Linear recurrences and asymptotic behavior of exponential sums of symmetric boolean functions

Transcription

1 Linear recurrences and asyptotic behavior of exponential sus of syetric boolean functions Francis N. Castro Departent of Matheatics University of Puerto Rico, San Juan, PR Luis A. Medina Departent of Matheatics University of Puerto Rico, San Juan, PR Subitted: Jan 28, 2011; Accepted: May 13, 2011; Published: May 25, 2011 Matheatics Subject Classification: 11T23, 05E05 Dedicated to Doron Zeilberger on the occasion of his 60th birthday Abstract In this paper we give an iproveent of the degree of the hoogeneous linear recurrence with integer coefficients that exponential sus of syetric Boolean functions satisfy. This iproveent is tight. We also copute the asyptotic behavior of syetric Boolean functions and provide a forula that allows us to deterine if a syetric boolean function is asyptotically not balanced. In particular, when the degree of the syetric function is a power of two, then the exponential su is uch saller than 2 n. Keywords: Exponential sus, recurrences, Cusick et al. Conjecture for eleentary balanced syetric boolean functions 1 Introduction Boolean functions are one of the ost studied objects in atheatics. They are iportant in any applications, for exaple, in the design of strea ciphers, block and hash functions. These functions also play a vital role in cryptography as they are used as filter and cobination generator of strea ciphers based on linear feed-back shift registers. The the electronic journal of cobinatorics 18(2 (2011, #P8 1

2 case of boolean functions of degree 2 has been intensively studied because of its relation to bent functions (see [11], [1]. One can find any papers and books discussing the properties of boolean functions (see [5], [9], [2] and [6]. The subject can be studied fro the point of view of coplexity theory or fro the algebraic point of view as we do in this paper, where we copute the asyptotic behavior of exponential sus of syetric boolean functions. The correlation between two Boolean functions of n inputs is defined as the nuber of ties the functions agree inus the nuber of ties they disagree all divided by 2 n, i.e., C(F 1, F 2 = 1 ( 1 F 1(x 1,...,x n+f 2 (x 1,...,x n. (1.1 2 n x 1,...,x n {0,1} In this paper we are interested in the case when F 1 and F 2 are syetric boolean functions. Without loss of generality, we write C(F instead of C(F 1, F 2, where F is a syetric boolean function. In [4], A. Canteaut and M. Videau studied in detail syetric boolean functions. They established a link between the periodicity of the siplified value vector of a syetric Boolean function and its degree. They also deterined all balanced syetric functions of degree less than or equal to 7. In [13], J. von zur Gathen and J. Rouche found all the balanced syetric boolean functions up to 128 variables. In[3], J. Cai et. al. coputed a closed forula for the correlation between any two syetric Boolean functions. This forula iplies that C(F satisfies a hoogeneous linear recurrence with integer coefficients and provides an upper bound for the degree of the inial recurrence of this type that C(F satisfies. In this paper we give an iproveent to the degree of the inial hoogeneous linear recurrence with integer coefficients satisfying by C(F. In particular, our lower and upper bounds are tight in any cases. Also, in the case of an eleentary syetric function we provide the inial hoogeneous linear recurrence. We also copute the asyptotic value of C(F. In particular, we give infinite failies of boolean functions that are asyptotically not balanced, i.e., li n C(F 0. In [7], T. Cusick et al. conjectured that there are no nonlinear balanced eleentary syetric polynoials except for the eleentary syetric boolean function of degree k = in l 1 variables, where r and l are any positive integers. In this paper, we prove that li C(σ n,k = 2w2(k 1 1, (1.2 n 2 w 2(k 1 where σ n,k is the eleentary syetric polynoial of degree k in n variables and w 2 (k is the su of the binary digits of k. Note that this iplies that Cusick et al. s conjecture holds for sufficiently large n. In particular, an eleentary syetric function is asyptotically not balanced if and only if its degree is not a power of 2. In [8], Cusick et al. presented soe progress on proving this conjecture. In particular, they presented the following stronger version of their conjecture: If n 2(k 1, where k is fixed and w 2 (k 6, then C(F > 1/2. Forula (1.2 iplies that this holds for sufficiently large n when w 2 (k 3. the electronic journal of cobinatorics 18(2 (2011, #P8 2

3 When the asyptotic value of C(F is zero, we copute the asyptotic values of 1 λ n x 1,...,x n {0,1} ( 1 F (x 1,...,x n, (1.3 where λ and λ are the roots with the biggest odulus of the characteristic polynoial associated to the exponential su of F. We prove that the coefficient of λ is not identically zero and obtain inforation about the spectru of F (X 1,..., X n. In particular, its liit is a periodic function in n. 2 Preliinaries Let F be the binary field, F n = {(x 1,..., x n x i F, i = 1,..., n}, and F (X = F (X 1,..., X n be a polynoial in n variables over F. The exponential su associated to F over F is: S(F = x F n ( 1 F (x. (2.1 Note that S(F = 2 n C(F. A boolean function F (X is called balanced if S(F = 0. This property is iportant for soe applications in cryptography. P. Sarkar and S. Maitra [12] found a lower bound for the nuber of syetric balanced boolean functions. In particular, in the case that n 13 is odd and n + 3 is a perfect square, this nuber is bigger than or equal to 2 (n+1/2 + 2 (n+1/2 3. In this paper we study exponential sus associated to syetric boolean functions F. Any syetric function is a linear cobination of eleentary syetric polynoials, thus we start with exponential sus of eleentary syetric polynoials. Let σ n,k be the eleentary syetric polynoial in n variables of degree k. For exaple, σ 4,3 = X 1 X 2 X 3 + X 1 X 4 X 3 + X 2 X 4 X 3 + X 1 X 2 X 4. (2.2 Fix k 2 and let n vary. Consider the sequence of exponential sus {S(σ n,k } n N where S(σ n,k = ( 1 σ n,k(x 1,,x n. (2.3 x 1,,x n F Define A j to be the set of all (x 1,, x n F n with exactly j entries equal to 1. Clearly, A j = ( n j and σn,k (x = ( j k for x Aj. Therefore, n ( S(σ n,k = ( 1 (j k n. (2.4 j j=0 In general, if 1 < k 2 < < k s are fixed integers, then n S(σ n,k1 + σ n,k2 + + σ n,ks = ( 1 ( j +( j k 2 + +( j ( n ks. (2.5 j j=0 the electronic journal of cobinatorics 18(2 (2011, #P8 3

4 Rear Note that the su on the right hand side of (2.5 akes sense for values of n less than k s, while S(σ n,k1 + + σ n,ks does not. However, throughout the paper we let S(σ n,k1 + + σ n,ks to be defined by the su in (2.5, even for values of n less than k s. 3 The recurrence Coputer experientation suggests that for fix 1 < < k s, the sequence {S(σ n,k1 + + σ n,ks } n N satisfies a hoogeneous linear recurrence with integer coefficients. For exaple, if we consider {S(σ n,7 } n N and type FindLinearRecurrence[Table[Su[((-1^Binoial[,7]* into Matheatica 7, then it returns Binoial[n,],{,0,n}],{n,1,30}]] {8,-28,56,-70,56,-28,8}. This suggests that {S(σ n,7 } n N satisfies the recurrence x n = 8x n 1 28x n x n 3 70x n x n 5 28x n 6 + 8x n 7. (3.1 If we continue with these experients, we arrive to the observation that if r = log 2 (k s + 1, then {S(σ n,k1 + + σ n,ks } n N sees to satisfy the recurrence ( 2 x n = ( 1 1 r x n. (3.2 =1 This result can be proved using eleentary achinery. The idea is to use the fact that if r = log 2 (k s + 1, then ( ( j + i2 r j (od 2 (3.3 k for all non-negative integers i and = 1, 2,, s, to show inductively that the faily of sequences a n,r,i = ( n ( n =, (3.4 j + i j j k j i (od i = 0, 1, 1 satisfies the sae recurrence (3.2. However, we should point out the fact that {S(σ n,k1 + +σ n,ks } n N satisfies (3.2 is a consequence of the following theore of J. Cai et al. [3]. Theore 3.1 Fix 1 < < k s and let r = log 2 (k s + 1. The value of the exponential su S(σ n,k1 + + σ n,ks is given by n S(σ n,k1 + + σ n,ks = ( 1 ( i + +( i ( n ks i i=0 = c 0 (,, k s 2 n + j=1 c j (,, k s (1 + ζ j n, (3.5 the electronic journal of cobinatorics 18(2 (2011, #P8 4

5 ( π where ζ j = exp 1 j and 1 c j (,, k s = 1 2r 1 ( 1 ( i i=0 + +( i The proofs of Cai et al. rely on linear algebra. They wrote i=0 ks ζ i j. (3.6 S(σ n,k1 + + σ n,ks = ( 1 ( i + +( i ks a n,r,i, (3.7 and used the eleentary identity ( ( n n 1 = + k k to find a recurrence for a n,r = a n,r,1 a n,r,2. a n,r,2 ( n 1, (3.8 k 1 (3.9 of the for a n,r = Ma n 1,r, for soe atrix M. Finding the eigenvalues and corresponding eigenvectors of M, they were able to solve this recurrence and prove Theore 3.1. See [3] for ore details. Fro Theore 3.1 it is now evident that {S(σ n,k1 + + σ n,ks } n N satisfies (3.2. Moreover, the roots of the characteristic polynoial associated to the linear recurrence (3.2 are all different and the polynoial is given by P r (x = ( 2 ( 1 r x 2 = (x 2Φ 4 (x 1Φ 8 (x 1 Φ (x 1, (3.10 where Φ (x represents the -th cyclotoic polynoial Φ (x = (x ζ. (3.11 ζ =1 priitive Even though {S(σ n,k1 + + σ n,ks } n N satisfies (3.2, in any instances (3.2 is not the inial hoogeneous linear recurrence with integer coefficients that {S(σ n,k1 + + σ n,ks } n N satisfies. For exaple, {S(σ n,3 + σ n,5 } n N satisfies (3.1, but its inial recurrence is x n = 6x n 1 14x n x n 3 10x n 4 + 4x n 5. (3.12 In the next section we use Theore 3.1 to give soe iproveents on the degree of the inial linear recurrence associated to {S(σ n,k1 + + σ n,ks } n N. the electronic journal of cobinatorics 18(2 (2011, #P8 5

6 4 On the degree of the recurrence relation Now that we are equipped with equation (3.6, we ove to the proble of reducing the degree of the recurrence relation that our sequences of exponential sus satisfy. The idea behind our approach is very siple. Consider all roots 1 + ζ s of Φ 2 t+1(x 1 where 1 t r 1. We know that (1 + ζ n appears in (3.5. If we show that the coefficient that corresponds to (1 + ζ n is zero for each 1 + ζ, then we reduce the degree of the characteristic polynoial, and therefore the degree of the recurrence, by 2 t. However, note that Φ 2 t+1(x 1 is irreducible over Q (according to Eisenstein s criterion on Φ 2 t+1(x 1 with Φ 2 t+1(x = x 2t + 1, see [10]. Therefore, the coefficients related to the roots of Φ 2 t+1(x 1 are either all zeros or all non-zeros. In view of (3.6, this can be deterined by checking whether or not the su + +( ( π 1 ks exp 2 t (4.1 is zero. First, we discuss the case of the exponential su of one eleentary syetric polynoial, i.e. {S(σ n,k } n N. We start with the following eleentary result. Lea 4.1 (Lucas theore Let n be a natural nuber with 2-adic expansion n = 2 a a a l. The binoial coefficient ( n k is odd if and only if k is either 0 or a su of soe of the 2 a i s. Proof: Recall that (1 + x x 2 (od 2 for all non-negative integer, thus (1 + x n (1 + x 2a 1 (1 + x 2a 2 (1 + x 2a l (od 2. Note that the coefficient of x k in (1 + x 2a 1 (1 + x 2a 2 (1 + x 2a l is 1 if and only if k = 0 or a su of soe of the 2 a i s. The next result is an iediate consequence of the above lea. Corollary 4.2 Fix a natural nuber k. Suppose its 2-adic expansion is k = 2 a a 2 ( a l. A natural nuber is such that k is odd if and only if has a 2-adic expansion of the for = k + δ i 2 i (4.2 where δ i {0, 1}. 2 i {2 a 1,2 a 2,,2 a l} Reark 2 Let k 1 be an integer with 2-adic expansion k = 2 a a l. Suppose {0, 1, 2, 3,, 1} is such that ( k is odd. Note that Corollary 4.2 iplies = k + δ 1 2 b 1 + δ 2 2 b δ t 2 b f, (4.3 where {2 b 1, 2 b 2,, 2 b f } = {1, 2, 2 2,, 1 }\{2 a 1, 2 a 2,, 2 a l }. We now proceed to show which coefficients c j (k are zero. We start with c 0 (k. the electronic journal of cobinatorics 18(2 (2011, #P8 6

7 Lea 4.3 Suppose k 2 is an integer. Then, c 0 (k = 2w 2(k w 2(k 1, (4.4 where w 2 (k is the su of the binary digits of k. In particular, c 0 (k = 0 if and only if k is a power of two. Proof: Recall fro Theore 3.1 that where r = log 2 (k + 1. Re-write c 0 (k as c 0 (k = 1 2r 1 ( 1 ( k, c 0 (k = 1 ( 2 N 1, (4.5 where N = { {0, 1, 2, 3,, 1} : ( } is odd. k (4.6 Note that (4.3 iplies that the cardinality of N is w2(k. A siple calculation yields the result. Reark 3 In [7], T. Cusick et al. conjectured that there are no nonlinear balanced eleentary syetric polynoials except for the eleentary syetric boolean function of degree k = in l 1 variables, where r and l are any positive integers. Note that Lea 4.3 iplies that Cusick et al. conjecture holds for sufficiently large n. In particular, Lea 4.3 shows that if k is not a power of two, then σ n,k is not balanced for sufficiently large n. We say that σ n,k1 + + σ n,ks is asyptotically not balanced if S(σ n,k1 + + σ n,ks li 0. (4.7 n 2 n In the case that S(σ n,k1 + + σ n,ks li = 0, (4.8 n 2 n we cannot conclude anything about whether or not S(σ n,k1 + + σ n,ks is balanced for soe values of n. For instance, but S(σ n,2 + σ n,5 0. S(σ n,2 + σ n,5 li = 0, (4.9 n 2 n the electronic journal of cobinatorics 18(2 (2011, #P8 7

8 Consider now the coefficients c j (k = 1 2r 1 ( ( 1 ( k π 1j exp with j > 0. Fro Theore 3.1 we know each c j (k is the coefficient of (1 + ζ j n where 1 + ζ j is a root of Φ 2 t+1(x 1 for soe t = 1, 2,, 1. Lea 4.4 Let k 2 be an integer with 2-adic expansion k = 2 a a l. Then c j (k = 0 if and only if it is the coefficient of (1 + ζ n, where 1 + ζ is a root of Φ 2 b+1(x 1 and b a i for all i = 1,, l, i.e. 2 b does not appear in the 2-adic expansion of k. Proof: Recall that to show that the coefficients of the roots of Φ 2 t+1(x 1 are zero is equivalent to show that 1 ( ( 1 ( k π 1 exp = 0. (4.10 If {2 b 1,, 2 b f } = {1, 2, 2 2,, 1 }\{2 a 1,, 2 a l }, then equation (4.3 iplies, ( ( 1 ( k π 1 exp = 2 2 t ( π 1k = 2 exp 2 t (δ 1,,δ f F f 2 2 t (δ 1,,δ f F f 2 ( π Thus, (4.10 holds if and only if exp 1 is a root of 2 t (δ 1,,δ f F f 2 ( π 1 exp (k + δ 2 t 1 2 b δ t 2 b f ( π 1 exp (δ 2 t 1 2 b δ t 2 b f. (4.11 x δ 12 b 1 + +δ t2 b f. (4.12 Consider first t = b 1. If we set δ 1 = 0 in the last su of (4.11, then we have ( π 1 exp (δ 2 2 b δ t 2 b f. (4.13 (δ 2,,δ f F f 1 2 (δ 2,,δ f F f b 1 However, if we set δ 1 = 1, then we have ( π 1 exp (δ 2 2 b δ t 2 b f. ( b 1 We conclude that (4.10 holds for t = b 1, i.e. the 2 b 1 coefficients related to the roots of Φ 2 b 1 +1(x 1 are zero. Repeat this arguent with t = b 2,, b f to conclude that the the electronic journal of cobinatorics 18(2 (2011, #P8 8

9 coefficients related to the roots of Φ 2 b i +1(x 1, i = 1,, f are zero. Since (4.12 is of degree d = 2 b b f, then only d of the coefficients cj (k can be zero. Since we already found d coefficients that are zero, then we conclude that these are all of the. The clai follows. Leas 4.3 and 4.4 are put together in the following theore. The function ɛ(n used in the theore is defined as { 0, if n is a power of 2, ɛ(n = (4.15 1, otherwise. Theore 4.5 Let k be a natural nuber and P k (x be the characteristic polynoial associated to the inial linear recurrence with integer coefficients that {S(σ n,k } n N satisfies. Let k = 2 k/ We know k has a 2-adic expansion of the for k = a a a l, (4.16 where the last exponent is given by a l = log 2 ( k. Then P k (x equals l (x 2 ɛ(k Φ 2 a j +1(x 1. (4.17 j=1 In particular, the degree of the inial linear recurrence that {S(σ n,k } n N satisfies is equal to 2 k/2 + ɛ(k. Theore 4.5 can be generalized to the case {S(σ n,k1 + + σ n,ks } n N. Define the OR operator on F 2 as 0 0 = 0 ( = = = 1. Extend to N by letting n be the natural nuber obtained by applying coordinatewise to the binary digits of n and. For exaple, and 4 6 = ( ( (4.19 = ( ( ( = = ( ( (4.20 = ( ( ( ( = 11. Next is a generalization of Theore 4.5. the electronic journal of cobinatorics 18(2 (2011, #P8 9

10 Theore 4.6 Let 1 < k 2 < < k s be fixed integers and P k1,,k s (x be the characteristic polynoial associated to the inial linear recurrence with integer coefficients that {S(σ n,k1 + + σ n,ks } n N satisfies. Let k = 2 ( k s / We know k has a 2-adic expansion of the for k = a a a l, (4.21 where the last exponent is given by a l = log 2 ( k. Then P k1,,k s (x divides the polynoial (x 2 l Φ 2 a j +1(x 1. (4.22 j=1 Proof: The proof is siilar to the one of Lea 4.4. Let k = 2 ( k s /2 + 1 and r = log 2 ( k + 1. Define { ( ( } N = {1, 2, 3,, 1} : + + is odd. (4.23 Suppose 2 b {2, 2 2,, 1 } is such that 2 b does not appear in the 2-adic expansion of k. We will show that + +( ( π 1 ks exp = 0, (4.24 which iplies that the coefficients related to the roots of x 2b + 1 are all zero. Observe that + +( ( π 1 ks exp = 2 ( π 1 exp. ( b 2 b N Suppose N is such that 2 b does not appear in the 2-adic expansion of. Note that equation (4.3 iplies + 2 b N. Thus, the sae arguent as in (4.13 and (4.14 iply that (4.24 is true. Hence, the clai follows. The following exaple presents a case when Theore 4.6 is tight. Exaple 4.7 Consider = 6 and k 2 = 17. Note that 2 (6 17/2 + 1 = 23 = In this case, the characteristic polynoial associated to {S(σ n,6 +σ n,17 } n N is P 6,17 (x = (x 2Φ 4 (x 1Φ 8 (x 1Φ 32 (x 1. This is the best case scenario of Theore 4.6, i.e we have equality rather than just divisibility. Also, note that in this case the recurrence given by Theore 3.1 is of degree 31, while the inial linear recurrence is of degree 23. The next exaple presents a case in which Theore 4.6 iproves the degree of the hoogeneous linear recurrence provided by Theore 3.1. However it did not provide the inial degree of the recurrence. 2 b k s the electronic journal of cobinatorics 18(2 (2011, #P8 10

11 Exaple 4.8 Consider = 3, k 2 = 5, and k 3 = 17. We have 2 (3 5 17/2 +1 = 23. In this case, the characteristic polynoial of the inial recurrence is P 3,5,17 (x = (x 2Φ 32 (x 1. It divides (x 2Φ 4 (x 1Φ 8 (x 1Φ 32 (x 1 as Theore 4.6 predicted, but are clearly not equal. The factors Φ 4 (x 1 and Φ 8 (x 1 do not appear in P 3,5,17 (x. This eans that the coefficients c j (3, 5, 17 related to the roots of Φ 4 (x = x and Φ 8 (x = x are zero. However, since 2 and 4 appear in the 2-adic expansion of 23, then Theore 4.6 cannot detect this. We now provide bounds on the degree of the inial linear recurrence that {S(σ n,k1 + + σ n,ks } n N satisfies. We start with the following theore. Theore 4.9 Suppose 1 < < k s are integers. Let r = log 2 (k s + 1. Then Φ (x 1 divides P k1,,k s (x, the characteristic polynoial associated to {S(σ n,k1 + + σ n,ks } n N. Proof: Note that the theore will follow if we show that + +( ( π 1 ks exp 0. ( This is equivalent to showing that x 2r does not divide + +( ks x. (4.27 We present the core of our proof with a particular exaple. The general case will follow in a siilar anner. Consider the case = 3, k 2 = 5, and k 3 = 10. Then (4.27 equals, 1 + x + x 2 x 3 + x 4 x 5 + x 6 + x 7 + x 8 + x 9 x 10 + x 11 + x 12 x 13 x 14 x 15. (4.28 Look at the sign of x j for j = 8, 9,, 15. If x j and x j 8 have the sae sign, then leave the sign of x j as it is. Otherwise, change the sign of x j. After doing this, we get 1 + x + x 2 x 3 + x 4 x 5 + x 6 + x 7 + x 8 + x 9 + x 10 x 11 + x 12 x 13 + x 14 + x 15, (4.29 which equals (1 + x 8 (1 + x + x 2 x 3 + x 4 x 5 + x 6 + x 7. (4.30 Of course, in order to get (4.28 back, we need to add to (4.30 two ties the ters for which we changed their signs: (1 + x 8 (1 + x + x 2 x 3 + x 4 x 5 + x 6 + x 7 2x x 11 2x 14 2x 15. (4.31 This last polynoial equals (1 + x 8 (1 + x + x 2 x 3 + x 4 x 5 + x 6 + x 7 + 2x 8 ( x 2 + x 3 x 6 x 7. (4.32 the electronic journal of cobinatorics 18(2 (2011, #P8 11

12 In general, + +( ks x = (x 2r ( ( ks x +2x 2r 1 q(x, (4.33 where q(x is a polynoial of degree at ost 1 1. We conclude that x 2r does not divide (4.27 and so the clai follows. Corollary 4.10 Let 1 < < k s be integers. Let D be the degree of the inial hoogeneous linear recurrence with integer coefficients that {S(σ n,k1 + + σ n,ks } n N satisfies. Then 2 log 2 (ks D 2 ( k s / Proof: Note that the upper bound follows fro Theore 4.6 while the lower bound is a consequence of Theore 4.9. Note that Corollary 4.10 is an iproveent of Theore 3.1 with respect to the degree D of the inial hoogeneous linear recurrence with integer coefficients that {S(σ n,k1 + + σ n,ks } n N satisfies. Fro Theore 3.1 we can only infer that D 1, where r = log 2 (k s + 1. However, now we know that 2 log 2 (ks D 2 ( k s /2 + 1 and 2 ( k s / Also, exaple 4.7 shows that the upper bound of Corollary 4.10 can be attained. In the next theore, we show that when k s (the highest degree is a power of two, then the lower bound is tight. Theore 4.11 Suppose 1 < k 2 < < k s are fixed integers with k s = 1 a power of two. Let P k1,k 2,,2r 1(x be the characteristic polynoial associated to the inial linear recurrence that {S(σ n,k1 + σ n,k2 + + σ n, 1} n N satisfies. Then 2r 1 P k1,k 2,, 1(x = Φ ( 2 2r(x 1 = 2 + ( 1 r 1 x. (4.34 In particular, deg(p k1,k 2,, 1(x = 2r 1 = 2 log 2 (ks. Proof: The theore will follow if we show that c 0 (, k 2,, 1 = 0, and +( k 2 + +( for each j = 1, 2,, r 2, and +( k 2 + +( 1 exp 1 exp =1 ( π 1 = 0, ( j ( π 1 0. ( the electronic journal of cobinatorics 18(2 (2011, #P8 12

13 Fro Theore 4.9 we know that (4.36 holds true. Now, the coefficient c 0 (,, k s 1, 1 is zero if and only if + +( k s 1 +( r 1 2 = 0. (4.37 Fro (3.3 we see that the period of + +( k s 1 is a proper divisor of, so However, ( k s 1 = = 1 ( 1 ( { ( 1 ( r 1 2 1, if 2 = r 1 1 1, if ( k s 1. (4.38 (4.39 Thus, (4.37 holds and therefore c 0 (,, k s 1, 1 = 0. Siilarly, the periodicity of ( 1 ( ( + +( k s 1 and exp π 1 iplies 2 j 1 1 ( 1 ( ( + +( k s 1 π 1 exp = 2 j = 1 ( 1 ( So, (4.39 and (4.40 iply (4.35. This concludes the proof. + +( k s 1 exp ( π 1 2 j. (4.40 We conclude this section with the following result, which shows that when k s is a power of two, then, as n increases, S(σ n,k1 + + σ n,ks is uch saller than 2 n. Corollary 4.12 Suppose 1 < k 2 < < k s are fixed integers with k s = 1 a power of two. Then, for 0 j 1, c j (,, k s 1, 1 0 if and only if j is odd. In particular, c 0 (,, k s 1, 1 = 0. 5 Asyptotic behavior In this section we discuss the asyptotic behavior of {S(σ n,k1 + + σ n,ks } n N. Note that Theore 3.1 iplies that S(σ n,k1 + + σ n,ks li = c n 2 n 0 (,, k s = 1 ( 1 ( + +( ks. (5.1 Thus, we study c 0 (,, k s first. We already discussed the case of one eleentary syetric polynoial {S(σ n,k } n N, see (4.4: c 0 (k = 2w 2(k w 2(k 1. (5.2 the electronic journal of cobinatorics 18(2 (2011, #P8 13

14 For instance, we know that c 0 (k 0, and the equality holds if and only if k is a power of two. The ethod of inclusion-exclusion can be used to get a forula in the case that we have ore than one syetric polynoial. For exaple, in the case of two eleentary syetric polynoials {S(σ n,k1 + σ n,k2 } n N, we have c 0 (, k 2 = w 2( 2 1 w 2(k w 2( k 2 (5.3 The reader can check that this forula iplies c 0 (, k 2 0, with equality if and only if w 2 ( k 2 = w( + w 2 (k 2 and w 2 (k i = 1, where i = 1 or i = 2. We start the general case with the following lea. Lea 5.1 Suppose that 1 < k 2 < < k s are integers. Define { ( ( } N(k i1, k ij = {1, 2, 3,, 1} : + + is odd. (5.4 k i1 k ij Then, #N(, k 2,, k s = s #N(k i 2 i=1 +4 i 1 <i 2 # (N(k i1 N(k i2 i 1 <i 2 <i 3 # (N(k i1 N(k i2 N(k i2 (5.5 +( 1 s 1 2 s 1 # (N( N(k 2 N(k s. Proof: Note that ( ( + + k s (5.6 is odd exactly when the aount of odd suands is itself an odd nuber (trivial. In ters of our sets N(k i, this iplies that N(,, k s is obtained by including all the intersections of an odd aount of the sets N(k i, while excluding all the intersections of an even aount of the. For exaple, the case of four k s can be represented by the Venn diagra in Figure 1. In this case, we want to include the shaded regions and exclude the white ones. We start by adding #N( + #N(k #N(k s. Then, we proceed to take out all the intersections of two sets N(k i N(k j, i j. In this case, each of the has been added twice in our previous su. Therefore, to take the out, we should add 2#(N( N(k 2 2#(N( N(k 3 2#(N(k s 1 N(k s to the previous su. So, now we have s #N(k i 2 i=1 i 1 <i 2 #(N(k i1 N(k i2. (5.7 This takes care of all intersections of two sets. Now, we need to add all intersections of three sets #(N(k i1 N(k i2 N(k i3. Each of the have been added three ties by the the electronic journal of cobinatorics 18(2 (2011, #P8 14

15 Figure 1: Representation of the case of four k s. first su and subtracted six ties by the second su. Thus, in order to add the into the equation, we have to add each of the four ties to (5.7. By doing this we have s #N(k i 2 i=1 Continue in this anner and use the identity to get the result. i 1 <i 2 #(N(k i1 N(k i2 ( i 1 <i 2 <i 3 #(N(k i1 N(k i2 N(k i3. j 1 ( { j 2 ( 1 i 1 2 i 1 = j 1, if j is even i 2 j 1 + 1, if j is odd, i=1 (5.9 Now that we have the above lea, we are ready to state our general forula for c 0 (,, k s. Theore 5.2 Suppose that 1 < k 2 < < k s are integers. Then s c 0 (,, k s = w 2(k i + i=1 Proof: Let r = log 2 (k s + 1. Note that i 1 <i w2(ki1 ki2 i 1 <i 2 <i w2(ki1 ki2 ki3 + + ( 1 s 2 s w2(k1 k2 ks. (5.10 c 0 (,, k s = 2r #N(,, k s. (5.11 Consider the integer k i. Suppose its 2-adic expansion is k i = 2 a a a l. Then, by Corollary 4.2 we know that 0 1 is such that ( k i is odd precisely when = k + δ 1 2 b 1 + δ 2 2 b δ t 2 bt, (5.12 the electronic journal of cobinatorics 18(2 (2011, #P8 15

16 where {2 b 1, 2 b 2,, 2 bt } = {1, 2, 2 2,, 1 }\{2 a 1, 2 a 2,, 2 a l } and δj is either 0 or 1. This iplies that #N(k i = w 2(k i. (5.13 Also, (5.12 iplies that if i j, then and in general, if 1 i 1 < i 2 < i t s, then #(N(k i N(k j = w 2(k i k j, (5.14 #(N(k i1 N(k i2 N(k t = w 2(k i1 k i2 k it. (5.15 Equations (5.11 and (5.15 and Lea 5.1 iply the theore. Exaple 5.3 Suppose = 7, k 2 = 9, k 3 = , and k 4 = , then c 0 (, k 2, k 3, k 4 = 1/4. In other words, S(σ n,k1 + σ n,k2 + σ n,k3 + σ n,k4 li = 1 n 2 n 4. (5.16 Exaple 5.4 Suppose = 31, k 2 = and k 3 = , then c 0 (, k 2, k 3 = 45/128. We now use the above theore to provide a faily of syetric polynoials that are not balanced for sufficiently large n. Suppose that and k 2 are two positive integers. We say that k 2 if each power of two appearing in the 2-adic expansion of also appears in the 2-adic expansion of k 2. For exaple, because 10 = and 14 = Corollary 5.5 Suppose that k 2 k s are positive integers. Then, s/2 c 0 (,, k s = 1 (s2 1 w 2(k s (2 w 2(k 2j k 2j w 2(k 2j. (5.17 Here (s equals 0 if s is even and 1 otherwise; in other words, (s = s od 2. j=1 Proof: This follows directly fro Theore 5.2 and the equality w 2 (k i = w 2 (k i k i 1 + w 2 (k i 1. Theore 5.6 Suppose that k 2 k s are positive integers. Then, c 0 (,, k s > 0. (5.18 In particular, {S(σ n,k1 + + σ n,ks } n N is asyptotically not balanced. the electronic journal of cobinatorics 18(2 (2011, #P8 16

17 Proof: Corollary 5.5 iplies that c 0 (,, k s > 0 if and only if s/2 (s w 2(k 2j k 2j 1 1 w 2(k s 2 w 2(k 2j < 1 2. (5.19 j=1 Since k 2 k s, then we have the inequality and the equality Note that (5.20 and (5.21 iply w 2 (k i 1 + w 2 (k i 1 (5.20 w 2 (k i = w 2 (k i k i 1 + w 2 (k i 1. (5.21 s/2 (s w 2(k 2j k 2j 1 1 w 2(k s 2 w 2(k 2j = (s w 2(k s 2 1 w 2( 2 + w 2(k 2 j=1 This finishes the proof. s/2 j=2 2 w 2(k 2j k 2j w 2(k 2j (s w 2(k s 2 1 w 2( s/2 1 w 2(k 2 2 w 2(k 2 2 2j 3 j=2 < 1 2 w 2( 1 2 w 2(k = 1 2 w 2( + ( w 2(k 2 j= j < 1 w 2(k 2 2. (5.22 We now turn our attention to the case when c 0 (,, k s = 0, which happens if and only if 2 is not a root of the characteristic polynoial associated to {S(σ n,k1 + + σ n,ks } n N. In this case S(σ n,k1 + + σ n,ks li = 0, (5.23 n 2 n because S(σ n,k1 + + σ n,ks is exponentially saller than 2 n. However, aside fro the size of S(σ n,k1 + + σ n,ks with respect to 2 n, knowing that c 0 (,, k s = 0 does not give a real sense of the behavior of S(σ n,k1 + + σ n,ks as n increases. Now, when c 0 (,, k s = 0, the biggest odulus of the roots of the characteristic polynoial associated to {S(σ n,k1 + +σ n,ks } n N is 2 cos(π/. This odulus is obtained at the roots 1 + exp ( π 1/( 1 and 1 + exp ( π 1/( 1. Thus, as n increases, approaches c 1 (,, k s S(σ n,k1 + + σ n,ks (2 cos(π/ n (5.24 ( ( π (1 + exp n ( ( n r 1 c2 (,, k s 1 + exp π 1 1. (5.25 (2 cos(π/ n the electronic journal of cobinatorics 18(2 (2011, #P8 17

18 ( π Let c i = c i (,, k s and ξ = 1 + exp 1 then ( 1 + exp ± π 1 ( π = 2 cos exp 1 c 1 ξ n + c ξn 1 = (2 cos(π/2r n + +( ks ( ( (n 2π 1 exp = (2 cos(π/2r n 1. Note that c 1 2 = c 1. Since ( ± π 1 ( + exp (n 2π 1, ( ( ( (n 2π ks cos Note that (5.27 is not the zero function, because + +( ( (n 2π ks cos is the real part of ( π 1n exp + +( ( ks exp π (5.27 (5.28 (5.29 and fro the proof of Theore 4.10 we know that the su in (5.29 is a non-zero constant. We conclude that S(σ n,k1 + + σ n,ks = (2 cos(π/2r n + +( ( (n 2π ks cos Reark 4 Note that + O 1 (( ( π n 2 cos. ( ( ( (n 2π ks cos, (5.31 is not identically zero, regardless if c 0 (,, k s is zero. Hence, the asyptotic expansion of S(σ n,k1 + + σ n,ks is S(σ n,k1 + + σ n,ks = c 0 2 n + (2 cos(π/2r n + +( ( (n 2π ks cos 1 (( ( π n + O 2 cos. ( the electronic journal of cobinatorics 18(2 (2011, #P8 18

19 Reark 5 If c 0 (,, k s = 0, then we define the function Error n (,, k s as Error n (,, k s = S(σ n, + + σ n,ks (2 cos(π/ n 1 ( 1 ( 1 + +( ks cos ( (n 2π. (5.33 We point out that (5.32 is a consequence of Theore 3.1. Moreover, it is not hard to continue refining this forula and re-write S(σ n,k1 + + σ n,ks copletely in ters of cosine (which will be a restateent of Theore 3.1. However, we stress that we are proving that + +( ( (n 2π ks cos (5.34 is not identically zero, so the second ter of the asyptotic expansion (5.32 is always present. By the discussion presented in section 4 we know that in any cases soe of the c i s, i 1, 1, are zero. Because of this, we write the asyptotic expansion of S(σ n,k1 + + σ n,ks as in (5.32 and decide not to continue with a refineent of it. Exaple 5.7 Consider = 5, k 2 = 9, and k 3 = 12. The reader can check that in this case c 0 (5, 9, 12 = 0. By (5.30 we know that as n increases, S(σ n,5 + σ n,9 + σ n,12 (2 cos(π/16 n (5.35 approaches 1 15 ( ( 1 ( 5+( 9+( 12 (n 2π cos = ( ( ( ( ( nπ 2 1 cos ( ( nπ sin Here, we siplified the expression using Matheatica. In Figure 2 you can see a graphical representation of this. The dots represents (5.35 and the curve is the one given by (5.36. In Table 1 you can see the error ter. Exaple 5.8 Siilarly, consider = 2, k 2 = 4, k 3 = 11, and k 4 = 35. In this case we also have c 0 (2, 4, 11, 35 = 0, so by (5.30 we know that as n increases, approaches S(σ n,2 + σ n,4 + σ n,11 + σ n,35 (2 cos(π/64 n (5.37 ( ( 1 ( 2+( 4+( 11+( 35 (n 2π cos 16. (5.38 In Figure 3 you can see a graphical representation of this, where the dots represents (5.37 and the curve is (5.38. In Table 2 you can see the error ter. the electronic journal of cobinatorics 18(2 (2011, #P8 19

20 Figure 2: Graphical representation of the asyptotic behavior when = 5, k 2 = 9, and k 3 = Table 1: The error ter for = 5, k 2 = 9, and k 3 = 12. n Error n (5, 9, Figure 3: Graphical representation of the asyptotic behavior when = 2, k 2 = 4, k 3 = 11, and k 4 = Table 2: The error ter for = 2, k 2 = 4, k 3 = 11, and k 3 = 35. n Error n (2, 4, 11, Acknowledgents. We would like to thank Claude Carlet and Doron Zeilberger for a careful reading of the anuscript and for all their helpful suggestions. A special thank goes to the referee for pointing out relevant references to this work and for iproving the presentation of this anuscript. the electronic journal of cobinatorics 18(2 (2011, #P8 20

21 References [1] C. Bey and G. M. Kyureghyan, On Boolean functions with the su of every two of the being bent, Des. Codes Cryptogr., 49, pp , [2] R. E. Canfield, Z. Gao, C. Greenhill, B. McKay and R. W. Robinson, Asyptotic enueration of correlation-iune Boolean functions, Cryptogr. Coun., 2, pp , 2010 [3] J. Y. Cai, F. Greeen, and T. Thierauf. On the Correlation of Syetric Functions. Theory of Coputing Systes, 29, pp , [4] A. Canteaut and M. Videau, Syetric Boolean Functions, IEEE Transactions on Inforation Theory, 51, pp , [5] C. Carlet, Boolean Functions for Cryptography and Error Correcting Codes, Boolean Models and Methods in Matheatics, Coputer Science, and Engineering, Eds. Yves Craa and Peter Haer, Cabridge University Press, [6] C. Carlet, X. Zeng, C. Li, and L. Hu, Further properties of several classes of Boolean functions with optiu algebraic iunity, Des. Codes Cryptogr., 52,pp , [7] T. Cusick, Y. Li, and P. Stănică, Balanced Syetric Functions over GF (p. IEEE Trans. on Inforation Theory, 54, pp , [8] T. Cusick, Y. Li, and P. Stănică, On a conjecture for balanced syetric Boolean functions, J. Math. Crypt., 3, pp. 1-18, [9] T. Cusick and P. Stănică, Cryptographic Boolean Functions and Applications, Acadeic Press, [10] D.J.H. Garling, A Course in Galois Theory, Cabridge University Press, [11] O. S. Rothaus, On Bent functions. J. Cobin. Theory Ser. A, 20, pp , [12] P. Sarkar and S. Maitra, Balancedness and correlation iunity of syetric Boolean functions. Discrete Matheatics, 307, pp , [13] J. von zur Gathen and J. Roche, Polynoial with Two Values, Cobinatorica, 17, pp , the electronic journal of cobinatorics 18(2 (2011, #P8 21