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Complexity of multiplication of polynomials of small degree over finite fields (Extended abstract) Shy Artzi? and Michael Kaminski Department of Computer Science Technion – Israel Institute of Technology Haifa 32000 Israel Abstract. Let µq (n) denote the number of multiplications required for computation of the product of two polynomials of degree n over a qelement field by means of a quadratic algorithm. It is known that if q ≥ 2n, then µq (n) = 2n + 1. This bound is achieved by evaluating the polynomials at 2n+1 different points (possibly, including ∞), multiplying the values, and interpolating the result. However, this method does not work when the number of the field elements is less than 2n. It is known from the literature that, for q/2 < n ≤ q + 1, µq (n) = 3n + 1 − bq/2c. In this paper we show that the same tight bound also holds in the extended range q/2 < n ≤ q + log2 q − 4.8, for an even q and q/2 < n ≤ q + log2 q − 5.8, for an odd q. 1 Introduction In infinite fields it is possible to compute the coefficients of the product of two polynomials of degree n in 2n+1 non-scalar multiplications. It is known from [31] that each algorithm for computing the above product in 2n + 1 non-scalar multiplications must evaluate the multiplicands at a minimum of 2n distinct points, multiply the samples, and interpolate the result. However in finite fields this method fails, if 2n exceeds the number of field elements. Thus, in general, the above bound cannot be achieved in finite fields. Let Fq denote the q-element field and let µq (n) denote the number of multiplications required to compute the coefficients of the product of two polynomials of degree n over Fq by means of quadratic algorithms. The best lower bound on µq (n) known from the literature, is 3.52n, if q = 2, see [6] and 3n−o(n) for q > 2, see [19]. Also, for small values of n, the following upper bound on µq (n) can be easily reached my an algorithm based on the Chinese Remainder Theorem. Proposition 1. ([19, Appendix B]) For n ≤ (q 2 + 1)/2, µq (n) ≤ 3n + 1 − bq/2c, ? Contact author. E-mail: [email protected], Fax: 972-4-8244798 On the other hand, it follows from [9] that µq ≤ 4(1 + √ 1 ) + o(n), if q is q−3 a square greater than or equal to 25. Thus, establishing the exact value of µq (n) should be of interest. It was shown in [19] that for q/2 < n ≤ q + 1, µq (n) = 3n + 1 − bq/2c. (1) As we have mentioned above, for n ≤ q/2, µq (n) = 2n + 1. In this paper we extend (1) to range q/2 < n ≤ q + log2 q − 4.8, for an even q and q/2 < n ≤ q + log2 q − 5.8, for an odd q. Namely, we shall prove the following theorem. Theorem 1. If q is even and q/2 < n ≤ q + log2 q − 4.8 or q is odd and q/2 < n ≤ q + log2 q − 5.8, then µq (n) = 3n + 1 − bq/2c. It is known from [28] that if a set of bilinear forms over an infinite field can be computed in t multiplications/divisions, then it can be computed in t multiplications by an algorithm without divisions whose total number of operations differs from that of the original one by a factor of a small constant. But it is unknown whether a similar result holds for finite fields, cf. [7]. However one can easily prove that quadratic algorithms for computing a set of bilinear forms are optimal within the algorithms without divisions. Also we would like to note that all the algorithms for polynomial multiplication over finite fields known from the literature are quadratic (and even bilinear), see [23] and [27]. The proofs are based on the theory of linear recurring sequences and an analysis of Hankel matrices1 representing bilinear forms defined by linear combinations of the coefficients of the product of two polynomials. This technique can was also applied in [19] to analysis of algorithms for multiplication of polynomials modulo a polynomial. The paper is organized as follows. In the next section we define the notion of a quadratic algorithm and in Section 3 we introduce some notation and definitions and state the major auxiliary technical lemmas. The proof of Theorem 1 is sketched in Section 4. Finally, the last section contains some concluding remarks. 2 Quadratic algorithms for polynomial multiplication In this paper we restrict ourselves to quadratic algorithms which are defined below. We denote by x = (x0 , x1 , . . . , xn )T and y = (y0 , y1 , . . . , yn )T column vectors of indeterminate. We remind the reader that a quadratic algorithm for computing a set of bilinear forms of x and y is a straight-line algorithm whose non-scalar multiplications are of the form L0 ∗ L00 , where L0 and L00 are linear forms in x and y and each bilinear form in the set is a linear P combination of these products. We have to compute zk = zk (x, y) = xi yj , k = 0, . . . , 2n. Let z = i+j=k (z0 , z1 , . . . , z2n )T . Assume that µq (n) = t, i.e., all the bilinear forms defined 1 The definition of Hankel matrices is given in Section 3 by the components of z can be computed in t multiplications. Namely, there exist t pairs linear forms L01 (x, y), L001 (x, y), . . . , L0t (x, y), L00t (x, y) of x and y such that each zk is a linear combination of products {L0i (x, y)L00i (x, y)}i=1,...,t . Let p = (L01 (x, y)L001 (x, y), . . . , L0t (x, y)L00t (x, y))T . It is known from [10] that t ≥ 2n + 1. By the definition of quadratic algorithms there exists a (2n + 1) × t matrix U whose entries are constants from F such that z = U p. We contend first that rankU = 2n + 1. Obviously, zk (x, y) = xT Ak y, where Ak = (ai,j,k ) is an (n + 1) × (n + 1) Hankel matrix defined by ½ 1, if i + j = k + 2 . (2) ai,j,k = 0, otherwise Since matrices A0 , A1 , . . . , A2n are linearly independent, the rows of U are independent as well. This proves our contention. Permuting the components of p, if necessary, we may assume that the first 2n + 1 columns of U are linearly independent. Hence there exist a non-singular (2n + 1) × (2n + 1) matrix W and a (2n + 1) × (t − 2n − 1) matrix V such that W z = (I2n+1 , V )p, where I2n+1 denotes the (2n + 1) × (2n + 1) identity matrix. That is, the first 2n + 1 columns of the product W U are those of I2n+1 . Let W z = (xT H1 y, xT H2 y, . . . , xT H2n+1 y)T . Let H = {H1 , H2 , . . . , H2n+1 }. We fix matrices H1 , H2 , . . . , H2n+1 till the end of this paper. Note that each matrix Hk ∈ H a linear combination of matrices A0 , A1 , . . . , A2n defined by (2): Hk = 2n X si Ai , s0 , s1 , . . . , s2n ∈ Fq . i=0 Therefore, Hk is of the form s0 s1 · · · sn s1 · · · sn s2 · · · sn+1 · · . · · · · sn+1 · · · s2n Matrices in such from are called Hankel matrices. They constitute a major tool in this paper. We conclude this section with an observation that if n ≤ (q 2 + 1)/2, then rankHk ≤ n, k = 1, 2, . . . , 2n + 1. Indeed, permuting rows and columns of (I2n+1 , V ), if necessary, we may assume that k = 2n + 1. Since computing of µq (xT H2n+1 y) requires rankH2n+1 multiplications, V has, at least, rankH2n+1 − 1 columns. Therefore µq (n) = t ≥ 2n + rankH2n+1 . Combining this with Proposition 1, we obtain the desired inequality. 3 Hankel matrices and linear recurring sequences In this section we introduce some notation and prove the major auxiliary lemmas needed for the proof of Theorems 1. Let k be a positive integer and let a0 , . . . , ak−1 be given elements of F. A sequence σ = s0 , s1 , . . . , s` of elements of Fq satisfying the relation sm+k = ak−1 sm+k−1 + ak−2 sm+k−2 + . . . + a0 sm , m = 0, 1, . . . , ` − k is called a (finite k-th-order homogeneous) linear recurring sequence in Fq . The terms s0 , s1 , . . . , sk−1 are referred as initial values. The polynomial f (α) = αk − ak−1 αk−1 − ak−2 αk−2 − . . . − a0 ∈ F [α] is called a characteristic polynomial of σ. Proposition 2 below shows that if a finite linear recurring sequence is “sufficiently long”, then it possesses an important property of infinite linear recurring sequences. Proposition 2. ([19, Proposition 1]) Let σ and f (α) be as above, and let fσ (α) be a characteristic polynomial of σ of the minimal degree. If deg fσ (α)+deg f (α) ≤ ` + 1, then fσ (α) divides f (α). A uniquely determined monic polynomial fσ (α) ∈ F [α] given by Proposition 2 is called the minimal polynomial of σ. For a sequence σ = s0 , s1 , . . . , s2n we define the (n + 1) × (n + 1) Hankel matrix H(σ) by s0 s1 · · · sn s1 s2 · · · sn+1 · · · . · · · · · · sn sn+1 · · · s2n Let H i (σ) denote the (i + 1)th row of H(σ). Let rankH(σ) ≤ n and let k be the minimal integer for which there exist a0 , a1 , . . . , ak−1 ∈ Fq such that k−1 X ai H i (σ) = H k (σ). i=0 We define sequence σ̃ = s̃0 , s̃1 , . . . , s̃2n by the linear recurrence s̃i+k = ak−1 s̃i+k−1 + ak−2 s̃i+k−2 + · · · + a0 s̃i , (3) with initial values s̃i = si , i = 0, 1, . . . , k − 1. Let σ̄ = σ − σ̃. We shall denote H(σ), H(σ̃), and H(σ̄) by H, H̃, and H̄, k−1 P respectively. Let fH (α) = αk − ai αi , i.e., fH (α) is a characteristic polynomial i=0 of the sequence defined by recurrence (3), and, in particular, of σ̃.2 Finally, we define the divisor f H (α) by f H (α) = fH (α)(α − ∞)rankH̄ . By [19, Proposition 2], rankH = deg f H (α). Let S be a set of (n + 1) × (n + 1) Hankel matrices of rank not exceeding n. Define fS (α) = lcm{f H (α) : H ∈ S}.3 In what follows we shall need Lemmas 1 and 2, below. Let V be a vector space over Fq , v1 , v2 , . . . , vm ∈ V . We denote the linear subspace of V spanned by v1 , v2 , . . . , vm by [v1 , v2 , . . . , vm ]. Lemma 1. ([19, Lemma 1]) Let S be a set of (n + 1) × (n + 1) Hankel matrices of rank not exceeding n. Then dim[S] ≤ deg f S (α). For Lemma 2 below we shall need the following notation. For a set S of (n + 1) × (n + 1) Hankel matrices we denote by µq (S) the quadratic complexity of S, i.e., the the minimal number of multiplications required to compute the set of bilinear forms {xXy T : H ∈ S} by means of a quadratic algorithm. Lemma 2. ([19, Lemmas 2 and 3]) Let S be a set of (n+1)×n Hankel matrices of rank not exceeding n. Then µq (S) ≥ min(deg f S (α), n + 1).4 Since matrices H1 , H2 , . . . , H2n+1 defined in the previous section are linearly independent, by Lemma 1, deg f H (α) ≥ 2n + 1. However, Lemma 2 bounds us to n + 1. In this paper we improve the lower bound given by Lemma 2 as follows. Lemma 3. Let S be a set of (n + 1) × (n + 1) Hankel matrices of rank not exceeding n such that for each H 0 , H 00 ∈ S divisors f H 0 (α) and f H 00 (α) are coprime. Let d = max{deg f H (α) : H ∈ S}. Then computing the set of bilinear forms of x and y defined by the elements of S requires at least min(deg f S (α), n+ blog2 ( n + 1)c − 1) multiplications. d Even though, using Lemma 3 instead of Lemma 2 in the proofs in [19] would improve the lower bound only by o(n), this lemma allows to extend the range of n for which the exact value of µq (n) can be calculated. To apply Lemma 3 we shall need the following results. Proposition 3. ([4, Proposition 1] Let S ⊆ H. then µq (n) ≥ 2n + 1 − |S| + µq (S).5 Corollary 1. Let n ≤ (q 2 + 1)/2. If for some S ⊆ H, µq (S) − |S| ≥ n − bq/2c, then µq (n) = 3n + 1 − bq/2c. Proof. The corollary follows from Propositions 1 and 3. 2 3 4 5 In fact, fH (α) is the minimal polynomial of those sequences, see [19, Section 3]. lcm is an abbreviation for “the least common multiple”. That is, computing the set of bilinear forms of x and y defined by the elements of S requires at least min(deg f S (α), n + 1) multiplications. We denote by |S| the cardinality of S. Let ½ cq = 0 if q is even . 1 if q is odd (4) In the next section for q/2 < n ≤ q + log2 q − cq we sketch a construction of a subset S of H such that µq (S) − |S| ≥ n − bq/2c. 4 Sketch of the proof of Theorem 1 Let H = {H1 , H2 , . . . , H2n+1 } be as in Section 2 and let f H (α) = Q̀ j=1 e pj j (α) be the decomposition of f H (α) into its irreducible factors. Let H ≥3 be a minimal subset of H such that e e – f H ≥3 (α) is divisible by all pj j (α) with deg pj j (α) ≥ 3 and – for each H 6∈ H ≥3 , deg f H ≥3 ∪{H} (α) ≤ deg f H ≥3 (α) + 2. Remark 1. By the definitions of H ≥3 , deg f H ≥3 (α) ≥ 3|H ≥3 |. Let H =2 be a minimal subset of H \ H ≥3 such that e – f H =2 (α) is divisible by all pj j (α) of degree 2 which are coprime with f H ≥3 (α). Remark 2. It follows from the definitions of H =2 that for each subset H ⊆ of H =2 , deg f H ≥3 ∪H ⊆ (α) = deg f H ≥3 (α) + 2|H ⊆ |. We proceed with a sequence of statements of properties of H ≥3 and H =2 . The proofs of these properties are similar each to other. We omit some of them and only sketch the others because of the space constraints. Lemma 4. We have |H =2 | + deg f H ≥3 (α) − |H ≥3 | ≥ n − bq/2c. From now on we assume that q ≥ 3 and q < n < 2q. We also fix a subset H ⊆ of H =2 . Lemma 5. If |H ⊆ | < 3(q − cq )/2 − n,6 then deg f H ≥3 (α) < n + 1 − 2|H ⊆ |. Lemma 6. If n < 3(q − cq )/2, then 1. µq (H ≥3 ) ≥ deg f H≥3 (α), 2. deg f H≥3 (α) − 2|H ≥3 | ≤ n − q + cq , and 3. |H ≥3 | ≤ n − q + cq . 6 See (4) for the definition of cq . Proof. Clause 1 of the Lemma follows from Lemma 5 with H ⊆ = ∅ and Lemma 2. The proof of clause 2 is a bit longer. By Lemma 1, deg f H (α) ≥ 2n + 1 and, by the definition of H ≥3 and H =2 , each H ∈ H \ {H ≥3 ∪ H =2 } contributes only a linear factor to f H (α). Therefore, deg f H ≥3 ∪H =2 (α) ≥ 2n − q (≥ n + 1). (5) Let H ⊆ be a minimal subset of H =2 such that deg f H ≥3 ∪H ⊆ (α) ≥ n + 1. Combining Proposition 3 with the 3n + 1 − bq/2c upper bound on µq (n) we obtain 3n + 1 − bq/2c ≥ µq (n) ≥ 2n + 1 − |H ≥3 ∪ H ⊆ | + µq (H ≥3 ∪ H ⊆ ). Therefore, by (5) and Lemma 2, |H ≥3 ∪ H ⊆ | = |H ≥3 | + |H ⊆ | ≥ bq/2c + 1. (6) By the definition of H ⊆ and Remark 2, |H ⊆ | = d(n + 1) − deg f H ≥3 (α))/2e which together with (6) implies the desired inequality. Finally, clause 3 of the lemma follows from clause 2 and Remark 1. Corollary 2. If n < 3(q − cq )/2, then deg f H ≥3 (α) ≤ 3(n − q + cq ). Proof. The inequality immediately follows from clauses 2 and 3 of Lemma 6. Lemma 7. Let H max be a maximal subset of [H ≥3 ∪H ⊆ ]\{0(n+1)×(n+1) } such that – divisors of the elements of H max are pairwise coprime, – divisors of the elements of H max are coprime with f H ≥3 (α), and – for each H ∈ H max , if jH is the number of elements of H ⊆ which participate in representation of H as a linear combination over H ≥3 ∪ H ⊆ , then jH < (n + 1)/2. Let n < (6(q − cq ) − 2)/5. (7) Then, for each H ∈ H max , 1. jH ≤ deg f H ≥3 (α) + 1 and 2. deg f H (α) = 2jH . It could happen that divisors of some matrices in H ⊆ do not divide f H max (α) or, in other words, there are matrices in H ⊆ which do not participate in representation of elements of H max as a linear combination over H ≥3 ∪ H ⊆ . We denote the set of such matrices by H − ⊆ , i.e., H− ⊆ = {H ∈ H ⊆ : gcd(f H (α), f Hmax (α)) = 1}. Some basic properties of H − ⊆ are summarized in Lemmas 8 and 9 below. Lemma 8. If (7), then 1. |H − ⊆ | + |H ≥3 | ≤ deg f H ≥3 (α) and 2. f H max (α) = f H ≥3 ∪(H ⊆ \H − ) (α)/f H ≥3 (α). ⊆ Lemma 9. If |H − ⊆ | < (4q − 3n − 3cq − 1)/2, (8) then there is a matrix in [H ≥3 ∪ H − ⊆ ] whose divisor is f H ≥3 ∪H − (α). ⊆ We denote a matrix whose existence is provided by Lemma 9 by HH ≥3 ∪H − ⊆ and we define a set of Hankel matrices H + ⊆ by H+ ⊆ = H max ∪ {HH ≥3 ∪H − }. (9) n < (8q − 7cq − 1)/7, (10) ⊆ Lemma 10. If then 1. the divisors of the elements of H + ⊆ are pairwise coprime, 2. f H + (α) = f H ≥3 ∪H ⊆ (α), and ⊆ 3. for each H ∈ H + ⊆ , 2 ≤ deg f H (α) ≤ max{2, 3 deg f H≥3 (α) − 2|H ≥3 |}. Proof. Since (10) implies (7), by clause 1 of Lemma 8, 2|H − ⊆ | ≤ 2 deg f H ≥3 (α) − 2|H ≥3 |. (11) Also, n < (8q − 7cq − 1)/7 implies n < 3(q − cq )/2. Thus, combining (11) with clause 2 of Lemma 6 and its corollary (Corollary 2), we obtain that |H − ⊆| ≤ 4(n−q+cq ), which, together with (10) implies (8). Thus, we may apply Lemmas 7 and 9 which together with (9) impliy the first clause of the lemma. The second clause of Lemma 10 follows from the second clause of Lemma 8 (and (9), of course). As for the last clause of the lemma, let H ∈ H + ⊆ . If H ∈ Hmax , then, by Lemma 7, 2 ≤ deg f H (α) ≤ 2(deg f H≥3 (α) + 1) ≤ max{2, 3 deg f H≥3 (α) − 2|H ≥3 |}, where the last inequality follows from Remark 1. If H = HH ≥3 ∪H − , then 2 ≤ deg f H (α). By Remark 2 and Lemma 9, ⊆ deg f H (α) = deg f H≥3 (α) + 2|H − ⊆ |, which together with (11) implies deg f H (α) ≤ 3 deg f H≥3 (α) − 2|H ≥3 |. Finally, Lemma 11 below provides a subset S of H that is the prerequisite of the corollary to Proposition 3 (Corollary 1). Lemma 11. If q < n ≤ q + log2 q − 4.8 − cq , (12) then there exists a subset S of H such that µq (S) − |S| ≥ n − bq/2c. Proof. Obviously, (12) implies (10). By clause 1 of Lemma 10, we may apply Lemma 3 to H + ⊆ and obtain µq (H + ⊆ ) ≥ min(deg f H + (α), n + blog2 ( ⊆ n + 1)c − 1), d (13) where d = max{deg f H (α) : H ∈ H + ⊆ }. We shall distinguish between the cases of deg f H + (α) < n+blog2 ( n +1)c−1 d =2 and deg f H + (α) ≥ n + blog2 ( n + 1)c − 1. d =2 If deg f H + (α) < n + blog2 ( n + 1)c − 1, by Remark 2, d =2 µq (H + =2 ) ≥ deg f H ≥3 (α) + 2|H =2 |. (14) + Let S = H ≥3 ∪ H =2 . Since H + =2 ⊆ [H ≥3 ∪ H =2 ], µq (S) ≥ µq (H =2 ) and, “subtracting |S| = |H ≥3 | + |H =2 | from (14)” results in µq (S) − |S| ≥ deg f H ≥3 (α) + |H =2 | − |H ≥3 |. (15) Now, the desired inequality follows from (15) and Lemma 4. The case of n deg f H + (α) ≥ n + blog2 ( + 1)c − 1 (16) =2 d is based on a long sequence of delicate calculations and we only sketch it.7 So, assume (16) and let H ⊆ be a minimal subset of H =2 such that deg f H + (α) ≥ n + blog2 ( ⊆ n + 1)c − 1. d (17) Then, by Remark 2, |H ⊆ | = d(n + blog2 ( n + 1)c − 1 − deg f H ≥3 (α))/2e. d (18) Let S = H ≥3 ∪ H ⊆ . Since H + ⊆ ⊆ [H ≥3 ∪ H ⊆ ], by (13), µq (S) ≥ µq (H + ⊆ ) ≥ n + blog2 ( n + 1)c − 1. d (19) We have to prove that µq (S) − |S| ≥ n − bq/2c. Since |S| = |H ≥3 | + |H ⊆ |, it suffices to show that n blog2 ( + 1)c + bq/2c ≥ |H ≥3 | + |H ⊆ | + 1. (20) d 7 In particular, constants 4.8 and 5.8 come from this case. By the lemma assumption, n − q < log2 q − 4.8 − cq ≤ log2 n − 4.8 − cq . (21) Using clause 3 of Lemma10, it can be shown that − log2 d + deg f H ≥3 (α) − 2|H ≥3 | ≥ 1 − log2 7 ≈ −1.8. 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