ANALYSIS OF COMPLEMENTED GROUP CA DERIVED FROM 90/150 GROUP CA

Abstract

In recent years, CA has been applied to image security due to its simple and regular structure, local interaction and random-like behavior. Since the initial state is regenerated after some iterations in the group CA, the receiver is able to decrypt by the same CA. Pries et al. showed that the all lengths of the cycles in the complemented group CA C with rules 195, 153, and 51 are equal to the order of C. Nandi et al. reported the encryption technique using C. These results can be made efficient use in cryptosystem by expanding the Nandi's key space. In this paper, we analyze the order of the complemented group CA derived from 90=150 group CA and show that all the lengths of the cycles in the complemented CA are equal to the order of the complemented CA.

1. Introduction

The concept of cellular automata(henceforth CA) was originally discovered in the 1940s by Ulam and von Neumann who suggested using a discrete system for creating a reductionist model of self-replication [10,13]. In 1960s, CA were studied as a particular type of dynamical system and the connection with the mathematical field of symbolic dynamics [6]. In 1980s, Wolfram engaged in a systematic study of one-dimensional CA and claimed that CA have applications in many fields of science. These include computer processors and cryptography [14]. Applications of CA in various fields have been proposed in [8,11].

In recent years, CA has also been applied to image security due to its simple and regular structure, local interaction and random-like behavior [1,7]. In the case of the group CA, information is preserved during the iteration. With this property, group CA can be made full use in cryptosystem. The CA rule is the key and the final configuration which is obtained by forward iteration of the CA for fixed time steps is an encrypted image.

In the group CA, since the initial state is regenerated after some iterations, the receiver is able to decrypt by the same CA. It was already known that the complemented CA derived from a group CA is also a group CA.

Pries et al.[11] showed that the all lengths of the cycles in the complemented group CA C with rules 195, 153, and 51 are equal to the order of C. And Nandi et al. reported the encryption technique using C in [9]. They also used F = (11 · · · 1)t as the complement vector to derive the complemented CA.

But if we use the rules 90 and 150 to generate the next state of each cell, the randomness is more strong since the dependency of the next state to its neighbor of the present state is higher than the dependency by the rules 60, 102, and 204. And we can use the results of Cho et al. to synthesize the CA according to the minimal polynomial [2,3,4].

In this paper, we analyze the order of the complemented group CA derived from 90=150 group CA and find the complement vectors F such that the all lengths of cycles are equal to the order of the complemented group CA derived from 90=150 group CA and F.

 

2. Preliminaries

CA consist of cells on a line where each cell has two possible values 0 or 1. Each configuration of CA evolves in discrete time steps and the next state is decided by the cell to its left, the cell itself, and the cell to its right, according to the combinational logic known as a rule. The next state transition function can be expressed as follows;

where xi(t) is the ith cell at the tth time step and f is a rule of the CA. Since there are 23 possible states for the three cells neighboring a given cell, there are 223 distinct mappings from all these neighborhood configurations to the next state, each of which can be indexed with an 8-bit binary number. For example,

The corresponding logic for rule 90 is xi(t + 1) = xi−1(t) ⊕ xi+1(t) and for rule 150 is xi(t + 1) = xi−1(t) ⊕ xi(t) ⊕ xi+1(t).

A CA having only XOR logic is called a linear CA and the corresponding rule is called a linear rule. In the case of the rules involving XNOR logic, the CA is called a complemented CA and the corresponding rule is called a complemented rule. In this paper, we will employ the rule 90, rule 150 and the null boundary conditions in which the boundary of the extreme cells is imposed as all 0(henceforth NBCA).

An n-cell linear CA is specified by n × n state transition matrix operating over GF(2) which can be represented as the following tridiagonal matrix[5].

In the matrix, the principal diagonal specifies the self-dependency if the next state of the ith cell depends on its present state. The other two diagonals specify the dependency of the corresponding cell on its left and right neighbors. Since all the entries in the other two diagonals of the state transition matrix are 1 for the rules 90 and 150, we abbreviate the matrix T as Tn = ⟨d1, d2, d3, · · ·, dn⟩, where each di is 0 or 1. That is, if the rule for the ith cell is rule 150, then di = 1. Similarly di = 0 represents the rule 90 for the rule of the ith cell. So Tn can also represent the rule vector for the CA. If X(t) stands for the state of the CA at the tth instant of time, then the state X(t + 1) at the next time instant can be represented as X(t + 1) = TX(t). Since the XNOR logic cannot be represented in the multiplicative notation, the state transition funcion for the complemented CA is symbolically represented as for the state transition matrix T of the corresponding CA with XOR logic only. Thus the next state of the complemented CA is X(t + 1) = TX(t) ⊕ F, where F is the complement vector which has significant entries in places of the cell positions where the inversion is required.

Lemma 2.1 ([5]). If denotes p times application of the complemented CA operator then where F is the complement vector.

The characteristic polynomial of a matrix T is given by |T ⊕ xI| and the minimal polynomial of T is the minimum degree factor of the characteristic polynomial that is annihilated by T. In general, the characteristic polynomial is different from the minimal polynomial for a matrix. However for the state transition matrix T of the rules 90 and 150, the two polynomials are identical [12].

Definition 2.2. A CA is called a group CA if det(T) = 1, where T is the state transition matrix for the CA and det(T) is the determinant of T. In a group CA, all states of the CA form cycles. And for a positive integer m, Tm = I where I is the identity matrix.

Das et al. [5] reported that the complement of a group CA is also a group CA. And Pries et al. [11] investigated the order of the complemented group CA derived from the group CA with rules 195, 153, and 51. We analyze the relation between the orders of the complemented CA and the corresponding noncomplemented 90/150 group NBCA. And we show the structure of the cycles in the complemented group CA.

Theorem 2.3 ([2]). For the 90/150 k-cell group CA, let Tk = ⟨d1, d2, · · ·, dk⟩, be the rule vector for the CA, where di = 0 for the rule 90 and di = 1 for the rule 150 at the ith cell. Then the followings hold :

 

3. The Structure of the Complemented CA

Lemma 3.1. For the state transition matrix T of the n-cell group NBCA C with the minimal polynomial mT (x) = (1 + x)n, 2r−1 < n ≤ 2r (r = 2, 3, 4, · · ·), let be the state transition function of the complemented group CA derived from C and the complement vector F = (f1f2· · ·fn)t, fi = 0 or 1 (1 ≤ i ≤ n). Then ord(T) ≤ ord(), where ord(T) is the order of T.

Proof. If ord(T) = p and ord() = k, then p = 2r from 2r-1 < n ≤ 2r. Suppose that then k = 2r-1 and for all the states X in C, So Then (I⊕T⊕T2⊕· · ·⊕Tk−1) F = (I⊕T)k−1 F = (I⊕Tk) X = (I⊕T)k X.

Since rank ((I⊕T)k) = n-k, dimN((I⊕T)k) = k and thus N((I⊕T)k-1) ⊂ N((I⊕T)k), where N(A) is the null space of A.

(i) If F ∈ N ((I⊕T)k-1), then (I⊕T)k-1 F = 0 but (I⊕T)k X ≠ 0 for X ∉ N ((I⊕T)k). This is a contradiction.

(ii) If F ∈ N ((I⊕T)k) \ N((I⊕T)k-1), then (I⊕T)k-1 F ≠ 0 but (I⊕T)k X = 0 for X = F. This is a contradiction.

(iii) If F ∉ N ((I⊕T)k), then (I⊕T)k-1 F ≠ 0 but (I⊕T)k X = 0 for X ∈ N ((I⊕T)k-1). This is a contradiction.

By (i), (ii) and (iii), (I⊕T)k-1 F ≠ (I⊕T)k X. So and thus □

The following lemma can be proved by Lemma 3.1 and the proof of Lemma 4.5.1 in [5].

Lemma 3.2. For the state transition matrix T of the n-cell group NBCA C with the minimal polynomial (1 + x)n, 2r−1 < n ≤ 2r (r = 2, 3, 4, · · ·), let be the state transition function of the complemented group CA derived from C and the complement vector F = (f1f2f3· · ·fn)t, fi = 0 or 1 (1 ≤ i ≤ n). If ord(T) = p,

The following theorem is very important for the results in this paper.

Theorem 3.3. For the state transition matrix Tn of the n-cell 90/150 group NBCA C with the minimal polynomial (1 + x)n, n = 2, 3, · · ·, let Sn = (I ⊕ Tn)n-1. Then the matrices S2n and S2n+1 can be obtained from Sn as followings;

where 0n is the n × 1 zero matrix.

Proof. For the case of n = 2, we can easily confirm the theorem.

For the case of n = k, let ⟨r1, r2, · · ·, rk⟩ be the rule vector of I ⊕ Tk. Then is the rule vector of I ⊕ T2k by Theorem 2.3.

Let where Ai (resp.Bi) is the ith row of Sk (resp.S2k) for 1 ≤ i ≤ k. Since (I ⊕ Tk) Sk = Ok and (I ⊕ T2k) S2k = O2k,

and

Since rank (I ⊕ Tk) = k − 1, the relation of Ai (i = 1, 2, · · · , k) is determined by the first (k − 1) equations of (3.1). Thus the relation of Bi (i = 1, 2, · · · , k) is determined by the first (k − 1) equations of (3.2).

Let Sk = (U1U2 · · ·Uk) and S2k = (V1V2 · · · V2k). Since Sk (I ⊕ Tk) = Ok and S2k (I ⊕ T2k) = O2k, we can show that the relation of Vi is equal to the relation of Ui by the similar method for i = 1, 2, · · · , k. Let where Wi is a k×k submatrix. Then W1 = Sk. By the same reason, we obtain W4 = Sk. From rank (Sk) = 1, we obtain W2 = W3 = Sk. Hence

In the case of S2k+1 = (I ⊕ T2k+1)2k, the rule vector of I⊕T2k+1 is ⟨r1, · · ·, rk-1, rk, 0, rk, rk-1, · · ·, r1⟩ from Theorem 2.3. If Yi is the ith row of S2k+1 for 1 ≤ i ≤ k, then Yl = Yk+l+1 for 1 ≤ l ≤ k, since ⟨r1, · · ·, rk-1, rk, 0, rk, rk-1, · · ·, r1⟩ is symmetric with respect to the center and (I⊕T2k+1) S2k+1 = O2k+1. Moreover the first k entries of Yi and the last k entries of Yk+1+i are equal to the first k entries of the ith row of Sk for 1 ≤ i ≤ k.

Since rank (I ⊕ Tk) = k −1, the kth row of I ⊕ T2k+1 is changed to by the elementary row operation. Thus the (k + 1)th row of (I ⊕ T2k+1) S2k+1 is Yk+1. Therefore Yk+1 is the zero vector. With the same reason, the (k + 1)th column of S2k+1 should be the zero column. Hence

Corollary 3.4. Let Tn be the state transition matrix of the n-cell 90/150 group NBCA with the minimal polynomial (1 + x)n, where n = 2r, r = 1, 2, · · ·. Then

Proof. Since T2 is ⟨0, 0⟩, By Theorem 3.3, we obtain (I ⊕ Tn)n−1 consisting of all columns with entries 1s. □

Example 3.5. For Then

where 03 is the 3 × 1 zero matrix.

The following two theorems are main results in this paper.

Theorem 3.6. Let C be the n-cell 90/150 group NBCA with the minimal polynomial (1 + x)n, 2r−1 < n ≤ 2r (r = 2, 3, 4, · · · ). And let be the complemented CA derived from C with the complement vector F = (f1f2 · · · fn)t with fi = 0 or 1, 1 ≤ i ≤ n. If ord(T) = p for the state transition matrix T of C, then ord() for the state transition function satisfies the following;

Proof. For ord(T) = p, let ord() = k for some positive integer k.

(i) In case of 2r−1< n< 2r :

From T2r = I, we obtain p = 2r and Since (I ⊕ T)n = O and n ≤ p−1, (I ⊕ T)p−1 F = 0. So Therefore k = p = 2r by Lemma 3.2.

(ii) In case of n = 2r :

Since and by Corollary 3.4, (I ⊕ T)p−1 F ≠ 0 for the complement vector F = (f1f2f3· · · fn)t with Therefore k > p and thus k = 2p = 2r+1 by Lemma 3.2. But (I ⊕ T)p-1 F = 0 for the complement vector F = (f1f2f3· · · fn)t with and thus k = p = 2r. □

Theorem 3.7. Let C be the n-cell 90=150 group NBCA with the minimal polynomial (1 + x)n, n = 2, 3, 4, · · · . And let C′ be the complemented CA derived from C with the complement vector F = (f1f2 · · · fn)t such that (I⊕T)n−1 F ≠ 0, where T is the state transition matrix of C and fi = 0 or 1, 1 ≤ i ≤ n. Then all the lengths of the cycles are equal to ord( ) in the state transition graph of C′.

Proof. Let Then k = 2r or k = 2r+1 by Lemma 3.2. Assume that there exists a cycle of length l such that l < k. Then l can be Since there exists a cycle of length l, there is a nonzero state X such that Then Thus

By multiplying (I ⊕ T)n−l to the both sides of (3.3), we obtain 0 ≠ (I ⊕ T)n-1 F = (I ⊕ T)n X = 0. This is a contradiction. So there does not exist any cycle with length l such that l < k. Hence all cycles in C′ have the same length □

To find the CA in which all the lengths of cycles are equal to it is sufficient to find F such that (I⊕Tn)n−1 F ≠ 0. F can be obtained from the form of (I⊕Tn)n−1. If we try to find the Sn = (I⊕Tn)n−1 from the state transition matrix Tn =< d1, d2, · · ·, dn >, the time complexity is O(n). By Theorem 3.3, we can easily derive (I⊕Tn)n−1 from And thus we can find the complement vector F such that (I⊕Tn)n−1 F ≠ 0 within the time complexity O(log2n).

 

4. Conclusion

In this paper, we analyzed the order of the complemented group CA derived from 90=150 group CA and showed that all the lengths of the cycles in the complemented CA are equal to the order of the complemented CA. Especially, the order of the complemented group CA derived from 90/150 group CA C is equal to or twice the order of C. And we showed that all the cycles in has the same length cycle with the order of . Also we showed that the time complexity to find the complement vector F such that (I⊕Tn)n−1 F ≠ 0 is O(log2n). So it is even more efficient than direct computation whose time complexity is O(n).