Complete characterization of generalized bent and 2 k -bent Boolean functions

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1 Complete characterization of generalized bent and k -bent Boolean functions Chunming Tang, Can Xiang, Yanfeng Qi, Keqin Feng 1 Abstract In this paper we investigate properties of generalized bent Boolean functions and k -bent (i.e., negabent, octabent, hexadecabent, et al.) Boolean functions in a uniform framework. We generalize the work of Stǎnicǎ et al., present necessary and sufficient conditions for generalized bent Boolean functions and k -bent Boolean functions in terms of classical bent functions, and completely characterize these functions in a combinatorial form. The result of this paper further shows that all generalized bent Boolean functions are regular. Index Terms Boolean functions, Walsh-Hadamard transforms, k -bent functions, generalized bent functions, cyclotomic fields I. INTRODUCTION Throughout this paper, let Z k be the ring of integers modulo k, Z n be the n-dimensional vector space over Z, and C the field of complex numbers, where k, n are positive integers. If x = (x 1,, x n ) and y = (y 1,, y n ) are two vectors in Z n, we define the scalar (or inner) product by x y = x 1 y x n y n. For a complex z = a + b 1, the absolute value of z is z = a + b and z = a b 1 denotes the complex conjugate of z, where a and b are real numbers. A function from Z n to Z k is called a generalized Boolean function on n variables [11], whose set is denote by GB k n. If k, a function in GB 1 n is a classical Boolean function on n variables. An important tool in the analysis of generalized Boolean function is the (generalized) Walsh-Hadamard transform, which is the function H g : Z n C, defined by H g (u) = n u x g(x), (1) k where g GB k n, u Z n, and k of such g is x Z n = e π 1 k is the complex k -primitive root of unity. The inverse Walsh-Hadamard transform g(x) k = n H g (u) u x. () u Z n The function g is said to be a generalized bent function if H g (u) for any u Z n. A generalized bent function g is regular if there exists some generalized Boolean function g satisfying H g (u) = g (u) for any u Z n k. Such function g is called the dual of g. From Equation (), the dual g of a regular generalized bent function g is also regular. When k, the generalized bent function g is just classical Boolean bent functions which have been introduced by Rothaus [8]. These bent functions only exist for n even. If n is odd, a function g GB 1 n is said to be a semibent function if and only if H g (u) 0, } for any u Z n. Such Boolean functions have been extensively studied, as they have important applications in cryptograph(stream ciphers [1]), sequences [6] and coding theory (Reed-Muller codes [3]). We refer to [], [4], [5], [16] for more on cryptographic Boolean function and bent functions. When k =, generalized Boolean functions in GB n were studied by Schmidt [10]. Solé and Tokareva [11] discussed the connection between generalized bent functions in GB n and bent functions in GB 1 n. Stǎnicǎ et al. [13] characterized generalized bent Boolean functions in GB n and GB 3 n. For Boolean functions f GB 1 n, the authors in [9], [7], [14], [15] introduced and investigated another transform which was called nega-hadamard transform. Later, Stǎnicǎ [1] generalized their results and proposed the k -Hadamard transform of f defined by H (k ) f (u) = n f(x)+u x wt(x), (3) k x Z n This work was supported by the National Natural Science Foundation of China (Grant No , ). C. Tang also acknowledges support from 14E013 and CXTD014-4 of China West Normal University. Y. Qi also acknowledges support from KSY of Hangzhou Dianzi University. The research of K. Feng was supported by NSFC No , and the Tsinghua National Lab. for Information Science and Technology. C. Tang is with School of Mathematics and Information, China West Normal University, Nanchong, Sichuan, 63700, China. tangchunmingmath@163.com C. Xiang is with the College of Mathematics and Information Science, Guangzhou University, Guangzhou , China. cxiangcxiang@hotmail.com Y. Qi is with School of Science, Hangzhou Dianzi University, Hangzhou, Zhejiang, , China. qiyanfeng07@163.com. K. Feng is with the Department of Mathematical Sciences, Tsinghua University, Beijing, , China. kfeng@math.tsinghua.edu.cn.

2 where u Z n and wt(x) = #i : 1 i n, x i 0} is the Hamming weight of x. A function f is called a k -bent function if the k -Hadamard transform are flat in absolute value, that is, H (k ) f (u), for any u Z n. We call a function f a strong k -bent function if and only if f is a l -bent function for any l k. Stǎnicǎ [1] completely characterized the octabent ( 3 -bent) and hexadecabent ( 4 -bent) functions in terms of bent functions. In this paper, we consider generalized bent Boolean functions in GB k n and k -bent functions in GB 1 n. Firstly, from the theory of cyclotomic fields we prove that all generalized bent Boolean functions are regular, i.e., their dual exist. Secondly, we completely characterize the generalized bent Boolean functions in GB k n (k 3) in terms of bent functions and also describe these functions by a combinatorial form. Finally, we associate every function in GB 1 n with a function in GB k n. Consequently, we completely characterize the k (k 3) bent Boolean functions in GB 1 n in terms of bent functions and also describe these functions by a combinatorial form. II. SOME RESULTS ON CYCLOTOMIC FIELDS AND THE REGULARITY OF GENERALIZED BENT BOOLEAN FUNCTIONS In this section we will give some results on cyclotomic fields, which will be used in the following sections. We also prove that any generalized bent Boolean function in GB k n (k 3) is always regular. Firstly, we state some basic facts on the cyclotomic field K = Q( k), which can be found in any book on algebraic number theory, such as [17] for example. Let O K be the ring of integers of K = Q( k). Any nonzero ideal A of O K can be uniquely(up to the order) expressed as A = P a1 1 P as s, where P 1,, P s are distinct(nonzero) prime ideals of O K and a i 1(1 i s). In other words, the set S(K) of all nonzero ideals of O K is a free multiplicative commutative semigroup with a basis B(K), the set of all nonzero prime ideals of O K. Such semigroup S(K) can be extended to the commutative group I(K), called the group of fractional ideals of K. Each element of I(K), called a fractional ideal, has the form AB 1, where A, B are ideals of O K. For each α K = K\0}, αo K is a fractional ideal, called a principal fractional ideal. And we have (αo K )(βo K ) = αβo K and (αo K ) 1 = (α 1 )O K. Therefore, the set P (K) of all principal fractional ideals is a subgroup of I(K). Some results on K are given in the following lemmas. Lemma.1: Let k and K = Q( k), then (i) The field extension K/Q is Galois of degree and the Galois group Gal(K/Q) = σ j : j Z, j 1 mod }, where the automorphism σ j of K is defined by k j. Moreover, σ k 1 (α) = α for any α K. (ii) The ring of integers in K is O K = Z[ k] and j : 0 j 1} is an integral basis of O k K. The group of roots of unity in O K is W K = j : 0 j k 1}. k (iii) Let ε OK (the units of O K). Then ε W K if and only if ε. (iv) The principal ideal (1 k)o K is a prime ideal of O K and the rational prime is totally ramified in O K, i.e., O K = ((1 k)o K ). Lemma.: Let n be a positive integer, k 3, α O K, and α α. Then W K. n n Proof: From the condition (i) of Lemma.1 and α, we have n From the condition (iv) of Lemma.1, we have ασ 1 (α) = αα = n. (αo K )σ 1 (αo K ) = ((1 k)o K ) n. According to the uniqueness of the decomposition of (αo K )σ 1 (αo K ) and the condition (iv) of Lemma.1, we have Note that k 3 and 1 We immediately have and αo K = σ 1 (αo K ) = ((1 k)o K ) n k. = O K. From the condition (iv) of Lemma.1, we have ( n OK ) = (O K ) n = ((1 k)o K ) n. αo K = n OK = ((1 k)o K ) n k α n O K. From the condition (iii) of Lemma.1, this lemma follows. Theorem.3: Let k 3. Then every generalized bent Boolean function in GB k n is always regular. Proof: From the definition of generalized bent Boolean functions and Lemma., this lemma follows.

3 3 Lemma.4: Let γ a = v Z a v j=1 vj j, where a Z k and v = (v 1, v,, v ). Then e k a Z u a γ a, where e = j=1 u j j, u = (u 1, u,, u ), and u j Z. Further, γ a : a Z } is a basis of K = Q( k) over Q. Proof: For simplicity, denote A u a γ a. Then we have Since a Z a (u+v) = This completes the proof. Lemma.5: Let γ a = v Z A a Z v Z u a 0, v u. This leads to, v = u a Z v Z j=1 vj j k v a j=1 vj j k a Z a (u+v). A j=1 uj j = e k k. a v j=1 vj j, where a Z k and v = (v 1, v,, v ).Then e + k e+k k k u a γ a, a Z,a 1=u 1 where e = j=1 u j j, u = (u 1, u,, u ), and u j Z. Proof: When u 1 = 0, from Lemma.4 we have e + k e+k k ( u a + a1+u a )γ a a Z a Z k (1 + a1 ) u a γ a a Z,a 1=u 1 When u 1, from e+k = e k and Lemma.4 we have u a γ a. e + k e+k = e k e k k k ( u a u a a1 )γ a a Z a Z k (1 a1 ) u a γ a a Z,a 1=u 1 u a γ a. This completes the proof. Lemma.6: Let a Z and γ a = v Z a v j=1 vj j. Then k γ a = (1 + aj j+1). j=1

4 4 Proof: For simplicity, denote Then we have This completes the proof. A = (1 + aj j+1). j=1 A = ( ajvj vj j j=1 v j Z = a v v Z k ) j=1 vj j. k III. COMPLETE CHARACTERIZATION OF GENERALIZED BENT BOOLEAN FUNCTIONS In this section, we present the complete characterization of generalized bent Boolean functions in terms classical Boolean bent functions. Let g : Z n Z k be a generalized Boolean function. It turns out that the generalized Walsh-Hadamard spectrum of g can be described (albeit, in a comlicated manner) in terms of the Walsh-Hadamard spectrum of its Boolean components g i. Theorem 3.1: Let g(x) = i=0 g i(x) i, where g GB k n and g i GB 1 n. Then H g (u) where γ a = v Z a v j=1 vj j. k Proof: According to the definition of H g (u), we have Note that Therefore, we have n Hg (u) = x Z n = x Z n = x Z n = x Z n = x Z n g(x) k u x a Z i=0 gi(x) i u x k i=0 gi(x) i u x k u x i=0 gi(x) i+1 g0(x)+u x x Z n x Z n gi(x) i+1 g0(x)+u x g0(x)+u x (1 + ai i+1) aigi(x) = n Hg (u) x Z n a Z g0(x)+u x H g0+ aigi(u)γ a, ((1 + gi(x) ) + (1 gi(x) ) i+1) ((1 + i+1) + (1 i+1) gi(x) ). 1 + i+1 if a i = 0, (1 i+1) gi(x) if a i. a Z (1 + ai i+1) x Z n aigi(x) (1 + ai i+1) g0(x)+u x+ aigi(x).

5 5 According to the definition of H g0+ aigi(u), we have H g (u) a Z H g0+ aigi(u) (1 + ai i+1). From Lemma.6, this theorem follows. Theorem 3.: Let k 3 and g(x) = i=0 g i(x) i, where g GB k n and g i GB 1 n. Then g is generalized bent if and only if condition (i), for n even, respectively, (ii), for n odd hold, where: (i) For any u Z n, there exist some v Z and some b 0 Z such that H g0+ aigi(u) = v a+b0, for all a Z. (ii) For any u Z n, there exist some v Z and some b 0 Z such that H g a1+v1 aigi(u) = ṽ ã+b0, for all a Z, where ṽ = (v,, v ) and ã = (a,, a ). Proof: If g is generalized bent, then for any u, H g (u). From Lemma., one has H g (u) W K. Thus, H g (u) = or e, where e = k e k When n is even, suppose H g (u) = e k j=1 v j j and v j Z. and e = j=1 v j j. According to Theorem 3.1 and Lemma.4, we have H g (u) a Z a Z H g0+ aigi(u)γ a v a γ a. From the uniqueness of the expansion of H g (u) by the basis γ a : a Z }, we have where v only depends on g and u. Suppose H g (u) = e k e = j=1 v j j, it can be verified that H g0+ aigi(u) = v a, for any a Z, and e = j=1 v j j, by the same technique used in the case H g0+ aigi(u) = v a+1, for any a Z. When n is odd, suppose H g (u) = e. Note that e k = e k ( k ) = + (e k 3 )+ k. If H e k 3 k k g (u) = e k 3 k and e k 3 = j=1 v j j, by Theorem 3.1 and Lemma.5, we have Hg (u) ( H g0+ aigi(u))γ a a Z a Z,a 1=v 1 From the uniqueness of the expansion of H g (u) by the basis γ a : a Z } and H g0+ aigi(u) Q, we have v a H g0+ aigi(u) =, a 1 = v 1, 0, a 1 v 1. Hence, we obtain H g0+ aigi(u) = v a γ a. ṽ ã+v 1, a 1 = v 1, 0, a 1 v 1, where ṽ = (v,, v ) and ã = (a,, a ). If H g (u) = e k 3 and e k 3 = k j=1 v j j, by the same technique used in the case H g (u) = e k 3 and k e k 3 = j=1 v j j, it can be verified that ṽ ã+v 1+1 H g0+ aigi(u) =, a 1 = v 1, 0, a 1 v 1,

6 6 where ṽ = (v,, v ) and ã = (a,, a ). If the condition (i) or (ii) in this theorem holds, then from the definition of generalized bent functions a simple computation shows that g is generalized bent. Hence, the theorem follows. Remark The cases k = and k = 3 in Theorem 3. are investigated by Stǎnicǎ, et al. in [13]. Corollary 3.3: Let k 3 and g(x) = i=0 g i(x) i be generalized bent, where g GB k n and g i GB 1 n. Then, conditions (i), for n even, respectively, (ii), for n odd hold, where (i) g 0 + a ig i is bent for all a Z ; (ii) g 0 + a ig i is semibent, for all a Z. Proof: From Theorem 3., this corollary follows. Corollary 3.4: Let k 3 and g(x) = i=0 g i(x) i be generalized bent, where g GB k n and g i GB 1 n. Then, g π is always generalized bent, where g π is defined as (i), for n even, respectively, (ii), for n odd, where (i) g π (x) = g 0 (x) + g π(i)(x) i for any permutation π of 1,,, k 1}; (ii) g π (x) = g 0 (x) + g 1 (x) k + i= g π(i)(x) i for any permutation π of, 3,, k 1}. Proof: From Theorem 3., this corollary follows. Corollary 3.5: Let k 3, l k, and g(x) = i=0 g i(x) i be generalized bent, where g GB k n and g i GB 1 n. Then, g I is always generalized bent in GB l n, where g I is defined as (i), for n even, respectively, (ii), for n odd, where (i) g I (x) = g 0 (x) l 1 + l 1 j=1 g i j (x) l 1 j for any subset I = i 1,, i l 1 } of 1,,, k 1}, where #I = l 1; (ii) g I (x) = g 0 (x) l 1 + g 1 (x) l + l 1 j= g i j (x) l 1 j for any subset I = i,, i l 1 } of, 3,, k 1}, where #I = l. Proof: From Theorem 3., this corollary follows. Theorem 3.6: Let k 3 and g(x) = i=0 g i(x) i, where g GB k n and g i GB 1 n. For any w Z, we define the set Γ w = x Z n : (g 1 (x),, g (x)) = w}. Then g is generalized bent if and only if conditions (i), for n even, respectively, (ii), for n odd hold, where (i) For any u Z n, there exist some v Z and some b 0 Z such that g0(x)+u x b0 n, w = v, = 0, otherwise. (ii) For any u Z n, there exist some v Z and some b 0 Z such that b 0 n, w = (0, ṽ), g0(x)+u x = b 0 +v 1 n, w = (1, ṽ), 0, otherwise, where v = (v 1, ṽ). Proof: Let A = n H g0+ aigi(u). According to the definition of H g 0+ aigi(u), one obtains A = g0(x)+u x+ gi(x)ai x Z n = x Z n = w Z g0(x)+u x gi(x)ai ( g0(x)+u x ) w a. Suppose g is generalized bent. When n is even, from Theorem 3., we have A = ( g0(x)+u x ) w a w Z = n b 0 v a, where a Z. Since w a (w Z ) as functions from Z to C (a w a ) are linear independent over C. Hence, we obtain g0(x)+u x b0 n, w = v, = 0, otherwise.

7 7 When n is odd, from Theorem 3. we have A = = n w Z ( g0(x)+u x ) w a 1 + a1+v1 ṽ ã+b0. where a Z, v = (v 1, ṽ), and a = (a 1, ã). Let c w = g0(x)+u x. Then one gets A = c w w a = w Z w 1 Z, w Z k = = n w Z k c w w1 a1 w ã (c (0, w) + c (1, w) a1 ) w ã 1 + a1+v1 b0 ṽ ã. Since w ã ( w Z k ) as functions from Z k to C (ã w ã ) are linear independent over C. Hence, we obtain 1+ a 1 +v 1 c (0, w) + c (1, w) a1 = b0 n, w = ṽ, 0, otherwise, where a 1 Z. This implies that 0 n, w = (0, ṽ), g0(x)+u x = b 0 +v 1 n, w = (1, ṽ), 0, otherwise. If the condition (i) or (ii) holds, from the definition of generalized bent functions, g is generalized bent. Hence, this theorem follows. IV. COMPLETE CHARACTERIZATION OF k -BENT BOOLEAN FUNCTIONS In this section, we consider the characterization of k -bent Boolean functions, connect k -bent Boolean functions with generalized bent functions, and characterize k -bent Boolean functions in terms of bent Boolean functions. For any f GB 1 n, we give a unique generalized boolean function g GB k n such that g(x) = f(x) + wt(x). (4) Suppose g(x) = i=0 g i(x) i, where g i GB 1 n. From Lemma 5 of [1], we have f(x) + s g i (x) = (x), i = 0, s i(x), 1 i k 1, where s i(x) are the elementary symmetric polynomials of degree i (we use the convention that s i(x) = 0, if x Z n, and i > n). The following theorem gives the relationship between f and g. Theorem 4.1: Let f and g defined as Equation (4). Then Further, f is k -bent if and only if g is generalized bent. Proof: According to the definition of H g (u), we have H g (u) = n = n = n H (k ) f (u) = H g (u). x Z n x Z n x Z n =H (k ) f (u). u x g(x) k u x f(x) +wt(x) k f(x)+u x wt(x) k (5)

8 8 Then f is k -bent if and only if g is generalized bent. Hence, this theorem follows. Theorem 4.: Let k 3 and f GB 1 n. Then f is k -bent if and only if conditions (i), for n even, respectively, (ii), for n odd hold, where (i) For any u Z n, there exist some v Z and some b 0 Z such that H f+s + ais i (u) = v a+b0, for any a Z. (ii) For any u Z n, there exist some v Z and some b 0 Z such that 1 + H f+s + a1+v1 ais (u) = ṽ ã+b0, i for any a Z, where ṽ = (v,, v ), and ã = (a,, a ). Proof: From Theorem 4.1, Theorem 3., and Equation (5), this theorem follows. Remark The case k = in Theorem 4. was considered by Stǎnicǎ et al. [14]. Stǎnicǎ [1] investegated cases k = 3, 4 in Theorem 4.. Corollary 4.3: Let k 3 and f GB 1 n be k -bent. Then, conditions (i), for n even, respectively, (ii), for n odd hold, where (i) f + s + k i=0 a is i is bent for all a Z ; (ii) f + s + k i=0 a is i is semibent for all a Z,. Proof: From Theorem 4., this corollary follows. Corollary 4.4: Let k 3 and f GB 1 n be a strong k -bent function. Then, conditions (i), for n even, respectively, (ii), for n odd hold, where (i) f + i=0 a is i is bent for all a Z k ; (ii) f + i=0 a is i is semibent for all a Z k. Proof: The conclusion follows from Corollary 4.3. Theorem 4.5: Let k 3 and f GB 1 n. For any w Z, denote the following set Γ w = x Z n : (s k (x),, s 0(x)) = w}. Then, f is k -bent if and only if conditions (i), for n even, respectively, (ii), for n odd hold, where (i) For any u Z n, there exist some v Z and some b 0 Z such that f(x)+s (x)+u x b0 n, w = v, = 0, otherwise. (ii) For any u Z n, there exist some v Z and some b 0 Z such that b 0 n, w = (0, ṽ), f(x)+s (x)+u x = b 0 +v 1 n, w = (1, ṽ), 0, otherwise, where v = (v 1, ṽ). Proof: From Theorem 4., this theorem follows. Corollary 4.6: Let n 3 and f GB 1 n be k -bent. Then k log n + 1. Proof: Suppose k > log n + 1. Then n = k i=0 w i i and w i 0, 1}. By the definition of Γ w, we have From Theorem 4.5 and n 3, we have This leads to a contradiction and completes the proof. Γ w = (1, 1,, 1)}. }} n 1 = #Γ w 0 mod. V. CONCLUDING REMARKS This paper generalized the work of Stǎnicǎ et al. on generalized Boolean bent functions and k -bent Boolean functions[1], [13], [14]. We showed that every generalized bent Boolean function was regular. A complete characterization for generalized Boolean bent functions and k -bent Boolean functions was presented in terms of bent functions. And we also completely characterized these functions in a combinatoric form. It would be interesting to characterize generalized p-ary bent functions.

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