Sensitivities and block sensitivities of elementary symmetric Boolean functions

Jing Zhang; Yuan Li; John O. Adeyeye

doi:10.1515/jmc-2020-0042

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Sensitivities and block sensitivities of elementary symmetric Boolean functions

Jing Zhang , Yuan Li and John O. Adeyeye

Published/Copyright: April 22, 2021

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From the journal Journal of Mathematical Cryptology Volume 15 Issue 1

Abstract

Boolean functions have important applications in molecular regulatory networks, engineering, cryptography, information technology, and computer science. Symmetric Boolean functions have received a lot of attention in several decades. Sensitivity and block sensitivity are important complexity measures of Boolean functions. In this paper, we study the sensitivity of elementary symmetric Boolean functions and obtain many explicit formulas. We also obtain a formula for the block sensitivity of symmetric Boolean functions and discuss its applications in elementary symmetric Boolean functions.

Keywords: Boolean functions; Lucas theorem

MSC 2010: 05A05; 05A15

1 Introduction

In 1938, Shannon [28] recognized that symmetric functions had efficient switch network implementation. Since then, a lot of research has been carried out on symmetric or partially symmetric Boolean functions, and detection of symmetry has become important in logic synthesis, technology mapping, binary decision diagram minimization, and testing [1,10,22].

For the applications of symmetric Boolean functions in cryptography, Canteaut and Videau [3] presented an extensive study in 2005, and more results on (totally) symmetric Boolean functions can be found in other papers [2,5,8,17,21,23,27].

It is clear that any symmetric Boolean function can be written as a sum of some elementary symmetric Boolean functions. Hence, it is a fundamental question to have a comprehensive understanding about elementary symmetric Boolean functions. In ref. [8], the authors studied the balancedness of elementary symmetric Boolean functions and they proposed a conjecture which has received a lot of attention [4,5,6,8,9,13,14,31].

In ref. [7], Cook et al. introduced the definition of sensitivity as a combinatorial measure for Boolean functions by providing lower bounds on the time needed by CREW PRAM (Concurrent Read Exclusive Write, Parallel Random Access Machine). The concept was extended by Nisan [24] to block sensitivity. The study of sensitivity and block sensitivity of Boolean functions has been an active research topic for many years [11,12,16,18,19,25,26,29,30,32].

Recently, Huang proved the long standing Sensitivity Conjecture [15]: for any Boolean function f , b s ( f ) ≤ 2 s ( f ) 4 , where b s ( f ) is the block sensitivity of f and s ( f ) is the sensitivity of f .

In Section 2 of this paper, we introduce the algebraic normal form (ANF) of Boolean functions and the definition of symmetric Boolean functions. In Section 3, we first recall definitions used in the paper, then obtain many explicit formulas of the sensitivities of elementary symmetric Boolean functions by using some elementary combinatorial results. The main idea of Section 3 is motivated by ref. [8,33]. In Section 4, we prove a formula for the block sensitivity of symmetric Boolean functions. Based on our knowledge, this is the first study about the block sensitivity of symmetric Boolean functions. We apply this formula to elementary symmetric Boolean functions and show that the block sensitivity can be strictly greater than the sensitivity for some elementary symmetric Boolean functions. The conclusion is included in Section 5.

2 Preliminaries

In this section, we introduce the definitions and notations. Let F = F 2 = { 0 , 1 } . If f : F n ⟶ F is a function with n variables and values in F , it is well known [20] that f can be expressed as a polynomial, called the ANF:

f ( x 1 , … , x n ) = ⊕ 0 ≤ k i ≤ 1 , i = 1 , … , n a k 1 ⋯ k n x 1 k 1 ⋯ x n k n ,

where each coefficient a k 1 ⋯ k n ∈ F . The symbol ⊕ stands for addition modulo 2. The number k 1 + ⋯ + k n is the multivariate degree of the term a k 1 ⋯ k n x 1 k 1 ⋯ x n k n with nonzero coefficient a k 1 ⋯ k n . The greatest degree of all the terms of f is called the algebraic degree, denoted by deg ( f ) .

The number of non-zeros in x = ( x 1 , … , x n ) , denoted by W ( x ) , is called the Hamming weight of x .

Let S n be the symmetric group of degree n , i.e., the set of all the permutations over the set { 1 , … , n } . If for any permutation π ∈ S n ,

f ( x 1 , … , x n ) = f ( x π ( 1 ) , … , x π ( n ) ) ,

then the function f ( x 1 , … , x n ) is called a symmetric function. Obviously, the value of symmetric functions f ( x 1 , … , x n ) depends only on the Hamming weight of ( x 1 , … , x n ) . In other words, f ( x 1 , … , x n ) is symmetric if and only if

W ( x ) = W ( y ) ⇒ f ( x ) = f ( y ) .

Let v k = f ( x ) with W ( x ) = k , we call v ( f ) = ⟨ v 0 , … , v n ⟩ the value vector of f ( x ) . It is clear that there are 2 n + 1 symmetric Boolean functions.

3 Sensitivity of elementary symmetric Boolean functions

In this section, we calculate the sensitivities of elementary symmetric Boolean functions. More precisely, we present explicit formulas for the sensitivities of X ( d , n ) when d is odd and d = 2 k , 2 k + t ± 2 t . We show that s ( X ( d , n ) ) and s ( X ( n − d , n ) ) are computable for fixed d and that the explicit formulas of s ( X ( d × 2 k , n ) ) can be obtained once we obtain the sequence 1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ .

Definition 3.1

For integers n and d , 1 ≤ d ≤ n , we define the elementary symmetric Boolean function by

X ( d , n ) = ⊕ 1 ≤ i 1 < ⋯ < i d ≤ n x i 1 ⋯ x i d .

Let x = ( x 1 , … , x n ) ∈ F n . We use x i to denote the word obtained by flipping the i -th bit of x .

Definition 3.2

[16,26] The sensitivity s ( f ; x ) of f at x is the number of indices i such that f ( x ) ≠ f ( x i ) . The sensitivity of f , denoted by s ( f ) , is max x s ( f ; x ) . The average sensitivity of f , denoted by s ¯ ( f ) , is 1 2 n ∑ x ∈ { 0 , 1 } n s ( f , x ) .

In the above definition,

s ( f ) = max x s ( f ; x ) = max x ∈ { 0 , 1 } n s ( f ; x ) .

Example 3.3

In Table 1, we list the sensitivities of X ( 1 , 3 ) , X ( 2 , 3 ) , and X ( 3 , 3 ) on every word. Their sensitivities are 3, 2, 3, respectively, and the average sensitivities are 3, 3 2 , 3 4 , respectively.

Table 1

Example 3.3, s ( X ( 1 , 3 ) ) = 3 , s ( X ( 2 , 3 ) ) = 2 , s ( X ( 3 , 3 ) ) = 3

x	X ( 1 , 3 ) ( x )	s ( X ( 1 , 3 ) , x )	X ( 2 , 3 ) ( x )	s ( X ( 2 , 3 ) , x )	X ( 3 , 3 ) ( x )	s ( X ( 3 , 3 ) , x )
(0,0,0)	0	3	0	0	0	0
(0,0,1)	1	3	0	2	0	0
(0,1,0)	1	3	0	2	0	0
(0,1,1)	0	3	1	2	0	1
(1,0,0)	1	3	0	2	0	0
(1,0,1)	0	3	1	2	0	1
(1,1,0)	0	3	1	2	0	1
(1,1,1)	1	3	1	0	1	3

Definition 3.4

The function f ( x 1 , … , x n ) is essential in the variable x i if there exist r , s ∈ F and x 1 ∗ , … , x i − 1 ∗ , x i + 1 ∗ , … , x n ∗ such that

f ( x 1 ∗ , … , x i − 1 ∗ , r , x i + 1 ∗ , … , x n ∗ ) ≠ f ( x 1 ∗ , … , x i − 1 ∗ , s , x i + 1 ∗ , … , x n ∗ ) .

First, we will calculate the sensitivities and average sensitivities of X ( 1 , n ) and X ( n , n ) .

Lemma 3.5

For every x in F n , s ( f ; x ) = n is equivalent to f ( x 1 , … , x n ) = x 1 ⊕ ⋯ ⊕ x n ⊕ a , where a ∈ F .

Proof

The sufficiency is obvious. We show the necessity in the following.

Since for every word x in F n , s ( f ; x ) = n , we know that every variable x i is essential in f . If deg ( f ) > 1 , without loss of generality, we may assume that the term with degree deg ( f ) contains variable x 1 . Hence, we can write f as

f ( x 1 , … , x n ) = x 1 g ( x 2 , … , x n ) ⊕ h ( x 2 , … , x n ) ,

where deg ( g ) = deg ( f ) − 1 ≥ 1 . Hence, function g is not a constant function. We can find ( b 2 , … , b n ) such that g ( b 2 , … , b n ) = 0 . Then, we have

f ( 1 , b 2 , … , b n ) = f ( 0 , b 2 , … , b n ) = h ( b 2 , … , b n ) ,

which means that the sensitivity of f over the word ( 1 , b 2 , … , b n ) is at most n − 1 . This contradiction shows deg ( f ) = 1 . Since every variable is essential, we have f ( x 1 , … , x n ) = x 1 ⊕ ⋯ ⊕ x n ⊕ a . For a ∈ F .□

From Lemma 3.5, we immediately have

Corollary 3.6

The sensitivity and average sensitivity of X ( 1 , n ) = x 1 + ⋯ + x n are both n . In other words, s ( X ( 1 , n ) ) = s ¯ ( X ( 1 , n ) ) = n .

For the sensitivity of X ( n , n ) = x 1 ⋯ x n , we have s ( X ( n , n ) , ( 1 , … , 1 ) ) = n , s ( X ( n , n ) , x ) = 1 with W ( x ) = n − 1 and s ( X ( n , n ) , x ) = 0 with W ( x ) ≤ n − 2 . So we have

Proposition 3.7

The sensitivity of X ( n , n ) is n , and the average sensitivity of X ( n , n ) is n 2 n − 1 . In other words, s ( X ( n , n ) ) = n and s ¯ ( X ( n , n ) ) = n 2 n − 1 .

Remark 3.8

Function X ( n , n ) is a nested canalizing function and the above proposition also appeared in [18, 19].

We need some lemmas to calculate the sensitivities of X ( d , n ) for 2 ≤ d ≤ n − 1 .

Lemma 3.9

[33] Let v ( f ) = ⟨ v 0 , v 1 , … , v n ⟩ be the value vector of symmetric Boolean function f ( x ) . If for some i , ( v i − 1 , v i , v i + 1 ) ∈ { ( 0 , 1 , 0 ) , ( 1 , 0 , 1 ) } , then the sensitivity of f is n . Otherwise, s ( f ) = max { k + 1 , n − k ∣ v k ≠ v k + 1 } .

Lemma 3.10

If v 0 ≠ v 1 or v n − 1 ≠ v n , then s ( f ) = n .

Proof

Because v 0 ≠ v 1 implies s ( f , ( 0 , … , 0 ) ) = n and v n − 1 ≠ v n implies s ( f , ( 1 , … , 1 ) ) = n , we have s ( f ) = n .□

Let C ( n , k ) = n ! k ! ( n − k ) ! if 0 ≤ k ≤ n and 0 otherwise, then we have

Lemma 3.11

[8] Let X ( d , n ) ( j ) be the value of X ( d , n ) when x has Hamming weight j , then

X ( d , n ) ( j ) = 1 − ( − 1 ) C ( j , d ) 2 .

Lemma 3.12

(Lucas theorem) Let n = a m p m + a m − 1 p m − 1 + ⋯ + a 1 p + a 0 and k = b m p m + b m − 1 p m − 1 + ⋯ + b 1 p + b 0 , where a i , b i ∈ N , 0 ≤ a i ≤ p − 1 , 0 ≤ b i ≤ p − 1 , and p is a prime, then

C ( n , k ) ≡ C ( a m , b m ) ⋯ C ( a 1 , b 1 ) ( mod p ) .

From the Lucas theorem, we immediately have

Lemma 3.13

For any prime p and natural number t , C ( n , k ) ≡ C ( p t n , p t k ) ( mod p ) . Particularly, we have C ( n , k ) ≡ C ( 2 t n , 2 t k ) ( mod 2 ) .

Lemma 3.14

[3, 8] For any integer k ≥ 2 , the sequence { ( − 1 ) C ( j , k ) } j = 0 ∞ is periodic and the least period is 2 ⌊ log 2 k ⌋ + 1 .

By Lemmas 3.12 and 3.14, we get the following computing results.

Lemma 3.15

[8] If we write the infinite periodic string a 1 a 2 ⋯ a k a 1 a 2 ⋯ a k ⋯ ⋯ as a 1 a 2 ⋯ a k ¯ , then

1 − ( − 1 ) C ( j , 2 ) 2 j = 0 ∞ = 0011 ¯

1 − ( − 1 ) C ( j , 3 ) 2 j = 0 ∞ = 0001 ¯

1 − ( − 1 ) C ( j , 4 ) 2 j = 0 ∞ = 00001111 ¯

1 − ( − 1 ) C ( j , 5 ) 2 j = 0 ∞ = 00000101 ¯

1 − ( − 1 ) C ( j , 6 ) 2 j = 0 ∞ = 00000011 ¯

1 − ( − 1 ) C ( j , 7 ) 2 j = 0 ∞ = 00000001 ¯

1 − ( − 1 ) C ( j , 8 ) 2 j = 0 ∞ = 0000000011111111 ¯

1 − ( − 1 ) C ( j , 9 ) 2 j = 0 ∞ = 0000000001010101 ¯

1 − ( − 1 ) C ( j , 10 ) 2 j = 0 ∞ = 0000000000110011 ¯

1 − ( − 1 ) C ( j , 11 ) 2 j = 0 ∞ = 0000000000010001 ¯

1 − ( − 1 ) C ( j , 12 ) 2 j = 0 ∞ = 0000000000001111 ¯

1 − ( − 1 ) C ( j , 13 ) 2 j = 0 ∞ = 0000000000000101 ¯

1 − ( − 1 ) C ( j , 14 ) 2 j = 0 ∞ = 0000000000000011 ¯

1 − ( − 1 ) C ( j , 15 ) 2 j = 0 ∞ = 0000000000000001 ¯

1 − ( − 1 ) C ( j , 16 ) 2 j = 0 ∞ = 00000000000000001111111111111111 ¯

1 − ( − 1 ) C ( j , 17 ) 2 j = 0 ∞ = 00000000000000000101010101010101 ¯

1 − ( − 1 ) C ( j , 18 ) 2 j = 0 ∞ = 00000000000000000011001100110011 ¯

1 − ( − 1 ) C ( j , 19 ) 2 j = 0 ∞ = 00000000000000000001000100010001 ¯

1 − ( − 1 ) C ( j , 20 ) 2 j = 0 ∞ = 00000000000000000000111100001111 ¯ .

Using Lemma 3.12 for p = 2 and Lemma 3.14, a straightforward calculation shows

Lemma 3.16

When d = 2 k , k ∈ N , we have

1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ = 1 − ( − 1 ) C ( j , 2 k ) 2 j = 0 ∞ = 0 ⋯ 0 ︷ 2 k 1 ⋯ 1 ︷ 2 k ¯ .

We first consider the sensitivities of X ( d , n ) for odd d .

Theorem 3.17

For d = 1 , 3 , 5 , … , 2 n − 1 2 + 1 , the sensitivity of X ( d , n ) is n :

s ( X ( d , n ) ) = n .

Proof

If n is odd, we already know s ( X ( n , n ) ) = n . For 0 < d < n and d odd, we have

X ( d , n ) ( d − 1 ) = v d − 1 = 1 − ( − 1 ) C ( d − 1 , d ) 2 = 0 ,

X ( d , n ) ( d ) = v d = 1 − ( − 1 ) C ( d , d ) 2 = 1 ,

and

X ( d , n ) ( d + 1 ) = v d + 1 = 1 − ( − 1 ) C ( d + 1 , d ) 2 = 0 .

Hence, ( v d − 1 , v d , v d + 1 ) = ( 0 , 1 , 0 ) and by Lemma 3.9, we have

s ( X ( d , n ) ) = n . □

In order to compute the sensitivities of X ( d , n ) for even d , we need to improve Lemma 3.9.

Lemma 3.18

Let v ( f ) = ⟨ v 0 , … , v n ⟩ be the value vector of X ( d , n ) for even d , k min = min { k ∣ v k ≠ v k + 1 } and k max = max { k ∣ v k ≠ v k + 1 } , then s ( X ( d , n ) ) = max { k max + 1 , n − k min } .

Proof

We first show that there is no i such that ( v i − 1 , v i , v i + 1 ) ∈ { ( 0 , 1 , 0 ) , ( 1 , 0 , 1 ) } .

Let d = 2 k , k ∈ N , we have

(3.1) ( i − 2 k ) C ( i , 2 k ) = i C ( i − 1 , 2 k ) ,

(3.2) ( i + 1 − 2 k ) C ( i + 1 , 2 k ) = ( i + 1 ) C ( i , 2 k ) .

Case 1

If there exists i such that

( v i − 1 , v i , v i + 1 ) = 1 − ( − 1 ) C ( i − 1 , 2 k ) 2 , 1 − ( − 1 ) C ( i , 2 k ) 2 , 1 − ( − 1 ) C ( i + 1 , 2 k ) 2 = ( 0 , 1 , 0 ) ,

then 2 ∣ C ( i − 1 , 2 k ) , 2 ∤ C ( i , 2 k ) , and 2 ∣ C ( i + 1 , 2 k ) . By equation (3.1), 2 ∣ ( i − 2 k ) , hence 2 ∣ i . But by equation (3.2), 2 ∣ ( i + 1 ) , which is a contradiction.

Case 2

If there exists i such that

( v i − 1 , v i , v i + 1 ) = 1 − ( − 1 ) C ( i − 1 , 2 k ) 2 , 1 − ( − 1 ) C ( i , 2 k ) 2 , 1 − ( − 1 ) C ( i + 1 , 2 k ) 2 = ( 1 , 0 , 1 ) ,

then 2 ∤ C ( i − 1 , 2 k ) , 2 ∣ C ( i , 2 k ) and 2 ∤ C ( i + 1 , 2 k ) . By equation (3.1), 2 ∣ i . Again by equation (3.2), 2 ∣ ( i + 1 − 2 k ) and 2 ∣ ( i + 1 ) , we receive a contradiction.

We show that there is no i such that

( v i − 1 , v i , v i + 1 ) ∈ { ( 0 , 1 , 0 ) , ( 1 , 0 , 1 ) } .

On the other hand, let

{ k ∣ v k ≠ v k + 1 } = { k min = k 1 , k 2 , … , k t = k max }

with k 1 < k 2 < ⋯ < k t . Since s ( X ( d , n ) ) = max { k + 1 , n − k ∣ v k ≠ v k + 1 } , we have

s ( X ( d , n ) ) = max { k 1 + 1 , n − k 1 , k 2 + 1 , n − k 2 , … , k t + 1 , n − k t } = max { k max + 1 , n − k min } . □

To simplify the notation, sometimes, we write the value vector ⟨ v 0 , … , v n ⟩ as ⟨ v 0 ⋯ v n ⟩ and word ( x 1 , … , x n ) as ( x 1 ⋯ x n ) .

Theorem 3.19

For n ≥ 2 k , k ∈ N , the sensitivity of X ( 2 k , n ) is

s ( X ( 2 k , n ) ) = 2 k n 2 k .

Proof

By Lemma 3.16, the value vector is the first n + 1 numbers of the sequence

1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ = 1 − ( − 1 ) C ( j , 2 k ) 2 j = 0 ∞ = 0 ⋯ 0 ︷ 2 k 1 ⋯ 1 ︷ 2 k ¯ .

Let n = 2 k q + r , 0 ≤ r ≤ 2 k − 1 , k ∈ N .

When q is even, we have

⟨ v 0 ⋯ v n ⟩ = ⟨ 0 ⋯ 0 ︷ 2 k 1 ⋯ 1 ︷ 2 k ⋯ 0 ⋯ 0 ︷ 2 k 1 ⋯ 1 ︷ 2 k ︷ q 0 ⋯ 0 ︷ r + 1 ⟩ .

It is clear that k min = 2 k − 1 and k max = 2 k q − 1 . By Lemma 3.18, we have

s ( X ( 2 k , n ) ) = max { k max + 1 , n − k min } = max { 2 k q , n − 2 k + 1 } = max { 2 k q , 2 k ( q − 1 ) + r + 1 } = 2 k q = 2 k n 2 k .

When q is odd, we have ⟨ v 0 ⋯ v n ⟩ = ⟨ 0 ⋯ 0 ︷ 2 k 1 ⋯ 1 ︷ 2 k ⋯ 0 ⋯ 0 ︷ 2 k ︷ q 1 ⋯ 1 ︷ r + 1 ⟩ . The calculation is identical to the case when q is even. We are done.□

A bound for the sensitivities of elementary symmetric Boolean functions can be received by Lemma 3.18.

Proposition 3.20

For any 1 ≤ d ≤ n , d ∈ N , we have n + 1 2 ≤ s ( X ( d , n ) ) ≤ n .

Proof

First, we have s ( X ( d , n ) ) ≤ n by definition. From Lemma 3.18, for even d , we have

s ( X ( d , n ) ) = max { k max + 1 , n − k min } ≥ max { k min + 1 , n − k min } ≥ n + 1 2 .

By Theorem 3.17, s ( X ( d , n ) ) = n for odd d , we are done.□

Remark 3.21

The upper bound in the Proposition 3.20 is tight. In Theorem 3.19, let n = 2 k + 1 − 1 , then s ( X ( 2 k , n ) ) = n + 1 2 . It shows that the lower bound can also be reached sometimes.

To understand and prove more general formulas, we will first calculate the sensitivities of X ( d , n ) for some small even d . The same techniques will be used to obtain more general formulas later.

Lemma 3.22

For n ≥ 6 , the sensitivity of X ( 6 , n ) is

s ( X ( 6 , n ) ) = 8 n 8 , n = 8 q + r , 0 ≤ r ≤ 5 n , n = 8 q + 6 n − 1 , n = 8 q + 7 .

Proof

By Lemma 3.15, the value vector ⟨ v 0 ⋯ v n ⟩ is the first n + 1 number of 1 − ( − 1 ) C ( j , 6 ) 2 j = 0 ∞ = 00000011 ¯ .

When n = 8 q + 6 , the value vector ⟨ v 0 ⋯ v n ⟩ ends with 01. By Lemma 3.10, s ( X ( 6 , n ) ) = n .

When n = 8 q + 7 , the value vector ⟨ v 0 ⋯ v n ⟩ is

⟨ 00000011 ︷ 8 ⋯ 00000011 ︷ 8 ︷ q + 1 ⟩ .

Since k min = 5 and k max = n − 2 , we have s ( X ( 6 , n ) ) = max { n − 2 + 1 , n − 5 } = n − 1 .

When n = 8 q + r , 0 ≤ r ≤ 5 ,

⟨ v 0 ⋯ v n ⟩ = ⟨ 00000011 ︷ 8 ⋯ 00000011 ︷ 8 ︷ q 0 ⋯ 0 ︷ r + 1 ⟩ .

Since k min = 5 and k max = 8 q − 1 , we have

s ( X ( 6 , n ) ) = max { 8 q − 1 + 1 , n − 5 } = max { 8 q , 8 q + r − 5 } = 8 q = 8 n 8 . □

For d = 10 , we have

Lemma 3.23

For n ≥ 10 , the sensitivity of X ( 10 , n ) is

s ( X ( 10 , n ) ) = 16 n 16 , n = 16 q + r , 0 ≤ r ≤ 9 n , n = 16 q + 10 , 16 q + 12 , 16 q + 14 n − 1 , n = 16 q + 11 , 16 q + 13 , 16 q + 15 .

Proof

By Lemma 3.15, the value vector ⟨ v 0 ⋯ v n ⟩ is the first n + 1 number of 1 − ( − 1 ) C ( j , 10 ) 2 j = 0 ∞ = 0000000000110011 ¯ .

When n = 16 q + 10 , 16 q + 12 , 16 q + 14 , the value vectors ⟨ v 0 ⋯ v n ⟩ end with 01, 10, and 01, respectively. By Lemma 3.10, s ( X ( 10 , n ) ) = n .

When n = 16 q + 11 , 16 q + 13 , 16 q + 15 , the value vectors ⟨ v 0 ⋯ v n ⟩ are

⟨ 0000000000110011 ︷ 16 ⋯ 0000000000110011 ︷ 16 ︷ q 000000000011 ︷ 12 ⟩ ,

⟨ 0000000000110011 ︷ 16 ⋯ 0000000000110011 ︷ 16 ︷ q 00000000001100 ︷ 14 ⟩ ,

and

⟨ 0000000000110011 ︷ 16 ⋯ 0000000000110011 ︷ 16 ︷ q + 1 ⟩ ,

respectively. In any case, we always have k min = 10 and k max = n − 2 , so

s ( X ( 10 , n ) ) = max { n − 2 + 1 , n − 10 } = n − 1 .

When n = 16 q + r , 0 ≤ r ≤ 9 , the value vector ⟨ v 0 ⋯ v n ⟩ is

⟨ 0000000000110011 ︷ 16 ⋯ 0000000000110011 ︷ 16 ︷ q 0 ⋯ 0 ︷ r + 1 ⟩ .

Since k min = 9 and k max = 16 q − 1 ,

s ( X ( 10 , n ) ) = max { 16 q − 1 + 1 , n − 9 } = max { 16 q , 16 q + r − 9 } = 16 q = 16 n 16 . □

For d = 12 , we have

Lemma 3.24

For n ≥ 12 , the sensitivity of X ( 12 , n ) is

s ( X ( 12 , n ) ) = 16 n 16 , n = 16 q + r , 0 ≤ r ≤ 11 16 n 16 + 12 , n = 16 q + r , 12 ≤ r ≤ 15 .

Proof

By Lemma 3.15, the value vector ⟨ v 0 ⋯ v n ⟩ is the first n + 1 number of 1 − ( − 1 ) C ( j , 12 ) 2 j = 0 ∞ = 0000000000001111 ¯ .

When n = 16 q + r , 0 ≤ r ≤ 11 , the value vectors ⟨ v 0 ⋯ v n ⟩ are

⟨ 0000000000001111 ︷ 16 ⋯ 0000000000001111 ︷ 16 ︷ q 0 ⋯ 0 ︷ r + 1 ⟩ .

We have k min = 11 and k max = 16 q − 1 , so

s ( X ( 12 , n ) ) = max { 16 q + r − 11 , 16 q } = 16 q = 16 n 16 .

When n = 16 q + r , 12 ≤ r ≤ 15 , the value vectors ⟨ v 0 ⋯ v n ⟩ are

⟨ 0000000000001111 ︷ 16 ⋯ 0000000000001111 ︷ 16 ︷ q 0 ⋯ 0 ︷ 12 1 ⋯ 1 ︷ r − 11 ⟩ .

Since k min = 11 and k max = 16 q + 11 , we have

s ( X ( 12 , n ) ) = max { 16 q + 12 , n − 11 } = max { 16 q + 12 , 16 q + r − 11 } = 16 q + 12 = 16 n 16 + 12 . □

It is clear that one can continue to compute the explicit formulas of X ( d , n ) for fixed d . In the following, we will consider the situation that d is very close to n . We already know s ( X ( n , n ) ) = n .

Lemma 3.25

Let n ≥ 3 , then s ( X ( n − 1 , n ) ) = 2 n 2 .

Proof

Since

X ( n − 1 , n ) = x 2 ⋯ x n ⊕ x 1 x 3 ⋯ x n ⊕ ⋯ ⊕ x 1 ⋯ x n − 1 ,

obviously, X ( n − 1 , n ) ( j ) = 0 for j = 0 , 1 , … , n − 2 , X ( n − 1 , n ) ( n − 1 ) = 1 , and X ( n − 1 , n ) ( n ) = 1 − ( − 1 ) n 2 . The value vector ⟨ v 0 , … , v n ⟩ is 0 , … , 0 , 1 , 1 − ( − 1 ) n 2 . The result follows from Lemmas 3.9 and 3.18.□

Lemma 3.26

For n ≥ 5 , the sensitivity of X ( n − 2 , n ) is

s ( X ( n − 2 , n ) ) = n , n = 4 q , 4 q + 1 , 4 q + 3 n − 2 , n = 4 q + 2 .

Proof

By Lemma 3.11, the value vector is

⟨ v 0 , … , v n ⟩ = 0 , … , 0 , 1 , 1 − ( − 1 ) n − 1 2 , 1 − ( − 1 ) n ( n − 1 ) 2 2 .

Case 1: n = 4 q + 0 , ⟨ v 0 , … , v n ⟩ = ⟨ 0 , … , 0 , 1 , 1 , 0 ⟩ .

Case 2: n = 4 q + 1 , ⟨ v 0 , … , v n ⟩ = ⟨ 0 , … , 0 , 1 , 0 , 0 ⟩ .

Case 3: n = 4 q + 2 , ⟨ v 0 , … , v n ⟩ = ⟨ 0 , … , 0 , 1 , 1 , 1 ⟩ .

Case 4: n = 4 q + 3 , ⟨ v 0 , … , v n ⟩ = ⟨ 0 , … , 0 , 1 , 0 , 1 ⟩ .

By Lemmas 3.9, 3.10, and 3.18, the formula of the sensitivity is obtained.□

It is clear that one can continue to calculate the explicit sensitivity formulas of X ( n − d , n ) for d = 3 , 4 , … .

Now we will discuss more properties of the sequence 1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ and generalize the above results.

Proposition 3.27

If 1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ = a 0 a 1 ⋯ a m − 1 ¯ , then

1 − ( − 1 ) C ( j , 2 d ) 2 j = 0 ∞ = a 0 a 0 a 1 a 1 ⋯ a m − 1 a m − 1 ¯ .

Proof

By Lemma 3.14, if the least period of 1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ = a 0 a 1 ⋯ a m − 1 ¯ is m = 2 ⌊ log 2 d ⌋ + 1 , then the least period of 1 − ( − 1 ) C ( j , 2 d ) 2 j = 0 ∞ is 2 m . Hence, we may assume

1 − ( − 1 ) C ( j , 2 d ) 2 j = 0 ∞ = b 0 b 1 ⋯ b 2 m − 1 ¯ .

We only need to show b 2 j = b 2 j + 1 = a j for j = 0 , … , m − 1 .

For all j = 0 , 1 , 2 , … , since a j ≡ C ( j , d ) ( mod 2 ) and b 2 j ≡ C ( 2 j , 2 d ) ( mod 2 ) , it is clear that C ( j , d ) ≡ C ( 2 j , 2 d ) ( mod 2 ) by Lemma 3.13. So a j = b 2 j . On the other hand, from

C ( 2 j + 1 , 2 d ) = C ( 2 j , 2 d ) + C ( 2 j , 2 d − 1 )

and

( 2 d − 1 ) C ( 2 j , 2 d − 1 ) = 2 j C ( 2 j − 1 , 2 d − 2 ) ,

we have 2 ∣ C ( 2 j , 2 d − 1 ) . Therefore,

b 2 j + 1 ≡ C ( 2 j + 1 , 2 d ) = C ( 2 j , 2 d ) + C ( 2 j , 2 d − 1 ) ≡ C ( 2 j , 2 d ) ≡ b 2 j ( mod 2 ) ,

and b 2 j + 1 = b 2 j .□

By a direct calculation and Lemma 3.14, we have

Lemma 3.28

For d = 2 k − 1 , 1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ = 0 ⋯ 01 ︸ 2 k ¯ .

From Proposition 3.27 and Lemma 3.28, we have

Lemma 3.29

For k ≥ 2 , d = 2 k + 1 − 2 , 2 k + 2 − 4 , … , 2 k + t − 2 t , and t ≥ 1 , we have

1 − ( − 1 ) C ( j , 2 k + 1 − 2 ) 2 j = 0 ∞ = 0 ⋯ 011 ︸ 2 k + 1 ¯ ,

1 − ( − 1 ) C ( j , 2 k + 2 − 4 ) 2 j = 0 ∞ = 0 ⋯ 01111 ︸ 2 k + 2 ¯ ,

⋯ ,

1 − ( − 1 ) C ( j , 2 k + t − 2 t ) 2 j = 0 ∞ = 0 ⋯ 0 1 ⋯ 1 ︸ 2 t ︸ 2 k + t ¯ .

Theorem 3.30

For k ≥ 2 , t ≥ 1 , d = 2 k + t − 2 t , and n ≥ d , the sensitivity of X ( d , n ) is

s ( X ( 2 k + t − 2 t , n ) ) = 2 k + t n 2 k + t , n = 2 k + t q + r , 0 ≤ r ≤ 2 k + t − 2 t − 1 2 k + t n 2 k + t + 2 k + t − 2 t , n = 2 k + t q + r , 2 k + t − 2 t ≤ r ≤ 2 k + t − 1 .

Proof

By Lemma 3.29, the value vector ⟨ v 0 ⋯ v n ⟩ is the first n + 1 number of 1 − ( − 1 ) C ( j , 2 k + t − 2 t ) 2 j = 0 ∞ = 0 ⋯ 0 1 ⋯ 1 ︸ 2 t ︸ 2 k + t ¯ .

When n = 2 k + t q + r , 0 ≤ r ≤ 2 k + t − 2 t − 1 , the value vector is

⟨ 0 ⋯ 0 1 ⋯ 1 ︷ 2 t ︷ 2 k + t ⋯ ⋯ 0 ⋯ 0 1 ⋯ 1 ︷ 2 t ︷ 2 k + t ︷ q 0 ⋯ 0 ︷ r + 1 ⟩ .

Since k min = 2 k + t − 2 t − 1 and k max = 2 k + t q − 1 , we have

s ( X ( 2 k + t − 2 t , n ) ) = max { 2 k + t q − 1 + 1 , n − 2 k + t + 2 t + 1 } = max { 2 k + t q , 2 k + t ( q − 1 ) + r + 2 t + 1 } = 2 k + t q = 2 k + t n 2 k + t .

When n = 2 k + t q + r , 2 k + t − 2 t ≤ r ≤ 2 k + t − 1 , the value vector ⟨ v 0 ⋯ v n ⟩ is

⟨ 0 ⋯ 0 1 ⋯ 1 ︷ 2 t ︷ 2 k + t ⋯ ⋯ 0 ⋯ 0 1 ⋯ 1 ︷ 2 t ︷ 2 k + t ︷ q 0 ⋯ 0 ︷ 2 k + t − 2 t 1 ⋯ ⋯ ⋯ 1 ︷ r − 2 k + t + 2 t + 1 ⟩ .

From k min = 2 k + t − 2 t − 1 and k max = 2 k + t q + 2 k + t − 2 t − 1 , we have

s ( X ( 2 k + t − 2 t , n ) ) = max { 2 k + t q + 2 k + t − 2 t , n − 2 k + t + 2 t + 1 } = 2 k + t q + 2 k + t − 2 t = 2 k + t n 2 k + t + 2 k + t − 2 t . □

Lemma 3.31

If d = 2 k + 1 , k ∈ N , then

1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ = 1 − ( − 1 ) C ( j , 2 k + 1 ) 2 j = 0 ∞ = 0 ⋯ 0 ︸ 2 k 0101 ⋯ 01 ︸ 2 k ¯ .

Proof

First, it is clear that C ( j , 2 k + 1 ) = 0 for j = 0 , 1 , … , 2 k − 1 . It is straight-forward to check C ( 2 k + 2 j , 2 k + 1 ) is even for j = 0 , 1 , … , 2 k − 1 − 1 (generally, C ( R , S ) is even for even R and odd S ). We only need to show that C ( 2 k + 2 j + 1 , 2 k + 1 ) is odd for j = 0 , 1 , … , 2 k − 1 − 1 .

Let 2 j = c r 2 r + ⋯ + c 1 2 . Then 2 k + 2 j + 1 = 2 k + c r 2 r + ⋯ + c 1 2 + 1 . By the Lucas theorem,

C ( 2 k + 2 j + 1 , 2 k + 1 ) , ≡ C ( 1 , 1 ) C ( c r , 0 ) ⋯ C ( c 1 , 0 ) C ( 1 , 1 ) ≡ 1 ( mod 2 ) .

We are done since the least period of 1 − ( − 1 ) C ( j , 2 k + 1 ) 2 j = 0 ∞ is 2 ⌊ log 2 ( 2 k + 1 ) ⌋ + 1 = 2 k + 1 by Lemma 3.14.□

From Proposition 3.27 and Lemma 3.31, we have

Lemma 3.32

For k ≥ 1 , d = 2 k + 1 + 2 , 2 k + 2 + 4 , … , 2 k + t + 2 t , and t ≥ 1 , we have

1 − ( − 1 ) C ( j , 2 k + 1 + 2 ) 2 j = 0 ∞ = 0 ⋯ 0 ︸ 2 k + 1 0011 ⋯ 0011 ︸ 2 k + 1 ¯ , 1 − ( − 1 ) C ( j , 2 k + 2 + 4 ) 2 j = 0 ∞ = 0 ⋯ 0 ︸ 2 k + 2 00001111 ⋯ 00001111 ︸ 2 k + 2 ¯ , ⋯ , 1 − ( − 1 ) C ( j , 2 k + t + 2 t ) 2 j = 0 ∞ = 0 ⋯ 0 ︸ 2 k + t 0 ⋯ 0 ︸ 2 t 1 ⋯ 1 ︸ 2 t ⋯ 0 ⋯ 0 ︸ 2 t 1 ⋯ 1 ︸ 2 t ︸ 2 k ¯ .

Theorem 3.33

Let k , t ∈ N , d = 2 k + t + 2 t , and n ≥ d .

If n = 2 k + t + 1 q + r , 0 ≤ r ≤ 2 k + t + 2 t − 1 , then

s ( X ( 2 k + t + 2 t , n ) ) = 2 k + t + 1 n 2 k + t + 1 .

If n = 2 k + t + 1 q + 2 k + t + ( j + 1 ) 2 t + i , 0 ≤ j ≤ 2 k − 2 , 0 ≤ i ≤ 2 t − 1 , then

s ( X ( 2 k + t + 2 t , n ) ) = 2 k + t + 1 n 2 k + t + 1 + 2 k + t + ( j + 1 ) 2 t .

Proof

By Lemma 3.32, the value vector ⟨ v 0 ⋯ v n ⟩ is the first n + 1 number of

1 − ( − 1 ) C ( j , 2 k + t + 2 t ) 2 j = 0 ∞ = 0 ⋯ 0 ︸ 2 k + t 0 ⋯ 0 ︸ 2 t 1 ⋯ 1 ︸ 2 t ⋯ 0 ⋯ 0 ︸ 2 t 1 ⋯ 1 ︸ 2 t ︸ 2 k ¯ = B ¯ .

Let n = 2 k + t + 1 q + r , 0 ≤ r ≤ 2 k + t + 1 − 1 .

Case 1

n = 2 k + t + 1 q + r , 0 ≤ r ≤ 2 k + t + 2 t − 1 .

The value vector of X ( 2 k + t + 2 t , n ) is

⟨ B ⋯ B ︷ q 0 ⋯ 0 ︷ r + 1 ⟩ .

Since k min = 2 k + t + 2 t − 1 and k max = 2 k + t + 1 q − 1 , we have

s ( X ( 2 k + t + 2 t , n ) ) = max { n − k min , k max + 1 } = max { 2 k + t + 1 q + r − 2 k + t − 2 t + 1 , 2 k + t + 1 q } = 2 k + t + 1 q = 2 k + t + 1 n 2 k + t + 1 .

Case 2

n = 2 k + t + 1 q + r , 2 k + t + 2 t ≤ r ≤ 2 k + t + 1 − 1 .

Let l = r − 2 k + t − 2 t , 0 ≤ l ≤ 2 k + t + 1 − 2 k + t − 2 t − 1 , l = 2 t j + i , 0 ≤ j ≤ 2 k − 2 , 0 ≤ i ≤ 2 t − 1 .

When n = 2 k + t + 1 q + 2 k + t + 2 t + 2 t j + i , j is even, the value vector is

⟨ B ⋯ B ︷ q 0 ⋯ 0 ︷ 2 k + t 0 ⋯ 0 ︷ 2 t 1 ⋯ 1 ︷ 2 t 0 ⋯ 0 ︷ 2 t ⋯ ⋯ 1 ⋯ 1 ︷ 2 t 0 ⋯ 0 ︷ 2 t ︷ j 1 ⋯ 1 ︷ i + 1 ⟩ .

From k min = 2 k + t + 2 t − 1 and k max = 2 k + t + 1 q + 2 k + t + 2 t + 2 t j − 1 , we have

s ( X ( 2 k + t + 2 t , n ) ) = max { n − k min , k max + 1 } = max { 2 k + t + 1 q + 2 t j + i + 1 , 2 k + t + 1 q + 2 k + t + 2 t + 2 t j } = 2 k + t + 1 q + 2 k + t + 2 t ( j + 1 ) = 2 k + t + 1 n 2 k + t + 1 + 2 k + t + ( j + 1 ) 2 t .

When n = 2 k + t + 1 q + 2 k + t + 2 t + 2 t j + i , j is odd, the value vector is

⟨ B ⋯ B ︷ q 0 ⋯ 0 ︷ 2 k + t 0 ⋯ 0 ︷ 2 t 1 ⋯ 1 ︷ 2 t 0 ⋯ 0 ︷ 2 t ⋯ ⋯ 1 ⋯ 1 ︷ 2 t ︷ j 0 ⋯ 0 ︷ i + 1 ⟩ .

The calculation is identical to the case of even j . The theorem is proved.□

Example 3.34

In Theorem 3.30, if k = 2 , t = 1 , then d = 6 . If k = 2 , t = 2 , then d = 12 .

In Theorem 3.33, if k = 1 , t = 1 , then d = 6 . If k = 2 , t = 1 , then d = 10 . If k = 1 , t = 2 , then d = 12 . One can check that these results are consistent with the previous lemmas.

If 1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ for an odd d can be calculated, then one can find the sequence

1 − ( − 1 ) C ( j , 2 k d ) 2 j = 0 ∞

by Proposition 3.27. Hence, the explicit formula of s ( X ( 2 k d , n ) ) can be obtained.

Let A = { 2 k + t − 2 t ∣ k ≥ 2 , t ≥ 1 } , B = { 2 k + t + 2 t ∣ k ≥ 1 , t ≥ 1 } , and C = { 2 k ∣ k ≥ 1 } . For A , B , C , any set is not contained in the other set. It is easy to check that

A ∩ B = { 2 i + 2 − 2 i = 2 i + 1 + 2 i ∣ i ≥ 1 } = { 3 × 2 i ∣ i ≥ 1 }

and

A ∪ B ∪ C = { 2 , 4 , 6 , 8 , 10 , 12 , 14 , 16 , 18 , 20 , 24 , 28 , 30 , 32 , 34 , 36 , 40 , 48 , 56 , 60 , 62 , 64 , 66 , 68 , 72 , 80 , 96 , 124 , … } .

4 The block sensitivities of symmetric Boolean functions

In this section, we will obtain a formula for the block sensitivity of symmetric Boolean function based on its value vector.

Let x = ( x 1 , … , x n ) ∈ F n , [ n ] = { 1 , … , n } . For any subset S of [ n ] , we form x S by complementing those bits in x indexed by elements of S .

Definition 4.1

[24] The block sensitivity b s ( f ; x ) of f at x is the maximum number of disjoint subsets B 1 , … , B r of [n] such that for all j , f ( x ) ≠ f ( x B j ) . We refer to such a set B j as a block. The block sensitivity of f , denoted b s ( f ) , is max x b s ( f ; x ) .

Obviously, we have 0 ≤ s ( f ; x ) ≤ b s ( f ; x ) ≤ n and 0 ≤ s ( f ) ≤ b s ( f ) ≤ n .

Example 4.2

Let n = 6 and ⟨ v 0 v 1 v 2 v 3 v 4 v 5 v 6 ⟩ = ⟨ 1100011 ⟩ be the value vector of a symmetric Boolean function f ( x ) . We calculate the block sensitivities of f over the words x 0 = ( 0 , 0 , 0 , 0 , 0 , 0 ) , x 3 = ( 1 , 1 , 1 , 0 , 0 , 0 ) , and x 5 = ( 1 , 1 , 1 , 1 , 1 , 0 ) .

For x 0 = ( 0 , 0 , 0 , 0 , 0 , 0 ) , f ( x 0 ) = v 0 = 1 . In order to change the value to 0, we have to change at least two 0s in x 0 = ( 0 , 0 , 0 , 0 , 0 , 0 ) since v 1 = 1 . We are looking for the maximal number of blocks such that the value of f will be changed when the bits in each of these blocks are changed. So, we just change exactly two zeros in x 0 = ( 0 , 0 , 0 , 0 , 0 , 0 ) . Hence, the maximal number of blocks is 6 2 = 3 . Therefore, b s ( f , ( 0 , 0 , 0 , 0 , 0 , 0 ) ) = 3 .

For x 3 = ( 1 , 1 , 1 , 0 , 0 , 0 ) , f ( x 3 ) = v 3 = 0 . In order to change its value to 1, we either change two 1s to 0s or change two 0s to 1s. There are 3 2 + 3 2 = 2 blocks. Hence, b s ( f , ( 1 , 1 , 1 , 0 , 0 , 0 ) ) = 2 .

For x 5 = ( 1 , 1 , 1 , 1 , 1 , 0 ) , f ( x 5 ) = v 5 = 1 . In order to change its value to 0, we change one 1 to 0 in x 5 = ( 1 , 1 , 1 , 1 , 1 , 0 ) , so, b s ( f , ( 1 , 1 , 1 , 1 , 1 , 0 ) ) = 5 .

Similarly, one can find b s ( f , ( 1 , 0 , 0 , 0 , 0 , 0 ) ) = 5 , b s ( f , ( 1 , 1 , 0 , 0 , 0 , 0 ) ) = b s ( f , ( 1 , 1 , 1 , 1 , 0 , 0 ) ) = b s ( f , ( 1 , 1 , 1 , 1 , 1 , 1 ) ) = 3 . Hence, the block sensitivity of the function f is b s ( f ) = 5 .

We have

Proposition 4.3

Let ⟨ v 0 v 1 ⋯ v n ⟩ = ⟨ u 1 ⋯ u 1 ︷ k 1 u 2 ⋯ u 2 ︷ k 2 ⋯ ⋯ u t ⋯ u t ︷ k t ⟩ be the value vector of symmetric Boolean function f ( x ) , where u 1 ≠ u 2 ≠ ⋯ ≠ u t , k i ≥ 1 , i ≥ 1 . If k i = 1 for some i , then s ( f ) = b s ( f ) = n .

Proof

This follows from Lemmas 3.9 and 3.10.□

Generally, we have

Theorem 4.4

Let ⟨ v 0 v 1 ⋯ v n ⟩ = ⟨ u 1 ⋯ u 1 ︷ k 1 u 2 ⋯ u 2 ︷ k 2 ⋯ ⋯ u t ⋯ u t ︷ k t ⟩ be the value vector of symmetric Boolean function f ( x ) , where u 1 ≠ u 2 ≠ ⋯ ≠ u t , k i ≥ 1 , i ≥ 1 , k 1 + k 2 + ⋯ + k t = n + 1 . v 0 = v 1 = ⋯ = v k 1 − 1 = u 1 , v k 1 = v k 1 + 1 = ⋯ = v k 1 + k 2 − 1 = u 2 , … , v k 1 + ⋯ + k t − 1 = ⋯ = v n = u t . Then the block sensitivity of f ( x ) is

(4.1) b s ( f ) = max ∑ i = 1 j k i + n − ∑ i = 1 j k i k j + 1 , n + 1 − ∑ i = 1 j k i + ∑ i = 1 j k i − 1 k j j = 1 , … , t − 1 .

Proof

If there exists j , 1 ≤ j ≤ t such that k j = 1 , by Proposition 4.3, b s ( f ) = n . It is clear the formula of equation (4.1) is also equal to n .

In the following, we assume k i ≥ 2 for i = 1 , … , t . Since f is symmetric, we only need to calculate the sensitivities of f over the n + 1 words x i = ( 1 ⋯ 1 ︷ i 0 ⋯ 0 ) , i = 0 , 1 , … , n and the greatest sensitivity will be b s ( f ) . We divide this n + 1 words into t groups. For each group, we do straightforward calculation as we did in Example 4.2. We list the results below. For easy notation, we write word ( x 1 , … , x n ) as ( x 1 ⋯ x n ) .

Group 1: x i = ( 1 ⋯ 1 ︷ i 0 ⋯ 0 ) , 0 ≤ i ≤ k 1 − 1 , f ( x i ) = u 1 .

For x 0 = ( 0 ⋯ 0 ) , b s ( f , x 0 ) = n k 1 .

For x 1 = ( 10 ⋯ 0 ) , b s ( f , x 1 ) = n − 1 k 1 − 1 .

⋯

For x k 1 − 1 = ( 1 ⋯ 1 ︷ k 1 − 1 0 ⋯ 0 ) , b s ( f , x k 1 − 1 ) = n − k 1 + 1 1 .

Group 2: x i = ( 1 ⋯ 1 ︷ i 0 ⋯ 0 ) , k 1 ≤ i ≤ k 1 + k 2 − 1 , f ( x i ) = u 2 .

For x k 1 = ( 1 ⋯ 1 ︷ k 1 0 ⋯ 0 ) , b s ( f , x k 1 ) = k 1 1 + n − k 1 k 2 .

For x k 1 + 1 = ( 1 ⋯ 1 ︷ k 1 + 1 0 ⋯ 0 ) , b s ( f , x k 1 + 1 ) = k 1 + 1 2 + n − k 1 − 1 k 2 − 1 .

⋯

For x k 1 + k 2 − 1 = ( 1 ⋯ 1 ︷ k 1 + k 2 − 1 0 ⋯ 0 ) , b s ( f , x k 1 + k 2 − 1 ) = k 1 + k 2 − 1 k 2 + n − k 1 − k 2 + 1 1 .

⋯

Group j : 2 ≤ j ≤ t − 1 , x i = ( 1 ⋯ 1 ︷ i 0 ⋯ 0 ) , k 1 + ⋯ + k j − 1 ≤ i ≤ k 1 + ⋯ + k j − 1 , f ( x i ) = u j .

For x k 1 + ⋯ + k j − 1 = ( 1 ⋯ 1 ︷ k 1 + ⋯ + k j − 1 0 ⋯ 0 ) , b s ( f , x k 1 + ⋯ + k j − 1 ) = k 1 + ⋯ + k j − 1 1 + n − k 1 − ⋯ − k j − 1 k j .

For x k 1 + ⋯ + k j − 1 + 1 = ( 1 ⋯ 1 ︷ k 1 + ⋯ + k j − 1 + 1 0 ⋯ 0 ) , b s ( f , x k 1 + ⋯ + k j − 1 + 1 ) = k 1 + ⋯ + k j − 1 + 1 2 + n − k 1 − ⋯ − k j − 1 − 1 k j − 1 .

⋯

For x k 1 + ⋯ + k j − 1 = ( 1 ⋯ 1 ︷ k 1 + ⋯ + k j − 1 0 ⋯ 0 ) , b s ( f , x k 1 + ⋯ + k j − 1 ) = k 1 + ⋯ + k j − 1 k j + n − k 1 − ⋯ − k j + 1 1 .

⋯

Group t : x i = ( 1 ⋯ 1 ︷ i 0 ⋯ 0 ) , k 1 + ⋯ + k t − 1 ≤ i ≤ n , f ( x i ) = u t .

For x k 1 + ⋯ + k t − 1 = ( 1 ⋯ 1 ︷ k 1 + ⋯ + k t − 1 0 ⋯ 0 ) , b s ( f , x k 1 + ⋯ + k t − 1 ) = k 1 + ⋯ + k t − 1 1 .

For x k 1 + ⋯ + k t − 1 + 1 = ( 1 ⋯ 1 ︷ k 1 + ⋯ + k t − 1 + 1 0 ⋯ 0 ) , b s ( f , x k 1 + ⋯ + k t − 1 + 1 ) = k 1 + ⋯ + k t − 1 + 1 2 .

⋯

For x k 1 + ⋯ + k t − 1 = x n = ( 1 ⋯ 1 ) , b s ( f , ( 1 ⋯ 1 ) ) = k 1 + ⋯ + k t − 1 k t = n k t .

We will first find the maximal sensitivity number in each group.

In Group 1, it is clear that b s ( f , x 0 ) ≤ b s ( f , x 1 ) ≤ ⋯ ≤ b s ( f , x k 1 − 1 ) = n − k 1 + 1 1 .

In Group j , 2 ≤ j ≤ t − 1 , we will show the maximal number will be the first or the last one. Namely, max k 1 + ⋯ + k j − 1 1 + n − k 1 − ⋯ − k j − 1 k j , k 1 + ⋯ + k j − 1 k j + n − k 1 − ⋯ − k j + 1 1 .

Let y m = x k 1 + ⋯ + k j − 1 + m , 0 ≤ m ≤ k j − 1 be the k j words in Group j . We already know

b s ( f , y m ) = k 1 + ⋯ + k j − 1 + m m + 1 + n − k 1 − ⋯ − k j − 1 − m k j − m , 0 ≤ m ≤ k j − 1 .

Now we consider the real variable function r ( x ) = k 1 + ⋯ + k j − 1 + x x + 1 + n − k 1 − ⋯ − k j − 1 − x k j − x over the closed interval [ 0 , k j − 1 ] . Let k 1 + ⋯ + k j − 1 − 1 = A > 0 and n − k 1 − ⋯ − k j = B > 0 , then r ( x ) = A x + 1 + B k j − x + 2 . Since d 2 r d x 2 = 2 A ( x + 1 ) 3 + 2 B ( k j − x ) 3 > 0 over closed interval [ 0 , k j − 1 ] , r ( x ) is concave up over [ 0 , k j − 1 ] . Hence, r ( x ) ≤ max { r ( 0 ) , r ( k j − 1 ) } for any x ∈ [ 0 , k j − 1 ] . For 0 ≤ m ≤ k j − 1 , we have

b s ( f , y m ) = k 1 + ⋯ + k j − 1 + m m + 1 + n − k 1 − ⋯ − k j − 1 − m k j − m ≤ k 1 + ⋯ + k j − 1 + m m + 1 + n − k 1 − ⋯ − k j − 1 − m k j − m = ⌊ r ( m ) ⌋ ≤ ⌊ max { r ( 0 ) , r ( k j − 1 ) } ⌋ = max { ⌊ r ( 0 ) ⌋ , ⌊ r ( k j − 1 ) ⌋ } = max k 1 + ⋯ + k j − 1 1 + n − k 1 − ⋯ − k j − 1 k j , k 1 + ⋯ + k j − 1 k j + n − k 1 − ⋯ − k j + 1 1 = max k 1 + ⋯ + k j − 1 + n − k 1 − ⋯ − k j − 1 k j , k 1 + ⋯ + k j − 1 k j + n − k 1 − ⋯ − k j + 1 = max { b s ( f , y 0 ) , b s ( f , y k j − 1 ) } .

In Group t , it is clear that k 1 + ⋯ + k t − 1 1 ≥ k 1 + ⋯ + k t − 1 + 1 2 ≥ ⋯ ≥ n k t .

Now we put all the maximal numbers or maximal candidates of each group together to form a set

S = n − k 1 + 1 , k 1 + ⋯ + k j − 1 + n − k 1 − ⋯ − k j − 1 k j , k 1 + ⋯ + k j − 1 k j + n − k 1 − ⋯ − k j + 1 , k 1 + ⋯ + k t − 1 2 ≤ j ≤ t − 1 = k 1 + ⋯ + k j − 1 + n − k 1 − ⋯ − k j − 1 k j , k 1 + ⋯ + k t − 1 , n − k 1 + 1 , k 1 + ⋯ + k j − 1 k j + n − k 1 − ⋯ − k j + 1 2 ≤ j ≤ t − 1 = k 1 + ⋯ + k j + n − k 1 − ⋯ − k j k j + 1 , k 1 + ⋯ + k t − 1 , 1 ≤ j ≤ t − 2 ⋃ n − k 1 + 1 , k 1 + ⋯ + k j − 1 k j + n − k 1 − ⋯ − k j + 1 , 2 ≤ j ≤ t − 1 = k 1 + ⋯ + k j + n − k 1 − ⋯ − k j k j + 1 , 1 ≤ j ≤ t − 1 ⋃ k 1 + ⋯ + k j − 1 k j + n − k 1 − ⋯ − k j + 1 , 1 ≤ j ≤ t − 1 = k 1 + ⋯ + k j + n − k 1 − ⋯ − k j k j + 1 , k 1 + ⋯ + k j − 1 k j + n − k 1 − ⋯ − k j + 1 , 1 ≤ j ≤ t − 1 = ∑ i = 1 j k i + n − ∑ i = 1 j k i k j + 1 , n + 1 − ∑ i = 1 j k i + ∑ i = 1 j k i − 1 k j j = 1 , … , t − 1 .

Take the maximal value of this set, we p r o v e the formula of b s ( f ) .□

Since s ( f ) ≤ b s ( f ) ≤ n for any Boolean function f with n variables, by definition, we have the following.

Theorem 4.5

For odd d and n ≥ d , the block sensitivity of X ( d , n ) is n .

Proof

This follows from Theorem 3.17.□

Theorem 4.6

If d = 2 k , k ∈ N and n ≥ d , then b s ( X ( d , n ) ) = b s ( X ( 2 k , n ) ) = 2 k n 2 k .

Proof

Let ⟨ v 0 v 1 ⋯ v n ⟩ = ⟨ u 1 ⋯ u 1 ︷ k 1 u 2 ⋯ u 2 ︷ k 2 ⋯ ⋯ u t ⋯ u t ︷ k t ⟩ be the value vector of X ( 2 k , n ) and n = 2 k q + r with 0 ≤ r ≤ 2 k − 1 , where u 1 ≠ u 2 ≠ ⋯ ≠ u t . From the proof of Theorem 3.19, we know

⟨ v 0 ⋯ v n ⟩ = ⟨ 0 ⋯ 0 ︷ 2 k 1 ⋯ 1 ︷ 2 k ⋯ 0 ⋯ 0 ︷ 2 k 1 ⋯ 1 ︷ 2 k ︷ q 0 ⋯ 0 ︷ r + 1 ⟩

when q is even and

⟨ v 0 ⋯ v n ⟩ = ⟨ 0 ⋯ 0 ︷ 2 k 1 ⋯ 1 ︷ 2 k ⋯ 0 ⋯ 0 ︷ 2 k ︷ q 1 ⋯ 1 ︷ r + 1 ⟩

when q is odd.

In either case, we always have t = q + 1 , k 1 = ⋯ = k t − 1 = 2 k and k t = r + 1 . After a routine simplification of equation (4.1), it is easy to find that no other number is greater than ∑ i = 1 t − 1 k i + n − ∑ i = 1 t − 1 k i k t = 2 k n 2 k in the set S . We obtain b s ( X ( 2 k , n ) ) = 2 k n 2 k .□

In the following, we see the sensitivity is strictly less than the block sensitivity for some elementary symmetric Boolean functions.

Proposition 4.7

For n ≥ 10 , we have

b s ( X ( 10 , n ) ) = 16 n 16 , n = 16 q + r , 0 ≤ r ≤ 3 16 n 16 + 1 , n = 16 q + 4 , 16 q + 5 16 n 16 + 2 , n = 16 q + 6 , 16 q + 7 16 n 16 + 3 , n = 16 q + 8 , 16 q + 9 n , n = 16 q + 10 , 16 q + 12 , 16 q + 14 n − 1 , n = 16 q + 11 , 16 q + 13 , 16 q + 15 .

Proof

When n = 16 q + r , 0 ≤ r ≤ 9 , by Lemma 3.23, the value vector

⟨ v 0 v 1 ⋯ v n ⟩ = ⟨ u 1 ⋯ u 1 ︷ k 1 u 2 ⋯ u 2 ︷ k 2 ⋯ ⋯ u t ⋯ u t ︷ k t ⟩ of X ( 10 , n ) is

⟨ 0000000000110011 ︷ 16 ⋯ 0000000000110011 ︷ 16 ︷ q 0 ⋯ 0 ︷ r + 1 ⟩ .

Therefore, we have t = 4 q + 1 ,

k i = 10 when i ≡ 1 ( mod 4 ) and 1 ≤ i ≤ 4 q ,

k i = 2 when i ≡ 2 , 3 , 4 ( mod 4 ) and 1 ≤ i ≤ 4 q ,

k t = k 4 q + 1 = r + 1 .

If j ≡ 1 and 1 ≤ j ≤ 4 q ( mod 4 ) , then K j = ∑ i = 1 j k i = j + 3 4 10 + 3 j − 3 4 2 = 4 j + 6 .

If j ≡ 2 and 1 ≤ j ≤ 4 q ( mod 4 ) , then K j = ∑ i = 1 j k i = j + 2 4 10 + 3 j − 2 4 2 = 4 j + 4 .

If j ≡ 3 and 1 ≤ j ≤ 4 q ( mod 4 ) , then K j = ∑ i = 1 j k i = j + 1 4 10 + 3 j − 1 4 2 = 4 j + 2 .

If j ≡ 4 and 1 ≤ j ≤ 4 q ( mod 4 ) , then K j = ∑ i = 1 j k i = j 4 10 + 3 j 4 2 = 4 j .

By Theorem 4.4, we have b s ( X ( 10 , n ) ) = max { A , B , C } , where A = max K j + n − K j k j + 1 j = 1 , … , t − 2 , B = K t − 1 + n − K t − 1 k t , and C = max n + 1 − K j + K j − 1 k j j = 1 , … , t − 1 .

Let A s = max K j + n − K j k j + 1 j = 1 , … , t − 2 , j ≡ s ( mod 4 ) , s = 1 , 2 , 3 , 4 .

Then

A 1 = max 4 j + 6 + n − 4 j − 6 2 1 ≤ j ≤ 4 q − 1 , j ≡ 1 ( mod 4 ) = max 2 j + 3 + n 2 1 ≤ j ≤ 4 q − 1 , j ≡ 1 ( mod 4 ) = 2 ( 4 q − 3 ) + 3 + n 2 = 16 q − 3 + r 2 ,

A 2 = max 4 j + 4 + n − 4 j − 4 2 1 ≤ j ≤ 4 q − 1 , j ≡ 2 ( mod 4 ) = max 2 j + 2 + n 2 1 ≤ j ≤ 4 q − 1 , j ≡ 2 ( mod 4 ) = 2 ( 4 q − 2 ) + 2 + n 2 = 16 q − 2 + r 2 ,

A 3 = max 4 j + 2 + n − 4 j − 2 2 1 ≤ j ≤ 4 q − 1 , j ≡ 3 ( mod 4 ) = max 2 j + 1 + n 2 1 ≤ j ≤ 4 q − 1 , j ≡ 3 ( mod 4 ) = 2 ( 4 q − 1 ) + 1 + n 2 = 16 q − 1 + r 2 ,

A 4 = max 4 j + n − 4 j 10 1 ≤ j ≤ 4 q − 1 , j ≡ 4 ( mod 4 ) = 4 ( 4 q − 4 ) + n − 4 ( 4 q − 4 ) 10 = 16 q − 16 + r + 16 10 ,

and A = max { A 1 , A 2 , A 3 , A 4 } = 16 q − 1 + r 2 . It is clear that B = K t − 1 + n − K t − 1 k t = K 4 q + n − K 4 q k 4 q + 1 = 16 q + r r + 1 = 16 q .

Let C s = max n + 1 − K j + K j − 1 k j 1 ≤ j ≤ t − 1 , j ≡ s ( mod 4 ) , s = 1 , 2 , 3 , 4 .

Then

C 1 = max n + 1 − ( 4 j + 6 ) + 4 j + 5 10 1 ≤ j ≤ t − 1 , j ≡ 1 ( mod 4 ) = n + 1 − ( 4 + 6 ) + 4 + 5 10 = n − 9 ,

C 2 = max n + 1 − ( 4 j + 4 ) + 4 j + 3 2 1 ≤ j ≤ t − 1 , j ≡ 2 ( mod 4 ) = n + 1 − ( 4 × 2 + 4 ) + 4 × 2 + 3 2 = n − 6 ,

C 3 = max n + 1 − ( 4 j + 2 ) + 4 j + 1 2 1 ≤ j ≤ t − 1 , j ≡ 3 ( mod 4 ) = n + 1 − ( 4 × 3 + 2 ) + 4 × 3 + 1 2 = n − 7 ,

C 4 = max n + 1 − 4 j + 4 j − 1 2 1 ≤ j ≤ t − 1 , j ≡ 4 ( mod 4 ) = n + 1 − 16 + 7 = n − 8 ,

and C = max { C 1 , C 2 , C 3 , C 4 } = n − 6 = 16 q + r − 6 .

In summary, for n = 16 q + r , 0 ≤ r ≤ 9 , we have

b s ( X ( 10 , n ) ) = max { A , B , C } = max 16 q − 1 + r 2 , 16 q , 16 q + r − 6 = 16 n 16 , n = 16 q + r , 0 ≤ r ≤ 3 16 n 16 + 1 , n = 16 q + 4 , 16 q + 5 16 n 16 + 2 , n = 16 q + 6 , 16 q + 7 16 n 16 + 3 , n = 16 q + 8 , 16 q + 9 .

When n = 16 q + 10 , 16 q + 12 , 16 q + 14 , by Lemma 3.23, we have b s ( X ( 10 , n ) ) = n .

When n = 16 q + 11 , 16 q + 13 , 16 q + 15 , the formula b s ( X ( 10 , n ) ) = n − 1 can be similarly proved.□

From Lemma 3.23 and Proposition 4.7, we know s ( X ( 10 , n ) ) < b s ( X ( 10 , n ) ) for n = 16 q + r , 4 ≤ r ≤ 9 .

After similar calculations, we can obtain the following.

Proposition 4.8

Let d = 12 , n − 2 , n − 1 . Then, s ( X ( d , n ) ) = b s ( X ( d , n ) ) .

Proposition 4.9

For n ≥ 6 , the block sensitivity of X ( 6 , n ) is

b s ( X ( 6 , n ) ) = 8 n 8 , n = 8 q + r , 0 ≤ r ≤ 3 8 n 8 + 1 , n = 8 q + 4 , 8 q + 5 n , n = 8 q + 6 n − 1 , n = 8 q + 7 .

It is clear that s ( X ( 6 , n ) ) < b s ( X ( 6 , n ) ) for n = 8 q + 4 , 8 q + 5 .

5 Conclusion

In this paper, we first improve a proposition of [33] and obtain some properties of the sequence 1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ discussed in ref. [8]. For elementary symmetric Boolean functions X ( d , n ) , we obtain explicit formulas for their sensitivities when d is odd and d = n − 2 , n − 1 , 2 k , 2 t + k ± 2 t for any natural numbers t and k . We also show that for the fixed value of d , the formulas of s ( X ( d , n ) ) and s ( X ( n − d , n ) ) can always be obtained. Further more, the explicit formulas of s ( X ( d × 2 k , n ) ) can be obtained from the formulas of 1 − ( − 1 ) C ( j , d ) 2 j = 0 ∞ . Based on the value vector of a symmetric Boolean function, we provide a formula for its block sensitivity. We apply this formula to elementary symmetric Boolean functions and obtain some block sensitivity formulas. We provide the tight upper and lower bounds for the sensitivities and block sensitivity for elementary symmetric Boolean functions. It is well known that sensitivity and block sensitivity of monotone Boolean functions are equal [24]. In ref. [18], the authors proved that sensitivity and block sensitivity of nested canalizing functions are also equal. Our results in this paper show that the block sensitivities for some elementary symmetric Boolean functions can be strictly greater than their sensitivities.

Acknowledgments

The authors sincerely thank the anonymous referees for their valuable suggestions and comments to improve the presentation quality of our manuscript.

Conflict of interest: Authors state no conflict of interest.

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Received: 2020-10-10

Revised: 2021-03-05

Accepted: 2021-03-05

Published Online: 2021-04-22

This work is licensed under the Creative Commons Attribution 4.0 International License.

Articles in the same Issue

https://doi.org/10.1515/jmc-2020-0042

Keywords for this article

Boolean functions; Lucas theorem

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