New Constructions of Quaternary Hadamard Matrices

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1 New Constructions of Quaternary Hadamard Matrices Ji-Woong Jang 1, Sang-Hyo Kim 1, Jong-Seon No 1, and Habong Chung 2 1 School of Electrical Engineering and Computer Science, Seoul National University, Seoul , Korea {stasera, shkim}@ccl.snu.ac.kr, jsno@snu.ac.kr 2 School of Electronics and Electrical Engineering, Hong-Ik University, Seoul , Korea habchung@hongik.ac.kr Abstract. In this paper, we propose two new construction methods for quaternary Hadamard matrices. By the first method, which is applicable for any positive integer n, we are able to construct a quaternary Hadamard matrix of order 2 n from a binary sequence with ideal autocorrelation. The second method also gives us a quaternary Hadamard matrix of order 2 n from a binary extended sequence of period 2 n 1, where n is a composite number. 1 Introduction A generalized Hadamard matrix H of order N is an N N matrix satisfying HH = NI N, where denotes the conjugate transpose and I N is the identity matrix of order N [3,8,13]. In other words, any two distinct rows of H are orthogonal. For this reason, Hadamard matrices have been studied for the applications in many areas such as wireless communication systems, coding theory, and signal design[1,,1,15,16]. Hadamard matrices have strong ties to sequences. Matsufuji and Suehiro proposed the complex Hadamard matrices related to bent sequences[9]. Popovic, Suehiro, and Fan[12] proposed orthogonal sets of quaternary sequences by using quadriphase sequence family A by Boztas, Hammons, and Kumar[2]. In this paper, we propose two new construction methods for quaternary Hadamard matrices. By the first method, which is applicable for any positive integer n, we are able to construct a quaternary Hadamard matrix of order 2 n from a binary sequence with ideal autocorrelation. The second method also gives us a quaternary Hadamard matrix of order 2 n from a binary extended sequence of period 2 n 1, where n is a composite number. Before we proceed to the next section, let us clarify some terms and notations used throughout this paper. Let F 2 n be the finite field with 2 n elements. Let F2 = F n 2n \{0} and s(x) be a mapping from F 2 n to F 2 or Z. If we restrict the mapping s(x) tof2 and n replace x by α t, where α is a primitive element in F 2 n, then we can obtain a sequence s(α t ), 0 t 2 n 2, of period 2 n 1. Hence, for convenience, we T. Helleseth et al. (Eds.): SETA 200, LNCS 386, pp , c Springer-Verlag Berlin Heidelberg 2005

2 362 J.-W. Jang et al. will use the expression a binary or quaternary sequence s(α t )ofperiod2 n 1 interchangeably with a mapping s(x) fromf 2 n to F 2 or Z. For δ F2 n, the crosscorrelation function between two quaternary sequences s i (x) ands j (x) is defined as R i,j (δ) = ω si(xδ) sj(x), x F2 n where w is a complex fourth root of unity. Let f(x) be a mapping from F 2 n onto F 2 m, where m n. The function f(x) is said to be balanced if each nonzero element of F 2 m appears 2 n m times and zero element 2 n m 1 times in the list {f(x) x F2 n}. A function f(x) is said to be difference-balanced if f(δx) f(x) is balanced for any δ F 2 n\{0, 1}. It is easy to see that the binary sequence with difference-balance property has the ideal autocorrelation property necessarily and sufficiently. It is not difficult to see that a variable v over Z can be expressed using two binary variables v 1 and v 2 as v = v 1 +2v 2, where addition is modulo. Let us define two maps φ and ψ as φ(v) =v 1, ψ(v) =v 2. It can be shown that φ(v w) andψ(v w) of the difference v w are expressed as φ(v w) =v 1 + w 1 ψ(v w) =v 1 w 1 + w 1 + w 2 + v 2. (1) 2 New Constructions of Quaternary Hadamard Matrices In this section, we propose two constructions for quaternary Hadamard matrices from binary sequences with ideal autocorrelation. Lemma 1. For a positive integer n, letg(t) be a binary sequence of period 2 n 1 with ideal autocorrelation. Then for any z, 1 z 2 n 2, the following sequence q z (t) given by q z (t) =g(t)+2g(t + z) is balanced over Z. Proof.LetN z (a, b), a,b {0, 1} be the number of t such that g(t) =a and g(t + z) = b. Since g(t) has the ideal autocorrelation property, it is balanced and difference-balanced. Thus we have N z (0, 0) + N z (0, 1) = 2 n 1 1 N z (0, 0) + N z (1, 0) = 2 n 1 1 N z (0, 0) + N z (1, 1) = 2 n 1 1.

3 Finally, from the facts that we can conclude that q z (t) is balanced. New Constructions of Quaternary Hadamard Matrices 363 N z (a, b) =2 n 1, a b Using the above lemma, we get the quaternary Hadamard matrices as in the following theorem. Theorem 1. Let n be an integer and g(t), 0 t 2 n 2, be a sequence of period 2 n 1 with ideal autocorrelation. Then the following matrix H Q is a 2 n 2 n quaternary Hadamard matrix. H Q =(h ij ), 0 i, j 2 n 1, where h ij is given as 1, for i =0orj =0 h ij = w 2g(j 1), for i =1and1 j 2 n 1 w g(j 1)+2g(i 1+j 1) = w qi 1(j 1), otherwise. Proof.Letu i be the ith row of H Q. It is clear that u i u i =2n, 0 i 2 n 1. In proving the orthogonality between u i and u k, we should consider the following three cases. Case 1) i =0and1 k 2 n 1: From Lemma 1 and balance property of g(t) andq k (t), it is clear that u 0 is orthogonal to u k, for any k, 1 k 2 n 1. Case 2) i =1,2 k 2 n 1: In this case, u 1 u k is given as 2 n 2 u 1 u k =1+ t=0 w 2g(t) g(t) 2g(t+k 1) 2 n 2 =1+ w g(t) 2g(t+k 1). t=0 From Lemma 1, it is straightforward that g(t) 2g(t + k 1) is also balanced and thus u 1 u k = 0, i.e., u 1 is orthogonal to u k.

4 36 J.-W. Jang et al. Case 3) 2 i<k 2 n 1: In this case, u i u k is given as 2 n 2 u i u k =1+ t=0 2 n 2 =1+ t=0 w {g(t)+2g(t+i 1)} {g(t)+2g(t+k 1)} w 2(g(t+i 1)+g(t+k 1)) 2 n 2 =1+ ( 1) g(t+i 1)+g(t+k 1). t=0 From the difference-balance property of g(t), u i u k =0. Here is an example of an 8 8 quaternary Hadamard matrix constructed from the above theorem. Example 1. Let α be a primitive element in F 2 3. Using the m-sequence tr 3 1(α t ) of period 7, we can construct the quaternary sequences of period 7 as s 0 (t) = 2tr 3 1(α t ) s i (t) =tr 3 1(α t ) + 2tr 3 1(α t+i ), 1 i 6, which gives us H Q ω 0 ω 0 ω 0 ω 0 ω 0 ω 0 ω 0 ω 0 ω 0 ω 2 ω 0 ω 0 ω 2 ω 0 ω 2 ω 2 ω 0 ω 1 ω 0 ω 2 ω 1 ω 2 ω 3 ω 3 H Q = ω 0 ω 1 ω 2 ω 0 ω 3 ω 2 ω 3 ω 1 ω 0 ω 3 ω 0 ω 2 ω 3 ω 2 ω 1 ω 1. ω 0 ω 1 ω 2 ω 2 ω 3 ω 0 ω 1 ω 3 ω 0 ω 3 ω 2 ω 2 ω 1 ω 0 ω 3 ω 1 ω 0 ω 3 ω 2 ω 0 ω 1 ω 2 ω 1 ω 3 No, Yang, Chung, and Song constructed extended sequences with ideal autocorrelation from sequences of shorter period with ideal autocorrelation[11]. Theorem 2 (No, Yang, Chung, and Song[11]). Let n and m be positive integers such that m n. Letf(y) be the function from F 2 m to F 2 with differencebalance property such that f(0) = 0. Let r be an integer such that gcd(r, 2 m 1) = 1 and 1 r 2 m 2. Then the sequence of period 2 n 1 defined by f([tr n m(x)] r ) has the ideal autocorrelation property.

5 New Constructions of Quaternary Hadamard Matrices 365 Using the extended sequences, we can construct the quaternary Hadamard matrix as in the following theorem. Theorem 3. Let n and m be integers such that m n, andr be an integer such that 1 r 2 m 2 and gcd(r, 2 m 1) = 1. Let T = 2n 1 2 m 1 and f(y) bethe sequence from F 2 m to F 2 which has the balance and difference-balance properties. Let s i (α t ) be defined as s 0 (α t )=2f([tr n m(α t )] r ) s i (α t )=f([tr n m(α t )] r +2f([tr n m(β i α t )] r ), 1 i 2 m 2, where β = α T is a primitive element in F 2 m. Then the following matrix H L is a 2 n 2 n quaternary Hadamard matrix. H L =(h ij ), where h ij is given as { 1, if i =0orj =0 h ij = w s (i 1)/T (j 1+i T ), otherwise, where x denotes the greatest integer not exceeding x and i T =(i 1) mod T. Proof of the above theorem requires following lemmas. Lemma 2. Let m, e, andn be positive integers such that n = em. Letq =2 m and A = {1,α,,α T 1 }, where α is a primitive element in F 2 n and T = qe 1 q 1. Let v(x) be a function from F q e onto F q with the balance and difference-balance properties. Further assume that v(x) satisfies v(yx) =yv(x) for any y F q and x F q e. For a given δ F q e \F q,letm δ (a, b) be the number of x 2 A satisfying Then, we have c F q v(δx 2 )=a and v(x 2 )=b, a, b F q. M δ (0, 0) = qe 2 1 q 1 M δ (c, 0) = c F q = 2n 2m 1 2 m 1 M δ (0,c)=q e 2 =2 n 2m M δ (cd, d) =q e 2 =2 n 2m, for any c Fq. d F q Proof. Let N δ (a, b) be the number of x F q e satisfying v(δx) =a and v(x) =b. Let x = x 1 x 2, where x 1 F q and x 2 A. Because v(x) is difference-balanced, v(δx) v(cx) =v(δx) cv(x) is balanced for any c F q and 0 occurs q e 1 1 times as x varies over F q e.thuswehave a F q N δ (ca, a) =q e 1 1. (2)

6 366 J.-W. Jang et al. Since v(x) is balanced, we have N δ (a, 0) = N δ (0,b)=q e 1 1. (3) a F q b F q Also, note that a F q b F q N δ (a, b) =q e 1. () Now, we have N δ (a, b) = N δ (a, 0) + N δ (0,b) a F q b F q a F q b F q N δ (0, 0) + N δ (a, ca) c Fq a Fq = a F q N δ (a, 0) + b F q N δ (0,b) N δ (0, 0) + N δ (a, ca) N δ (0, 0). (5) a F q c F q Plugging (2), (3), and () into (5), we have N δ (0, 0) = q e 2 1. (6) From (2) and (6), we also have N δ (ca, a) = N δ (ca, a) N δ (0, 0) = q e 2 (q 1). a F q a F q Let β = α T. For a given x 2 such that v(δx 2 )=cv(x 2 ), the ordered pair (v(δx),v(x)) = (x 1 v(δx 2 ),x 1 v(x 2 )) takes each value in the list (c, 1), (cβ, β),, (cβ q 2,β q 2 ) exactly once as x 1 varies over Fq. Therefore we have N δ (ca, a) =(q 1) M δ (ca, a), a F q which, in turn, tells us that a F q M δ (ca, a) =q e 2. a F q

7 Similarly, we have New Constructions of Quaternary Hadamard Matrices 367 M δ (0, 0) = N δ(0, 0) = qe 2 1 q 1 q 1 c F N q M δ (c, 0) = δ (c, 0) = q e 2 q 1 c F N q M δ (0,c)= δ (0,c) = q e 2. q 1 c F q c F q Lemma 3. Let s(x) be a function from any domain B to Z, where s(0) = 0. Define two Boolean constituent functions of s(x) as φ s (x) =φ(s(x)), ψ s (x) =ψ(s(x)) and their modulo-2 sum as µ s (x) =φ s (x)+ψ s (x). (7) Let N f (c) denote the number of occurrences of f(x) =c as x varies over B. Then, we have x B ω s(x) =(N ψs (0) N µs (1)) + j(n µs (1) N ψs (1)). Proof. It is clear that ω s(x) =(N s (0) N s (2)) + j(n s (1) N s (3)) and x B From (8), (9), and (10), we have N ψs (1) = N s (2) + N s (3) (8) N ψs (0) = 2 n N ψs (1) = N s (0) + N s (1) (9) N µs (1) = N s (1) + N s (2). (10) N s (0) N s (2) = N ψs (0) N µs (1) N s (1) N s (3) = N µs (1) N ψs (1). Thus we prove the lemma.

8 368 J.-W. Jang et al. Corollary 1. Let s(x) be a function from F 2 n to Z. Then, ω s(x) =0 x F 2 n if and only if the functions ψ s (x) andµ s (x) are balanced. Now we are ready to prove Theorem 3. Proof of Theorem 3. Let v i be the ith row of H L,0 i 2 n 1. We have to show that v i v k = 0 for all i k. The case when i = 0 is simple. Since v 0 is an all one sequence, we need to show that the row sum is zero for each row v k,k 0. From the structure of H L, it is manifest that the rows v 1+lT through v T +lt, 0 l 2 m 2, are the cyclic shifts of s l (x). Also note that s 0 (x) is balanced since it is in fact the binary extended sequence, and s l (x),l 0, is also balanced from Lemma 1. Thus we have v 0 v k = 0 for all k 0. Now, for any nonzero i and k, i k, v i v k canbeexpressedas 2 n 2 v i v k =1+ t=0 =1+ w s (i 1)/T (t+i T ) s (k 1)/T (t+k T ) w si (δx) sk (x), x F2 n where δ = α it kt, i = (i 1)/T, andk = (k 1)/T. Forδ = α it kt, showing that v i v k = 0 is equivalent to showing that the crosscorrelation R i,k (δ) between s i (x) ands k (x) is 1. For a, b F 2 m \F 2, define two quaternary sequences u a (x)andu b (x) of period 2 m 1as u a (x) =f(x)+2f(ax) u b (x) =f(x)+2f(bx) and let d(x, η) =u a (ηx) u b (x). Define S ψd and S µd as S ψd = S µd = ( 1) ψ(d(x,η)) x F2 m η F2 m x F 2 m η F 2 m ( 1) µ(d(x,η)). Then from (1) and (7), S ψd and S µd canbeexpressedas S ψd = S µd = ( 1) f(ηx)f(x)+f(x)+f(bx)+f(aηx) x F2 m η F2 m x F 2 m (11) η F 2 m ( 1) f(ηx)f(x)+f(ηx)+f(bx)+f(aηx). (12)

9 New Constructions of Quaternary Hadamard Matrices 369 Now, let I 1 (x) andi 2 (x) be the inner summation in (11), η F 2 m ( 1) f(ηx)f(x)+f(x)+f(bx)+f(aηx) for the cases when f(x) =0andf(x) = 1, respectively, i.e., I 1 (x) = η F 2 m ( 1) f(bx)+f(aηx) and I 2 (x) = ( 1) f(ηx)+1+f(bx)+f(aηx). η F2 m Then S ψd canbeexpressedas S ψd = I 1 (x)+ x {x f(x)=0,x F 2 m } The first term in (13) is computed as I 1 (x) = x {x f(x)=0,x F 2 m } = x {x f(x)=0,x F 2 m } x {x f(x)=0,x F 2 m } since f(x) is balanced. The second term in (13) is computed as I 2 (x) = x {x f(x)=1,x F 2 m } = x {x f(x)=1,x F 2 m } x {x f(x)=1,x F 2 m } since f(x) is difference-balanced. Thus, we have S ψd = ( 1) f(bx) x {x f(x)=1,x F 2 m } x {x f(x)=1,x F 2 m } ( 1) f(bx) ( 1) f(bx)+1, ( 1) f(bx)+1 ( 1) f(bx) I 2 (x). (13) η F 2 m ( 1) f(aηx) x {x f(x)=0,x F 2 m } Finally, from the difference-balance property, we have (0, 0), 2 m 2 1 times (1, 0), 2 m 2 times (f(x),f(bx)) = (0, 1), 2 m 2 times (1, 1), 2 m 2 times, η F 2 m ( 1) f(ηx)+f(aηx) ( 1) f(bx).

10 370 J.-W. Jang et al. as x varies over F2 m. Therefore, we have S ψd =1. In the similar way, we get S µd =1. Now consider two sequences s i (x) =f([tr n m(x)] r )+2f(a r [tr n m(x)] r ) s k (x) =f([tr n m(x)] r )+2f(b r [tr n m(x)] r ), where a = β i and b = β k for nonzero i and k. Then R i,k (δ) is given by R i,k (δ) = ω si (δx) sk (x) x F2 n = ω {f(x x 2 A x 1 F2 m r 1 [trn m (δx2)]r )+2f(x r 1 ar [tr n m (δx2)]r )} ω {f(xr 1 [trn m (x2)]r )+2f(x r 1 br [tr n m (x2)]r )}. Case 1) i k for nonzero i and k : For δ/ F 2 m, with the replacement of tr n m(δx 2 )bycd and tr n m(x 2 )byd and also from Lemma 2, R i,k (δ) is rewritten as R i,k (δ) =M δ (cd, d) d F2 m +M δ (0, 0) + ω {f([x1cd] r )+2f([x 1acd] r )} {f([x 1d] r )+2f([x 1bd] r )} c F2 m x 1 F2 m x 1 F 2 m ω 0 M δ (c, 0) c F2 m + M δ (0,c) c F2 m =2 n 2m ω {f([x1c] r )+2f([x 1ac] r )} x 1 F2 m ω {f([x1c] r )+2f([x 1bc] r )} x 1 F2 m ω {f([x1c] r )+2f([x 1ac] r )} {f(x r 1 )+2f([x1b]r )} c F2 m x 1 F2 m + 2n 2m 1 2 m 1 +2 n 2m +2 n 2m x 1 F 2 m ω 0 ω {f([x1c] r )+2f([x 1ac] r )} x 1 F2 m ω {f([x1c] r )+2f([x 1bc] r )}. x 1 F2 m From Lemma 3 and the facts that S ψd =1andS µd =1,R i,k (δ) canbe computed as R i,k (δ) =2n 2m +2 n 2m n 2m ( 1) = 1.

11 For δ =1,wehave R i,k (1) = New Constructions of Quaternary Hadamard Matrices 371 ω (f([trn x F2 n = 1 m (x)]r )+2f(a r [tr n m (x)]r )) (f([tr n m (x)]r )+2f(b r [tr n m (x)]r )) from the difference-balance property of f(x). Case 2) i = k for nonzero i and k : Obviously, R i,i (1) = 2n 1. When δ/ F 2 m, the correlation function is given as R i,i (δ) r )+2f([x 1ac] r )} {f(x r 1 =2n 2m )+2f([x1a]r )} ω {f([x1c] c F2 m x 1 F2 m + 2n 2m 1 2 m 1 +2 n 2m +2 n 2m x 1 F 2 m ω 0 ω {f([x1c] r )+2f([x 1ac] r )} x 1 F2 m ω {f(x r 1 cr )+2f([x 1ac] r )} x 1 F2 m =2 n 2m +2 n 2m n 2m ( 1) = 1. Case 3) i =0ork =0: Inthiscase,itiseasytoshowthatR i,0(δ) =R 0,i (δ) = 1 forδ/ F 2 m and R 0,0 (δ) = 1 forδ 1. Here is an example of 6 6 quaternary Hadamard matrix constructed from the Theorem 3. Example 2. Let α be a primitive element in F 2 6.LetT = =9andr =5. Using the GMW-sequence tr 3 1([tr 6 3(α t )] r ) of period 63, we can construct quaternary sequences of period 63 as s 0 (t) = 2tr 3 1([tr 6 3(α t )] 5 ) s i (t) =tr 3 1([tr 6 3(α t )] r ) + 2tr 3 1([tr 6 3(α t+9i )] 5 ), 1 i 8. These sequences make a quaternary Hadamard matrix as H L =(h ij ), where h ij is given as w 0 if i =0orj =0 h ij = w 2tr3 1 ([tr6 3 (αj 1+i 9 )] 5 ) if 1 i T and j 0 w tr3 1 ([tr6 3 (αj 1+i 9 )] r )+2tr 3 1 ([tr6 3 (αj 1+i 9 +9 (i 1)/9 )] 5 ) otherwise, where i 9 =(i 1) mod 9.

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