A Quadratic Lower Bound for the Permanent and Determinant Problem over any Characteristic 6= 2
Jin-Yi Cai Xi Chen Dong Li Computer Sciences School of Mathematics School of Mathematics Department, University of Institute for Advanced Study Institute for Advanced Study Wisconsin, Madison U.S.A. U.S.A. and Radcliffe Institute [email protected] [email protected] Harvard University, U.S.A. [email protected]
ABSTRACT is also well-studied, especially in combinatorics [12]. For In Valiant’s theory of arithmetic complexity, the classes VP example, if A is a 0-1 matrix then per(A) counts the number and VNP are analogs of P and NP. A fundamental problem of perfect matchings in a bipartite graph with adjacency A concerning these classes is the Permanent and Determinant matrix . Problem: Given a field F of characteristic = 2, and an inte- These well-known functions took on important new mean- ger n, what is the minimum m such that the6 permanent of ings when viewed from the computational complexity per- spective. It is well known that the determinant can be com- an n n matrix X =(xij ) can be expressed as a determinant of an×m m matrix, where the entries of the determinant puted in polynomial time. In fact it can be computed in the × complexity class NC2. By contrast, Valiant [22, 21] showed matrix are affine linear functions of xij ’s, and the equal- ity is in F[X]. Mignon and Ressayre (2004) [11] proved a that computing the permanent is #P-complete. quadratic lower bound m = Ω(n2) for fields of characteristic In fact, Valiant [21] (see also [4, 5]) has developed a sub- 0. We extend the Mignon-Ressayre quadratic lower bound stantial theory. The complexity classes VPF and VNPF are to all fields of characteristic = 2. the analogs of P and NP in this theory of arithmetic com- 6 plexity, and det and per functions are the central objects in the two classes, respectively. It was shown that the com- Categories and Subject Descriptors plexity of computing the permanent characterizes the class F.2.1 [Analysis of Algorithms and Problem Complex- VNPF and the complexity of computing the determinant (al- ity]: Numerical Algorithms and Problems—Computations most) characterizes the class VPF. in finite fields, Computations on matrices, Computations on More precisely, a family of polynomials fn is in VPF if O(1) { } polynomials deg(fn)= n and there is a family of arithmetic circuits O(1) of size n computing fn . A family of polynomials gn { O}(1) { } General Terms is in VNPF if deg(gn) = n , and there exists a family of polynomials fn VPF such that Theory { } ∈ gn(x1, ..., xn)= fn+m(x1, ..., xn,y1, ..., ym),
Keywords y ,...,ym∈{0,1} 1 X O(1) Permanent, determinant, arithmetic complexity, finite field where m = n . We say that fn is a projection of gm { } { } if there are some α1, α2,...,αm F x1,...,xn , such ∈ ∪ { } 1. INTRODUCTION that fn(x1,...,xn)= gm(α1,...,αm). It isa p-projection if 2 O(1) Given a set of n indeterminates X =(xi,j )i,j=1,...,n over m = n . A projection is a particularly simple reduction. a field F, we can define It is a special case of an affine linear reduction, where each αi is an affine linear function of xi’s. Valiant proved that det(X)= sign(π) n x , and π∈Sn i=1 i,π(i) n For any field F, F. per(X)= x . Theorem 1 (Valiant). per VNP P π∈Sn i=1Q i,π(i) ∈ Moreover, for any F with char F = 2, any fn VNPF is 6 { } ∈ The determinant functionP is certainlyQ one of the most well- a p-projection of per. studied functions in mathematics. The permanent function It is also known that det is in VPF [23]. More exact char- acterizations of det were given in terms of polynomial-sized arithmetic branching programs [7, 19, 24]. Permission to make digital or hard copies of all or part of this work for personal or classroom use is granted without fee provided that copies are Theorem 2 (Valiant). Any polynomial fn is a pro- not made or distributed for profit or commercial advantage and that copies jection of detm of an m m matrix, where m is linear in × bear this notice and the full citation on the first page. To copy otherwise, to the formula size of fn. In particular, if fn has polyno- republish, to post on servers or to redistribute to lists, requires prior specific { } mial formula size, then fn is a p-projection of det. Also permission and/or a fee. { } STOC’08, if fn VPF, then fn is the projection of detm for some May 17–20, 2008, Victoria, British Columbia, Canada. { }O ∈(log n) Copyright 2008 ACM 978-1-60558-047-0/08/05 ...$5.00. m = n .
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2 n By Ryser’s formula [12], pern has formula size O(n 2 ). and the matrix H becomes 0. In this paper we overcome Thus by Valiant’s theorem it is the projection of detm, where this difficulty by considering another explicit construction 2 n m = O(n 2 ). Furthermore if we view Ryser’s formula as of matrix X0. on the truncated linear row sums directly (instead of on the We mention some other related results. In [9], Jerrum and variables), then Valiant’s theorem implies that Snir showed that any monotone arithmetic circuit family that computes permanent must have exponential size. For Theorem 3. For every n, there exists a collection A of 2 depth-three arithmetic circuits over fields of characteristic 0, affine linear functions Ak,l(X) over n variables, where 1 n ≤ Shpilka and Wigderson [17] proved that the permanent (and k,l m = O(2 ), such that per (X) = detm(A(X)). 2 ≤ n determinant) requires circuit size Ω(n ). For depth-three It is remarkable that this is the best general upper bound arithmetic circuits over finite characteristic, Grigoriev and known for this. Razborov [8] showed an exponential lower bound for both determinant and permanent. Raz [15] proved a lower bound O(log n) Definition 1. The determinantal complexity dc of fn is of n on the size of families of multilinear formulas the minimum integer m, such that there exist affine linear computing permanent and determinant. For syntactically functions Ak,l(X), where 1 k,l m, such that fn(X) = multilinear arithmetic circuits, Raz, Shpilka and Yehuday- ≤ ≤ 4/3 2 detm(A(X)). off [16] proved a Ω(n / log n) lower bound for an explicit multilinear function. A survey of some work on the Perma- dc The question addressed in this paper is about (pern). nent and Determinant Problem can be found in [1], where Valiant’s analog of P = NP will follow if one can show a it also discusses an algebraic geometry approach by Mulmu- dc 6 ω(log n) lower bound (pern)= n . ley and Sohoni [13] and connections to the pseudorandom In some sense this problem has a longer history. P´olya [14] generator used in the AKS proof for primality [2, 3]. was the first to ask a question on when one can express a This paper is organized as follows. In Section 2, we discuss permanent as a modified determinant. He noticed that the general approach by Mignon and Ressayre, and state our 2 a b a b result. In Section 3, we prove an Ω(n ) lower bound valid per = det , c d c− d for all characteristic = 2. In Section 4 we indicate how to improve the leading constant6 in Ω(n2) to match the Mignon- and asked if there are any similar equations, by affixing 1 to Ressayre bound. the variables, for n 3. This was answered in the negative± ≥ by Szeg¨o [18]. This line of inquiry culminated in 2. THE APPROACH AND THE THEOREM Theorem 4 (Marcus, Minc). If char F = 0 and n ≥ 3, then there are no homogeneous linear functions fk,ℓ in the 2.1 The Proof by Mignon and Ressayre indeterminates xi,j (1 i,j,k,ℓ n) such that per(xi,j )= Given an n n matrix X = (xi,j )i,j=1,2,...,n over a field ≤ ≤ × det(fk,ℓ). F, the determinant det(X) and the permanent per(X) are 2 dc both polynomials of degree n over n variables. Their partial In terms of (pern), this celebrated theorem is equivalent to dc(per ) n + 1, over fields of char F = 0 (note that if derivatives of all orders are defined formally. n H X Hes- the permanental≥ matrix is also n n, then clearly constant We use ( )=(Hij,kl)i,j,k,l=1,2,...,n to denote the sian X terms in affine linear equations do× not help, as seen by the matrix of per( ): homogeneous part.). ∂2per(X) dc F X The first non-trivial lower bound for (pern) is by von zur Hij,kl = [ ], for all 1 i,j,k,l n. ∂xi,j ∂xk,l ∈ ≤ ≤ dc Gathen [25], who showed that (pern) 8/7n. (This was proved for p-projections.) Von zur Gathen’s≥ result was then Similarly, we can define the Hessian matrix of det(X), and p improved independently by Babai and Seress (as reported denote it by Hdet(X). in [26]), by Cai [6], and by Meshulam [10]. Their results Now suppose that there exists a collection A of m2 affine were (ignoring lower order terms) dc(per ) √2n. linear functions, where A = Ak,l(x1,1,x1,2,...,xn,n), k,l : n { This rather weak lower bound stood as≥ the best bound 1 k,l m , such that in the polynomial ring F[X], ≤ ≤ } until 2004, when Mignon and Ressayre [11] proved that 2 dc(per ) n /2, over any field of char F = 0. Over a field of pern(X) = detm (Ak,l(X)) . (1) n ≥ 1≤k,l≤m char F = p = 2, the best bound is a very recent unpublished result by Valiant6 [20], which is Ω(n5/4) for projections. The first step in the proof by Mignon and Ressayre [11] is to transform A to a normal form. Consider a fixed ma- More important than the bound 8/7n, von zur Ga- n×n trix X0 F such that per(X0) = 0. We expand the then [25] introduced a method of taking derivatives and ∈ then comparing appropriate dimensions/ranks.p The follow- affine linear functions Ak,l(X) at X0, and write (Ak,l(X)) = √ (Lk,l(X X0))+ Y0 for some homogeneous linear functions up improvements to 2n all use this approach. − m×m Lk,l and some matrix Y0 F . It follows from (1) that The Mignon-Ressayre breakthrough [11] uses a new idea: ∈ Take second-order derivatives. det(Y0) = per(X0) = 0. Let C and D be two non-singular The key step in their proof [11] is to lower bound the rank matrices such that CY0D is a diagonal matrix of the second derivative matrix H of the permanent at a 0 0 certain matrix X0. However, their proof encounters a major , where s 492 of CY0D are both zero, we may multiply diagonal matrices Theorem 6. Let p> 2 be a prime, then diag(det(C)−1, 1,..., 1) and diag(det(D)−1, 1,..., 1) to the 1. If p = 23, then for any n > 2 that satisfies p (n + 1), left and right, so we may just assume det(C) = det(D) = 1. 6 there exists an (n + 1) (n + 1) matrix X0 over finite It follows that, by (multiplying matrices C and D to the left × field Fp, such that per(X0) 0 (mod p) and and right, and) renaming Lk,l and Y0, we may assume (1) ≡ takes the form rank(H(X0)) (n 2)(n 3); ≥ − − per(X) = det Lk,l (X X0) + Y0 , − 2. If p = 3, 5, then for any n> 1 that satisfies p (n + 2), 6 there exists an (n + 1) (n + 1) matrix X0 over finite where Y0 = diag(0, 1,..., 1). × field Fp, such that per(X0) 0 (mod p) and Now we can take second-order derivatives, and evaluate ≡ them at X0. By the chain rule, rank(H(X0)) (n 2)(n 3). ≥ − − T H(X0)= L Hdet(Y0) L , dc · · This implies the lower bound for (pern) over field Fp. (We 2 2 where L is an n m matrix over F. It follows immediately remark that a lower bound for Fp is also valid over Q.) that rank(H(X ))× rank(H (Y )). 0 det 0 Corollary 1. For any prime p = 2, dc(per ) (n It is relatively easy≤ to derive a O(m) upper bound for n 2)(n 3)/2 over a field of char F = p6 . ≥ − the rank of Hdet(Y0). Notice that when one takes a partial − derivative ∂/∂xij on the determinant (as well as on the per- To prove the theorem, we introduce, for any v Fp, and ∈ n manent), one simply gets the minor after striking out row i integer n 1, the following (n + 1) (n + 1) matrix Mv = 2 ≥ × and column j. Second order derivative ∂ /∂xij ∂xkl simply (Mi,j ): M(n+1),(n+1) = v, Mi,i = M(n+1),i = Mi,(n+1) = 1 strikes out rows i, k and columns j, l . By the form of for all i : 1 i n, and Mi,j = 0 otherwise. For example, Y { } { } H Y 3 ≤ ≤ 0, to get a non-zero value for an entry (ij, kl) in det( 0), M2 is given by it must be that 1 i, k and 1 j, l . In fact the only 1 0 0 1 ∈ { } ∈ { } non-zero entries are 0 1 0 1 M3 = . 2 0 0 1 1 (ij, kl) = (11,tt), (tt, 11), (1t,t1) or (t1, 1t), 1 1 1 2 for all t> 1. This immediately gives a 2m upper bound for n n We will prove the two cases of Theorem 6 using M1 and M2 rank(Hdet(Y0)). (If we did not assume s = m 1, then it − respectively. Given v and n, the following lemma essentially would have been even more difficult to get a non-zero entry n n defines the Hessian matrix H(Mv ) of Mv . in Hdet(Y0). If s = m 2, there could be at most O(1) − 2 2 many non-zero entries. If s 493 T (ti), where 1 t n and t = i, j. As a result, Hij Hij = Now we can write RR as an S S block matrix: (v + n 2)2 +≤ 2(n≤ 2). For the6 third case, the only non-zero· | | × | | A B B B entries− in H H −are (v + n 2) for indices (ji) and (ki), ij ik B A B · · · B and 1 for indices· (ti), where 1− t n and t = i, j, k, thus T · · · H H = 2(v + n 2)+(n ≤3).≤ 6 RR = B B A B , where ij ik . . . ·. · · . · − − ...... . . . . We also need the following lemma concerning the det of B B B A matrices of a specific form. · · · a b b b b c c c · · · · · · Lemma 3. Let A = (Ai,j )i,j=1,...,n be an n n matrix b a b b c b c c × · · · · · · over Fp satisfying Ai,i = α for all 1 i n and Ai,j = β A = b b a b , B = c c b c , otherwise. Then, ≤ ≤ . . . ·. · · . . . . ·. · · . ...... ...... . . . . . . . . A n−1 det( )= α +(n 1)β (α β) . b b b a c c c b − − · · · · · · are both T T matrices with a = 2, b = 8 and c = 2. 3. A WEAKER THEOREM Next we| apply| × | the| following operations− to RR− T : subtract In this section, we prove the following weaker version of the second last column from the last column of RRT (here Theorem 6. what we mean by “a column” is a whole block column of RRT ). Then subtract the third last column from the second Lemma 4. Let p> 2 be a prime, then for any sufficiently n last column . . . till subtract the first column from the second large n satisfying p (n + 1), we have per(M1 ) 0 (mod p) n | 2 ≡ column. We end up with and rank(H(M1 )) = Ω(n ). n A B A Proof. In the proof, we denote matrix M by M. Clearly 0 0 1 B A − B B A · · · per(M)= n + 1 0 (mod p), so we only need to prove the 0 ≡ B −0 A − B · · · 0 second part. . . −. ·. · · . . Let S be a maximal subset of integers i : 1 i < n/2 ...... { ≤ } with S 2 (mod p), and T be a maximal subset of j : B 0 0 B A | | ≡ { n/2 j n with T 2 (mod p). Both S and T are B 0 0 · · · A − B ≤ ≤ } | | ≡ | | | | Ω(n). · · · − We will show that there exists a sub-matrix R of H(M) Then we add the first row to the second row. Add the second with S T rows, such that, det(RRT ) is non-zero. As a row to the third row, etc. Finally, we get | | · | | result, we have A B A 0 0 0 − · · · rank H(M) rank(R) A + B 0 B A 0 0 ≥ A + 2B 0− 0 B A · · · 0 rank(RRT )= S T = Ω(n2), . . . −. ·. · · . . ...... ≥ | | · | | . . . . . and the lemma follows. A +( S 2)B 0 0 0 B A To get the matrix R, we choose the following subset of A +(|S|− 1)B 0 0 0 · · · −0 rows and columns of H(M): rows (ij), where i S and | |− · · · j T ; and columns (kl), where 1 k = l n. So∈R is an Clearly, all these operations do not change its determinant. ∈ 2 ≤ 6 ≤ By using Lemma 3, we have (here we use s and t to denote ( S T ) (n n) matrix. Let S = i1, i2,...,i and | | · | | × − { |S| } S 1 and T 1, respectively) T = j1, j2,...,j , then we can write R as { |T | } | |− | |− det(RRT ) = det(A + sB) (det(B A))s Hi1j1 ± · − t Hi1j2 = (a + sb + t (b + sc))(a + sb (b + sc)) ± − s . (b a + t (c b))(b a (c b))t . − − − − − H t t s R = i1j|T | , ( 16)( 4) (4)( 16) 0 (mod p), ≡ ± − − − 6≡ Hi2j1 T . since p> 2. Therefore, we have rank(RR )= S T , and . | | · | | . the lemma is proven. Hi j |S| |T | 2 where Hij is the (n n)-dimensional vector truncated from 4. PROOF OF THE MAIN THEOREM th − the (ij) row of H(M). In this section, we prove Theorem 6. As already men- n n Consider the inner products of arbitrary two rows of R. tioned in Section 2.2, we will use M1 and M2 to prove the By Lemma 2, we have for i S and j T , two cases, respectively. The idea behind the proof is similar ∈ ∈ 2 to the previous one. However, the sub-matrix R we pick 1. Hij Hij =(v +n 2) +2(n 2) 2 (mod p), since · − − ≡− this time is a square matrix with n2 n rows. By showing v = 1, n 1 (mod p); T − ≡− that the rank of RR is almost full, the theorem follows. ′ ′ 2. when j = j and j T , Hij Hij′ = 2(v + n 2) + (n 3) 6 8 0 (mod∈ p); · − Proof of Theorem 6. Let v = 1 in the first case and − ≡− 6≡ ′ ′ v = 2 in the second case. Note that in both cases, we have 3. when i = i and i S, Hij H ′ 8 0 (mod p); 6 ∈ · i j ≡− 6≡ n v (mod p). ′ ′ ′ ′ ≡− 4. when i = i, j = j, i S and j T , Hij Hi′j′ 2 Let S = (i, j) : 1 i = j n , then we use Rv to denote 6 6 ∈ ∈ · ≡ { ≤ 6 ≤ } n (mod p). the following sub-matrix of H(Mv ): Row (or column) (ij) 494 of H(Mv) is selected if and only if (i, j) S. Thus, Rv is Proof. To prove the lemma it suffices to show that, for 2 2 ∈ th T an (n n) (n n) matrix. Again, we write Rv as all 1 i 495 12 13 14 15 16 21 23 24 25 26 31 32 34 35 36 41 42 43 45 46 51 52 53 54 56 61 62 63 64 65 12 abbbb 01111 1 b 222 1 b 222 1 b 222 1 b 2 2 2 13 babbb 1 b 222 01111 12 b 22 12 b 22 12 b 2 2 14 bbabb 1 2 b 22 12 b 22 01111 122 b 2 122 b 2 15 bbbab 1 2 2 b 2 122 b 2 122 b 2 01111 1222 b 16 bbbba 1222 b 1222 b 1222 b 1222 b 01111 21 01111 abbbb b 1222 b 1222 b 1222 b 1222 23 1 b 2 2 2 babbb 10111 21 b 22 21 b 22 21 b 2 2 24 1 2 b 2 2 bbabb 2 1 b 22 10111 212 b 2 212 b 2 25 1 2 2 b 2 bbbab 2 1 2 b 2 212 b 2 10111 2122 b 26 1222 b bbbba 2122 b 2122 b 2122 b 10111 31 10111 b 1222 abbbb b 2122 b 2122 b 2122 32 b 1222 10111 babbb 2 b 122 2 b 122 2 b 1 2 2 34 2 1 b 22 21 b 2 2 bbabb 11011 221 b 2 221 b 2 35 2 1 2 b 2 212 b 2 bbbab 2 2 1 b 2 11011 2212 b 36 2122 b 2122 b bbbba 2212 b 2212 b 11011 41 11011 b 2122 b 2122 abbbb b 2212 b 2212 42 b 2122 11011 2 b 1 2 2 babbb 2 b 212 2 b 2 1 2 43 2 b 122 2 b 122 11011 bbabb 2 2 b 12 22 b 1 2 45 2 2 1 b 2 221 b 2 221 b 2 bbbab 11101 2221 b 46 2212 b 2212 b 2212 b bbbba 2221 b 11101 51 11101 b 2212 b 2212 b 2212 abbbb b 2221 52 b 2212 11101 2 b 212 2 b 2 1 2 babbb 2 b 2 2 1 53 2 b 212 2 b 212 11101 22 b 1 2 bbabb 2 2 b 2 1 54 2 2 b 12 22 b 12 22 b 12 11101 bbbab 2 2 2 b 1 56 2221 b 2221 b 2221 b 2221 b bbbba 11110 61 11110 b 2221 b 2221 b 2221 b 2221 abbbb 62 b 2221 11110 2 b 221 2 b 221 2 b 2 2 1 babbb 63 2 b 221 2 b 221 11110 22 b 21 22 b 2 1 bbabb 64 2 2 b 21 22 b 21 22 b 21 11110 222 b 1 bbbab 65 2 2 2 b 1 222 b 1 222 b 1 222 b 1 11110 bbbba T Figure 1: An example of matrix Rv Rv when n = 6 Lemma 6. For all 1 i < j n such that i + 1 < j and by P∗. It directly follows from Lemma 6 that P∗ is a lower j + 1 n, we have ≤ ≤ triangular block matrix, and the block matrices along the ≤ T T T T diagonal are: (Cj Cj−1)BCi+1 =(Cj Cj−1)BCi. − − T T T th Ci (B A)Ci (Ci Ci−1)BCi−1 , i = 2, ..., n 1. Proof. For 1 k n 1, we use Bk to denote the k − − − − ≤ ≤ − ′ T T row vector of B. We also use B to denote (Cj Cj−1)B, On the other hand, as implied by the proof of Lemma 6, the ′ th ′ − T T and Bk to denote the k row of B . It is not hard to check rank of (Ci Ci−1)BCi−1 is exactly 1, so ′ ′ ′ − that B = Bj B1, B = B1 Bj , and B = 0 for all j−1 − j − k T ∗ k = j 1, j. rank(Rv Rv ) rank(P ) 6 − ≥ On the other hand, all the entries of Bj B1 are equal n−1 th − T T T to 1 except the j entry which is equal to b 1. As we rank Ci (B A)Ci (Ci Ci−1)BCi−1 ′ ′ − ′ ≥ − − − assumed that i + 1 < j, we have B C = B C = B , and i=2 i+1 i X the lemma is proven. (n 2) rank(B A) 1 . ≥ − − − T We apply the following operations to RvRv : subtract the Finally, by Lemma 3, the determinant of the lower right T second last column from the last column of RvRv , then (n 2) (n 2) sub-matrix of B A is subtract the third last column from the second last column − × − − . . . till subtract the first column from the second column. Let (b a)+(n 3)(2 b) (b a) (2 b) n−3 P denote the upper right sub-matrix, after the operations, − − − − − − T ( 46)( 16)n−3 (mod p) when v = 1; of RvRv containing (n 1) (n 1) blocks (see Fig.2). − × − − − n−3 Next, we transform P as follows: Subtract the second last ≡ (( 60)( 16) (mod p) when v = 2. row from the last row, then subtract the third last row from − − n T the second last row . . . till subtract the first row from the As a result, we have rank(H(M1 )) rank(R1R1 ) (n ≥n ≥T − second row. We only need to focus on the lower right part 2)(n 3) when p = 23; and rank(H(M2 )) rank(R2R2 ) of P containing (n 2) (n 2) blocks, which we denote (n −2)(n 3) when6 p = 3, 5. ≥ ≥ − × − − − 6 496 T T T T T T T C (B A)C1 (C C )BC1 (C C )BC1 (C C )BC1 1 − 2 − 1 3 − 2 · · · n−1 − n−2 T T T T T C (B A)C2 (C C )BC2 (C C )BC2 ∗ 2 − 3 − 2 · · · n−1 − n−2 P = T T T C (B A)C3 (C C )BC3 ∗ ∗ 3 − · · · n−1 − n−2 ...... . . . . . T C (B A)Cn−1 ∗ ∗ ∗ ··· n−1 − T Figure 2: Matrix P transformed from Rv Rv A natural question is what makes this sequence of matrices [10] R. Meshulam. Two extremal matrix problems. Linear works for the proof. We can only offer our take on this. We Alg. and its Applications, 114/115:261–271, 1989. believe that probably most matrices X, where per(X) = 0, [11] T. Mignon and N. Ressayre. A quadratic bound for the will work, i.e., its Hessian will have a quadratic rank. The determinant and permanent problem. International problem is rather how to prove this. Over characteristic 0, Mathematics Research Notices, pages 4241–4253, 2004. Mignon and Ressayre gave a construction which is essentially [12] H. Minc. Permanents. Encyclopedia of Mathematics the all 1 matrix (except the (1, 1) entry to make per(X) = 0). and its Applications, vol 6. Addison-Wesley, 1978. This makes most second derivatives in the Hessian of the [13] K. Mulmulay and M. Sohoni. Geometric complexity permanent a constant (but involving a large factorial). The theory, P vs. NP, and explicit obstructions. SIAM key to our matrix is to choose it sufficiently uniform so that Journal on Computing, 31(2):496–526, 2002. we can still prove its rank analytically, but not so uniform [14] G. P´olya. Aufgabe 424. Arch. Math. Phys., 20:271, so as to involve large factorials. 1913. [15] R. Raz. Multi-linear formulas for permanent and Acknowledgement determinant are of super-polynomial size. J. ACM, to We wish to thank Les Valiant for many interesting discus- appear. sions on the topic. This work would not have been possible [16] R. Raz, A. Shpilka, and A. Yehudayoff. A lower bound for us without the discussions and comments from Les. for the size of syntactically multilinear arithmetic Jin-Yi Cai’s work was supported by NSF Grant NSF CCR- circuits. In Proceedings of the 48th Annual IEEE 0511679. Xi Chen and Dong Li were both supported by NSF Symposium on Foundations of Computer Science, Grant DMS-0635607. pages 438–448, 2007. [17] A. Shpilka and A. Wigderson. Depth-3 arithmetic 5. REFERENCES circuits over fields of characteristic zero. Computational Complexity, 10(1):1–27, 2001. [1] M. Agrawal. Determinant versus permanent. [18] G. Szeg¨o. Zu aufgabe 424. Arch. Math. Phys., manuscript. 21:291–292, 1913. [2] M. Agrawal. Proving lower bounds via pesudo-random [19] S. Toda. Counting problems computationally generators. In Proceedings of the FST and TCS, pages equivalent to the determinant. manuscript, 1991. 96–105, 2005. [20] L. Valiant. Private communication. [3] M. 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