Divisibility of Polynomials over Finite Fields and Combinatorial Applications

Transcription

1 Designs, Codes and Cryptography anuscript No. (will be inserted by the editor) Divisibility of Polynoials over Finite Fields and Cobinatorial Applications Daniel Panario Olga Sosnovski Brett Stevens Qiang Wang Septeber 1, 211 Abstract Consider a axiu-length shift-register sequence generated by a priitive polynoial f over a finite field. The set of its subintervals is a linear code whose dual code is fored by all polynoials divisible by f. Since the iniu weight of dual codes is directly related to the strength of the corresponding orthogonal arrays, we can produce orthogonal arrays by studying divisibility of polynoials. Muneasa (Finite Fields Appl., 4(3):252-26, 1998) uses trinoials over F 2 to construct orthogonal arrays of guaranteed strength 2 (and alost strength 3). That result was extended by Dewar et al. (Des. Codes Cryptogr., 45:1-17, 27) to construct orthogonal arrays of guaranteed strength 3 by considering divisibility of trinoials by pentanoials over F 2. Here we first siplify the requireent in Muneasa s approach that the characteristic polynoial of the sequence ust be priitive: we show that the ethod applies even to the uch broader class of polynoials with no repeated roots. Then we give characterizations of divisibility for binoials and trinoials over F 3. Soe of our results apply to any finite field F q with q eleents. Keywords Polynoials over finite fields divisibility of polynoials orthogonal arrays. Matheatics Subject Classification (2) 12E2, 94A55, 5B15 1 Introduction Maxiu-length shift-register sequences are widely used in pseudo-rando nuber generation and several engineering applications [9, 1]. The fewer nonzero ters in the characteristic polynoial of the shift-register sequence, the faster is the generation of the sequence. However, the nuber of nonzero ters in ultiples of the characteristic polynoial deterines the statistical bias in the sequence, fewer ters iplying ore bias [12, 16]. The authors are supported in part by NSERC of Canada. School of Matheatics and Statistics, Carleton University 1125 Colonel By Drive, Ottawa, ON, K1S 5B6

2 2 A classical result [7] relates the iniu weight of dual codes to the strength of the corresponding orthogonal arrays. In this paper we follow a ethod introduced by Muneasa [23] that constructs orthogonal arrays. The procedure first constructs a code with subintervals of a shift-register sequence generated by a polynoial f. Its dual code is characterized by all polynoials that are divisible by f. Hence, by studying the divisibility of polynoials we can produce dual codes, thus deterining the strength of coverage in the arrays. This suggests studying the divisibility of polynoials over finite fields, and this is the focus of this paper. This procedure was used by Muneasa [23] for the case trinoials (polynoials with three nonzero ters) dividing trinoials over F 2 to produce orthogonal arrays with guaranteed strength 2 (and alost strength 3). Then, Dewar et al. [8] extended this to pentanoials (polynoials with five nonzero ters) dividing trinoials over F 2 to give orthogonal arrays with guaranteed strength 3. Dewar et al. [8] suggests extending the results to finite fields other than F 2. Once we are not in F 2 we can consider binoials (there are no irreducible binoials over F 2 ). We focus on low weight polynoials since we can produce precise results in these cases. However, the general proble of when polynoials over a finite field of given weight divide polynoials of another given weight is interesting and not well understood. As concrete applications of this divisibility proble for ranges on the weight of the polynoials that exceed the results in this paper, see the cryptosyste TCHo [1, 13] and the turbo codes applications [24]. Also of interest are soe general results on the weights of ultiples of polynoials over F 2 which do not depend on the low weight of f [11,14,17,21]. This paper contains results siilar to [8] and [23] but for binoials and trinoials over non-binary fields. In Section 2, we define notation and give previous results as well as we outline the general ethodology. The ost iportant result in this section is a siplification of Muneasa s conditions: we only require irreducible polynoials (or even reducible ones under the condition of no repeated roots) for the inial polynoial of the LFSR instead of the priitive polynoial condition in previous results [8, 23]. We also give soe cobinatorial applications for our results. The results in this section are valid for any finite field F q. Our ain results are obtained for finite field F 3. That is why throughout this paper, unless otherwise stated, we use the finite field F 3. Moving fro F 2 to F 3 has coplicated the proofs considerably. Although in principle one could perhaps extend soe of our results to other finite fields, the level of added coplications would be even greater with our ethodology. We believe that other techniques are needed to obtain siilar results over larger finite fields, ideally techniques independent of the base fields. Section 3 focuses on divisibility of binoials by trinoials over F 3. Section 4 deals with the case of trinoials dividing trinoials over F 3. In Section 5, we conclude with soe questions for further studies and research. 2 Background and preliinaries 2.1 Definitions and previous results We give next soe required definitions. A polynoial f of degree is called priitive over F q if k = q 1 is the sallest positive integer such that f divides x k 1.

3 3 A shift-register sequence with characteristic polynoial f(x) = x P 1 i= c ix i is the sequence a = (a, a 1,...) defined by X 1 a n+ = c i a i+n for n. i= If f is priitive over F q the sequence has period q 1. A subset C of F n q is called an orthogonal array of strength t if for any t-subset T = {i 1, i 2,..., i t } of {1, 2,..., n} and any t-tuple (b 1, b 2,..., b t ) F t q there exists exactly C /q t eleents c = (c 1, c 2,..., c n) of C such that c ij = b j for all 1 j t. Fro the definition, if C is an orthogonal array of strength t, then it is also an orthogonal array of strength s for all 1 s t. Orthogonal arrays and codes are related by the next theore. Theore 1 ([2]) Let C be a linear code over F q. Then, C is an orthogonal array of axial strength t if and only if C, its dual code, has iniu weight t + 1. Delsarte was able to generalize this result to non-linear codes [7]. The following theore describes the dual code of the code generated by shift-register sequences in ters of ultiples of its characteristic polynoial. Theore 2 ([23]) Let f be a priitive polynoial of degree over F q and let 2 n q 1. Let Cn f be the set of all subintervals of the shift-register sequence with length n generated by f, together with the zero vector of length n. The dual code of Cn f is given by (Cn) f n 1 = {(b 1,..., b n) X : b i+1 x i is divisible by f}. i= Previous studies on divisibility of polynoials and cobinatorial applications were done for polynoials over the binary field F 2 [8,23]. Let f be a priitive polynoial of degree over F 2 and let a = (a, a 1,...) be a shift-register sequence with characteristic polynoial f. As in [8], we denote by C f n the set of all subintervals of this sequence with length n, where < n 2, together with the zero vector of length n. Muneasa [23] investigates the shift-register sequences when f is a trinoial, that is, a polynoial with three ters over F 2. Theore 3 ([23]) Let f(x) = x +x l +1 be a trinoial over F 2 such that gcd(, l) = 1. If g is a trinoial over F 2 of degree at ost 2 that is divisible by f, then g(x) = x deg g f(x), g(x) = f(x) 2, or g(x) = x 5 + x = (x 2 + x + 1)(x 3 + x + 1) or, its reciprocal, g(x) = x 5 + x + 1 = (x 2 + x + 1)(x 3 + x 2 + 1). The ain result in [23] iplies that, in the case of a priitive trinoial f satisfying certain properties, C f n is an orthogonal array of strength 2 having the property of being very close to an orthogonal array of strength 3. Muneasa [23] shows that for ost 3-tuples of {1, 2,..., n}, the orthogonal property is satisfied, exception to this are the triples of coordinates corresponding to the exponents of trinoials of the for x i f and f 2. Muneasa [23] suggested the extension of his results to polynoials f with ore than three ters. Dewar et al. [8] extended Muneasa s result to shift-register sequences generated by priitive pentanoials, polynoials with five ters, over F 2.

4 4 No. f(x) h(x) type 1 x 5 + x 4 + x 3 + x x 3 + x p 2 x 5 + x 3 + x 2 + x + 1 x 3 + x + 1 p 3 x 5 + x 3 + x 2 + x + 1 x 4 + x + 1 p 4 x 5 + x 4 + x 3 + x + 1 x 2 + x + 1 p 5 x 6 + x 5 + x 4 + x x 4 + x r 6 x 6 + x 4 + x 2 + x + 1 x 3 + x + 1 i 7 x 6 + x 4 + x 3 + x + 1 x 2 + x + 1 p 8 x 6 + x 5 + x 2 + x + 1 x 5 + x 4 + x 3 + x + 1 p 9 x 6 + x 5 + x 3 + x + 1 x 2 + x + 1 r 1 x 7 + x 4 + x 2 + x + 1 x 3 + x + 1 r 11 x 7 + x 4 + x 3 + x x 3 + x p 12 x 7 + x 5 + x 2 + x + 1 x 7 + x 6 + x 5 + x 4 + x 3 + x + 1 p 13 x 7 + x 5 + x 3 + x x 5 + x 4 + x 3 + x r 14 x 8 + x 5 + x 3 + x + 1 x 5 + x 4 + x 2 + x + 1 p 15 x 8 + x 5 + x 3 + x x 8 + x 7 + x 5 + x 4 + x 3 + x p 16 x 8 + x 6 + x 3 + x + 1 x 6 + x 4 + x 2 + x + 1 r 17 x 8 + x 7 + x 5 + x x 6 + x 5 + x 4 + x r 18 x 9 + x 6 + x 5 + x x 8 + x 5 + x 4 + x i 19 x 9 + x 7 + x 4 + x x 8 + x 6 + x 4 + x i 2 x 9 + x 8 + x 5 + x x 6 + x 5 + x 4 + x r 21 x 1 + x 4 + x 3 + x x 8 + x 7 + x 4 + x i 22 x 1 + x 7 + x 2 + x + 1 x 6 + x 4 + x 3 + x + 1 r 23 x 11 + x 7 + x 6 + x x 8 + x 7 + x 4 + x r 24 x 13 + x 1 + x 2 + x + 1 x 9 + x 7 + x 6 + x 4 + x 3 + x + 1 r 25 x 13 + x 1 + x 9 + x x 12 + x 9 + x 8 + x 6 + x 4 + x p Table 1 Table of binary polynoial exceptions in the ain theore of [8]: p in type indicates that the given polynoial f(x) is priitive, i indicates that f(x) is irreducible and r indicates that f(x) is reducible. Theore 4 ([8]) Let f(x) = x + x l + x k + x j + 1 be a pentanoial over F 2 such that gcd(, l, k, j) = 1. If g is a trinoial of degree at ost 2 divisible by f, with g = fh, then 1. f is one of the polynoial exceptions given in Table 1; 2. 1 od 3 and f, g, h are as follows f(x) = 1 + x + x 2 + x 3 + x = (1 + x + x 2 )(1 + x 3 + x 2 ), h(x) = (1 + x) + (x 3 + x 4 ) + + (x 7 + x 6 ) + x 4, f(x)h(x) = g(x) = 1 + x x 2 4 ; or 3. f is the reciprocal of one of the polynoials listed in the previous ites. The result in [8] constructs orthogonal arrays of guaranteed strength at least Reoving the priitivity condition In this subsection, we generalize Theore 2 by reoving the priitivity condition for the characteristic polynoial f. Theore 5 Let a = (a, a 1, a 2,...) be a shift-register sequence over F q with inial polynoial f F q[x], and suppose that f has degree with distinct roots. Let ρ be

5 5 the period of f and 2 n ρ. Let C f n be the set of all subintervals of the shift-register sequence a with length n. Then the dual code of C f n is given by X (Cn) f n 1 = {(b 1,..., b n) : b i+1 x i is divisible by f}. i= Proof. Since a = (a, a 1, a 2,...) is the sequence with inial polynoial f, the least period per(a) of sequence a equals per(f) which is denoted by t. Let P α 1,..., α be all distinct roots of f. Then by Theore 8.21 in [15], we have a n = j=1 β jαj n for n =, 1,.... First we assue that all β j are nonzero. Hence an eleent w = (b 1,..., b n) F n q belongs to (Cn) f if and only if n 1 X b i+1@ X i= j=1 β j α i+k j 1 A =, (1) for k =, 1,..., t 1. Equation (1) can be rewritten as X n 1 X! b i+1 αj i β j αj k =, k =,..., t 1. (2) j=1 i= P n 1 Consider the above syste of equations with unknown variables i= b i+1αj i and coefficients β j αj k. There is at least 1 solution (the zero solution). Moreover, we have t >. However, because all β j and all α j are distinct, the first equations in (2) give a unique solution as the coefficient atrix is an invertible P Vanderonde atrix. n 1 Therefore, there is only one solution in Equation (2). Hence i= b i+1αj i =, for all j = 1,...,. This iplies that the polynoial w(x) = P n 1 i= b i+1x i is divisible by f. If not all β j are nonzero, then, without loss of generality, we assue that a n = P j=1 β jα n j for n =, 1,... and <. Naely, we assue β j = for j = + 1,...,. We show that each α j where j = + 1,..., ust be a conjugate of one of α j where j = 1,...,. Otherwise, if there exists one α j such that + 1 j and α j is not a conjugate of any α j where j = 1,...,, then all conjugates of α j ust be in the set {α +1,..., α }. Let g be P the inial polynoial with roots α j and its conjugates. Then g f. Because a n = j=1 β jαj n for n =, 1,..., we have that sequence a is derived by the polynoial f/g, which has degree <, contradicting that f is the inial polynoial of sequence a. Hence each α j where j = + 1,..., is a conjugate of one of α j where j = 1,...,. Now an eleent w = (b 1,..., b n) F n q belongs to (C f n) if and only if n 1 X i= b i+1@ X j=1 β j α i+k j 1 A =, (3) for k =, 1,..., t 1. Equation (3) can be rewritten as X n 1 X! b i+1 αj i β j αj k =, k =,..., t 1. (4) j=1 i=

6 6 P n 1 Consider the above syste of equations with unknown variables i= b i+1αj i and coefficients β j αj k. There is at least 1 solution (the zero solution). Moreover, we have t > >. However, because all β j for all 1 j and all α j are distinct, the first equations in (4) give a unique solution as the coefficient atrix is an invertible Vanderonde atrix. Therefore, there is only one solution in Equation (4). Hence i= b i+1(α q j )i = P n 1 i= b i+1α i j =, for all j = 1,...,. Furtherore, we obtain P n 1 P n 1 ( i= b i+1αj i )q = because b i+1 F q. This iplies that a conjugate of α j where j = 1,..., is still a root of w. Because each α j satisfies that + 1 j is a conjugate of one of α j such that 1 j, we therefore obtain f w. Suppose f = f 1 f k such that f 1,..., f k are irreducible polynoials and f satisfies the conditions in Theore 5. Let (Cn) f be the code generated by f and Cn fi be the code generated by each f i as in Theore 5. Let C be the union of Cn f and Cn fi, together with the zero vector of length n. The dual code C of C is contained in the dual code (Cn) f of Cn. f Therefore, if (Cn) f has iniu weight t + 1, then C has iniu weight at least t + 1 as well. If C can be generated by a priitive polynoial of degree as in Theore 2, then C is an orthogonal array of strength at least t. Rearks about cobinatorial applications. One of the priary applications suggested by Muneasa [23] was the construction of orthogonal arrays when f is priitive. Theore 5 shows that orthogonal arrays can be constructed fro a far larger class of polynoials. However, the characterization of the dual code in ters of the ultiples of the polynoial f deonstrates that other interesting cobinatorial objects can be built fro these ethods. Arrays which have orthogonal or covering properties for a selected collection of subsets of coluns, rather than all subsets of coluns of a fixed size t, have recently been studied. Indeed, the investigation of partial orthogonal arrays [22] was otivated directly by Muneasa s original paper. Another well-known exaple of arrays which are orthogonal for selected subsets of coluns are the (t,, s)-nets [18]. A covering array is a generalization of orthogonal arrays which in its siplest definition requires that for any given set of t coluns, every t-tuple appears at least λ ties [5]. These are used extensively in reliability testing where each colun corresponds to an input or paraeter of the syste under test and each row corresponds to the settings for one test. This is typically used in a black-box anner assuing nothing about the internal structure of the syste. But when internal knowledge of the syste is known we can relax the requireent that all t-subsets of coluns should be covered and only require coverage for those known to interact. This setting can even be generalized to collections of colun subsets of various sizes. There has been a substantial aount of recent work in this area, for exaples and ore references see [3, 4, 19, 2]. Arrays constructed fro shift-register sequences offer a eans to construct such objects with coverage over algebraically deterined collections of colun subsets. 2.3 Notation In this section, we introduce soe notations that are widely used in the reainder of the paper. Also, in the next sections we use the sae terinology as in [8]. In particular, when the su of coefficients in the sae colun of our figures is we write that corresponding ters x i cancel. Any use of the terinology up, down, left, above, lower, etc., is with respect to the layout in the figure.

7 7 Notation 1 Let F be a field, f, g, h F[x]. Let f(x) = a + bx k + x P divide g(x) = c + dx l + x n d 1 such that n 3, a, c and fh = g, where h(x) = i= h ix i. We say that l is the left-over of g. Moreover, in the expansion of fh = g we get: d = h l + bh l k + ah l, h i = c. i [, 2], If only ah l, then the left-over is of Type. If only bh l k, then the left-over is of Type K. If only h l, then the left-over is of Type M. If only h l = and bh l k + ah l, then the left-over is of Type K. If only bh l k = and h l + ah l, then the left-over is of Type M. If only ah l = and h l + bh l k, then the left-over is of Type KM. If ah l, bh l k, h l and h l + bh l k + ah l, then the left-over is of Type KM. The above notation is given for the case when f is trinoial. If f is binoial we do not have bh l k ter when calculating dx l. Thus, we do not have left-overs of Type K, K, KM and KM. In order to better understand all the definitions and notations let us consider the following exaple in F 3 : f(x) = 2+x 2, h(x) = 2+2x 2 +x 4, g(x) = 1+x 4 +x 6. The exaple illustrates the case when f is binoial and the ter bh l k is oitted. The product fh corresponds to the left box diagra fro Fig. 1, where a colun indicates the degree of the onoial in the expansion fh. The boxes in row i correspond to the nonzero coefficients in f(x) (h i x i ), i < d, and the entries in the boxes coe fro the exponents of the onoials fro f yielding this ter in the expanded product. The box diagra on the right of Fig. 1 gives the explanation of the figure on the left but using the variable x. Throughout this paper the notation A b is used to refer to a box in colun b containing the label A. For exaple, in Fig. 1 we say that 2 is canceled up with 2 2, since the su of coefficients is. We can see it on the right diagra, 2x 2 + x 2 =. On the diagra, 2 4 is not canceled down with 4 since the su of coefficients in this colun is not. In this case we have left-over l = 4 of Type M x 2 2 x 2 2x 4 2 2x 4 x 6 1 x 4 x 6 Fig. 1 An illustration of the notation in equation g(x) = h(x)f(x) = ( P a i x i )f(x) with f(x) = 2 + x 2, h(x) = 2 + 2x 2 + x 4 and g(x) = 1 + x 4 + x 6 over F 3.

8 8 2.4 Reductions of the proble In this section, we introduce a result that is used to reduce the probles in the next sections. Let w(f) denote the weight of f, that is, the nuber of nonzero ters of f. Theore 6 Let f, g, h F[x], fh = g, w(f) = n > 1 and w(g) =. If there exists an f F[x] such that f(x) = f (x k ) for k > 1 then there exist g i F[x], w(g i ) = i for i < k such that X k 1 g(x) = g i (x k )x i k 1, = i, and i 1. (5) i= P P d Proof. Suppose that h(x) = i= a ix i d/k and define h i (x) = j= a jk+i x j, for i < k. Thus we have that X X i= k 1 h(x) = h i (x k )x i. (6) i= Let g i = f h i and define i to be the weight of g i. Equation (5) now follows fro g = fh and Equation (6). Since the powers of x in each g i (x k )x i are disjoint sets, is partitioned into i and since every g i is a ultiple of f whose weight is ore than 1, we have that i 1. When f is a binoial divisibility by x a is equivalent to a being a root. We can use this to derive an instance of Theore 6 for the case when f has weight 2. Corollary 1 Let f(x) = x k + a F[x]. Then f divides g with w(g) = if and only if there exist g i F[x] with weights w(g i ) = i 1 such that g i (a) = and X k 1 g(x) = g i (x k )x i k 1, = i. i= Corollary 2 Let f, g, h F[x], fh = g, w(f) = n and w(g) 3. If there exists f F[x] such that f(x) = f (x k ) for k > 1 then there exists g F[x] such that g(x) = g (x k ). Proof. There are no integer partitions of 2 or 3 that contain ore than one nonzero part and that do not contain any part of size 1. Thus, considering w(g) 3, we ay assue that gcd of the exponents of f is 1. This will be used frequently in our proofs. For exaple in the case that w(f) = 2 and w(g) 3, Corollaries 1 and 2 give for binoials dividing binoials: X i= f(x) = x + a, g(x) = x d ( a) d ; and for binoials dividing trinoials k f(x) = x + a, g(x) = x d + ( a) d k c( a) x k + c, 1 k d 1.

9 9 3 Trinoials dividing binoials 3.1 Polynoials dividing binoials There are soe general coents which can be given about polynoials dividing binoials. Let f, g, h F[x], where F is a field, fh = g, w(f) = n > 1, ρ is the period of f, and w(g) = 2. If the degree of g is greater than ρ then Sadjadpour et al. [24] describe all possible binoial g in ters of binoial ultiples of f whose degree is less than ρ. We start with a couple siple facts. First, if f divides a polynoial of degree d and weight w then f ust divide a onic polynoial of the sae degree and weight. If f divides two onic polynoials of the sae degree then it ust divide their difference. These cobined with the fact that a non-onoial cannot divide a onoial gives the following fact. Fact 1 Let f F[x] and w(f) > 1. The polynoial f can divide at ost one onic binoial with nonzero constant of any fixed degree. A siple induction gives the next fact. Fact 2 In F[x] the following divisibility always holds for any i 1: x k a x ik a i. Proposition 1 Let f F[x], w(f) > 1 and the period of f be ρ. If f divides a binoial of degree k < ρ with a nonzero constant ter, then it ust divide a binoial of degree gcd(ρ, k). Proof. If k is not a divisor of ρ then we have ρ = sk + r where < r < k. Fact 2 now gives that f divides a onic binoial h of degree sk with a nonzero constant ter. Now x ρ 1 x r h results in a binoial of degree r with a nonzero constant ter. An induction and the Euclidean algorith coplete the proof. We can now conclude that to know everything about f dividing binoials it is sufficient to restrict the attention to onic binoials with nonzero constant ters and degrees that are divisors of the period ρ of f. Fact 2 also gives constraints as to what the constant ters can be as a function of roots of unity in the field F. On this topic there is uch known [15]. Sadjadpour et al. [24] and Fact 2 construct all others with nonzero constant ters and ultiplication by non-trivial onoials yields the rest. Of course these are quite general stateents. In this paper we give ore precise stateents about binoials as ultiples of a polynoial f, of degree, for a sall range of degrees, naely those not ore than char(f). 3.2 Trinoials dividing binoials over F 3 This section focuses on the divisibility of binoials by trinoials over F 3. The ain result is given in Theore 7. As in previous sections, results are derived for onic polynoials f and g such that f divides g. Since we work with F 3, results in Theore 7 can be extended to that f divides 2g, 2f divides g and 2f divides 2g. If a polynoial f F 3 [x] with f() = 1 divides polynoial g(x) = x n 1 then the sallest such n is the period of f and periods of polynoials are well studied [15]. All

10 1 other binoials g with f g and deg(g) greater than the period of f are characterized by Sadjadpour et al. [24]. Next we give results for trinoials dividing binoials over F 3. To enhance readability of the paper we use K s and M s although they are k s and s. Theore 7 Let f(x) = a + bx k + x (a, b ) be a onic trinoial over F 3. If g(x) = c + x n (c ) is a onic binoial over F 3 with degree at ost 3 divisible by f, with g = fh, then f and g are as given in Table 2. Case f(x) g(x) bx /2 + x b + x 3/ bx /2 + x 1 + x bx /2 + x 1 + x a + x /3 + x 1 + x 8/3 1.5 b + bx 2/3 + x 1 + x 8/3 Table 2 Polynoials over F 3 such that g = fh for onic trinoial f and onic binoial g. Proof. The idea is siilar to the previous section. If there exists f and f(x) = f (x k ) for soe integer value k, the proble is reduced to finding binoials g divisible by trinoials f. Thus, we assue that f (x) = a + bx k + x and gcd(k, ) = 1. First we consider the case that 4 and 2k >. Consider the box diagra in Fig. 2. In order to get the binoial g the left-ost and the right-ost ters ust reain and the rest of the ters ust be canceled. So we ust have row [1] and row [2] to cancel K k down with k. Since (k, ) = 1, M can only be canceled down with and we ust have row [3]. Also row [4] ust occur because K 2k cancels down with 2k. Since we have row [4] we have to cancel M 2 and K 3k. After we cancel the, we get deg(g) + 2k. Therefore, we need row [5] to cancel M +2k down with K +2k. Cancelation with +2k gives deg(g) 3. Also row [6] ust exist to cancel M 2 and M 2+k as in Fig. 2. If 2 3k, there are ore rows on Fig. 2 to coplete all cancelations to get the binoial. If we cancel K 3k up, we get a left-over ter of Type to the left of M. If we cancel K 3k down, we get M +3k ter that is not canceled. If we cancel M +3k down with +3k, we get deg(g) > 3. If we cancel M +3k down with K +3k, we get +2k to the right of M 2 and deg(g) > 3. Therefore, there is no solution in this case. If 2 = 3k, and (k, ) = 1, we have = 3 and k = 2. This contradicts to 4. Therefore, there is no solution in this case. We have thus proved that for 4, and 2k > there are no trinoials f dividing binoials g. Using reciprocity we can state that there are no trinoials f dividing binoials g when 4. The proble is now reduced to finding those trinoials f of degree at ost 3 dividing binoials of degree at ost 9. This is a finite proble and can be exhaustively coputed. The results are shown in Table 3. The generalization of Table 3 to arbitrary such that gcd(k, ) is not necessarily 1 is given in Table 2.

11 11 k 2k + k 3k 2 + 2k 2 + k 3 [1] k [2] k [3] k [4] k [5] k [6] k Fig. 2 An illustration of equation g (x) = ( P a i x i )f (x) with trinoial f and binoial g when deg(g) [ + 1, 3] and 2k > over F 3. f (x) g (x) Case in Table bx + x 2 b + x bx + x x bx + x x a + x + x x b + bx 2 + x x Table 3 Polynoials over F 3 with g = f h for onic trinoial f (x) = a + bx k + x, such that gcd(k, ) = 1, and onic binoial g. 4 Trinoials dividing trinoials over F 3 This section focuses on the divisibility of trinoials by trinoials over F 3. The ain results are shown in Theore 8. Results are derived for onic polynoials f and g such that f divides g. As before, since we work with F 3, results in Theore 8 can be extended to f divides 2g, 2f divides g and 2f divides 2g. Again, to enhance readability of the paper we use K s and M s as though they are k s and s. Definition 1 Let f(x) = a + bx k + x divide g(x) = c + dx l + x n such that fh = g in F 3 [x], and n 3. Let N k denote the nuber of K s to the right of 2 and N denote the nuber of M s to the right of 2. For exaple, Fig. 2 shows the case when N k = 2 and N = 3. Lea 1 Let f, g, h be as above. If deg(g) > 2, N k and N as defined, then: N = N k + 1, if l (, 2] of any Type or l (2, 3) and of Type MK; N = N k, if l (2, 3) and of Type K; N = N k + 2, if l (2, 3) and of Type M. Proof. All the M s to the right of 2 can only be canceled with K s, since canceling with a iplies deg(g) > 3. If l (, 2] of any type or l (2, 3) of Type MK, all but one M to the right of 2 can only be canceled (or paired) with K s. The other one corresponds

12 12 to the leading ter of g. Therefore, for the given condition on the left-over l, we get N = N k + 1. If l (2, 3) of Type K, all the M s to the right of 2 can only be canceled with K s. We have one K for the left-over and one M for the leading ter of trinoial g. Therefore, in this case we get N = N k. If l (2, 3) of Type M, all but two M s to the right of 2 can only be canceled with K s: two reain for left-over and for the leading ter of trinoial g. Therefore, in this case we get N = N k + 2. Lea 2 Let f, g, h be as above and N k and N as defined. If deg(g) > 2 then: N k 1, if k = 1; N k 1, if l (, 2] of any Type or l (2, 3) of Type MK; N k 2, if l (2, 3) of Type M; N k = 1, if l (2, 3) of Type K i 2 + j 2 + x 2 + y [1] [2] [3] [4] k [5] k Fig. 3 An illustration of relationship of N k and left-over in g. Proof. Let > 2k, otherwise we use the reciprocal of f. Consider the portion of the box diagra for g = fh in Fig. 3. Since we only work with N k and N as defined before, we need to check only what happens to the right of 2. l (, 2] of any type or l (2, 3) of Type MK or k = 1 We assue that N k 1, and therefore row [5] exists and it is the last row in Fig. 3. If l (, 2] of any type or l (2, 3) of Type MK we ust have row [3] to cancel K 2+j up with M 2+j. If N k > 1 (l (2, 3) of Type MK) and [4] is the second to last row, we have ore M s that are not canceled. If k = 1 and l is of any type then the existence of row [5] and the fact that deg(g) 3 iply that j = 1 and row [4] cannot exist. Therefore, N k 1. l (2, 3) of Type M Assue [5] is the last row on Fig. 3. Then we have row [3] to cancel K 2+j up with M 2+j. If we have row [4] then we have [2] to cancel K and M in colun

13 i. If we have ore rows with K s to the right of 2 it gives ore M s to the right of 2 + j that are not canceled. Therefore, N k 2. l (2, 3) of Type K Assue [5] is the last row on Fig. 3. If N k > 1 then we have ore rows above [5] and it gives M s that are not canceled, given that left-over is of Type K. Therefore, N k = 1. Theore 8 Let f(x) = a + bx k + x (a, b ) be a onic trinoial over F 3. If g(x) = c + dx l + x n (c, d ) is a onic trinoial over F 3 with degree at ost 3 divisible by f, with g = fh, then 1. g = f 3 ; 2. f and g are as in Table 4; or 3. f and g are reciprocals of polynoials listed in Table 4. Case f(x) g(x) bx /2 + x 1 bx /2 + x bx /2 + x b + x /2 + x 5/ bx /2 + x b bx + x 5/ bx /2 + x b x 3/2 + x 5/ bx /2 + x b + bx 4/2 + x 5/ bx /2 + x 1 + x + x bx /2 + x b + x + x 3/ bx /2 + x b bx + x 3/2 1.9 a x /3 + x a x /3 + x a x /3 + x 1 + x 2/3 + x 8/ a + x /3 + x a + ax 2/3 + x 7/ a x /3 + x a ax 4/3 + x 7/ a x /3 + x a + x 5/3 + x 7/ a + x /3 + x 1 + ax 5/3 + x a x /3 + x a + ax 4/3 + x 5/ bx /4 + x b + bx 6/4 + x 11/ bx /4 + x 1 + bx 9/4 + x 1/4 Table 4 Table of polynoials such that g = fh with f and g onic trinoials over F 3. Proof. Let f be a trinoial dividing a trinoial g, where f and g are as given in the stateent of the theore. Let h(x) = a + a 1 x + a 2 x 2 + (a ) such that g = fh. The proof is split into parts. The idea is siilar to the proof in the previous section. If there exists f (x) and f(x) = f (x k ) for soe integer value k, the proble is reduced to finding trinoials g divisible by trinoials f. Thus, we assue that gcd(, k) = 1. We get two cases: 1. (k, ) = 1, k 1; 2. k = (k, ) = 1, k 1 In this part we only consider the case when > 2k; using reciprocity we can extend the results to < 2k. The proof in this part is divided into subcases according to the

14 14 type and position of the left-over, l. We assue for the whole case 1 that = jk + i and i < k. (a) l (, ) of Type K Consider the box diagra for g = fh fro Fig. 4. In order to get the trinoial g, the left-ost, the right-ost and one of the iddle ters ust reain and the rest of the ters ust cancel. On the diagra we always have row [1]. Since left- k 2k 3k + k + 2k 2 k k 3 k 3 [1] k [2] k [3] k [4] k [5] k [6] k [7] k Fig. 4 An illustration of g(x) = ( P a i x i )f(x) with f, g trinoials over F 3 for case 1(a). over is of Type K and gcd(k, ) = 1 and k 1, we need row [4] to cancel M (corresponds to M in colun ) with. Assue that K +k is canceled up only by M +k, thus we have row [2]. We get the following syste of equations: Rearranging the above equations we get ba + a k =, colun +k, aa + a =, colun, aa k + ba =, colun k. aba =, contradicting that a, b and a, since we always have row [4]. We have shown that K +k cannot be canceled up only, and we have row [5]. According to Leas 1 and 2, N k 1 and N 2. Therefore, we can only have at ost 1 ore row after [5]. We ust have row [7] to cancel M 2 down by 2 and cancel M 2+k down by K 2+k. There are no ore rows between [5] and [7]. Thus, K +2k cancels up with M +2k in row [3]. Next, row [2] ust be present to cancel 2k up with K 2k and k up with K k. There are no rows between [3] and [4] because otherwise we get extra M s between and 2. They cannot be canceled since no ore rows are possible between [5] and [7]. The solution in this case is f(x) = a + bx k + x, g(x) = a + bx 3k + x 3,

15 15 corresponding to the trivial case g(x) = f 3 (x). (b) l (, ) of Type K or Consider the box diagra for g = fh fro Fig. 5. In order to get the trinoial g, the left-ost, the right-ost and one of the iddle ters ust reain and the rest of the ters ust cancel. k 2k = jk + i + k + 2k jk (j + 1)k (j + 2)k k 3 i + jk + (j + 1)k 3 k 3 [1] k [2] k [3] k [4] k [5] k [6] k [7] k [8] k [9] k Fig. 5 An illustration of g(x) = ( P a i x i )f(x) with f, g trinoials over F 3 for cases 1(b)-1(g). On the diagra we always have rows [1]-[3], since > 2k and left-over is of Type K or, all the K s ust be canceled with s or atched with s (in case left-over is of Type K). There are soe rows possible between [1]-[2] and [2]-[3]. For now the type of l does not atter, since we have to cancel or atch all K s to the left of. According to Lea 2, N k 1. According to Lea 1, N = N k + 1, i.e. at ost 2 rows are possible below [4]. Assue N k =. If we have row [4] then M +jk cancels down with K +jk (observe that if M +jk cancels down with a, then there is a k to the right of 2) and M 2 cancels down with row [7]. We have at least one M between 2 and 3 k that is not canceled. If we do not have row [4] then row [6] ust exist to cancel M +k down. Also M cancels down with K and it gives one ore M between + k and + jk. We still have to cancel at least two M s between + k and + jk and cancel M 2+k. However, only one ore row is allowed below [4]. Therefore, there is no solution for this case when N k =. Assue N k = 1. If we have row [8] then row [5] ust exist and there is no row [4]. Therefore, M is not canceled. If we have row [9] then rows [4] and [6] ust exist. No ore rows are allowed below [4] and therefore K (j+1)k ust be canceled

16 16 up. Therefore, there will be ore rows between [2] and [3] to cancel all extra K s which will give ore M s between + k and 2 that are not canceled. Therefore, there is no solution in this case when l (, ) of Type K or. (c) l = Consider the box diagra for g = fh fro Fig. 5. On the diagra we always have row [1]-[3], since > 2k. There are soe possible rows between [2] and [3]. Then we have row [5] to cancel K (j+1)k down and row [8] to cancel M +(j+1)k down. Thus, we have N 2 and Leas 1 and 2 give N = 2, N k = 1 and no other rows below [4] other than [5] and [8]. Therefore, K (j+2)k is not canceled (it cannot be canceled up since (k, ) = 1) and there is no solution in this case. (d) l (, 2] Consider the box diagra for g = fh fro Fig. 5. On the diagra, as before, we always have row [1]-[3], since > 2k. There are soe possible rows between [2] and [3]. Since (k, ) = 1 and k 1 we have row [4]. Let = jk + i, j 2, 1 i < k. This holds for reaining cases when k 1. According to Leas 1 and 2, N k 1, N = N k + 1 and at ost 2 rows are possible below [4]. If l (j + 1)k then we have [5] to cancel K (j+1)k down and row [8] to cancel M +(j+1)k down, since only one row is possible between [4] and [8] and we have [5]. Therefore K (j+2)k and M 2 are not canceled and there is no solution in this case. Let l = (j + 1)k. If l is of Type K then again we have [5] and [8] and therefore K (j+2)k and M 2 are not canceled and there is no solution in this case. If l is of Type K then we do not have [5] and [8]. Now we need to cancel M +jk and we get deg(g) > 2. Therefore, we also need to cancel M 2. Since we do not have [8], we can only cancel M +jk down with K +jk. This iplies the existence of a row beneath row [4] which, because of the gcd condition, cannot be row [7]. If we also have row [7] to cancel M 2 down with K 2, we get two M s to the right of M 2 that are not canceled. Therefore, [7] does not exist and we have [9] to cancel M 2 down with 2. The row [6] ust exist to cancel K 2+k up with M 2+k and M +jk down with K +2k. We have + 2k = + jk and j 2 gives j = 2. Therefore, the solution in this case is f(x) = a + bx k + x, g(x) = a + bx 3k + x 3, corresponding to the trivial case g(x) = f 3 (x). (e) l (2, 3] of Type K According to Leas 1 and 2, N = N k = 1. Therefore, no rows are possible between [4] and [8]. Therefore, K (j+1)k is not canceled and there is no solution in this case. (f) l (2, 3] of Type M According to Lea 1, N = N k + 2. Therefore, we have at ost two rows between [4] and [8]. If we have [9], then we ust have [6] to cancel K 2+k up. We have [5] to cancel K (j+1)k down. We already have two rows between [4] and [8] and K (j+2)k is not canceled down. Also K (j+2)k cannot cancel up with M (j+2)k because gcd(k, ) = 1. Therefore, there is no solution in this case. If we do not have [9], then we have [7] to cancel M 2 down. We have [5] to cancel K (j+1)k down. We already have two rows between [4] and [8] and K (j+2)k has to be canceled. We can cancel it only with 2 k and get (j + 2)k = 2 k,

17 17 Solving for (k, ) = 1, = jk + i and j 2 we get = 5, k = j = 2 and i = 1. We should have row [8] to cancel M +jk and M +(j+1)k down. We can write the following syste of equations: a 9 = 1, colun 3-i, a 8, colun 3-k, ba 9 + a 6 =, colun +(j+1)k, aa 9 + a 4 =, colun +jk, ba 8 + a 5 =, colun 2, aa 8 + ba 6 =, colun (j+2)k, aa 6 + ba 4 =, colun (j+1)k, ba 5 + a 2 =, colun +k, aa 5 + a =, colun =jk+i, aa 4 + ba 2 =, colun 2k=jk, aa 2 + ba =, colun k. Rearranging the above equations we get ab =, contradicting that a, b. Therefore, there is no solution in this case. (g) l (2, 3] of Type MK According to Lea 1, N = N k + 1. Therefore, we have at ost two rows below [4]. Again we cannot have [8] and [9] at the sae tie. Assue we have [9]. Then we have [6] and this is the only row between [4] and [9]. Therefore, K (j+1)k is not canceled and there is no solution in this case. Assue we have [8]. Then we have [5] and this is the only row between [4] and [8]. Therefore, M 2 is not canceled and there is no solution in this case. 2. k = 1. First suppose 6. According to Lea 1 and Lea 2 we have - if l 2, then N k 1 and N 2; - if l > 2, then. l is Type M: N k 1 and N 3;. l is Type K: N k = 1 and N = 1;. l is Type MK: N k 1 and N 2. This eans that, for k = 1, N 3. Therefore, we can consider the following cases according to N (the nuber of nonzero coefficients a i of polynoial h(x), i [ + 1, 2], where fh = g): (a) N = 3 According to the above stateent, this case is possible only if l > 2 and l is of Type M. Consider the box diagra for g = fh fro Fig. 6. We always have rows [1]-[4] to cancel all s and 1 s to the left of M. In general, the total of rows ust exist between [1] and [4] inclusive. Since N = 3, one of the M s to the right of 2 has to be canceled down with 1. Since N k 1, we have row [7] and row [6] to cancel up with M 2+1 and only one ore row ust exist between [6] and [7] inclusive to give left-over of Type M. Since 6, there are at least three M s between + 3 and 2 1 inclusive that are not canceled. All three of the cannot be canceled with a single row. Therefore, there is no solution in this case. (b) N = 2 This case is possible if either l 2 of any type, or l > 2 and l of Type MK or M.

18 [1] [2] [3] [4] [5] [6] [7] 1 Fig. 6 An illustration of g(x) = ( P a i x i )f(x) with f, g trinoials over F 3 and k = 1. i. l 2 If l or l < and of Type K = 1 then consider the box diagra for g = fh fro Fig. 6. Here we have a siilar situation as in 2(a). There are at least two M s between + 3 and 2 2 (inclusive) that are not canceled. One of the can be part of left-over, but others are not canceled. Therefore, there is no solution in this case. Consider l < of Type K = 1. Here we have a siilar situation as in 1(a). The flow of the proof is exactly the sae and we get trivial solution g(x) = f 3 (x). The case when l < of Type K = cannot happen because k = 1 and either all 1 s canceled with s or there are 2 left-overs and 1. ii. l > 2 and l of Type MK We have a siilar situation as in 2(b)i. and the sae box diagra for g = fh fro Fig. 6 that contains rows as discussed earlier. The difference is that there are no ore rows allowed between [6] and [7]. This eans that for given 6, there are at least two M s between + 3 and 2 2 (inclusive) that are not canceled. Therefore, there is no solution in this case. iii. l > 2 and l of Type M Here we have a siilar situation as in previous part. Again there are at least five M s between + 2 and 2 that need to be canceled by at ost two rows below [5]. Therefore, there is no solution in this case. (c) N = 1 This case is possible if l > 2 and l is of Type K or l 2. i. l > 2 and l is of Type K Here we have a siilar situation as in 2(a). We have the sae box diagra for g = fh fro Fig. 6 that contains rows as discussed earlier. The difference is that there are no ore rows allowed between [5] and [7] (row [7] is required in order to get left-over of Type K). This eans that for given 6, there are at least four M s between + 2 and 2 1 (inclusive) that are not canceled. Therefore, there is no solution in this case.

19 19 ii. l 2 If l or l < of Type 1 then consider the box diagra for g = fh fro Fig. 6. We always have rows [1]-[4] to cancel or atch all s to the left of M. All rows between [1] and [4] inclusive are present. Even if there is row [5] to cancel M +1 down with 1 +1, for given 6 there are at least five M s between +2 and 2 or +1 and 2 1 (inclusive). Only one row is allowed below row [5] that can cancel two M s, one M is left-over and there are still M s that are not canceled. Thus, there is no solution in this case. Consider l < of Type K = 1 and the box diagra for g = fh fro Fig. 6. Since l <, we have row [1] and [5] to cancel M down with. If row [6] is present then N = 1 shows this is the last row. Also M 2 ust cancel down and this cannot be done. Therefore, row [6] is absent and we ust have row [2] to cancel 1 +1 up with M +1 and 1 1 down with 1. Having one ore row below [5] leads to cancelation of M 2 and existence of row [4] and therefore we have to cancel all s to the left of M. This contradicts that the left-over is of Type 1. Therefore, there is no solution in this case. The case l < of Type is not possible since it gives K to the left of M that is not canceled. (d) N = In this case we get that no rows exist below [5]. In 1(a) we proved that the case with only rows [1], [2] and [5] is not possible. If we have other rows between [2] and [5] we get M s between M +1 and M 2 and K s to the left of M that are not canceled. Therefore, there are no solutions in this case. According to the above results, given 6, there is only one case possible for trinoial f diving trinoial g, when g = f 3. This eans the proble is reduced to 5 which is a finite proble and can be coputed with our progra. Running the progra, the results in Table 5 are obtained. f (x) g (x) Case in Table bx + x 2 1 bx + x bx + x 2 b + x + x bx + x 2 b bx 2 + x bx + x 2 b x 3 + x bx + x 2 b + bx 4 + x bx + x x 2 + x bx + x 2 b + x 2 + x bx + x 2 b bx 2 + x a x + x 3 a x + x a x + x x 2 + x a + x + x 3 a + ax 2 + x a x + x 3 a ax 4 + x a x + x 3 a + x 5 + x a + x + x ax 5 + x a x + x 3 a + ax 4 + x bx + x 4 b + bx 6 + x bx + x bx 9 + x Table 5 Polynoials over F 3 such that g = f h with onic trinoial f and onic trinoial g.

20 2 5 Conclusion Divisibility of polynoials over finite fields with control of their weights is not well understood, even though it has any concrete practical applications. In this paper we study the divisibility of binoials and trinoials by binoials and trinoials over finite fields. We ostly focus in F 3. Natural extensions are to consider polynoials with ore onoials and larger finite fields. The overall goal of the area is to copletely characterize when polynoials of certain weight divide polynoials of another weight. Unfortunately, this sees to be out of reach with the known ethods. We cannot use our ethods for an arbitrary field as the field grows, the nuber of ways of cancelling explodes; this causes the syste of equations to have any solutions. New techniques are required to go beyond fewnoials (polynoials with few onoials) dividing fewnoials as the ones presented in this paper. We feel it is iportant that the next steps in the research should be to find a ethod which works over an arbitrary field and not just a ethod to solve the next largest field or the next highest weight. References 1. J. Ph. Auasson, M. Finiasz, W. Meier and S. Vaudenay, TCHo: a Hardware-Oriented Trapdoor Cipher, Proc. ACISP 7, LNCS 4586, , R. C. Bose, On soe connections between the design of experients and inforation theory, Bull. Inst. Internat. Statist., 38: , C. T. Cheng, The test suite generation proble: optial instances and their iplications, Discrete Applied Matheatics, 155: , M. B. Cohen, C. J. Colbourn, J. S. Collofello, P. B. Gibbons and W. B. Mugridge, Variable strength interaction testing of coponents, Proc. 27th Internat. Cop. Softw. and Applications, , C. J. Colbourn, Covering arrays, Handbook of Cobinatorial Designs, Chapter VI.1, , C. J. Colbourn and J. H. Dinitz (eds.), Handbook of Cobinatorial Designs, Discrete Matheatics and its Applications, Chapan & Hall/CRC, second edition, P. Delsarte, Four fundaental paraeters of a code and their significance, Inforation and Control, 23:47 438, M. Dewar, L. Moura, D. Panario, B. Stevens and Q. Wang, Division of trinoials by pentanoials and orthogonal arrays, Designs, Codes and Cryptography, 45:1 17, S. Golob, Shift Register Sequences, Aegean Park Press, S. Golob and G. Gong, Signal Design for Good Correlation, Cabridge University Press, K. C. Gupta and S. Maitra, Multiples of priitive polynoials over GF(2), In Progress in cryptology INDOCRYPT 21 (Chennai), volue 2247 of Lecture Notes in Coput. Sci., pages Springer, Berlin, H. F. Jordan and D. C. M. Wood, On the distribution of sus of successive bits of shiftregister sequences, IEEE Transactions on Coputers, C-22:4 48, M. Herrann and G. Leander, A practical key recovery attack on basic TCHo, Proc. PKC 29, LNCS 5443, , K. Jabunathan, On choice of connection-polynoials for LFSR-based strea ciphers, In Progress in cryptology INDOCRYPT 2 (Calcutta), volue 1977 of Lecture Notes in Coput. Sci., pages Springer, Berlin, R. Lidl and H. Niederreiter, Introduction to Finite Fields and their Applications, Cabridge University Press, J. H. Lindhol, An analysis of the pseudo-randoness properties of subsequences of long -sequences, IEEE Transactions on Inforation Theory, IT-14, , S. Maitra, K. C. Gupta, and A. Venkateswarlu, Results on ultiples of priitive polynoials and their products over GF(2), Theoretical Coputer Science, 341(1-3): , 25.

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