COUNTING VALUE SETS: ALGORITHM AND COMPLEXITY

Transcription

1 COUNTING VALUE SETS: ALGORITHM AND COMPLEXITY QI CHENG, JOSHUA E. HILL, AND DAQING WAN Abstract. Let p be a prime. Given a polynomial in F p m[x] of egree over the finite fiel F p m, one can view it as a map from F p m to F p m, an examine the image of this map, also known as the value set. In this paper, we present the first non-trivial algorithm an the first complexity result on explicitly computing the carinality of this value set. We show an elementary connection between this carinality an the number of points on a family of varieties in affine space. We then apply Lauer an Wan s p-aic point-counting algorithm to count these points, resulting in a non-trivial algorithm for calculating the carinality of the value set. The running time of our algorithm is pm) O). In particular, this is a polynomial-time algorithm for fixe if p is reasonably small. We also show that the problem is #P-har when the polynomial is given in a sparse representation, p = 2, an m is allowe to vary, or when the polynomial is given as a straight-line program, m = 1 an p is allowe to vary. Aitionally, we prove that it is NP-har to ecie whether a polynomial represente by a straight-line program has a root in a prime-orer finite fiel, thus resolving an open problem propose by Kaltofen an Koiran in [5, 6]. 1. Introuction In a finite fiel with q = p m p prime) elements, F q, take a polynomial, f F q [x] with egree > 0. Denote the image set of this polynomial as V f = {f α) α F q } an enote the carinality of this set as # V f ). There are a few trivial bouns that can be immeiately establishe. There are only q elements in the fiel, so # V f ) q. Aitionally, any polynomial of egree can have at most roots, thus for all a V f, fx) = a is satisfie at most times. This is true for every element in V f, so # V f ) q, whence q # V f ) q where is the ceiling function). Both of these bouns can be achieve: if # V f ) = q, then f is calle a permutation polynomial an if # V f ) = q, then f is sai to have a minimal value set. The problem of establishing # V f ) has been stuie in various forms for at least the last 115 years, but exact formulations for # V f ) are known only for polynomials in very specific forms. Results that apply to general polynomials are asymptotic in nature, or provie estimates whose errors have reasonable bouns only on average [11]. All three authors are partially supporte by the NSF. 1

2 2 QI CHENG, JOSHUA E. HILL, AND DAQING WAN The funamental problem of counting the value set carinality # V f ) can be thought of as a much more general version of the problem of etermining if a particular polynomial is a permutation polynomial. Shparlinski [14] provie a baby-step giant-step type test that etermines if a given polynomial is a permutation polynomial by extening [17] to an algorithm that runs in Õq)6/7 ). This is still fully exponential in log q. Ma an von zur Gathen [10] provie a ZPP zero-error probabilistic polynomial-time) algorithm for testing if a given polynomial is a permutation polynomial. Accoring to [7], the first eterministic polynomial-time algorithm for testing permutation polynomials is obtaine by Lenstra using the classification of exceptional polynomials which in turn epens on the classification of finite simple groups. Subsequently, an elementary approach base on the Gao-Kaltofen-Lauer factorization algorithm is given by Kayal [7]. For the more general problem of exactly computing # V f ), essentially nothing is known about this problem s complexity an no non-trivial algorithms are known. For instance, no baby-step giant-step type algorithm is known in computing # V f ). No probabilistic polynomial-time algorithm is known. Fining a non-trivial algorithm an proving a non-trivial complexity result for the value counting were raise as open problems in [10], where a probabilistic approximation algorithm is given. In this paper, we provie the first non-trivial algorithm an the first non-trivial complexity result for the exact counting of the value set problem Our results. Perhaps the most obvious metho to calculate # V f ) is to evaluate the polynomial at each point in F q an count how many istinct images result. This algorithm has a time an space complexity q) O1). One can also approach this problem by operating on points in the co-omain. One has fx) = a for some x F q if an only if f a X) = fx) a has a zero in F q ; this algorithm again has a time complexity q) O1), but the space complexity is improve consierably to log q) O1). In this paper we present several results on etermining the carinality of value sets. On the algorithmic sie, we show an elementary connection between this carinality an the number of points on a family of varieties in affine space. We then apply Lauer an Wan s p-aic point-counting algorithm [9], resulting in a non-trivial algorithm for calculating the image set carinality in the case that p is sufficiently small i.e., p = O log q) C ) for some positive constant C). Precisely, we have Theorem 6. There exists an explicit eterministic algorithm an an explicit polynomial R such that for any f F q [x] of egree, where q = p m p prime), the algorithm computes the carinality of the image set, # V f ), in a number of bit operations boune by R m p ). The running time of this algorithm is polynomial in both p an m, but is exponential in. In particular, this is a polynomial-time algorithm for fixe if the characteristic p is small q = p m can be large, but p = O log q) C ). On the complexity sie, we have several harness results on the value set problem. We frame these results using some stanar classes in complexity theory, which we outline here. NP is the complexity class of ecision problem whose positive solutions can be verifie in polynomial time. NP-har is the computational class of ecision problems that all NP problems can be reuce to using a polynomial time reuction. NP-complete is the complexity class of all NP-har problem whose

3 COUNTING VALUE SETS: ALGORITHM AND COMPLEXITY 3 solution can be verifie in polynomial time that is, NP-complete is the intersection of NP-har an NP). Co-NP-complete is the complexity class of problems where answering the logical complement of the ecision problem is NP-complete. The corresponing counting complexity theory classes that we use are as follows. #P rea sharp-p ) is the set of counting problems whose corresponing ecision problem is in NP. #P-har is the computational class of counting problems that all #P problems can be reuce to using a polynomial time counting reuction. #P-complete is the intersection of #P-har an #P. With a fiel of characteristic p = 2, we have Theorem 3. The problem of counting the value set of a sparse polynomial over a finite fiel of characteristic p = 2 is #P-har. The central approach in our proof of this theorem is to reuce the problem of counting satisfying assignments for a 3SAT formula to the problem of value set counting. Over a prime-orer finite fiel, we have Theorem 5. Over a prime-orer finite fiel F p, the problem of counting the value set is #P-har uner RP-reuction ranomize polynomial time reuction) if the polynomial is given as a straight-line program. Aitionally, we prove that it is NP-har to ecie whether a polynomial in Z[x] represente by a straight-line program has a root in a prime-orer finite fiel, thus resolving an open problem propose in [5, 6]. We accomplish the complexity results over prime-orer finite fiels by reucing the prime-orer finite fiel subset sum problem PFFSSP) to these problems. In the PFFSSP, given a prime p, an integer b an a set of integers S = {a 1, a 2,, a t }, we want to ecie the solvability of the equation a 1 x 1 + a 2 x a t x t b mo p) with x i {0, 1} for 1 i t. The main iea comes from the observation that if t < log p/3, there is a sparse polynomial αx) F p [x] such that as x runs over F p, the vector αx), αx + 1),, αx + t 1)) runs over all the elements in {0, 1} t. In fact, a lightly moifie version of the quaratic character αx) = x p 1)/2 + x p 1 )/2 suffices. So the PFFSSP can be reuce to eciing whether the sparse shift polynomial t 1 i=0 a i+1αx + i) b = 0 has a solution in F p. 2. Backgroun 2.1. The subset sum problem. To prove the complexity results, we use the subset sum problem SSP) extensively. The SSP is a well-known problem in computer science. In one instance of the SSP, given an integer b an a set of positive integers S = {a 1, a 2,, a t }, 1) Decision version) the goal is to ecie whether there exists a subset T S such that the sum of all the integers in T equals b, 2) Search version) the goal is to fin a subset T S such that the sum of all the integers in T equals b,

4 4 QI CHENG, JOSHUA E. HILL, AND DAQING WAN 3) Counting version) the goal is to count the number of subsets T S such that the sum of all the integers in T equals b. The ecision version of the SSP is a classical NP-complete problem. The counting version of the SSP is #P-complete, which can be easily erive from proofs of the NP-completeness of the ecision version, e.g. [3, Theorem 34.15]. One can view the SSP as a problem of solving the linear equation a 1 x 1 + a 2 x a t x t = b with x i {0, 1} for 1 i t. The prime-orer finite fiel subset sum problem is a similar problem where in aition to b an S, one is given a prime p, an the goal is to ecie the solvability of the equation with x i {0, 1} for 1 i t. a 1 x 1 + a 2 x a t x t b mo p) Proposition 1. The prime-orer finite fiel subset sum problem is NP-har uner RP-reuction. Proof. To reuce the subset sum problem to the prime-orer finite fiel subset sum problem, one fins a prime p > t i=1 a i, which can be one in ranomize polynomial time. Remark 1. To make the reuction eterministic, one nees to e-ranomize the problem of fining a large prime, which appears to be ifficult [15] Polynomial representations. There are ifferent ways to represent a polynomial over a fiel F. The ense representation lists all the coefficients of a polynomial, incluing the zero coefficients. The sparse representation lists only the nonzero coefficients, along with the egrees of the corresponing terms. If most of the coefficients of a polynomial are zero, then the sparse representation is much shorter than the ense representation. A sparse shift representation of a polynomial in F[x] is a list of n triples a i, b i, e i ) F F Z 0 which represents the polynomial a i x + b i ) e i. 1 i n More generally, a straight-line program for a univariate polynomial in Z[x] or F p [x] is a sequence of assignments, starting from x 1 = 1 an x 2 = x. After that, the i-th assignment has the form x i = x j x k where 0 j, k < i an is one of the three operations +,,. We first let α be an element in F p m such that F p m = F p [α]. A straight-line program for a univariate polynomial in F p m[x] can be efine similarly, except that the sequence starts from x 1 = α an x 2 = x. One can verify that a straight-line program computes a univariate polynomial, an that sparse polynomials an sparse shift polynomials have short straight-line programs. A polynomial prouce by a short straight-line program may have very high egree, an most of its coefficients may be nonzero, so it may be costly to write it in either a ense form or a sparse form.

5 COUNTING VALUE SETS: ALGORITHM AND COMPLEXITY 5 3. Harness of solving straight-line polynomials It is known that eciing whether there is a root in a finite fiel for a sparse polynomial is NP-har [8]. In a relate work, it was shown that eciing whether there is a p-aic rational root for a sparse polynomial is NP-har [1]. However, the complexity of eciing the solvability of a straight-line polynomial in Z[x] within a prime-orer finite fiel was not known. This open problem was propose in [5] an [6]. We resolve this problem within this section, an this same iea will be use later on to prove the harness result of the value set counting problem. Let p be an o prime. Let χ be the quaratic character moulo p, namely χx) equals 1, 1, or 0, epening on whether x is a quaratic resiue, a quaratic non-resiue, or is congruent to 0 moulo p. For x F p, χx) = x p 1)/2. Consier the list 1) χ1), χ2),, χp 1). It is a sequence in {1, 1} p 1. The following boun is a stanar consequence of the celebrate Weil boun for character sums, see [13] for a etaile proof. Proposition 2. Let b 1, b 2,, b t ) be a sequence in {1, 1} t. Then the number of x F p such that χx) = b 1, χx + 1) = b 2,, χx + t 1) = b t is in the range p/2 t t3 + p), p/2 t + t3 + p)). The proposition implies that if t < log p/3, then every possible sequence in { 1, 1} t occurs as a consecutive sub-sequence in expression 1). In many situations it is more convenient to use binary 0/1 sequences, which suggests instea using the polynomial x p 1)/2 + 1)/2, but this results in a small problem at x = 0. We instea use the sparse polynomial 2) αx) = x p 1)/2 + x p 1 )/2. αx) takes value in {0, 1} if x F p an αx) = 1 iff χx) = 1. Corollary 1. If t < log p/3, then for any binary sequence b 1, b 2,, b t ) {0, 1} t, there exists a x F p such that αx) = b 1, αx + 1) = b 2,, αx + t 1) = b t. In other wors, if t < log p/3, the map x αx), αx + 1),, αx + t 1)) is an onto map from F p to {0, 1} t ; this map thus sens an algebraic object to a combinatorial object. Given a straight-line polynomial fx) Z[x] an a prime p, how har is it to ecie whether the polynomial has a solution in F p? We now prove that this problem is NP-har. Theorem 1. Given a sparse shift polynomial fx) Z[x], an a large prime p, it is NP-har to ecie whether fx) has a root in F p uner RP-reuction. Proof. We reuce the ecision version of the) subset sum problem to this problem. Given b Z 0 an S = {a 1, a 2,, a t } Z 0, one can then fin a prime, p, such

6 6 QI CHENG, JOSHUA E. HILL, AND DAQING WAN that p > max2 3t, t i=1 a i) an construct a sparse shift polynomial t 1 3) βx) = a i αx + i) b. i=0 If the polynomial has a solution moulo p, then the answer to the subset sum problem is yes, since for any x F p, αx + i) {0, 1}. In the other irection, if the answer to the subset sum problem is yes, then accoring to Corollary 1, the polynomial has a solution in F p. Note that the reuction can be compute in ranomize polynomial time. 4. Complexity of the value set counting problem In this section, we prove several results about the complexity of the value set counting problem Finite fiels of characteristic 2. We will use a problem about NC 0 5 circuits to prove that counting the value set of a sparse polynomial in a fiel of characteristic 2 is #P-har. A Boolean circuit is in NC 0 5 if every output bit of the circuit epens only on at most 5 input bits. We can view a circuit with n input bits an m output bits as a map from {0, 1} n to {0, 1} m an call the image of the map the value set of the circuit. The following proposition is implie in [4]; we provie a sketch of the proof. Proposition 3. Given a 3SAT formula with n variables an m clauses, one can construct in polynomial time an NC 0 5 circuit with n+m input bits an n+m outputs bits, such that if there are M satisfying assignments for the 3SAT formula, then the carinality of the value set of the NC 0 5 circuit is 2 n+m 2 m 1 M. In particular, if the 3SAT formula can not be satisfie, then the circuit computes a permutation from {0, 1} n+m to {0, 1} n+m. Proof. Denote the variables of the 3SAT formula by x 1, x 2,, x n, an the clauses of the 3SAT formula by C 1, C 2,, C m. Buil a circuit with n + m input bits an n + m output bits as follows. The input bits will be enote by x 1, x 2,, x n an y 1, y 2,, y m, an output bits will be enote by z 1, z 2,, z n, w 1, w 2,, w m. Set z i = x i for 1 i n. An set w i = C i y i y i+1 mo m)) )) C i y i ) for 1 i m. In other wors, if C i is evaluate to be TRUE, then output y i y i+1 mo m)) as w i, an otherwise output y i as w i. Note that C i epens only on 3 variables from {x 1, x 2,, x n }, thus we obtain an NC 0 5 circuit. After fixing an assignment to x i s, z i s are also fixe, an the transformation from y 1, y 2,, y m ) to w 1, w 2,, w m ) is linear over F 2. One can verify that the linear transformation has rank m 1 if the assignment satisfies all the clauses, an it has rank m namely it has full rank) if some of the clauses are not satisfie. So the carinality of the value set of the circuit is M2 m n M)2 m = 2 n+m 2 m 1 M. If we replace the Boolean gates in the NC 0 5 circuit by algebraic gates over F 2, we obtain an algebraic circuit that computes a polynomial map from F n+m 2 to itself, where each polynomial epens only on 5 variables an has egree equal to or

7 COUNTING VALUE SETS: ALGORITHM AND COMPLEXITY 7 less than 5. There is an F 2 -basis for F 2 n+m, say ω 1, ω 2,, ω n+m which inuces a bijection from F n+m 2 to F 2 n+m given by x 1, x 2,, x n+m ) x = n+m which has an inverse that can be represente by sparse polynomials in F 2 n+m[x]. Using this fact, we can replace the input bits of the algebraic circuit by sparse polynomials, an collect the output bits together using the base to form a single element in F 2 n+m. We thus obtain a sparse univariate polynomial in F 2 n+m[x] from the NC 0 5 circuit such that their value sets have the same carinality. We thus have the following theorem: Theorem 2. Given a 3SAT formula with n variables an m clauses, one can construct in polynomial time a sparse polynomial γx) over F 2 n+m such that the value set of γx) has carinality 2 n+m 2 m 1 M, where M is the number of satisfying assignments of the 3SAT formula. Since counting the number of satisfying assignments for a 3SAT formula is known to be #P-complete, we have our main theorem: Theorem 3. The problem of counting the value set of a sparse polynomial over a finite fiel of characteristic p = 2 is #P-har. Corollary 2. The set of sparse permutation polynomials over finite fiels of characteristic p = 2 is co-np-complete Prime-orer finite fiels. The construction in Theorem 2 relies on builing extensions over F 2. The technique cannot be aopte easily to the prime-orer finite fiel case. We will prove that counting the value set of a straight-line polynomial over prime-orer finite fiel is #P-har. We reuce the counting version of the subset sum problem to the value set counting problem. Theorem 4. Given access to an oracle that solves the value set counting problem for straight-line polynomials over prime-orer finite fiels, there is a ranomize polynomial-time algorithm solving the counting version of the SSP. Proof. Given an instance of the counting subset sum problem, say b with the set S = {a 1, a 2,, a n }, if b > n i=1 a i, we answer 0; if b = 0, then we answer 1. Otherwise we fin a prime p > max2 3t, 2 n i=1 a i) an ask the oracle to count the value set of the sparse shift polynomial i=0 i=1 x i ω i t 1 fx) := 1 βx) p 1 ) αx + i)2 i ) over the prime-orer fiel F p, where αx) an βx) are as efine in equations 2) an 3), respectively. We output the answer # V f ) 1, which is easily seen to be exactly the number of subsets of {a 1,, a n } which sum to b. Since the counting version of the SSP is #P-complete, this theorem yiels Theorem 5. Over a prime-orer finite fiel F p, the problem of counting the value set is #P-har uner RP-reuction, if the polynomial is given as a straight-line program.

8 8 QI CHENG, JOSHUA E. HILL, AND DAQING WAN 5. The Image Set an Point Counting Proposition 4. If f F q [x] is a polynomial of egree > 0, then the carinality of its image set is 4) # V f ) = 1) i 1 N i σ i 1, 1 2,..., 1 ) i=1 where N k = # { x 1,..., x k ) F k q fx 1 ) = = fx k ) }) an σ i enotes the ith elementary symmetric function on elements. Proof. For any y V f, efine Ñ k,y = { x 1,..., x k ) F k q fx 1 ) = = fx k ) = y } an enote the corresponing carinality of these sets as N k,y = # an finally, note that 5) N k = Ñk,y ) y V f N k,y. Let us refer to the right han sie of 4) as η; plugging 5) into this expression an rearranging, we get η = 1) i 1 N i,y σ i 1, 1 2,..., 1 ). y V f i=1 Let us call the inner sum ω y, that is: ω y = 1) i 1 N i,y σ i 1, 1 2,..., 1 ). i=1 If we can show that for all y V f we have ω y = 1, then we clearly have η = # V f ). Let y V f be fixe. Let k = # f 1 y) ). It is clear that 1 k an N i,y = k i for 0 i. Substituting this in, our expression mercifully becomes somewhat nicer: 6) 7) ω y = 1 = 1 = 1 1) i k i σ i 1, 1 2,..., 1 ) 1) i σ i k1, k 1 2,..., k 1 ) i=0 i=0 [ 1 k1) 1 k 1 ) 1 k 1 )] 2 = 1. From step 6) to step 7), we are using the ientity n n λ X j ) = 1) j λ n j σ j X 1,..., X n ). j=1 j=0

9 COUNTING VALUE SETS: ALGORITHM AND COMPLEXITY 9 Note that the brackete term of 7) is 0, as k must be an integer such that 1 k, so one term in the prouct will be 0. Thus, we have η = # V f ), as esire. Proposition 4 gives us a way to express # V f ) in terms of the numbers of rational points on a sequence of curves over F q. If we ha a way of getting N k for 1 k, then it woul be easy to calculate # V f ). We procee by examining a family of relate spaces, Ñ k = { x 1,..., x k ) F k q fx 1 ) = = fx k ) }. ) We immeiately note that N k = # Ñk. Spaces similar to our Ñk have been use several times[16, 2] to establish various asymptotic results for # V f ). The spaces use in these works require that x i x j for i j. We will see that our work woul have been ramatically harer if we ha impose these aitional restrictions. The spaces Ñk aren t of any nice form in particular, we cannot assume they are non-singular projective, abelian varieties, etc.), so we procee by using the p- aic point counting metho escribe in [9], which runs in polynomial time for any variety over a fiel of small characteristic i.e., p = O log q) C ) for some positive constant C). Theorem 6. There exists an explicit eterministic algorithm an an explicit polynomial R such that for any f F q [x] of egree, where q = p m p prime), the algorithm computes the carinality of the image set, # V f ), in a number of bit operations boune by R m p ). Proof. We first note that: Ñ k = { x 1,..., x k ) F k q fx 1 ) = = fx k ) } fx 1 ) fx 2 ) = 0 = x 1,..., x k ) F k fx 1 ) fx 3 ) = 0 q.. fx 1 ) fx k ) = 0 For reasons soon to become clear, we nee to represent this as the solution set of a single polynomial. Let us introuce aitional variables z 1 to z k 1, an enote x = x 1,..., x k ) an z = z 1,..., z k 1 ). Now examine the auxiliary function 8) F k x, z) = z 1 fx 1 ) fx 2 )) + + z k 1 fx 1 ) fx k )). Clearly, if γ Ñk, then F k γ, z) is the zero function. If γ F k q \ Ñk, then the solutions of F k γ, z) = 0 specify a k 2)-imensional F q -linear subspace of F k 1 q. Thus, if we enote the carinality of the solution set to F k x, z) = 0 as # F k ), then we see that # F k ) = q k 1 N k + q k 2 q k ) N k Solving for N k, we fin that = N k q k 2 q 1) + q 2k 2. 9) N k = # F k) q 2k 2 q k 2. q 1)

10 10 QI CHENG, JOSHUA E. HILL, AND DAQING WAN Thus we have an easy way to etermine what N k is epening on the number of points on this hypersurface efine by the single polynomial equation F k = 0. The main theorem in [9] yiels an algorithm for toric point counting in F q l that is polynomial time when the characteristic is small i.e., p = O log q) C ) for some positive constant C) that works for general varieties. In [9, 6.4], this theorem is aapte to be a generic point counting algorithm. Aapting this result to our problem, we see that F k has a total egree of + 1, is in 2k 1 variables, an that we only care about the case where l = 1. Thus, the runtime for this algorithm is Õ28k+1 m 6k+4 k 6k+2 6k 3 p 4k+2 ) bit operations. In orer to calculate # V f ) using equation 4), we calculate N k for 1 k, scale by an elementary symmetric polynomial. All of the necessary elementary symmetric polynomials can be evaluate using Newton s ientity see [12]) in less than O 2 log ) multiplications. As such, the entire calculation has a runtime of Õ2 8+1 m p 4+2 ) bit operations. For consistency with [9], we can then note that as > 1, we can write = log 2)8+1). Thus, there is a polynomial, R, in one variable such that the runtime of this algorithm is boune by Rm p ) bit operations. In the ense polynomial moel, the polynomial f has input size O log q), so this algorithm oes not have polynomial runtime with respect to the input length. This algorithm has runtime that is exponential in the egree of the polynomial,, an polynomial in m an p. As a note, ha we aopte the spaces constructe in prior works[16, 2], we woul have then require x i x j for i j. The stanar approach to representing such inequalities is the Rabinovich trick. To use this trick, we woul have introuce an aitional variable, say y, an the aitional equation y i<jx j x i ) = 1. This is a egree k 2) + 1 polynomial, which woul have le to an equation corresponing to 8) of at least egree k 2) + 2 with 2k + 1 variables, which woul have increase the work factor of the algorithm significantly. 6. Open Problems The algorithm presente relies on Lauer-Wan, which is intene to calculate the number of F q -rational points associate with general varieties. We use this on a polynomial of a very special form. As such, it may be possible to get a consierably more efficient algorithm by exploiting symmetry in the resulting Newton polytope. Though value sets of polynomials appear to be closely relate to zero sets, they are not as well stuie. There are many interesting open problems about value sets. The most important one is to fin a counting algorithm with running time log q) O1), that is, a eterministic polynomial-time algorithm in the ense moel. It is not clear if this is always possible. Our result affirmatively solves this problem for fixe if characteristic p is reasonably small. We conjecture that the same result is true for fixe an all characteristic p. For the complexity sie, can one prove that the counting problem for sparse polynomials in prime-orer finite fiels is har? Can one prove that the counting problem for ense input moel is har for general egree? Acknowlegment: We thank Dr. Tsuyoshi Ito for pointing out reference [4] to us.

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