Technion - Computer Science Department - M.Sc. Thesis MSC Constrained Codes for Two-Dimensional Channels.

Transcription

1 Technion - Computer Science Department - M.Sc. Thesis MSC Constraine Coes for Two-Dimensional Channels Keren Censor

2 Technion - Computer Science Department - M.Sc. Thesis MSC

3 Technion - Computer Science Department - M.Sc. Thesis MSC Constraine Coes for Two-Dimensional Channels Research Thesis Submitte in Partial Fulfillment of the Requirements for the Degree of Master of Science in Computer Science Keren Censor Submitte to the Senate of the Technion - Israel Institute of Technology ar 5766 Haifa March 2006

4 Technion - Computer Science Department - M.Sc. Thesis MSC The research thesis was one uner the supervision of Prof. Tuvi Etzion in the Department of Computer Science. I thank Prof. Tuvi Etzion for his evote guiance. I thank little Ofek, Noga, Neta an lmog, for the light they shine. The generous financial help of the Technion is gratefully acknowlege. I am also grateful to the Jeanette an Samuel Lubell Founation.

5 Technion - Computer Science Department - M.Sc. Thesis MSC Contents bstract bbreviations an Notations 3 Introuction 4. Physical Constraints in Digital Storage Systems (,k)-rll constraints Encoing One-Dimensional Constraine Coing Two-Dimensional Constraine Coing Connectivity Moels Previous Work Description of the Work asic Techniques 3 2. Positive Capacity Zero Capacity - The Scanning Metho symmetric Run-length Constraine Channels 9 3. Constructions for Proving Positive Capacity Proving Zero Capacity Summary of Results for the Diamon Moel The Square Moel 3 4. Proving Zero Capacity Summary of Results for the Square Moel

6 Technion - Computer Science Department - M.Sc. Thesis MSC The Triangular Moel Construction for Proving Positive Capacity Proving Zero Capacity The Capacity for Small Values of Summary of Results for the Triangular Moel Discussion an Open Problems The Scanning Metho ouning the Capacity The Connectivity Moels The Diamon Moel The Square Moel The Triangular Moel ibliography 64

7 Technion - Computer Science Department - M.Sc. Thesis MSC List of Figures. n encoer with rate p n graph for the (,k)-rll constraint Neighbors of position (i,j) in the: (a) iamon moel, (b) square moel, (c) hexagonal moel Neighbors of positions (i,j, 0) an (i,j, ) in the triangular moel [7 2, 3 5] skeleton tile Scanning of a (, + ) array Scanning of ρ positions in a row The tree T when no constraints are impose The tree T. Every subtree which oes not inclue vertices on the leftmost path, oes not have vertices that represent positions that are in state (s3) The array T The skeleton tile for the (, 2 +, 2, 2 + ) constraint Two skew tetrominoes for substitution in the skeleton tile Tiling the plane with skeleton tiles reas crossing two tiles for the (, 2 +, 2, 2 + ) constraint Relative locations of T + arrays skeleton array for the (, 2 + 2, 2 +, 2 + 2) constraint The array H 8, The skeleton tile for (, 2 +, + r, + r + ) constraint reas crossing two tiles for the (, 2 +, + r, + r + ) constraint Relative locations of H,r arrays Labels of the array in Proposition

8 Technion - Computer Science Department - M.Sc. Thesis MSC Proving C (, 4) = Proving C (2, 5) = Scanning of a (, + 3) array Case of Theorem Case 2 of Theorem Case 3 of Theorem triangular array Three 2 2 exchangeable triangular arrays The triangular array T The pattern PEven Labels implie by the pattern PEven The possible orientations of a scanne position Case of Lemma Case a of Lemma Case b of Lemma Case c of Lemma Case 2 of Lemma Case 2a of Lemma Case 2b of Lemma Case 2c of Lemma The pattern PO Labels implie by the pattern PO Case of Lemma Case a of Lemma Case b of Lemma Case c of Lemma Case 2 of Lemma Case 2a of Lemma Case 2b of Lemma Case 2c of Lemma Case of Lemma Case 2 of Lemma The array for the proof that C (, 3) > Two triangular tiles to prove that C (, 3) > force run of 4 zeroes in a (2,k) triangular array Two triangular arrays to prove that C (2, 4) > Case of Lemma

9 Technion - Computer Science Department - M.Sc. Thesis MSC Case a of Lemma Case b of Lemma Case c of Lemma Case 2 of Lemma Case 2a of Lemma Case 2b of Lemma Case 2c of Lemma The array for the proof that C (3, 7) > Four triangular tiles to prove that C (3, 7) > Proving C (4, 8) =

10 Technion - Computer Science Department - M.Sc. Thesis MSC bstract In igital ata storage systems, such as magnetic an optical storage evices, the recore ata has to satisfy certain constraints that are impose by the physical structure of the meia. One of the most frequently investigate type of constraints are the (,k) run-length limite (RLL) constraints. binary sequence satisfies a one-imensional (,k) constraint if every run of zeroes has length at least an at most k. Recent evelopments in optical storage, especially in holographic memory, regar the recore ata as two-imensional. one-imensional constraint has to be satisfie in each of the array irections. Similarly to the oneimensional case, the capacity of a two-imensional constraint Θ is efine as: C(Θ) = lim n,m log 2 N(n,m Θ), nm where N(n,m Θ) is the number of arrays of size n m that satisfy Θ. Few connectivity moels have been propose in the literature to hanle twoimensional ata: the iamon moel, the square moel, the hexagonal moel, an the triangular moel. The constraints may be asymmetric, i.e. vary among the ifferent irections. In this work, we erive some new methos for etermining zero an positive capacity. We generalize a technique for proving zero capacity, which is base on scanning a Θ-constraine array whose labels are partially known, an counting the number of possible ways to label the rest of the array. This metho provies an upper boun for the number of constraine arrays of size n m, which is small enough to etermine that C(Θ) = 0. For proving positive capacity of some constraints, we efine shapes which can tile the plane. Given such a shape, we fin two ifferent vali ways to label it. We then show that tiling the plane with copies of the shape, where each copy can have either one of the two labels, results in a Θ-constraine

11 Technion - Computer Science Department - M.Sc. Thesis MSC array. This provies a lower boun for the number of constraine arrays of size n m, which is large enough to etermine that C(Θ) > 0. We apply the above methos to the ifferent connectivity moels in orer tocharacterize their zero/positive capacity regions. We consier asymmetric constraints in the iamon moel, an provie an almost complete characterization of the positive capacity region. In the triangular moel, we show that C(, + 3) = 0 for every 3. For (mo 4), 5, we show a tight characterization: C(,k) > 0 if an only if k + 4. Together with the former result, it implies that for other values of, the gaps between the known zero an positive capacity regions are relatively small. Finally, in the square moel we show that C(,+3) = 0 for every. 2

12 Technion - Computer Science Department - M.Sc. Thesis MSC bbreviations an Notations N(n Θ) N(n,m Θ) C(Θ) (,k)-rll number of Θ-constraine sequences of length n number of Θ-constraine arrays of size n m capacity of the constraint Θ constraint in which the number of zeroes between every pair of consecutive ones is at least an at most k (,k, 2,k 2 )-RLL two-imensional constraint where (,k ) is the horizontal constraint an ( 2,k 2 ) is the vertical constraint C (,k) C (,k) C (,k) C (,k) capacity of the (,k) constraint in the iamon moel capacity of the (,k) constraint in the square moel capacity of the (,k) constraint in the hexagonal moel capacity of the (,k) constraint in the triangular moel [n m,k l] tile n m array from which a k l array was remove from the upper right corner 3

13 Technion - Computer Science Department - M.Sc. Thesis MSC Chapter Introuction Constraint coing is wiely use in igital storage applications, particularly magnetic an optical storage evices [0, ]. In such systems, the physical structure of the storage evice imposes constraints on the recore ata. This chapter introuces the fiel of constraine coing, an escribes our main lines of research.. Physical Constraints in Digital Storage Systems Magnetic storage evices consist of tracks of magnets. When the ata is recore, a bit one is represente by a reversal of the magnetic polarity along the ata track, an no reversal of the polarity represents a zero. While reaing the ata, the hea which reas respons to a polarity change by an inuce voltage. When no change occurs, no voltage is prouce. sufficiently high voltage is consiere as a one, an otherwise the bit is consiere to be a zero. On one han, if successive ones are too close, the voltage levels rea by them might interfere with each other. Hence there is a lower boun on the number of zeroes between successive ones that are allowe in the recore ata. On the other han, the clock of the evice is ajuste when high voltages are rea an a one is etecte. To avoi clock rifting, that might cause erroneous recovery of ata, there is an upper boun on the number of zeroes between successive ones that are allowe in the recore ata. When recoring ata on an optical evice such as a CD, the bit one is 4

14 Technion - Computer Science Department - M.Sc. Thesis MSC representeas a peak on the surface. In orer to rea the ata, alaser beam is projecte. The light is reflectefrom the surface, an whenreaing apeak, a estructive interference occurs. Therefore the etector sees arkness an interprets it as the bit one, an otherwise the bit is a zero. On one han, in orer for the etector not to miss the peak, the peak has to be wie enough, which implies a lower boun on the number of zeroes between successive ones that are allowe in the recore ata. On the other han, reaing peaks allows the etector to ajust the spee of rotation of the CD accoring to the istance of the track from the center. Hence, there is an upper boun on the number of zeroes between successive ones that are allowe in the recore ata..2 (, k)-rll constraints The constraints that are implie from the iscussion above are calle (,k)- RLL constraints. Formally, a binary sequence satisfies a (,k)-rll constraint (or a (,k) constraint), if every run of zeroes between successive ones has length at least an at most k. t the beginning an en of the sequence, the runs are only require to be of length at most k. Example The sequence is (2, 5)-constraine. Inee, there are many stanar storage evices that use (,k)-rll constraints. Example 2 Floppy-isks are (, 7) or (2, 7)-constraine. DVDs are (2, 0)-constraine..3 Encoing The user of a storage system may wish to recor any binary ata on the evice, an therefore it has to be change in orer to comply with the constraints. This is calle encoing. n encoer (see Fig..) is require to transform any binary sequence of length p into a constraine sequence of length n (a coewor). Usually n p, since not all sequences are vali. The 5

15 Technion - Computer Science Department - M.Sc. Thesis MSC encoing proceure has to be reversible in orer to later rea the recore ata correctly, hence a ecoer is require, which converts the constraine sequences of length n back into the original sequences of length p. p Encoer Figure.: n encoer with rate p n. The ratio p is the rate of the encoer. higher rate implies that fewer n bits are written per one input bit, which ecreases the amount of space neee to recor the ata. Given a constraint Θ, we are intereste in fining the maximal rate possible for an encoer. The capacity of a one-imensional constraint Θ is efine as: log C(Θ) = lim 2 N(n Θ), n n where N(n Θ) is the number of coewors of length n that satisfy Θ. Given N(n Θ) output wors of length n, the maximum length p of input sequences can be at most log 2 N(n Θ). Hence, the capacity C(Θ) upper bouns the rate of any encoer for the constraint Θ. Therefore given a constraint Θ, we are intereste in fining the capacity C(Θ)..4 One-Dimensional Constraine Coing Given a one-imensional (,k) constraint we construct the following graph (see Fig..2): The set of noes is {0,,k}. For 0 i k, there is an ege from noe i to noe i +, that is labelle by 0. n 6

16 Technion - Computer Science Department - M.Sc. Thesis MSC k Figure.2: graph for the (,k)-rll constraint. For i k, there is an ege from noe i to noe 0, that is labelle by. The graph escribes the one-imensional (,k) constraint in the following sense: any walk in the graph prouces a (,k) coewor by the sequence of labels on the eges of the walk, an any (,k) coewor has a corresponing walk. The capacity of a (,k) constraint is known to be equal to log 2 λ, where λ is the Perron eigenvalue of the ajacency matrix of the graph..5 Two-Dimensional Constraine Coing Recent evelopments in optical storage especially in the area of holographic memory increase recoring ensityby exploiting the fact that the recoring evice is a surface. In this new moel, the recore ata is regare as two-imensional, as oppose to the track-oriente one-imensional recoring paraigm. This new approach, however, necessitates the introuction of new types of constraints which are two-imensional rather than one-imensional. While the one-imensional case has been wiely explore, results in the twoimensional case have been slower to arrive. This is mainly ue to the fact that imposing constraints in a few irections makes the coing problem much more ifficult. Nevertheless, in the last ecae there has been a consierable progress in the stuy of two-imensional constraints. Similarly to the one-imensional efinition, a two-imensional surface is sai to satisfy a (,k) constraint, if each irection efinebyits connectivity moel satisfies a one-imensional (,k) constraint. The capacity of a twoimensional constraint Θ is efine by: 7

17 Technion - Computer Science Department - M.Sc. Thesis MSC log C(Θ) = lim 2 N(n,m Θ) n,m nm where N(n,m Θ) is the number of n m arrays satisfying the constraint Θ. n array which satisfies the constraint Θ is calle Θ-constraine or a Θ array. In general, the algebraic tools use to compute the capacity of oneimensional constraints, cannot be similarly use in the two-imensional case. This work focuses on the reuce task of characterizing the region of parameters,k for which C(,k) > 0. We escribe the ifferent connectivity moels in the following section..5. Connectivity Moels Data shoul be organize on a two-imensional surface in some orer. This orer will be efine by the way in which the ata is rea. For this purpose four connectivity moels are efine. The iamon moel, the square moel, an the hexagonal moel are frequently consiere in the literature, e.g., for constraine coes they were consiere first by Weeks an lahut [2]. The triangular moel was consiere by [9] for constraine coes an for other applications in [6]. Some other papers which consier capacities of constraints in such moels are [7, 3, 4, 8, 20]. The first connectivity moel is the iamon moel. In this moel, a point (i,j) Z 2 has the following four neighbors: {(i +,j), (i,j), (i,j + ), (i,j )}. When (i,j) is a bounary point, the neighbor set is reuce to points within the array. In this moel the ata is organize in the two-imensional rectangular gri, an it is rea horizontally an vertically. The secon moel is calle the square moel, in which each point (i,j) Z 2 has eight neighbors: {(i +,j), (i,j), (i,j + ), (i,j ), (i +,j + ), (i,j + ), (i +,j ), (i,j )}. In this moel the ata is organize in the two-imensional rectangular gri an it is rea horizontally, vertically, an in the two iagonal irections., 8

18 Technion - Computer Science Department - M.Sc. Thesis MSC The thir moel is calle the hexagonal moel. Instea of the rectangular gri we have use up to now, we efine the following graph. We start by tiling the plane R 2 with regular hexagons. The vertices of the graph are the center points of the hexagons. These points efine the hexagonal lattice [5]. We connect two vertices if an only if their respective hexagons are ajacent. In this way, each vertex has exactly six neighboring vertices. We will use an isomorphic representation of the moel. This representation inclues Z 2 as the set of vertices. Each point (i,j) Z 2 has the following neighboring vertices: {(i +,j), (i,j), (i,j + ), (i,j ), (i,j ), (i +,j + )}. It can be shown that the two moels are isomorphic [2]. From now on, by abuse of notation, we will also call the last moel the hexagonal moel. In this isomorphic moel the ata is organize in the two-imensional rectangular gri an it is rea horizontally, vertically, an in one of the iagonals irection calle right iagonal. The neighbor sets of the three ifferent moels are summarize in Fig..3. square with a ot is the point (i,j). In all moels, rows an columns of the arrays will be inexe in ascening orer, bottom to top an left to right. (i +, j) (i, j ) (i, j + ) (i, j) (a) (b) Figure.3: Neighbors of position (i,j) in the: (a) iamon moel, (b) square moel, (c) hexagonal moel. The fourth moel is calle the triangular moel. gain, we start by tiling the plane R 2 with regular hexagons. The vertices of the graph are now the vertices of the hexagons, rather than their centers. The eges between the vertices are the sies of the hexagons. Hence, each vertex has exactly three (c) 9

19 Technion - Computer Science Department - M.Sc. Thesis MSC neighboring vertices. If we connect the centers of the hexagons with lines we will obtain a tiling of the R 2 with equilateral triangles. The vertices of the graph are the center points of the equilateral triangles. The set of vertices is also a union of two translates of the hexagonal lattice. Clearly, a point in this moel can be represente by a triple (i,j,s) Z 2 {0, }. Each point (i,j, 0) Z 2 {0} has the following neighboring vertices: {(i,j, ), (i,j, ), (i,j, )}. Each point (i,j, ) Z 2 {} has the following neighboring vertices: {(i,j, 0), (i +,j, 0), (i,j +, 0)}. The neighbor sets in this moel are illustrate in Fig..4. (i,j,) (a) (i,j,) (i,j,) (i+,j,0) (i,j,0) (i,j+,0) Figure.4: Neighbors of positions (i,j, 0) an (i,j, ) in the triangular moel. s the vertices are two translates of the hexagonal lattice, one can consier the moel as having six irections. We will consier it slightly ifferently. Instea of ata store in the centers of the triangles, the ata will occupy the whole area of the triangle. Therefore, in this interpretation there are three irections in this moel. Finally, we note that in the triangular moel an n m array has 2nm points. Therefore the efinition of the capacity in this moel is accoringly ajuste to be: (b) log C(Θ) = lim 2 N(n,m Θ) n,m 2nm. 0

20 Technion - Computer Science Department - M.Sc. Thesis MSC Previous Work Let C (,k) enote the capacity of the (,k) two-imensional constraint in the iamon moel. The value of C (, ) has been investigate in many works. Calkin an Wilf [3] showe that C (, ) Weeks an lahut improve these results in [2], showing that C (, ) Then they use a numerical convergence-speeing technique calle Richarson Extrapolation to estimate that C (, ) an that this approximation is correct up to 2 igits. For, Siegel an Wolf [22], an Halevy, Chen, Roth, Siegel an Wolf [9], boune C (, ) by stuying bit-stuffing encoers. Kato an Zeger [3] also showe bouns for these capacities. For k, the value of C (0,k) was investigate by Talyansky [23], an by Kato an Zeger [3]. For other values of the capacity of C (,k) is generally unknown. Kato an Zeger [3] characterize the positive capacity region of (,k) constraints in the iamon moel, by proving that C (,k) > 0 if an only if k + 2. We are intereste in asymmetric constraints in this moel, in which there can be ifferent constraints for rows an for columns. C (,k, 2,k 2 ) enotes the capacity of the asymmetric (,k, 2,k 2 ) constraint in the iamon moel, i.e., horizontally the constraint is (,k ) an vertically the constraint is ( 2,k 2 ). These constraints were hanle in [4]. C (,k), C (,k), an C (,k) enote the capacity of the (,k) constraint in the square moel, hexagonal moel, an triangular moel, respectively. In the hexagonal moel, the exact value of C (, ) was given by axter []. The positive capacity region of hexagonal constraints has been stuie by Kukorelly an Zeger in [6, 5]. Finally, the capacity of the har-triangle constraint (isolate ones) was shown in [9] to be boune by C (, )

21 Technion - Computer Science Department - M.Sc. Thesis MSC Description of the Work The rest of the chapters are organize as follows. In Chapter 2 we present the known basic techniques to prove zero or positive capacity. We generalize these techniques, so that they coul be applie to more complicate cases which we will have in succeeing chapters. In Chapter 3 we examine asymmetric constraints in the iamon moel an provie an almost complete solution for the zero/positive capacity region problem. In Chapters 4, an 5 we examine capacities of constraints in the square moel an the triangular moel, respectively. Discussion an open problems are in Chapter 6. 2

22 Technion - Computer Science Department - M.Sc. Thesis MSC Chapter 2 asic Techniques In this chapter we will survey the known techniques, except for a-hoc methos, use to prove zero capacity, an those use to prove positive capacity. We will generalize these techniques in a way that will enable them to hanle more complicate scenarios. The first lemma which appeare in [4] is an immeiate consequence of the efinition of the (,k) constraint. Lemma Let Θ be a constraint with minimum runlength an maximum runlength k in irection. Let Θ be a constraint with minimum runlength an maximum runlength k k in irection, an the same constraints as in Θ in the other irections. Then C(Θ) C( Θ). 2. Positive Capacity n [n m,k l] skeleton tile is a tile which consists of an n m array from which a k l array was remove from the upper right corner. If l = we simply have an [n m,k] skeleton tile. n example of a [7 2, 3 5] skeleton tile is given in Fig. 2.. Figure 2.: [7 2, 3 5] skeleton tile. 3

23 Technion - Computer Science Department - M.Sc. Thesis MSC For two points z = (x,y ) an z 2 = (x 2,y 2 ), z,z 2 Z 2, let L(z,z 2 ) = {(ix + jx 2,iy + jy 2 ) : i,j Z} be the set of points spanne by z,z 2. This is the lattice efine by z an z 2 (see [5, 8]). Note, that by abuse of notation, the first coorinate is for the row inex an the secon is for the column inex. The following lemma can be easily verifie. Lemma 2 Let be an [n m,k l] skeleton tile. If we place the bottom leftmost point of on the points of L((n k,m l), (n, l)), then a tiling of Z 2 with copies of is obtaine. The tiling obtaine by Lemma 2 will be calle the stanar tiling. If is an n m array (a skeleton array) then the stanar tiling is obtaine by substituting k = 0 an l = 0 in the skeleton tile of lemma 2. Clearly, we can also use a parallelogram instea of a rectangle. stanar tiling can use a few tiles with the same shape an ifferent labels. In this case each one of the tiles can have any one of the labels. The next lemma is a straightforwar generalization of similar lemmas for skeleton arrays, given in [7, 4]. Lemma 3 Let an be two ientical tiles with ifferent labels, an Θ a two-imensional constraint. If any stanar tiling with an yiels a two-imensional array which is Θ-constraine, then C(Θ) > 0. Moreover, if we can use t ientical tiles with ifferent labels,, t, an the number of points in i is N, then C(Θ) N log 2t. 2.2 Zero Capacity - The Scanning Metho The most effective metho to prove zero capacity was given by lackburn [2] for specific constraints. However, this metho can be formulate to hanle general two-imensional constraints. ssume we want to show that the capacity of a two-imensional constraint Θ is zero. We consier an (n + r + r 2 ) (m + t + t 2 ) array which is Θ-constraine, where t, t 2, r, an r 2 are constants which might epen on the runlength constraints, but o not epen on n an m. ssume further that the labels at positions of the first r rows, the last r 2 rows, the first t columns, an the last t 2 columns, are known. We now scan the other positions of. We scan the other n rows from bottom to top, an the m positions in a row are scanne from left to right. We assume that all positions in the array are scanne, i.e. we omit arrays in which not all positions can 4

24 Technion - Computer Science Department - M.Sc. Thesis MSC be labelle. If each position is etermine by the known labels an the positions which are alreay scanne, then the capacity of the constraint Θ is zero. We will not give a proof to the claim, since we will prove a much stronger result. This technique will be calle scanning. The strength of scanning is emonstrate by proviing a very short proof to the following theorem by Kato an Zeger [3]. Theorem C (, + ) = 0 for every. Proof. Consier an n m array which is (, + )-constraine. We will show that the labels of are etermine by the labels at positions (i,j), where 0 i or 0 j or j = m. X. Y. C D Figure 2.2: Scanning of a (, + ) array. We show that for every + i, j m 2, the label of the position marke by X (see Fig. 2.2) is etermine by the labels to the left of it an the labels below it. ssume the contrary that X can be a zero an can be a one. It implies that all the positions marke by are zeroes an either X or Y is a one. Since Y can be a one, it follows that all positions marke by are zeroes. Since X can be a zero it follows by the vertical constraint that C is a one. Similarly, since Y can be a zero, it follows that D is a one, a contraiction to the horizontal constraint. Hence, C (, + ) = 0. We strengthen the technique as follows: 5

25 Technion - Computer Science Department - M.Sc. Thesis MSC Theorem 2 ssume the scanning metho is applie to a two-imensional constraint Θ. If for the label in each scanne position (i,j), one of the following three states hols: (s) The label in position (i,j) is completely etermine; (s2) The label can be either zero or one, but with one of these labels the suffix of the row is completely etermine; (s3) The label can be either zero or one, but the prefix of the row before position (i,j) is a given sequence P(i,j); then C(Θ) = 0. Proof. ssume ρ positions, numbere by 0,,,ρ, are scanne in a row, as epicte in Fig Figure 2.3: Scanning of ρ positions in a row. Let T be a irecte tree with ρ + levels efine as follows. The root of T (level 0) represents position 0. For l < ρ, the vertices in level l represent position l. The vertices in level ρ represent all the vali labels of all the ρ positions in the row. vertex v which is not a leaf has out-egree one or two epening whether the label of the corresponing position is completely etermine or not, respectively. The ege which connects a vertex v in level l to vertex u in level l + is labelle with one of the possible labels of the position represente by v. If the out-egree of v is two then one ege is labelle by a zero an one ege is labelle by a one. Each vertex v is labelle with the orere labels of the path from the root to v. Hence, each leaf correspons to a vali sequence of labels for the ρ positions. We now boun the number of leaves in the tree, which gives an upper boun to the number of possible rows in the scanning. ρ 6

26 Technion - Computer Science Department - M.Sc. Thesis MSC n example of a tree T that represents the scanning of ρ positions with no constraints is illustrate in Fig When no constraints are impose level 0 level. level ρ level ρ ρ 0 0 ρ 0 ρ ρ Figure 2.4: The tree T when no constraints are impose. every row is vali, therefore the tree is a complete binary tree. The number of leaves equals the number of all rows of length ρ, which is 2 ρ. We now boun the number of leaves in any tree T. First, we note that the label on a vertex v, which represents position (i,j), represents the labels of the positions before position (i,j). vertex representing a position in which state (s) hols has exactly one son. vertex representing a position in which state (s2) hols has two sons, but one of them is a chain of vertices representing positions in state (s). Hence, the number of leaves of a subtree whose root is in level l, an oes not have vertices which represent positions in state (s3), is at most ρ l +. If state (s3) hols in position (i,j) represente by v, then the label on v must be P(i,j). Therefore, in each level there is at most one vertex which represents a position in which state (s3) hols. Now, we construct a tree T from T by swapping subtrees of T, with roots on the same level. Clearly, the number of leaves in T is equal to the number of leaves in T. T will be constructe in a way that all vertices which 7

27 Technion - Computer Science Department - M.Sc. Thesis MSC correspon to positions in which state (s3) hols, are on the same path (see Fig. 2.5). The total number of leaves of T, which are not on this path, is at most ρ (ρ+)ρ l= (ρ l + ) = 2. (s3) (s3) (s3) (s3) Figure 2.5: The tree T. Every subtree which oes not inclue vertices on the leftmost path, oes not have vertices that represent positions that are in state (s3). The number of leaves in T is equal to the number of ifferent labels for a row in the (n + r + r 2 ) (m + t + t 2 ) array which is Θ-constraine (for ρ = m). We now have that the total number of possible ifferent labels for an (n+r +r 2 ) (m+t +t 2 ) array which is Θ-constraine is at most ( (m+)m +) n, 2 which implies that C(Θ) = 0. 8

28 Technion - Computer Science Department - M.Sc. Thesis MSC Chapter 3 symmetric Run-length Constraine Channels The positive capacity region of (,k) constraints in the iamon moel has been etermine by Kato an Zeger in [3]: for every, C (,k) > 0 if an only if k + 2. In this chapter we investigate asymmetric constraints in the iamon moel. The zero/positive region of asymmetric constraints, enote by C (,k, 2,k 2 ), has been stuie by Kato an Zeger in [4]. They have summarize their results in which seven cases remaine unsolve: (u) =, k = 3, 2 = 2, k 2 = 3. (u2) 2, k = +, 2 =, k (u3) 2, + 2 k 2, 2 =, k 2 = 2 +. (u4) 2, + 2 k 2, < 2 < k, k 2 = 2 +. (u5) 2, + 2 k 2, 2 = k, k (u6) 2, 2 < k, < 2 < k, k 2 = 2 +. (u7) 2, 2 < k, 2 = k, k In this chapter we solve most of these cases. 9

29 Technion - Computer Science Department - M.Sc. Thesis MSC Constructions for Proving Positive Capacity Lemma 4 C (, 2 +, 2, 2 + ) > 0 for every. Proof. Let T n be the following (2n 2) (2n) array. T n (, 2n 2) = an T n (0,n 2) = ; if T n (i,j) = then T n (i + 2,j ) = provie that i + 2 2n 3. ll other positions of T n are zeroes. T 4 is illustrate in Fig. 3.. Figure 3.: The array T 4. Consier the [(4 + 4) (2 + 3), 2 + 3] skeleton tile shown in Fig Let an be the two [(4 + 4) (2 + 3), 2 + 3] tiles obtaine from T T + Figure 3.2: The skeleton tile for the (, 2 +, 2, 2 + ) constraint. 20

30 Technion - Computer Science Department - M.Sc. Thesis MSC this skeleton tile by substituting the two skew tetrominoes shown in Fig. 3.3 instea of the four asterisks. We claim that any stanar tiling with the Figure 3.3: Two skew tetrominoes for substitution in the skeleton tile. arrays an yiels a (, 2 +, 2, 2 + )-constraine array. One can easily verify that it is sufficient to prove that the [(4 + 4) (2 + 3), 2 + 3] skeleton tile is a (, 2 +, 2, 2 + ) tile, an that the constraint is not violate on rows an columns crossing two ifferent skeleton tiles, on the positions marke in bol in Fig Figure 3.4: Tiling the plane with skeleton tiles. Now, consier the portions of the rows that cross two skeleton tiles. The scenario is epicte in Fig First note that in Figures 3.2 an 3.5 all the gaps between ones, in which at least one of the ones is not in T + are calculate an written. Therefore, we only have to calculate the gaps between ones in the rectangles epicte in Fig In each one of the two figures Fig. 3.6(a),(b), let α be the leftmost copy of T +, an β the rightmost copy. In each one of the two figures Fig. 3.6(c),(), let α be the upper copy of T +, an β the lower copy. 2

31 Technion - Computer Science Department - M.Sc. Thesis MSC T T T T + T T Figure 3.5: reas crossing two tiles for the (, 2 +, 2, 2 + ) constraint. 22

32 Technion - Computer Science Department - M.Sc. Thesis MSC We start with the ones of Fig. 3.6(a). We calculate the gaps between ones of α an β. α an β are separate by one column, an since the with of T + is 2 + 2, an β is shifte own by two rows, the gaps between ones are In Fig. 3.6(b), the gaps between ones of α an β are also 2 +, since the with of T + is In Fig. 3.6(c), α an β are separate by three rows, an since the height of T + is 2, an β is shifte to the right by one column, the vertical gaps between ones are In Fig. 3.6(), the vertical gaps between ones are also 2, since the height of T + is 2, an α an β are separate by one row. 2 + α (a) β 2 + α (b) β 3 2 (c) Figure 3.6: Relative locations of T + arrays. Hence, any stanar tiling with an is a (, 2 +, 2, 2 + ) array. Therefore, by Lemma 3 we have C (, 2+, 2, 2+) (4+4)(2+3) (2+3) = Lemma 5 C (, 2 + 2, 2 +, 2 + 2) > 0 for every. Proof. Consier the (4+5) (2+3) skeleton array of Fig Let an be the two (4 + 5) (2 + 3) arrays, obtaine from the skeleton array by substituting a one instea of one of the asterisks an a zero instea of the other. 2 () 23

33 Technion - Computer Science Department - M.Sc. Thesis MSC I I + + I I Figure 3.7: skeleton array for the (, 2 + 2, 2 +, 2 + 2) constraint. Consier any stanar tiling of the plane with an. Every row that oes not contain an asterisk, has the pattern (0 2+2 ), which is (, 2 + 2)- constraine. The rows that contain asterisks have the pattern (0 + 0 ) or (0 0 + ), which will be vali in any substitution of zeros an ones instea of the asterisks. Similarly, every column that oes not contain an asterisk, has the pattern ( ), which is (2 +, 2 + 2)-constraine. The columns that contain asterisks have the pattern ( ), which will be vali in any substitution of one zero an one one instea of each consecutive asterisks. Hence, any stanar tiling of the plane with an yiels a twoimensional (, 2+2, 2+, 2+2)-constraine array. Therefore, by Lemma 3 C (, 2 + 2, 2 +, 2 + 2) =. (4+5)(2+3) Lemma 6 If, k > 2, 2 = k an k k then C (,k, 2,k 2 ) > 0. Proof. ssume, k = 2 + t, t > 0, 2 = k, an k 2 = k. We istinguish between two cases: Case : t = 2r +, r 0. y Lemma 4 we have C ( + r, 2 + 2r +, 2 + 2r, 2 + 2r + ) > 0. Therefore, by Lemma we have C (, 2 +2r+, 2 +2r, 2 +2r+) > 0. Case 2: t = 2r + 2, r 0. y Lemma 5 we have C ( + r, 2 + 2r + 2, 2 + 2r +, 2 + 2r + 2) > 0. Therefore, Lemma implies C (, 2 +2r+2, 2 +2r+, 2 +2r+2) > 0. 24

34 Technion - Computer Science Department - M.Sc. Thesis MSC Hence, C (, 2 + t, 2 + t, 2 + t) > 0 an thus by Lemma we have that if, k > 2, 2 = k an k k then C (,k, 2,k 2 ) > 0. Lemma 7 If 2 an r, then C (, 2+,+r,+r+) > 0. Proof. We begin by recursively efining a ( + r ) array, H,r, as follows. For ρ let, H δ,2ρ = H δ,2ρ , H δ,2ρ+ = 0 where H δ, = I δ. H 8,6 is illustrate in Fig Figure 3.8: The array H 8, H δ,2ρ, Next, we efine the ( + r ) array H,r, by rotation of H,r by 80. Note that H,r = H,r if an only if r is o. lso, in the center of H,r (H,r ) there is the ientity matrix I r+. This part of the array will be calle center. Consier the [(2+2r +4) (3+2), 2+2r +] skeleton tile of Fig Let an be the two [(2 + 2r + 4) (3 + 2), 2 + 2r + ] tiles obtaine from the skeleton tile by substituting the two skew tetrominoes of Fig. 3.3 instea of the four asterisks. s in the proof of Lemma 4 we have to prove that any stanar tiling with an is a (, 2+,+r,+r+)-constraine array. One can easily 25

35 Technion - Computer Science Department - M.Sc. Thesis MSC H,r H,r H,r + r + r + r r 2 H,r H,r H,r + r 2 + r 3 Figure 3.9: The skeleton tile for (, 2 +, + r, + r + ) constraint. verify that it is sufficient to prove that the [(2+2r+4) (3+2), 2+2r+] skeleton tiles are (, 2 +, + r, + r + ) tiles, an that the constraint is not violate on rows an columns crossing two ifferentskeletontiles on the positions marke in bol in Fig The scenario is epicte in Fig First note that rotating the plane by 80, aroun any of the tetrominoes (while the tetrominoes are still labelle with the asterisks) leaves the plane with exactly the same labels. Note also that in Figures 3.9 an 3.0 all the gaps between ones, in which at least one of the ones is not in H,r or H,r are calculate an written. Therefore, we only have to calculate the gaps between ones in the rectangles epicte in Fig. 3.. In each one of the three figures (Fig. 3.(a),(b),(c)), let α be the leftmost copy of H,r, β the mile copy, an γ the rightmost copy of H,r.. We start with the ones of Fig. 3.(a). We calculate the gaps between ones, where one of the ones is in α. If the secon one is in β then both ones belong to the center of H,r, an hence the gap between them is. If the corresponing row in β consists only of zeroes, then the corresponing row in γ contains a one as epicte in Fig. 3.(a). The gap between these two ones is 2. The gaps between ones of β an γ are the same as the gaps between the ones of α an β. 2. The gaps between the ones of α an β in Fig. 3.(b) are the same 26

36 Technion - Computer Science Department - M.Sc. Thesis MSC H,r H,r H,r H,r H,r H,r H,r H,r H,r H,r H,r H,r H,r H,r H,r Figure 3.0: reas crossing two tiles for the (, 2 +, + r, + r + ) constraint. H,r H,r H,r 27

37 Technion - Computer Science Department - M.Sc. Thesis MSC α β (a) γ α β γ (b) α β γ (c) 2 () Figure 3.: Relative locations of H,r arrays. as the gaps between the ones of α an β in Fig. 3.(a). The gaps between the ones of α an γ in Fig. 3.(b), where the corresponing row of β has zeroes are greater by one than the gaps between the ones of α an γ in Fig. 3.(a), an hence these gaps have length 2 +. Similarly, the gaps between β an γ are The gaps between the ones of α an β in Fig. 3.(c) are greater by one than gaps between the ones of α an β in Fig. 3.(a), an hence these gaps have length +. The gaps between the ones of α an γ in Fig. 3.(c), where the corresponing row of β has zeroes are the same as the gaps between the ones of α an γ in Fig. 3.(a). Similarly, the gaps between β an γ are Since the height of H,r is + r, it follows that the vertical gaps between ones in Fig. 3.() is + r if r is o. If r is even, then the gap between two ones is +r if at least one of them is not in the center of its shape, an + r + between the other ones. 5. The vertical gaps between ones in Fig. 3.(e) is + r if r is even. If r is o, then the gap between two ones is + r if at least one of them is not in the center of its shape, an + r + between the other ones. This completes the proof that any stanar tiling with an is a (, 2+,+r,+r+)-constraine array. Therefore, by Lemma 3 we have C (, 2+, + r, + r + ) =. (2+2r+4)(3+2) (2+2r+) r+4+2r+7 4 (e) 28

38 Technion - Computer Science Department - M.Sc. Thesis MSC Lemma 8 If 2, k > 2, < 2 < k an k 2 = 2 +, then C (,k, 2,k 2 ) > 0. Proof. ssume 2, k > 2, < 2 < k, an k 2 = 2 +. We istinguish between two cases: Case : < 2 < 2. y Lemma 7 we have C (, 2 +, 2, 2 + ) > 0. Since k 2 +, by Lemma we have C (,k, 2, 2 + ) > 0. Case 2: 2 2 < k. Since 2 2 then by Lemma 6 we have C (, 2 +, 2, 2 +) > 0. In this case k > 2 +, an therefore Lemma implies C (,k, 2, 2 +) > Proving Zero Capacity Proposition If 2, k 2, 2 k, an k 2 = 2 + then C (,k, 2,k 2 ) = 0. Proof. Consier an array which is (,k, 2,k 2 )-constraine. We will show that the label X at position (i,j) is etermine by the labels to the left of it, an the labels of the ( 2 +) ( +) array below it (see Fig. 3.2). 2 X. F E E. E C D F E E. E D Figure 3.2: Labels of the array in Proposition. ssume the contrary that X can be labelle by a zero an can be labelle by a one. It implies that all the positions marke by are zeroes. If any of them was labelle with a one, it woul imply that X is a zero, in orer to 29

39 Technion - Computer Science Department - M.Sc. Thesis MSC avoi a pattern which violates the horizontal constraint. The same argument vertically implies that all the positions marke by are zeroes. If the position marke by C is a zero, then the positions marke by or C form a run of 2 + = k 2 zeroes, which implies that X is a one. Hence, C is a one an all the positions marke by D are zeroes, in orer to satisfy the horizontal constraint. Consier the 2 positions marke by E in one of the corresponing columns. If all these 2 positions are zeroes, then the position marke by F in the same column shoul be labelle with a one by the vertical constraint, an X is a zero by the horizontal constraint. Therefore, in each column with positions marke by E, one of these positions is a one which implies that all the positions marke by F are labelle by zeroes. Since all positions marke by are also zeroes, it follows that X is a one, which contraicts our assumption. Thus, by Theorem 2 we have C (,k, 2,k 2 ) = Summary of Results for the Diamon Moel The results in this chapter prouce solutions to most of the seven unsolve cases of [4]: (u) is solve in Lemma 4, (u2), (u3), an (u4) in Proposition, (u6) in Lemma 8, (u7) in Lemma 6, an (u5) was solve when k 2 = 2 +, in Proposition. The only case which remains unsolve is when 2, +2 k 2, 2 = k, k

40 Technion - Computer Science Department - M.Sc. Thesis MSC Chapter 4 The Square Moel In this moel the ata is organize in the two-imensional rectangular gri, an is rea horizontally, vertically, an in the two iagonal irections. 4. Proving Zero Capacity Recall that the hexagonal moel can be represente as a rectangular gri with 3 irections. Therefore, any (,k)-constraine array in the square moel is also a (,k)-constraine array in the hexagonal moel, which implies the following lemma: Lemma 9 For every,k, C (,k) C (,k). In particular, Lemma 9 implies that if C (,k) = 0 then C (,k) = 0. We will use this in proving zero capacity for some constraints in the square moel. Theorem 3 C (, + 3) = 0 for every. Proof. We beginbyprovingfor =. Kukorelly an Zeger [6] showe that C (, + 2) = 0 for every. In particular, C (2, 4) = 0 an therefore by Lemma 9 we have C (2, 4) = 0. Hence, if C (, 4) > 0 then there exists a (, 4) array that has a run of exactly zero (see Fig. 4.). Each one implies zeroes in each of its 8 neighbors, therefore all the positions marke by are zeros. This creates a run of 5 zeros horizontally, which is a contraiction. Hence, C (, 4) = 0. 3

41 Technion - Computer Science Department - M.Sc. Thesis MSC Figure 4.: Proving C (, 4) = 0. For = 2, the proof is similar. gain we have by Lemma 9 that C (3, 5) = 0, hence if C (2, 5) > 0 then there exists a (2, 5) array that has a run of exactly 2 zeroes (see Fig. 4.2). Each one implies zeroes in each of its Figure 4.2: Proving C (2, 5) = 0. neighbors, therefore all the positions marke by are zeros. This creates a run of 6 zeros horizontally, which is a contraiction. Hence, C (2, 5) = 0. In [5], Kukorelly an Zeger prove that C (, + 3) = 0 for = 3, 4, 5. Therefore by Lemma 9, we have that C (, + 3) = 0 for = 3, 4, 5. The rest of the proof assumes 6. Consier an array which is (, + 3)-constraine. We will show that the label X at position (i,j) is etermine by the labels to the left of it an labels below it (see Figure 4.3). ssume the contrary, i.e. that X can be labelle by a zero an can be labelle X Y Y 2 Y 3 Figure 4.3: Scanning of a (, + 3) array. by a one. It implies that all the positions marke by are zeroes an either X or one of the three positions to the right of X is a one. Therefore, at least one of the following three cases must be vali. 32

42 Technion - Computer Science Department - M.Sc. Thesis MSC Case : X can be a one an Y can be a one (see Fig. 4.4). Clearly, all X Y.. C D D C 2 Figure 4.4: Case of Theorem 3. positions marke by are zeroes. X can be a zero, an therefore by the vertical constraint either C or C 2 is a one. This implies that both positions marke by D are zeroes, which will create a vertical run of +4 zeroes when Y will be a zero, which is a contraiction. Case 2: X can be a one an Y 2 can be a one (see Fig. 4.5). s in case E F 2 E E D E E E E C 2 X Y 2. E D. F C E Figure 4.5: Case 2 of Theorem 3. E, the positions marke by are zeroes. lso, if X will be a zero then the 33

43 Technion - Computer Science Department - M.Sc. Thesis MSC positions marke by C an C 2 will be ones, an if Y 2 will be a zero then the positions marke by D will be ones. Therefore, the positions marke by E must be zeroes. This implies, by the iagonals constraints, that if C 2 will be a zero then both F an F 2 will be ones, a contraiction to the horizontal constraint since the gap between them is of length 5. Case 3: X can be a one an Y 3 can be a one (see Fig. 4.6). Clearly, all C 2 E D 2 C X Y 3. F F D F F. F. F F G G G G E. Figure 4.6: Case 3 of Theorem 3. positions marke by are zeroes. If X will be a zero, then by the vertical constraint either C or C 2 will be a one, an by the right iagonal constraint either D or D 2 will be a one, which implies that C an D 2 will be ones. Similarly, if Y 3 will be a zero, then by the vertical constraint an the left iagonal constraint, the positions marke by E will be ones. This implies that D an all positions marke by F must be zeroes. Hence, similarly to case, in orer to avoi a vertical run of +4 zeroes, two of the four positions marke by G must be ones, which is clearly impossible. y Theorem 2, this completes the proof that C (, + 3) = 0, for every. 34

44 Technion - Computer Science Department - M.Sc. Thesis MSC Summary of Results for the Square Moel The following positive capacity results in the square moel appear in [4]: C (, + 6) > 0, for every, 2(mo 30) C (, + 8) > 0, for every 2, 30(mo 42) C (, + 6) > 0, for every 2, 46(mo 66) C (, + 8) > 0, for every 3, 55(mo 78) Theorem 3 proves that C (, + 3) = 0, for every. Thus, there is still a gap between the known zero an positive capacity regions. 35

45 Technion - Computer Science Department - M.Sc. Thesis MSC Chapter 5 The Triangular Moel In this chapter we investigate the positive capacity region of the triangular moel. Let be an n n triangular array. We say that has n rows, n right columns, an n left columns. (i,j,s) belongs to row i, right column j, an left column [i + j + s] n (see Fig. 5.). row left column right column * * * * * * * * * * * * * * * * * * * * * * * Figure 5.: triangular array 5. Construction for Proving Positive Capacity n n n triangular array is calle a oubly perioic non-attacking triangle queens array if each row, right column, an left column has exactly one one. 36

46 Technion - Computer Science Department - M.Sc. Thesis MSC Lemma 0 n n n oubly perioic non-attacking triangle queens array exists if an only if a (2n, 2n ) triangular array exists. Proof. Let be an n n oubly perioic non-attacking triangle queens array. Consier the following 2n 2n triangular array: = Clearly, each row (right column) of has two ones separate by 2n zeroes. Now, consier the bottom right an the upper left copies of. Each left column which has a one in these arrays has two ones on the corresponing left column of. They are separate by 2n zeroes as the other two copies of cannot have a one on the same left column of. Note that any run of 2n symbols in the tiling has a representation in. Therefore, ones in each row of the tiling are separate by 2n zeroes, an the same is true for right an left columns. Lemma If is an n n (,) triangular array then any exchanges of copies of the patterns shown in Fig. 5.2 in isjoint positions of will result in a ( 2, + 2) array. Figure 5.2: Three 2 2 exchangeable triangular arrays. Proof. The ones in all three triangular arrays occupy the same rows, an right an left columns. In each irection, the ifference between the arrays, is a change of at most 2 positions for the label one. In a (,) array, any two ajacent copies of the patterns above must be ientical. Therefore, by exchanges of copies of the above patterns in isjoint 37

47 Technion - Computer Science Department - M.Sc. Thesis MSC positions of a (,) array, the length of any given run of zeroes may increase or ecrease by at most 2. This results in a ( 2, + 2) array. We now make use of Lemmas 0 an, to construct tiles that imply positive capacity of some constraints in the triangular moel. Lemma 2 If (mo 4) then C (, + 4) 2(+3) log 23. Proof. For even n we construct an n n oubly perioic non-attacking triangle queens array T n, where T n (i,i,s) = if s i (mo 2), for every 0 i n (T 6 is illustrate in Fig. 5.3). y Lemma 0, the stanar Figure 5.3: The triangular array T 6. tiling with T n is a (2n, 2n ) array. y Lemma, any exchanges of copies of the pattern shown in Fig. 5.2 in isjoint positions of will result in a (2n 3, 2n+) array. The total number of ifferent (2n 3, 2n+) arrays use in the tiling is 3 n 2. Hence, by Lemma 3 we have that C (2n 3, 2n+) 4n log 23. The following lemmashows thatwhen 3(mo 4), asimilar construction to the one above oes not exist. Lemma 3 If n is o then there is no n n oubly perioic non-attacking triangle queens array which contains an appearance of any of the patterns shown in Fig Proof. ssume that n is o an an n n oubly perioic non-attacking triangle queens array exists. We write as a sequence a 0,a,,a n, where a i = (j i,s i ) if (i,j i,s i ) =. Since is a oubly perioic nonattacking triangle queens array, it follows that for every 0 r < l n, we have j r j l because there cannot be 2 ones in the same right column, 38

48 Technion - Computer Science Department - M.Sc. Thesis MSC an j r + r + s r j l + l + s l (mo n) because there cannot be 2 ones in the same left column. Therefore, j 0,j,,j n an [j s 0 ] n, [j + + s ] n,, [j n +(n )+s n ] n are permutations of 0,,,n. For any given permutation p 0,p,,p n of 0,,,n we have n p i = i=0 since n is o. Therefore, (n )n 2 0(mo n), n n n n s i = (j i + i + s i ) j i i 0(mo n). i=0 i=0 Hence, either s i = 0 for each 0 i n, or s i = for each 0 i n, implying that all the positions that are ones have the same orientation. Thus, there is no oubly perioic n n non-attacking triangle queens array which contains an appearance of any 2 2 array shown in Fig i=0 5.2 Proving Zero Capacity In this section we prove that C (, + 3) = 0 for every 5. For technical reasons, the proof is ivie into two parts, one proof for even values of, an another for o ones. The following lemma will be use in Lemma 5, when proving that C (, + 3) > 0 for even 6. Lemma 4 Let 6 be an even integer, h = +6, an let be an infinite 2 (, + 3) array. If contains an r h sub-array whose first two rows form the pattern PEven (see Fig. 5.4), then the first two an the last two right columns of are substrings of (0 + ). +2 i=0 Figure 5.4: The pattern PEven. 39

49 Technion - Computer Science Department - M.Sc. Thesis MSC Proof. Let C be an r (h + ) sub-array of with the pattern PEven as epicte in Fig Clearly the positions marke by are zeroes. y the left column constraint either or 2 will be a one an hence all positions marke by C are zeroes. ssume the position marke by D is a one. Then, E E E E C D C 2 +2 C F Figure 5.5: Labels implie by the pattern PEven. all positions marke by E will be zeroes which will create a run of + 7 zeroes in the right column, a contraiction. Hence, D is a zero, F is a one, is a zero, an 2 is a one. The four ones in the left columns of 2 an F form the pattern PEven an hence by the same arguments the two positions marke by G are ones. The positions marke by 2, F, an G form again the pattern PEven. The claim of the lemma is prove now by inuction. Lemma 5 If 6 is even then C (, + 3) = 0. Proof. We use the scanning technique again, an show that one of the three states of Theorem 2 occurs in each scanne position. ssume we have to label the next scanne position marke by X. We have to istinguish between two ifferent types of orientations of the position as epicte in Fig X Figure 5.6: The possible orientations of a scanne position. G G X 40

50 Technion - Computer Science Department - M.Sc. Thesis MSC Case : ssume that X, as epicte in Figure 5.7 (to simplify the picture, the array is rawn in a ifferent orientation), is not uniquely etermine, i.e., it can be labelle by a zero an it can be labelle by a one. It implies that all the positions marke by are zeroes, either X or one of the three positions to the right of X is a one, an either or 2 is a one. Therefore, the positions marke by C are zeroes an at least one of the following three cases must be vali. X Y Y Y C C Figure 5.7: Case of Lemma 5. Case a: X can be a one an Y can be a one (see Fig. 5.8). Clearly, all positions marke by D are zeroes. Y can be a zero, therefore E is a one, an hence 2 is a zero an is a one. Therefore, the seven positions marke by F are zeroes. I G I L 2 D X Y D J J D H H J J H H D F F D E F 2 C F C 7 F F F M M K L 2 Figure 5.8: Case a of Lemma 5. ssume G is a one. Then the positions below it (ening with the 5 positions marke by H) are zeroes, creating a run of + 7 zeroes, a contraiction. Hence, G is a zero. 4

51 Technion - Computer Science Department - M.Sc. Thesis MSC If X will be a zero then either I or I 2 will be a one. ssume I will be a one. Then, all the positions marke by J are zeroes. If X will be a one then I an K will be zeroes, Y will be a zero, either L or L 2 will be a one an the two positions marke by M will be labelle by zeroes. Therefore, there is a run of + 4 zeroes is the right column of K, a contraiction. Hence, if X will be a zero then I will be a zero, I 2 an Y will be ones. E,, I 2 an Y will form the pattern Peven, an hence by Lemma 4 the suffix of the current row is completely etermine, an we are in state (s2). Case b: X can be a one an Y 2 can be a one (see Fig. 5.9). If Y 2 will be a one then all positions marke by D are zeroes. If X will be a one then Y 2 will be a zero, an hence there is a run of + 5 zeroes in the left column of Y 2, a contraiction. Thus, Y 2 cannot be a one. D D X Y2 D D D D D D 2 C C Figure 5.9: Case b of Lemma 5. Case c: X can be a one an Y 3 can be a one (see Fig. 5.0). Clearly, all F F F D E Y D 3 X D D 2 C C Figure 5.0: Case c of Lemma 5. positions marke by D are zeroes. If X will be a zero, then Y 3 will be a one 42

52 Technion - Computer Science Department - M.Sc. Thesis MSC an the position marke by E will be a one, an hence all positions marke by F are zeroes. Therefore, there is a run of length + 4 in left column of X, a contraiction. Thus, X cannot be a zero. Case 2: ssume that X, as epicte in Fig. 5., is not uniquely etermine, i.e., it can be labelle by a zero an it can be labelle by a one. It implies that all the positions marke by are zeroes, either X or one of the three positions to the right of X is a one, an at least one of the following three cases must be vali. X Y Y 2 Y3 Figure 5.: Case 2 of Lemma 5. Case 2a: X can be a one an Y can be a one. Clearly, all positions marke by are zeroes (see Fig. 5.2). X can be a zero an hence either C or C 2 X Y H K H K 5 J J K H H H J J 5 H H C E E E F E H F2 H C 2 D D2 E E E E L 2 L M I Figure 5.2: Case 2a of Lemma 5. is a one. Y can be a zero an therefore either D or D 2 is a one. It implies that C an D are ones. Hence, all positions marke by E are zeroes. y the horizontal constraint either F or F 2 is a one. Since Y can be a zero, it G G 2 43

53 Technion - Computer Science Department - M.Sc. Thesis MSC follows that either G or G 2 is a one. Hence, F is a one an all positions marke by H are zeroes. ssume I will be a one. Then all positions marke by J will be zeroes, creating a run of +4 zeroes in their right column, a contraiction. Therefore, I is labelle by a zero. ssume all the 5 positions marke by K are zeroes. Then L is labelle by a one an Y cannot be a one, a contraiction. Hence, one of the positions marke by K is a one, L an L 2 are labelle by zeroes. Therefore, if X will be a one then Y will be a zero an by its right column constraint M will be a one. M, X, C, an D will form the pattern PEven, an hence by Lemma 4 all the prefix of the row before X is a given sequence P(i,j), an we are in state (s3). Case 2b: X can be a one an Y 2 can be a one. Clearly, all positions marke by are zeroes (see Fig. 5.3). X can be a zero an hence exactly one of the C i s is a one, an exactly one of the D i s is a one. Y 2 can be a zero an therefore exactly one of the E i s is a one, an exactly one of the F i s is a one. Clearly, D 3 an E 3 cannot be ones. C C3 C 2 H G X Y 2 E D F E E 2 3 D 2 F D 2 3 F3 Figure 5.3: Case 2b of Lemma 5. If E 2 is a one then C is a one. If X will be a one then Y 2 will be a zero an by its left column constraint G will be a one. E 2, C, X, an G will form the pattern PEven, an hence by Lemma 4 all the prefix of the row before X is a given sequence P(i,j), an we are in state (s3). If D 2 is a one then F is a one. If Y 2 will be a one then X will be a zero an by its right column constraint H will be a one. D 2, F, Y 2, an H 44

54 Technion - Computer Science Department - M.Sc. Thesis MSC will form the pattern PEven, an hence by Lemma 4 the suffix of the current row is completely etermine, an we are in state (s2). If D 2, D 3, E 2, an E 3 are zeroes then D an E are ones which is impossible since the gap between them is an the horizontal constraint will be violate. Case 2c: X can be a one an Y 3 can be a one. Clearly, all positions marke D D C D X Y 3 Figure 5.4: Case 2c of Lemma 5. by are zeroes (see Fig. 5.4). If X will be a one then Y 3 will be a zero an by its left column constraint C will be a one. Hence, all the positions marke by D will be labelle by zeroes, creating a run of + 4 zeroes in the right column of Y 3, a contraiction. Thus, by Theorem 2, C (, + 3) = 0 for every even 6. Similarly tolemma 4, the following lemma will be usein Lemma 7, when proving that C (, + 3) > 0 for o 5. Lemma 6 Let 5 be an o integer, h = +7, an let be an infinite 2 (, + 3) array. If contains an r h sub-array whose first two rows form the pattern PO (see Fig. 5.5), then the first two an the last two right columns of are substrings of (0 +2 ). +3 Figure 5.5: The pattern PO + 45

55 Technion - Computer Science Department - M.Sc. Thesis MSC Proof. Let C be an (r + 2) h right sub-array of with the pattern PO as epicte in Fig Clearly the positions marke by are zeroes. ssume the position marke by is a one. Then the 4 positions marke by C will be zeroes, creating a run of + 4 zeroes in their right column, a contraiction. Therefore, is a zero an either D or D 2 is a one. E C F C E D 2 G G D + +3 E 2 E Figure 5.6: Labels implie by the pattern PO. ssume D is a one. Then D 2 an all positions marke by E or E 2 will be zeroes. Hence, by the right column constraint, F will be a one an the two positions marke by G will be zeroes, an it will create a run of + 4 zeroes in their left column, a contraiction. Therefore, D is a zero an D 2 is a one. It implies that all positions marke by E or G are zeroes, an hence E 2 is a one. The four ones in the left columns of D 2 an E 2 form the pattern PO an hence by the same arguments the two positions marke by H are ones. The positions marke by D 2, E 2, an H form again the pattern PO. The claim of the lemma is prove now by inuction. Similarly to Lemma 5 we have the following lemma. Lemma 7 If 5 is o then C (, + 3) = 0. Proof. We will use the scanning technique again. ssume we have to label the next position marke by X. We have to istinguish between two ifferent types of positions as epicte in Figure 5.6. H H 46

56 Technion - Computer Science Department - M.Sc. Thesis MSC Case : ssume that X, as epicte in Figure 5.7, is not uniquely etermine, i.e., it can be labelle by a zero an it can be labelle by a one. It implies that all the positions marke by are zeroes, either X or one of the three positions to the right of X is a one, an either or 2 is a one. Therefore, the positions marke by C are zeroes an at least one of the following three cases must be vali. X Y Y Y 3 2 C C 2 Figure 5.7: Case of Lemma 7. Case a: X can be a one an Y can be a one (see Fig. 5.8). Clearly, all positions marke by D are zeroes, E is a one, an hence is a zero an 2 is a one. If X will be a zero then Y an F will be ones. E, 2, Y an F will form the pattern Po, an hence by Lemma 6 the suffix of the current row is completely etermine, an we are in state (s2). F D X Y D D D D D C C E Figure 5.8: Case a of Lemma 7. Case b: X can be a one an Y 2 can be a one (see Fig. 5.9). If Y 2 will be a one then all positions marke by D are zeroes, an hence E will be a one. Therefore, the positions marke by F are zeroes, an since also X will be a 2 47

57 Technion - Computer Science Department - M.Sc. Thesis MSC zero, it follows that there is a run of + 4 zeroes in the left column of X, a contraiction. F F D E D X Y 2 D D D D C C 2 Figure 5.9: Case b of Lemma 7. Case c: X can be a one an Y 3 can be a one (see Fig. 5.20). Clearly, all G D E G D Y X 3 D D F D F D F D F D C D C Figure 5.20: Case c of Lemma 7. positions marke by D are zeroes. If X will be a zero, then Y 3 will be a one an the position marke by E will be a one, an hence all positions marke by F are zeroes. If X will be a one, then Y 3 will be a zero an E will be a zero, an to avoi a run of length + 4 in the left columns we must have ones in the positions marke by G, which is impossible. Case 2: ssume that X, as epicte in Fig. 5.2, is not uniquely etermine, i.e., it can be labelle by a zero an it can be labelle by a one. It implies that all the positions marke by are zeroes, either X or one of the three positions to the right of X is a one, an at least one of the following three cases must be vali. D 2 48

58 Technion - Computer Science Department - M.Sc. Thesis MSC X Y Y 2 Y3 Figure 5.2: Case 2 of Lemma 7. Case 2a: X can be a one an Y can be a one (see Fig. 5.22). Clearly, all positions marke by are zeroes. If Y will be a one then X will be a zero, an therefore either D or D 2 is a one. If X will be a one then Y will be a zero, an therefore either E or E 2 is a one. Hence, D 2 an E 2 are ones. If X will be a one then F will be a one. D 2, E 2, X an F will form the pattern Po, an hence by Lemma 4 all the prefix of the row before X is completely etermine, an we are in state (s3). D2 D E E 2 F X Y Figure 5.22: Case 2a of Lemma 7. Case 2b: X can be a one an Y 2 can be a one (see Fig. 5.23). Clearly, all positions marke by are zeroes. X can be a zero an hence exactly one of the C i s is a one, an exactly one of the D i s is a one. Y can be a zero an hence exactly one of the E i s is a one, an exactly one of the F i s is a one. Clearly, C an F cannot be ones. If C 2 is a one then E 3 is a one. If X will be a one, then by the left column constraint G will be a one. C 2, E 3, X an G will form the 49

59 Technion - Computer Science Department - M.Sc. Thesis MSC pattern Po, an hence by Lemma 6 all the prefix of the row before X is completely etermine, an we are in state (s3). If F 2 is a one then D 3 is a one. If Y 2 will be a one then by the right column constraint H will be a one. F 2, D 3, Y 2 an H will form the pattern Po, an hence by Lemma 6 the suffix the current row is completely etermine, an we are in state (s2). If C 2 an F 2 are zeroes then C 3 an F 3 are ones which is impossible since the gap between them is + 4 an the horizontal constraint will be violate. H G X Y 2 C E C2 D E F D2 F2 2 C 3 E 3 D 3 F 3 Figure 5.23: Case 2b of Lemma 7. Case 2c: X can be a one an Y 3 can be a one (see Fig. 5.24). Clearly, all G G C X Y 3 D D D F F E E Figure 5.24: Case 2c of Lemma 7. positions marke by are zeroes. If X will be a one then C will be a one 50

60 Technion - Computer Science Department - M.Sc. Thesis MSC by the left column constraint, an hence all the positions marke by D are zeroes; Y 3 will be a zero an hence one of the positions marke by E is a one an the positions marke by F are zeroes. If Y 3 will be a one then C will be a zero an hence by the right columns constraint both positions marke by G shoul be ones which is impossible by the horizontal constraint. Thus, by Theorem 2, C (, + 3) = 0 for every o 5. Corollary C (, + 3) = 0 for The Capacity for Small Values of For small values of the zero/positive capacity region is slightly ifferent, an is escribe in this section. Lemma 8 C (,k) > 0 if an only if k 3. Proof. We first show that C (, 2) = 0. This is one by a simple scanning argument. ssume we have to label the next scanne position marke by X. We have to istinguish between the two ifferent types of orientations of the position as epicte in Fig Case : ssume that X, as epicte in Figure 5.25, is not uniquely etermine, i.e., it can be labelle by a zero an it can be labelle by a one. It implies that the positions marke by are zeroes. If X will be a zero, there will be a run of 3 zeroes in the left column, a contraiction. X Figure 5.25: Case of Lemma 8. Case 2: ssume that X, as epicte in Figure 5.26, is not uniquely etermine, i.e., it can be labelle by a zero an it can be labelle by a one. It implies that the position marke by is a zero. If X will be a zero, Y will be a one by the horizontal constraint, an therefore is a zero. Moreover, if X will be a zero there will be 2 zeroes in the right column, hence C is a one. Similarly, Y can be a zero, which implies that D is a one. C an D are ajacent ones, a contraiction. 5

61 Technion - Computer Science Department - M.Sc. Thesis MSC X Y C D Figure 5.26: Case 2 of Lemma 8. This completes the proof that C (, 2) = 0. This proof can be trivially generalize to show that C (, + ) = 0 for every o, but our results for the triangle moel are stronger, an hence the generalization is omitte. We now show that C (, 3) > 0. Consier the (, ) array of size n n, where (i,j, 0) = (see Fig. 5.27). Clearly, any change of nonconsecutive ones into Figure 5.27: The array for the proof that C (, 3) > 0. zeroes, results in a (, 3) array. ny tiling of the plane with the lattice points {(x,y) : x = i, y = i + 3j, i,j Z}, using the two triangular tiles of Fig. 5.28, correspons to some array constructe in the above manner. y Figure 5.28: Two triangular tiles to prove that C (, 3) > 0. Lemma 3, this tiling implies that C (, 3) 6. In this proof, we make use of the fact that any change of nonconsecutive ones into zeroes in the (, ) array, results in a (, 3) array. ut the lower boun on the capacity that is achieve by the corresponing tiling, can be much improve by noticing the following. The ones in the (, ) array form a hexagonal lattice, where two consecutive ones correspon to ajacent hexagons. ny set of nonconsecutive ones is an inepenent set in the 52

62 Technion - Computer Science Department - M.Sc. Thesis MSC hexagonal lattice. The exact number of inepenent sets in the hexagonal moel, also known as the number of arrays in the har hexagonal moel, has been given by axter in []. Since the hexagonal lattice inuce by the ones is a hexagonal n m array, we have the following boun on the capacity: log C (, 3) = lim 2 N(n,m (, 3)) n,m 2nm 2 C (, ) , 2 which is better than the boun of given by the tiling. 6 Lemma 9 C (2,k) > 0 if an only if k 4. Proof. We first show that C (2, 3) = 0. Clearly C (3, 3) = 0, hence if C (2, 3) > 0, then there exists a (2, 3) array that has a run of zeroes whose length is exactly 2. We analyze such an array, an show that a run of zeroes whose length is 4 must exists. Let be an n n array with a run of zeroes of length 2 as epicte in Fig The leftmost one implies that the position 0 0 Figure 5.29: force run of 4 zeroes in a (2,k) triangular array. marke by is a zero, an the rightmost one implies that the positions marke by are zeroes, which creates a run of 4 zeroes horizontally. Hence, C (2, 3) = 0. We now show that C (2, 4) > 0. ny tiling of the plane with the lattice points {(x,y) : x = 3i, y = 3i + 9j, i,j Z}, using the two triangular arrays of Fig is a vali (2, 4) array. y Lemma 3, this tiling implies Figure 5.30: Two triangular arrays to prove that C (2, 4) > 0. that C (2, 4)

63 Technion - Computer Science Department - M.Sc. Thesis MSC Lemma 20 C (3,k) > 0 if an only if k 7. Proof. First, we use the scanning technique again to prove that C (3, 6) = 0. ssume we have to label the next scanne position marke by X. We have to istinguish between the two ifferent types of orientations of the position as epicte in Fig Case : ssume that X, as epicte in Figure 5.3, is not uniquely etermine, i.e., it can be labelle by a zero an it can be labelle by a one. It implies that all the positions marke by are zeroes, either X or one of the three positions to the right of X is a one, therefore at least one of the following three cases must be vali. Y Y X Y2 3 Figure 5.3: Case of Lemma 20. Case a: X can be a one an Y can be a one (see Fig. 5.32). Clearly, all positions marke by are zeroes. X can be a zero, therefore either C or C 2 is a one, an D is a zero. Y can be a zero, therefore E is a one, C is a zero, C 2 is a one, an the positions marke by F are zeroes. Hence, by the right column constraint G is a one, which implies that H is a zero. This implies that either I or I 2 is a one, an the positions marke by J are zeroes, which creates a run of 8 zeroes in the left column when X is a zero, a contraiction. I I 2 J J X Y H C D G E F C F 2 Figure 5.32: Case a of Lemma 20. Case b: X can be a one an Y 2 can be a one (see Fig. 5.33). Clearly, all positions marke by are zeroes. X can be a zero, therefore C is a one, 54

64 Technion - Computer Science Department - M.Sc. Thesis MSC an similarly Y 2 can be a zero, therefore D is a one. This implies that all positions marke by E are zeroes. y the horizontal constraint F is a one, an therefore G is a zero, which creates a run of 7 zeroes in the left column when X is a zero, a contraiction. G X E F E E C E Y 2 E Figure 5.33: Case b of Lemma 20. Case c: X can be a one an Y 3 can be a one (see Fig. 5.34). Clearly, all positions marke by are zeroes, therefore C is a one, an all positions marke by D are zeroes, which creates a run of 7 zeroes in the left column when X is a zero, a contraiction. Y 3 X C D D D Figure 5.34: Case c of Lemma 20. Case 2: ssume that X, as epicte in Figure 5.35, is not uniquely etermine, i.e., it can be labelle by a zero an it can be labelle by a one. It implies that all the positions marke by are zeroes, either X or one of the three positions to the right of X is a one, therefore at least one of the following three cases must be vali. X Y Y 2 Y 3 Figure 5.35: Case 2 of Lemma 20. D 55

65 Technion - Computer Science Department - M.Sc. Thesis MSC Case 2a: X can be a one an Y can be a one (see Fig. 5.36). Clearly, all positions marke by are zeroes. X can be a zero, therefore either C or C 2 is a one. Y can be a zero, therefore either D or D 2 is a one. This implies that C 2 an D 2 are ones, an the positions marke by E are zeroes. y the horizontal constraint F is a one, an the positions marke by G are zeroes. y the left column constraint H is a one, an the positions marke by I are zeroes. X can by a zero, therefore by the horizontal constraint J is a one, an K is a zero. If X will be a one, Y will be a zero, hence L will be a one an the positions marke by M will be zero, which will create a horizontal run of 7 zeroes, a contraiction. K I I I M M M I H M I G G L J G X Y I E F G E E C C2 D Figure 5.36: Case 2a of Lemma 20. Case 2b: X can be a one an Y 2 can be a one (see Fig. 5.37). Clearly, all positions marke by are zeroes, which creates a horizontal run of 7 zeroes, a contraiction. D 2 X Y 2 Figure 5.37: Case 2b of Lemma 20. Case 2c: X can be a one an Y 3 can be a one (see Fig. 5.38). Clearly, all positions marke by are zeroes, therefore C is a one, an all positions marke by D are zeroes. Y 3 can be a zero, therefore by the right column constraint, E is a one, an all positions marke by F are zeroes. X can be a zero, therefore by the right column constraint G is a one, all positions marke by H are zeroes, an either I or I 2 is a one. This implies that J is 56

66 Technion - Computer Science Department - M.Sc. Thesis MSC a zero, which creates a run of 7 zeroes in the right column when X is a zero, a contraiction. This completes the proof that C (3, 6) = 0. H H I I 2 J X Y 3 D C F D G D F F D D D E Figure 5.38: Case 2c of Lemma 20. We now show that C (3, 7) > 0. Consier the (3, 3) array of size n n, where (2i, 2j, ) = an (2i +, 2j +, 0) = (see Fig. 5.39). Clearly, any change Figure 5.39: The array for the proof that C (3, 7) > 0. of nonconsecutive ones into zeroes, results in a (3, 7) array. ny tiling of the plane with the lattice points {(x,y) : x = 2i, y = 2i + 6j, i,j Z}, using the four triangular tiles of Fig. 5.40, correspons to some array constructe in the above manner. y Lemma 3, this tiling implies that C (3, 7) 2. Figure 5.40: Four triangular tiles to prove that C (3, 7) > 0. 57

67 Technion - Computer Science Department - M.Sc. Thesis MSC s in Lemma 20, the lower boun on the capacity that is achieve by the corresponing tiling, can be much improve by noticing that the ones in the (3, 3) array form two hexagonal lattices, where two consecutive ones correspon to ajacent hexagons. Since each hexagonal lattice inuce by the ones is a hexagonal n m array, we have the following boun on the 2 2 capacity: log C (3, 7) = lim 2 N(n,m (3, 7)) n,m 2nm [ 2 4 C (, ) + ] 4 C (, ) = = 4 C (, ) , 4 which is better than the boun of 2 Lemma 2 C (4,k) > 0 if an only if k 9. given by the tiling. Proof. y Lemma 7 we have that C (5, 8) = 0. Hence if C (4, 8) > 0, then there exists a (4, 8) array that has a run of zeroes whose length is exactly 4. We analyze such an array, an show that a run of zeroes whose length is 9 must exists. Let be an n n array with a run of zeroes of length 4 as epicte in Fig Clearly the positions marke by are zeroes. ssume J H K J J L H H H I F F H H J H G F F C E D Figure 5.4: Proving C (4, 8) = 0. the position marke by is a zero. Then, by the horizontal constraint C is a one, an by the left an right columns, D an E are zeroes, which creates a run of 9 zeroes horizontally. Hence, is a one, all the positions marke 58

68 Technion - Computer Science Department - M.Sc. Thesis MSC by F are zeroes, C an E are zeroes, an D is a one. This implies that G must be a one by the right column constraint, an all the positions marke by H are zeroes. I is a one by the left column constraint, an hence all the positions marke by J are zeroes. Therefore, K is a one by the right column constraint, an L is a zero, which creates a run of 9 zeroes in that left column. Hence, C (4, 8) = 0. y using = 5 in Lemma 2 we have that C (5, 9) > 0. Therefore Lemma implies that C (4, 9) > 0, which completes the proof. 5.4 Summary of Results for the Triangular Moel This chapter shows a tight characterization for C (,k) when (mo 4), given by Lemmas 2 an 7: Corollary 2 For every (mo 4), 5 we have: C (,k) > 0 if an only if k 4. For other values of, by Lemmas 2 an, we have: Corollary 3 C (, + 5) > 0 if 0 (mo 4) C (, + 6) > 0 if 3 (mo 4) C (, + 7) > 0 if 2 (mo 4) y Corollary we have that C (, + 3) = 0 for all 5, hence the remaining gaps are relatively small. 59

69 Technion - Computer Science Department - M.Sc. Thesis MSC Chapter 6 Discussion an Open Problems In this work we consiere the positive capacity region of two-imensional run-length constraine channels in a few connectivity moels the iamon, square, an triangular moels. We have manage to fin some regions where the capacity is positive an some in which the capacity is zero, by using generalizations an moifications of known techniques. 6. The Scanning Metho The main contribution regaring techniques for proving zero capacity, is the generalization of the scanning metho of [2] in Theorem 2. Previous techniques for proving zerocapacity, strongly epeneon the specific constraint they were applie to. The proofs were much longer an require the consieration of many ifferent cases. The alternative proof in chapter 2 for the result of Kato an Zeger that C (, + ) = 0, shows the efficiency of the scanning metho. Perhaps more important, is that the generalization of scanning metho allows to etermine zero capacity for constraints Θ that have larger values of N(n,m Θ). n interesting path for further research is generalizing the scanning metho to hanle constraints in which the number of constraine arrays is much larger. 6.2 ouning the Capacity When proving positive capacity, we fin tiles with ifferent labels, an show that tiling the plane with them inuces vali arrays. This implies a boun 60

70 Technion - Computer Science Department - M.Sc. Thesis MSC on the capacity as escribe in Lemma 3. The proofs of Lemmas 8 an 20 show that the boun inuce by the tiling coul be far from the actual capacity. How goo are these bouns for other constraints? For example, can the boun of Lemma 2, C (, + 4) log 2(+3) 23 for (mo 4), be improve? 6.3 The Connectivity Moels 6.3. The Diamon Moel We consiere asymmetric constraints in the iamon moel in Chapter 3. We solve most of the open cases of [4], using the techniques presente in Chpater 2, an showe a characterization of the zero/positive capacity region in which only one case remains unsolve. We woul like to see the capacity of the last case etermine: for 2, + 2 k 2, 2 = k, k 2 2 2, is C (,k, 2,k 2 ) = 0 or C (,k, 2,k 2 ) > 0? The Square Moel The gaps between the known zero an positive capacity regions in the square moel are relatively large. In Chapter 4 we prove that C (, + 3) = 0 for every, but the known positive capacities are much farther. Further research shoul attempt to fin an infinite set S of positive integers, an an integer r, such that C (, + r) = 0 an C (, + r + ) > 0 for each S The Triangular Moel We consiere the triangular moel in Chapter 5 an showe a tight characterization of the positive capacity region for many values of. We showe that C (,k) > 0 if an only if k + 4, for every (mo 4). Together with the proof that C (,+3) = 0 for every 3, it implies that the gaps between the zero an positive capacity regions in this moel are relatively small. full characterization in the triangular moel is yet to be etermine. 6

71 Technion - Computer Science Department - M.Sc. Thesis MSC ibliography [] R. J. axter. Har hexagons: Exact solution. J. Phys. : Math. Gen., 3:L6 L70, 980. [2] S. R. lackburn. Two-imensional runlength constraint arrays with equal horizontal an vertical constraints. IEEE Transactions on Information Theory, submitte, [3] N. J. Calkin an H. S. Wilf. The number of inepenent sets in a gri graph. SIM J. Discret. Math., ():54 60, 998. [4] K. Censor an T. Etzion. The positive capacity region of twoimensional run length constraine channels. IEEE Symposium on Information Theory, Seattle W, submitte, [5] J. H. Conway an N. J.. Sloane. Sphere Packings, Lattices an Groups. New York, Springer-Verlag, 988. [6] S.I.R.Costa, M.Muniz,E.gustini, an R.Palazzo. Graphs, tessellations, an perfect coes on flat tori. IEEE Transactions on Information Theory, 50(0): , [7] T. Etzion an K. G. Paterson. Zero/positive capacities of twoimensional runlength-constraine arrays. IEEE Transactions on Information Theory, 5(9): , [8] T. Etzion an. Vary. Two-imensional interleaving schemes with repetitions: Constructions an bouns. IEEE Transactions on Information Theory, 48(2): , [9] S. Halevy, J. Chen, R. M. Roth, P. H. Siegel, an J. K. Wolf. Improve bit-stuffing bouns on two-imensional constraints. IEEE Transactions on Information Theory, 50(5): ,

72 Technion - Computer Science Department - M.Sc. Thesis MSC [0] K.. S. Immink. Coing Techniques for Digital Recorers. New York, Prentice Hall, 99. [] K.. S. Immink. Coes for Mass Data Storage Systems. The Netherlans, Shannon Founation Publishers, 999. [2] W. Weeks IV an R. E. lahut. The capacity an coing gain of certain checkerboar coes. IEEE Transactions on Information Theory, 44(3):93 203, 998. [3]. Kato an K. Zeger. On the capacity of two-imensional runlength constraine channels. IEEE Transactions on Information Theory, 45(5): , 999. [4]. Kato an K. Zeger. Partial characterization of the positive capacity region of two-imensional asymmetric run length constraine channels. IEEE Transactions on Information Theory, 46(7): , [5] Zs. Kukorelly an K. Zeger. utomate theorem proving for hexagonal run length constrine capacity computation. IEEE Symposium on Information Theory, Seattle W, submitte, [6] Zs. Kukorelly an K. Zeger. The capacity of some hexagonal (,k) constraints. IEEE International Symposium on Information Theory, Washington DC, page 263, June 200. [7]. H. Marcus, R. M. Roth, an P. H. Siegel. n introuction to coing for constraine systems. Lecture Notes, 200. [8] Zs. Nagy an K. Zeger. symptotic capacity of two-imensional channels with checkerboar constraints. IEEE Transactions on Information Theory, 49(9):25 225, [9] Zs. Nagy an K. Zeger. Capacity bouns for the har triangle moel. IEEE International Symposium on Information Theory, Chicago Illinois, page 62, June [20] R. M. Roth, P. H. Siegel, an J. K. Wolf. Efficient coing schemes for the har-square moel. IEEE Transactions on Information Theory, 47(3):66 76,

73 Technion - Computer Science Department - M.Sc. Thesis MSC [2] M. Schwartz an T. Etzion. Two-imensional cluster-correcting coes. IEEE Transactions on Information Theory, 5(6):22 232, [22] P. H. Siegel an J. K. Wolf. it-stuffing bouns on the capacity of 2-imensional constraine arrays. IEEE International Symposium on Information Theory, Cambrige M, page 323, ug [23] R. Talyansky. Coing for two-imensional constraints. M.Sc. Thesis (in Hebrew), Computer Science Department, Technion, Haifa, Israel,

74 Technion - Computer Science Department - M.Sc. Thesis MSC צפנים עם אילוצים לערוצים דו-מימדיים קרן צנזור

75 Technion - Computer Science Department - M.Sc. Thesis MSC

76 Technion - Computer Science Department - M.Sc. Thesis MSC צפנים עם אילוצים לערוצים דו-מימדיים קרן צנזור

77 Technion - Computer Science Department - M.Sc. Thesis MSC

78 R - R h. g + ; ; - - I I J I I I I p x H Technion - Computer Science Department - M.Sc. Thesis MSC , ! " # $ % & ' ( ) ' * "! / ' * "! - +. < : ' * "! ' : = - +, ' * "! ' : -? ' % >! / ' " # : # & %! # G H C D E F K R J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J M P L M N O LP M Q JK K V J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J S T U O LP M Q GI bg W X X ^ _ ` X a \ W X Y Z [ X \ W ] W X Y Z ^ ] I h J J J J J J J J J J J J J J J J J J J J J J J J J J J M P L M N O LP M Q c N L e O c L M f P R JK J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J S T U O LP M Q c N L e R J J J J J J J J J J J J J J J J J J J J j M f k e c l P m U O n L o m LP l c L U i L c e j L M S R JR H G R K R w H y R z ~ ~ W X q r s t u q Z ` s [ v Z ` J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J S T U O LP M Q c N L e V JK J J J J J J J J J J J J J J J J J J J j M f k e c f L o k O n L o m LP l c L U i L c e j L M S V J W X s [ Z s ` t [ v Z ` J J J J J J J J J J J J J J J J J J J J J J J J J J J M P L M N O LP M Q c N L e O c L M f P w JK J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J J S T U O LP M Q c N L e w J } } } } } } } } } } } } } } } } } } } } } } } } O k j M f { Q j M m l m LP l O LP M Q e w JR } } } } } } } } } } } } } } } } } } } } } ƒ ƒ ˆ Š Œ Ž Œ ~ }

79 * Technion - Computer Science Department - M.Sc. Thesis MSC ! " # $ %! # $ % $ # # & ' ( )

80 $ i i g g Ç õ? â > > f f Ž Ž Ž Ž Æ > ; t ; 9 4 Ò Ç Ç Ç. Å Å X f Þ È Ç [ G G > G G > R Gi { fig { f { f { { f { { É Æ È É Æ Ç É Æ É É É Þ È Ç Technion - Computer Science Department - M.Sc. Thesis MSC Y g h g z i i i i «Ç È Ç È Ç È Ç Ç Ç Ç Ç É Ç è õ ö! " # # / = 3 * +, - ) ' &, 9 : ) ( 8 9 : / < 3 * +, - ) ' & ( 8 9 : / 5 3 / & ' ( ) * +, - ) > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > > * + ) 9 ) : ' ( 8 9 : / C E 3 / C D 3 > > > > > > > > > > > > > > * + ) ( 9 ) : ' ( 8 9 : 4 * +. & ' ( ) * +, - ) K L M E N O M P Q > > > > > > > > > > > > > > > > > > > > > > > > > > > > H 9 I : ' 7 9 J / U U V E 3 > > > > > > > > > > > > > > > > > > > > > > > > > > S 7 6 : ( ) ' T + 7 I > > > > > > > > > > > > > > > > > > > > > > > > > > > > > ' 7 9 ) 4 * +. & ( ) ' T + 7 I f f f f f f f f f f f f f f f f f f f f f f f f f f f f f \ ] ^ _ ` ] a b ] a c a e Z 6 ' \ ] p q ^ b ] a c p _ ] o ] ` a q m ` _ ` r q m q \ ] p q ^ ` ] e q _ s ] a k l p p ` e o [ n k l m v w x y f f f f f f f f f f f f f f f f f f f f f f f o ^ q o \ ] a ^ q s m \ ] a p \ ] u ^ ] ] q } ~ f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f f c l q m v ƒ ƒ y f f f f f f f f f f f f f f f f f f f k _ ` ] a m c _ o l m c _ ^ m f f f f f f f f f f f f f f f f f f f f f f m c _ ^ o m o ^ m ` \ ] a p p _ s o p _ c _ ^ p f f f f f f f f f f f f f f f f f f f f f f f f f f f f m c _ ^ p c l o c _ ] q m _ ^ ] c v ƒ ƒ y f f f f f f f f k _ ` ] a m c _ o l p _ c _ ^ ] p \ ] ^ _ ˆ \ ] c _ a } Œ Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž u _ r q \ ] e c l q ` \ ] ] r ˆ ] \ ] q _ Š ] q œ ž Ÿ Ÿ ž Ÿ ž Ÿ Ÿ Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž š ª Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž š œ ž Ÿ ž ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž š œ ž Ÿ ž ž Ž Ž Ž Ž Ž š ± ± ² ³ ² ³ µ º Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž ³ š ² ³ ³ ¹ Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž Ž» š ¼ š ± ± ½ ¾ À Á Â Ã Ä Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ š Ñ ¾ Ò Á Ó Ã Ä Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Ê Ë Ì Í Î Ï Ð Ë Î Û Á Û Ü Ý Ã Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Ô Ê Í Õ Ö Î Ø Ù Ê Ú Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ ß à Õ Ì Î Ê Ø Õ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ ß à Õ Ì Î Ê Ø Õ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ ß à Õ Ì Î Ê Ø Õ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ ã Ù Ö Ë Õ Î Ö ä Ë Õ Ì Ô Ê Í Õ å Ò Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Î à Ö Ï Î Ö ã Ù æ ç Ù æ Î ã Ù Ð Ê Í Õ Î Ë Ö é ê ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ë ì í î ï í ð ñ î ò ï ð ó Ô Ê Í Õ Î û ü ý þ ÿ ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ô ù ï ú ò ñ % > > f j f j f Ž Ž Ž Ž Ž Ž Ž Æ á Æâ á Æ á Æ á Æ ôø

81 " " " M M M M W W W W! # " " "! " #! # " " "! " # " X X \ X ] X ^ Z Z \ Z ] Z ^ N M N M N M N M N M N M N M N M N M N M N M N M Technion - Computer Science Department - M.Sc. Thesis MSC " " $ $ N O N O N [ N [ N N N N N N N _ N _ N ` N ` N ` N a % & & % & & * +, -. / 0 2 ' ( ) * +, -. / 0 2 ' ( 3 C D E F G M M M M M M M M M M M M H I = K J = > : ; < = > S T C D E U G V M M M M M M M M M M M M J K Q K J H I = K J = > H I R ; < = I Z P M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M K = J > K ; Y = Z P M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M K = J > K ; Y = Z P M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M K = J > K ; Y = Z P M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M K = J > K ; Y = Z P M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M K = J > K ; Y = Z P M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M K = J > K ; Y = Z P M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M K = J > K ; Y = Z P M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M M K = J > K ; Y = S T C c E G V M M M M M M M M M M M M M M M M M M M M M M M M M J K Q K J : ; < = K S T C c E G V M M M M M M M M M M M M J K Q K J H I = K J = > < > ; f S T C U E g G h M M M M M M M M M M M M M M M M M M M M M M M M M M M M > < K Q K " $ " " " " " $ " " " " " " O P O X O Z O O O [ O N O _ O ` O a O b [ P [ X

82 ««º 5 s o «ª a ª «j «i t ª U ƒ a _ f w ª ª ª I Technion - Computer Science Department - M.Sc. Thesis MSC ! " # $ % & ' ( ) * ' +, - )./., *! 0 $ 7 : V G D ] C \ U > ; > ; < ; = ; >? V D C K W J G D K X G W J Y C D N H O P Z O [ P O W [ O O C P G Q Q R S T C D E F E G H I J K L M N K O G E P ` C O G K I J K J H G G [ H V D C K W J G D K X G W J Y C D N H Z C O ^ O D CW F G D ^ K I G O H G M [ H I W ] [ I C I H K C ] H O K G I I D Z H I M N G I K C I E E C [ H O b c S n o p q r n k l m G E H G H E N ` C O G K O g R b h S al C J X ` C O G K O E E C [ H ƒ w x y z { } ~ ~ y u v ž Ÿ š œ Š Œ Ž ˆ ž Ÿ Ÿ Ÿ œ š Ž Ÿ š Ÿ œ Ÿ š š š Ÿ Ÿ š Ÿ œ Ÿ Ÿ Ÿ ² ± µ ² µ ± ± ± ² ³ Ÿ œ š Ÿ š š š š Ž ž Ÿ Ÿ œ š Ÿ Ÿ ¹ š Ÿ ¹ š Ž ž Ÿ ž Ÿ Ÿ Ÿ «š Ÿ

83 » 3 Ä Ä ] ] «ˆ ˆ 3 V 7 U V Ç ˆ 3 ] ] ˆ Technion - Computer Science Department - M.Sc. Thesis MSC $ " ( + 2 ' ( & ( ( 0 & * $ ( 4 " 2 $ ( " " 2 $ 8 ( 2 $, $ # ( " 9 ( ( #, / " # $ % & ' ( ( ) ( * + ' (, & + -. & $ ( &, & + $ 5 * $ 0 4 ) ' ( ( # & ( & " $ * ( 0 + ' + +, ( * 0 " 2 & 5 & $ 5 & + + (, * * 2 0. $, $ : + # $ *, [ W E I Y N F E I Œ x y P I E N X { x { x O H W X Y Z [ G \ P M N F E I O H H P Q I H K M z { } ~ z { w y v ~ { u v w x Œ } y v ~ Œ } y { x { x p q r s ' ( ) $ + ' ( " # $ & * H F I H I [ E H Z N O H N E X H O H ^ E N H M Z t u v w x y p q r s t p q r s ƒ t { } Ž { } y { Š Ž { } y p r t z { { Š x z { } ~ w y v Œ x y u v w x v ~ Œ } y { x { x µ «± ¹ ± º ¼ ½ à À Á à  Œ { p r t v x Œ } ' ( ( # & ( & ( * $ ( 4 $. - $ + $ + " ( # & ( & 7! # 4 + ' ( 5 $ " ( " # $ % ( ) * $ # ' ( 5 $ " ( " " $ * ( 0 + $ # $ " ( ; < < = < ; < C E F G F H I J E K L M N O H H P Q I E M H R O H N F E I P S K I O H I H H T O H Y N E Y I Z N F E I E N F E I Z N Y O H W E Y Z O H W E E H V M [ z { { Š x ª µ ª µ ª ª ¼ ½ À Á ¼ ½ ¾ À Á à  «É µ ª º «µ ª ««µ º N X N n o V p q r s t H F H N G p q ƒ P M E [ I r s t O H Y E Y I Z N F E I " O H W X Y Z X Z J H \ [ E W [ Y Z N O H N E X H _ `a b c e f g hi j k lm z { { } { ~ y { ~ w w { x { v Š Œ x y z p r t Œ { z { { Š x z { } ~ z { { v x { z { Œ { y z Š w Œ x y Œ { z { { Š x z à  ² ³ ¼ ½ À Á ¾  ¼ ½ à À Á ¾ Â Æ È «ª ««µ º x w Œ } y p r r š r š t { x { x ª «ª «± ¼ ½ à À Á  ¼ ½ ¾ À Á ¾  ««µ º ª º ¼ ½ ¾ Œ { p š r š t w y v x w Œ { œ ž Ÿ Ÿ À Á  Šº ± ± ª Æ

84 $ 8 / M M M! $ Technion - Computer Science Department - M.Sc. Thesis MSC % & - ( ) * - ( +.. " # % & ' ( ) * ' ( : ; 4 < = H I C L E V. % & ) *, ( + 0 % Y F K F % & ) * ' ( + % &, ( ) * + % & ' ( ) * Y D F Y D V 8 T V I W I F T S G N 4 O Q < 4 < 7 G N 4 O R < 4 ; H I J D K L F I G N P < 4 O 4 < H I J D G N R < 4 O 4 ; H I J L H I J L L I D K L F 8. G N 4 O 4 < K L F G N 4 O 4 ; G ; 7 I C E F G < 7 C D C E F L E T U 8 E V T W I V K E C I X H I F L V K F W L V 8 C E F D F H I J D K

85 u y ~ b Š 2 ª ª! " I _ y G V ª P ~ Technion - Computer Science Department - M.Sc. Thesis MSC G N F C & ) # * +, - J 3 : K L G E 6 C M F C ] c V e Y a V Z [ a [ ~ { z v Œ Ž V a _ w ~ y z Q R S T U ` V Y [ ^ 4 G _ ^ ] _ Z i w ~ } ~ y x y # $ %. & / ' 0 ( ) = Y V Z [ \ ƒ ˆ ] ^ / 0 H 5 3 F W X Q R S T U 5 C 3 D E F ; < = ; Q R S T U : ; O = f [ e Y V e _ Y a V Z V f [ e V e _ Y a V Z h ~ y r v z x š œ ƒ ˆ œ Š v { ~ w ~ x r v ~ ~ { z s q r j k o l k p : f [ c g e _ h j k l k m n q r s t u v w x y z { } ~ y ~ z r u š «š š š œ ƒ ˆ š œ ± Œ ² ³ Ÿ µ ² Œ Œ r z { w ~ z ¹ º z ~ y x r ~ } { ~ v Œ ž ž Ž Ÿ { z s v w ~ z } u w ~ s t w v w { v r w r t w z x y { ~ ƒ ˆ š «š š z x v v { z s v t t r t ~ { z v v } ~ s v Š