MDS Array Codes for Correcting a Single Criss-Cross Error

Transcription

1 1068 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 3, MAY 2000 MDS Array Codes for Correctng a Sngle Crss-Cross Error Maro Blaum, Fellow, IEEE, and Jehoshua Bruck, Senor Member, IEEE Abstract We present a famly of Maxmum-Dstance Separable (MDS) array codes of sze ( 1) ( 1), a prme number, and mnmum crss-cross dstance 3,.e., the code s capable of correctng any row or column n error, wthout a pror knowledge of what type of error occurred. The complexty of the encodng and decodng algorthms s lower than that of known codes wth the same error-correctng power, snce our algorthms are based on exclusve OR operatons over lnes of dfferent slopes, as opposed to algebrac operatons over a fnte feld. We also provde effcent encodng and decodng algorthms for errors and erasures. Index Terms Array codes, crss-cross errors, error-correctng codes, MDS codes, rank errors. I. INTRODUCTION In ths correspondence we descrbe two-dmensonal array codes that can correct errors gven by ether a row or a column n error (wthout a pror knowledge of whch one occurred). There exst codes that can do so. Moreover, the known codes are stronger n the sense that they can correct the rank of an array. The dea of usng the rank as a metrc comes from Delsarte [4]. See also Gabduln [6] and Roth [12] [14]. However, these constructons are based on fnte-feld arthmetc, as Reed Solomon codes. Therefore, for very large arrays, they may become mpractcal, snce they may need a very large lookup table. In ths correspondence, we wll present array codes that have the same error-correctng capablty n terms of rows and columns (although sometmes they cannot correct the rank) as the ones n [4], [6], and [12], but they have less complexty. The new codes are based on smple party along lnes of dfferent slopes, n the sprt of [3]. There are applcatons n whch nformaton bts are stored n n 2 n bt arrays. The error patterns are such that all corrupted bts are confned to at most some prespecfed number t of rows or columns (or both). We wll refer to such errors as crss-cross errors. Crss-cross errors can be found n memory chp arrays, where row or column falures occur due to the malfunctonng of row drvers, or column amplfers (see, for nstance [5], [8], and [9]). Another applcaton of codes correctng crss-cross errors occurs n multtrack magnetc tapes, where the errors usually occur along the tracks, whereas the nformaton unts (bytes) are recorded across the tracks. Computaton of check bts s equvalent to decodng of erasures at the check bt locatons, and n ths case these erasures are perpendcular to the erroneous tracks. There exst codes for multtrack magnetc recordng [2], [10], [11]. We need some defntons. Let E [e j ] n01 ; be an n 2 n matrx over a feld F.Acover of E s a par (X; Y ) of sets X; Y f0; 1; 111;n 0 1g, such that e j 6 0 ) (( 2 X) or (j 2 Y )) for all 0 ; j n 0 1. The sze of a cover (X; Y ) s defned Manuscrpt receved March 29, 1998; revsed August 6, Ths work was supported n part by an IBM Unversty Partnershp Award, an NSF Young Investgator Award CCR , and by a Sloan Research Fellowshp. The materal n ths correspondence was presented n part at the IEEE Internatonal Symposum on Informaton Theory, Ulm, Germany, M. Blaum s wth IBM Research Dvson, Almaden Research Center, San Jose, CA USA (e-mal: blaum@almaden.bm.com). J. Bruck s wth the Calforna Insttute of Technology, Mal Stop , Pasadena, CA USA (e-mal: bruck@paradse.caltech.edu). Communcated by E. Soljann, Assocate Edtor for Codng Technques. Publsher Item Identfer S (00) by j(x; Y )j jxj + jy j. The crss-cross weght of E, denoted by w(e), s the mnmum sze j(x; Y )j over all possble covers (X; Y ) of E. Note that a mnmum-sze cover of a gven matrx E s not always unque. The rank of E over F s never greater than ts crss-cross weght. A well-known result by Köng (see [7, Theorem 5.1.4]) states that the mnmum sze of a cover of a f0; 1g-matrx s equal to the maxmum number of 1's that can be chosen n that matrx wth no two on the same row of column. The crss-cross dstance d(a; B) between two n 2 n matrces A and B over F s defned by d(a; B) w(a 0 B): Example 1.1: Consder the array E over GF (2). It s easy to verfy that E has two covers of sze 3, namely, (f0; 2g; f1g) and (f2g; f0; 1g). Furthermore, snce the three nonzero elements on the man dagonal of E belong to dstnct rows and columns, the crss-cross weght of E must be at least 3. Therefore, w(e) 3. Let 0[c n01 j] ; be an n 2 n matrx over F, denotng the correct array to be stored, and let 0 8 E denote the array actually recorded, wth E [e j ] n01 ; standng for the error array. The crss-cross error model assumes that w(e) t for some prespecfed t. An [n 2 n; k; d] lnear array code C over a feld F s a k-dmensonal lnear space of n 2 n matrces over F wth d beng the mnmum of all crss-cross dstances between pars of dstnct matrces n C. Adoptng the termnology of conventonal lnear codes, we call d the mnmum crss-cross dstance of C. As wth regular block codes, d equals the mnmum crss-cross weght of any nonzero matrx n C. An [n 2 n; k; d] array code C can correct any pattern of s crss-cross errors together wth t crss-cross erasures f and only f 2s + t d01. The proof s agan dentcal to the proof for block codes. In ths correspondence, we present array codes wth mnmum crsscross dstance d 3. The constructons n [4], [6], and [12] operate over a feld GF (2 n ). When n s a large number, lke n holographc storage applcatons, the resultng complexty may be prohbtve. Thus we want to construct codes wth low complexty but stll havng mnmum crss-cross dstance 3. To ths end, we wll consder codes over the rng of polynomals modulo 1+x + x x, p a prme number, as n [3]. We have the followng verson of the Sngleton bound [12]. Theorem 1.1: For any [n 2 n; k; d] array code over a feld F k n(n 0 d +1): Codes meetng the Sngleton bound are called Maxmum-Dstance Separable (MDS). In the next secton, we wll construct [(p 0 1) 2 (p 0 1); (p 0 1)(p 0 3); 3] array codes, p a prme number. Accordng to Theorem 1.1, these codes are MDS. In Secton III, we prove the man propertes of the codes, /00$ IEEE

2 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 3, MAY manly, the condtons under whch they are MDS wth respect to the crss-cross dstance. We also show that, n general, our crss-cross dstance s not equvalent to the rank dstance, as wth the codes n [4], [6], and [12]. We also brefly dscuss possble generalzatons to multple partes. In Secton IV we present effcent decodng algorthms n the case of errors and erasures. We end the correspondence by drawng some conclusons. II. CONSTRUCTION OF THE CODES We gve two descrptons of the codes, one algebrac, the other geometrc. In the sequel, p denotes a prme number and l a number such that 2 l p 0 2. Let us start wth the algebrac descrpton. The entres of the elements of the code are polynomals modulo 1+x+111+x. Let be a root of 1+x x and, moreover, assume that 1+x x s the mnmal polynomal of. Then, a party-check matrx of code C(p; l) s gven by H(p; l) l 2l 111 ()l : (1) Now, assume that where (c 0();c 1(); 111;c ()) 2C(p; l) c j () c j : The codewords may be nterpreted as (p 0 1) 2 (p 0 1) arrays (c j) 0;j such that each symbol n a codeword s gven by a column n the array. We denote by C(p; l) the bnary code of (p 0 1) 2 (p 0 1) arrays derved from C(p; l). Normally, and n order to smplfy notaton, we wll add an magnary 0-row and an magnary 0-column to the arrays n C(p; l). So, the codewords may be nterpreted as p 2 p arrays (c j ) 0;j, such that c ;j c ; 0 for 0 ; j p 0 1. Also, from now on, we take all the subndces modulo p. We apologze for ths abuse, but the notaton s somewhat awkward f we want to denote the modulo p subndces every tme. We wll see that a geometrc nterpretaton of code C(p; l) as derved from code C(p; l) defned by (1), s as the set of arrays havng ether even or odd party along lnes of slope 1 and l. Ths wll be made clear by the followng lemma. Lemma 2.1: Vector (c 0();c 1(); 111;c ()) belongs n C(p; l), where c j () f and only f, for each 0 p 0 1 where b 2 GF (2). c j c 0j;j b (2) c 0lj;j b; (3) The geometrc meanng of (2) and (3) s the followng: we have party n the array along lnes of slope 1 and l, respectvely. Ths party can be ether even or odd: t s even when b 0, and odd when b 1. Before provng Lemma 2.1, let us gve an example. Example 2.1: Let us consder p 5and l 2,.e., code C(5; 2) or C(5; 2) as bnary arrays. Accordng to (1), a party-check matrx of the code s gven by H(5; 2) where Now, consder the followng codeword n C(5; 2): c() ( + 3 ; ; ; ): The reader can easly verfy that c()(h(5; 2)) T 0(H T denotes the transpose matrx of H), therefore, c() 2C(5; 2). Wrtng c() as the array n C(5; 2) n whch the columns correspond to the entres of c(), we obtan (remember that we are addng an extra 0-row and an extra 0-column) The reader can verfy that we have odd party along the lnes of slope 1 and of slope 2, as predcted by (2) and (3). We are ready now to prove Lemma 2.1. Proof of Lemma 2.1: Remember that a 0-row and a 0-column have been added to the (c ;j) array,.e., c ;j 0and c ; 0.If c() (c 0 ();c 1 (); 111;c ()) 2C(p; l) takng the nner product of c() wth the second row of H(p; l) and rememberng that the subndces are taken modulo p, we obtan 0 c j () lj t0 +ljt t0 c ;j c ;j +lj c ;j lj t c t0lj;j t (4) where (4) s obtaned usng the fact that p 1and groupng together the terms correspondng to the same power t, 0 t p 0 1. Reducng modulo t0 c t0lj;j t 0

3 1070 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 3, MAY 2000 f and only f c t0lj;j b l ; b l 2 GF (2) for each 0 t p 0 1, whch nearly concdes wth (3). Smlarly, by repeatng ths procedure wth l 1, we obtan c t0j;j b 1 ; b 1 2 GF (2) for each 0 t p 0 1, whch nearly concdes wth (2). In order to complete the proof, we need to show that b 1 b l. Notce that, takng the XOR of all the elements n the array n two dfferent ways, we obtan completng the proof. b 1 pb 1 t0 t0 pb l b l ; c t0j;j c t0lj;j We next consder a code C 0 (p; l) whose elements are the transposes of the arrays n C(p; l). The followng lemma connects the two codes. Lemma 2.2: Consder the code C(p; l) of bnary () 2() arrays defned by (2) and (3), and let C 0 (p; l) be the code whose elements are the transposes of the elements n C(p; l). Then, C 0 (p; l) C(p; 1l),.e., the arrays n C 0 (p; l) have even or odd party along lnes of slope 1 and 1l. Algebracally, a party-check matrx for the correspondng code C 0 (p; l) s gven by H 0 (p; l) (5) 1 1l 2l 111 ()l Proof: Let (c ;j) 2 C(p; l) and (c;j) 0 2 C 0 (p; l) such that c 0 ;j c j;. Accordng to (4) and Lemma 2.1 +ljt c ;j b; b 2 GF (2); for 0 t p 0 1: Dvdng the subndces by l, ths occurs f and only f j+(l)tl f and only f j+(l)t c ;j b; b 2 GF (2); for 0 t p 0 1 c ;j b; b 2 GF (2); for 0 t p 0 1 snce dvdng by l each 0 t p 0 1 s a 1-1 functon. Ths occurs f and only f f and only f c ;t0(l) b; b 2 GF (2); for 0 t p 0 1 c 0 t0(l); b; b 2 GF (2); for 0 t p 0 1: By Lemma 2.1, (c 0 ;j) 2 C(p; 1l) and the party-check matrx correspondng to C 0 (p; l) s gven by (5). The next example llustrates Lemma 2.2. Example 2.2: Consder code C(5; 2) as n Example 2.1. Accordng to Lemma 2.2, snce 12 3modulo5, a party-check matrx of the code C 0 (5; 2) s gven by H 0 (5; 2) : The transpose of the array gven n Example 2.1 s We can see that the array above has odd party along lnes of slope 1 and 3. An easy observaton s, code C(p; l) s MDS,.e., f we erase any two columns n an array (c ;j ) 2 C(p; l), regardng these two columns as elements modulo 1+x x, they wll be recovered by the code. Of course, the same s true for the correspondng code C 0 (p; l) n whch we dentfy the rows of the arrays wth elements modulo 1+ x x : any par of erased rows wll be recovered. Let us prove these facts n the next lemma. Lemma 2.3: Code C(p; l) s MDS. Proof: We have to show that any two columns n H(p; l) are lnearly ndependent,.e., any determnant n H(p; l) as gven by (1) s nvertble. Notce that det j +lj + j+l l lj j+l (l01)(j0) +1 ; and snce 1 l 0 1 p 0 3 and 1 j 0 p 0 2, (l 0 1)(j 0 ) 6 0 (mod p), and (l01)(j0) 61. Moreover, gcd(x t (x s +1); 1+x x )1; for s 6 0 (mod p) [3], thus j+l ( (l01)(j0) +1)s nvertble. However, our goal s to show that the bnary code C(p; l) of (p 0 1) 2 (p 0 1) arrays s MDS wth respect to the crss-cross dstance. To ths end, we have to show that any erased row together wth any erased column wll be unquely recovered. Ths wll not happen for every code C(p; l). Actually, t wll occur f and only f l s prmtve n GF (p),.e., the powers of l generate all the nonzero elements n GF (p). For nstance, 2 s prmtve n GF (5), but not n GF (7). However, 3 s prmtve n GF (7). Thus C(7; 2) s not MDS wth respect to the crss-cross dstance, but C(7; 3) s. We wll prove these propertes n the next secton. III. MAIN PROPERTIES Before provng our man theorem, let us gve some examples of codes C(p; l) for whch 2 s not prmtve n GF (p).

4 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 3, MAY Example 3.1: Consder the followng array n C(7; 2) (to whch a 0-row and a 0-column have been added): We can see that the array above has even party on lnes of slope 1 and 2, therefore, t belongs n C(7; 2). Ths array has crss-cross weght 2,soC(7;2) s not MDS wth respect to the crss-cross dstance. Example 3.2: Consder the array n C(17;2) (to whch a 0-row and a 0-column have been added) shown at the top of ths page. As n Example 3.1, we can see that the array above has even party on lnes of slope 1 and 2, therefore, t belongs n C(17;2). Ths array has crss-cross weght 2, soc(17;2) s not MDS wth respect to the crss-cross dstance. Notce that 2 s not prmtve nether n GF (7) nor n GF (17). The next theorem s our man result. Its proof s based on the generalzaton of Examples 3.1 and 3.2. Moreover, we'll refer to these two examples n the proof as an llustraton. Theorem 3.1: Code C(p; l) has mnmum crss-cross dstance 3 f and only f l s prmtve n GF (p). Proof: )) Assume that l s not prmtve n GF (p). We wll exhbt an array wth a cover of sze 2. Consder the set S(p; l) f1 l 0 ;l

5 1072 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 3, MAY 2000 l 1 ;l 2 ; 111;l s01 g such that these powers are taken modulo p and l s 1. Snce l s not prmtve n GF (p); S(p; l) 6 GF (p) 0f0g. There are two cases: p S(p; l) and p S(p; l). We defne a set U 6 ; as follows: U S(p; l); f p S(p; l) U GF (p) 0 S(p; l) 0f0g; f p S(p; l): Next, consder the array (c ;j ) 0;j of crss-cross weght 2 such that c ;0 1; f 2 U c 0;j 1; f j 2 U c ;j 0; elsewhere. As an example of the case p S(p; l), consder p 7and l 2. Then, S(7; 2) U f1; 2; 4g (notce that 6 62 S(7; 2)). The array (c ;j) n ths case s depcted n Example 3.1. As an example of the case p S(p; l), consder p 17and l 2. Then and S(17; 2) f1; 2; 4; 8; 9; 13; 15; 16g U f3; 5; 6; 7; 10; 11; 12; 14g: The array (c ;j ) n ths case s depcted n Example 3.2. In order to reach a contradcton, we need to prove that the array (c ;j ) s n C(p; l). Frst we make the followng observaton: snce U consst of all the powers of l modulo p, the functon! l s closed n U and s 1-1. Next we show that all the lnes of slopes 1 and l have party 0 and use Lemma 2.1 to complete the proof. Notce that the only nonzero entres n (c ;j) are as defned by U n the frst row and frst column. Namely, for every 0 p 0 1 c 0j;j 0 because the frst row and frst column are dentcal and the entres not n the unon of the frst row and the frst column are zero. Moreover, for every 0 p 0 1 c 0lj;j c ;0 + c 0;t where 0 lt 0. But c ;0 + c 0;t 0f and only f c ;0 c 0;t, whch, by the defnton of the array (c ;j ), s equvalent to the followng statement: 2 U, t 2 U: Snce 0 lt 0, then lt. Ift 2 U, snce the functon j! lj s closed n U, then 2 U. Conversely, f 2 U, then t l s01, and agan, snce j! lj s closed n U, then t 2 U. Hence, by Lemma 2.1, (c ;j) s n C(p; l). () Assume now that l s prmtve n GF (p). We have already proven n Lemma 2.3 that codes C(p; l) and C 0 (p; l) are MDS, thus any par of columns or of rows n an array n C(p; l) can be unquely retreved. We need to show now that the same s true for any row and column. Let (c s;t ) 0s;t be an array n C(p; l), and assume that c s;t 0 whenever s 6 and t 6 j. As usual, assume that the last row s an magnary 0-column as well as the last column. We wll show that also c ;t 0and c s;j 0. Snce c ;j 0, then c ;j0(+1) b, snce they belong n the same dagonal, and b 0or b 1, accordng to the party of the dagonals and the lnes of slope l. Snce c ;j0(+1) b, then c 0l(+1);j 0, snce they belong n the same lne of slope l. By nducton, assume that c 0for 1 r. Then, c 0l (+1);j ;j0l (+1) b, snce c 0l and c (+1);j ;j0l (+1) belong n the same dagonal. Ths mples that c 0l (+1);j 0, snce c ;j0l (+1) and c 0l (+1);j belong n the same lne of slope l. Therefore, c 0l (+1);j 0 and c ;j0l (+1) b for 0 r. Snce l s prmtve n GF (p), then there s an r such that l r (j +1)( +1). For that r; j 0 l r ( +1) p 0 1, butc ; 0b, thus c ;j0l (+1) 0for 0 r p 0 2. Agan, usng the fact that l s prmtve n GF (p), we conclude that c s;j 0for s 6 and c ;t 0for t 6 j. Fnally, c ;j 0snce the dagonals and lnes of slope l have even party. Ths completes the proof. In [4], [6], and [12], the authors prove that ther constructon can correct the rank of an array when the rank s used as a metrc. We have seen that the rank s a more powerful metrc than the crss-cross dstance consdered here. Therefore, a legtmate queston s: can the codes C(p; l) also correct the rank? The answer s no, n general. For nstance, consder the followng array n C(7; 3): Ths array has crss-cross weght 3 but rank 2. So, the queston s, under whch condtons the codes C(p; l) can correct the rank? The answer s, whenever the polynomal 1+x x s rreducble,.e., when 2 s prmtve n GF (p), then the nonzero arrays n C(p; l) have rank at least 3. In ths case, the rng of polynomals modulo 1+x x s a feld. Explctly Theorem 3.2: Every nonzero array n code C(p; l), 2, and l prmtve n GF (p), has rank at least 3. Proof: Let 0 be a nonzero array n C(p; l) wth rank 2, therefore, we can wrte where s a (p 0 1) 2 2 array and 0UD U (u ;j ) D (d ;j ) s a 2 2 (p 0 1) array, and both U and D have rank 2. Lookng at each column of 0 as an element modulo 1+x x, we obtan that the jth column of 0 s gven by 1 t0 u ;td t;j 1 t0 d t;j u ;t :

6 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 3, MAY Usng the party-check matrx H(p; l) defned by (1), we obtan that 1 d t;j u ;t j 0 t0 1 d t;j t0 whch can be rearranged as 1 t0 1 t0 d t;j j d t;j lj u ;t lj 0 u ;t 0 (6) u ;t 0: (7) Snce 2 s prmtve n GF (p), then l s a power of 2, and thus d t;j lj so (6) and (7) can be wrtten as d0;j j d0;j j Let D t become l u ;0 + u ;0 + dt;jj and U t d t;j j d1;j j d1;j j D0U0 + D1U1 0 D0U0 l + D1U1 l 0: l l u ;1 0 (8) u ;1 0: (9) u;t, then (8) and (9) Snce we are n a feld, the system above has a nontrval soluton f and only f det D0 D1 D0 l D1 l 0 and, snce D0 6 0and D1 6 0, ths can only occur f D l D l01 1 0(D0 + D1) l01 : Thus D0 D1, a contradcton, snce we are assumng that matrx D has rank 2. We have not found an adequate generalzaton of the codes C(p; l) to more than two partes. It s an open problem f such a generalzaton exsts. However, f l 2and 2 s prmtve n GF (p), then the code wth r partes defned by the party-check matrx ()2 H () (2 ) 111 ()2 s MDS wth respect to the rank. Ths code s a partcular case of the ones descrbed n [4], [6], and [12]. IV. ENCODING AND DECODING In ths secton we gve encodng and decodng algorthms for errors and erasures. If there are no ndcatons of erased rows or columns n a receved array, the decoder attempts to correct ether a column or a row. In the case of erasures, the decoder can correct ether two erased columns, two erased rows, or an erased column together wth an erased row. The encodng s a partcular case of the decodng of two erased columns. We examne all these cases separately. Let us start wth errors. Assume that (r ;j ) s a receved array, possbly a nosy verson of an orgnally stored array (c ;j) 2 C(p; l). Moreover, assume that ether a column or a row n (r ;j ) are n error. The frst step s fndng the column syndromes usng the party-check matrx H(p; l) gven by (1). To ths end, we defne r() (r0();r1(); 111;r ()) as r j () Estmatng r()(h(p; l)) T, we obtan r ;j : r j () j S1() (10) r j () jl S l () (11) Let us pont out that multplyng a vector of length p 0 1 by a power t s equvalent to rotatng the vector t tmes modulo 1+x+111+x. For nstance, let p 5, and consder the vector (a0;a1;a2;a3). Asa polynomal n, ths vector s wrtten as a0 + a1 + a2 2 + a3 3.If we multply ths polynomal by 2, snce 5 1and , we obtan (a3 + a2)+a2 +(a0 + a2) 2 +(a1 + a2) 3, whch n vector form s (a3 +a2;a2;a0 +a2;a1 +a2). Ths s what we call rotatng the vector (a0;a1;a2;a3) two tmes modulo 1+x + x 2 + x 3 + x 4. Alternatvely, we could have taken the vector and added a 0 to t to obtan (a0;a1;a2;a3; 0). Rotatng ths extended vector twce to the rght (.e., rotaton modulo 1+x 5 ), we obtan (a3; 0;a0;a1;a2).To reduce t modulo 1+x+x 2 +x 3 +x 4 we need to add the last coordnate to each of the frst four, gvng the vector (a3+a2;a2;a0+a2;a1+a2). Ths provdes a computatonally smple method for multplyng by. For detals, we refer the reader to [3]. If there was an error E n, say, column t, and all the other columns are correct, (10) and (11) gve E t S1() (12) E tl S l (): (13) We need to fnd the error locaton t and the error tself E. Solvng (12) and (13), we obtan (l01)t S1() S l (): (14) So, the decoder apples repeatedly the operaton (l01)j S1() for 0 j p 0 2 untl t fnds a j t satsfyng (14). If there s such a t, then the decoder declares an error n column t, and the value E of the error, from (12), s gven by E 0t S1(). The fnal step s addng E to column t, completng the decodng. However, f there s no t satsfyng (14), the decoder wll assume that there was a row error, and wll repeat the procedure but ths tme for rows. Specfcally, the decoder now consders r 0 () (r 0 0();r 0 1(); 111;r 0 ()) as r 0 () r ;j j :

7 1074 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 3, MAY 2000 Estmatng r 0 ()(H(p; 1l) T, we obtan the row syndromes Notce that r() 0 S1() 0 (15) r() 0 l Sl(): 0 (16) S1() 0 r() 0 r ;j j r ;j j r j() j S1() so S 0 1() does not need to be calculated once S1() s known. If there was an error E 0 n, say, row s, and all the other rows are correct, (15) and (16) gve E 0 s S 0 1() (17) E 0 sl S 0 l(): (18) Solvng (17) and (18), we obtan ((1l)01)s S1() 0 Sl(): 0 (19) Now, the decoder apples repeatedly the operaton ((1l)01)j S1() 0 for 0 j p 0 2 untl t fnds a j s satsfyng (19). If there s such an s, then the decoder declares an error n row s, and the value E 0 of the error, from (12), s gven by E 0 0s S1(). 0 The fnal step s addng E 0 to row s, completng the decodng. If there s no s satsfyng (19), then the decoder declares an uncorrectable error. Let us llustrate the decodng procedure wth an example. Example 4.1: Consder code C(7; 3) and assume that the followng array s receved: The values S1(); S3(); S 0 1() and S 0 3() are gven by and (S1();S3()) r()(h(7; 3)) T (S 0 1();S 0 3()) r 0 ()(H(7; 5)) T : Performng these operatons, we obtan S 0 1() S1() S3() S 0 3() : Frst we check f there s a column error. Usng (14), we have to check f there s a t such that 2t S1() S3(). We can verfy that there s no such t, so we concentrate next on rows. Usng (19), we have to check f there s an s such that 4s S1() 0 S3(). 0 We can verfy that for s 3 12 S1() 0 5 ( ) S 0 3() so there s an error n row 3. From (17), ths error s gven by E 0 0s S 0 1() 4 ( ) : Addng ths error value to locaton 3 of r 0 (), we obtan c 0 () (1+ 4 ; 2 ; ; ; ; ): Each of the entres of c 0 () represents a row n the array, so the corrected array s gven by The correspondng r() and r 0 () are gven by r() (1; ;; ; ; ) r 0 () (1+ 4 ; 2 ; ;+ 3 ; ; ): Notce that 13 5modulo7,so H(7; 3) H 0 (7; 3) H(7; 5) and Let us formally wrte the algorthm descrbed above. Algorthm 4.1 (Decodng Algorthm for a Row or a Column n Error): Assume that (r ;j ) s a receved array, possbly a nosy verson of an orgnally stored array n C(p; l), where l s prmtve n GF (p). Assume that ether a column or a row n (r ;j) are n error. Defne r() (r0();r1(); 111;r ()), where r j () r ;j and r 0 () (r 0 0();r 0 1(); 111;r 0 ()), where r 0 () r ;j j :

8 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 3, MAY Fnd S 1() S1() 0 accordng to (10), S l () accordng to (11), and () accordng to (16). Then S 0 l If there s a t such that S ()S (), then: Let E S ();c () r () for j 6 t and c () r ()+E, where c () c, output (c ) and stop. Else, f there s no t such that If there s an s such that S ()S (), then: S ()S (), then: Let E S ();c ()r () for 6 s and c () r ()+E ; where c () c, output (c ) and stop. S ()S (), then declare an uncor- Else, f there s no s such that rectable error and stop. Next we concentrate on erasures. Frst, assume that two erasures have occurred n columns s and t, 0 s < t p 0 2. In order to compute the column syndromes accordng to (10) and (11), we assume that r s () r t () 0. Then, we have to fnd the mssng elements E s and E t. In ths case, (10) and (11) gve E s s + E t t S 1() E s ls + E t lt S l (): Solvng the lnear system above, we obtan (l01)(t0s) +1 E s l(t0s)0t S 1() + 0sl S l () (20) (l01)(t0s) +1 E t 0t S 1 () + 0(l01)s0t S l (): (21) Solvng effcently recursons of the type ( j +1)A B modulo 1+x x was done n detal n [3]. Certanly, a soluton s guaranteed to occur snce snce, as proved n [3] gcd(1 + x j ; 1+x x )1 therefore, j +1s nvertble. Let us llustrate the case of two erased columns wth an example. Example 4.2: As n Example 4.1, consder code C(7; 3) and assume that the followng array s receved (the? sgns denote erased bts): 1 0? 0? ? 0? ? 0? ? 0? ? 0? ? 1? 1 0 Therefore, columns 2 and 4 have been erased. The receved vector r() can be wrtten as r() ( ; ; 0; 5 ; 0; ): Performng (S 1();S 3()) r()(h(7; 3)) T as n Example 4.1, we obtan S 1 () S 3() : Applyng (20) and (21) wth l 3;s2;t4; and the values of S 1 () and S 3 () above, we obtan ( 4 +1)E ( 4 +1)E : For the sake of completeness, let us solve the recurson ( 4 +1)E : For detals of the method, see [3]. Let Then E 4 x 0 + x 1 + x x x x 5 5 : (1 + 4 )E 4 (x 3 + x 2 + x 0 )+(x 4 + x 2 + x 1 ) + x 5 2 +(x 2 + x 3) 3 +(x 0 + x 2 + x 4) 4 +(x 1 + x 2 + x 5 ) : Solvng ths system recursvely, we obtan therefore, Smlarly, f x 5 1 x 1 + x 2 1 x 4 1 x 0 + x 2 1 x 3 0 x 2 1 x 1 0 x 0 0 E : E 2 x 0 + x 1 + x x x x 5 5 we have to solve the recurson (1 + 4 )E 2 (x 3 + x 2 + x 0 )+(x 4 + x 2 + x 1 ) + x 5 2 +(x 2 + x 3) 3 +(x 0 + x 2 + x 4) 4 +(x 1 + x 2 + x 5 ) : Proceedng lke n the prevous case, ths gves Therefore, the decoded array s E 2 2 : Notce that the lnes of slope 1 and 3 have even party.

9 1076 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 3, MAY 2000 Let us pont out once more that the encodng s a partcular case of the erasure decodng descrbed above: we choose two columns for the redundancy, say, columns p 0 3 and p 0 2 (the last two columns n the array), and usng the nformaton n the frst p 0 3 columns, we reconstruct the two redundant columns. Let us wrte down formally the algorthm for decodng of two erased columns. Algorthm 4.2 (Decodng Algorthm for Two Erased Columns): Assume that (r ;j ) s a receved array from an orgnally stored array n C(p; l), where two columns, 0 s<t p 0 2,have been erased. Defne r() (r 0 ();r 1 (); 111;r ()), where r j () r ;j ; for j 6 s; t and r s () r t () 0: Fnd S 1() accordng to (10) and S l () accordng to (11). Then, let E s be the soluton of the recurson gven by (20) and E t the soluton of the recurson gven by (21). Defne c j () r j() for j 6 s; t; c s () E s and c t () E t, where c j() c ;j output (c ;j ) and stop. If two rows are erased, the procedure s analogous, except that we have to consder now the syndromes S1() 0 and S 0 l () and replace l by 1l. Formally Algorthm 4.3 (Decodng Algorthm for Two Erased Rows): Assume that (r ;j) s a receved array from an orgnally stored array n C(p; l), where two rows, 0 s < t p 0 2, have been erased. Defne r 0 () (r0();r 0 1(); 0 111;r()), 0 where r 0 () r ;j ; for j 6 s; t and r 0 s() r 0 t() 0: Fnd S1() 0 accordng to (15) and S 0 l () accordng to (16). Then, let E 0 s and E 0 t be the soluton of the followng recursons: ((1l)01)(t0s) +1 E 0 s (1l)(t0s)0t S 0 1() + 0s(1l) S 0 l() ((1l)01)(t0s) +1 E 0 t 0t S 0 1() + 0((1l)01)s0t S 0 l(l): Defne c 0 () r 0 j () for 6 s; t; c 0 s() E 0 s and c 0 t() E 0 t, where c() 0 c ;j output (c ;j ) and stop. The last case we need to consder n order to complete the decodng of two erasures, s the case n whch a row and a column have been erased. In ths case, we need to assume that l s prmtve n GF (p), an assumpton that was not necessary n the decodng of two erased columns or two erased rows. Assume then that (r ;j) s a receved array where row s and column t have been erased, 0 s; t p 0 2. We want to fnd the values r s;j and r ;t. As usual, we assume ntally that those values are 0 n order to calculate the syndromes, and we also assume that a 0-row and a 0-column have been added to the array. From (10) and (11), let us start by estmatng, for each 0, the p syndromes of slope 1 and l, respectvely, as follows: S (l) r 0j;j (22) r 0lj;j: (23) The reader may ask, what s the relatonshp between these syndromes, and the column syndromes gven by (10) and (11)? Proceedng lke n Lemma 2.1, we easly fnd out that and S 1 () ( + ) S l () (S (l) + S (l) ) : Therefore, (22) and (23) provde a computatonally effcent method to fnd (10) and (11). Let b be the unknown party of lnes of slope 1 and l. Proceedng nductvely lke n the f part of the proof of Theorem 3.1, for each 0 j, we obtan the followng recurson: r s;t0l (s+1) S(1) s+t0l (s+1) + r s0l (s+1);t + b (24) r s0l (s+1);t S(l) s+lt0l (s+1) + r s;t0l (s+1) + b (25) Therefore, applyng the recurson repeatedly, for each 0 j p 0 2, we obtan r s;t0l (s+1) s+t0l (s+1) + S (l) s+lt0l (s+1) j1 In partcular, let be the unque value n GF (p) such that + b: (26) l t +1 s +1 : (27) Snce l s prmtve n GF (p), we know that there exsts such an. Replacng (27) n (26), snce r s;t01 0;bs gven by b s+t0l (s+1) + S (l) (28) s+lt0l (s+1) j1 Thus the recurson gven by (24) and (25) provdes the soluton, snce b s now known. Fnally, r s;t + b (29) s+t completes the decodng. Let us llustrate the procedure wth an example. Example 4.3: As n Example 4.2, consder agan C(7; 3). Assume that we receve the followng array: 1 0? ? ? ? ?????? 0 1 0?

10 IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 46, NO. 3, MAY n whch row s 4and column t 2have been erased. The frst step s fndng the syndromes accordng to (22) and (23). Takng as zero the symbols denoted by?, ths gves Accordng to (27) 0 0 S (3) S (3) S (3) S (3) S (3) S (3) S (3) 6 0: l 3 (t +1)(s +1)35 2 snce 15 3n GF (7). Solvng for 3 2, we conclude that 2. From (28), replacng l 3;2;s4; and t 2; we obtan b S (3) 2 + S (3) : We are ready now for the fnal recurson gven by (29), (24), and (25) r 4;2 6 0 r 4;4 1 + r 6;2 1 (j 0) r3;2 S (3) 2 + r4;4 0 (j 0) r 4;1 5 + r 3;2 1 (j 1) r 1;2 S (3) 0 + r 4;1 0 (j 1) r4;6 3 + r1;2 0 (j 2) r2;2 S (3) 1 + r4;6 1 (j 2) r 4;0 4 + r 2;2 0 (j 3) r 5;2 S (3) 4 + r 4;0 0 (j 3) r4;3 0 + r5;2 0 (j 4) r 0;2 S (3) 6 + r 4;3 0 (j 4) r 4;5 2 + r 0;2 0 (j 5) r 6;2 S (3) 5 + r 4;5 0 (j 5) Notce that, for j 2, we already know that r4;6 0, snce ths value s n the magnary seventh 0-column that has been added. However, ths fact was exploted n the calculaton of the party bt b gven by (28). The fnal decoded array s thus gven by Let us end ths secton by wrtng formally the decodng algorthm for an erased row together wth an erased column. Algorthm 4.4 (Decodng Algorthm for an Erased Row and an Erased Column): Assume that (r ;j ) s a receved array from an orgnally stored array n C(p; l); lprmtve n GF (p), where row s and column t; 0 s; t p 0 2; have been erased. For each 0 p 0 1, fnd the syndromes gven by (22) and (23). Then, f s such that l (t +1)(s +1), determne b accordng to (28). Fnally, fnd r s;t accordng to (29) and the rest of the values accordng to the recurson gven by (24) and (25). V. CONCLUSION We presented a famly of ()2() array codes, p a prme, that can correct any row or any column n error. The constructon s based on takng all the arrays wth even or odd party along lnes of slope 1 and of slope l, l prmtve n GF (p). Known codes n the lterature dffer from our codes n the sense that they can correct errors defned by the rank of an array. Our codes can also correct the errors defned by the rank when 2 s prmtve n GF (p). However, the man new feature of our codes s ther lower encodng/decodng complexty, as ther encodng/decodng algorthms are based on smple XOR operatons, n contrast to known codes that requre operatons over fnte felds. Although we presented our results for bnary codes, they may be trvally extended to any feld. REFERENCES [1] M. Blaum and P. G. Farrell, Array codes for cluster-error correcton, Electron. Lett., vol. 30, no. 21, pp , [2] M. Blaum and R. J. McElece, Codng protecton for magnetc tapes: A generalzaton of the Patel-Hong code, IEEE Trans. Inform. Theory, vol. IT-31, pp , Sept [3] M. Blaum and R. M. Roth, New array codes for multple phased burst correcton, IEEE Trans. Inform. Theory, vol. 39, pp , Jan [4] P. Delsarte, Blnear forms over a fnte feld, wth applcatons to codng theory, J. Comb. Theory Ser. A, vol. 25, pp , [5] S. A. Elknd and D. P. Seworek, Relablty and performance of errorcorrectng memory and regster codes, IEEE Trans. Comput., vol. C-29, pp , [6] E. M. Gabduln, Theory of codes wth maxmum rank dstance, Probl. Inform. Transm., vol. 21, no. 1, pp. 3 16, [7] M. Hall Jr., Combnatoral Theory. Waltham, MA: Blasdell, [8] L. Levne and W. Meyers, Semconductor memory relablty wth error detectng and correctng codes, Computer, vol. 9, pp , Oct [9] W. F. Mkhal, R. W. Bartoldus, and R. A. Rutledge, The relablty of memory wth sngle-error correcton, IEEE Trans. Comput., vol. C-31, pp , [10] A. M. Patel and S. J. Hong, Optmal rectangular code for hgh densty magnetc tape, IBM J. Res. Develop., vol. 18, pp , [11] P. Prusnkewcz and S. Budkowsk, A double-track error-correcton code for magnetc tape, IEEE Trans. Comput., vol. C-19, pp , [12] R. M. Roth, Maxmum rank array codes and ther applcaton to crsscross error correcton, IEEE Trans. Inform. Theory, vol. 37, pp , Mar [13], Tensor codes for the rank metrc, IEEE Trans. Inform. Theory, vol. 42, pp , Nov [14], Probablstc crsscross error correcton, IEEE Trans. Inform. Theory, vol. 43, pp , Sept whch concdes wth the one n Example 4.2.