#A48 INTEGERS 12 (2012) ON A COMBINATORIAL CONJECTURE OF TU AND DENG

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1 #A48 INTEGERS 12 (2012) ON A COMBINATORIAL CONJECTURE OF TU AND DENG Guixin Deng Schoo of Mathematica Sciences, Guangxi Teachers Education University, Nanning, P.R.China dengguixin@ive.com Pingzhi Yuan Schoo of Mathematics, South China Norma University, Guangzhou, P.R.China mcsypz@mai.sysu.edu.cn Received: 12/22/11, Revised: 4/2/12, Accepted: 8/19/12, Pubished: 9/3/12 Abstract Recenty, Tu and Deng obtained two casses of Booean functions with nice properties based on a combinatoria conjecture about binary strings. In this paper, using different approaches, we prove this conjecture is true in severa cases. 1. Introduction Let x be a nonnegative integer. If the binary expansion of x is x i x i2 i, then the Hamming weight of x is w(x) i x i. In [7] Tu and Deng proposed the foowing conjecture. Conjecture 1. Let S t {(a, b) a, b Z 2n 1, a+b t (mod 2 n 1 ), w(a)+w(b) n 1}, where 1 t 2 n 2, n 2. Then S t 2 n 1. Based on this conjecture, the authors in [7] constructed some casses of Booean functions with many nice cryptographic properties. In this paper we make use of the foowing bijection from Z 2n 1 onto X n, where X n is the set of binary strings of ength n except the string consisting of n copies of 1: n 1 Z 2n 1 X n, x i 2 i x 0 x 1... x n 1. We use t to denote the ength of a binary string t t 0 t 1... t n 1. Let t t 0 t 1... t n 1, where t i 1 t i. We aso use the notation 1 k 0 m : k times m times In [7], Tu and Deng construct an agorithm which they used it to show that the conjecture above is true when n 29. Cusick, Li and Stanica [2] show that Con-

2 INTEGERS: 12 (2012) 2 jecture 1 is true when w(t) 2, or w(t) t 4. In this paper, we wi consider the foowing conjecture, which is equivaent to Conjecture 1. Conjecture 2. Let 1 t 2 n 2, n 2. Let S(t) {a a Z 2n 1, w(x) w(a) + 1, t + a x (mod 2 n 1 )}. Then S(t) 2 n 1. Lemma 3. Let t t 0 t 1... t n 1. The foowing statements are true: (i) S(t) S(t i t i+1... t n 1 t 0... t i 1 ) for any i; (ii) w(t) + w( t) t ; (iii) The map ϕ : S t S( t), ϕ((a, b)) a, is bijective. Hence S t S( t). Proof. The assertions (i) and (ii) are trivia. If (a, b) S t, then w( t + a) w( b) n w(b) w(a) + 1. Hence a S( t). The map ϕ is we-defined and injective. Conversey, if a S( t), then w( t + a) w(a) + 1 so that w(a) + w(t a) w(a) + n w( t + a) w(a) + n w(a) 1 n 1. Thus (a, t a) S t, and therefore ϕ is aso surjective. Note that the authors in [3] actuay showed that Conjecture 2 is true when w(t) t 2, or w(t) 4 according to Lemma 3. The remainder of this paper is organized as foows. In Section 2 we construct a partition on the set of binary strings of fixed ength. The partition guarantees us to compute a cass of S(t), which reaches the bound of Conjecture 2. In Section 3 we compare S(t) and S(t0 m ) based on the partition, and we show that Conjecture 2 is true for some other casses. 2. The Partition t of X n We begin with a emma. Lemma 4. Let t t 0 t 1... t n 1 X n, a a 0 a 1... a n 1 X n. Suppose that I {j 0 j n 1, t j a j } {i 1, i 2,..., i }, where 0 i 1 < i 2 < < i n 1. Assume that t + a 0 n. Then w(t + a) w(a) + w(t) (i s+1 i s ), where we set i +1 i 1 + n. s I,t is 1 Proof. It is cear that I if and ony if t + a 0 n. Suppose that t + a x 0 x 1... x n 1. If j I, then j < i 1 or there exists an s 1 such that i s < j < i s+1.

3 INTEGERS: 12 (2012) 3 If i s < j < i s+1, by a simpe computation we obtain x j 1 t is. If j < i 1, then x j 1 t i. On the other hand, if j i s I, then x j t is 1 if s > 1, and x j t i if s 1. Let I j {i s 1 s, t is a is j}, j 0, 1. Then w(t) + w(a) w(t + a) (n + I 1 I 0 ) ( (i s+1 i s 1) + I 1 ) i s I 0 n (i s+1 i s ) i s I 0 (i s+1 i s ) (i s+1 i s ) i s I i s I 0 (i s+1 i s ). i s I 1 Let S i (t) {a X n w(t) + w(a) w(t + a) i} for any t X n. Then S(t) w(t) 1 S i (t) and X n n 1 S i(t) are both disjoint unions. We construct a partition t on X n according to Lemma 4. Definition 5. Let t t 0 t 1... t n 1 be a binary string of ength n. Suppose that w(t) r, and t m1 t m2 t mr 1, where 0 m 1 < m 2 < < m r n 1. Let a a 0 a 1... a n 1 be a binary string. Suppose that I a {0 j n 1 : a j t j } {i 1, i 2,..., i }, where 0 i 1 < i 2 < < i n 1 and set i +1 i 1 + n. We define (x 1, x 2,..., x r ) t to be the subset of X n such that a (x 1, x 2,..., x r ) t if and ony if the foowing two conditions hod (i) x j i s+1 i s if m j i s I a ; (ii) x j 0 if m j I a. Moreover, we wi use the notation a t : (x 1, x 2,..., x r ) t if a (x 1, x 2,..., x r ) t. Remark 6. The idea of this partition comes from the carries in [5]. It is a refinement of the partition X n i S i (t). So we can obtain more information about the structure of S(t). Definition 7. Let t t 0 t 1... t n 1, a a 0 a 1... a n 1 be two given binary strings of ength n. For any 0 m n 1, we set a t m i, if b m i for a b a t, i 0, 1. Moreover, we say that a t m is free if there exist two strings b and b in a t such that b m 0 and b m 1. Exampe 8. Let t , a , b , and c Then t 1 t 4 t 7 1, and I a {0, 1, 3, 5, 6, 7}, I b {1, 2, 4, 5, 7, 8}, and I c {2, 4, 5, 7}. So a (2, 0, 2) t, b (1, 1, 1) t, and c (0, 1, 4) t. Moreover, by definition 5 a a t if and ony if a , where 0 or 1. That is, a t 5 and a t 6 are both free and a t 4. Moreover, b b t if and ony if b , c c t if and ony if c , where 0 or 1.

4 INTEGERS: 12 (2012) 4 It foows that a t (x i ) t S r i1 xi(t) from Definition 5 and Lemma 4. Moreover, if k is the number of indices such that a t i is free, then at 2 k. Let a i (j) {(x 1, x 2,..., x j ) x j N, j i + j 1 x k i}. i k1 The proof of the foowing emma is easy so we omit it. Lemma 9. For any r 1, c i a i (r) (c i ji+1 c j )a i (r + i + 1). In particuar, 2 i a i (r) 2 a i(r + i + 1). The resuts of Theorem 10 and Lemma 11 have been proved in [5]. Here we give another proof. Theorem 10. Let t 10 s1 10 s sr, where s i r 1 and w(t) r. Then S(t) 2 t 1. Proof. Suppose that t n, a (x 1, x 2,..., x r ) t and r i1 x i r 1. Then a is of the form r , k 1 k 2 k r where i 1 and k i x i if x i > 0, and i 0, k i 0 if x i 0. Observe that there are exacty n r r i1 x i free indices of (x 1, x 2,..., x r ) t if x i s i. Since S(t) S i(t) is a disjoint union, one has S(t) S i (t) r j1 xji (x 1, x 2,..., x r ) t 2 n r i a i (r) 2r 1 2 n 2r+1 i 2 n 2r n 1. Lemma 11. Let t 1 k 0 m. Then S j (t) b j 2 m j, where 1, j 0 4 b j j 1 3, 1 j k 4 k 1 3, k < j m

5 INTEGERS: 12 (2012) 5 Proof. The case j 0 is trivia. In fact for any 1 < j m, S j (t) (x 1, x 2,..., x k ) t k i1 xij k (0,..., 0, x i, x i+1,..., x k ) t ik j+1 x i>0, i xij k 2 k i 2 m j+k i ik j+1 2 m j k b j 2 m j. ik j+1 2 2(k i) Theorem 12. Let t 1 k 0 s1 10 s s r k+1, where w(t) r, k 1 and s i r 1. Then S(t) 2 t 1. Proof. Suppose that t 1 k 0 s1, t 10 s s r k+1 such that t n 1, t n 2, and t n. Let b j satisfy S j (t ) b j 2 n1 k j. Then S(t) S i (t) i1 r i 1 j0 2 n r [ S i j (t ) S j (t ) r i 1 j0 2 n r [ c i a i (r k)] b j 2 i j a i (r k)] r k 1 2 n r [ 2 k i 1 a i (r k + 1) r+1 a (r k + 1)], ir k 2 i 2 a i (r k + 1) where c i 2 k i 2 r+2k+1 2 r+1 3 if 0 i r k 1, and c i 2i +2 r+1 3 if r k i r 1.

6 INTEGERS: 12 (2012) 6 Let T 1 k 1 0 s1 10 s s r k+1+1. Then T n and w(t ) r 1. Simiary, r k 1 S(T ) 2 n r+1 [ 2 k i 2 a i (r k + 1) + Therefore, + 2 r+2 a (r k + 1)] r k 1 2 n r [ 2 k i 1 a i (r k + 1) r+3 a (r k + 1)]. r 3 ir k r 3 ir k 2 i 3 a i (r k + 1) 2 i 2 a i (r k + 1) S(T ) S(t) 2 n 2r+1 [2a (r k + 1) a (r k + 1)] 2r k 2 2r k 1 2 n 2r+1 [2 ] r 2 r 1 k 1 2r k 2 2 n 2r+1 > 0. r 1 r 2 It foows by induction that S(t) S(10 s1+k 1 10 s s r k+1+k 1 ) 2 n A Comparison Lemma and its Appication It was conjectured in [5] that the converse of Theorem 10 is true. Let t t 0 t 1... t n 1 and T t0 N n be two binary strings, where N n w(t) 1. Another question is whether S(T ) 2 N n S(t). We cannot prove that, but we have the foowing resut based on the partition on X n. We consider what happens if we add some 0 s in the string. Lemma 13. Let t 10 s su 10 s sr, T 10 s su s r such that t T n, w(t) w(t ) r. Suppose that s i r 1 for any i u + 2, s r r 1. Let m i (s i + 1) for any 1 u + 1. Then S(T ) S(t) 1 j0 i1 xij 2 r+j+1 [a r j 1 (r u 1) a r j 2 (r u)] (x 1,..., x 1, x m, 0,..., 0) T. Proof. Suppose that u + 1 and x < m. By comparing the free indices the cardinaity of (x 1,..., x, 0,..., 0, x u+2, x u+3,..., x r ) t is equa to the cardinaity of (x 1,..., x, 0,..., 0, x u+2, x u+3,..., x r ) T. Let I 1 1 x m, i xi (x 1,..., x, 0,..., 0, x u+2, x u+3,..., x r ) T,

7 INTEGERS: 12 (2012) 7 I 2 1 x m, i xi (x 1,..., x, 0,..., 0, x u+2, x u+3,..., x r ) t. Then S(T ) S(t) I 1 I 2. For any given (x 1, x 2,..., x, 0,..., 0, x u+2, x u+3,..., x r ) t such that x m, one has x u+2 0 if x > m, x u+2 > 0 if x m. Moreover, if x > m, the cardinaity of (x 1,..., x, 0,..., 0, x u+2, x u+3,..., x r ) t is equa to the cardinaity of (x 1,..., x 1, m, 0,..., 0, x u+2 x m, x u+3,..., x r ) t. So I x m,x u+2>0, r i1 xi (x 1,..., x, 0,..., 0, x u+2,..., x r ) t (x 1,..., x m, 0,..., 0, x u+2,..., x r ) T x u+2>0, r i1 xi 1 j0 i1 xij r iu+2 xi r j 1,xu+2>0 4 Simiary, I 1 Therefore, 1 j0 i1 xij 1 r j 2 k0 (x 1,..., x m, 0,..., 0, x u+2,..., x r ) T 2 1 k a k (r u 1) (x 1,..., x m, 0,..., 0) T. (x 1,..., x, 0,..., 0, x u+2,..., x r ) T x m, i1 xi 1 j0 1 i1 xi+m j 1 j0 i1 xij I 1 I 2 r j 1 k0 y+ r iu+2 r j 1 (x 1,..., m + y, 0,..., 0, x u+2,..., x r ) T 2 k a k (r u) (x 1,..., x 1, x m, 0,..., 0) T. r j 1 [ 1 j0 i1 xij k0 r j 2 2 k a k (r u) (x 1,..., x 1, x m, 0,..., 0) T 1 j0 k0 2 1 k a k (r u 1)] i1 xij 2 r+j+1 [a r j 1 (r u 1) a r j 2 (r u)] (x 1,..., x 1, x m, 0,..., 0) T.

8 INTEGERS: 12 (2012) 8 This finishes the proof. Theorem 14. Let t 10 s1 10 s s 10 sr, where s i i 2. Then S(t) 2 t 1. In particuar, S(t) 2 t 1 if s 1 < s 2 < < s r. Proof. It is cear that 2 m S(t) S(t0 m ) 2 m 1 S(t0) for any m 1 since s r r 2. So we can assume that s r is sufficienty arge. We set t (i) 10 s1 10 s si sr,i, such that t (i) n, w(t i ) r, and s r,i r 1. Let m, i (s i + 1) for any u + 1 r 1. By Lemma 13, u S(t (u) ) S(t () ) 1 j0 i1 xij 2 r+j+1 [a r j 1 (r u 1) a r j 2 (r u)] (x 1,..., x 1, x m,, 0,..., 0) t(u). Suppose that (x 1,..., x 1, x m,, 0,..., 0) t(u), where i1 x i j and m, s +1. Then r j 1 r s 2 r u 1 since s u 1. It foows that a r j 1 (r u 1) a r j 2 (r u) 0 and u 0. By induction S(t) S(t () ) S(t (r 3) ) S(t (0) ) 2 n 1. Theorem 15. Let t 10 s1 10 s s 10 sr. Suppose that s i + s j r 2 for any 1 i, j r. Then S(t) 2 t 1. Proof. We can assume that s r < r 1. Let T 10 s1 10 s s 10. Suppose that T N, t n. It suffice to show that S(T ) 2 N n S(t). Simiar to the proof of Lemma 13, one has S(T ) 2 N n S(t) (x 1, x 2,..., x r ) T xi,x r s r+1 2 N n+1 xi,x 1>0 (x 1, x 2,..., x r s r + 1) t. But I 1 xi (x 1, x 2,..., x r s r + 1) T r s j0 2 N r j s a j (r), I 2 2 xi,x 1>0 (x 1, x 2,..., x r s r + 1) t r s r 3 j0 2 N r sr j a j (r 1).

9 INTEGERS: 12 (2012) 9 Thus, S(T ) 2 N n S(t) I 1 I 2 r s r s r 3 2 N r s [ a j (r) 2 1 j a j (r 1)] The proof is competed. Coroary 16. S(t) 2 t 1 if w(t) 6. j0 j0 2 N r s [a r s(r 1) a r sr 3(r)] 0. Proof. We ony treat the case w(t) 6. Suppose that t 10 s1 10 s2 10 s3 10 s4 10 s5 10 s6. Let t 10 r1 10 r2 10 r3 10 r4 10 r5 10 r6, where r i min{s i, 5}. Then (x 1, x 2, x 3, x 4, x 5, x 6 ) t (x 1, x 2, x 3, x 4, x 5, x 6 ) t is a bijection between S(t) t and S(t ). Moreover, by t comparing the number of free indices (x 1, x 2, x 3, x 4, x 5, x 6 ) t 2 t t (x 1, x 2, x 3, x 4, x 5, x 6 ) t. Therefore, S(t) 2 t t S(t ). It remains to show that S(t ) 2 n 1. We can assume that t 30. If min 1 i 6 {r i } 2, then r i +r j w(t) 2 4 for any 1 i, j 6. If min 1 i 6 {r i } r 1 1, then min 2 i 6 {r i } 3. If min 1 i 6 {r i } r 1 0, then min 2 i 6 {r i } 4. We have r i + r j w(t) 2 4 for any 1 i, j 6. The coroary foows from Theorem 15. References [1] C. Caret, On a weakness of the Tu-Deng functions and its repair. IACR Crypotoogy eprint Archive 606 (2009). [2] G. Cohen, Jean-P. Fori, On a generaized combinatoria conjecture invoving addition mod 2 k 1. IACR Cryptoogy eprint Archive 400 (2011). [3] T. W. Cusick, Y. Li, and P. Stănică, On a combinatoria conjecture. Integers 11 (2011), [4] Jean-P. Fori, H. Randriam, G. Cohen, and S. Mesnager, On a conjecture about binary strings distribution. Sequences and Their Appications SETA 2010, LNCS 6338 (2010), [5] Jean-P. Fori, H. Randriam, G. Cohen, and S. Mesnager, On a conjecture about binary strings distribution. IACR Cryptoogy eprint Archive 170 (2010). [6] Jean-P. Fori, H. Randriam, On the number of carries occuring in an addition mod 2 k 1. Integers 12 (2012). [7] Z. Tu and Y. Deng, A conjecture on binary string and its appication on constructing Booean Functions of optima agebraic immunity. Designs, Codes and Cryptography 60 (2010), [8] Z. Tu and Y. Deng, A cass of 1 Resiient function with high noninearity and agebraic immunity. IACR Cryptoogy eprint Archive 179 (2010).