# On the 4-Adic Complexity of Quaternary Sequences with Ideal Autocorrelation

In this paper, we determine the 4-adic complexity of the balanced quaternary sequences of period 2p and 2(2^n-1) with ideal autocorrelation defined by Kim et al. (ISIT, pp. 282-285, 2009) and Jang et al. (ISIT, pp. 278-281, 2009), respectively. Our results show that the 4-adic complexity of the quaternary sequences defined in these two papers is large enough to resist the attack of the rational approximation algorithm.

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## I Introduction

With the development of correlation attack and algebraic attack, it is becoming the main trend to use the nonlinear feedback shift register sequences with pseudorandom property as the driving sequences in stream cipher design. The feedback with carry shift register (FCSR) proposed by [6] and [7] is a kind of generator which can produce nonlinear sequences quickly.

Balanced binary and quaternary sequences with good autocorrelation play important roles in communication and cryptography systems. The -adic complexity measures the smallest length of FCSR which generates the sequence over . Sequences over with low -adic complexity are susceptibly decoded by the rational approximation algorithm, see [6], [8-9]. Particularly, a quaternary sequence can be decoded by the rational approximation algorithm with consecutive bits. Hence, the 4-adic complexity of a safe sequence with period should exceed . There are numerous results about the 2-adic complexity of binary sequences with good autocorrelation, see [2-3], [11-14], for example. However, the 4-adic complexity of quaternary sequences with good autocorrelation has not been studied so fully and there are few quaternary sequences with good autocorrelation whose 4-adic complexity is known, see [10]. This may pose risk to communication and cryptography system.

In this paper, we determine the 4-adic complexity of the balanced quaternary sequences of even period and with ideal autocorrelation defined in [5] and [4], respectively. Our results show that the 4-adic complexity of the quaternary sequences with period and defined in these two papers is larger than and respectively. Hence they are safe enough to resist the attack of the rational approximation algorithm.

## Ii Preliminaries

In the application of communication and cryptography, balanced sequences with good autocorrelation property are preferred.

For a sequence over with period , it is said to be balanced if for any pair of with ,where

 Ak={t|gt=k,0≤t

The autocorrelation function of a sequence over with period is defined by

 Cs(τ)=N−1∑i=0ζsi−si+τd,   0≤τ

where is a complex -th primitive root of unity.

The maximal out-of-phase autocorrelation magnitude should be as small as possible and the number of the occurrences of the maximal out-of-phase autocorrelation magnitude should be minimized. A sequence with the possible minimum value of the maximal out-of-phase autocorrelation magnitude and the minimum number of occurrences of the maximal out-of-phase autocorrelation magnitude is said to have the ideal autocorrelation property.

For a binary sequence with period , it is well known that if

 Cs(τ)=−1  for all 0<τ

then is an ideal autocorrelation sequence.

The autocorrelation distribution of a quaternary sequence of even period with ideal autocorrelation and balance property is given by

 Cs(τ)=⎧⎪ ⎪ ⎪⎨⎪ ⎪ ⎪⎩N,1 times,0,N2−1 times,−2,N2 times.

in [5].

By using the Legendre sequences and the Gray mapping, two classes of balanced quaternary sequences of even period with ideal autocorrelation were constructed in [5]. Balanced quaternary sequences of period with ideal autocorrelation were constructed in [4] by using the binary sequences of period with ideal autocorrelation and the Gray mapping.

For an odd prime

, let and be the set of quadratic residues and quadratic non-residues in the set , respectively. Two classes of Legendre sequences and of period are defined by

 bt=⎧⎪⎨⎪⎩0,for t=00,for t∈QR1,for t∈QNR
 ct=⎧⎪⎨⎪⎩1,for t=00,for t∈QR1,for t∈QNR

respectively.

The Gray mapping is defined by

 ϕ(0,0)=0, ϕ(0,1)=1, ϕ(1,1)=2, ϕ(1,0)=3.

According to the definition of the Gray mapping, we can get

 ϕ(a,e)=2a−a(e−1)−(a−1)e (2)

where and .

The following two classes of quaternary sequences and of even period defined by using the Gray mapping and the Legendre sequences were shown to have ideal autocorrelation and balance property in [5].

###### Definition 1.

([5]) For an odd prime with , let and be two binary sequences of the same period defined by

 s0t={bt,for t≡0mod2ct,for t≡1mod2
 s1t={bt,for t≡0mod21−ct,for t≡1mod2.

The quaternary sequence of period is defined by .

###### Definition 2.

([5]) For an odd prime with , let and be two binary sequences of the same period defined by

 s2t={bt,for 0≤t
 s3t={ct,for t≡0mod21−ct,for t≡1mod2.

The quaternary sequence of period is defined by .

Let . Assume that is a binary sequence of period with ideal autocorrelation. Let be the characteristic set of defined by

 D0={t|st=1,0≤t≤2n−2}

and . By the Chinese remainder theorem, we have the isomorphism

 ϕ:Z2×(2n−1)≃Z2×Z2n−1,h↦(hmod2,hmod2n−1).

The following class of quaternary sequences of even period defined by using the Gray mapping and the ideal autocorrelation sequences with period were shown to have ideal autocorrelation and balance property in [4].

###### Definition 3.

([4]) Let be binary sequence of period with ideal autocorrelation and a characteristic set of . Let be the quaternary sequence defined by

 g3t=ϕ(ut,vt),

where and are the binary sequences of period defined by

 ut={1,if t∈{0,1}×D00,if t∈{0,1}×¯¯¯¯¯D0
 vt={1,if t∈{0}×D0 ⋃ {1}×¯¯¯¯¯D00,if t∈{0}×¯¯¯¯¯D0 ⋃ {1}×D0.

The definition about the -adic complexity of quaternary sequences with period is defined as follows.

###### Definition 4.

([6, 9]) For a quaternary sequence with period , let . The 4-adic complexity is defined by where denotes the greatest common divisor of and . (The exact value of the smallest length of FCSR which generates the quaternary sequence is .

According to Definition 4, determining the -adic complexity of quaternary sequences is equivalent to determining .

## Iii Main result

In this section, we study the 4-adic complexity of the quaternary sequences of period and with ideal autocorrelation in Section II.

For , the Legendre symbol is defined by

 (ip)={1,if % i∈QR −1,otherwise.

The following four lemmas are useful in the sequel.

###### Lemma 1.

([1], Theorem 7.3) If is a periodic binary sequence of odd period with ideal autocorrelation, then the number of nonzero bits in one period of is .

The proof of the lemma is similar to that of Lemma 2(1) in [14]. For the completeness of the paper, we give a simple proof.

###### Lemma 2.

Let be an odd prime. Then

 (p−1∑i=1(ip)4i)2≡−(−1p)4p−13+(−1p)p(mod4p−1).
###### Proof.

Since

 (p−1∑i=1(ip)4i)2 =p−1∑i=1(ip)4ip−1∑j=1(jp)4j =p−1∑i,j=1(ijp)4i+j  (% let j=ik) =p−1∑i,k=1(kp)4i(k+1) =p−2∑k=1(kp)p−1∑i=14i(k+1)+p−1∑i=14ip(p−1p).

Then from and

 p−1∑i=14i(k+1)≡p−1∑i=14i(mod4p−1) (1≤k≤p−2),

we get

 (p−1∑i=1(ip)4i)2 ≡−(−1p)4p−13+(−1p)p(mod4p−1).

###### Lemma 3.

For a prime , if , then we have .

###### Proof.

Since and has only one solution in the set , then from , we get which implies . Hence we get . ∎

###### Lemma 4.

For an odd prime , we have

 p−1∑t=1(2tp)42t+p−1∑t=0t≠p−12(2t+1p)42t+1≡⎧⎪⎨⎪⎩2p−1∑t=1(tp)4t(mod4p−1)0(mod4p+1)
###### Proof.

From

 p−1∑t=1(2tp)42t≡p−1∑t=1(tp)4t≡p−1∑t=0t≠p−12(2t+1p)42t+1(mod4p−1)

we get

 p−1∑t=1(2tp)42t+p−1∑t=0t≠p−12(2t+1p)42t+1≡2p−1∑t=1(tp)4t(mod4p−1).

Since

 p−12∑t=1(2tp)42t≡−p−1∑t=p−32+2(2t+1p)42t+1(mod4p+1)
 p−1∑t=p−12+1(2tp)42t≡−p−32∑t=0(2t+1p)42t+1(mod4p+1)

then the rest result follows from

 p−1∑t=1(2tp)42t =p−12∑t=1(2tp)42t+p−1∑t=p−12+1(2tp)42t

and

 p−1∑t=0t≠p−12(2t+1p)42t+1=p−32∑t=0(2t+1p)42t+1+p−1∑t=p−32+2(2t+1p)42t+1.

Now we study the 4-adic complexity of the quaternary sequence in Definition 1.

###### Theorem 5.

For the quaternary sequence in Definition 1, we have

 Φ4(g1)=⎧⎨⎩log442p−115,if 5|(p+2) log442p−13,else.
###### Proof.

(i) Firstly, we prove

 gcd(g1(4),4p−1)=3.

Let the symbols be the same as before. Then we get

 g1(4) =2p−1∑t=0ϕ(s0t,s1t)4t =p−1∑t=0ϕ(b2t,b2t)42t+p−1∑t=0ϕ(c2t+1,1−c2t+1)42t+1 =p−1∑t=0[2b2t−b2t(b2t−1)−(b2t−1)b2t]42t+p−1∑t=0[2c2t+1−c2t+1(−c2t+1) −(c2t+1−1)(1−c2t+1)]42t+1  (by (???)) =p−1∑t=0[2b2t−(b2t)2+b2t−(b2t)2+b2t]42t+p−1∑t=0[2c2t+1+(c2t+1)2+1−2c2t+1+(c2t+1)2]42t+1 =p−1∑t=02b2t42t+p−1∑t=0(2c2t+1+1)42t+1  (since a2=a (0≤a≤1)) =2b040+p−1∑t=12b2t42t+(2cp+1)4p+p−1∑t=0t≠p−12(2c2t+1+1)42t+1 =p−1∑t=12b2t42t+3⋅4p+p−1∑t=0t≠p−12(2c2t+1+1)42t+1  (since% b0=0, c0=cp=1) =p−1∑t=0t≠p−1242t+1+3⋅4p+p−1∑t=12b2t42t+p−1∑t=0t≠p−122c2t+142t+1 =4p−1∑t=042t−4p+3⋅4p+2p−1∑t=11−(2tp)242t+2p−1∑t=0t≠p−121−(2t+1p)242t+1 ≡⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩9p−1∑t=042t−2p−1∑t=1(tp)4t(mod4p−1)9p−1∑t=042t−2(mod4p+1).    (by Lemma ???) (3)

Since ), then from (3) we know

 g1(4)≡p−1∑t=1(tp)4t≡0(mod3) \bigg{(}since p−1∑t=1(tp)=0\bigg{)}.

It then follows that . If , then from and , we get which contradicts with . Therefore .

Assume that is a prime divisor of such that . By (3) we get

 g1(4)≡−2p−1∑t=1(tp)4t(modd1).

Then we have . Combining with Lemma 2, we have which implies . Hence, we have . By Fermat’s little Theorem, we get . Then we have which is a contradiction. Hence, we know .

Therefore we get

 gcd(g1(4),4p−1)=3. (4)

(ii) Next, we prove

 gcd(g1(4),4p+1)={5,if 5|(p+2) 1,else.

By (3) we have

 g1(4)≡−p−2(mod5).

Then we get only when .

Assume that and , then by Lemma 3 we get which contradicts with . It then follows that .

Let be a divisor of such that . Then from (3) we have Thus which implies Therefore

 gcd(g1(4),4p+1)={5,if 5|(p+2) 1,else. (5)

Combining with (4-5) and the definition of the 4-adic complexity, the result is proven. ∎

The 4-adic complexity of the sequence in Definition 2 is given by the following theorem.

###### Theorem 6.

For the quaternary sequence in Definition 2, we have

 Φ4(g2)={log442p−15,if 5|(p−2) log4(42p−1),else.
###### Proof.

(i) Firstly, we determine the exact value of .

Assume that the symbols are the same as before. Then we have

 g2(4) =2p−1∑t=0ϕ(s2t,s3t)4t =p−1∑t=0ϕ(b2t,c2t)42t+p−1∑t=0ϕ(b2t+1,1−c2t+1)42t+1 =p−1∑t=0[2b2t−b2t(c2t−1)−(b2t−1)c2t]42t+p−1∑t=0[2b2t+1+b2t+1c2t+1 −(b2t+1−1)(1−c2t+1)]42t+1  (by (???)) =(2×0−0×(1−1)−(0−1)×1)×40+p−1∑t=1[2b2t−b2tc2t+b2t−b2tc2t+c2t]42t +[2bp+bpcp−(bp−1)(1−cp)]4p+p−1∑t=0t≠p−12[2b2t+1+b2t+1c2t+1+(b2t+1)2+1−2b2t+1]42t+1 =1+p−1∑t=12b2t42t+p−1∑t=0t≠p−12(2b2t+1+1)42t+1  (since% a2=a (0≤a≤1)) =1+p−1∑t=12b2t42t+4p−1∑t=042t−42⋅p−12+1+p−1∑t=0t≠p−122b2t+142t+1 =2p−1∑t=11−(2tp)242t+2p−1∑t=0t≠p−121−(2t+1p)242t+1+1−4p+4p−1∑t=042t ≡⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩9p−1∑t=042t−2p−1∑t=1(tp)4t−2(mod4p−1)9p−1∑t=042t+2(mod4p+1).(by % Lemma ???) (6)

Since , then by (6) we know

 g2(4)≡1(mod3).

Then we get .

Let be a prime divisor of . From (6) and we have

 g2(4)≡−2(p−1∑t=1(tp)4t+1)(modd3).

Then by Lemma 2 and , we have

 1≡(p−1∑t=1(tp)4t)2≡−p+4p−13≡−p(modd3).

Hence, we have . From and we get which is a contradiction. Therefore

 gcd(g2(4),4p−1)=1. (7)

(ii) Now we determine .

Since , then by (6) we get

 g2(4)≡−p−1∑t=042t+2≡−p+2(mod5).

Hence we get only when .

Assume that and , then from Lemma 3 we have which contradicts with . It then follows that . Assume that is a prime divisor of and such that . Then by (6) we get

 g2(4)≡2(modd4)

which implies . Hence we have

 gcd(g2(4),4p+1)={5,if 5|(p−2) 1,otherwise. (8)

Combining with (7-8) and the definition of the 4-adic complexity, the result is proven. ∎

The 4-adic complexity of the sequence with period in Definition 3 is given as follows.

###### Theorem 7.

For the quaternary sequence with period in Definition 3, we have

 Φ(g3)=log45⋅(42n−1−1).
###### Proof.

With the symbols the same as before, we have

 g3(4) =2n+1−3∑t=0ϕ(ut,vt)4t =2n+1−3∑t=0t∈{1}×¯¯¯¯D04t+2n+1−3∑t=0t∈{0}×D02⋅4t+2n+1−3∑t=0t∈{1}×D03⋅4t =2n+1−3∑t=02∤t,st=04t+2⋅2n+1−3∑t=02|t,st=14t+3⋅2n+1−3∑t=02∤t,st=14t =2n+1−3∑t=02∤t(1−st)4t+2⋅2n+1−3∑ct=02|tst⋅4t+3⋅2n+1−3∑t=02∤tst⋅4t =2n+1−3∑t=02∤t4t+22n+1−3∑t=0st⋅4t <