Tilings of Binary Spaces Gérard Cohen Département Informatique ENST, 46 rue Barrault 75634 Paris, France Simon Litsyn Department of Electrical Engineering Tel-Aviv University Ramat-Aviv 69978, Israel Alexander Vardy Coordinated Science Laboratory University of Illinois Urbana, Illinois 61801, USA Gilles Zémor Département Réseaux ENST, 46 rue Barrault 75634 Paris, France Abstract We study partitions of the space IF 2 n of all the binary n-tuples into disjoint sets, where each set is an additive coset of a given set V. Such a partition is called a tiling of IF 2 n and denoted (V, A), where A is the set of coset representatives. We give a sufficient condition for a set V to be a tile, in terms of the cardinality of V +V. We then employ this condition to classify all tilings with sets of small cardinality. Further, periodicity of tilings in IF 2 n is discussed, and a simple construction of nonperiodic tilings of IF 2 n is presented for all n 6. It is also shown that the nonperiodic tiling of IF 2 6 is unique. A tiling (V, A) is said to be proper if V generates IF 2 n ; it is said to be full-rank if both V and A generate IF 2 n. We show that, in general, the classification of tilings can be reduced to the study of proper tilings. We then prove that any tiling may be decomposed into smaller tilings that are either trivial or have full rank. Existence of full-rank tilings is exhibited by showing that each tiling is uniquely associated with a perfect binary code. Moreover, it is shown that periodic full-rank tilings may be further decomposed into smaller tilings, and then the existence of nonperiodic full-rank tilings is deduced. Finally, we generalize the well-known Lloyd theorem, originally stated for tilings by spheres, for the case of arbitrary tilings. The work of Alexander Vardy was supported in part by the NSF under Grant NCR 9501345.
1. Introduction Given a body in an n-dimensional metric space, is it possible to tile the space with translations of this body? This problem has been extensively studied for the Euclidean space IR n, see [29, 32] and references therein. The case of Euclidean spaces of small dimension has received a considerable attention, and a substantial progress was achieved for tilings of the Euclidean plane [11, 9]. In particular, Penrose [24] demonstrated the existence of nonperiodic tilings of IR 2. Another related topic is tilings of the plane with a finite collection of specially defined bodies. See, for example, publications on polyomino tilings and related problems [12]. In another setting, one may define tilings of finite abelian groups G, where by a tiling we mean a decomposition of the form G = A + V, such that V A = G. In other words, tiling is a partition of G into additive cosets of a tile set V. The problem of describing all possible tilings of abelian groups was first brought up by Hajós [13] and is largely unsolved. Hajós suggested a first effort in this direction by considering tilings of a certain type, which will be called periodic in this paper. For a precise definition of such tilings see Section 5. If all the tilings of a given group G are periodic, then they can be described accurately with the help of a recursive decomposition. After a substantial amount of effort, the problem of determining all the abelian groups that admit only periodic tilings was finally solved by Sands [30], following ingenious ideas of de Bruijn [3] and Rédei [28]. In this work we are interested in tiling the Hamming space IF 2 n consisting of all binary n-tuples. This problem falls into both the above-mentioned categories of tilings, since IF 2 n is at the same time a metric space and a finite abelian group. We will show that other features of tilings besides periodicity, and in particular their rank, may be employed for the recursive decomposition of tilings. More generally, we shall be interested in describing those subsets of IF 2 n that are tiles. Results on the problem of tiling IF 2 n have applications in coding theory. Thus, certain particular cases of this problem have been already studied in the coding theory literature [20, 22]. For example, tilings of IF 2 n with Hamming spheres are known as perfect binary codes [19, 20]. Although the parameters of all such tilings have been determined in [19, 34], a complete classification remains an open problem. Another example is tilings of IF 2 n by the so-called L-spheres, which are unions of some of the shells of the Hamming sphere. For this case all possible parameters are also known [4, 15], whenever L {0, 1,..., n/2}. In general, tilings of IF 2 n with arbitrary bodies correspond to perfect codes correcting an arbitrary set of errors. Conditions for the existence of codes correcting an arbitrary set of errors and estimates of their cardinality were investigated in [6, 7, 8, 16, 17]. Such codes are useful, for instance, for correcting errors at the output of logic networks the set of errors depends on the structure of the network and a single error in an element of the network may lead to an error of greater multiplicity at the network output [8]. Other examples where the problem of correcting a given set of errors arises are concatenated coding schemes [33] and artificial noise channels, which occur if the process of data transmission is considered as a game situation [6]. Tilings of IF 2 n were also found useful in [35] for the design of soft-decision decoders for BCH codes. This paper is organized as follows. We start with some notation and definitions in Section 2. In Section 3 we present a sufficient condition for a given set V to be a tile in terms of the cardinality of the difference set of V. In Section 4 we provide a complete classification of tiles of cardinality 8. Tilings with sets of large rank are also considered in Section 4. In Section 5 we define periodicity of tilings in IF n 2, which is analogous to periodicity of tilings in the Euclidean plane. 1
We prove that nonperiodic tilings of IF 2 n do not exist for n 5, and that the nonperiodic tiling of IF 2 6 is unique up to coordinate transformations. An explicit construction of nonperiodic tilings of IF 2 n for all n 6 is also presented. In Section 6 we introduce the concept of proper tilings and show that the classification of tilings of IF 2 n may be reduced to the study of proper tilings. This leads to a recursive decomposition of tilings into tilings of smaller and smaller size, and ultimately shows that any tiling of IF 2 n may be constructed from tilings that are either trivial or have full rank. In Section 7 we show that a tiling (V, A) is uniquely associated with a perfect binary code of length V 1. This allows us to deduce the existence of full-rank tilings from the existence of full-rank perfect codes established in [10]. A construction of tilings from perfect codes is also presented in Section 7. In Section 8 we describe a technique for further decomposing a periodic full-rank tiling into tilings of smaller size. This technique is then employed to show that nonperiodic full-rank tilings exist. In Section 9 we generalize the well-known Lloyd theorem [20, Ch. 7], originally stated for tilings by spheres, for the case of arbitrary tilings. We give examples where such generalization may be used to prove the nonexistence of certain tilings. 2. Preliminaries Let IF 2 n denote the vector space of dimension n over GF(2). The Hamming distance between vectors x, y IF 2 n, denoted d(x, y), is the number of positions where x and y differ. The Hamming weight of x is wt(x) = d(x, 0), where 0 denotes the all-zero vector. Given a subset C IF 2 n and x IF 2 n we denote d(x, C) = min y C d(x, y). If C = M and min x y C d(x, y) = d we shall say that C is an (n, M, d) binary code. For any x IF 2 n and a nonnegative integer R n, the Hamming sphere of radius R about x is given by B n (x, R) = { y IF 2 n : d(x, y) R }. Given x = (x 1, x 2,..., x n ) IF 2 n, we denote by supp(x) the subset of {1, 2,..., n} consisting of all i such that x i 0. Given a subset V IF 2 n with 0 V, we denote iv def = V + V + + V }{{} i times where + stands for the direct sum. Thus iv is the set of all vectors in IF n 2 that may be represented as a sum of some i elements of V. The linear span of V, denoted V, is the subspace of IF n 2 generated by V. The rank of V is given by rank(v ) = dim V. We let ρ(v ) denote the minimum number of elements of V required to generate any vector in its linear span, namely ρ(v ) def = min j { j : jv = V } It is easy to see that ρ(v ) is the covering radius of the linear code defined by a parity-check matrix H(V ), having the elements of V \ {0} as its columns. We shall say that a given set V is a tile of IF 2 n if it is possible to partition IF 2 n into disjoint additive cosets of V. Note that the set of coset representatives A is also a tile of IF 2 n. Without loss of generality we assume that both V and A contain the 0 element, unless stated otherwise. Evidently, each x IF 2 n has a unique representation of the form x = v + a, where v V and a A. Thus we have the following definition. Definition 2.1. The pair (V, A) is a tiling of IF 2 n if V + A = IF 2 n and 2V 2A = {0}. 2
A trivial necessary condition for (V, A) to be a tiling of IF 2 n is that V = 2 k and A = 2 n k, for some 0 k n. Thus, hereafter, when denoting a subset of IF 2 n by V we assume that V = 2 k. If both V and A are linear subspaces of IF 2 n, then (V, A) is a tiling iff A = IF 2 n /V. Hence, in what follows we mainly focus on those tilings where at least one of the sets V, A is not linear. However, we shall say that V IF 2 n is a linear tile if there is a tiling (V, A) such that A is a linear subspace of IF 2 n. 3. A sufficient condition A well-known example of a linear tile is the Hamming sphere, in which case the set of coset representatives A is a perfect binary code. More precisely, a sphere of radius R < (n 1)/2 is a (linear) tile iff R = 1, n = 2 m 1 or R = 3, n = 23 (cf. [22]). In this section we present a sufficient condition for a set V to be a linear tile, which shows that many more such tiles exist. Theorem 3.1. If 2V < 2 V then V is a linear tile. Proof. Let V = 2 k and suppose that 2V < 2 k+1. Then either k = n, in which case V = IF 2 n is a trivial linear tile, or 2V < 2 n and there exists an a 1 IF 2 n \2V. Set V 1 = V (a 1 +V ). Since a 1 2V, we have V (a 1 + V ) = and V 1 = 2 V = 2 k+1. If V 1 = IF 2 n then (V, {0, a 1 }) is a tiling of IF 2 n and we are done. Otherwise, since 2V 1 = 2V (a 1 + 2V ) < 2 V 1, there exists an a 2 IF 2 n \2V 1 and we set V 2 = V 1 (a 2 +V 1 ) with V 1 (a 2 +V 1 ) = and V 2 = 2 V 1. Continuing in this manner we construct V n k = IF 2 n and a tiling (V, A) of IF 2 n, where A is a subspace of IF 2 n generated by a 1, a 2,..., a n k. Using the results of Zémor [38] it is, in principle, possible to characterize all the sets V which satisfy the condition of Theorem 3.1. Here we give an example of construction that produces such a set. Fix a subspace S IF 2 n and a nonzero element s S. Partition S into cosets modulo the two-element subspace {0, s}, and construct V by choosing one and only one element from each coset. By assumption, we need 0 V and therefore s V. But then, by construction, s is not in V + V and hence 2V < S = 2 V. In the following sections we will be particularly interested in those tiles V which satisfy V = IF 2 n. It can be shown that for these tiles 2V 7 4 V. More bounds on the cardinality of V + V for a given set V may be found in [38]. The condition 2V < 2 V of Theorem 3.1 may be somewhat relaxed if additional information pertaining to V is available. For instance, we have the following corollary to Theorem 3.1. Corollary 3.2. If 2V = 2 V then V is a tile if and only if 2V is not linear. Proof. ( ) If 2V is not linear then 2V 4V and there is an a 1 4V \2V. Set V 1 = V (a 1 +V ). Since a 1 2V, we have V (a 1 + V ) = and V 1 = 2 V as before. Now 2V 1 = 2V (a 1 + 2V ), and since a 1 4V it follows that 2V (a 1 + 2V ). Thus 2V 1 < 2 2V = 2 V 1. Applying Theorem 3.1 to V 1, we conclude that there exists a subspace A 1 IF 2 n such that (V 1, A 1 ) is a tiling. But then so is (V, A 1 + {0, a 1 }). ( ) If 2V is linear then V = 2V, and therefore V does not tile its linear span. The claim now follows directly from Proposition 6.1. 3
Note that 2V is not linear iff ρ(v ) > 2. In fact, the argument of Corollary 3.2 may be pushed a little further, provided that ρ(v ) is large enough. Corollary 3.3. If either of the following holds a. 2V 2 V + 2 and ρ(v ) > 2 b. 2V 2 V + 4 and ρ(v ) > 5 c. 2V 2 V + 5 and ρ(v ) > 20 then V is a linear tile. Proof. (a). If ρ(v ) > 2 there exists an a 1 3V \ 2V, say a 1 = v 1 + v 2 + v 3. As before let V 1 = V (a 1 + V ) with V 1 = 2 V and 2V 1 = 2V (a 1 + 2V ). Note that 2V (a 1 + 2V ) must contain the six vectors v 1, v 2, v 3, v 1 + v 2, v 1 + v 3, v 2 + v 3. Thus 2V 1 2 2V 6 2 V 1 2. Again, applying Theorem 3.1 to V 1, we conclude that V 1 is a linear tile. Hence so is V. (b). A similar argument yields 2V 1 2 V 1 + 2 in this case. Since ρ(v ) > 5 it follows that 2V 1 = 2V (a 1 + 2V ) V, or in other words ρ(v 1 ) > 2. Hence, the proof of case (a) may be used to establish that V 1, and therefore also V, is a linear tile. (c). Similarly, 2V 1 2 V 1 + 4, and since ρ(v ) > 20 we have ρ(v 1 ) > 5. Thus, the proof of case (b) applies. 4. Classification of small tiles In this section we completely characterize all tiles of size 8. In addition, we also consider tilings with sets of large rank. First, we have the simple Proposition 4.1. If V 4 then V is a tile. Proof. Notice that 2V ( V ) ( 2 + 1, and for V 4 we have V ) 2 + 1 < 2 V. Hence V is a linear tile by Theorem 3.1. Next we classify tiles of size 8. Let V = {0, v 1,..., v 7 }. When is V a tile? The answer to this question is greatly facilitated by the following simple fact: V is a tile of IF 2 n if and only if it is a tile of its own linear span V. The proof of this statement is postponed until Proposition 6.1. Here, we distinguish between several cases according to the rank of V. Claim 4.2. If rank(v ) = 7 then V is a tile. Proof. Let C be the perfect binary Hamming code of length 7. Define A = { c 1 v 1 + c 2 v 2 + + c 7 v 7 : (c 1, c 2,..., c 7 ) C } We claim that (V, A) is a tiling of V. Since V = 2 3, A = 2 4 and V = 2 7, it would suffice to show that 2V 2A = {0}. If a 1 + a 2 2V for some distinct a 1, a 2 A, then there are two distinct codewords in C at distance 2 from each other. However, since the minimum distance of C is 3, this is a contradiction. Note that the proof of Claim 4.2 amounts to choosing the appropriate coordinates for V, so that the set V of rank 7 becomes the Hamming sphere B 7 (0, 1). In general, we may always assume w.l.o.g. that B r (0, 1) V, where r = rank(v ). 4
Consider the case where rank(v ) = 6. After a suitable choice of coordinates for V, we have V = B 6 (0, 1) {x}, where x IF 6 2 is of weight 2. In this case, V is a tile if and only if wt(x) 4, 5. The proof of this follows from Proposition 4.5 below, and is therefore postponed. Now let rank(v ) = 5. Again, with a suitable choice of coordinates, V = B 5 (0, 1) {x, y}. Claim 4.3. V is a tile if and only if one of the following holds: a. wt(x) = 2, wt(y) = 2 and wt(x + y) = 2; b. wt(x) = 3, wt(y) = 2 and wt(x + y) = 1; c. wt(x) = 3, wt(y) = 2 and wt(x + y) = 5; d. wt(x) = 3, wt(y) = 3 and wt(x + y) = 2 Proof. Note that 2V may be written as B 5 (0, 2) B 5 (x, 1) B 5 (y, 1) {x + y}. Hence, V is a tile of V if and only if there exists a set A IF 5 2 of cardinality 4, such that 2A B 5 (0, 2) = {0} (1) 2A B 5 (x, 1) = 2A B 5 (y, 1) = 2A {x + y} = (2) Condition (1) means that the Hamming distance between any two vectors in A is at least 3. Since the (5,4,3) binary code is unique [22], we have that A = { 00000, 11100, 10011, 01111 } (3) up to a permutation of coordinates. Note that A is linear, and therefore 2A = A. Hence, condition (2) translates into d(x, A) 2 d(y, A) 2 x + y A (4) The cases (a),(b),(c),(d) may be now derived by inspection from (3) and (4). For rank(v ) = 4 we have V = B 4 (0, 1) {x, y, z}. In this case it is obvious (from Corollary 3.2) that V is a tile iff 2V V. Claim 4.4. V is a tile if and only if one of the following holds: a. wt(x) = wt(y) = wt(z) = 2 and wt(x + y) = wt(x + z) = wt(y + z) = 2; b. wt(x) = wt(y) = 2, wt(z) = 3 and wt(x + y) = 2, wt(x + z) = wt(y + z) = 1; c. wt(x) = 2, wt(y) = wt(z) = 3 and either wt(x + y) = 1 or wt(x + z) = 1; d. wt(x) = wt(y) = wt(z) = 3 Proof. All we need to show is that there exists a vector in IF 4 2 which is not a sum of 2 elements from B 4 (0, 1) {x, y, z}. This is easily done by inspection. For instance, (1111) is such a vector in case (a), and so A = {0000, 1111}. In all other cases A has the form {0000, 1110}. Finally, if rank(v ) = 3 then V is linear, and hence is a trivial tile. Note that the foregoing classification shows that in all cases where V 8 and V is a tile, it is also a linear tile. This also follows directly from Corollary 7.3. We now consider tiles of large rank, that is, those tiles for which rank(v ) is close to the upper bound V 1. For instance, if rank(v ) = V 1 then w.l.o.g. V = B r (0, 1) and is therefore a tile. If rank(v ) = V 2 then w.l.o.g. V = B r (0, 1) {x} for some x IF r 2 of weight 2. Proposition 4.5. If V = B r (0, 1) {x} then V is a tile, provided wt(x) r 2, r 1. 5
Proof. We have 2V = B r (0, 2) B r (x, 1) with r = 2 k 2. Thus, we may take A to be an (r, 2 r k, 3) linear code, obtained by shortening the (r + 1, 2 r k+1, 3) Hamming code C, provided d(x, 2A) = d(x, A) 2. It is well-known [10, 22] that for any 3 w r 2 or w = r + 1 there exists a codeword c C of weight w. Hence, if wt(x) r 2, r 1, we may always find a permutation of C such that c coincides with x in the first r coordinates and has a 1 in the last coordinate. Upon shortening (that is, taking all codewords with a zero) in the last coordinate, we obtain A with d(x, A) 2. As a consequence of Proposition 4.5, we have Claim 4.6. If V = 8 and rank(v ) = 6, that is V = B 6 (0, 1) {x}, then V is a tile if and only if wt(x) 4, 5. It follows from the proof of Proposition 4.5 that if V = B r (0, 1) {x} and a tiling (V, A) exists, then A must have the parameters (n, M, d) of a shortened Hamming code. Further, it is easy to show that there cannot be a vector of weight n 2 or n 1 at distance 2 from any shortened perfect code C of length n = 2 k 2 (either linear or nonlinear). Assume to the contrary that x is such a vector, and extend C to a perfect code C e = C 0 C 1 of length n + 1. Here C 0 = {(c 0) : c C} and C 1 = {(c 1) : c C } for some C, where ( ) denotes concatenation. Since d((x 0), C e ) 1 while d((x 0), C 0 ) 2, it follows that d((x 0), C 1 ) 1. Therefore x C and hence C e contains a codeword of weight n 1 or n, which is a contradiction. Thus, if any code with the parameters (2 k 2, 2 2k k 2, 3) can be obtained by shortening a perfect code, then no tiling is possible for wt(x) = r 2 or wt(x) = r 1. This is certainly true for k = 3, since the (6, 8, 3) shortened Hamming code is unique, which establishes the only if part of Claim 4.6. 5. Periodicity of tilings A tiling T of the Euclidean space IR n is said to be periodic if there exists a translation mapping from IR n to IR n which takes T into itself. In other words, T is periodic if t + T = T for some nonzero t IR n. Similarly, we shall say that a tile V of IF 2 n is periodic if there exists a nonzero v IF 2 n, such that v + V = V. Note that since 0 V by assumption, we must have v V. We call v a periodic point of V. To keep the notation rigorous, we shall consider 0 to be a periodic point of any set. A periodic point is also called a stabilizer of the set in some texts [14]. Given a tiling (V, A) of IF 2 n, we shall say that it is nonperiodic if neither V nor A contain a nonzero periodic point. The existence of nonperiodic tilings of IR 2 is a well-known problem in tessellation theory [24]. The kites and darts tiling of Penrose [24] is a notable example of a nonperiodic tiling of the plane with two distinct polygons. It is still unknown, however, whether there exists a single (nonconvex) body which tiles the plane only nonperiodically. In this section we consider the periodicity of tilings in IF 2 n. First, we construct a nonperiodic tiling of IF 2 6 and then show that IF 2 6 is the smallest binary space which admits a nonperiodic tiling. We also present without proof a simple general construction of tilings in IF 2 n, which establishes the existence of nonperiodic tilings for all n 6. Although this also follows from the general classification of groups that admit nonperiodic tilings, concluded by Sands [30], we hope these constructions will prove insightful. Furthermore, we show that the nonperiodic tiling of IF 2 6 is unique up to coordinate transformations. Proposition 5.1. Let 6
V = {0, v 1,..., v 7 } = A = {0, a 1,..., a 7 } = 0 1 0 0 0 0 1 0 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 1 0 0 0 0 0 0 0 0 0 1 1 0 0 0 1 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 0 0 1 1 0 1 0 1 0 1 0 1 0 0 1 1 0 0 0 0 1 1 1 1 (5) (6) where the elements of V, A IF 6 2 are represented as column vectors. Then (V, A) is a nonperiodic tiling of IF 6 2. Proof. To see that (V, A) is a tiling define A 0 = {0, a 1, a 2, a 3 } and A 1 = {a 4, a 5, a 6, a 7 }. Then obviously 2V (A 0 +A 1 ) = (consider the last row in (5),(6) above), and it remains to show that 2V 2A 0 = 2V 2A 1 = {0}. Note that the first five rows of (5) correspond to B 5 (0, 1) {x, y} with wt(x + y) = 5, while the first five rows in both A 0 and A 1 are isomorphic to the tile set in (3). Thus, 2V 2A 0 = 2V 2A 1 = {0} follows from Claim 4.3. To see that (V, A) is nonperiodic, observe that rank(v ) = rank(a) = 5. It is obvious that a set of cardinality 8 and rank 5 cannot be periodic. We now show that the nonperiodic tiling of Proposition 5.1 is the smallest possible. Proposition 5.2. If (V, A) is a nonperiodic tiling, then V 8 and A 8. Proof. Obviously, any tile of cardinality 2 is linear and, hence, periodic. Now let (V, A) be a tiling of IF n 2 with A = 4. We claim that either A is linear or V is periodic. Indeed, let A = {0, a, b, c} and define A = {0, a, b, a + b}. Then 2A 2A which implies 2V 2A = {0}. Hence, both (V, A) and (V, A ) are tilings of IF n 2, and therefore we must have c+v = (a+b)+v. This is only possible if c = a + b, in which case A is linear, or if c = (a + b) + v, where v is a nonzero periodic point of V. Furthermore, it can be shown that the tiling of Proposition 5.1 is unique. Namely, in any nonperiodic tiling of IF 2 6 both tiles must be of the form (5), up to coordinate transformations. For example, under the coordinate transformation corresponding to taking {a 1, a 2, a 4, a 5, a 6, 1}, where 1 denotes the all-one vector, as the basis of IF 2 6, the tile A given in (6) maps into (5). In view of Proposition 5.2, in order to establish the uniqueness of (5),(6) we only need to consider tiles of cardinality 8. All such tiles have been classified in Section 4. Furthermore, if V = 8 and V is a nonperiodic tile of IF 2 6, then obviously 4 rank(v ) 6. If rank(v ) = 6 then V can tile IF 2 6 = V only linearly, in view of Proposition 4.5 and the fact that a (6, 8, 3) code must be linear. Hence rank(v ) = 4 or rank(v ) = 5. Following the various cases in claims 4.3 and 4.4, it may be readily verified that none of them produces a nonperiodic tiling, except case (c) in Claim 4.3. This is precisely the case corresponding to (5). We now describe a general construction of tilings in IF 2 n which shows that nonperiodic tilings exist for all odd n 7. Let ν = 2 m 1, where m 3, and let h 1, h 2,..., h ν be the distinct nonzero elements of IF 2 m, arranged in some definite, lexicographic say, order. Fix a permutation π on the set {1, 2,..., ν} and consider A 0, A 1, V IF 2 2m+1 given by 7
A 0 = 0 h 1 h 2 h ν (7) A 1 = 0 h 1 h 2 h ν 1 1 1 1 (8) V = 0 h 1 h 2 h ν 0 h π(1) h π(2) h π(ν) (9) where the elements of A 0, A 1, V are again represented as column vectors, and (..) denotes concatenation. Let A = A 0 A 1. Then it is easy to see that (V, A) is a tiling of IF 2 2m+1. Further, the set A = A 0 A 1 is clearly nonperiodic. It may be also shown that if the permutation π is given by, for instance, π = (1, 4, 2)(3, 6)(5, 7)(8, 9)(10, 11) (ν 1, ν) then the set V in (9) is nonperiodic as well. Thus, we have constructed nonperiodic tilings of IF 2 n for all odd n 7. To exhibit nonperiodic tilings of IF 2 n for even n we use a variant of the same construction. As before, let ν = 2 m 1. Now let π be any derangement of the set {1, 2,..., ν}. Define A, V IF 2 2m+2 as follows A = 0 h 1 h 2 h ν 0 h 1 h 2 h ν 1 1 1 1 (10) V = 0 h 1 h 2 h ν 0 h π(1) h π(2) h π(ν) 0 h 1 h 2 h ν 0 h 1 h 2 h ν 1 1 1 1 (11) Then (V, A) is a nonperiodic tiling of IF 2m+2 2 for all m 2. We omit the proof of (7) (11). 6. Recursive decomposition of tilings In this section we present a recursive decomposition (construction) of tilings, which shows that any tiling of IF n 2 may be decomposed into (constructed from) smaller tilings of certain particular type. First, we prove the following proposition which was used in the classification of Section 4. 8
Proposition 6.1. A set V IF n 2 is a tile of IF n 2 if and only if it is a tile of V. Proof. ( ) Since V is linear it is a tile of IF 2 n. Thus, if ( V, A 1 ) is a tiling of IF 2 n and (V, A 0 ) is a tiling of V, then evidently (V, A 0 + A 1 ) is a tiling of IF 2 n. ( ) Let (V, A) be a tiling of IF 2 n and define A 0 = A V. We claim that (V, A 0 ) is a tiling of V. Indeed, since A 0 A and 2V 2A = {0}, it follows that 2V 2A 0 = {0}. Further, since V IF 2 n = V + A, any w V can be written as w = v + a, where v V and a A. However, since V is linear we have a = v + w V, so that a A 0. It follows that V V + A 0. The converse inclusion V + A 0 V is obvious from V, A 0 V. In view of Proposition 6.1, we shall be particularly interested in those tilings (V, A) for which IF 2 n = V. Note that until now the order of the sets in the tiling pair (V, A) was of no importance, since the roles of V and A were completely symmetric. However, this is no longer true if we require V = IF 2 n. Definition 6.1. The ordered pair (V, A) is a proper tiling of IF 2 n if (V, A) is a tiling of IF 2 n and V = IF 2 n. The following proposition shows that the classification of all tilings in IF 2 n may be, in principle, reduced to the study of proper tilings. Let V be a tile of IF 2 n with V = IF 2 n, or equivalently rank(v ) < n. Denote m = 2 n r 1, where r = rank(v ). Theorem 6.2. The pair (V, A) is a tiling of IF n 2 if and only if A has the following form: 1. For i = 0, 1,..., m let A i be a subset of V, such that (V, A i ) is a tiling of V. 2. Let c 0 = 0, c 1,..., c m be a set of representatives for IF n 2 / V. 3. For i = 0, 1,..., m let v i be any element of V. Then A = A 0 (v 1 + c 1 + A 1 ) (v m + c m + A m ) (12) Proof. ( ) Let A be as in (12), and denote V = 2 k. Then A i = 2 r k for all i, and A = (m+1)2 r k = 2 n k. Hence, it remains to show that 2V 2A = {0}. We have 2A = U W, where U = m i=0 2A i and W = 0 i<j m (v i + v j + c i + c j + A i + A j ). Now, 2V U = {0} since 2V 2A i = {0} for all i. Since c i + c j V for all 0 i < j m, it follows that V and W are disjoint, and hence 2V W =. ( ) Let (V, A) be a tiling of IF 2 n. Pick any set of representatives c 0 = 0, c 1, c 2,..., c m for IF 2 n / V and define ( ) A i = v i + c i + A (c i + V ) where v i is any element of V such that 0 A i. To see that such v i exists note that c i + (A c i + V ) V. We have v i + c i + A i = A (c i + V ) and, hence, m (v i + c i + A i ) = i=0 m A (c i + V ) = A (13) i=0 We need to show that (V, A i ) is a tiling of V for all i. Clearly 2A i = (A (c i + V ))+(A (c i + V )) 2A, and therefore 2V 2A i = {0}. Note that A i V, which implies V + A i V. Thus, to establish that (V, A i ) is a tiling of V, it remains to show that A i = 2 r k. Since 2V 2A i = {0}, we obviously have A i 2 r k. However, (m+1)2 r k = 2 n k = A m i=0 A i, where the last inequality is from (13). This implies A i = 2 r k and completes the proof. Remark. Note that the vectors v 1, v 2,... v m and the sets A 0, A 1,..., A m in Theorem 6.2 are not necessarily distinct. It is assumed that 0 A i for all i = 0, 1,..., m. 9
The analysis of Theorem 6.2 shows that if all the proper tilings of IF r 2 are known for r = 1, 2,..., n, we can construct all the tilings of IF n 2. However, as will be now shown, the set of all the proper tilings is not the smallest class of tilings which permits complete classification of all tilings of IF n 2. Indeed, let (V, A) be a proper tiling of V and consider the tiling (A, V ). Unless rank(a) = rank(v ), this tiling is not proper and, hence, by Theorem 6.2 V = V 0 (a 1 + c 1 + V 1 ) (a m + c m + V m ) (14) where (A, V i ) is a proper tiling of A for all i, 0, c 1,..., c m are representatives of V / A, and a 1, a 2,..., a m A. Thus, using (14), each of the tilings (V, A i ) of V in Theorem 6.2 may be decomposed into yet smaller tilings, unless A i = V. This process may be iterated until a complete decomposition of the original tiling (V, A) is obtained. Example. Consider the tiling (V, A) of IF 2 6 given in (5) and (6). This tiling is not proper since rank(v ) = rank(a) = 5 < 6. If we take c 0 = 0 and c 1 = (000001) as the representatives of IF 2 6 / V, we can write A = A 0 (c 1 + A 1 ) with A 0 = 0 1 1 0 0 0 1 1 0 1 0 1 0 0 1 1 0 1 0 1 0 0 0 0 A 1 = 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 0 0 1 1 0 0 0 0 Note that (V, A 0 ) and (V, A 1 ) are proper tilings of V. Since rank(a 0 ) = rank(a 1 ) < rank(v ), we may further decompose (V, A 0 ) and (V, A 1 ) as follows. Let 0, c 0,1,..., c 0,7 and 0, c 1,1,..., c 1,7 be the representatives of V / A 0 and V / A 1, respectively. We may take, for instance, {c 0,1, c 0,2,..., c 0,7 } = {c 1,1, c 1,2,..., c 1,7 } = Then the corresponding decompositions of V are given by where and 1 0 1 0 1 0 1 0 1 1 0 0 1 1 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 V = V 0,0 (a 0,1 + c 0,1 + V 0,1 ) (a 0,7 + c 0,7 + V 0,7 ) V = V 1,0 (a 1,1 + c 1,1 + V 1,1 ) (a 1,7 + c 1,7 + V 1,7 ) a 0,1, a 0,2,..., a 0,7 = 0, 0, (110100), 0, (101010), (011110), 0 A 0 a 1,1, a 1,2,..., a 1,7 = 0, 0, (110010), 0, (101110), (011100), 0 A 1 V 0,0 = = V 0,7 = V 1,0 = = V 1,7 = {0} Indeed, since A 0 and A 1 are linear, both (A 0, V 0,i ) = (A 0, {0}) and (A 1, V 1,i ) = (A 1, {0}) are trivial tilings of A 0 = A 0 and A 1 = A 1, respectively. As illustrated in the foregoing example, the recursive decomposition of (12) and (14) terminates when trivial tilings, of the form (V, {0}) with V linear, are obtained. The only other case where the recursion of (12),(14) stops is when a tiling (V, A) with V = A is encountered. Definition 6.2. A tiling (V, A) of IF 2 n is said to be of full rank if V = A. 10
It is easy to see that a tiling (V, A) of IF 2 n is of full rank iff rank(v ) = rank(a) = n. In the sense described in this section, any tiling of IF 2 n can be constructed in a unique way from the trivial tilings and tilings of full rank. This clearly leads to the following question: do full-rank tilings exist? We shall settle this question affirmatively in the next section. 7. Tilings and perfect binary codes A perfect binary code of length n and Hamming distance 2R + 1 is a tiling of IF n 2 with Hamming spheres of radius R. Perfect codes have been extensively studied, see for instance [10, 18, 23, 25, 26, 34, 36]. It is known [18, 34] that perfect codes exist only for R=0, R=n, R = (n 1)/2 with n odd, R =1 with n = 2 m 1, and R =3 with n=23. Since the first three cases are trivial while the last case corresponds to the well-known binary Golay code [22], we shall henceforth use the word perfect to refer to the perfect codes with R=1 and n = 2 m 1. It can be shown that each tiling (V, A) of IF 2 n is uniquely associated with a perfect binary code of length ν = V 1. A similar result in the context of coverings was established by Blokhuis and Lam [2]. Let H(V ) be an n ν matrix having the elements of V \ {0}, arranged in some fixed order, as its columns. For x IF ν 2 let s(x) = H(V )x t denote the syndrome of x with respect to H(V ). Proposition 7.1. Let C = { c IF ν 2 : s(c) = H(V )c t A } (15) Then C is a perfect code of length ν. Proof. We first show that d(c) = min c1,c 2 C d(c 1, c 2 ) 3. Denote a 1 = s(c 1 ) and a 2 = s(c 2 ), where a 1, a 2 A by (15). Suppose that a 1 = a 2, or equivalently s(c 1 + c 2 ) = 0. Then d(c 1, c 2 ) = wt(c 1 + c 2 ) 3 since the columns of H(V ) are distinct. Now suppose that a 1 a 2. Note that a 1 + a 2 = s(c 1 + c 2 ) = i supp(c 1 +c 2 ) v i. Further note that a 1 + a 2 2V, since 2V 2A = {0}. Hence, again d(c 1, c 2 ) = supp(c 1 + c 2 ) 3. It remains to show that d(x, C) 1 for all x IF ν 2. Since V + A = IF n 2, we have s(x) = v + a for some v V and a A. If v = 0 then x C by (15) and we are done. Otherwise, let c IF ν 2 be a vector which coincides with x in all positions except one, which is the position corresponding to the location of v in H(V ). Then d(x, c) = 1, and s(c) = a which implies that c C by (15). Proposition 7.1 provides a means for constructing a perfect code C from any given tiling (V, A). We shall say that this code C is the perfect code associated with (V, A). It should be pointed out that the correspondence between sets V, A such that V +A = IF n 2 and coverings by spheres of radius R = 1 has been initially noticed by Blokhuis and Lam in [2]. The relevance of their rank, however, seems to have been overlooked in [2]. We now elaborate on this issue. Proposition 7.2. If C is the perfect code associated with a proper tiling (V, A), then rank(c) = ν rank(v ) + rank(a) Proof. Define C 0 = {c IF ν 2 : s(c) = 0}. Clearly, C 0 is a linear code and dim C 0 = ν rank(v ). Now let A = {0, a 1,..., a µ } where µ = A 1. Since V = IF n 2 we can always find a set C 1 = {0, c 1,..., c µ } IF ν 2 such that s(c i ) = a i for all i, rank(c 1 ) = rank(a), and C 1 C 0 = {0}. Then C = c C 1 (c + C 0 ) and rank(c) = dim C 0 + rank(c 1 ) = ν rank(v ) + rank(a). 11
Remark. If (V, A) is not a proper tiling then Proposition 7.2 does not apply. This is so because for rank(v ) < n there exist elements a A which cannot be represented in the form H(V )x t for any x IF ν 2. In this case we have where A 0 = A V. rank(c) = ν rank(v ) + rank(a 0 ) < ν rank(v ) + rank(a) Note that C 0 is a linear subcode of C (regardless of whether (V, A) is proper or not), and C itself is linear, i.e. equivalent to the Hamming code of length ν, if and only if A 0 = A V is linear. The following corollary is an immediate consequence of this fact. Corollary 7.3. If V 8 and (V, A) is a proper tiling, then A is linear. Proof. The perfect code C associated with (V, A) has length at most 7 and, hence, is equivalent (see e.g. [25]) to the Hamming (7, 16, 3) code or to the (3, 2, 3) repetition code. We now employ the foregoing results on the relation between tilings and perfect codes to establish the existence of full-rank tilings. Indeed, a full-rank perfect code (that is, a perfect code of length ν and rank ν) together with the Hamming sphere B ν (0, 1) constitute an example of a full-rank tiling. Furthermore, we have shown that full-rank tilings exist if and only if there exist full-rank perfect codes, for if (V, A) is a full-rank tiling then the associated perfect code must have full rank by Proposition 7.2. Unfortunately, none of the classical constructions of nonlinear perfect codes given in [1, 23, 25, 26, 36] produces perfect codes of full rank. However, such codes have been recently constructed by Etzion and Vardy in [10]. Theorem 7.4 ([10]). For all m 4, there exists a full-rank perfect code of length 2 m 1. Thus, full-rank tilings of IF 2 n exist for all n = 2 m 1 with m 4. It is shown in the next section that they also exist for n = 2 m 2 with m 4, and in particular for n = 14. As a consequence of Corollary 7.3 full-rank tilings do not exist for n 7. It is still an open question whether full-rank tilings exist for n = 8, 9,..., 13, and many other values of n. However, we have the following theorem. Theorem 7.5. Full-rank tilings of IF 2 n exist for all sufficiently large n. Proof. Let (V 1, A 1 ) and (V 2, A 2 ) be tilings of IF n 1 2 and IF n 2, respectively. Define V = { ( v 1 v 2 ) : v 1 V 1, v 2 V 2 } A = { ( a 1 a 2 ) : a 1 A 1, a 2 A 2 } Then clearly 2V 2A = {0} and V A = 2 n 1+n 2, which means that (V, A) is a tiling of IF n 1+n 2 2. Now rank(v ) = rank(v 1 ) + rank(v 2 ) and rank(a) = rank(a 1 ) + rank(a 2 ). Thus, if (V 1, A 1 ) and (V 2, A 2 ) are full-rank tilings then so is (V, A). By Theorem 7.4 there exist full-rank tilings of IF 2 n for n = 15, 31, 63,... Since 15 and 31 are relatively prime, by the conductor theorem of Frobenius [31, p.376], we have full-rank tilings of IF 2 n for all n n 0, where n 0 is sufficiently large. 2 Remark. In fact, using the results of the next section in the proof of Theorem 7.5 it is possible to show that full-rank tilings of IF 2 n exist for all n 112. This leaves exactly 53 values of n for which we do not know whether full-rank tilings exist. We conclude this section with a construction of proper tilings from perfect binary codes. This construction is, in a sense, the converse of Proposition 7.1. Let C be a perfect binary code 12
of length ν, and let Γ be a linear subcode of C, such that Γ + C = C. For instance, by Proposition 8.1 of the next section, we may take as Γ any linear subspace of the set of all the periodic points of C (the kernel of C in the terminology of [1, 27]). Denote γ = dim Γ, and let H(Γ) be a (ν γ) ν parity-check matrix of Γ. Note that the matrix H(Γ) is full-rank by assumption. Take V = {0} { the columns of H(Γ) } and define A = {H(Γ)c t : c C}. Proposition 7.6. With V and A as defined above, (V, A) is a proper tiling of IF ν γ Proof. For x IF ν 2 let s(x) = H(Γ)x t. We claim that 2. C = { c IF ν 2 : s(c) = H(Γ)c t A } (16) Indeed, if c C then s(c) A by the definition of A. To see the converse inclusion, consider a vector x IF 2 ν such that s(x) = a A. By the definition of A, there exists c C such that s(c) = a. Hence s(x + c) = 0, which implies that x + c Γ. But since Γ + C = C it follows that x = (x + c) + c C. It is now easy to see that d(c) = 3 implies 2V 2A = {0}. Suppose to the contrary that v 1 + v 2 = a 1 + a 2 for some v 1 v 2 V and a 1 a 2 A. By (16) and the fact that V = IF ν γ 2, there exists a codeword c 1 C such that s(c 1 ) = a 1. Let c 2 IF 2 ν be a vector which coincides with c 1 in all but the two positions corresponding to the locations of v 1 and v 2 in H(Γ). Then s(c 2 ) = a 1 + v 1 + v 2 = a 2 A, and therefore c 2 C by (16). This is a contradiction, since d(c 1, c 2 ) = 2. The fact that V + A = IF ν γ 2 also follows from (16). Let x IF ν γ 2. Since V = IF ν γ 2, there exists y IF ν 2 such that x = s(y). If y C then x A by (16), and we are done. Otherwise, there exists a codeword c C at distance 1 from y. Let s(c) = a A. Then x = a + v, where v is the column of H(Γ) located at the position where c and y differ. Note that γ is possibly 0, in which case V is a sphere and A = C. However, as will be shown in the next section, given any perfect code C we may always find a linear subcode Γ {0}, such that Γ + C = C. 8. Decomposition of full-rank tilings In this section we show that a periodic full-rank tiling may be further decomposed into smaller tilings, in a way similar to Theorem 6.2. Analysis of the rank and periodicity of tilings resulting from such recursive decomposition is presented. Using this analysis we deduce the existence of nonperiodic full-rank tilings. Let A IF 2 n and let A 0 denote the set of all the periodic points of A. The set A 0 is sometimes called the kernel of A [1, 27]. The following proposition was first established in [1]. We include the proof herein for completeness. Proposition 8.1. The set A 0 is a linear subcode of A. Furthermore, A is the union of disjoint additive cosets of A 0. Proof. The proposition hold vacuously if A is either nonperiodic or linear. Hence assume that A 0 {0} and A 0 A. To see that A 0 is linear, note that (a 1 + a 2 ) + A = a 1 + (a 2 + A) = a 1 + A = A for any a 1, a 2 A 0. To see that A 0 tiles A, let a A \ A 0. Then (a + A 0 ) A 0 = 13
since A 0 is linear, and a + A 0 A since A 0 consists of periodic points of A. If A = A 0 (a + A 0 ) we are done; otherwise continue in this manner until A is exhausted. It follows from Proposition 8.1 that there exists A A, such that A + A 0 = A and 2A 0 2A = {0}. We shall write A = A/A 0. More generally, for any set V IF 2 n we define V = V/A 0 as follows. Fix a basis a 1, a 2,..., a m for A 0, where m = dim A 0, and complete this to a basis a 1, a 2,..., a m, b 1, b 2,..., b n m for IF 2 n. Then each vector v = m i=1 α i a i + n m i=1 β ib i in V is mapped onto the vector v = n m i=1 β ib i in V/A 0. Thus V/A 0 is just the projection of V onto IF 2 n /A 0. Note that the mapping from V to V/A 0 is not necessarily one-to-one. Thus, for instance, the mapping from A to A = A/A 0 is A 0 to one, and it is easy to see that the two definitions of A/A 0 coincide. With the above notation we have the following theorem. Theorem 8.2. Let (V, A) be a tiling of IF 2 n and let A 0 be the set of periodic points of A. Then A has the following form: A = A (c 1 + A ) (c 2 m 1 + A ) (17) where (V/A 0, A ) is a tiling of IF n 2 /A 0, and 0, c 1,..., c 2 m 1 are representatives for IF n 2 /A 0 in IF n 2. Proof. Set A = A/A 0 and {0, c 1, c 2,..., c 2 m 1} = A 0. Then (17) holds by Proposition 8.1, and all we need to show is that (V/A 0, A/A 0 ) is a tiling of IF n 2 /A 0. First, we claim that the mapping from V to V/A 0 is one-to-one in this case. Indeed, suppose to the contrary that v 1, v 2 V which map onto the same vector v V/A 0. Then v 1 = m i=1 α i a i + v and v 2 = m i=1 β i a i + v which implies that v 1 + v 2 = m i=1 (α i + β i )a i A 0. This is a contradiction, since A 0 A and 2V 2A = {0}. Thus V/A 0 = V, and therefore V/A 0 A/A 0 = IF n 2 /A 0. It remains to show that 2(V/A 0 ) 2(A/A 0 ) = {0}. Again, suppose to the contrary that v 1 + v 2 = a 1 + a 2 for some v 1 v 2 V/A 0 and a 1 a 2 A/A 0. Let v 1, v 2 be the two vectors in V which map onto v 1, v 2, and let a 1, a 2 be some two vectors in A that map onto a 1, a 2. Then v 1 + v 2 = a 1 + a 2 = m m α i a i + β i a i + v 1 + v 2 i=1 i=1 m m γ i a i + δ i a i + a 1 + a 2 i=1 i=1 which implies, by linearity of A 0, that v 1 + v 2 = a 0 + a 1 + a 2 for some a 0 A 0. But, since A 0 consists of periodic points of A, we have a 0 + a 1 A which contradicts 2V 2A = {0}. Remark. Note that Proposition 8.1 and Theorem 8.2 hold without change if A 0 is any linear subspace of the set of all the periodic points of A. Using Theorem 8.2 a full-rank tiling (V, A) of IF 2 n may be further decomposed into smaller tilings, provided that at least one of the sets V, A is periodic. We have the following sufficient condition for periodicity of proper tilings. Proposition 8.3. Let (V, A) be a proper tiling of IF 2 n and let ṽ = v V v. Then ṽ is a periodic point of A. Proof. Let C be the perfect binary code of length ν = V 1 associated with (V, A). Since the weight distribution of any perfect code containing 0 is uniquely determined [22], the vector 1 of weight ν belongs to c+c for all c C. Thus, 1 is a periodic point of C. Now, ṽ = H(V )1 t = s(1) which shows that ṽ A. Furthermore, since the tiling (V, A) is proper, for any a A there exists c C such that s(c) = a. But then 1 + c C, and therefore s(1 + c) = s(1) + s(c) = ṽ + a belongs to A. 14
By Proposition 8.3 a full-rank tiling (V, A) is nonperiodic only if v = a = 0 a A v V In particular, a full-rank tiling consisting of the Hamming sphere and a full-rank perfect code is necessarily periodic. Example. Using theorems 7.4 and 8.2 it is possible to construct a full-rank tiling of IF 2 14. Let V = B 15 (0, 1) and let A be the full-rank perfect code of length 15 constructed in [10]. Since 1 is a periodic point of A, we may take A 0 = {0, 1} and let u 1, u 2,..., u 14, 1 be the corresponding basis of IF 2 15, where u j denotes a vector of weight 1 with the nonzero entry in the j-th position. Then V/A 0 = {0, u 1,..., u 14 } {(111111111111110)} and A/A 0 is the set of all codewords of A with a zero in the last position. Upon puncturing in the last coordinate we obtain a full-rank tiling (V, A ) of IF 2 14. The following two propositions are concerned with the periodicity of A/A 0 and V/A 0. Proposition 8.4. If A 0 contains all the periodic points of A, then A/A 0 is nonperiodic. Proof. Let a be a periodic point of A/A 0. Then a + A = a + (A/A 0 ) + A 0 = (A/A 0 ) + A 0 = A. Hence a A 0. But (A/A 0 ) A 0 = {0}, and therefore a = 0. Proposition 8.5. The mapping from V to V/A 0 takes periodic points of V into periodic points of V/A 0. Proof. Let v 0 = m i=1 α i a i + n m i=1 β ib i be a periodic point of V, and let v 0 = n m i=1 β ib i be its image in V/A 0. Clearly, for any v = m i=1 γ i a i + n m i=1 δ ib i in V, the fact that v 0 + v V implies that v 0 + v = n m i=1 (β i + δ i )b i is in V/A 0. By Proposition 8.5, if V is periodic then so is V/A 0 and, hence, Theorem 8.2 can be applied to the tiling (V/A 0, A/A 0 ). In general, this process may be iterated until we obtain a decomposition of the original tiling (V, A) into nonperiodic tilings. At each iteration one of the two tile sets loses all its periodic points, by Proposition 8.4. However, the recursion of Theorem 8.2 does not necessarily terminate after two iterations, since the other tile set may acquire new periodic points that is, V/A 0 can be periodic even if the original set V is not. Such a situation is illustrated in the following example. Example. Let V = B 7 (0, 1), and let A be the linear Hamming code generated by 1 1 1 0 1 0 0 0 0 1 0 0 {a 1, a 2, a 3, a 4 } = 0 0 1 0 1 0 0 1 0 1 0 1 0 0 1 1 Take, for instance, A 0 = a 1, a 2, a 3. An appropriate basis for IF 2 7 is a 1, a 2, a 3, u 1, u 2, u 3, u 4. Then, upon puncturing in the last three coordinates, V/A 0 = 0 1 0 0 0 1 1 1 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 0 0 0 0 0 1 0 0 1 (18) 15
Note that V/A 0 in (18) is periodic (with (1000) t being the unique nonzero periodic point), even though V = B 7 (0, 1) is not. Using the iterative decomposition described above, we can deduce the existence of nonperiodic full-rank tilings. First, note that rank(a) m rank(a/a 0 ) n m (19) rank(v ) m rank(v/a 0 ) n m (20) where m = dim A 0. The inequalities (19),(20) show that if (V, A) is a full-rank (proper) tiling of IF 2 n then (V/A 0, A/A 0 ) is a full-rank (proper) tiling of IF 2 n /A 0. Thus, starting with any fullrank tiling (V, A), iterative application of Theorem 8.2 produces a decomposition of (V, A) into nonperiodic full-rank tilings. In fact, if V = B 15 (0, 1) and A = C, where C is a full-rank perfect code of length 15, we need only one iteration. This is so because B 15 (0, 1) = 16 and by Corollary 7.3 full-rank tilings (V, A) do not exist for V 8. Thus, if C 0 is the set of all the periodic points of C, then the tiling (B 15 (0, 1)/C 0, C/C 0 ) is full-rank and nonperiodic. We conclude this section by observing that, as a consequence of theorems 6.2 and 8.2, the classification of tiles reduces to the classification of nonperiodic full-rank tiles. That is, if we can determine whether any given nonperiodic full-rank set is a tile, then using theorems 6.2 and 8.2, we can determine whether any given set is a tile. The next section provides a means for showing that certain sets cannot be tiles. 9. Nonexistence results: generalized Lloyd theorem In this section we derive several necessary conditions for the existence of tilings. Our main result herein is a generalization of the Lloyd theorem [21], originally stated for tilings with Hamming spheres, for arbitrary tiles. The Lloyd theorem plays a crucial role in the classification of perfect binary codes [18, 34], for more details see [20, Ch. 7]. Other generalizations of the Lloyd theorem may be found in [19]. As in [22], we represent a vector v = (v 1, v 2,..., v n ) IF 2 n by its image z v = z v 1 1 zv 2 2 zn vn in the group algebra QG, where G is isomorphic to IF 2 n. Then for any u, v IF 2 n and V IF 2 n, the characters χ u ( ) are defined in the standard way (cf. [22, Ch. 5]), χ u (z v ) = ( 1) <u,v> χ u (V ) = v V χ u (z v ) where <, > stands for the inner product modulo 2 in IF n 2. The Krawtchouk polynomial of degree k in the indeterminate x is defined by P k (x) = ( )( ) k x n x ( 1) i i k i i=0 16
for k = 0, 1,..., n. The distance distribution of a subset A IF n 2 D i (A) = 1 W i (a + A) A a A is given by where W i (a + A) is the number of vectors of weight i in a + A. The MacWilliams transform of D i (A) is given by D i(a) = 1 n D j (A)P i (j) A Now let (V, A) be a tiling of IF n 2. Define the sets N(A), N (A) {0, 1,..., n} as follows j=0 N(A) = { j : D j (A) 0 } N (A) = { j : D j(a) 0 } Thus, N(A) is the set of distances occurring in A while N (A) is the set of distances occurring in the formal dual A = {a : χ a (A) 0} of A. Further, let U = { u : χ u (V ) = 0 } Q(U) = { j : u U such that wt(u) = j } With this notation we have the following theorem. Theorem 9.1. N (A) Q(U) {0}. Proof. Since A + V = IF n 2, in the group algebra QG we have χ u (V ) χ u (A) = χ u (IF n 2 ). Since χ u (IF n 2 ) = 0 for all u 0, either χ u (V ) = 0 or χ u (A) = 0 or both χ u (V ) = χ u (A) = 0, unless u = 0. Now, consider the following sum [22, p.139] S j (A) def = 1 A 2 wt(u)=j χ u (A + A) = 1 A 2 wt(u)=j Furthermore, since χ u (A + A) = i N(A) wt(w)=i χ u (z w ), we have w A+A S j (A) = 1 A 2 wt(u)=j i N(A) wt(w)=i w A+A χ u (z w ) = 1 A i N(A) χ u (A) 2 0 (21) D i (A) wt(u)=j χ u (z x ) Here x is an arbitrary vector of weight i, and we have used the fact that wt(u)=j χ u(z x ) depends only on the weight of x. In fact [22, p.135], wt(u)=j χ u(z x ) = P j (i), and therefore S j (A) = 1 A i N(A) D i (A)P j (i) = D j(a) (22) Now let j {1, 2,..., n}. Clearly, j N (A) iff D j (A) 0. From (21) and (22) it follows that if D j (A) 0 then there exists a vector u of weight j, such that χ u(a) 0. Since χ u (V ) χ u (A) = 0 unless u = 0, for this vector u we must have χ u (V ) = 0. Hence u U and j Q(U). If V is the Hamming sphere of radius R and u IF n 2 is a vector of weight j then 17