f - Semantic Scholar

On the Capacity of Two-Dimensional Run Length Constrained Channels Akiko Kato 

Kenneth Zeger y

IEEE Trans. on Information Theory Submitted: March 10, 1998 Revised: November 11, 1998 Abstract Two-dimensional binary patterns that satisfy one-dimensional (d; k) run length constraints both horizontally and vertically are considered. For a given d and k, the d;k =mn, where N d;k denotes capacity Cd;k is de ned as Cd;k = limm;n!1 log Nm;n m;n the number of m  n rectangular patterns that satisfy the two-dimensional (d; k) run length constraint. Bounds on Cd;k are given and it is proven for every d  1 and every k > d that Cd;k = 0 if and only if k = d + 1. Encoding algorithms are also discussed. (

)

(

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2

A. Kato is with the Department of Mathematical Engineering and Information Physics, Graduate School of Engineering, University of Tokyo, Tokyo 113-8656, Japan, email: [email protected]. This work was supported in part by a JSPS Fellowship for Young Scientists and was performed during her stay at the Department of Electrical and Computer Engineering, University of California, San Diego, CA 92103-0407. y K. Zeger is with the Department of Electrical and Computer Engineering, University of California, San Diego, CA 92103-0407, email: [email protected]. This work was supported in part by the National Science Foundation. 

1

1 Introduction and Main Results A one-dimensional binary sequence is said to satisfy a (d; k)-constraint if the number of 0's between any pair of consecutive 1's is at least d and at most k. The one-dimensional capacity is de ned as Ed;k = limm!1 log Nmd;k =m, where Nmd;k is the number of patterns of length m on a line that satisfy the (d; k)-constraint. The one-dimensional capacity Ed;k is known to be the logarithm (base 2) of the largest real root of the equation X k ?X k?d ? X k?d? ?


? X ? 1 = 0 for 0  d  k < 1, and it is known that Ed;1 = Ed? ; d? for d  1 (see, e.g., [1, 10]). Therefore, for every nonnegative integer d, the one-dimensional capacity Ed;k is positive for all k > d. A two-dimensional binary pattern of 0's and 1's arranged in an m  n rectangle is said to satisfy a two-dimensional (d; k)-constraint if it satis es a one-dimensional (d; k)-constraint both horizontally and vertically. We call such patterns valid. The two-dimensional (d; k)-capacity is de ned as (

)

(

)

2

+1

1

1 2

1

d;k log Nm;n mn ; d;k denotes the number of valid patterns on an m  n rectangle. It is trivial where Nm;n to see that Cd;d = 0 for all d  0, and hence we assume k > d throughout this paper. Note that the de nition of (d; k)-constraints implies monotonicity of the capacity in each variable, namely, Cd;j  Cd;k for j  k; (1) Cj;k  Cd;k for j  d: (2) Thus, in particular, limk!1 Cd;k = Cd;1. The two-dimensional capacity is important for certain digital recording applications, and has recently become the focus of increased study. In this paper we derive various upper and lower bounds on Cd;k , and in particular demonstrate the curious result that for every d  1, the two-dimensional capacity equals zero if and only if k = d +1. The two-dimensional capacity has been mentioned previously in the literature, but a concise and complete proof of its existence appears to be lacking. For the sake of completeness we provide such a proof in the Appendix. While there have been numerous studies of one-dimensional constrained codes, far fewer results have appeared concerning two-dimensional codes. Marcellin and Weber introduced multitrack (d; k)-constrained binary codes in [9]. In an n-track (d; k)-constrained binary code, the d-constraint is required to be satis ed one-dimensionally on each track, but the k-constraint is only required to be satis ed only by the bitwise logical \or" of

Cd;k = m;nlim !1

(

)

2

(

2

)

n consecutive tracks. Orcutt and Marcellin [15] computed capacities of redundant multitrack (d; k)-constrained binary codes, which allow only some xed size subset of the tracks (redundant tracks) to be faulty at every time instant. For the case of d > k, those capacity bounds were derived by Vasic [22]. Erxleben and Marcellin [4] examined error-correcting one-dimensional (d; k)-constrained binary codes for multitrack (d; k)constrained codes; they constructed multitrack (1; 3)-constrained codes having better rates and better error-correcting capabilities than those previously known. Etzion [5] obtained results on mergings of two-dimensional patterns that satisfy both a (d ; k )constraint horizontally and a (d ; k )-constraint vertically, and discussed the Hamming distances of such two-dimensional patterns. Weeks and Blahut [23] calculated numerical bounds on the capacity of various two-dimensional non-\run-length" constrained systems and used a Richardson extrapolation to obtain conjectures for tighter bounds. In contrast to the one-dimensional capacity Ed;k , there is little known about the twodimensional capacity Cd;k . It was shown by Calkin and Wilf [3] that C ;1 exists and is bounded as :587891  C ;1  :588339. Siegel and Wolf [18] used \bit stung" techniques to map one-dimensional sequences onto diagonals in the plane in order to create two-dimensional (d; 1) and (0; k) constrained codes. Ashley and Marcus [2] recently discovered the surprising result that C ; = 0. That is, for the (1; 2)-constraint, eectively no positive amount of information can be stored per bit written in two dimensions. In the present paper, we generalize this result and show that Cd;k = 0 if and only if k = d + 1, for all d  1. Numerous bounds are also given. We are con dent that some of the bounds in this paper can be improved upon by future researchers. Our motivations for presenting these bounds are that they are analytically aesthetic, the derivations are interesting, and in most cases no previous bounds were published or known. The main results of this paper are Theorems 1-8 and Corollaries 1-4, which are stated below. Their proofs are given in Section 2. Theorem 1 For every positive d, Cd;d = 0: Theorem 2 If d < k, then 1 + d log j ! + log (r!) j ; (3) Cd;k  maxk?d (j + d) j  1

2

2

1

1

1 2

+1

%

8$ > > > >
> > > :

where r = d mod j .

3

9 > > > > = > > > > ;

1

It can be seen from (3) (by taking j = 2) that Cd;k > 0 for all (d; k) such that k  d + 2, since 8 1 > > > ; d even > > > < 2(d + 2) Cd;d  > > > d + 1 ; d odd > > > : 2(d + 2) +2

2

which is positive for all d  0. Combining this fact with Theorem 1 gives a characterization of which (d; k)-constraints induce nonzero capacities.

Corollary 1 For every d  1 and every k > d, Cd;k = 0 , k = d + 1: Fact 1 C ; = C ;1. Fact 1 holds since the two-dimensional (1; 1)-constraint is equivalent to the two-dimensional 0 1

1

(0; 1)-constraint, by interchanging the roles of 0 and 1. From Fact 1 and the monotonicity in (1), it immediately follows that C ;k  C ;1, for all k  2. Theorem 3 gives a stronger lower bound on C ;k ; this lower bound approaches 1 as k ! 1. The bounds in Theorems 3 { 5 are given in terms of the quantity C ;1, whose value was determined to within :0002 by Calkin and Wilf [3]. 0

1

0

1

Theorem 3 For every positive integer k,

C ;k  1 ? 1 d?k=C2e;1 :

(4)

1

0

In [20, 21], Talyansky, Etzion, and Roth provided an encoding algorithm for generating \conservative arrays". As a special case, their algorithm generates two-dimensional binary patterns that do not contain more than k consecutive 0's or 1's, which yields the lower bound   (5) C ;k  1 + (bk=2c1 + 1) log 1 ? (bk=2c + 1)
2? bk= c? : The lower bound in (5) appeared in [19] (in Hebrew) for k  8 and is stronger than the lower bound in (4) for all k  8. The proof technique of Theorem 3, is however, interesting in its own right and may lead to future ideas for improving bounds. Theorem 4 If d and k are positive integers such that kd ++ 11 is an even integer, then (6) Cd;k  d +1 1 ? k +2 1 (1 ? C ;1): (

0

2

2

1

4

2

1)

The inequality in (6) is valid whenever k  ?1 mod 2(d + 1). The right hand side of (6) gives a lower bound on Cd;k0 for all k0 2 fk + 1; k + 2; : : : ; k + 2d + 1g by the monotonicity in (1). Thus, Theorem 4 actually gives a lower bound on Cd;k for all d and k. Note that as k ! 1 the lower bound in (6) approaches Cd;1  1=(d + 1). The following theorem gives a tighter lower bound for Cd;1 than the limiting inequality of (6).

Theorem 5 For every d  2,

1 : (7) Cd;1  1 +Cb;d= 2c Theorem 6 below tightens the lower bound in Theorem 5 if and only if d 6= 3, but is less analytically attractive. 1

Theorem 6 For every d  2, Cd;1  max sd 1

s s 1 r r 1 r ! + log 1 + ds log s! i i! i i! i i (s + d) %

8$ > > > >
> > > :

!9 > > > > = > > > > ;

;

(8)

where r = d mod s.

Corollary 2 The lower bound in Theorem 6 is stronger than the lower bound in Theorem 5 if and only if d = 6 3. Theorem 7 For every positive integer k,

(9) C ;k  1 ? k +1 1 log 1 ? 21? k : Theorem 8 For every positive integer d, d d 1 Cd;1  d1 log d! i i! i p  d1 log 2ed + d1 log 2d + 121d log e: The rst upper bound in Theorem 8 is twice the value of the term inside the maximization in (8) when s = d, and becomes the trivial upper bound C ;1  1 for d = 1. Note that since 1 1 n ? ln(1 ? x) = xn < x + x2 + xn = x + x2 + 1 x? x < x + (ln 2)x n n 





0



2

!

X

( +1)

!

2

2

=0

!

2

2

2

3

2

1

X

=1

2

2

X

3

2

=3

5

whenever 0  x  (2 ln 2 ? 1)=(2 ln 2 + 1), the lower bound in (5) implies that log e 4 log e + 16
2? bk= c  + 16
2?k? 1 ? C ;k  b k= c b k= c? (bk=2c + 1)2 (k + 1)2 2

0

2(

(

2

2 +1)

2 +1)

1

(

2

2)

for all suciently large k, which was seen in [19]. Combining this with Theorem7, gives asymptotic bounds on how fast (as k grows) the capacity C ;k approaches 1 for the (0; k)-constraint. 0

Corollary 3 For suciently large k,

p

log e < 1 ? C ;k  4 2
logk=e + 8k : (10) k (k + 1)2 (k + 1)2 2 It is interesting to note that the one-dimensional capacity E ;k is known to converge to one (as k grows) at the rate ( log e)=2k . Corollary 4 follows from Theorems 6 and 8, and it shows that Cd;1 decays to zero (as d grows) exactly at the rate (log d)=d. The one-dimensional capacity Ed;1 is known to decay to zero (as d grows) exactly at the same rate (log d)=d. 1

2

2

2

0

2

0

1

2

4

2

2

Corollary 4

d
C = 1: lim d;1 d!1 log d !

2

2 Proofs of Results The set of integers is denoted by Z , and Z denotes the 2-dimensional integer lattice. A two-dimensional binary code F on Z is a set of distinct mappings f : Z ! f0; 1g, and each mapping is called a codeword. Given a codeword f , for each point (x; y) 2 Z we call the value f (x; y) the label of (x; y) (under f ), and for any set S  Z the set of labels of the points of S is called the label of S (under f ) and is denoted by S^. When no confusion results, the label of Z may be also referred to as a codeword. If all the codewords in F satisfy the (d; k)-constraint, we say that F is a (d; k)-constrained binary code on Z . A subset of Z is called a rectangle if it can be written in the form f(x; y) 2 Z j a  x  a + n ? 1; b  y  b + m ? 1g for some integers a; b; n; m, and we n;m . A rectangle of the form S n;n is called a square and is denoted denote this set by S a;b a;b n . by S a;b Note that, given a two-dimensional binary code F on Z , the label S^ of any square S  Z under f 2 F can be viewed as a binary square matrix. Let 0j denote j consecutive 0's. If the pattern 10d1 occurs as a label of a horizontal line segment f(x; b) 2 Z j a  2

2

2

2

2

2

2

2

2

(

(

(

(

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(

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(

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2

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x  a + d + 1g of length d + 2 for some integers a and b, then we call the binary d  d square matrices S^ ad ;b and S^ ad ;b?d the adjacent matrices of the 10d1 pattern. Figure 1 shows an example of the two adjacent matrices of an occurrence of 10d1 in a (3; 5)-constrained binary code. Similarly, if the pattern 10d1 occurs as a label of a vertical line segment f(a; y) 2 Z j b  y  b + d + 1g of length d + 2 for some integers a and b, we call the binary d  d square matrices S^ ad ;b and S^ ad?d;b the adjacent matrices of the 10d1 pattern. A square matrix is a permutation matrix if there is exactly one 1 on each row and also on each column, and all other components are 0's. If all the anti-diagonal components ai;d?i (i = 1; 2; : : : ; d) are 1's in a d  d permutation matrix A = (ai;j )i;j then A is called an anti-identity matrix. A subset of Z is called a diagonal of width w if it can be written in the form f(x; y) 2 Z j a  x + y  a + w ? 1g for some integer a. Similarly, a subset of Z is called an anti-diagonal of width w if it can be written in the form f(x; y) 2 Z j a  x ? y  a + w ? 1g for some integer a. For nonnegative integers a; b; c, we use the notation a  b mod c to indicate that cj(a ? b) and we use the notation a = b mod c to mean a  b mod c and 0  a < c (i.e. a = b ? b c). Finally, for any given collection of V valid codewords on an m  n c rectangle, we call the quantity log V=(mn) the coding rate of the collection. ( )

( )

( +1

+1)

( +1

)

2

( )

( +1

( )

+1)

(

+1)

+1

2

2

2

2

$

%

2

2.1 Proof of Theorem 1

It is already known that C ; = 0 [2]. Hence we will prove Theorem 1 for d  2 in what follows. (Our proof of Theorem 1 does not directly specialize to d = 1, but a slight modi cation does.) Before giving the formal proof of Theorem 1 we give a brief intuitive description of the proof in order to facilitate an understanding of the rigorous details. The main idea in showing that the capacity Cd;k is zero when k = d + 1 is to show that the number of valid patterns in a rectangle grows sub-exponentially as a function of the area of the rectangle. That is, the ratio of the growth exponent to the area of the rectangle tends to zero as the rectangle's area grows without bound. As an example, the capacity is zero if every bit of information stored in a large square requires, for example, an amount of storage space that is linear in the side length of the square, instead of constant in the side length. Our proof of Theorem 1 rst looks for any occurrence of the pattern 10d1 in the plane and then inspects the two corresponding adjacent matrices. First it is shown that these adjacent matrices must equal each other and must be permutation matrices. Then two 1 2

7

cases are considered: (a) the adjacent matrices are neither the identity matrix nor the anti-identity matrix; or (b) the adjacent matrices are either the identity matrix or the anti-identity matrix. In case (a) it is shown that the (d; d + 1)-constraint forces the label of all of Z to be completely determined so that there is no freedom for choosing any bits beyond the choice of the permutation matrix. In case (b) it is shown that the bits that appear on any horizontal or vertical line in Z completely determine the rest of the choice of bits in Z , since every occurrence of 10d1 or 10d 1 forces the existence of an in nite diagonal or anti-diagonal of width at least d. Hence, each bit of stored information occupies an amount of area in a square that grows linearly, instead of constant, with the length of the side of the square. We conclude that the combined number of patterns that can be stored in a rectangle due to cases (a) and (b) is not enough to achieve positive capacity. Conversely, to prove that the capacity is nonzero for k  d + 2, we demonstrate codes that achieve nonzero coding rates. 2

2

2

+1

Lemma 1 Let d  2. For any (d; d + 1)-constrained binary code on Z , if the pattern 10d1 occurs either horizontally or vertically in a codeword then its d  d adjacent matrices 2

are permutation matrices and are equal to each other.

Proof. Without loss of generality, we can assume that the pattern 10d 1 occurs horizontally as the label of the line segment f(x; 0) j 0  x  d + 1g. Let f : Z ! f0; 1g be a codeword in a (d; d + 1)-constrained binary code on Z such that f (0; 0) = f (d + 1; 0) = 1 (and thus f (0; y) = f (d + 1; y) = 0 for y = 1; 2; : : : ; d). Therefore, for each z 2 f1; 2; : : : ; dg there must exist an x 2 f1; 2; : : : ; dg such that f (x; z) = 1, for otherwise 0d would occur on the horizontal line y = z. Also, for each w 2 f1; 2; : : : ; dg there can be at most one y 2 f1; 2; : : : ; dg such that f (w; y) = 1. Hence the adjacent matrix S^ d; is a d  d permutation matrix. Let  be the unique permutation of f1; 2; : : : ; dg such that f (? (y); y) = 1 for y = 1; 2; : : : ; d. We will show that f (? (y); y ? (d + 1)) = 1 for all y = 1; 2; : : : ; d. We have f (? (d); ?1) = 1 since f (? (y); ?1) = 0 for y = 1; 2; : : : ; d ? 1 and f (0; ?1) = f (d + 1; ?1) = 0; the former follows from f (? (y); y) = 1 for y = 1; 2; : : : ; d ? 1 and the latter from f (0; 0) = f (d + 1; 0) = 1. Thus, the statement is true for y = d, and a straightforward induction argument shows that it is also true for y = d ? 1; d ? 2; : : : ; 1, which completes the proof. 2 2

2

+2

( )

(1 1)

1

1

1

1

1

Lemma 2 Let d  2. Any (d; d + 1)-constrained binary code on Z has at most (d + 2)! distinct codewords that contain the pattern 10d 1, and whose d  d adjacent matrices are 2

neither the identity matrix nor the anti-identity matrix.

8

Figure 2 shows an example of the statement of the lemma, and Figure 3 is useful for following the steps in the proof. In Figure 2, the set S d; and the sets whose labels are the same as S^ d; are shown as square areas surrounded by thick lines (including the boundaries), where  indicates that the label of the point is 1; otherwise the label of the point is 0. (We adopt this convention in all the gures in this paper.) ( +1)

(0 0)

( +1)

(0 0)

Proof. Given a (d; d + 1)-constrained binary code on Z , assume that f : Z ! f0; 1g is a codeword such that f (0; 0) = f (d + 1; 0) = 1 and S^ d; is neither the identity matrix nor the anti-identity matrix. It suces to prove that f (0; d +1) = f (d +1; d +1) = 1, for then Lemma 1 forces the remainder of Z to be labeled in repeated patterns of adjacent matrices, i.e., the label of the whole space Z is uniquely determined by the label S^ d; of the square S d; . Let  be a permutation of f1; 2; : : : ; dg such that f (x; (x)) = 1 for x = 1; 2; : : : ; d, (as given in the proof of Lemma 1). Either (1) 6= d or (d) 6= d, so assume without loss of generality that (d) 6= d. For all y = 1; 2; : : : ; d such that y 6= (d), we have f (?1; y) = 0, since f (? (y); y) = 1 and ? (y) < d for such y's. Also, we have f (?1; ?1) = 0, since f (? (d); d) = f (? (d); ?1) = 1 and ? (d) < d, and we have f (?1; 0) = 0, since f (0; 0) = 1. Therefore, f (?1; (d)) = 1 (for otherwise 0d occurs vertically) and hence f (?1; d + 1) = 0 because (d)  1. Together with the fact that f (x; d + 1) = 0 for all x 2 f1; 2; : : : ; dg, this implies that f (0; d +1) = 1 (for otherwise 0d occurs horizontally). It thus remains to be shown that f (d + 1; d + 1) = 1. Let r = minfx  1 j (x) 6= d + 1 ? xg, i.e., the rth row is the rst row from the top of the adjacent matrix S^ d; that diers from the identity matrix. If r = 1 then f (d + 1; d + 1) = 1 follows by symmetry from an analogous argument to the proof in the preceding paragraph that showed f (0; d + 1) = 1. So assume r  2. First we show that 2

2

( )

(1 1)

2

2

( +1)

( +1)

(0 0)

(0 0)

1

1

1

1

1

+2

+2

( )

(1 1)

f (d + 1 + j; ?j ) = 1 for j = 1; 2; : : : ; r ? 1

(11)

by induction on j . (These points are indicated in Figure 3 as the circled black discs below the x-axis.) For j = 1 we have f (d + 2; ?1) = 1 since f (d + 2; y) = 0 for y = 0; 1; : : : ; d ? 1 and also for y = ?2, because ? (y)  2 for these y's, and because f (d + 1; 0) = 1 and f (? (d ? 1); ?2) = f (? (d ? 1); d ? 1) = 1 (the last equality follows from Lemma 1 combined with the assumption f (0; 0) = f (d + 1; 0) = 1). Now assume the induction statement is true up to and including j (1  j < r ? 1). Then f (d + 1 + (j + 1); y) = 0 for y = d ? j ? 1; d ? j ? 2; : : : ; ?j and for y = ?j ? 2. More precisely, for y = d ? j ? 1; d ? j ? 2; : : : ; 1 the equality follows since ? (y)  j + 2 1

1

1

1

9

for these y's by the assumption j < r ? 1 and the de nition of r; for y = 0 it follows since f (d + 1; 0) = 1; for y = ?1; ?2; : : : ; ?j it follows from the induction hypothesis; for y = ?j ? 2 it follows since f (? (d ? j ? 1); ?j ? 2) = f (? (d ? j ? 1); d ? j ? 1) = 1 by Lemma 1 and ? (d ? j ? 1)  j + 2. Hence we have f (d + 1 + (j + 1); ?j ? 1) = 1 (for otherwise 0d appears vertically), completing the induction argument for (11). From (11) and the de nition of r we have f (d + 1 + r; y) = 0 for y = d ? (r ? 1); d ? r; d ? (r + 1); : : : ; ?r such that y 6= (r). More precisely, for y = d ? (r ? 1); d ? r; : : : ; (r) + 1; (r) ? 1; : : : ; 1, the equality follows since ? (y)  r + 1 for these y's by the de nition of r; for y = 0 it follows since f (d + 1; 0) = 1; for y = ?1; ?2; : : : ; ?(r ? 1) it follows from (11), indicated by the third type circle in Figure 3; for y = ?r it follows since f (r +1; ?r) = f (r +1; d +1 ? r) = 1 by Lemma 1. Therefore, since (r) 6= d +1 ? r by de nition, we must have f (d + 1 + r; (r)) = 1; (12) for otherwise 0d appears vertically. Now we will show that 1

1

1

+2

1

+2

f (d + 1 + r ? j; d + 1 ? r + j ) = 1 for j = 1; 2; : : : ; r ? 1

(13)

by induction on j . (These points are indicated in Figure 3 as the circled black discs in the upper-right corner.) For j = 1 we have f (d + r; d ? r + 2) = 1 for otherwise 0d would appear horizontally, since f (x; d ? r + 2) = 0 for x = r; r + 1; : : : ; d + r ? 1 (because f (r ? 1; d ? r + 2) = 1) and for x = d + r + 1 by (12). Now assume the statement is true up to and including j ? 1, for j  2. Then f (x; d ? r + j + 1) = 0 for x = r ? j + 1; r ? j + 2; : : : ; d + r ? j since f (r ? j; d ? r + j + 1) = 1, and f (d + r ? j + 2; d ? r + j + 1) = 0 since f (d + r ? j + 2; d ? r + j ) = 1 (by the induction hypothesis for j ? 1). Therefore, to avoid 0d occurring horizontally, we must have f (d + r ? j + 1; d ? r + j + 1) = 1, completing the induction argument. In particular, we have f (d + 2; d) = 1 so that f (d + 2; d + 1) = 0. Thus, since f (x; d + 1) = 0 for x = 1; 2; : : : ; d, we must have f (d + 1; d + 1) = 1 to avoid a horizontal 0d . It is concluded that in a (d; d + 1)-constrained binary code on Z , the number of distinct codewords f : Z ! f0; 1g such that f (0; 0) = f (d + 1; 0) = 1 and the adjacent matrix S^ d; is neither the identity matrix nor the anti-identity matrix is at most d! ? 2. There are d +1 choices of x 2 f0; 1; : : : ; dg for which a codeword f : Z ! f0; 1g satis es f (x; 0) = f (x + d + 1; 0) = 1, and there are d + 1 choices of y 2 f0; 1; : : : ; dg for which f (0; y) = f (0; y + d + 1) = 1. Therefore a (d; d + 1)-constrained binary code on Z can have at most (d + 1) (d! ? 2)  (d + 2)! distinct codewords that contain the pattern 10d1, +2

+2

+2

2

2

( )

(1 1)

2

2

2

10

and whose d  d adjacent matrices are neither the identity matrix nor the anti-identity matrix. 2 Note that the proof of Lemma 2 implies that if a (d; d+1)-constrained binary codeword on Z contains the pattern 10d1, and whose adjacent matrices are neither the identity matrix nor the anti-identity matrix, then it is forced to be a (d; d)-constrained codeword. In other words, such a codeword cannot contain 10d 1 horizontally nor vertically, in spite of the (d; d + 1)-constraint. 2

+1

To determine the capacity Cd;k we need an analogue of Lemma 2 for two-dimensional (d; d + 1)-constrained binary codes on a bounded rectangle rather than all of Z . Given integers n; m  1, a two-dimensional (d; d + 1)-constrained binary code F on S n;m ; is a n;m set of distinct mappings f : S ; ! f0; 1g that satisfy the (d; d + 1)-constraint. That is, every codeword (i.e., the label of S n;m ; under every f 2 F ) satis es the (d; d + 1)-constraint. Note that a (d; d + 1)-constrained binary code on S n;m ; might not be extendible to Z . Figure 4 shows an example of (3; 4)-constrained binary code on S ;; that cannot be extended to Z . In [11] related nonextendable patterns are discussed. Corollary 5 Let d  2 and n  m  3d + 3. Any (d; d + 1)-constrained binary code on d n m? d? distinct codewords such that the pattern 10d 1 S n;m ; has at most (d + 2)!2 is contained in S dn? ;dd? ;m? d? and its d  d adjacent matrices are neither the identity matrix nor the anti-identity matrix. Proof. The number of points contained in the m  n rectangle S n;m ; but not   n ? d ? ;m? d? is contained in the m ? 2( d + 1)  n ? 2( d + 1) inner rectangle S d ;d    mn ? m ? 2(d + 1) n ? 2(d + 1) = 2(d + 1)(n + m ? 2d ? 2). The corollary follows from Lemma 2 since there are at most 2 d n m? d? labels of these points and there are at most (d + 2)! valid labels of the inner rectangle. 2 2

(

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(

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(0 0)

(

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(0 0)

(6 4)

2

(0 0)

2

(

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2( +1)(

+

2

2)

(0 0)

(

2

( +1

2

2

2)

+1)

(

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(0 0)

(

2

( +1

2( +1)(

+

2

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+1)

2)

Figure 5 shows an example of Corollary 5 for n = 23; m = 19; d = 3.

Lemma 3 Let d  2. For any (d; d + 1)-constrained binary code on Z , if the pattern 10d1 occurs in a codeword and its d  d adjacent matrices are identity (resp. anti-identity) 2

matrices then the 10d1 pattern is contained in an in nite diagonal (resp. anti-diagonal) of width d + 2. Proof. Let f : Z ! f0; 1g be a codeword in a (d; d +1)-constrained binary code on Z such that f (0; 0) = f (d +1; 0) = 1 and S^ d; is the identity matrix. It suces to show 2

2

( )

(1 1)

11

that f (?1; 1) = f (0; d + 1) = 1. Since f (?1; y) = 0 for y = 2; 3; : : : ; d and also for y = 0 and y = ?1, we have f (?1; 1) = 1. Therefore f (0; d + 1) = 1 because f (?1; d + 1) = 0 and f (x; d + 1) = 0 for x = 1; 2; : : : ; d. 2 Figure 6 shows an example of Lemma 3 for d = 7, and the proof is also illustrated in Figure 7.

Corollary 6 Let d  2. For any (d; d + 1)-constrained binary code on Z , a codeword 2

cannot have occurrences of both the identity and the anti-identity matrices as adjacent matrices of dierent 10d 1 patterns.

Proof. In light of Lemma 3, if the pattern 10d 1 occurs on Z with its adjacent matrices being the identity matrix, and the pattern 10d1 also occurs somewhere else on Z with its adjacent matrices being the anti-identity matrix, then the diagonal strip of width d+2 that contains the former 10d1 pattern and the anti-diagonal strip of width d+2 that contains the latter 10d1 pattern intersect somewhere on Z . This is a contradiction. 2

2

2

2

Remark 1 Let d  2. In view of the proof of Lemma 2, for any (d; d + 1)-constrained binary code on Z , a codeword cannot have occurrences of both the identity matrix and 2

any other matrix as adjacent matrices of dierent 10d 1 patterns. Similarly, a codeword cannot have occurrences of both the anti-identity matrix and any other matrix as adjacent matrices of dierent 10d 1 patterns.

Lemma 4 Let d  2. Suppose that a codeword in a (d; d + 1)-constrained binary code on Z has the property that every 10d1 pattern has identity (anti-identity) matrices as its 2

adjacent matrices. Then, given the label of any in nite horizontal strip of width 1, the codeword is uniquely determined.

Proof. Without loss of generality assume we are given the label of the integers on the x-axis and that every occurrence of 10d1 induces (via Lemma 3) a diagonal of width d + 2. Further assume without loss of generality that the sequence 10d10d 1 occurs on the horizontal line segment from (0; 0) to (2d + 3; 0). The sequence 10d1 induces a diagonal of width d + 2, and therefore the sequence 10d 1 induces a diagonal of width d + 3. This follows because each 1 on the diagonal f(y; ?y) : y 2 Zg forces d + 1 0's to the right (if only d 0's followed then another diagonal of width d + 2 would result, forcing a 1 in position (2d + 2; 0)). This argument can easily be extended using induction to show that all occurrences of 10d 1 on the x-axis induce diagonals of width d+3. 2 +1

+1

+1

12

The following corollary is an analogue of Lemma 4 for (d; d +1)-constrained binary codes on S n;m ; . Corollary 7 Let d  2 and n  m  3d + 3. Given a (d; d + 1)-constrained binary d code on S n;m ; , suppose that a codeword has the property that every 10 1 pattern in S dn? ;dd? ;m? d? has identity (anti-identity) matrices as its adjacent matrices. Then, given the labels of the horizontal line segments L = f(x; d + 1) 2 S n;m ; j 0  x  n ? 1g n;m n and L = f(x; m ? d ? 2) 2 S ; j 0  x  n ? 1g, the label of S d ? ;dd? ;m? d? in the codeword is uniquely determined. Proof. In view of Lemma 4, if the label of L is given then the label of the set S = f(x; y) 2 S dn? ;dd? ;m? d? j d + 1  x + y  n + dg is uniquely determined. Furthermore, if the label of the set L n S is given and is consistent with the label of S under the (d; d + 1)-constraint, then it additionally determines the label of the set S = f(x; y) 2 S dn? ;dd? ;m? d? j n + d  x + y  m + n ? d ? 3g: This completes the proof. 2 Figure 9 shows an example of Corollary 7 for n = 23; m = 19; d = 3. Corollary 8 Let d n 2m and n  m  3d + 3. Any (d; d + 1)-constrained binary code on d n m? d? distinct codewords in which every 10d 1 pattern S n;m ; has at most 2 d in S dn? ;dd? ;m? d? has the identity (anti-identity) matrix as its adjacent matrices. Proof. Immediately follows from Corollary 7, since we have (n + m)=(d + 1) choices for the labels of the horizontal line segments L and L , and also in S n;m ; , there are at most 2(d + 1)(n + m ? 2d ? 2) grid points in the rectangular annulus of width d, as in the proof of Corollary 5. 2 (

)

(0 0)

(

)

(0 0)

(

2

( +1

2

2

2)

+1)

(

1

(

2

)

(0 0)

)

(

(0 0)

2

( +1

2

2

2)

+1)

1

(

2

2

( +1

2

2)

+1)

2

(

2

2

( +1

(

+ +2( +1)( + +1

)

(0 0)

(

2

2

( +1

2

2

2)

+1)

2

2)

2)

+1)

(

1

2

)

(0 0)

Lemma 5 Let d  2. Any (d; d + 1)-constrained binary code on Z has at most (d + 2)! 2

codewords that do not contain the pattern 10d 1.

Proof. Any codeword in a (d; d + 1)-constrained binary code that does not contain the pattern 10d1 on Z is uniquely determined by the label S^ d; of a (d + 2)  (d + 2) square, which must be a permutation matrix. There are (d + 2)! such matrices. 2 ( +2)

2

(0 0)

The following corollary immediately follows from the proof of Lemma 5 for (d; d + 1)constrained binary codes on S n;m ; , since 0 or 1 can be chosen arbitrarily for at most 2(d + 1)(n + m ? 2d ? 2) grids in S n;m ; . (

)

(0 0)

(

)

(0 0)

13

Corollary 9 Let d  2 and n  m  3d + 3. Any (d; d + 1)-constrained binary code

d on S n;m ; has at most (d + 2)!2 pattern 10d1 in S dn? ;dd? ;m? d? . (

)

n m? d?

2( +1)(

(0 0)

(

2

( +1

2

2

+

2

2)

distinct codewords that do not contain the

2)

+1)

Proof of Theorem 1. For suciently large n and m, Remark 1 holds for any (d; d + 1)-constrained binary code on the bounded rectangle S n;m ; . Therefore, combining Corollaries 5, 8, and 9, any(d; d + 1)-constrained binary code on S n;m ; has at n m d n m ? d ? 2 distinct codewords; this number is most (d + 2)! + (d + 2)! + 2 d smaller than 2 d n m for suciently large n; m. Hence we have d;d log Nm;n 3(d + 1)(n + m) = 0:  m;nlim Cd;d = m;nlim !1 !1 (

)

(0 0)

(

+ +1 +1

3( +1)(

+

)

(0 0)

2( +1)(

+

2

2)

)

(

2

+1)

mn

+1

mn

2

2.2 Proof of Theorem 2

Lemma 60 Let d and k0 be0 nonnegative integers such that k0 ? d is positive and even. k + d + 2 , b = k ? d + 2 , r = a (mod b), and let V 0 be the set of all block Let a = d;k 2 2 diagonal a  a matrices of the form 2 6 6 6 6 6 6 6 6 6 4

A

1

3

A

2

...

7 7 7 7 7 7 7 7 7 5

Aba=bc

;

A0 where each Ai is a b  b permutation matrix, A0 is an r  r permutation matrix, and all a 2V 0 the elements not speci ed are 0's. If a mapping f : Z ! f0; 1g satis es S^ x;y d;k whenever x  y  0 (mod a), then f is a (d; k0)-constrained codeword. Proof. By symmetry, it suces to show that if S^ a; 2 Vd;k0 then S^ a;a can be any element of Vd;k0 without violating the (d; k0)-constraint. If S^ a; ; S^ a;a 2 Vd;k0 , then for 1  i  l the label of the rectangle S ;aa;b?bi (i.e., rows b(i ? 1) + 1 through bi of S^ ;a;a ), is of the form ( )

2

(

(2

( )

(

( )

( )

(0 0)

(

0)

(2

)

(0

)

)

=4

)

(0 0)

2

a;b ;a?bi

(2

0)

)

(0

S^

( )

(0 0)

)

3

Ob i? (

1)

P

Od

P0

Od?b i? (

1)

5

;

where Oj is the b  j rectangular matrix, all of whose elements are 0's, and P and P 0 are b  b permutation matrices. Therefore the number of consecutive 0's between any pair of 14

two horizontally consecutive 1's in S ;aa;b?ib is at least d and at most (b?1)+d+(b?1) = k0. Similarly, the label of the rectangle S ;a;r is of the form (2

)

(0

)

(2

)

(0 0)

S^

3

2

a;r ;

(2

)

(0 0)

=4

Obb ab c

P

P0

Oa?r

5

;

where P and P 0 are r  r permutation matrices. Therefore, the number of consecutive 0's between any pair of two horizontally consecutive 1's in S ;a;r is at least a ? r  a ? (b ? 1) = d + 1 and at most (r ? 1) + (a ? r) + (r ? 1) = a + r ? 2  a + b ? 3 = k0 ? 1. This completes the proof. 2 (2

)

(0 0)

Proof of Theorem 2. Let F be a (d; k0 )-constrained binary code on Z consisting of all mappings f given in the statement of Lemma 6. Given integers n and m, we de ne a n;m n;m (d; k0)-constrained binary code F n;m ; on S ; as the restriction of F to S ; ; i.e., each n;m codeword in F n;m ; is the restriction of a codeword in F to S ; . Then, since a ? b = d, we have 2

(

(

)

(0 0)

(

)

(

(0 0)

)

(0 0)

)

(

(0 0)

)

(0 0)

bc log jF n;m ; j log jVd;k0 j log (b!) b (r!) = = = Cd;k0  m;nlim !1 mn a a (

2

a



)

(0 0)



j

2

2

2

b d b +

2

k

log (b!) + log (r!) : (b + d) 2

2

2

Since Cd;k  Cd;k0 for all k0  k, we have j

Cd;k  =

max d k0 k; k0 ?d

b d b +

k

2

b

2

j

maxk?d 1+

(14)

2

2

+2

even

log (b!) + log (r!) (b + d)

b d b +

k

log (b!) + log (r!) : (b + d) 2

2

2

2

The range of k0 in (14) is restricted to k0  d + 2 instead of k0  d because Cd;d = 0 and k0 ? d must be even. 2 n;m Recall that the coding rate of any (d; k)-constrained binary code F n;m ; on S ; is de ned as log jF n;m ; j mn ; (

(0 0)

(

2

)

(

)

(0 0)

)

(0 0)

n;m where jF n;m ; j is the number of codewords in F ; . We will see in the following sections that the limits of the coding rates of certain sequences of binary codes will also establish the lower bounds in Theorems 3{6, as was done in the proof of Theorem 2. (

)

(

(0 0)

(0 0)

15

)

2.3 Proof of Theorem 3

We construct (0; k)-constrained binary codes on m  n rectangles, whose coding rates approach the lower bound in (4) as n; m ! 1. Figure 10 illustrates this construction for a (0; 5)-constrained binary codeword on S ; ; , i.e., k = 5, m = 9, and n = 12. Intuitively, if k is odd then such a sequence of (0; k)-constrained binary codes is constructed as follows. Let n and m be positive integers that are divisible by k and assume n  m. Let n0 = k n = and m0 = k m = . First, we construct (0; 1)0 0 constrained binary codewords f ; f ; : : : ; f k? = on S n; ;m . Then, for each codeword ft,
{z

X} we replace its bits by 1  k rectangles containing the bit patterns X
{z

X} 0 X | | (12 9)

(0 0)

( +1) 2

( +1) 2

( +1) 2 (

0

1

(

1) 2

)

(0 0)

( +1) 2

t

k?1) ?t

(

and X
{z

X} 1 X
{z

X}, where each X is arbitrarily chosen from f0; 1g. That is, each such | | t

2

k?1) ?t

(

rectangle has (k ? 1)=2 \free bits" and one xed 0 or one xed 1. Finally, we merge these 0 (k + 1)=2 codewords on S n; ;m into a single codeword on the original m  n rectangle k i? + t + 1th row S n;m , by interlacing; i.e., regarding the i th row of f as the t ; n;m of the resulting codeword on S ; , for i 2 f1; 2; : : : ; m0g. (The rows are ordered from bottom to top.) 0 0 Formally, let F n0;m0 be a (0; 1)-constrained binary code on S n; ;m , let H = f f , f ,. . . , f k? = g be a set of k codewords in F n0;m0 , and let gH n;m : S n;m ; ! f0; 1g k = ; be any mapping such that the label of the rectangle S x;y under gH n;m satis es 2

(

)

(0 0)

(

)

( +1)(

(0 0)

1)

2

(

)

(0 0)

(

(

)

)

0

(0 0)

+1

1

(

1) 2

(

)

(

(

=;

(( +1) 2 1)

(

)

(

)

)

x ; y?t =0 if ft (k + 1)=2 (k + 1)=2

8 > > > > > > >
> > > > > :

X t1X k? ?t if ft (k +x1)=2 ; (k y+?1)t=2 = 1

=>

)

(0 0)

2

(( +1) 2 1)

S^ x;yk

(

)

!

(15)

!

2

1

for (x; y)  (0; t) (mod k ), for each t = 0; 1; : : : ; (k ? 1)=2, and where each X is an arbitrary choice of either 0 or 1. The resulting codeword gH n;m is shown to satisfy the (0; k)-constraint on S n;m ; as follows. It is straightforward from the de nition that gH n;m satis es the (0; k)-constraint horizontally; the number of consecutive 0's between any pair of two horizontally consec   utive 1's is at most (k ? 1)=2 ? t + (k + 1)=2 + t = k since each ft satis es the (0; 1)-constraint horizontally (this is achieved if 101 maps to 0t10 k? ?t 0 k 0t 10 k? ?t). A similar argument shows that gH n;m satis es the (0; k)-constraint vertically. +1 2

(

(

)

(0 0)

(

)

2

(

)

16

1

+1 2

2

1

)

Proof of Theorem 3. First, suppose that k is odd. Let n; m; n0 ; m0 be positive integers such that k n, k m, and n  m, and let n0 = k n = and m0 = k m = . Let GH n;m be the set of all mappings gH n;m de ned above, and let (

+1

+1

2

2

( +1) 2

)

(

G n;m = (

( +1) 2

)

[

)

GH n;m ; (

H F (n0 ;m0 ) jH j k+1 2 :

)

=

n;m a (0; k)-constrained binary code on S n;m ; . We will show that the coding rate of G approaches the lower bound in (4) as n; m ! 1, for a particular choice of a (0; 1)0 0 constrained code F n0;m0 on S n; ;m . Let fF n0;m0 g1 n0 ;m0 be a sequence of (0; 1)-constrained 0 ;m0 n binary codes on S ; such that the coding rate log jF n0;m0 j=n0m0 of F n0;m0 approaches C ; as n0 ; m0 ! 1 (such codes were shown to exist in [3]). k patterns are interlaced, each chosen from among jF n0;m0 j possible patterns. Each resulting interlaced m  n rectangular pattern has kk? mn free bits. Thus, the total number of valid m  n patterns created with this construction is (

)

(

)

(0 0)

(

(

)

)

(

)

=1

(0 0)

(

)

(

)

(

)

2

(0 0)

+1

0 1

2

(

)

1

+1

jG n;m j = jF n0;m0 j k
2 kk? mn; (

)

(

)

1 +1

+1 2

and hence

n0 ;m0 j m0 n0 n;m j k?1 k + 1 log jF log jG + = lim
lim 0 0 m;n !1 m;n !1 mn 2 mn mn k+1 1: = k +2 1 C ; + kk ? +1 Together with Fact 1, this completes the proof of Theorem 3 for odd k. For even k, Theorem 3 immediately follows from the monotonicity in (1), and Theorem 3 for k ? 1 odd. 2 (

(

)

)

!

2

2

0 1

2.4 Proof of Theorem 4 Theorem 4 follows directly from Theorem 3 and (16) in Lemma 7 below (using k0 odd).

Lemma 7 Let d, k, d0, k0 be nonnegative integers such that d+1 = k+1 d0 + 1 k0 + 1 is an integer. Then,

0 Cd;k  dd ++ 11 Cd0 ;k0 :

17

In particular, when d0 = 0 this implies

Cd;k  d +1 1 C ;k0 ;

(16)

0

where

+ 1 ? 1: k0 = kd + 1

Proof of Lemma 7. Let

s = dd0 ++ 11 = kk0 ++ 11 : To establish Lemma 7, we construct (d; k)-constrained binary codes on m  n rectangles, whose coding rates approach 1 Cd0;k0 as n; m ! 1. Figure 11 illustrates this construction s for a (1; 7)-constrained binary codeword on S ; ; from two (0; 3)-constrained binary codeword on S ;; , i.e., d = 1, k = 1, d0 = 0, k0 = 3, s = 2, m = 10, and n = 12. Let n and m be positive integers that are divisible by s, assume n  m, and let n0 = n=s and m0 = m=s. A (d; k)-constrained binary codeword on S n;m ; is constructed 0 ;m0 n 0 0 from a set of s (d ; k )-constrained binary codewords on S ; as follows. Let F n0;m0 0 0 be a (d0; k0)-constrained binary code on S n; ;m , and let H = ff ; f ; : : : ; fs? g be a set of s codewords in F n0 ;m0 . For each codeword ft, we replace every occurrence of 0 and 1 by a 1  s rectangle containing the bit patterns 0s and 0t10s? ?t, respectively. Then, we merge these s codewords into a new large codeword, on the original m  n rectangle S n;m ; , by regarding the ith row of ft as the (s(i ? 1) + t + 1)th row of the resulting codeword for i 2 f1; 2; : : : ; m0g (the rows are ordered from bottom to top). Formally, s; n;m let gH n;m : S n;m ; ! f0; 1g be any mapping such that the label of S x;y under gH satis es t =0 0s if ft xs ; y ? s s; = S^ x;y t =1 0t10s? ?t if ft xs ; y ? s for (x; y)  (0; t) (mod s), for each t = 0; 1; : : : ; s ? 1. The resulting codeword gH n;m is shown to satisfy the (d; k)-constraint on S n;m ; 0 0 as follows. Since each ft satis es the (d ; k )-constraint horizontally by de nition, the number of consecutive 0's between any pair of two horizontally consecutive 1's in gH n;m is at least s ? 1 ? t + d0s + t = d and at most s ? 1 ? t + k0s + t = k; the former is achieved since the pattern 10d0 1 maps to 0t 10s? ?t 0s


0s 0t 10s? ?t, and the latter is achieved (12 10)

(0 0)

(6 5)

(0 0)

(

)

(0 0)

(

)

(

)

(0 0)

(

)

0

(0 0)

(

1

1

)

1

(

)

(0 0)

(

)

(

)

(0 0)

(

1)

(

)

8 > > > > > < > > > > > :

(









(

1)

(

)

(

)

(

)

1

)

(0 0)

(

1

1

|

18

{z

d0

}

)

since the pattern 10k0 1 maps to 0t10s? ?t |0s
{z

0}s 0t10s? ?t. A similar argument shows 1

1

k0 n;m that gH satis es the (d; k)-constraint vertically. n;m Let GH be the set of all mappings gH n;m , and (

)

(

)

(

G n;m = (

)

[

)

let

GH n;m ; (

H F (n0 ;m0 ) jH j s :

)

=

n;m a (d; k)-constrained binary code on S n;m ; . We show that the coding rate of G approaches the lower bound in Lemma 7 as n; m ! 1, for a particular choice of a 0 0 (d0; k0)-constrained binary code F n0 ;m0 on S n; ;m . Let fF n0;m0 g1 n0 ;m0 be a sequence of 0 ;m0 n 0 0 (d ; k )-constrained binary codes on S ; such that the coding rate log jF n0;m0 j=n0m0 of F n0;m0 approaches Cd0 ;k0 as n0; m0 ! 1. Note that such a sequence exists because of the de nition of the capacity Cd0 ;k0 . Then, we have (

)

(

)

(0 0)

(

(

)

)

(

)

=1

(0 0)

(

)

(

(

)

2

(0 0)

)

n0 ;m0 js n;m j log jF log jG = m;nlim lim !1 m;n !1 mn mn 0 0 n ;m j m0 n0 log jF
mn = s m;nlim !1 m0 n0 = 1s Cd0 ;k0 : (

(

)

)

2

2

(

)

2

2

2.5 Proof of Theorem 5

The following Corollary is Lemma 7 with k = k0 = 1 (the proof is unchanged).

Corollary 10 Let d and d0 be nonnegative integers such that (d0 + 1) (d + 1). Then, 0 Cd;1  dd ++ 11 Cd0 ;1:

Setting d0 = 1, Corollary 10 implies Theorem 5 for odd d. For even d, Theorem 5 immediately follows from the monotonicity Cd;1  Cd ;1, and Theorem 5 for d +1 odd. Also, when d0 = 0, Corollary 10 implies that (17) Cd;1  d +1 1 for all positive integers d, since C ;1 = 1 is the unconstrained capacity. It is easy to check that the lower bound in (7) in Theorem 5 is tighter than that in (17), unless d = 2 or d = 4. +1

0

19

Corollary 10 ensures that a (d; 1)-constrained binary code can be constructed from a (1; 1)-constrained binary code in the manner of the previous section for odd d, by setting s = d and k = k0 = 1. We next demonstrate this construction. Let n and m be positive integers that are divisible by d , assume n  m, and let n0 = n;m m n 0 d = and m = d = . A (d; 1)-constrained binary codeword on S ; is constructed 0 0 from a set of d (1; 1)-constrained binary codewords on S n; ;m as follows. Let F n0;m0 0 0 be a (1; 1)-constrained binary code on S n; ;m , and let H = ff ; f ; : : : ; f d? = g be a set of d codewords in F n0;m0 . For each codeword ft , we replace every occurrence of 0 and 1 by a 1  (d + 1)=2 rectangle containing the bit patterns 0 d and 0t10 d? ?t, respectively. Then, we merge these (d + 1)=2 codewords into a new large codeword, on the original d i? + t + 1)th row of m  n rectangle S n;m ; , by regarding the ith row of ft as the ( the resulting codeword for i 2 f1; 2; : : : ; m0g (the rows are ordered from bottom to top). Formally, a (d; 1)-constrained binary codeword gH n;m : S n;m ; ! f0; 1g is de ned as d = ; any mapping such that the label of S x;y under gH n;m satis es +1 2

+1 2

(

( +1) 2

+1

(

2

(0 0)

(

)

(

(

0

1

(

1) 2

)

2

+1 2

(

)

)

(0 0)

+1

)

(0 0)

( +1) 2

2

)

( +1)(

(0 0)

1

1)

2

(

(

)

)

(0 0)

(( +1) 2 1)

(

S^ x;yd

=;

(( +1) 2 1)

(

)

)

if ft (d +x1)=2 ; (dy+?1)t=2 = 0

8 > > > > > > >
> > > > > :

0t10 d?2 1 ?t

=>

(

)

!

d+1 2

if ft (d +x1)=2 ; (dy+?1)t=2 = 1 !

for (x; y)  (0; t) (mod d ), and for each t = 0; 1; : : : ; (d ? 1)=2 (for odd d). Figure 12 illustrates this construction for a (5; 1)-constrained binary codeword on S ; ; , i.e., d = 5, m = 9, and n = 12. +1 2

(12 9)

(0 0)

2.6 Proof of Theorem 6

Let s be an integer such that 1  s  d, let r = d (mod s), and let Vd;s be the set of all block diagonal (s + d)  (s + d) binary matrices of the form 2 6 6 6 6 6 6 6 6 6 4

A

3 1

A

2

...

A

bd=sc

1+

A0

7 7 7 7 7 7 7 7 7 5

;

where each Ai is an s  s matrix and A0 is an r  r matrix, and each row and each column of Ai and A0 has at most one 1, and all the non-block-diagonal elements are 0's. If a 20

s d 2 V whenever x  y  0 (mod s + d), then f mapping f : Z ! f0; 1g satis es S^ x;y d;s is a (d; 1)-constrained codeword, since the number of 0's between consecutive 1's is at least d. Let F be the set of all such mappings f . Given integers n and m, we de ne a n;m is the (d; 1)-constrained binary code F n;m on S n;m ; such that each codeword in F restriction of a codeword in F to S n;m ; . Then, we have log jF n;m j = log jVd;sj : (18) Cd;1  m;nlim !1 mn (s + d) Since each Ai and A0 has at most one 1 in each row and in each column, the number of choices for each Ai is (enumerating over nonzero rows and columns) s s! 1 s s! X X i i! = s! i i i! ; i and the number of choices for A0 is r r! 1 r r! X X i i! = r! i i i! : i Hence r r! 1 ! s s! 1 ! b ds c X X r! jVd;sj = s! i i! i i! : ( + )

2

(

(

)

(

)

)

(

)

(0 0)

(

)

(0 0)

(

)

2

2

2

2

=0

=0

2

=0

=0

1+

i

i

=0

=0

The maximization in (8) follows since (18) holds for every s  d.

2

2.7 Proof of Corollary 2

For d  13, Corollary 2 is veri ed by direct calculation. Note that the right hand side of (8) is greater than or equal to log d! 2

d i

P

 

d i i

=0

!

1 !

2d (by setting s = d in the maximization of (8)), and Stirling's inequality [6] implies that 2

log d! 2

But for all d  14,

 

d i

P

d i i

=0

2d

1 !

2

 log2d(d!) > 2

log

2



p

!

 2

2d 2d

  

d d e

2

2

(19)



log de 2C ;1 > C ;1 C ;1 :   2d d 1 + ((d ? 1)=2) 1 + bd=2c 2

 

log de > 2d :

1

1

21

1

2

2.8 Proof of Theorem 7

Let n and m be divisible by d and n  m, and let F n;m be any (0; k)-constrained n;m k binary code on the n  m rectangle S n;m ; . We break S ; into squares S x;y of size (k + 1)  (k + 1), where x  y  0 (mod k + 1). Since every codeword in F n;m must k , we have have at least one 1 in each row and in each column of each S^ x;y (

(

)

)

(

(0 0)

)

( +1)

(0 0)

(

)

(

)

( +1)

(

(

C ;k

)

2

2

?

kX

k + 1 (2k ? 1)k i (k + 1) !

+1

i

=1

2

0



log

n;m j log jF   m;nlim !1 mn

2k

( +1)

)

+1

?i

!

+1

;

2



where k i (2k ? 1)k ?i indicates the number of (k + 1)  (k + 1) binary squares consisting of i zero rows and k ? i nonzero rows. Eq. (9) follows since +1

+1

kX

+1

k + 1 (2k ? 1)k i !

+1

+1

i

=1

?i

+1

=



=

2k

2k

( +1)

kX

k + 1 (2k ? 1)?i ?1 i i 1 ? (1 ? 2? k )k : k



+1

2

+1

!

+1

+1

=1



( +1)

+1



2

2.9 Proof of Theorem 8

Let n and m be divisible by d and n  m, and let F n;m be any (d; 1)-constrained n;m d binary code on the n  m rectangle S n;m ; . We break S ; into squares S x;y of size d  d, where x  y  0 (mod d). Since every codeword in F n;m can have at most one d , we have 1 in each row and in each column of each S^ x;y (

(

)

)

(

(0 0)

)

( )

(0 0)

(

(

)

)

( )

(

log jF n;m j  Cd;1  m;nlim !1 mn Since Pdi

 

=0

Cd;1 

log



d i i

1 !

p 2

d i

 

P

=0

d i

d d 2d e e = d !

d

1 (12 )

2

log

d

d i! i !2

X

2

!

log d!

d

X

2

d 1 i i!

i = d d = 2d , Stirling's upper bound [6] implies that (

2

)

)


2d

i

=0

=0 2

2

!

!

:

! ! ! p 2 d 1 1 = d log 2d + d log e + 12d log e : 2

2

2

2

2

22

2.10 Proof of Corollary 3 The right hand inequality in (10) is straightforward from Theorem 3. The left hand inequality in (10) is derived from (9) as follows:   1 ? C ;k  k?+11 log 1 ? 2? k > k +1 1 (log e)2? k (as k ! 1) = (k +log1)2ek ; xn 2 where we used ? ln(1 ? x) = P1 n n > x, for all x. ( +1)

0

2

( +1)

2

2

+1

=1

2.11 Proof of Corollary 4 The lower bound in (8) is greater than or equal to s s! 1 ! d log s! X i i! s i : (20) (s + d) s s! 1 X Since i i!  1, the quantity in (20) is larger than i d log p2s  s s d log (s!) > s(s + d) s(s + d) e > (s +d d) log es for all d (using Stirling's lower bound [6]). Substituting s = "d (for any " 2 (0; 1)), d log s = 1 log " + log d : (s + d) e (1 + ") d e Thus for any " 2 (0; 1), it follows that ! 1 ; d
C  lim d; 1 d!1 log d (1 + ") and therefore ! d
Cd;1  1: lim d!1 log d But Theorem 8 immediately implies that ! d
Cd;1  1: lim d!1 log d 2

=0

2

=0

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

23

Appendix A Existence of the Two-Dimensional Capacity Theorem 9 The two-dimensional (d; k)-capacity Cd;k exists. Theorem 9 is stated in [8, 12] without a complete proof. It is a special case of a complicated proof in the preprint [7]. We give here a concise proof using an extension of [8, Lemma 4.1.7] to double sequences. Note that both the proof in [7] and our proof below essentially depend only on the two-dimensional sub-additivity.

Lemma 8 Let fam;ng1m;n be a double sequence of nonnegative reals such that =1

am m ;n  am ;n + am ;n am;n n  am;n + am;n : 1+

2

1

1+ 2

(21) (22)

2

1

2

Then limm;n!1 am;n =(mn) exists and equals inf m;n am;n =(mn). 1

Proof.

It follows from (21) and (22) that

am

m2 ;n1 n2

1+

+

 am ;n + am ;n + am ;n + am ;n : 1

1

2

1

1

2

2

2

(23)

Also, by induction we get apm;n  pam;n and am;pn  pam;n for all m; n, and p. Let  = inf m;n am;n=(mn). By de nition, am;n=(mn)   for every m; n  1. Fix an arbitrarily small " > 0. It suces to show that am;n=(mn) <  + " for all m; n suciently large. Since  is the largest number less than or equal to all of the am;n=(mn), there exist  and  such that a; =( ) <  + " . Let p and q be positive integers such that a ; =p < " , a ; =q < " , and let i and j be integers satisfying 0  i <  and 0  j <  . Then ap;q ap;j ap i;q j  + (p + i)(q + j ) (p + i)(q + j ) (p + i)(q + j ) ai;j i;q + (p + ai)( + q + j ) (p + i)(q + j ) a a ai;q + ai;j p;q p;j  pq + pq + pq pq pja iqa pqa ;  pq; + pq; + pq; + ija pq   " <  + 4 + aq; + ap; + apq; <  + ": (24) 1

4

1 1

4

1 1

4

+

+

1

1 1

24

1

1 1

1 1

1 1

For any m  p and any n  q we can always write m = p0 + i and n = q0 + j , where p0  p, q0  q, 0  i <  and 0  j <  , and thus the inequalities above show that am;n=(mn) <  + " for all m  p and n  q . 2 Proof of Theorem 9. Let

d;k : am;n = log Nm;n (

)

2

Then, we have

am m ;n  am ;n + am ;n; since Nmd;k m ;n  Nmd;k;nNmd;k;n, because the right-hand side is the number of patterns on an (m + m )  n rectangle created by concatenating valid patterns on an m  n rectangle and valid patterns on an m  n rectangle. Similarly, we have 1+

(

)

1+

1

(

2

)

1

(

2

1

2

)

2

2

1

2

am;n

n

1+ 2

 am;n + am;n : 1

2

Hence, Lemma 8 can be applied to this double sequence fam;ng1 m;n . =1

2

Acknowledgment This paper was motived in large part by Paul Siegel's presentation of [17] and by interesting discussions with Ron Roth at the 1997 Information Theory Workshop in Svalbard, Norway. The authors would like to thank Dan Goldstein for a helpful suggestion in Theorem 3. Also, thanks to Brian Marcus and Bill Weeks for providing preprints [2] and [23], respectively, and thanks to Tuvi Etzion and Ron Roth for providing preprint [21].

References [1] J. J. Ashley and P. H. Siegel, \A Note on the Shannon Capacity of Run-LengthLimited Codes," IEEE Trans. Inform. Theory, vol. 33, no. 4, July 1987, pp. 601{605. [2] J. J. Ashley and B. H. Marcus, \Two-Dimensional Lowpass Filtering Codes," IBM Research Division, Almaden Research Center, IBM Research Report RJ 10045 (90541), Oct. 1996. [3] N. J. Calkin and H. S. Wilf, \The Number of Independent Sets in a Grid Graph," SIAM J. on Discrete Mathematics, vol. 11, February 1998, pp. 54-60. 25

[4] W. H. Erxleben and M. W. Marcellin, \Error-Correcting Two-Dimensional Modulation Codes," IEEE Trans. Inform. Theory, vol. 41, no. 4, July 1995, pp. 1116{1126. [5] T. Etzion, \Cascading Methods for Runlength-Limited Arrays," IEEE Trans. Inform. Theory, vol. 43, no. 1, January 1997, pp. 319{324. [6] W. Feller, An Introduction to Probability Theory and its Applications, vol. 1, John Wiley and Sons, 3rd Ed. New York, 1968, p. 54. [7] J. Justesen and Y. M. Shtarkov, \The Combinatorial Entropy of Images," preprint. [8] D. Lind and B. H. Marcus, An Introduction to Symbolic Dynamics and Coding, Cambridge Univ. Press, New York, 1995. [9] M. W. Marcellin and H. J. Weber, \Two-Dimensional Modulation Codes," IEEE Journal on Selected Areas in Communications, vol. 10, no. 1, January 1992, pp. 254{ 266. [10] B. H. Marcus, P. H. Siegel, and J. K. Wolf, \Finite-State Modulation Codes for Data Storage," IEEE Journal on Selected Areas in Communications, vol. 10, no. 1, January 1992, pp. 5{37. [11] N. Markley and M. Paul, Matrix Subshifts for Z  Symbolic Dynamics, Proceedings of the London Mathematical Society, vol. 43, no.3, 1981, pp. 251{272. [12] N. Markley and M. Paul, Maximal Measures and Entropy for Z  Subshifts of Finite Type, Classical Mechanics and Dynamical Systems, Dekker Notes in Mathematics, ed. R. Devaney and Z. Nitecki, 70 (1981), pp. 135{157. [13] S. W. McLaughlin, \The Construction of M -ary (d; 1) Codes that Achieve Capacity and Have the Fewest Number of Encoder States," IEEE Trans. Inform. Theory, vol. 43, no. 2, March 1997, pp. 699{703. [14] S. W. McLaughlin, Jian Luo, and Qun Xie, \On the Capacity of M -ary RunlengthLimited Codes," IEEE Trans. Inform. Theory, vol. 41, no. 5, September 1995, pp. 1508{1511. [15] E. K. Orcutt and M. W. Marcellin, \Redundant Multitrack (d; k) Codes," IEEE Trans. Inform. Theory, vol. 39, no. 5, September 1993, pp. 1744{1750.

26

[16] P. N. Perry, \Runlength-Limited Codes for Single Error Detection in the Magnetic Recording Channel," IEEE Trans. Inform. Theory, vol. 41, no. 3, May 1995, pp. 809{815. [17] P. H. Siegel, \Problems in Coding for Multi-Dimensional Storage Devices," IEEE Information Theory Workshop, Longyearbyen, Norway, June, 1997. [18] P. H. Siegel and J. K. Wolf, \Bit Stung Bounds On the Capacity of 2-Dimensional Constrained Arrays," Proceedings of 1998 IEEE International Symposium on Information Theory, Boston, MA, August, 1998, p. 323. [19] R. Talyansky, Coding for Two-Dimensional Constraints, M.Sc. thesis, Computer Science Department, Technion, Haifa, Israel, 1997 (in Hebrew). [20] R. Talyansky, T. Etzion, and R. M. Roth, \Ecient Code Constructions for Certain Two-Dimensional Constraints," Proceedings of 1997 IEEE International Symposium on Information Theory, Ulm, Germany, June, 1997, p. 387. [21] R. Talyansky, T. Etzion, and R. M. Roth, \Ecient Code Constructions for Certain Two-Dimensional Constraints," preprint. [22] B. V. Vasic, \Capacity of Channels with Redundant Multitrack (d; k) constraints: The k < d Case", IEEE Trans. Inform. Theory, vol. 42, no. 5, September 1996, pp. 1546{1548. [23] W. Weeks IV and R. E. Blahut, \The Capacity and Coding Gain of Certain Checkerboard Codes," IEEE Trans. Inform. Theory, vol. 44, no. 3, May 1998, pp. 1193{1203.

27

List of Figures Example of 3  3 adjacent matrices of (3; 5)-constrained code on S ;; . 0 and 1 are considered to be placed on grids. . . . . . . . . . . . . . . . . . 2 Example of Lemma 2 (d = 7): a (7; 8)-constrained binary codeword whose 7  7 adjacent matrices of any 10 1 pattern are neither the identity nor the anti-identity matrix. . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Illustration of the proof of Lemma 2 for the case d = 7; r = 4. . . . . . . 4 Example of (3; 4)-constrained binary code on S ;; that cannot be extended to Z . If one more row is appended below the rectangle, it must be a binary pattern of the form 100X 00, but both X = 0 and X = 1 violate the horizontal (3; 4)-constraint. . . . . . . . . . . . . . . . . . . . 5 Example of Corollary 5 (n = 23; m = 19; d = 3). . . . . . . . . . . . . . . 6 Example of Lemma 3 (d = 7). . . . . . . . . . . . . . . . . . . . . . . . . 7 Illustration of the proof of Lemma 3 for the case d = 7. . . . . . . . . . . 8 Illustration of the proof of Lemma 4 for the case d = 5; a = ?1. . . . . . 9 Example of Corollary 7 (n = 23; m = 19; d = 3). . . . . . . . . . . . . . . 10 A (0; 5)-constrained binary codeword gH ; on S ; ; constructed from three (0; 1)-constrained binary codewords f , f , and f on S ;; . See Theorem 3. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11 A (1; 7)-constrained binary codeword gH ; on S ; ; constructed from two (0; 3)-constrained binary codewords on S ;; . See Theorem 4 and Lemma 7. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12 A (5; 1)-constrained binary codeword gH ; on S ; ; constructed from three (1; 1)-constrained binary codewords f ; f ; f on S ;; . See Theorem 5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

1

(7 11)

(0 0)

ii

7

iii iv

(6 4)

(0 0)

2

v vi vii vii viii ix

(12 9)

(12 9)

(0 0)

(4 3)

0

1

2

(0 0)

x

(12 10)

(12 10)

(0 0)

(6 5)

(0 0)

xi

(12 9)

(12 9)

(0 0)

(4 3)

0

i

1

2

(0 0)

xii

y

0

1

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

1

0

0

0

0

0

1

0

0

1

0

0

0

0

0

1

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

1

0

0

0

0

0

1

0

0

0

0

0

1

0

0

0

1

0

0

0

0

Figure 1: Example of 3  3 adjacent matrices of (3; 5)-constrained code on S 1 are considered to be placed on grids.

ii

x

; ;

(7 11) (0 0)

. 0 and

y

x

Figure 2: Example of Lemma 2 (d = 7): a (7; 8)-constrained binary codeword whose 7  7 adjacent matrices of any 10 1 pattern are neither the identity nor the anti-identity matrix. 7

iii

f(0,d+1)=1 f(-1,d+1)=0

f( σ -1 (d), d)=1

y(0,8)

f(d+1,d+1)=1

(8,8)

y=d

f(d+2,d)=1 f(d+r,d-r+2)=1

f(r-1,d-r+2)=1

r=4 (12,2) f(d+1+r, σ (r))=1

f(-1, σ (d))=1

x

o

f(d+2,-1)=1

f( σ -1 (d), -1)=1

f(d+r,-r+1)=1

y=-d x=1

x=d

: the original 0’s : the new 0 by f(-1, σ (d))=1 : the new 0’s by Eq. (11) : the new 0’s by Eqs. (12) and (13) : f(x,y)=1

Figure 3: Illustration of the proof of Lemma 2 for the case d = 7; r = 4.

iv

y

0

0

0

1

0

0

0

0

0

0

1

0

0

0

1

0

0

0

0

1

0

0

0

1

x

Figure 4: Example of (3; 4)-constrained binary code on S ;; that cannot be extended to Z . If one more row is appended below the rectangle, it must be a binary pattern of the form 100X 00, but both X = 0 and X = 1 violate the horizontal (3; 4)-constraint. (6 4)

(0 0)

2

v

y

(n-d-2,m-d-2) (n-1,m-1)

(d+1,m-d-2)

(0,m-1)

x o

(n-d-2,d+1)

(d+1,d+1)

(n-1,0)

:the labels of the grid points bordering these squares might not be determined

Figure 5: Example of Corollary 5 (n = 23; m = 19; d = 3).

vi

y

o

y

(8,0)

x

o (8,0)

Figure 6: Example of Lemma 3 (d = 7).

y (0,8)

(-1,1)

x

o : the original 0’s : the new 0 by f(-1,1)=1

Figure 7: Illustration of the proof of Lemma 3 for the case d = 7. vii

x

y

x

o

: the original 0’s by f(x,-x-1)=1 : the new 0’s

Figure 8: Illustration of the proof of Lemma 4 for the case d = 5; a = ?1.

viii

y (0,m-1)

(n-d-2,m-d-2) (n-1,m-1)

(d+1, m-d-2)

y=14

y=4

x

o (d+1,d+1)

(n-d-2,d+1)

: the labels of the grid points bordering these squares might not be determined : the labels of the grid points bordering these squares or triangles are determined by the label of y=4 : the labels of the grid points bordering these squares or triangles are determined by the label of y=14

Figure 9: Example of Corollary 7 (n = 23; m = 19; d = 3). ix

(n-1,0)

f0

f1

f2

0 1 1 0 1 0 1 1 0 1 1 0

0 1 1 1 1 1 0 1 1 0 1 0

1 0 1 0 0 1 0 1 1 1 1 0

XX1 XX0 XX1 XX0 X0X X1X X1X X1X 0XX 1XX 1XX 0XX XX0 XX1 XX0 XX1 X1X X1X X0X X1X 1XX 0XX 1XX 1XX XX1 XX1 XX1 XX0 X1X X0X X1X X0X 0XX 1XX 1XX 0XX y

y X X 0 X X 1 X X 0

X 0 X X 1 X X 1 X

1 X X 0 X X 1 X X

X X 1 X X 0 X X 1

X 1 X X 1 X X 0 X

0 X X 1 X X 1 X X

X X 1 X X 1 X X 1

X 1 X X 0 X X 1 X

1 X X 0 X X 1 X X

X X 0 X X 1 X X 0

X 1 X X 1 X X 0 X

0 X X 1 X X 0 X x X

0 1 0 1 0 1 0 0 0

0 0 1 1 1 0 0 1 0

1 1 1 0 0 0 1 1 0

0 1 1 0 1 0 0 1 1

0 1 0 0 1 0 0 0 0

0 1 0 1 0 0 1 1 0

0 0 1 0 0 1 1 1 1

0 1 1 0 0 0 1 1 0

1 0 0 0 0 0 1 0 0

1 0 0 0 1 1 0 0 0

1 1 0 0 1 0 1 0 1

One example of a complete (0,5)-constrained binary code with particular choices of X ’s.

All free bits X can be arbitrarily chosen from {0,1 }.

Figure 10: A (0; 5)-constrained binary codeword gH ; on S ; ; constructed from three (0; 1)-constrained binary codewords f , f , and f on S ;; . See Theorem 3. (12 9)

(12 9)

(0 0)

(4 3)

0

0 1 0 1 0 0 0 1 0

1

2

x

(0 0)

x

f0 1 0 0 0 1

10 00 00 00 10

00 10 00 10 00

0 1 0 1 0

0 1 1 0 1

00 10 10 00 10

f1 0 1 0 0 0

1 0 0 1 1

1 0 0 0 1

00

0 0 0 1 1

10

10

00

00

00

00

10

00

0 1 0 1 0

1 0 1 0 0

0 0 1 0 0

1 1 0 0 1

0 0 1 1 0

00

00

01

00

01

00

00

01

00

00

01

00

00

00

01

01

00

01

01

01

00

00

00

01

01

00

00

00

01

00

10 00 00 00

10

10

0 1 1 0 0 0 0 0 0 0 1

0 0 1 0 1 0 0 0 1 0 0

y 0 0 0 0 0 0 1 0 1 0

0 0 11 0 0 0 1 0 0 0

0 0 0 1 1 0 1 0 0 0 1

1 0 0 1 0 1 0 0 0 0 0

0 0 0 1 0 0 0 0 0 0

0 0 1 0 0 1 0 0 0 0 0

0 1 00 0 0 0 0 1 0 1

1 0 1 0 0 0 0 0 1 0

0 1 0 0 0 0 0 0 0 1

0 0 0 0 1 0 1 0 0 0

x

Figure 11: A (1; 7)-constrained binary codeword gH ; on S ; ; constructed from two (0; 3)-constrained binary codewords on S ;; . See Theorem 4 and Lemma 7. xi (12 10)

(12 10)

(0 0)

(6 5)

(0 0)

f0

f1

1 0 0 1 0 1 0 0 1 0 0 1

f2

1 0 0 0 0 0 1 0 0 1 0 1

0 1 0 1 1 0 1 0 0 0 0 1

00 0 001 00 0 001 0 10 0 00 0 00 0 00 100 000 000 100 00 1 000 00 1 000 0 00 0 00 0 10 0 00 000 100 000 000 00 0 000 000 001 0 00 0 10 0 00 0 10 100 000 000 100 y 0 0 1 0 0 0 0 0 1

0 1 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0

0 0 0 0 0 1 0 0 0

0 0 0 0 0 0 0 1 0

1 0 0 0 0 0 0 0 0

0 0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0

0 0 0 1 0 0 0 0 0

0 0 1 0 0 0 0 0 1

0 0 0 0 0 0 0 1 0

1 0 0 0 0 0 1 0 0

x

Figure 12: A (5; 1)-constrained binary codeword gH ; on S ; ; constructed from three (1; 1)-constrained binary codewords f ; f ; f on S ;; . See Theorem 5. (12 9)

(12 9)

(0 0)

(4 3)

0

xii

1

2

(0 0)