing EΣ,n (x) Suppose permanent family can be computed by a circuit family of size s(2o(n) ) over F . Then, the polynomial family 2c log n Mn can be computed by a circuit family of size s(no(1) ). Let circuit C compute 2c log n Mn . j Modify C by replacing its input zj by x 2 . This adds O(log n) multiplication gates to C . Multiply the resulting circuit by 2−c log n in F (since char(F ) > 2, it always exists). The final circuit computes EΣ,n (x) and has size s(no(1) ), a contradiction. Manindra Agrawal (IITK) CSR 2011 16 / 33
  29. 29. Computing EΣ,n (x) Suppose permanent family can be computed by a circuit family of size s(2o(n) ) over F . Then, the polynomial family 2c log n Mn can be computed by a circuit family of size s(no(1) ). Let circuit C compute 2c log n Mn . j Modify C by replacing its input zj by x 2 . This adds O(log n) multiplication gates to C . Multiply the resulting circuit by 2−c log n in F (since char(F ) > 2, it always exists). The final circuit computes EΣ,n (x) and has size s(no(1) ), a contradiction. Manindra Agrawal (IITK) CSR 2011 16 / 33
  30. 30. Computing EΣ,n (x) Suppose permanent family can be computed by a circuit family of size s(2o(n) ) over F . Then, the polynomial family 2c log n Mn can be computed by a circuit family of size s(no(1) ). Let circuit C compute 2c log n Mn . j Modify C by replacing its input zj by x 2 . This adds O(log n) multiplication gates to C . Multiply the resulting circuit by 2−c log n in F (since char(F ) > 2, it always exists). The final circuit computes EΣ,n (x) and has size s(no(1) ), a contradiction. Manindra Agrawal (IITK) CSR 2011 16 / 33
  31. 31. Outline1 Euler Function and Permanent Polynomial2 Computing Euler Function3 Proof of First Theorem4 Proof of Second Theorem5 Black-box Derandomization of Identity Testing6 Open Questions and a Conjecture Manindra Agrawal (IITK) CSR 2011 17 / 33
  32. 32. Setup Assume that there is a circuit family of size s(2o(n) ) computing permanent polynomial over Z. Let P(x) = EΠ,n (x) for some n > 1. Degree of P(x) equals 1 n(n + 1) < n2 . 2 Let char(F ) = p. Since coefficients of P(x) are in Fp , we can assume F = Fp . ˆ ˆ Let F be an extension of F with n2 ≤ q = |F | = O(n2 ). Manindra Agrawal (IITK) CSR 2011 18 / 33
  33. 33. Setup Assume that there is a circuit family of size s(2o(n) ) computing permanent polynomial over Z. Let P(x) = EΠ,n (x) for some n > 1. Degree of P(x) equals 1 n(n + 1) < n2 . 2 Let char(F ) = p. Since coefficients of P(x) are in Fp , we can assume F = Fp . ˆ ˆ Let F be an extension of F with n2 ≤ q = |F | = O(n2 ). Manindra Agrawal (IITK) CSR 2011 18 / 33
  34. 34. Setup Assume that there is a circuit family of size s(2o(n) ) computing permanent polynomial over Z. Let P(x) = EΠ,n (x) for some n > 1. Degree of P(x) equals 1 n(n + 1) < n2 . 2 Let char(F ) = p. Since coefficients of P(x) are in Fp , we can assume F = Fp . ˆ ˆ Let F be an extension of F with n2 ≤ q = |F | = O(n2 ). Manindra Agrawal (IITK) CSR 2011 18 / 33
  35. 35. An Alternative Expression for P(x) By Langrange’s interpolation formula, we have: β∈F ,β=α (x ˆ − β) P(x) = P(α) · . β∈F ,β=α (α ˆ − β) ˆ α∈F Observe that (α − β) = β = −1, ˆ β∈F ,β=α ˆ β∈F ∗ and β∈F (x ˆ − β) (x − β) = x −α ˆ β∈F ,β=α q−1 xq − x = = αj−1 x q−j . x −α j=1 Manindra Agrawal (IITK) CSR 2011 19 / 33
  36. 36. An Alternative Expression for P(x) By Langrange’s interpolation formula, we have: β∈F ,β=α (x ˆ − β) P(x) = P(α) · . β∈F ,β=α (α ˆ − β) ˆ α∈F Observe that (α − β) = β = −1, ˆ β∈F ,β=α ˆ β∈F ∗ and β∈F (x ˆ − β) (x − β) = x −α ˆ β∈F ,β=α q−1 xq − x = = αj−1 x q−j . x −α j=1 Manindra Agrawal (IITK) CSR 2011 19 / 33
  37. 37. An Alternative Expression for P(x) Therefore, q−1 P(x) = − P(α) αj−1 x q−j ˆ α∈F j=1 q−1 = − ( P(α)αj−1 )x q−j . j=1 α∈F ˆ Now if we can compute P(α) efficiently, we can compute P(x) as permanent of a small size matrix. However, as n P(α) = (1 − αm ), m=1 we cannot compute it directly. Manindra Agrawal (IITK) CSR 2011 20 / 33
  38. 38. An Alternative Expression for P(x) Therefore, q−1 P(x) = − P(α) αj−1 x q−j ˆ α∈F j=1 q−1 = − ( P(α)αj−1 )x q−j . j=1 α∈F ˆ Now if we can compute P(α) efficiently, we can compute P(x) as permanent of a small size matrix. However, as n P(α) = (1 − αm ), m=1 we cannot compute it directly. Manindra Agrawal (IITK) CSR 2011 20 / 33
  39. 39. Computing P(α) ˆ Let g be a generator of F ∗ . Define NTM N as: on input α, guess t and m with 0 ≤ t < q and 1 ≤ m ≤ n. Check if g t = 1 − αm . If yes, output t on the part, else output 0. N is a polynomial time TM, and n #N(α) = tm , m=1 where g tm = 1 − αm . Hence, g #N(α) = P(α). Therefore, P(α) is computable in P#P . Manindra Agrawal (IITK) CSR 2011 21 / 33
  40. 40. Computing P(α) ˆ Let g be a generator of F ∗ . Define NTM N as: on input α, guess t and m with 0 ≤ t < q and 1 ≤ m ≤ n. Check if g t = 1 − αm . If yes, output t on the part, else output 0. N is a polynomial time TM, and n #N(α) = tm , m=1 where g tm = 1 − αm . Hence, g #N(α) = P(α). Therefore, P(α) is computable in P#P . Manindra Agrawal (IITK) CSR 2011 21 / 33
  41. 41. Computing P(α) ˆ Let g be a generator of F ∗ . Define NTM N as: on input α, guess t and m with 0 ≤ t < q and 1 ≤ m ≤ n. Check if g t = 1 − αm . If yes, output t on the part, else output 0. N is a polynomial time TM, and n #N(α) = tm , m=1 where g tm = 1 − αm . Hence, g #N(α) = P(α). Therefore, P(α) is computable in P#P . Manindra Agrawal (IITK) CSR 2011 21 / 33
  42. 42. Computing P(x) Since permanent is complete for #P, we get that P(α) can be computed by boolean circuits of size s(no(1) ). Therefore, P(x) can be computed by arithmetic circuits of size o(1) s(2o(s(n )) ) over F . A Contradiction. Manindra Agrawal (IITK) CSR 2011 22 / 33
  43. 43. Computing P(x) Since permanent is complete for #P, we get that P(α) can be computed by boolean circuits of size s(no(1) ). Therefore, P(x) can be computed by arithmetic circuits of size o(1) s(2o(s(n )) ) over F . A Contradiction. Manindra Agrawal (IITK) CSR 2011 22 / 33
  44. 44. Outline1 Euler Function and Permanent Polynomial2 Computing Euler Function3 Proof of First Theorem4 Proof of Second Theorem5 Black-box Derandomization of Identity Testing6 Open Questions and a Conjecture Manindra Agrawal (IITK) CSR 2011 23 / 33
  45. 45. Polynomial Identity Testing ProblemPIT over FGiven an arithmetic circuit over field F , determine if the polynomialcomputed by the circuit is identically zero. Admits a number of randomized polynomial time algorithms but no deterministic one is known. Has an interesting connection with hardness of computing EΠ,n (x). Manindra Agrawal (IITK) CSR 2011 24 / 33
  46. 46. Polynomial Identity Testing ProblemPIT over FGiven an arithmetic circuit over field F , determine if the polynomialcomputed by the circuit is identically zero. Admits a number of randomized polynomial time algorithms but no deterministic one is known. Has an interesting connection with hardness of computing EΠ,n (x). Manindra Agrawal (IITK) CSR 2011 24 / 33
  47. 47. Computing Multiples of EΠ,n (x)Let Pm (x) be a family of polynomials with Pm (x) of degree mO(1) . Thefamily is an n(m)-multiple of the family EΠ,n (x) if for every m, EΠ,n(m) (x)divides Pm (x). It is possible that EΠ,n (x) requires circuit of size nΩ(1) to compute. Does it also mean that every n(m)-multiple of EΠ,n (x) also requires circuits of size (n(m))Ω(1) to compute? If yes, we get a black-box derandomization of PIT. Manindra Agrawal (IITK) CSR 2011 25 / 33
  48. 48. Computing Multiples of EΠ,n (x)Let Pm (x) be a family of polynomials with Pm (x) of degree mO(1) . Thefamily is an n(m)-multiple of the family EΠ,n (x) if for every m, EΠ,n(m) (x)divides Pm (x). It is possible that EΠ,n (x) requires circuit of size nΩ(1) to compute. Does it also mean that every n(m)-multiple of EΠ,n (x) also requires circuits of size (n(m))Ω(1) to compute? If yes, we get a black-box derandomization of PIT. Manindra Agrawal (IITK) CSR 2011 25 / 33
  49. 49. Derandomization of PITTheoremIf every n(m)-multiple of EΠ,n (x), for every n(m) = mO(1) , requirescircuits of size (n(m))Ω(1) to compute over field F , then there exists apolynomial-time black-box derandomization of PIT over F . Manindra Agrawal (IITK) CSR 2011 26 / 33
  50. 50. Proof Assume that every n(m)-multiple of EΠ,n (x) requires circuits of size (n(m))δ for some δ > 0. Let C be an arithmetic circuit of size m computing a polynomial Q(y1 , . . . , ym ) over F . The degree of Q is bounded by 2m . We give a polynomial time algorithm for checking if Q is identically zero. Manindra Agrawal (IITK) CSR 2011 27 / 33
  51. 51. Proof Assume that every n(m)-multiple of EΠ,n (x) requires circuits of size (n(m))δ for some δ > 0. Let C be an arithmetic circuit of size m computing a polynomial Q(y1 , . . . , ym ) over F . The degree of Q is bounded by 2m . We give a polynomial time algorithm for checking if Q is identically zero. Manindra Agrawal (IITK) CSR 2011 27 / 33
  52. 52. Proof Assume that every n(m)-multiple of EΠ,n (x) requires circuits of size (n(m))δ for some δ > 0. Let C be an arithmetic circuit of size m computing a polynomial Q(y1 , . . . , ym ) over F . The degree of Q is bounded by 2m . We give a polynomial time algorithm for checking if Q is identically zero. Manindra Agrawal (IITK) CSR 2011 27 / 33
  53. 53. The Algorithm i−1 Let D = 2m + 1 and replace y i by x D as input to C . This requires an additional O(m2 ) wires at the bottom of C . ˆ Let the resulting circuit be C , and R(x) be the polynomial computed by it. 2 ˆ The size of C is O(m2 ) and the degree of R(x) is at most 2m . It is easy to see that R(x) is non-zero iff Q(y1 , . . . , ym ) is. Test if R(x) = 0 (mod (x − 1)k ) for 1 ≤ ≤ n = m3/δ and k is the largest number such that (x − 1)k divides EΠ,n (x). Output ZERO iff all the tests succeed. Manindra Agrawal (IITK) CSR 2011 28 / 33
  54. 54. The Algorithm i−1 Let D = 2m + 1 and replace y i by x D as input to C . This requires an additional O(m2 ) wires at the bottom of C . ˆ Let the resulting circuit be C , and R(x) be the polynomial computed by it. 2 ˆ The size of C is O(m2 ) and the degree of R(x) is at most 2m . It is easy to see that R(x) is non-zero iff Q(y1 , . . . , ym ) is. Test if R(x) = 0 (mod (x − 1)k ) for 1 ≤ ≤ n = m3/δ and k is the largest number such that (x − 1)k divides EΠ,n (x). Output ZERO iff all the tests succeed. Manindra Agrawal (IITK) CSR 2011 28 / 33
  55. 55. The Algorithm i−1 Let D = 2m + 1 and replace y i by x D as input to C . This requires an additional O(m2 ) wires at the bottom of C . ˆ Let the resulting circuit be C , and R(x) be the polynomial computed by it. 2 ˆ The size of C is O(m2 ) and the degree of R(x) is at most 2m . It is easy to see that R(x) is non-zero iff Q(y1 , . . . , ym ) is. Test if R(x) = 0 (mod (x − 1)k ) for 1 ≤ ≤ n = m3/δ and k is the largest number such that (x − 1)k divides EΠ,n (x). Output ZERO iff all the tests succeed. Manindra Agrawal (IITK) CSR 2011 28 / 33
  56. 56. The Algorithm i−1 Let D = 2m + 1 and replace y i by x D as input to C . This requires an additional O(m2 ) wires at the bottom of C . ˆ Let the resulting circuit be C , and R(x) be the polynomial computed by it. 2 ˆ The size of C is O(m2 ) and the degree of R(x) is at most 2m . It is easy to see that R(x) is non-zero iff Q(y1 , . . . , ym ) is. Test if R(x) = 0 (mod (x − 1)k ) for 1 ≤ ≤ n = m3/δ and k is the largest number such that (x − 1)k divides EΠ,n (x). Output ZERO iff all the tests succeed. Manindra Agrawal (IITK) CSR 2011 28 / 33
  57. 57. Correctness The algorithm is clearly deterministic, polynomial-time, and black-box. Observe that if all the tests succeed, it implies that EΠ,n (x) divides R(x). If R(x) is non-zero then, by our assumption on n-multiples of EΠ,n (x), R(x) requires a circuit of size nδ = m3 to compute. ˆ However, circuit C , of size O(m2 ), computes R(x). A contradiction. Manindra Agrawal (IITK) CSR 2011 29 / 33
  58. 58. Correctness The algorithm is clearly deterministic, polynomial-time, and black-box. Observe that if all the tests succeed, it implies that EΠ,n (x) divides R(x). If R(x) is non-zero then, by our assumption on n-multiples of EΠ,n (x), R(x) requires a circuit of size nδ = m3 to compute. ˆ However, circuit C , of size O(m2 ), computes R(x). A contradiction. Manindra Agrawal (IITK) CSR 2011 29 / 33
  59. 59. Correctness The algorithm is clearly deterministic, polynomial-time, and black-box. Observe that if all the tests succeed, it implies that EΠ,n (x) divides R(x). If R(x) is non-zero then, by our assumption on n-multiples of EΠ,n (x), R(x) requires a circuit of size nδ = m3 to compute. ˆ However, circuit C , of size O(m2 ), computes R(x). A contradiction. Manindra Agrawal (IITK) CSR 2011 29 / 33
  60. 60. Correctness The algorithm is clearly deterministic, polynomial-time, and black-box. Observe that if all the tests succeed, it implies that EΠ,n (x) divides R(x). If R(x) is non-zero then, by our assumption on n-multiples of EΠ,n (x), R(x) requires a circuit of size nδ = m3 to compute. ˆ However, circuit C , of size O(m2 ), computes R(x). A contradiction. Manindra Agrawal (IITK) CSR 2011 29 / 33
  61. 61. Outline1 Euler Function and Permanent Polynomial2 Computing Euler Function3 Proof of First Theorem4 Proof of Second Theorem5 Black-box Derandomization of Identity Testing6 Open Questions and a Conjecture Manindra Agrawal (IITK) CSR 2011 30 / 33
  62. 62. Open QuestionsSeveral questions remain open: 1 Is the polynomial EΠ,n (x) computable over Fp in Modp P? 2 Does EΠ,n (x) require circuits of size nΩ(1) ? 3 Does every n-multiple of EΠ,n (x) requires circuits of size nΩ(1) ? Manindra Agrawal (IITK) CSR 2011 31 / 33
  63. 63. Open QuestionsSeveral questions remain open: 1 Is the polynomial EΠ,n (x) computable over Fp in Modp P? 2 Does EΠ,n (x) require circuits of size nΩ(1) ? 3 Does every n-multiple of EΠ,n (x) requires circuits of size nΩ(1) ? Manindra Agrawal (IITK) CSR 2011 31 / 33
  64. 64. Open QuestionsSeveral questions remain open: 1 Is the polynomial EΠ,n (x) computable over Fp in Modp P? 2 Does EΠ,n (x) require circuits of size nΩ(1) ? 3 Does every n-multiple of EΠ,n (x) requires circuits of size nΩ(1) ? Manindra Agrawal (IITK) CSR 2011 31 / 33
  65. 65. A ConjectureConjectureAnswer to all three questions above is yes.If the conjecture is true then an exponential lower bound on permanentpolynomial follows. Manindra Agrawal (IITK) CSR 2011 32 / 33
  66. 66. A ConjectureConjectureAnswer to all three questions above is yes.If the conjecture is true then an exponential lower bound on permanentpolynomial follows. Manindra Agrawal (IITK) CSR 2011 32 / 33
  67. 67. Thoughts on the Conjecture The conjecture relates the size of an arithmetic circuit computing a polynomial to the number of distinct small roots of unity that the polynomial can have. It is similar in spirit to τ -conjecture of Shub-Smale that relates the size of an arithmetic circuit computing a polynomial to the number of integer roots the polynomial can have. Manindra Agrawal (IITK) CSR 2011 33 / 33
  68. 68. Thoughts on the Conjecture The conjecture relates the size of an arithmetic circuit computing a polynomial to the number of distinct small roots of unity that the polynomial can have. It is similar in spirit to τ -conjecture of Shub-Smale that relates the size of an arithmetic circuit computing a polynomial to the number of integer roots the polynomial can have. Manindra Agrawal (IITK) CSR 2011 33 / 33

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