On the representation of bent functions by bent rectangles

We propose a representation of boolean bent functions by bent rectangles, that is, by special matrices with restrictions on rows and columns. Using this representation, we exhibit new classes of bent functions, give an algorithm to construct bent functions, improve a lower bound for the number of bent functions. 1 Preliminaries Let Vn be an n-dimensional vector space over the field F2 = {0, 1} and Fn be the set of all Boolean functions Vn → F2. We identify a function f ∈ Fn of x = (x1, . . . , xn) with its algebraic normal form, that is, a polynomial of the ring F2[x1, . . . , xn] reduced modulo the ideal (x1 − x1, . . . , x 2 n − xn). Denote by deg f the degree of such polynomial. Write ◦ γ = (−1) for γ ∈ F2. The Walsh–Hadamard transform of f is defined as ∧ f (λ) = ∑ x∈Vn ◦ f (x)(−1), λ ∈ Vn, where 〈λ, x〉 = λ1x1 + . . . + λnxn. The symbol + denotes both the addition in F2 and the addition of integers or real numbers. The way of the symbol usage depends on operands. A function f ∈ F2n is called a bent function if | ∧ f (λ)| = 2 for any λ ∈ V2n. Let B2n be the set of all bent functions of 2n variables. The Sylvester–Hadamard matrix Hn of order 2 n is defined by the recursive rule Hn = ( 1 1 1 −1 ) ⊗Hn−1, H0 = (1) , ∗Appeared in Probabilistic Methods in Discrete Mathematics: Proceedings of the Fifth International Petrozavodsk Conference (Petrozavodsk, June 1–6, 2000). Utrecht, Boston: VSP, pp. 121–135, 2002. 1 where ⊗ denotes the Kronecker product. Note that Hn is symmetric and HnHn = 2 E, where E is the identity 2 × 2 matrix. Let An = {l ∈ Fn : deg l ≤ 1} denote the set of affine functions. A function l ∈ An is of the form l(x) = α1x1 + . . .+ αnxn + γ = 〈α, x〉+ γ, α ∈ Vn, γ ∈ F2, and is a linear function if γ = 0. Let v1 = 0, v2, . . . , v2n be the lexicographically ordered vectors of Vn. The sequence and the spectral sequence of f are defined, respectively, as the row vectors ◦ f = ( ◦ f (v1), . . . , ◦ f (v2n)) and ∧ f = ( ∧ f (v1), . . . , ∧ f (v2n)). It is known that the ith row of Hn is a sequence of the linear function 〈α, x〉, where α = vi (α is the ith vector of Vn) and the Walsh–Hadamard transform can be written as ∧ f = ◦ f Hn. Therefore, (i) ∑ λ∈Vn ∧ f (λ) = 2 (Parseval’s identity); (ii) if l ∈ An, l(x) = 〈α, x〉+ γ, then ∧ l (α) = ◦ γ · 2 and ∧ l (λ) = 0, λ 6= α. Lemma 1. Let f1, f2, f3, f4 ∈ Fn. The vector 1 2 ( ∧ f 1 + ∧ f 2 + ∧ f 3 + ∧ f 4) is a spectral sequence of some function g ∈ Fn if and only if f1(x) + f2(x) + f3(x) + f4(x) = 1 for all x ∈ Vn. Moreover, g(x) = f1(x)f2(x) + f1(x)f3(x) + f2(x)f3(x). Proof. If ∧ g = 1 2 ( ∧ f 1 + ∧ f 2 + ∧ f 3 + ∧ f 4), then ◦ g = 1 2 ( ◦ f 1 + ◦ f 2 + ◦ f 3 + ◦ f 4) or ◦ g (x) = 1 2 ( ◦ f 1(x) + ◦ f 2(x) + ◦ f 3(x) + ◦ f 4(x)), x ∈ Vn. But ◦ f 1(x) + ◦ f 2(x) + ◦ f 3(x) + ◦ f 4(x) = ±2 if and only if f1(x) + f2(x) + f3(x) + f4(x) = 1. Let g be the function defined in the statement of the lemma. For all x, among f1(x), f2(x), f3(x) there are two identical values. Without loss of generality we can assume that f1(x) = f2(x). Then g(x) = f1(x), f4(x) = f3(x) + 1, and 1 2 ( ◦ f 1(x) + ◦ f 2(x) + ◦ f 3(x) + ◦ f 4(x)) = 1 2 ( ◦ f 1(x) + ◦ f 1(x) + ◦ f 3(x)− ◦ f 3(x)) = ◦ f 1(x) = ◦ g (x). Thus ◦ g = 1 2 ( ◦ f 1 + ◦ f 2 + ◦ f 3 + ◦ f 4) and ∧ g = 1 2 ( ∧ f 1 + ∧ f 2 + ∧ f 3 + ∧ f 4). 2 Further we will use functions from the set Qn = {q ∈ Fn : w( ∧ q ) = 4}, where w( ∧ q ) is the number of nonzero values of ∧ q . The following lemma completely characterizes the elements of Qn. Lemma 2. Let q ∈ Qn be such that ∧ q (λ) 6= 0 only if λ ∈ {λ(1), λ(2), λ(3), λ(4)} ⊆ Vn, n ≥ 2. Then ∑ i λ(i) = 0 and ∧ q (λ(i)) = ◦ γ i ·2 n−1 for some γi ∈ F2 such that ∑ i γi = 1 or ∏ i ◦ γ i = −1, i = 1, 2, 3, 4. Moreover, q(x) = l1(x)l2(x) + l1(x)l3(x) + l2(x)l3(x), where li(x) = 〈λ(i), x〉+ γi. Proof. By induction on n it is easy to show that the equation ∑ i z 2 i = 2 2n has an unique solution zi = 2 n−1 in positive integers (here and throughout the proof, i varies from 1 up to 4). According to the Parseval identity