European Journal of Combinatorics

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1 European Journal of Combnatorcs 0 (009) Contents lsts avalable at ScenceDrect European Journal of Combnatorcs journal homepage: Tlngs n Lee metrc P. Horak 1 Unversty of Washngton, Tacoma, Tacoma, WA 98465, USA a r t c l e n f o a b s t r a c t Artcle hstory: Receved 6 February 007 Accepted 8 Aprl 008 Avalable onlne 4 June 008 Graver et al. proved [S. Graver, M. Mollard, Ch. Payan, On the exstence of three-dmensonal tlng n the Lee metrc, European J. Combn. 19 (1998) ] that there s no tlng of the three-dmensonal space R wth Lee spheres of radus at least. In partcular, ths verfes the Golomb Welch conjecture for n =. Špacapan, [S. Špacapan, Non-exstence of face-to-face four-dmensonal tlng n the Lee metrc, European J. Combn. 8 (007) 17 1], usng a computer-based proof, showed that the statement s true for R 4 as well. In ths paper we ntroduce a new method that wll allow us not only to provde a short proof for the four-dmensonal case but also to extend the result to R 5. In addton, we provde a new proof for the three-dmensonal case, just to show the power of our method, although the orgnal one s more elegant. The man ngredent of our proof s the non-exstence of the perfect Lee -error correctng code over Z of block sze n =, 4, Elsever Ltd. All rghts reserved. 1. Introducton Let (C, d) be a metrc space. Then a code s any subset M of C, M. The elements of C wll be called words, whle elements of M wll be referred to as codewords. The most common metrc n codng theory s the Hammng metrc. In ths paper we deal wth another frequently used metrc, the so-called Lee metrc ( =the zg zag metrc, the Manhattan metrc). The Lee metrc d L n R n s gven by d L (U, V) = n =1 u v, where U = (u 1, u,..., u n ), V = (v 1, v,..., v n ). As usual Z wll stand for the set of ntegers. The perfect Lee t-error correctng code over Z of block sze n, denoted PL(n, t), s a set M Z n of codewords so that each word A Z n s at Lee dstance at most t from exactly one codeword n M. Snce PL(n, t) code can be seen as a partton of Z n nto spheres E-mal address: horak@u.washngton.edu. 1 Tel.: /$ see front matter 008 Elsever Ltd. All rghts reserved. do: /j.ejc

2 P. Horak / European Journal of Combnatorcs 0 (009) Fg. 1. wth radus t centered at codewords, only a small step s needed to get a geometrcal nterpretaton of PL(n, t) codes. Consder the space R n. The n-cube centered at X = (x 1,..., x n ) R n s the set: C(X) = {Y = (y 1,..., y n ), y = x + α, where 1 α 1 }. By a Lee sphere of radus r n Rn, L(n, r), centered at O we understand the unon of n-cubes centered at Y, where d L (O, Y) r, and Y O has nteger coordnates. Fnally, a Lee sphere of radus r n R n centered at X R n s a translaton of L(n, r) centered at O along the coordnate axes so that O s mapped on X. Clearly, a PL(n, t) code exsts f and only f there s a tlng of R n by Lee spheres of radus t. The Lee spheres L(, 1), L(, ), L(, 1), and L(, ) are depcted n Fg. 1. The most famous and ntensvely studed problem n the area of Lee codes s the Golomb Welch conjecture. In [] t s shown that PL(n, 1) code exsts for all n 1, and PL(, t) code exsts for all t 1. In addton, t s proved there that there s no P(, ) code, and that there are no PL(, a n ) codes, where a n s not explctly specfed. The authors conjectured: Conjecture 1. Golomb Welch: There are no PL(n, t) codes for n > and t > 1. There are many results supportng the conjecture. The strongest one was proved by Post [8]: Theorem. PL(n, t) codes do not exst for n = and t ; for 4 n 5 and t n ; and for n 6 and t n 1 4 ( ). In the fnal remark Post states that, by usng a computer to evaluate coeffcents of the Taylor seres of a sutable functon, t s possble to show that there are no perfect t-error correctng codes for 6 n 10 and t 1 (9n 15); and for 11 n 05 and t 1 (9n 14). The reader nterested n the non-exstence results for Lee codes over fnte sets s referred to [1,,7], and also to [9] for the sze of the largest Lee codes over a fnte set. It s speculated n [6] that the most dffcult cases to prove n the Golomb Welch conjecture are those for t = because they are the threshold cases (PL(n, 1) codes do exst). The Golomb Welch conjecture has been verfed there for the two smallest opened cases: Theorem. There s no PL(n, ) code for n = 5 and 6.

3 48 P. Horak / European Journal of Combnatorcs 0 (009) Thus, the Golomb Welch conjecture has been verfed for all pars (n, t) where n 6. In [4,5], the authors prove, for the three-dmensonal case, a result even stronger than conjectured by Golomb and Welch. They formulate t n terms of tlngs of R by Lee spheres. It s shown n [4] that: Theorem 4. There s no tlng of R wth Lee spheres of rad at least two, even wth dfferent rad. Thus, as a specal case, they showed that there s no PL(, t) code for any t. The authors provde a very elegant a pcture says t all proof. Yet, a stronger result s proved n [5], a sequel to [4], where t s shown that there s no tlng of R wth Lee spheres f radus of at least one sphere s greater than one. Recently, Špacapan [10] extended Theorem 4 to the four-dmensonal case. Theorem 5. There s no tlng of R 4 wth Lee spheres of rad at least two, even wth dfferent rad. The both proofs n [4] and n [10] have one feature n common; they are from scratch, they do not use any known result. On the other hand, the proof n [10] dffers essentally from that one n [4]. It requres checkng a large amount of cases and therefore t s computer-based. In ths paper we ntroduce a new method whch provdes a relatvely short proof, not aded by a computer, for Theorem 5, but also for the fve-dmensonal case. We wll gve a new, short proof for Theorem 4 as well, although the orgnal one gven n [4] s more elegant, just to show the power of our method. The proof does not splt nto cases for n =, and consders only two case for n = 4, and three cases for n = 5. Unlke the proofs n [4] and n [10], our method s based on a known result, namely on the non-exstence of the perfect Lee -error correctng codes over Z of block sze n =, 4, and 5. Thus, as a by-product, our method provdes some evdence that the most dffcult cases n the Golomb Welch conjecture are those for t =, because they mply, as a specal case, the nonexstence of PL(n, t) for n 5, and t. Our proof s algebrac n nature. Therefore we wll frst generalze the noton of the perfect Lee t-error correctng code. As usual, by a sphere S = (W, r W ), centered at W and of radus r W, we understand the set of all words V Z n so that d L (W, V) r W. For V S, we wll also say that S covers V. The perfect Lee code over Z of block sze n, denoted PL(n), s a set P of spheres (W, r W ), W Z n, r W, so that each word n Z n s covered by exactly one sphere n P. The man theorem of the paper reads as follows: Theorem 6. There s no PL(n) code for n 5. We beleve that a further refnement of the method should provde a proof of the non-exstence of PL(n) at least for n = 6. At the end of ths ntroducton we menton a result whch s related to the topc of ths paper. A tlng of R n by Lee spheres s called regular f neghborng spheres meet along entre (n 1)-dmensonal faces of the orgnal cubes. It s shown n [4] and [5] that the results stated there hold even n the case f we admt non-regular tlngs. At the frst glance t seems obvous that there are no non-regular tlngs of R n by Lee spheres. However, n [11] Szabo proved the followng surprsng result: Theorem 7. There s a non-regular tlng of R n f and only f n + 1 s not a prme.. PL(n) codes for n 5. In ths secton we prove the man result of the paper. Throughout the proof words n Z n wll be denoted by upper case block letters, and ther coordnates by the same lower case letter endowed wth an ndex, e.g., a word W wll have coordnates (w 1,..., w n ). Further, we drop subscrpt L when dealng wth Lee metrc, so the Lee dstance wll be denoted smply by d. The statement wll be proved by contradcton. Suppose that there s a PL(n) code P, where n 5. By Theorems and, there s no perfect Lee -error correctng code PL(n, ) for n 5 (note that Theorems 4 and 5 mply the statement for n = and n = 4, respectvely, as well). Thus, there s a sphere S 0 = (A, r A ) P so that r A. By a sutable translaton of P we

4 P. Horak / European Journal of Combnatorcs 0 (009) may assume that A = ( r A +, 0,..., 0). Consder the set V of words V wth d(v, A) = r A + 1, and v 1 0. Clearly, V V ff n v =. =1 Indeed, d(v, A) = n =1 v a = v 1 ( r A ) + n = v = v 1 +r A + n = v = r A +1, and (1) follows. Therefore, each word V n V s ether of type [±], or of type [±, ±1], or [±1 ]. To prove the non-exstence of PL(n) code for n 5, we show that t s mpossble to cover all words n V, that s, we show that there s no set of parwse dsjont spheres (and dsjont from S 0 ), each of radus at least, coverng all words n V. To ths extent, let S P be the set of all spheres n P whch cover at least one word n V. The words W so that (W, r W ) P wll be called codewords, the words W so that (W, r W ) S wll be called codewords n S. Moreover, f a word V belongs to a sphere S = (W, r W ) P, we wll abuse slghtly the language and sometmes nstead of sayng S covers V we wll say that the codeword W covers V. Now we state a seres of statements whch are rather smple but wll be appled over and over n ths proof, although not always explctly referred to. By defnton of PL(n) code we get Clam 8. If W, Z are codewords then the spheres (W, r W ) and (Z, r Z ) are dsjont, that s, d(w, Z) r W + r Z + 1. For any two words U, V, ther Lee dstance d(v, W) s nvarant wth respect to addng the same nteger to a coordnate, multplyng a coordnate by 1, or swappng the order of coordnates. Therefore: Clam 9. If P s a PL(n) code, then () translatng all codewords of P, () multplyng a coordnate of each codeword of P by 1, () swappng the order of coordnates n all codewords of P, results n a new PL(n) code. The followng clam plays a crucal role n the descrpton of words n V covered by a codeword n S. Clam 10. Let W, Z be codewords (not necessarly n S), V be a word covered by W, and d(z, V) r Z +. Then, for each coordnate, t s ether z v w, or z v w. Proof. By Clam 9, we may assume that Z = O = (0, 0,..., 0). Then n fact we need to prove that v w 0, and w v for all. Snce W covers V we have r W d(w, V) = n w =1 v. The spheres (W, r W ) and (O, r O ) are dsjont, therefore d(w, O) = n w =1 r W + r O + 1 n w =1 v + (d(o, V) ) + 1 (by assumpton d(o, V) r O + ). After smple rearrangements n we get w =1 n v =1 n w =1 v 1. Trvally a b a b for all a, b, wth equalty ff ab 0, and a b. To complete the proof t suffces to note that for ab < 0 t s a b < a b 1. Thus v w 0, and w v for all n. The clam follows. As an mmedate corollary of the above clam we get: Clam 11. If W s a codeword n S then n =1 w 5, and w 1 0. Proof. By defnton of S there s a word V V covered by W. For each word V n V we have d(a, V) = r A + 1. Snce v 1 0 for all words n V, by Clam 10, w 1 0 as well. In addton, d(w, A) = (w 1 ( r A ))+ n w = r A + r W + 1 r A + + n 1, that s, w =1 5. The clam follows. The followng corollary s the most frequently used statement of all clams gven here. It provdes a smple but very useful descrpton of all words V n V covered by a codeword W n S. We pont out that the descrpton nvolves only coordnates of W but not the radus r W of the sphere S = (W, r W ) S. Clam 1. Let W be a codeword n S. Then W covers a word V n V f and only f v w 0, w v for all = 1,..., n. (1)

5 484 P. Horak / European Journal of Combnatorcs 0 (009) Proof. The necessary part follows from Clam 10. Indeed, for each word V n V we have d(a, V) = r A + 1, therefore, for, ether 0 v w, or 0 v w,.e., v w > 0, and w v. By defnton, v 1 0, and by Clam 10, w 1 v 1. As to the suffcency part, t suffces to note that W beng a codeword n S mples that W covers at least one word n V, and that f V, V V are two words fulfllng the condton n the clam then d(w, V) = d(w, V ). Example. Let W = (1, 7, 1, ) be a codeword n S. Then Clam 1 mples that W covers n V only one word of type [±], namely (0,, 0, 0), the followng words of type [±, ±1] : (1,, 0, 0), (0,, 1, 0), (0,, 0, 1), (1, 0, 0, ), (0, 1, 0, ), and (0, 0, 1, ), and four words of type [±1 ] : (1, 1, 1, 0), (1, 1, 0, 1), (1, 0, 1, 1), and (0, 1, 1, 1). The followng three relatons are essental for the proof of our theorem. Let C(t) be the set of codewords n S havng t non-zero coordnates. Further, let C (t) be a subset of C(t) that contans codewords W so that {j, w j > 1} =. Set C(t) = c(t) and C (t) = c (t). There are n 1 words of type [±] n V (note that for V n V t s v 1 0). By Clam 11, for each codeword W n C 1 () there s, 1 n, so that w, and consequently, by Clam 1, W covers one of those words of type [±]; thus c 1 () n 1. ( n Further, there are 8 ) 4(n 1) words of type [±, ±1] n V. Hence ( n ) 4c() c 1 () + c(4) 8 4(n 1) () as, by Clam 1, each codeword n C 1 () covers exactly two of those words, each codeword n C() C 1 () at least 4 of them, and each codeword n C(4) at least of them (for each codeword W n C(4) there s at least one so that w, ( ) ( ) n see Clam 11). Fnally, there are 8 n 1 4 words of type [±1 ] n V, whch leads to n ( ) ( n ) ( ) n 1 C() = 8 4 = ( because, by Clam 1, each codeword n C() covers exactly ) words of type [±1 ]. Clearly, there are many solutons of Eqs. () (4) n the non-negatve ntegers. A soluton, whch corresponds to a perfect Lee code over Z, wll be called a feasble soluton. So our am wll be to show that there s no feasble soluton of these equatons..1. n =, 4 Here we provde alternatve proofs to Theorems 4 and 5. We start wth a statement, whch plays an essental role n the closng argument of those proofs. () (4) Theorem 1. For n =, there s no codeword W C () so that w 1 codeword W C (), so that w 1 =. = 0. For n = 4, there s no Proof. The statement wll be proved by contradcton. We start wth the case n = 4. Suppose that W C (), where w 1 =. Then there s, 4, so that w, (see Clam 11). By Clam 9 we assume, wthout loss of generalty, that w, and w = 0, for 4. Let, for j = 1,,, M j be a set of words gven by M 1 {V, v 1 = 1, v =, 4 v = = 1}, M = {V, v 1 = 1, v = 1, 4 v = = 1}, M = {V, v 1 = 0, v =, 4 v = = 1}. Clearly, for all j, Mj = 4. The followng clam s crucal for the proof of ths theorem. Clam 14. Let Z be a codeword coverng a word V M j, 1 j. Then z 1 = v 1, and z = v.

6 P. Horak / European Journal of Combnatorcs 0 (009) Proof of Clam. We recall that A = ( r A, 0, 0, 0), r A, s a codeword. Let V M j, 1 j. Then d(a, V) r A +, and therefore, by Clam 10, z 1 v 1 0 > r A, and z v 0. On the other hand, W s a codeword n S, so, by Clam 1, W covers both words (, 1, 0, 0) and (1,, 0, 0). Hence r W = d(w, (, 1, 0, 0)) = w 1 (note that r W cannot be > w 1 as then the spheres (A, r A ) and (W, r W ) would ntersect). Therefore, d(w, V) = 4 =1 w v = (w 1 v 1 ) (w v ) w + 1 = w + 1 = r w + and, agan by Clam 10, z 1 v 1 w 1, and z v w. The proof follows. Now we wll classfy codewords Z coverng words n M =1. Set, for j = 1,, A j = {Z, Z s a codeword coverng a word n M j=1 j, and {,, z 0} = j. Put a = A. We prove two nequaltes on a. It s clear that each codeword from A covers words n M j=1 j. Indeed, snce M j= j V, the statement n ths case follows from Clam 1. For a codeword coverng a word n M 1, the assumpton that U = (1,, d, 0), d 0, covers a word (1,, 0, c), c = 1, mples that spheres (U, r U ) and (W, r W ) ntersect. The other part s obvous. As = 1, we get a + a 1 = 1. To fnsh the proof of the theorem we prove that j=1 M j a 1 + a 4 (6) whch contradcts (5). Frst we pont out that f Z covers a word V j=1 M j then z + z 4. Indeed, f V s n M M, then the nequalty follows from Clam 11; for a word V n M 1 t can be routnely checked that z + z 4 < mples that the spheres (Z, r Z ) and (W, r W ) ntersect because r W w 1 and d(z, W) w + 1 n ths case. Let Z, Z be a codeword n A 1 A coverng a word V, V M =1, respectvely. To prove (6) t s suffcent to show, wth respect to (7), that f Z, Z A 1 A, and z j z j > 0, j, then ether zj = 1, or z j = 1. Assume wlog j =. Suppose by contradcton that z > 1, and z > 1. We have z z > 0, and mn{ z, z }, whch yelds z z z + z 4. (8) Further, the spheres (Z, r Z ) and (Z, r Z ) are dsjont, thus we get d(z, Z ) = 4 =1 z z r Z + r Z + 1 d(z, V) + d(z, V ) + 1 = 4 =1 z v + 4 =1 z v + 1. Applyng Clam 14 nto 4 =1 z z 4 =1 z v + 4 =1 z v +1 provdes v 1 v 1 + v v + 4 = z z 4 = z v + 4 = z v + 1; now usng (8) and obvous z4 z 4 z 4 + z 4 yelds v 1 v 1 + v v. However, v 1 v 1 1, and v v 1, a contradcton, and (6) follows. Fnally, let n =. Assume wlog that W = (0, a, b) C(), where a, b. Consder the words V 1 = (1, 1, 1), and V = (1, 1, ). By the same argument as n the proof of Clam 14, we get that f a codeword Z covers V, = 1,, then Z 1 = (c, 1, 1), Z = (d, 1, ), where, by Clam 11, c, and d (otherwse the spheres (A, r A ) and (Z, r Z ) would ntersect). However, then the spheres (Z 1, r Z1 ) and (Z, r Z ) ntersect. Indeed, r Z1 c 1 whle r Z d 1, but d(z 1, Z ) = c d + 1 < (c 1) + (d 1) + 1 = r Z1 + r Z + 1. The proof of the theorem s complete. Theorem 15. There s no PL() code. Proof. Puttng n = nto Eq. (4), and takng nto account that n ths case C(4) = C(5) = 0 we get c() = 4. Frst of all we show that there s a codeword W C 1 () so that w 1 = 1. Indeed, otherwse c() c 1 (), and the codewords n C() C 1 () would have to cover twce some word n A = {V, (5) (7)

7 486 P. Horak / European Journal of Combnatorcs 0 (009) V s of type [±, ±1], v 1 = 1}. So, suppose wlog that W = (1, d, 1), where d. Let Z = (a, b, c) be a codeword coverng the word (0,, 1). Then, by Clam 1, b and c 1; n addton, a = 0, otherwse the word (1,, 0) would be covered by both W and Z. If b > then Z would cover (0,, 0), whch leads to a contradcton because (0,, 0) would be covered by both W and V. Hence, b =, and by Clam 11, c, that s, Z C () wth z 1 = 0, contradctng Theorem 1. Theorem 16. There s no PL(4) code. Proof. For the reader s convenence we state that, for n = 4, the equatons () (4) turn nto c 1 () 7 4c() c 1 () + c(4) 6 c() + 4c(4) = 0. Frst of all we show that f there were a feasble soluton of Eqs. () (4) then c(4) 4. Let A = {V, V s a word of type [±1 ], v 1 = 0}. Clearly, A = 8. By Clam 1, each word n C() C(4) covers at most one word n A, thus c() + c(4) 8. Therefore, from (4), c(4) 4. On the other hand, t s easy to see that, for c(4), m = mn(4c() c 1 () + c(4)) s attaned when c 1 () and c(4) are maxmum possble,.e. c 1 () = 7, and c(4) =. Thus, m = = 40 > 6, contradctng (). We consder two cases. I. c(4) = 4. By Clam 9 we may assume that there s a codeword Z C(4) wth z > 0 for all = 1,..., 4. Then the coordnates of the four codewords n C(4) have the followng sgns: (+, +, +, +), (+, +,, ), (+,, +, ), and (+,,, +). Let B = {V, V s of type [±, ±1], v 1 = }. Clearly, B = 6. At most one codeword W n C(4) has the property that w 1 > 1, otherwse, by Clam 1, a word V of type [±, ±1] wth v 1 = and v j = 1, for some j, j 4, would be covered by two codewords from C(4). Thus, codewords n C(4) cover at most three words n B. There are twelve words V of type [±1 ] wth v 1 0. Each codeword n C(4) covers three of them. As c(4) = 4, all of those twelve words are covered by codewords n C(4). Hence, by Clam 1, for each codeword W C() we have w 1 = 0. Ths n turn mples that no word n B s covered by a codeword n C(). Hence, at least three words n B have to be covered by codewords Z n C(), wth z 1 > 1. By Clam 1, each codeword n C() covers at most one word from B, so there are at least three codewords n C() coverng a word n B. Snce at most one of them has ts frst coordnate (otherwse the word (, 0, 0, 0) would be covered by more than one codeword), there s a codeword U n C() wth u 1 =. However, ths contradcts Theorem 1. II. c() =. Assume wlog that the coordnates of codewords n C(4) have the followng sgns: (+, +, +, +), (+, +,, ), and (+,, +, ). These three codewords n C(4) cover nne out of twelve codewords V of type [±1 ] wth v 1 = 1. To cover the three remanng words V, there have to be, by Clam 1, codewords U, = 1,,, n C() wth coordnates (+, 0,, +), (+,, 0, +), and (+,,, 0), respectvely. Moreover, there has to be a codeword U 4 n C() wth coordnates (0,,, +) to cover the word (0, 1, 1, 1). It s not dffcult to check that, to avod some word of type [±, ±1] to be covered by two codewords U, that U C 1 () for = 1,..., 4. Thus, there s, 1 4, so that the frst coordnate of U s. Now we focus on the set of words B = {V, V s of type [±, ±1], v 1 = }. As n the case c(4) = 4, there s n C(4) at most one codeword W wth w 1. In addton, there s exactly one codeword U n C() wth u 1 (n fact for ths codeword ts frst coordnate, as mentoned above). Therefore, at most 5 words from B are covered by codewords n C() C(4); hence there has to be a codeword Z C(), coverng a word n B. To avod coverng the word (, 0, 0, 0) twce, t has to be z 1 =, contradctng Theorem 1... n = 5 In order to facltate our dscusson we ntroduce more notons and notaton. Two words U and V are sad to be sgn equvalent n the th coordnate f u v > 0. We wll deal very often wth a set of words that are sgn equvalent n some coordnate. For each coordnate we have two such sets. To smplfy the language we wll ntroduce the noton of the sgned coordnate. For the rest of the paper

8 P. Horak / European Journal of Combnatorcs 0 (009) by the set of sgned coordnates we wll understand the set I = {+1,..., +5,,..., 5} (we recall that by defnton t s v 1 0 for each word V n V; and by Clam 1 t s w 1 0 for each codeword W n S). Let A be a set of words, and, j, j I. Then A (A j ) s the set of all words n A so that.w > 0 (w > 0, jw j > 0). In other words, A s the set of words n Z n that are parwse sgn equvalent n the th coordnate, and ther common sgn n the th coordnate concde wth the sgn of. We note that no two codewords n S are sgn equvalent n three coordnates because they would cover the same word of type [±1 ], see Clam 1. The Eqs. () (4) descrbe global propertes of parameters c(t). The followng three theorems descrbe ther local propertes. ( 5 Theorem 17. It s t= 18. Consequently, C() 0 (mod ) for all I. ) t 1 C(t) +1 = 4, and for each I, ( 5 +1, we have t= ) t 1 C(t) = ( ) Proof. Let T be the set of all words n V of type [±1 4 ]. Clearly, T +1 = 4 = 4, and for I, ( ) 4 +1, t s T = 4 6 = 18. Further, each word n T s covered by exactly one word ( ) 5 from C(t) t 1 t=, and, by Clam 1, each codeword from C(t) covers of them. We are done ( ) wth the frst part of the statement. The second part follows trvally from the fact that for t = 4, 5. Theorem 18. For each, j I, +1 {, j}, 5 j, t= (t )C(t) j = 6; and for each, j I, +1 {, 5 j}, t= (t )C(t) j = 5. Consequently, C()j and C(5),j have the same party for +1 {, j}, and the opposte party for +1 {, j}. Proof. As n the prevous proof, let T be the set of all words of type [±1 ]. Then, for j, Tj = (5 ) for +1 {, j}, otherwse Tj = 5. Each word n Tj s covered by exactly one 5 codeword from C(t) t=. As, by Clam 1, each codeword from C(t) j covers t words n T j, the man part of the statement follows. The second part s obvous. We state one more theorem that wll sgnfcantly decrease the number of feasble solutons of () (4). t 1 Theorem 19. It s C() For I, +1, C() when C(5) = 0, and C() 6 when C(5) = 1. Proof. Let I. Further, set A = {V, v = ±1}, and B = {V, v = ±}. Clearly, A = B = 7 for each I, +1, and A +1 = B +1 = 8. Let a be the number of words n A covered by codewords n C(). It s easy to see that f W C() does not cover any word n A then W C 1 () and w. However, there s at most one such codeword n C() ; thus a C() 1. Ths n turn mples, C() +1 9 as A +1 = 8. Let I, +1. Put D = {W, W 5 C(t) t=, W covers a word n B }. Now we state some bounds on D. Let U C(t), t 5, u > 1 (.e., U C). Then U covers t 1 words n B. Thus, B = 7 mples D, and consequently D = mples 5 C(t) t=4 D 1. Further, f 5 C(t) t=4 D = then C() D =. Clearly, f W C(4) but not n B then W covers at least one word n A (by Clam 11 there s an ndex j so that w j ). Therefore, A a + C(4) C(4) D C() + C(4). Snce A = 7, we get C() + C(4) 9. By Theorem 17, C() + C(4) = 18 6 C(5), and C() 0 (mod ). Hence C() for C(5) = 0, and C() 6 for C(5) = 1. The proof s complete.

9 488 P. Horak / European Journal of Combnatorcs 0 (009) We are ready to prove the man result of the paper: Theorem 0. There s no PL(5) code. Proof. As n the prevous cases, for the reader s convenence we state that, for n = 5, the Eqs. () (4) have the form c 1 () 9 4c() c 1 () + c(4) 64 c() + 4c(4) + 10c(5) = 56. We pont out that c(5). Indeed, by Clam 11, for each W C(5) t s w 1 0, whch n turn mples C(5) +1,.e. c(5), otherwse some word of type [±1 ] would be covered by two codewords from C(5) +1. We wll consder three cases. I. c(5) =. Let W, Z C(5). By Clam 9 we assume wlog w > 0 for all. We wll consder two subcases wth respect to the number of coordnates n whch W and Z are sgn equvalent. Ia. Assume frst that z < 0 for 5. Then C(4) +1 = 0, otherwse there would be a word of type [±1 ] covered by two codewords. Thus, by Theorem 17, C() +1 = 1, whch contradcts Clam 11. Ib. There s, 5, so that z > 0. By Clam 9, we assume z > 0, and z < 0 for 5. It s easy to see that for each U C(4) t s u < 0. Set A = {W, W C(4), w 1 0}, and B = {W, W C(4), w 1 = 0}. By Theorem 19, C(), whch n turn mples C(4) 5. However, c(4) = C(4), thus A + B 5. Further, for, j, C(4)j 1, otherwse some word of type [±1 ] would be covered by two codewords. There are 6 pars of ndces, j, wth, j, and j. Each codeword U A belongs to C(4) j for one of those sx pars of sgned coordnates, and each codeword n B to two of those sx pars; thus A + B 6. Fnally, for, t s A 1, otherwse some word of type [±1 ] would be covered by two codewords n A ; hence A. However, ths contradcts A + B 5 because A + B 6. II. c(5) = 1. Let W C(5). As n case I, we assume w > 0 for all. By Theorem 18, C()+1, s odd for every {+, +, +4, +5}, hence C() +1 6, as C() 0 (mod ), see Theorem 17. Further, agan by Theorem 18, C(),j s odd f +1 {, j} and at least one of and j s n {,..., 5}. Therefore, C() for all sgned ndces I. Moreover, for {,..., 5}, f C()+1,, > 0 then C() 6 as then there are at least seven ndces j I for whch C(),j > 0. No codeword n C() +1 has all ts coordnates non-negatve. Hence, C()+1 6 mples that there are at least three ndces n {,..., 5} for whch C() 6, contradctng Theorem 19 because c (5) = 0. III. c(5) = 0. By Theorem 18, C()j s odd for each, j I, j, +1 {, j}. Thus C(), and by Theorem 19, C() =. By (4), we get c() 0 (mod 4). Hence C() +1 0 (mod 4). As C() +1 = 1 contradcts Theorem 19, we get C() +1 = 0. Thus c() = 8, and by (4), c(4) = 1. Moreover, C() +1 = 0 mples C(4) +1 = 8. Therefore there are four codewords U C(4) wth u 1 = 0. It s easy to see that there have to be two codewords U 1, U among those four codewords whch are sgn equvalent n two coordnates, say, j. (In fact ths would be true even f there were only three codewords U C(4) wth u 1 = 0}. As C() +1 = 0, there has to be a codeword W C(4) +1 that s sgn equvalent n the th and jth coordnate wth both U 1 and U. Let k, k 5, k {, j} be a coordnate so that w k 0. Clearly, U 1 and U are not sgn equvalent n k, as they would be sgn equvalent n three coordnates, thus one of them s sgn equvalent wth W n k as well, whch agan mples that there are two codewords sgn equvalent n three coordnates, a contradcton. References [1] J. Astola, On perfect codes n the Lee metrc, Ann. Unv. Turku (A)1 (176) (1978) 56pp. [] J. Astola, On perfect Lee-codes over small alphabets of odd cardnalty, Dscrete Appl. Math. 4 (198) 7 8. [] S.W. Golomb, L.R. Welsh, Algebrac codng and the Lee metrc, n: Error Correctng Codes, Wley, New York, 1968, pp [4] S. Graver, M. Mollard, Ch. Payan, On the exstence of three-dmensonal tlng n the Lee metrc, European J. Combn. 19 (1998)

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