On the minimum distance of ternary cyclic codes - Information Theory ...
409
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL 3Y. NO. 2. MARCH 1YY3
On the Minimum Distance of Ternary Cyclic Codes Marijn van Eupen and Jacobus H. van Lint
Abstract-There are many ways to find lower bounds for the minium distance of a cyclic code, based on investigation of the defining set. Some new theorems are derived. These and earlier techniques are applied to find lower bounds for the minimum distance of ternary cyclic codes. Furthermore, the exact minimum distance of ternary cyclic codes of length less than 40 is computed numerically. A table is given containing all ternary cyclic codes of length less than 40 and having a minimum distance exceeding the BCH bound. It seems that almost all lower bounds are equal to the minimum distance. Especially shifting, which is also done by computer, seems to be very powerful. For length 40 5 11 5 50, only lower bounds are computed. In many cases (deriven theoretically), however, these lower bounds are equal to the minimum distance.
of C is
where R = ( ; j l . , j 2 . . . . .,it)is the defining set of C (we do not require that a parity check matrix has independent rows). For purposes that will become clear soon, we will define, for I c (0.1 :... n - l}, M ( R ) I as the submatrix of M ( R ) consisting of the ith columns of M ( R ) , where i E I . One can see immediately that if R contains a consecutive set of Index Terms-Ternary cyclic code, minimum distance, shifting, length s, then M ( R ) rcontains a Vandermonde .s x s submatrix selforthogonal, contraction. and so r a n k ( M ( R ) l ) = s, whenever 111 = s . So if we define for every codeword c E C its support I as the set of I. INTRODUCTION positions where c is nonzero, then a codeword c with support CYCLIC CODE C of length 71, over the alphabet GF(q) I , 111 5 s (i.e., the weight of c(= wt(c) :=IT[)is at most (gcd(n,,q) = 1) can be characterized as an ideal in the s) cannot occur since M ( R ) I has full rank. So the minimum distance d = t t ( C ) : =niin,--c wt(c) is at least s 1. We call ring GF(~)[x:]/(:~:'~ - 1 ) with generator g ( . r ) (say), which is a divisor of :I:" - 1. A codeword of C will be written either as this bound the BCH bound (cf. [ 3 ] ,[4]), and we will write c ( x ) E GF(q)[:c]/(z" - 1) or as the vector c of length 71, having ~ B C H= s + 1, i f s is the largest integer having the property that as ith entry the coefficient of :I;' in c(.r;). If (k is a primitive there is a code equivalent to C with a defining set containing 7bth root of unity in some extension field of GF(q), then all a consecutive subset of size .s (notice that R depends on the zeros of xT1- 1 can be written as cuJ(0 5 j 5 71 - 1).If g ( r ) is choice of tr; we call two codes equivalent if their defining not constant, then g ( x ) has some zero oJrJsince g(.c) divides sets are the same up to multiplication with some integer coprime to 72). The first generalization of the BCH bound z7L - 1. But because g(x) is a polynomial over GF(q), it also was given by Hartmann and Tzeng [5]. They prove that if d 3 0 . . . . as its zeros (we will say: m j , ( . I ; ) divides has (Po, g(z), where n ~ ~ , ( .isc )the minimal polynomial of do, i.e., the the consecutive sets { I ,in. 'i 1 j u . . . , i 6 - 2 ,ja} . .). (0 5 ,j 5 s ) are contained in R, and if ( d . n ) < 6, then monic polynomial that only has zeros ( Y ~ O .c u q J ( ] . (@.lo. So we can characterize g(z) (and also C) by the set G := the minimum distance of the code with defining set R is (;joln~,,,(x) divides ,y(x)} (here we use that gcd(n, ti) = I , at least h + .s. Roos [6] generalized this by proving that if and so every zero of :E'' - 1 (and so every zero of !](.I;)) has the statement is true for sufficiently many (say s') values s' - 1. A multiplicity one). Mostly we will characterize C by its defining of , j , then the minimum distance is at least 6 last generalization can be found in [l] and is called the ABset R : { j l d is a zero of g ( : x ) } (and sometimes by R , which is the set of integers modulo 71 that are not in R). We define method. It says that if A, I3 c (0.1. . . ' , 71. - l} are such that the check polynomial h ~ ( . i : )as the reciprocal polynomial of A B:={ a b rriod n ) n E A. b E I?} is a subset of R, then h,(z):= (x" - l)/,q(:c) ( h l ( n . ) is the generator polynomial of the code with defining set R has minimum distance at least 6, the dual code). Of course every ( ( 2 ) E C has zeros d, ,j E R, if ~ r t ~ d ( M ( A+)r~a n) k : ( M ( B ) , )> III for every subset I of and this is the same as saying: c has inner product zero with ( 0 . 1 . ' . . . ? I , - l } for which 111 < h . If COis a subcode of C , then we will write CO5 C . If Ro is the vector (1nJ (Y2J . . o ( ~ - ' ) J ) . So a parity check matrix the defining set of CO,then we must have R c Ro. Moreover we have: d ( C ) 5 d(C0). If CO 5 C and CO # C , then we write CO< C' (and we say: COis a proper subcode of C). A Manuscript received July X, 1991. This work was presented in part at Coding and Information Theory Conference, Essen, Germany, December 15- 17. cyclic code C is called minimal if C # ( 0 ) (0 denotes the 1991. zero word) and if {0} 5 CO< C' implies that CO= (0).The M. van Eupen is with Eindhoven University of Technology, Eindhoven, generator of a minimal code is (.rT1- l ) / m j ( : c )for some .j. The Netherlands, HG 987. J. H. van Lint is with Eindhoven University of Technology, Eindhoven, The In Section 11, we first give some theorems, that give good Netherlands, BG 335. He is also with Philips Research Laboratories, P.O. Box lower bounds for the minimum distance. The first one is XOOOO, 5600 JA Eindhoven, The Netherlands. IEEE Log Number 9203861, called shifting and turned out to be very powerful. We used
A
+
+
+ +
+
>
+
+
00 lX-'~44X/93$0?1.00 0 1993 IEEE
+
+
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
410
a computer to compute the shifting bounds and for length less than 40 this bound was for approximately 95% of the codes equal to the minimum distance. Where the shifting bound did not equal the minimum distance we were often (but unfortunately not always) able to find a bound that did equal the minimum distance. Of course for many codes it is possible to find a lower bound more quickly by hand using other theorems than shifting (including the AB-method). But to avoid a long list of tedious examples, we only used other theorems where it was necessary (i.e., where the shifting bound did not equal the minimum distance). The results can be found in Table I. In Section 111, we treat the lengths 40 5 n 1. 50. Because shifting is in the worst case exponential in n, the computer could in some cases not find the shifting bound in reasonable time. Fortunately other theorems were powerful enough to find almost all minimum distances. The results can be found in Table I1 and Table 111. 11. LOWERBOUNDS
The technique of shifting was introduced in 1986 by Van Lint and Wilson [l]. It gives very good lower bounds for the minimum distance of cyclic codes. We now give a definition of the shifting bound ~ S H I F T (of C )a cyclic code C , that is slightly different from the definition in [l], but easier to implement on a computer. Definition 1: Suppose C is a cyclic code of length n with defining set R. We define the shifting bound dsHIFT(C) inductively. a) If C is a minimal cyclic code and q ,7-2, . . . , r , is the longest sequence of different integers (mod n) such that 1) {7-17-2,...,7-,} c R, 2 ) There is a sequence a l , a2, . . , a, such that for all i : { a ; + r l , . . . , a i + r t - l } c R a n d a ; + r ; 6 R(al1 Cw) 1. additions taken modulo n), then ~ S H I F T (= b) If C is not a minimal code and +
+
& H I F T ( ~ ):= m i n c o < c ~ S H I F T ( ~ O ) ,
and r l , r2,. . . , T , is the longest sequence of different integers with I dsHIFT(C) - satisfying conditions 1) and 2) in a), then ~ S H I F T (= Cw ) 1.
+
Theorem 1 (Shifting): Suppose C is a cyclic code with minimum distance d. Then, ~ S H I F T (I C d. ) Proof: By induction w.r.t. the number of irreducible factors in the check polynomial. 1) Let C be a minimal code (i.e., there is one irreducible factor in the check polynomial) and T I , 1-2, . . . , rw(w > 0) a sequence satisfying conditions 1) and 2) in Definition a). Suppose e(.) E C is of weight w and has support I. Then, we have for all 1 5 i 5 w : rank(M(rl,7-2,. . . , T = rank(M(ai
+ = rank(M(ai +
; ) ~ ) T I , ai
+
TZ,.
. . ,ai+ r i ) I )
+ ~ i - 1 ) ~+) 1 = rank(M(r1, ~ 2 , .. . , ~ i - 1 ) ~+)1. TI,.
. .,a,
The first equality follows from the fact that the rank of a matrix stays invariant under multiplication with ~ ) . the second diag( 1,sat, crzaa, . . . , ~ & ~ - l )To~ prove equality, we recall that
and see that c (the vector of length n corresponding to e(.)) is orthogonal to the first i - 1 rows of this matrix, but is not orthogonal to the last row (since otherwise c(z) would have aaa+rl as a zero, but a; T; R and C is a minimal code, so this would G 0). So the last row is linearly imply that e(.) independent from the other rows and the equality follows (notice that we may restrict ourselves to the columns corresponding to support I ) . The third equality is as obvious as the first one. So rank(M(r1, r z , . * * ,r w ) I ) = r a n k ( M ( r l , r 2 , . . . , r w - l ) I ) + l = ... = rank(M(rl),) w - 1 = w . But this means that M(r1,~ 2 , ..* ,r w ) I has full rank and so c(z) cannot be a codeword. This implies that ~ S H I F T (5C d. ) 2) Suppose C is not a minimal code. By induction, we know that &FT(C) Id(C) := m i n c o < c d ( C o ) . Suppose T I , 7-2, . . . , rw satisfy conditions 1) and 2) and w 5 &HIFT(C) - 1 5 d(C) - 1. Again let c(z) be a codeword of weight w ( > 0). We see that if is a zero of c(z), then c(z) is in a proper subcode of C, since ai r; 6 R. This contradicts the fact that 0 < w 5 a(C) - 1. So again the last row of M ( ai+ T I , a, + 7-2, . . . , ai + ri) is linearly independent of the others and in the same way as in 1, we have r a n k ( M ( r l , r 2 , . . . , r , ) I ) = w and c(z) cannot be a codeword. 0
+
+
+
In Some caseS shifting is not powerful enough and we have to improve the lower bound, using some other theorems. The first theorem is very useful and is also a consequence of a theorem of McEliece . 171. .
Theorem 2: Suppose C is a ternary cyclic code of length n with defining set R. Suppose that if i @ R, then n - z E R. Then, C is selforthogonal (or equivalently: wt(c(x)) E 0 mod 3 for all c(z) E C). Proof: Suppose c(z), c’(z) E C and a: is a primitive nth root of unity. If .(ai) # 0, then i @ R SO by assumption n - i E R and ~ ’ ( a - ~=) 0. This 0 mod (P- 1). is true for all i , so c(z)c’(z-’) Looking at the coefficient of zn-l in c(z)c’(s-’), we get (c(z), c’(z-l)) 5 0 mod 3 ((c(z), c’(z)) denotes the inner product of c(z) and c’(z) in GF(3)). So C is selforthogonal (c(z), c(x)) (c’(z), c’(z)) (and because (c(z), e’($)) 2(c(z) e’(.), c(z) c’(z)), this is equivalent to saying that 0 all weights in C are divisible by 3).
+
+
=
+
+
~
VAN EUPEN AND VAN LINT ON THE MINIMUM DISTANCE OF TERNARY CYCLIC CODES
Example11.1: n = 11, R = (0.2.6.7.8. lo}. This code satisfies the condition in Theorem 2. Here ~ ~ H I F=T 5, but because weights are divisible by 3, we have d 2 6. Example 22.1: n = 22, R = {2,6. 8.10.18). Here, ~ ~ H I F=T10 and by Theorem 2 we have d 2 12. Example 22.2: n = 22, R = (0.4.7.11.12.13. 14.16,17. 19.20.21). Here, ~ ~ ~ H I=F T7 and by Theorem 2 we have d 2 9. Example 22.3: n = 22, R = {0,2.6.7.8.10.11.13.17,18. 19.21}. Here, SHIFT = 5 and by Theorem 2 we have d 2 6. Example 23.1: n = 23, R = {0.5.7.10.11.14,15,17.19. 20,21,22). Here. ~ ~ H I F=T 7 and by Theorem 2 we have d 2 9. Example26.1: it = 26, R = (1.2.3.6.9.18). Here, ~ ~ H I F=T 1 3 and by Theorem 2 we have (1 2 15. Example 26.2: I I = 26, &' = { 1.2.3.4.%.6.9,10,12.15.18. 19}. Here, ~ ~ H I F=T8 and by Theorem 2 we have d 2 9. Example 26.3: 11 = 26, R = { 1 , 2 , 3 . 4 , 6 . 7 . 9 . 1 0 . I 1 . 1 2 . 1 8 . 21). Here, ~ ~ H I F=T8 and by Theorem 2 we have (1 2 9.
411
ProoJ C,) is generated by go(.) = ( 2 - l)g(z). The condition in Theorem 3 is satisfied. We see that Cl is generated by (:eTL- 1)/(x - 1) = :rTL-' d-' . . . 1. So all nonzero codewords in C1 are of weight n. Applying Theorem 3 and using the fact that wt,(co(x)) E 0 mod 3 (by 0 mod 3, assumption), we get a). Moreover, if wt(c(z)) then by Theorem 3 also wt(cl(:c)) 0 mod 3, i.e., q ( z ) 0 (since 71, $ 0 mod 3). This proves statement b). 0
+
+
+
=
Example 11.2: 71 = 11, R = { 2,6.7,8.10). The conditions in Corollary 1 are satisfied (see Example 11.1). Using shifting we get d 2 4. But because wt(c(,x)) = 0 or wt(c(x)) 2 for all codewords c ( x ) , we have d 2 5 . Example 22.4: rL = 22, R = {0,2,6,8,10,18}. The conditions in Corollary 1 are satisfied (see Example 22.1) and SHIFT = 8. By Corollary 1 a) we have d 2 9. But Corollary 1 b) says that if a codeword c ( x ) of weight 9 occurs, then c(z) is an element of the code in Example 22.1, which has d 2 12. This is a contradiction and so d 2 10. Example 22.5: 71 = 22, R = {4.7.11,12,13.14,16.17.19. Theorem 3 (Orthogonal Suhcode Representation): Suppose 20.21}. The conditions in Corrollary 1 are satisfied (see C and COare ternary cyclic codes of length 71 generated by Example 22.2) and ~ ~ H I F=T6. Using Corollary 1 b) and the fact that the code in Example 22.2 has d 2 9, we get d 2 7. Example 22.6: 71, = 22, R = {2.6,7,8.10.11.13: 17,18, 19%21}.The conditions in Corollary 1 are satisfied (see respectively, where t > s . Suppose that if y is a zero of Example 22.3) and ~ ~ H I F=T5 . Using Corollary 1 a) we get go(x)/g(x),then 7-l is a zero of go(:r). Then every c(x) E C d 2 6. Example 22.7: 7) = 22, R = (4.7.12. 13,14,16, 17,19,20, can be written as C ( X ) = c o ( : E ) c1(x), where cg(:~;) E CO 21). Here &HIFT = 6. We wish to apply Theorem 3 to and c ~ ( : eis) an element of the code C1, that is generated by prove that (1 2 7. Let the code C with defining set R :Cn - 1 be generated by g(:7:). Let CO be generated by g(z)(z l ) ( . r 1). Then the condition in Theorem 3 is satisfied. CO is the code of Example 22.2 and is selforthogonal and has d 2 9. Furthermore, C1 is generated by ( x ' ~- 1)/(x2 - 1) Moreover, wt(c(x)) = wt(cO(:i:)) w t ( q ( . c ) ) ~rioc-l3 . and q ( : r ) E C1 can only have a weight divisible by 3 if Proof: It is easy to see that gcd(go(.c),gl(.c)) = ,q(x). cl(.r) e 0. Now suppose c ( r ) E G has weight 6. Write So g ( z ) = o,(:c)gO(x) b ( x ) g l ( : c ) for certain polynomials C ( T ) = t ~ ~ ] ( . t : ) q ( z ) ,where q,(:r) E Go and q ( z ) E C1. Then by the congruence relation wt(c(x)) wt(co(x)) o,(:c), b ( x ) E GF(3)[.c]/(.c'L- 1). So also C ( J ) = c ~ ( : c ) w t ( q ( . I . ) ) iiiotl 3, and from the observations previously made q ( x ) for certain C O ( : I ; ) E CO and r1(.1;) E C1. Further0. So .(:I;) E CO,which is a contradiction, more ,qn(x)gl(:Y1) = 0 mod (:eT' - l ) , because if y is not we have (:I(:/-) a zero of go(:.), then by assumption it is not a zero of since COhas no codewords of weight 6. So d 2 7. Example 23.2: 71 = 23, R = {5,7,10,11,14,15,17,19,20. go(:~;-')/g(x-') (i.e., y is a zero of , q ~ ( : r ; - ' ) ) . But now we have c " ( : L ) ~ ( : I ; - ' ) = 0 rnod (:r:" - l ) , and this means that 21-22}. The conditions of Corrolary 1 are satisfied (see Example 23.1) and SHIFT = 6. Suppose a codeword of (co(.T:),cl(.c)) = 0 mod 3 . SO weight 6 occurs, then by Corollary 1 b) it is in the code of wt(c(:r)) ( . ( . E ) . C ( . X ) ) Example 23.1, which has (1 2 9. A contradiction, so d 2 7. Applying Corollary 1 a), we get d 2 8 (this is also shown (co(:c) q ( z ) eo(..) , q(x)) in [SI). (c(l(:c).co(:r)) 2(c0(z). q ( : r ) ) ( q ( x ) ,e1(:1;)) Example 26.4: 'rt, = 26, R = { 0 , 1 . 2 , 3 , 6 , 9 ,IS}. The (cg(z). Co(Z)) (q(:1;). q ( : r ) ) conditions of Corollary 1 are satisfied (see Example 26.1). wt(co(x)) w t ( c l ( x ) ) iriotl 3 . 0 Here SHIFT = 12, but by Corollary 1 b) we must have d 2 13 (see Example 26.1). And by Corollary 1 a) we get d 2 14. Corollary I : Suppose C is a ternary cyclic code of length Example 26.5: 7) = 26, R = {7.8.11.13,14,16,17,20,21, 11, with defining set R not containing 0. Also suppose that the 22,23,24.25}. Here ~ ~ H I F=T 7, and by Corollary 1 a) we code COcorresponding to R U (0) is selforthogonal. Then, the have r l 2 8 (see Example 26.2). following holds for all c ( x ) E C: Since our codes are ternary, 3 is not a divisor of n. So if 71, a) wt(c(z)) mod 3 or wt(c(z)) 71, mod 3, is not too big (say n 5 SO), then 2 is a divisor of n, unless b) if wt(c(z)) c 0 mod 3, then c(x) E Co. TI, is a prime or n = 25, 35, or 49. But if 2 divides n, then
=
+
+
+
+
+
+
+ +
+
+
+
+
+
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
412
we can use knowledge of the cyclic codes of length n/2 to find better lower bounds. Definition 2: Suppose C is a ternary cyclic code of length m . Then the code of length 2m defined by:
c2:= {c(z2) + zc’(z2) Ic(x), c’(x) E c }
Theorem 4: A ternary cyclic code D of length n = 2m is a square, if and only if R = m R, where R is the defining set of D and m R:= {m T mod n)T E R}.
+
+
Proof: j: Suppose D = C 2 , where C is a cyclic code of length m . Suppose a is a primitive nth root of unity. Then obviously am = -1 and so am+i = -ai for all i. Let i E R. Because D is a square of C , we know that c(a22) aicJ(azi)= 0, for all c ( x ) , c’(x) E C. But then also - ( ~ ~ c ’ ( a ’= ~ ) 0, for all c ( x ) , c’(x) E C , since c’(z) E C implies -c’(x) E C. So we also have c(( -ai)2> (-aZ)c’( ( -ai)2) = 0, for all c(x), c’(x) E C, and so -a2= a m f i is a zero of all elements of D. This means that also m i E R. e:Again a is a primitive nth root of unity. We write p:= a2 and /3 is a primitive mth root of unity. We shall prove that if R = m R, then D = C 2 ,where C has defining set R’ = R mod m. Take d(x) E D and write d(z) d ( - x ) = a ( z 2 ) ,d ( z ) - d ( - x ) = xb(z2),where u ( z ) , b(x) E GF(3)[z]/(xm - 1). We have d(a2) = d(-ai) = 0 for all i E R. So also .(a2;) = b(aZi)= 0 for all i E R. But this means that a ( @ ) = b(p2) = 0 for all z E €2’ d(-2)) so a ( z ) , b(z) E C and so d(x) = 2((d(z) (d(z) - d ( - 2 ) ) ) = 2(u(x2) z b ( x 2 ) )E C2. On the other hand, if c(@) = c’(T) = 0 for all i E R’ (i.e., c(z), c‘(a) E C), then c(aZi) f aic’(aZi) = 0 for all i E R’. But then c ( x 2 ) xc’(z2) has zeros aiand -ai = am+i for 0 all i E RI. So c(x2) xc’(x2) E D.
+
,+ +
+
+
+
+
+
e
0
is called the square of C. It is easy to see that C2 is a ternary cyclic code again, which has the same minimum distance as C. Moreover the following holds.
+
are squares. Of course CL 5 C 5 Cu. Suppose c(x) E C and ~ ( z ) CL. Then c(z) is nonzero both on the even positions and on the odd positions, since otherwise .(ai) = 0 would imply c(a-’) = c(am+2)= 0, and c(z) would be an element of CL.But c(x) is also in the square Cu. So wt(c(x)) 2 2du.
+
Corollary2: If dL 2 2du, then d = dL. Proof: From Theorem 5 we have: d 2 dl;. But CL 5 C,
so we also have: d 5
Example 22.9: n = 22, R = {2,4,6,7,8,10,12,13,14,16, 17,18,19,20,21}. Cu is the code of Example 22.8 and so dU 2 5. CL has defining set {0,11} and so d L = 11. Using Theorem 5, we get: d 2 10. Theorem 6: Suppose C is a ternary cyclic code of length n = 2m with defining set R. Let CE be the cyclic code of length n with defining set R U {jlj 0 mod 2) and CO the cyclic code of length n with defining set R U { j [ j G 1 mod 2). Then every element of C can be written as (a b, -a b), where (a,-a) E C E , and ( b , b ) E CO (a and b are of length m), and a) w t ( ( a + b , - a + b ) ) = 3(m-loo)-wt(a)-wt(b), where 100 := I{ilai = bi = O}l, b) w t ( ( a + b , - a + b ) ) 2 2max{wt(a), wt(b)} -min{wt. (a),wt(b)}, c) wt((a b,-a b ) ) 2 min{dE, d o , max{d~/2,d0/2}}, where d E and d o are the minimum distances of CE and CO, respectively, and (a b, -a b) # 0. Proof: By Theorem 3, we can write every codeword of C as the sum of a codeword of CE and a codeword of CO. But every codeword of CE is of the form (a,-a) and every codeword of CO is of the form (b,b) (since xm - 1 and xm 1 are divisors of their respective generators). Now, define for given a and b in GF(3)” : l,, := I{ilut = T and bi = s}I(r,s = 0,1,2). Then, we have
Theorem 5 (“Paris by Night”): Suppose C is a ternary cyclic code of length n = 2m and defining set R. Let CL be the cyclic code of length n with defining set R U (m R) (the so called lower square of C ) and Cu the cyclic code of length n with defining set R n ( m R) (the upper square of C). Call d L and dU the minimum distances of CL and Cu, respectively. Then, for the minimum distance d of C, we have d 2 min{d~,2du}.
+
+
Proof: Notice that CL and C” are ternary cyclic codes. Moreover, using Theorem 4, it is easy to see that CL and Cu
+
+
+
+
+
+
+
+
Notice that the code of Example 22.3 is the square of the code of Example 11.1 Example 22.8: n = 22, R = {2,6,7,8,10,13,17,18,19, 21). By Theorem 4, this code is the square of the code of Example 11.2. So d 2 5. This also explains why the code has no words of weight 7 or 21. In the explanation of Table 111, the following theorem will be quite useful. We dedicate it to Paris, since it was proved (at night) in a hbtel room in that city.
0
dL.
wt(a)
=
+ wt(b) + wt(a + b) + wt(-a + b)
Erf0 E, +
=
+
j,,
E,c,,,
E,
&,
+
E,E,+,
Er#o .
(E, lr,
+
1,s
Esf0 = Crfo (E, +3
+
E,#-,
+
c,,,
IT,)
10s
1,s
+ 2 E, 1rs) + 3
= 3 ( x . # 8 ~ ’ . .+xs#olos =3(Cr
1,s
E, -
= 3(m - loo).
1,s
100)
10s
__
VAN EUPEN AND VAN LINT: ON THE MINIMUM DISTANCE OF TERNARY CYCLIC CODES
And so wt((a
+ b, -a + b ) ) = wt(a + b)
+ wt(-a + b)
-
loo) - wt(a) - wt(b)
3("
-
L 3 max{wt(a). wt(b)) - wt(a) - wt(b) - 2max{wt(a), wt(b)} -
min{wt(a), wt(b))
, ( b )) . L max{ wt ( a ) wt
413
so wt(c) 2 13) or wt(c') = 0. So, if wt(c) < 13, then wt(c) is even. This is also obvious if we use Theorem 6 and observe that CO has defining set (0). Suppose C is a ternary cyclic code of length n = 2m and c = ( a b, -a b ) , where ( a ,- a ) E CE and ( b , b ) E CO, is a codeword of C. Then obviously c' = 2b (see Lemma 1) and so wt(c') = wt(b). So actually d o / 2 equals the minimum distance of C'. One could also define C": = { ( C O - e,, -c1+ cm+l. . . *rm-l c ~ ~ - ~ ) I c E C} and derive something similar for dE/2, but unfortunately C" need not be cyclic. We shall not go into detail about this. The reader can verify easily that in the ternary case Theorem 6 is always at least as powerful as contraction to length n/2. A ternary code of length n = 2m is called double if { , j I , j 1 mod 2) c R, and it is easy to see that the minimum distance of a double code is twice the minimum distance of its contraction.
+
+
So a) and b) have been proved. To prove c), we observe that there are three possibilities: 1) wt(a) = 0; then w t ( ( a + b . - a + b ) ) 2 d o ; Special Cases 2) wt(b) = 0; then wt((a b, -a b ) ) 2 d ~ ; 3) wt(a) # 0, wt(b) # 0; then max{wt(a).wt(b)} 2 Example 26.6: n = 26, R = { 1,2,3,6.8,9,17,18,20,23, max{dE/2, d0/2}. 0 24.25). ~ ~ H I F=T6. We wish to prove that d 2 7. Suppose Example 22.10: n = 22, R = {0.1.3.5,9,15). CE is c(.x) is a codeword of weight 6. Obviously we have equivalent to the code of Example 22.1 and so d E 2 12. CO (e(.) c(-.z)j2 - (.(:Ti) = c(x)c(-x). (1) has defining set (0) and so d o = 22. Applying Theorem 6c) we get d 2 11. But a codeword of weight 11 cannot occur by Notice that c ( x ) c - x ) is in the code with defining set {0,13}. Corollary 1 a). So d 2 12. Moreover, gcd(c(z).z2 - 1) = gcd(c(-z),x2 - 1) = 1 (since Theorem 5 and Theorem 6 will be used more optimally in the codes corresponding to R U (1) and R U {-I) have the next section. We shall now discuss the relationship between d 2 8), and so c(.c)c(-x) is equal to (.zz6- 1)/(z2- I), up to Theorem 6 and contraction. The method of contraction was cyclic shift or multiplication by 2. We wish to prove that this is introduced in [l].For the parameters that we shall treat in this not possible for any W O , 1111, where 7110: = wt(c(z) c ( - x ) ) paper, we could actually do without the method. However, one and 7/11: = wt,(c(:r) - c(-x)) = fi - 7110. First, we have to special case is so often quite useful that we mention it here. make some preparations. We apply Theorem 3 to get e(.) = Lemma 1: Let C be a cyclic code of length 2m with c g ( z ) c ~ ( x )where , C " ( : I : ) is in the code with defining set defining set R, containing as even integers the set 2R' (where RU (0, 13) and c1(.c) is in the code with defining set {0,13}. R' c {0,1;...m - 1)). Define the contraction C' as We do this because looking at the zeros of co(z) and ~ ( zwe ) 0 and c o ( ~ ; ) c ~ ( - x - ' E ) 0, and so have co(:e)co(--~c-~) (co(:I;),c"(-.c)) = (co(.c).c1(-2)) = ( c " ( - x ) , c l ( z ) ) = 0. s o wg = (e(.) C ( - 3 ; ) . e(.) e(-.)) = Z ( c ( x ) ,C(-.)) = and write e::= e, e,+, (0 5 i < m). Then C' is a cyclic Z(co(:c) c ~ ( x ) , c ~ ( - x + C ( - I C ) ) ~and ( C ( Z ) - C ( - I C ) ) ~ , of their sum must be 13. But also by (l),we must then have that all nonzero coefficients in ( c ( x ) C ( - X ) ) ~ have the same sign. This is a contradiction, since the nonzero coefficients in C(Z) +e( - E ) have the same sign. So c ( x ) cannot have weight 6 and so d 2 7. Example 26.7: n = 26, R = (0,13,14,16,17,22,23,25}. Here, d s H I F T = 5 and the Roos bound equals 6 (which is rather remarkable). To prove this, observe that the consecutive sets (13 3 j , 14 3 j ) are contained in R if j E (0, 1 , 3 , 4 } . By Roos [6],this means that d 2 3 4 - 1 = 6.
+
+
+
+
Explanation of Table I By computer we calculated the minimum distance of all ternary cyclic codes with length < 40. In many cases this minimum distance is equal to the BCH bound. In Table I we give the minimum distance of all ternary cyclic codes of length < 40, and minimum distance not equal to the BCH bound. So the minimum distances of all other ternary cyclic codes of length < 40 can easily be found by computing the BCH bound. Also by computer we calculated the SHIFT (see Definition 1) of the codes listed in Table 1. We see from Table I that in many cases the minimum distance equals the &HIFT. In the other cases, some minimum distances are equal to bounds given by other theorems in this section (or belong to the special cases). In these cases we refer to an example in this section. Unfortunately some minimum distances are left, that could not be derived theoretically by us. These are indicated by a question mark. By the shifting certificate we mean the sequence r l , rz, . . . ,rdsHrFT-l, which has been put in the order of Definition 1. Of course we did not mention codes that were equivalent to a code that was already in the list. We saw that we could call two codes equivalent, if their defining sets were the same up to multiplication with an integer coprime to n. But if n = 2m, we can also call two codes equivalent, if their defining sets are the same up to a shift over m (this corresponds to a substitution of IC by --z in the codewords c(z), and so the minimum distance stays invariant). So we call two codes equivalent if their defining sets can be obtained from each other by some combination of multiplying with an integer coprime to n and shifting over m (where n = 2m). To find the defining sets of the codes in Table I and to find equivalent codes, we refer to the Appendix.
m.
LONGER CODES
In this section, ternary cyclic codes will be studied with 40 5 n 5 50. Computing exact minimum distances for these
codes is very time consuming, so we only computed lower bounds. Of course, first of all we used shifting, since this method seems to be very powerful as we saw in Section 11. But, for some codes, shifting also took too much time and so we had to compute a lower bound using one of the other theorems in Section 11. Of course, we also tried to improve the dsHIFT using one of the other theorems.
n=40 There are too many ternary cyclic codes of length 40 to give a list of all of them. So for each dimension k we computed the best of the lower bounds for the minimum distance of all codes of dimension IC, and only listed the corresponding codes (Table 11). For some codes in Table I1 we were able to find an upper bound for the minimum distance (just by computing some codewords), that equaled the lower bound for its minimum distance. This is indicated by a boldface entry in the column “d,, 2.” In the last column of Table 11, an explanation is given for the lower bound. If this lower bound is not the BCH bound and is not explained in some example, then the bound equals the d s H I F T and the sequence rl,rz,..‘,rdSHIFT-lis given. Example 40.1: n = 40, fi = (1,3,9,27}. We will use Theorem 5 to prove that d 2 24. Notice that CL has as defining set all integers modulo 40, and so we may take d L infinite. Cu is the square of the code of length 20 with defining set (0,2,4,5,6,8,10,11,12,13,14,15,16,17,18,19}, which has minimum distance at least 12 by the BCH bound. So by Theorem 5 we have d 2 24. Example 40.2: n = 40, R = (0,1,3,9,27}. Again we will use “Paris by night” to prove that d 2 22. Also again d L is infinite. Cu is the square of the code of length 20 with defining set {0,1,3,7,9}, which has minimum distance at least 11 by the BCH bound. So by Theorem 5, we have d 2 22.
40<ns50 In Table 111, we give a complete list of all ternary cyclic codes of length 40 < n 5 50, having a lower bound for the minimum distance more than the BCH bound. Again we also computed upper bounds and if the lower bound equals the upper bound, we indicate this with a boldface entry in the column “d 2.”In the last column one can find an explanation for the lower bound for d. If shifting gives the best bound, then a shifting certificate is given. Otherwise we either refer to a theorem (“Paris” means Theorem 5) or to an example. Sometimes we also give an explanation for the upper bound, by giving a subcode with a known minimum distance contained in our code (e.g., “>nr.13”). The “-”-sign denotes “a code equivalent to.” If the code is double or a square, then the minimum distance can be calculated from Table I and there is nothing else to explain. Notice that not for all codes the shifting bound is given in Table 111. Here the computer spent too much time to compute the shifting bound. Because we need lower bounds for the minimum distance of all proper subcodes of a code to compute the shifting bound, we have put other lower bounds in the computer where shifting bound was not known. Notice that this means that the entries in the column
415
VAN EUPEN AND VAN L I N T ON THE MINIMUM DISTANCE OF TERNARY CYCLIC CODES
I, r 1 1
13
- ir , > > , >
13 16
> I
> 3 >
, 3 ,
3
1 > 1
$
I1 I1
3 4 5 6
TABLE I (PART 1)
k 1
L6
1 > 1 >
8
16
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
16
>
16
3
1 > 1
20 20 20 20 20
5
3
5
7
5
8
8
3 10 12 12 13 13 13 14 14 I5 5 6 6 7 10 10 11 11 12 12 15 16 11 12 6
5
9 B 5
41
42 43 44 45 46 47 48 49 50
51 52 53 54 55
56 57 58 59
60 61 62 63 64 65 66
67 68 69
TO 71
72 i3
74 i5 i6 77
20 20 20
20 20 20 20 20 22 22 22 22 22 22 22 22 22 22 22 22 23 23 26 26
6
26 26 26 26 26 26 26
6 7 7 7 8 8 9
26
9 9 9 10 10 10 10
26 26 26 26 26 26 26 26 26 26 26 26 26 26 26
IO 10 10 11
11 11 11
12 12
2€
12
2z 2f 2f 2f 2f 2f 2f 2t 2t 2f 2t 2f 2t
12 12 12 12
2t
- -
13
13 13 13 13
13 13 13 13 13
5 4 4 4 4
I 4
4 4 12 12 10 10 9 6 7 6 7 5 4 4 9 8
15 12 12 14 12 12
11 10 9 9 9 11 9 9 8
9 10 10 3 8 6 7
I 4 4 4 4 4 4
10 10 8 8 7 5 6 5 6 4 4 4 7 6 13 12 12
12 12 12 11
10 9 9 9 10 9 9
8 9 10 10 9 8
6 7
2
1,4,7 1~2.7
5 6 5 3 4 3
5,10,11
3 3 3 3 3
4.5,lO 2,5,10 Q,10.11 4.5 10,11 0.1,4,7.11 2.4,7.11 1,4,7,11 Z34,7
,
8
7 5 4 5 4 4 4
3 3 6
5
,
i 6
8 9 7 7 5 6
6 6 6 8 8 6 8 7 6 8
6 6 6
6 7 6 5 5
8
4.5,7,8,13.14,17 2,4,7,8,13,14,17
8 7 9 7
8 8
8
11.12.13,22.10,19,20,21,14,15,4,0 19,21,13,23,20,22,14,24,7,0.1
6 8
8
6
4,7,11 ’2.7,ll 4!7 2.7 0,7,11 4,11 0,5 5
10 10
5
- -
10,16,11,1i,12,13,0 10,11,16,12,17,13,2 11,4,16,12,15.10.0 11.15.10,4.5 11.12.19.10.0 10,19.0 4,15,0 11,17,5 4.15,5 14.5,2 10,19.0 4,15.5 11,17,0 13.19.14,20.11,12,3.0,1 10.17,12,13,11,6,16,7,2 11,12,13,14,3,4,1 12.6,17.16,18,7,2 11,12,20,19,13.0 7,10,6,0 11,12.19,13,4 6.17.7,2 16,13,12,19.4 6.7,2 19.1 1,o 12,14,4 14,10,21.19,20,0 14,10.21.19,5
11
9 10 8 8
5 6 5 5
6 4 7 5 5 5
Expl.
Ex 11.1 Ex 11.2
k,i,5
3
i
6
L.5,8,10
3,4.5,10,11 4,5,10,11 3,4,10,11 3.5,10.11 3,4.5,10
8
..
7,10.6.0 5.7,~ 10.11,7,12.0 5,0,11,5,2 7.13.8.14,3,0,1 4.i.12.13.5 4,7,0 7,14,13,5
1,1,5.8.10
5 3 3
8
7 8 6 7 7 6 8 6 6
t.i.0
5
9 9 8 9
8
Shifting certificate
1,4,5 ),2 4
6 6 5 5
6 4
c
2,i,8,13,14,17 1.i,8.13.14.17
4~5,1,l3~14~17 1.4.7.13.14.17 2,i,8,14,17 2.4.8.14.li 1,4,8,14,17 1,2.8,14.17 O,i.8.13,14.17 0,5.8.13,14.17
0,1,7,13,14,17 0,1,4,13,14,17 0,4,7,8.13,17 7,8,13,14,17 5.8,13,14.17 4,8.13.14.17 1.8.13.14.17 4.i.13.14.17 1,7,13,14,1i 4,5,13,14,17 2,4.13.14,17 1.4,13.14.17 1.2.13.14.17
SHIFT" need not equal the ~ ~ H I FofTDefinition 1 (but it is easy to see that the d s H l F T from Table I11 is still at most d). Example44.1: n = 44, R = {2,6,10,11,18,30,33}. We shall use Theorem 5 to prove that d 2 22. CL has defining set {11,33} and so d L = 22 CLI is the square of the code of length 22 with defining set {0,1,3,4,5,7,9,12,13,14,15,16,17,19,20,21},which has minimum distance 12, since it is equivalent to the code of
10,11,12,13,21,20,19.14,15,4.5 16,6,12,20.10.21,22,11,23,7,2 1,9,3,13.17,24,23,12,8,2,4 19,22,10,23,20.14,15,21,4,5
E x . 22.1 E x . 22.10
Ex.
22.4
E x . 22.9 E x . 22.2
Ex. Ex. Ex. Ex. Ex.
22.3 22.5 22.6 22.7 22.8
Ex. 23.1 E x 23.2 E x . 26.1
E x . 26.4
1,9.17,22.3,24.23,2,4 10,22,21,20,11,12.7,0 20,21,22,16,24.6,7,0 22,12,8,25,23,24,0,1 22,24.25,0,16.1,23,2.3 10.11.22,21,20,12,7,4 20,16.6,22,21.23,7,2 7.23.21.20,22,8.1 23,16,9,13,12,8,3,1 2 2 , 1i,8,23.1.13,24,2.3 12,13,21,15,11,10,14,4,5
10.3,16,22,12.11.9.1 20,16,6,23.21,7,2 22.16,23,17,2 22,9.24,3,8.1 22.23,1,16.2.3 20.21,22,23.24,7,0 13,22,14.23.15.16.0 23.24,13.0.22,8,1 11,12,16,25,21,10,0 22,13.0.16,1 9,0.12,3,1 10,11.7,12.0 20.21,22,23.i,8 13.14.22.15,23,16.5 22.12.23.13.4 23,16,13,24,8.1 12,21.10,11.13.4 22,16,21,13.1 13.14,22.16,12,4.5 12.13,17.16.2 9.13,22,3,1 1.17.13,2.16.22.3
E x . 26.2 E x . 26.3 7
Ex. 26.5
Example 22.10. So also d~ = 12. By Theorem 5, we have d 2 22. By Corollary 2, we have d = 22. Example 44.2: n = 44,R = {0,2,6,10,11,18,30,33}. We wish to apply Theorem 6 to prove that 21 is the best lower bound for the minimum distance that we can find (notice that “Paris by Night” gives d 2 20). Suppose c = (a b, -a b) is a nonzero element of the code of weight less than 21, where (a,-a) E CE and ( b , b ) E Co. If c E CE or c E CO, then
+
+
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
416
TABLE I (PART2)
- - - nr - - 6 78 3 L
79 80 81 82
83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107
108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 I4e 141 14f 14: 15C 151 152 15: 154 15: 15f
-
6 6 6 6 6 6 6 6 6
6 6 6 6 6 6 ,6 26 26 76 26 26 26 26 26 26 26 26 26 26 28 28 28 28 28 28 28 28 32
3 4 4 4 4 4 4 4 4 5 5 5 5 6 L6 16 I6 16 16 16 16 16 17
I7 18 18 19 19 20 6 7 8 10
18
32 32 32 32 32
19 19 20 4 7 8 9 9 13 14 14 15 15 16 16 16 17 17 17 17 18 18
32
18
32 32 32 32
18 18 19 19 19 19 20 2c 21 21 1: 1:
32 32 32 32 32
32 32 32 32
32 32
32 32
32
32 32 32 32
32 3: 3:
3: 35
1f
3: 3:
1f 2: 2:
3! 3: 3: 3' 3;
-
2:
11 1: -
ELK
b
>
>
,
> >
i
>
i >
>
, > , 4 5 4 4 12 12 12 7 4 4 4 4 16 12
8 10 8
6 6 6 6 6 4 6 6 4 6 6 6 4 4 4 5 5 4 4 4 5 4 4 4 4 li 12 8
li
2A
>
, I > 4 4 12 I2 12 7 4 4
I
I
I I
I
IO IO i
, 3 3
6 4
3
16 12 B 10 B 6
12
6
5 5 5 5
6 6 6 4 6 6 4 6 6 6
4 4 4 5 5 4 4 4 5 4 4 4 4 12 11 8 7 7 6
5 4 4 8 8
3
IO 3
3 5 3
3 5 5
3 5 5 3 3 3 3 4 1 3 3 3 4 3 3 3 3 10 10 6 6 5 4 4 3 3 5 5
ihifting certificate .i,8,13,17 .7.8,13,17 .8,14.17 ?8,14,17 ',8,14,17 -8*14,17 1?7,14,17 ,7,14,17 ,7,8,17 ,2,8,li l,7,13,14,17 1,4,13,14,17 1,7,8,13.17 l,4,8,13,17 ',13,14,17 l,13,14,17 2,13,14,17 l.13,14.17 r,8.13.17 4,8,13,17 l,8,13,17 4.7,13,17 2.4,13.17 4.14.17 2,8,1? ),13,14,17 ).7,13,17 13,14,17 ),7,17 14,17 D,2,4,5,7,14 274v5,7>14 2.4A7 2,4,5 D,4,7,14 4,7>14 2,i,14 4>7 0,1,2,5,10,16,2b 1,5,8,10,16,20 0,1.5.8,10,16 1.5,10,16,20 1,5,8,10,16 2,5,10,16,20 0,1,4,10,16,20 4.5.8.10.20 5,8,10,16,20 1,4,10,16.20 0,4.5.8,16,20 0,1,10,16.20 5,8,10,20 4,5.8,16,20 5.10.16,20 1.10.16,20 2,5,16,20 0,5.8.16,20 0.4,5,16.20 4,5,8,20 0,1,10,16 5,8.10 5.8.16,20 1.8.16,20 4,5,16,20 1.10.16 0,5.16.20 5,8,20 5,16,20 1,16,20 0.2,5,7 2.5,7 0.2.5 2,s 0.2.7 0,2 2 0.5.7 5.7 0.2 2
wt(c) 2 22 (C, has defining set { 11,33}and CO is the code number 7 in Table 111). If wt(a) = 22, then by Theorem 6b) we also have wt(c) 2 22. So we can assume that wt(a) = 11. Now we can say immediately that wt(c) e 2wt(a)+2wt(b) e 1 2wt(b) $ 2 mod 3 (since b is an element of the code of Example 22.10 and so wt(b) f 2 mod 3) and so wt(c) # 20 so d 2 21. We shall show that we cannot improve this bound
+
0,23.11,7,4 !0.8,24.7,1 4.15,23.22,16,5 6,24,22.23.4 !2,16,23,17,2 !3,16,8,24,22.1 .1.16,10,21,4 !1,23,7,1,3 !0,24,8,7,1 !3,24,2,1,3 !3,22,21,16,0 !2,13,12,16,0 23.24,7.20,0 12,25.23,24,0 22,16,21.13,7 12.10,23,14,4 22,17,16.13,2 14.1,25,23,3 7.24.23.20,8 3,20.4.13,10 23.13.9,20,1 10,23.21.11,4 10,4,2,23,6 12,10.14.4 17,24.23,2 13,22,16,0 23.11,O 13,22.16,14 23,21,0 22,16,14 4,14,5,15,10,11,12,13.6,7.0 4,14,10,12,20,11,13,5,6,7,2 10,5,11,6.16,15,4,20,21,7.2 11,4.22.10,5,2 20,7,0 20,7.4 6.21,2 20,7,4 13,25.14,26,15,27,16,28,17,18,19,0,1,2,3 7.13,23,25,27.24.26.8,9,10,1 7,13,8,14,9,0,1 13,27.25,14,28.26,9,10,1 7,23,13,8,9,10,1 13,20,5.6,2 9,11,0,3,1 4,7,12.13,5 13,28.14,7,5 9.25,10,3,1 4.7,O 25,9,26.0,1 13,29,28.7.5 4.7.5 13,28.14,15,5 25.9,26,10,1 21,5,20,6,2 20.7,O 4.15,O 4,7,5 9,10,0.1 i,14.13.5 20,7,5 8.19,l 4.15,5 9,25,10,1 20.15,O 20,7,5 20.15,5 16,19,1 18,21.30,24,22,6,19,20,5,7,0 18,19,20.21,23.5.22,6,7.2 18,24,23.30.22.5,0 18,30,22.23,5,2 18.23,21.31.6,0 34.23.31.18.0 31,23,18,2 20.14,O 20.14.5 13.18.14.19,31.5.0 13.18.14,19.31,5,2
:xpl
Ex. 26.6
I
Ex. 26.7
, ?
, 7
by using Theorem 6. Let b(z)be the polynomial corresponding to b. Notice that b(z) dz b(-x) is an element of the code of Example 22.8 (minimum distance 5 ) and cannot be the zero word (since otherwise wt(c) 2 22f10 = 32 or c E CL,which has minimum distance 22). Since a is either zero on the even or zero on the odd positions, we have loo 5 11 - 5 = 6. So 22 - loo 2 22 - 6 = 16 and, by Theorem 6a), we have
VAN EUPEN AND VAN LINT ON THE MINIMUM DISTANCE OF TERNARY CYCLIC CODES
417
TABLE I1
k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
18 19 20 21 22 23 24
25 26 27 28 29 30 31 32 33 34 35
-
dmaz 2
40 30 25 24 22 24 22 20 18 18
16 16 14 14 14 12 12 11 11 10 9 8 8 8 8 7 6 6 6 5 5
4 4 4 3
G
Explanation
1,?.4,5,7,8,10.11,13,~0,22,?5 0,1,2,4,7,8,10,11,13,20,22,25 1,2.4.7,8,10,11,13.20,22,25 0,2,4,5,7,8,10.11,13.20,22,25 2,4,5,7,8,10,11.13,20,2?,25
BCH BCH BCH
0,1,4,7,8,10,11,13,?0,22,25
BCH BCH BCH
1,4,7,8,10,11,13,20,22,25 1,4,7,8,10,11,13,22,25 4,5,7,8,10,11,13,20,22,25 0,4,7,8,10,11,13,20,22,25 2.7,8,10,11,13,20,22,25 0,4,7,8,11,13,20,22,25 1,4,5,8,10,13,20,?2,25 0,7,8,10,11,13,20,22.25 1,2,5,8,13,20,22,25 0,5.7,8,10,13.20.22,25 5,7,8,10,13,20,22,25 0,7,8,10,13,20,22,25 7,8,10.11,20,22,25 0,7,8,13,20,22,25 8,11,13,20,22,25 0,10,11,13,20,22,25 10,11,13,20,22,25 0,1,10,13,20,22 7,11,20,22,25 4,13,22,25 5,7,20,22,25 0,13,20,?2,25 7,20,22,25 0,13.20,22 7,20,?? 5,22,25 20.22,25 22,25 13,20
wt(c) 2 3 * 16 - wt(a) - wt(6) = 37 - wt(6). We have three cases. 1) wt(b) 5 15: wt(C) 2 37 - 15 = 22. 2) wt(b) = 16: wt(c) 2 37 - 16 = 21, and also by Theorem 6b): wt(c) 2 2 * 16 - 11 = 21. 3) wt(b) > 16: wt(c) > 21 by Theorem 6b). Example 44.3: n = 44, R = {2,4,G. 10.11.12,16>18.20. 30,33,36}. We wish to apply Theorem 6 to prove that d 2 16. Suppose c = ( a b, -a b ) is a nonzero codeword of weight less than 16, where ( a ,- a ) E CE (having defining set {11,33}) and ( b , b ) E CO (code number 14). As in Example 44.2, we may assume that w t ( a ) = 11. If b(.r) is the polynomial corresponding to b, then b(.r) f b(-.r) is an element of the code of length 22 with defining set {0. ll} and we may assume that b ( r )fb( -.r) $ 0 (since otherwise wt ( b ) = 0 and so wt(c) = 22). So lo0 5 11-2 = 9 and 22-10" 2 13. So by Theorem 6a), we have wt(c) 2 3 * 13 - wt ( a )- wt(b) = 28 - wt(b). But wt(b) E 0 mod 3 (by Theorem 2 ) and we must be in one of the following two cases. 1) wt(b) 5 12: wt(c) 2 28 - 12 = 16. 2) wt(b) 2 15: wt(c) 2 2 * 15 - 11 = 19 by Theorem 6b). So d 2 16. Example 44.4: n = 44, R = { 1 , 2 , 3 , 4 , 5 . 6 . 9 , 1 0 . 1 2 , 1 5 , 16,18,20,23,25,27.30.31,36,37}. This code satisfies the condition of Theorem 2, and so d 5 0 mod 3. ~ ~ H I F=T 10 and so d 2 12. Moreover d = 12, since this code contains number 13 in Table 111. Example 44.5: n = 44, R = {0,1,2,3,4.5,6,9,10,12.15. 16,18,20,23,25,27,30,31,36,37}. This code satisfies the conditions of Corollary 1 a) (see Example 44.4). Here
+
+
Example 40.1 Example 40.2
19,20,33,32,~9,30,21,34.28,31,24,15,11,22,10,4,5 28,31.32,19,33,29,20,34,30,21,35.22,36,10,11,7,0 30,6,24.32.33,29.?1,19,37,18,17,16,22,7,2 31,32,19,35,36,23,37,34,11,24,33,20,21,7,0 8,27,4,12,37,31,35,30.26.?4,9,3,1 29,?0,23,24,38,30,37,33,32,21,19,7,0
BCH 20,21,22,30,38,29,37,24,23,7,0 23,24,15,31,7,29,21,30,22,20,5 37,24,38,30,21,22,20,23,7,0 19,20,23,32,22,33,?4,21,7,8 22,23,31,7,39,20,25,?1,0 34,16,37,38,19,32,24,0 19,10,37,34,30,25,0 30,10,37,33,19.25,11 26,30,9,39,37,0,1 21,25,34,22,19.?0,7
BCH 21,34,22,20,5 37.38,25,39,0 ?1.34,21,?0,7
BCH BCH 34,25,5 34,25,20 34,25,22
BCH
SHIFT
= 10, so we have by Corollary 1 c) that d 2 11. Example 44.6: rt = 44, R = {0.7,8,13,14,17,19,21,22. 24,26,28,29.32.34.35.38,39,40,41,42,43}. We wish to apply Theorem 6 to prove that d 2 10. Suppose c = (a b, -a+b) is a nonzero codeword of weight less than 10, where ( a ,- a ) E CE (number 19) and ( b , b ) E CO (number 14). Notice that CE is a square with minimum distance 10, i.e., if wt(a, - a ) < 20, then ( a ,- a ) (and also a) is either zero on the even or on the odd positions (and if it is zero both on the even and on the odd positions, then wt(c) 2 18). Let b ( r ) be the polynomial corresponding to b, then b ( s ) f b ( - s ) has zeros 1 and -1 and so b must have weight at least 2 on both the even and the odd positions (if b is zero on either the even or on the odd positions, then b = 0 and wt(c) 2 lo). We have three cases. 1) w t ( a ) > 9 or wt(b) > 9. Then, wt(c) 2 10 by Theorem 6b). 2 ) wt(a) < 9 and wt(b) = 9. Then, wt(c) 2 10 by Theorem 6b). 3) w t ( a ) = 9 and wt(b) = 9. Then by the observations just made loo 5 2 (11 - 2) = 11 and so by Theorem 6a) we have: wt(c) 5 3 * 11 - 9 - 9 = 15. So d 2 10, and since code nr.19 in Table 111 is contained in this code, we have d = 10. Example 44.7: n = 44, R = { 1.3,5,8,9,11,15,22.23,24, 25.27.28,31,32,33,37,40}. We wish to apply Theorem 6 to prove that d 2 8. Suppose c = ( a b. -a b) is a nonzero codeword of weight < 8, where ( a .- a ) E C:E(equivalent to number 13) and ( b . 6 ) E CO (number 38). CE is selforthogonal and has minimum distance 12, so wt(c) can only be
+
+
+
+
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
418
n
-
ii-
P
2
41
3 4 5
41 41 41 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44
3 17 14
nr. 1
6 7 8 9 10 11
12 13 14 15 16 17 18 19
20 21
22 23 24 25 28
27
28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58
59
60 61
62 63 64 65 66 67 68 69 70
71
72 73 74 75 76 77
-
k
-
!L -rpr 0 8
8
2
2
5 6
6 7 7 8
8 10 10 10 11 11 11 12 12 12 12 12
12 13 13 13 14 14 14 15 15 15
--
F
I
25
I2
IO
L 3
6 4
6 3 4
12 L8
La 14 L2 LO 16
La La 14 LO LO L4
La LO
I1 LO
L2 12 B
1 IO
LO LO
I
I IO
LO
I
3
I LO
LO LO
r
5
3
14 10 8 11 10
LO
10
10 8
P 5 6
7 3 3
7 7 S T
16
12
10
7
16 16 16 16 17 17 17 17 17 18
11
10 10 10 8 8 8 8
7 5
10
7 7
12
ia 8
10
10 B
18
12 11 10 10
18 18
10
18 19 I9
20 20 20 20 21 21 21 21
22 22 22 22 22
22 22 23 23 23 23 24 24 24 25 25 26
28 21
-
I)
6 8 8 6
7 8
10 8 8 8 8 8 8 8 10 7 8 5 10 7 5 8 8 6 8
4
8
5 8
6 5 8 6
5 4
10 8
10 10
ia
8 8
e 11 Q
e 8
10
e 7
e. 8
I e. 8
8 7 K 8
e e.
8
7 -
Explanation
L2.4.7.8
~8,34,30,36,17,5,23,19,~,25,15,10,12,18,20,16,4,6,0
44.7.8 L.73
L,33,21,5,34,22.23.17,16,15,20.10,18,19.6.7,2 r,21 p , i i,20,15,25.29.4.i0,5 1.33,31,39,8,6,2 r,21.16,30,19,10,8 iouble iouble iouble Example 44.1 double Example 44.2 ‘Paris‘,3 n r . 8 iquare double double ‘Paris’,>nr.l3 double double square Example 44.3 square square double double 1Paris’,3nr.19 21,22,24,25,27,13,41.3t,39,23,7,8,l.>nr.17 ‘Paris’, 30,121 rquare 23,37,24,38.25,39,26,7,8~ 1 square ‘Paris’, 311.13 ‘Paris’, 3 n r . 1 3 double ‘Parid’, 311.13 ‘Paris’,3 n r . 1 3 ‘Paris’, 3 ? m r . 1 3 ‘ParL’,>~nr.l3 double ‘Paris’,Dnr.lS ‘Paris’83 n r . 1 9
42.8
I 14 I4 IO I2 IO ’1
La
TABLE I11 (PART 1)
5
8 8 6 4 7 5 5 7 8
7
7 8 8
6 7 7 8 3 5
4
7 5
4 5
8 5 5 7
7
5 4 5 7 4 4 5 4 4 5 5
r,a
1,1,4,7,8,11.14,22 1,4,7,8,1l.I4,22 L.2.7.8.11,14.22 ).1,4,7,8,14,22 1,4,7,8,11,14 L.4,7,8.14,22 1,2,7,0,14.22 1,2,4,7,8,11,14,22 1,1,7,8,11,14,22 1,1,4,7,11,14,22 2,4,7,8,11,14,22 1,7,8,11,14,22 1,4,7,11,14,22 D.2.4.7,8,14p22 D,1,7,8,14,22 0,1,4,7,14,22 2,4,7,8,11,14 1,7,0,11,14 1,4,,7,11,14 2,4,7,8,14,22 1,7,8,14,22 1.4,7B14.22 2,4.7,8,14 1,7,8,14 1,4,7,14 0,4,7,8,11,14,22 o,a,7,8,ii,i4,22 0.1.7,11,14,22 4,7,8,11,14,22 2,7,8,11.14,22 1,4,8,11,14,22 1,2,8.11.14.22 1,7,8,11,22 0,4,7,8,14,22 0,2.7,8,14,22 0,1,7,14,22 4,7,8,11,14 2,7,8,11,14 4.7,8,14,22 2.7,0,14,22 1,4,8,14,22 1,2,8,14,22 1,7,8,22 4.7.8,14 2,7,8,14 0,7,8,11,14,22 0,4,7,11,14,22 0,2,7.11,14,22 0,1,4,11,14,22 7,8,11,14,22 4,7,11,14,22 1,4,11,14,22 4.7,8,11,22 0,7.8,14,22 0#4,7,14,22 0,2,7,14,22 0,1,4,14,22 7.8,11,14 4.7,11,14 1,4,11,14 7,8.14.22 4.7,14,22 1,4.14.22 4,7,8.22 7.8,14 4,7,14 1,4,14 0,7,11,14,22 0,1,11,14,22 7,8.11.22 1,8,11,22 0.7,14.22
37,21,13,25,42.0,1,3nr.33 ‘Paris’,3111.22 <Paris’,3 n r . 2 2 ’Paris’.>nr. 19 ‘Paris’,3 n r . 1 9 ‘Paris’,>zznr.lS ‘Paris’,3 z n r . 19 21,39,23,27,7,8,1,>nr.38 ‘Paris’,3 n r . 2 8 ‘Paris,, 3 n r . 2 8 E x a m p l e 44.4 square 10,18.38,43,41.8,0 square E x a m p l e 44.5 ‘Paris’,3nr.52 ‘Paris’,3 n r . 5 4 32,21,40,16,19,7,4 E x a m p l e 44.6 square 38,13,34.17.6,21,0,3nr.53 square 38.39,32,40.33,7,8 square square 38,39,13,28,34.7,8 ’Paris’, 3nr.60 ‘Paris’. 3111.62 16,35,12.20,39,7,4, Jnr.58 38,39,13,28,34.7.8 square square ‘Paris’, 311.33 ‘Paris’, 3111.54 Theorem 6 ( i i i ) , 3 z n r . 5 4 E x a m p l e 44.7 ‘Paris’, 3111.60
2) m,- loa 2 7. Then, by Theorem 6a), we have: wt(c) 2 smaller than 8 if wt(a) = 6 (if wt(a) = 0, then c E CO, 21-6-6=9. which has minimum distance 8). But if wt(a) = 6, then a is zero on either the even or the odd positions. By Theorem 6b) So d 2 8, and since code number 38 is contained in it, we wt(c) can only be smaller than 8 if wt(b) 5 6. So we have have d = 8. two cases. Example 44.8: n = 44, R = {7,8,13,17,19,21,22,24,28, 1) m - ZOO = 6. Then c is in the lower square CL by the 29,32,35,39,40,41,43}. We wish to apply Theorem 6 to b) is a nonzero observations just made (c is zero on either the even or prove that d 2 6. Suppose c = ( a b, -a codeword of weight < 6, where ( a , - a ) E CE (number 19) the odd positions). But d L = 9, so this is impossible.
+
+
419
VAN EUPEN AND VAN LINT: ON THE MINIMUM DISTANCE OF TERNARY CYCLIC CODES
TABLE 111 (PART2)
- - -
-
nr n k 78 44 27 79 80 81
44 44 44 44 44 44 44 44 44 44 44 44 44 44 14 44 46 46 46 46 46 46 46 46 46 46 46 46 47 47
82
83 84 85 86 87 88
89 90 91 92 93 94 95 96 97 98 99 100 101
102 103 104 105 106 107 108
-
G 0,1.14.22 7.8.22 1.8.22 0 , 4 , 1 1, l 4 . 2 2 0.2,11.14.22
d >
0
28
28 30 30 30 31 31 31 31 32 32 32 33 35 36 37 11 12 12 13
5
6
5
7
6 4 4 4 4 4 4 4 4 4 4 4 4 4 4
4 4 4 4 4 4 4 4 4 4 4 4 4 4 18
18 10
4 5
3 3 3 3 3 3
3 3 3 3 3
18
le 9 9 9 9
22 22
23 23 24 24
8
8 4 4
33
34 23 24
-
3 3 3 12 12 10 10
7 9 7
6 6
8 6 8
6
9
4 4 9
8
8
0,7,11.22
4,11,14,22 2,11,14,22 4,8,11,22 7,11,22 0,7,22 4,11,14 2,11,14 7,22 0,11,14,22 8,11,22 11,14 0,2.5.10,23 2.5,10.23 1.5,10.23 2.5,lO 0,5.10.23 0.1,10.23 5,10.23 1.10,23 5,lO 1.10 0,10,23 10,23 0.5 5
6
5 6 3 3 6 5
Explanation ' P a r i s ' , > n r 62 Example 4 4 8 Example 44 9 square 10.33.0 square 11.33.4 10.33.2 32.11.4 19.11,7 square square 10,33,2 19,17,7 38.22.0 28,24,8 34,38.11 -.double 16,10.6,18,14.4,20,22,32,36,42,5,17,15,19,7,2,~nr.95
double 'Paris'. 3 r r n r . 9 7 square 38,27,39,22.41.0.9,1 'Paris', 3 n r . 9 9 E x a m p l e 46.1 square 38,27,28,39,29.9.1 20,28,0
20.28.10 38,44.30,19.43,10,22,0 43.29.44.38.39.10.5
TABLE IV (PART1 ) n
4
5 i
8
IO
m i n i m a l polvnomialr = + 2 1 2 2 + 1 3 r + 1 trivial trivial 1 r + 2
zeros e x p o n e n t s 0 1,3
2 3 4 5
r 2 + r + ?
1.3
2 + 1 r + l
2.6
13
1
X + 2
2
r 3 + r 2 t r + 2 r 3 t r 2 + 2 r 3 t 21' 21 2 r 3 + 2 r t 2
1 2
3
14
4 5 1 2
3 4 16
0
3
3 4 11
2
r2+2r+2 r + 2 r4 t 2r3 t r 2 + 2 z + 1 r 4 + r 3 + x 2 t r t 1 r+1 r t 2 r5+2r3+r2+2rt2 r5 r4 zr3 t r2 t 2
1 2
1 2
3
-+
+
+
14-2
+
+
r6 + 2 r 5 r4 + 2 r 3 + r2 + 2 1 + 1 s6 r5 t r 4 s3 r2 t r t 1
+
+
+
r+1
X t 2 r 4 + r 2 + 2 r2+r+2 ,=+I s4 2r2 + 2
4 5 r+1 6 7 r2+2r+2 trivial trivial 1 r + 2 2 r1+r3+2r+1 zr3 r2 2r t 1 3 4 s 4 + r 3 t 2 t r + 1 s ,'+I 6 r+1
+
17 19 20
+
7
+
permutations 1 2 3 3 2 1
+
r 4 + 2 2 + r + l
4 5*7 0 1.3,7,9 2,4,6,8 5 0 1,3,4.5,9 2,6,7,8,10 0 1,3,9 2.5.6 4,10,12 7.8.11 0 1,3,5,9,11,13 2.4.6.8,10.12 7 0 1,3,9,11 2,6 4,12 5,7,13,15
1 1 4
2 5 5
3 3 3
4 1 1
5 2 2
1 4
2 3
3 2
4 1
1 1
2 3
3 2
1 1 1 1
2 3 4 5
3 1 5 2
4 5 2 3
1 4
2
3
3
2
4 1
1 1 6
2 5 2
3 7 7
4 4 4
5 2 5
6 6 1
7 3 3
I 1 6
2 7 7
3 3 4
4 4 3
5 5 5
6 6 1
7 2 2
5 2 3 4
8
10.14
0 1,3,7,9 2,6,14,18 4,8.12,16 5,15 10
11.13,17,19
and (b, b) E CO (number 38). The only way to get wt(c) < 6, is to take wt(a) = 5 and wt(b) = 4 or 5. But CE is a square and so wt(a) = 5 implies that a is zero on either the even or
the odd positions. We have two cases. 1) m,- loo = 5. Then c is an element of the lower square, which has minimum distance 6,
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
420
TABLE IV (PART2) minimal polynomials 1 s+z z 2 zt4 zS3 2 2 I 3 2 z t 3 2 zS 2 4 .5 t4 2s3 t2 2 s ,5 2r3 2 2 t 2t 1 t + l 6 1 s+Z 2 2 1 2 2 2x6 s4 s3+ 21 2 +2s2 3 tll-+tlo+ t 9+ 2 s 8 + 2 s 7 +
n
22
zeros exponents 0 1,3,5.9.15 2,6,8,10,18 4.12,14.16,20 7.13.17.19,21 11 0
+ + + + + + + + + + + + + + +
23
+
25 I
28
+
1
2 3 4 5 6 7 8 9 10
I
1
2
+
+ + +
+
+
+
+
+
+
4 5 zS5 zS3 Z+ 6 t 2 + 1 7 s+1 trivial trivial 1 s+2 2 4 + s 4 + 2 3 s 4 + 2 + 2 4 s2+.+2 5 2 +zS4 + Z 6 s2+l 7 s4 2x2 + 2 8 S+l 9 z2+21+2 1 s + 2 2 ,I6 2J5 s14 ,I5 zl‘ 3 t16 4 t+i 1 s + 2
+
+ +I
+ +1
+
34
35
+
I
+
2 3
37
4 5 1 2
3
+
1
2 3 4
+ 2, + 1 + I
+1
4,8,12.16.20,24 5,11,13,15,17,23 7,21 14
0 1.3.9.11,17.19,25.27 2.6,18,22 4,lZ 5,7.13,15,21,23,23,31 8,24 10.14.26,30 I6 20.28 0 1.3,5,7. ,33 2.4.6.8 32 17 0
+=7
33
t2
S4+S3+.2+S+l s + 2 2 8 2,16 zt14 2 P + +P ,7 zS5 zS4+ 1 + 2 2 S18+s17+2z16+2s15 +s14+ 2s” 2s” Zs9+ +22 2 2 zS5 t4 zS3+ s 1 +22 t + Z Zt17 tl‘ . . . +It I *I7 SI5 . . .s 1 SI8
+
+ +
+ +
+d3 +
+
+
t+l
+
+
+
+
+
+
+
+
+
+
+
+
6 6 1
3 2
1 1 1 i 8
2 . 6 1 9
3 4 5 6 7 8 9 1 5 4 7 1 0 2 9 8 3 9 3 ’ 2 1 0 4 8 7 0 7 9 6 5 3 8 4 5 1 0 3 7 6 1 2
I 1 7
2 5 5
3 3 4
4 4 3
5 2 2
1 1 8
2 5 2
3 7 3
1 9 9
5 2 5
1 4
2 3
3 2
4 1
6 6 6
0 6 5 2 1
7 7 1
1
1.3,4,9,11.12.13.16.17,27,29.
+ + 2 2 + +Z + 1 s 1 2 + 210 + 2 2 + s 7 + s 5 + +zr4 + ,3 + 2 2 + zS + 1 z6 + z5 + z4 + s3 + +s+1
5 2 3
I none
s12_+2t~’1+2r10+_t9+2s8+
+ + + +
38
+ +
+
2 3
4 3 2
1
(1,2,3,4) (0,5,1O,I5,20} 5,10,15.20 0 1.3.9 2.6,18 4,10,12 5.15.19 7,11,21 8.20.24 13 14,16,22 17,23.25 0 1,3,9,19,25,27 2,6.10.18,22,26
+
+
5.7.10.11,14,15,17.19.20,21,
22
1 0
t 4 + . 3 + 2 + s + 1 t + Z z3+21+1 ,3+,2+.+2 s 3 + 2 + 2 S 3 + 2 2 + S + 1 z3+s2+2z+1 13 212 2s 2 S+l t3+21+2 t 3 t 2x2 1 s+2 S 6 + S ~ + S 3 + S + l zS5 ,4 zt3 2+ + I = +1 t6 2’ z4 s3 t2 s
+
1 1
1.2,3,4.6.8,3.12,13,16,18
+ 23 + 2 220 + S I 5 + 2 0 + 2 5 + 1
3
29 31 32
+
+t5 r + 2
1 2 3
_I
+
+ +
permutations 1 2 3 1 5 4 6 4 5
1
2
3
4
5
- 1
3
2
4
5
1 1
2 3
3 2
6 6 6
7 3 7
8 8 1
9 4 4
2,6,8,18,19,22,23,24,26,31,
32,34 5,10,15.20.25.30 7,14,21,28 0 1,3.4.~.9,10.11.12.16.21.
25,26.27.28.30.33,34,36 2.5.6.8.13.14,15.17,18,19,
20,22.23,24,29.31.32,35
0 1 . 3 , 5 , 7 , . . . .37 2,4,6,8,. . . .36 19
2) m - loo >_ 6. Then, by Theorem 6a), we have wt(c) 2 1 8 - 5 - 5 = 8. So d 2 6, and since the lower square has minimum distance 6, we have d = 6. Example 44.9: n = 44, R = {1,3,5,8,9,15,22,23,24,25, 27,28,31,32,37,40}. We wish to apply Theorem 6 to prove that d 2 7. Suppose c = (a+b, -a+b) is a nonzero codeword of weight < 7, where (a,-a) E CE (equivalent to number 19)
1234 4
3
2
1
and (b, b) E CO (number 38). The only way to get wt(c) < 7 is by taking wt(a) = 5 and wt(b) = 4 or 5 or wt(a) = 6 and wt(b) = 6. But CE is a square and so wt(a) < 10 implies that a is zero on either the even or the odd positions. We have two cases.
1) m - loo = wt(a). Then, c is an element of the lower square, which has minimum distance 7. 2) m - loo 2 wt(a) 1. Then, by Theorem 6a), we have
+
VAN
EUPEN
AND VAN LINT:
ON THE
42 1
MINIMUM DISTANCE OF TERNARY CYCLIC CODES
TABLE IV (PART3 )
n 40
~~~~
~~~~
minimal polynomials s+2
1
2 3 4 5 6 7 8 9 IO
+
+
+
11
41
s 4 + s 3 + r 2 t 1 s4 + r 3 +IS+ 1 2r3 s2 2s I r4 s 2 + s + 2 r4 2 s 3 r2 1 z 4 + r 3 + r 2 + s + ~ ,'+I r 4 + 2 + 2 1 + 1 r 4 + s 2 + r + 1 r + 1 r4+2z3+s+1 s2 2s+ 2 s + 2 28 216 s5 r3+ 1 +2s2 2 s7 zS4 s2+ +s 1 s8 s7 216 2 2 +
12 13 1
2
+
+
+
+
4
+
+
+
+ + +
+
+
+
3
+
+
+
+
43 44
46
aero$ exponents 0 1,3,9.27 2,6,14,18 4,12,28,36 5,151 7,21,23.29 8.16,24,32 10,30 11,17,19,33 13,31,37.39 20 22,26,34.38 25.35 0 1.3.9.14.27,32.38,40
permutations 1 2 3 1 6 3 4 1 9 1 2 0
12
11
6
12
4 1 4 4 7
1 1 1 1 1
2 3 4 5 6
3 4 6 2 5
4 6 5 3 2
5 2 3 6 4
6 5 2 4 3
0 1,3,5,9,15,23,25,27, 31,37 2,6,10,18,30 4.12,16,20,36 7,13,17.19,21,29,35, 39,41,43 8,24.28.32,40 11.33 14,26,34,38,42 22
1 1 9
2 5 2
3 8 6
4 6 8
5 2 5
6 4 3
0 1,3,9,13,25,27,29, 31,35,39,41 2,4.6.8,12,16,18,24, 26,32,36 5,7,11,15,19,21,33, 37,43,45 10,14,20,22,28,30, 34.38,40,42.44 23 0
1 1 6
2 4 3
3 5 2
4 2 5
5 3 4
6 6 1
1 1
2 3
3 2
2 3
3 2
4 5
5 4
1
1
2.6.13.18.23,28.35, 39 4,5,12,15,26,29,36. 37 7,16.19,20,21,22,25, 34 8,10,11,17,24,30,31,
1
5 3 2 5 1 13 13
6 7 0 9 2
7 8 7 7 4
8 1 8
9 0 2
8
6 10
8 3 4
9 9 1
8
10 9 6
1 1 2 9
11 12 1 1 2 1 3 1 11 3 1 3
13 5 3 5
5
33
9 1 2
3
r+l r + 2 S I 1 + 2 2 0 + r 9 + sa+ +zS7 ,5 ,3 I SI1 + r 1 0 + s 9 + 2 s 8 + +zS7 r 5 s 3 2
+ +
+ +
+ +
47
1,2.3.4,6,7,8,9,12, 14.16.17,18,21,24. 25,27,28,32,34,36, 37,42 5,10,11,13,15,19,20, 22,23,26,29,30,31, 33,35,38,39,40,41, 43,44.45.46 0 [1.2.3.4.5,6} IO, 7.14.21,28,35,42} 7,14,21.28,35,42
49 50
7 7 7
none
0
1
(1,3,7,9}*(0,10.20,30,40}~6
6
( 2 , 4 , 6 . 8 } I [O, 1 0 . 2 0 , 3 0 , 4 0 ) 5.15.35.45 10,20,30,40 25
wt(c) wt(b)
2
S(wt(a) 2 8.
+3
+ 1)
-
wt(a)
-
wt(b) = 2wt(a)
-
So d 2 7, and since the lower cquare has minimum distance 7, we have (1! = 7 . Example 46.1: 71 = 46, R = { 1 . 3 . 9 . 10.13.13.20.22.23. 25.27.28.29.30.31.34.35.38.39.40.41. 42.44). ~ ~ H I F=T 8. The conditions in Corollary 1 are satisfied. By Corollary 1 a) weights are congruent to 0 or 1 mot1 3. So d 2 9.
APPENDIX
In Table IV, we give a complete list of all irreducible factors of .r" - 1, 71 5 50, over GF(3) (a list for n 5 100 can be found in [9]). We also give the exponents of the zeros of these polynomials for some chosen nth root of unity. As we saw, a cyclic code of length 71 can be represented by a set of irreducible factors of .I.'~ - 1. In the last column, we give the permutations on the minimal polynomials, that permute
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
422
+
+
+
codes into equivalent codes. If xn-’ xn-2 . . . + x 1 is an irreducible factor of xn - 1, this will be denoted by “trivial.”
ACKNOWLEDGMENT The authors are very grateful to F. Bussemaker, Department of Mathematics, Eindhoven University of Technology, for writing the programs.
REFERENCES [l] J.H. van Lint and R.M. Wilson, “On the minimum distance of cyclic codes,” IEEE Trans. Inform. Theory, vol. IT-32, pp. 23-40; Jan. 1986.
M. van Eupen and C. van Eijl, “Minimum afstanden van temaire cyclische codes,” rep., Eindhoven Univ. of Technol., 1989. J.H. van Lint, Introduction to Coding Theory. New York: SpringerVerlag, 1982. F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes. Amsterdam: North Holland, 1977. C. R. P. Hartmann and K. K. Tzeng, “Generalizations of the BCH bound,” Inform. Contr., vol. 20, pp. 489-498, 1972. C. Roos, “A new lower bound for the minimum distance of a cyclic code,” IEEE Trans. Inform. Theory, vol. IT-29, pp. 330-332, May 1983. R. J. McEliece, “Weight congruences for p-ary cyclic codes,” Discrete Math., vol. 3, pp. 117-192, 1972. E. F. Assmus, Jr. and H. F. Mattson, Jr., “On weights in quadratic residue codes,” Discrete Math., vol. 3, pp. 1-20, 1972. R. J. McEliece, “Factorization of polynomials over finite fields,” Math. Computat., vol. 23, pp. 861-867, 1969.
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL 3Y. NO. 2. MARCH 1YY3
On the Minimum Distance of Ternary Cyclic Codes Marijn van Eupen and Jacobus H. van Lint
Abstract-There are many ways to find lower bounds for the minium distance of a cyclic code, based on investigation of the defining set. Some new theorems are derived. These and earlier techniques are applied to find lower bounds for the minimum distance of ternary cyclic codes. Furthermore, the exact minimum distance of ternary cyclic codes of length less than 40 is computed numerically. A table is given containing all ternary cyclic codes of length less than 40 and having a minimum distance exceeding the BCH bound. It seems that almost all lower bounds are equal to the minimum distance. Especially shifting, which is also done by computer, seems to be very powerful. For length 40 5 11 5 50, only lower bounds are computed. In many cases (deriven theoretically), however, these lower bounds are equal to the minimum distance.
of C is
where R = ( ; j l . , j 2 . . . . .,it)is the defining set of C (we do not require that a parity check matrix has independent rows). For purposes that will become clear soon, we will define, for I c (0.1 :... n - l}, M ( R ) I as the submatrix of M ( R ) consisting of the ith columns of M ( R ) , where i E I . One can see immediately that if R contains a consecutive set of Index Terms-Ternary cyclic code, minimum distance, shifting, length s, then M ( R ) rcontains a Vandermonde .s x s submatrix selforthogonal, contraction. and so r a n k ( M ( R ) l ) = s, whenever 111 = s . So if we define for every codeword c E C its support I as the set of I. INTRODUCTION positions where c is nonzero, then a codeword c with support CYCLIC CODE C of length 71, over the alphabet GF(q) I , 111 5 s (i.e., the weight of c(= wt(c) :=IT[)is at most (gcd(n,,q) = 1) can be characterized as an ideal in the s) cannot occur since M ( R ) I has full rank. So the minimum distance d = t t ( C ) : =niin,--c wt(c) is at least s 1. We call ring GF(~)[x:]/(:~:'~ - 1 ) with generator g ( . r ) (say), which is a divisor of :I:" - 1. A codeword of C will be written either as this bound the BCH bound (cf. [ 3 ] ,[4]), and we will write c ( x ) E GF(q)[:c]/(z" - 1) or as the vector c of length 71, having ~ B C H= s + 1, i f s is the largest integer having the property that as ith entry the coefficient of :I;' in c(.r;). If (k is a primitive there is a code equivalent to C with a defining set containing 7bth root of unity in some extension field of GF(q), then all a consecutive subset of size .s (notice that R depends on the zeros of xT1- 1 can be written as cuJ(0 5 j 5 71 - 1).If g ( r ) is choice of tr; we call two codes equivalent if their defining not constant, then g ( x ) has some zero oJrJsince g(.c) divides sets are the same up to multiplication with some integer coprime to 72). The first generalization of the BCH bound z7L - 1. But because g(x) is a polynomial over GF(q), it also was given by Hartmann and Tzeng [5]. They prove that if d 3 0 . . . . as its zeros (we will say: m j , ( . I ; ) divides has (Po, g(z), where n ~ ~ , ( .isc )the minimal polynomial of do, i.e., the the consecutive sets { I ,in. 'i 1 j u . . . , i 6 - 2 ,ja} . .). (0 5 ,j 5 s ) are contained in R, and if ( d . n ) < 6, then monic polynomial that only has zeros ( Y ~ O .c u q J ( ] . (@.lo. So we can characterize g(z) (and also C) by the set G := the minimum distance of the code with defining set R is (;joln~,,,(x) divides ,y(x)} (here we use that gcd(n, ti) = I , at least h + .s. Roos [6] generalized this by proving that if and so every zero of :E'' - 1 (and so every zero of !](.I;)) has the statement is true for sufficiently many (say s') values s' - 1. A multiplicity one). Mostly we will characterize C by its defining of , j , then the minimum distance is at least 6 last generalization can be found in [l] and is called the ABset R : { j l d is a zero of g ( : x ) } (and sometimes by R , which is the set of integers modulo 71 that are not in R). We define method. It says that if A, I3 c (0.1. . . ' , 71. - l} are such that the check polynomial h ~ ( . i : )as the reciprocal polynomial of A B:={ a b rriod n ) n E A. b E I?} is a subset of R, then h,(z):= (x" - l)/,q(:c) ( h l ( n . ) is the generator polynomial of the code with defining set R has minimum distance at least 6, the dual code). Of course every ( ( 2 ) E C has zeros d, ,j E R, if ~ r t ~ d ( M ( A+)r~a n) k : ( M ( B ) , )> III for every subset I of and this is the same as saying: c has inner product zero with ( 0 . 1 . ' . . . ? I , - l } for which 111 < h . If COis a subcode of C , then we will write CO5 C . If Ro is the vector (1nJ (Y2J . . o ( ~ - ' ) J ) . So a parity check matrix the defining set of CO,then we must have R c Ro. Moreover we have: d ( C ) 5 d(C0). If CO 5 C and CO # C , then we write CO< C' (and we say: COis a proper subcode of C). A Manuscript received July X, 1991. This work was presented in part at Coding and Information Theory Conference, Essen, Germany, December 15- 17. cyclic code C is called minimal if C # ( 0 ) (0 denotes the 1991. zero word) and if {0} 5 CO< C' implies that CO= (0).The M. van Eupen is with Eindhoven University of Technology, Eindhoven, generator of a minimal code is (.rT1- l ) / m j ( : c )for some .j. The Netherlands, HG 987. J. H. van Lint is with Eindhoven University of Technology, Eindhoven, The In Section 11, we first give some theorems, that give good Netherlands, BG 335. He is also with Philips Research Laboratories, P.O. Box lower bounds for the minimum distance. The first one is XOOOO, 5600 JA Eindhoven, The Netherlands. IEEE Log Number 9203861, called shifting and turned out to be very powerful. We used
A
+
+
+ +
+
>
+
+
00 lX-'~44X/93$0?1.00 0 1993 IEEE
+
+
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
410
a computer to compute the shifting bounds and for length less than 40 this bound was for approximately 95% of the codes equal to the minimum distance. Where the shifting bound did not equal the minimum distance we were often (but unfortunately not always) able to find a bound that did equal the minimum distance. Of course for many codes it is possible to find a lower bound more quickly by hand using other theorems than shifting (including the AB-method). But to avoid a long list of tedious examples, we only used other theorems where it was necessary (i.e., where the shifting bound did not equal the minimum distance). The results can be found in Table I. In Section 111, we treat the lengths 40 5 n 1. 50. Because shifting is in the worst case exponential in n, the computer could in some cases not find the shifting bound in reasonable time. Fortunately other theorems were powerful enough to find almost all minimum distances. The results can be found in Table I1 and Table 111. 11. LOWERBOUNDS
The technique of shifting was introduced in 1986 by Van Lint and Wilson [l]. It gives very good lower bounds for the minimum distance of cyclic codes. We now give a definition of the shifting bound ~ S H I F T (of C )a cyclic code C , that is slightly different from the definition in [l], but easier to implement on a computer. Definition 1: Suppose C is a cyclic code of length n with defining set R. We define the shifting bound dsHIFT(C) inductively. a) If C is a minimal cyclic code and q ,7-2, . . . , r , is the longest sequence of different integers (mod n) such that 1) {7-17-2,...,7-,} c R, 2 ) There is a sequence a l , a2, . . , a, such that for all i : { a ; + r l , . . . , a i + r t - l } c R a n d a ; + r ; 6 R(al1 Cw) 1. additions taken modulo n), then ~ S H I F T (= b) If C is not a minimal code and +
+
& H I F T ( ~ ):= m i n c o < c ~ S H I F T ( ~ O ) ,
and r l , r2,. . . , T , is the longest sequence of different integers with I dsHIFT(C) - satisfying conditions 1) and 2) in a), then ~ S H I F T (= Cw ) 1.
+
Theorem 1 (Shifting): Suppose C is a cyclic code with minimum distance d. Then, ~ S H I F T (I C d. ) Proof: By induction w.r.t. the number of irreducible factors in the check polynomial. 1) Let C be a minimal code (i.e., there is one irreducible factor in the check polynomial) and T I , 1-2, . . . , rw(w > 0) a sequence satisfying conditions 1) and 2) in Definition a). Suppose e(.) E C is of weight w and has support I. Then, we have for all 1 5 i 5 w : rank(M(rl,7-2,. . . , T = rank(M(ai
+ = rank(M(ai +
; ) ~ ) T I , ai
+
TZ,.
. . ,ai+ r i ) I )
+ ~ i - 1 ) ~+) 1 = rank(M(r1, ~ 2 , .. . , ~ i - 1 ) ~+)1. TI,.
. .,a,
The first equality follows from the fact that the rank of a matrix stays invariant under multiplication with ~ ) . the second diag( 1,sat, crzaa, . . . , ~ & ~ - l )To~ prove equality, we recall that
and see that c (the vector of length n corresponding to e(.)) is orthogonal to the first i - 1 rows of this matrix, but is not orthogonal to the last row (since otherwise c(z) would have aaa+rl as a zero, but a; T; R and C is a minimal code, so this would G 0). So the last row is linearly imply that e(.) independent from the other rows and the equality follows (notice that we may restrict ourselves to the columns corresponding to support I ) . The third equality is as obvious as the first one. So rank(M(r1, r z , . * * ,r w ) I ) = r a n k ( M ( r l , r 2 , . . . , r w - l ) I ) + l = ... = rank(M(rl),) w - 1 = w . But this means that M(r1,~ 2 , ..* ,r w ) I has full rank and so c(z) cannot be a codeword. This implies that ~ S H I F T (5C d. ) 2) Suppose C is not a minimal code. By induction, we know that &FT(C) Id(C) := m i n c o < c d ( C o ) . Suppose T I , 7-2, . . . , rw satisfy conditions 1) and 2) and w 5 &HIFT(C) - 1 5 d(C) - 1. Again let c(z) be a codeword of weight w ( > 0). We see that if is a zero of c(z), then c(z) is in a proper subcode of C, since ai r; 6 R. This contradicts the fact that 0 < w 5 a(C) - 1. So again the last row of M ( ai+ T I , a, + 7-2, . . . , ai + ri) is linearly independent of the others and in the same way as in 1, we have r a n k ( M ( r l , r 2 , . . . , r , ) I ) = w and c(z) cannot be a codeword. 0
+
+
+
In Some caseS shifting is not powerful enough and we have to improve the lower bound, using some other theorems. The first theorem is very useful and is also a consequence of a theorem of McEliece . 171. .
Theorem 2: Suppose C is a ternary cyclic code of length n with defining set R. Suppose that if i @ R, then n - z E R. Then, C is selforthogonal (or equivalently: wt(c(x)) E 0 mod 3 for all c(z) E C). Proof: Suppose c(z), c’(z) E C and a: is a primitive nth root of unity. If .(ai) # 0, then i @ R SO by assumption n - i E R and ~ ’ ( a - ~=) 0. This 0 mod (P- 1). is true for all i , so c(z)c’(z-’) Looking at the coefficient of zn-l in c(z)c’(s-’), we get (c(z), c’(z-l)) 5 0 mod 3 ((c(z), c’(z)) denotes the inner product of c(z) and c’(z) in GF(3)). So C is selforthogonal (c(z), c(x)) (c’(z), c’(z)) (and because (c(z), e’($)) 2(c(z) e’(.), c(z) c’(z)), this is equivalent to saying that 0 all weights in C are divisible by 3).
+
+
=
+
+
~
VAN EUPEN AND VAN LINT ON THE MINIMUM DISTANCE OF TERNARY CYCLIC CODES
Example11.1: n = 11, R = (0.2.6.7.8. lo}. This code satisfies the condition in Theorem 2. Here ~ ~ H I F=T 5, but because weights are divisible by 3, we have d 2 6. Example 22.1: n = 22, R = {2,6. 8.10.18). Here, ~ ~ H I F=T10 and by Theorem 2 we have d 2 12. Example 22.2: n = 22, R = (0.4.7.11.12.13. 14.16,17. 19.20.21). Here, ~ ~ ~ H I=F T7 and by Theorem 2 we have d 2 9. Example 22.3: n = 22, R = {0,2.6.7.8.10.11.13.17,18. 19.21}. Here, SHIFT = 5 and by Theorem 2 we have d 2 6. Example 23.1: n = 23, R = {0.5.7.10.11.14,15,17.19. 20,21,22). Here. ~ ~ H I F=T 7 and by Theorem 2 we have d 2 9. Example26.1: it = 26, R = (1.2.3.6.9.18). Here, ~ ~ H I F=T 1 3 and by Theorem 2 we have (1 2 15. Example 26.2: I I = 26, &' = { 1.2.3.4.%.6.9,10,12.15.18. 19}. Here, ~ ~ H I F=T8 and by Theorem 2 we have d 2 9. Example 26.3: 11 = 26, R = { 1 , 2 , 3 . 4 , 6 . 7 . 9 . 1 0 . I 1 . 1 2 . 1 8 . 21). Here, ~ ~ H I F=T8 and by Theorem 2 we have (1 2 9.
411
ProoJ C,) is generated by go(.) = ( 2 - l)g(z). The condition in Theorem 3 is satisfied. We see that Cl is generated by (:eTL- 1)/(x - 1) = :rTL-' d-' . . . 1. So all nonzero codewords in C1 are of weight n. Applying Theorem 3 and using the fact that wt,(co(x)) E 0 mod 3 (by 0 mod 3, assumption), we get a). Moreover, if wt(c(z)) then by Theorem 3 also wt(cl(:c)) 0 mod 3, i.e., q ( z ) 0 (since 71, $ 0 mod 3). This proves statement b). 0
+
+
+
=
Example 11.2: 71 = 11, R = { 2,6.7,8.10). The conditions in Corollary 1 are satisfied (see Example 11.1). Using shifting we get d 2 4. But because wt(c(,x)) = 0 or wt(c(x)) 2 for all codewords c ( x ) , we have d 2 5 . Example 22.4: rL = 22, R = {0,2,6,8,10,18}. The conditions in Corollary 1 are satisfied (see Example 22.1) and SHIFT = 8. By Corollary 1 a) we have d 2 9. But Corollary 1 b) says that if a codeword c ( x ) of weight 9 occurs, then c(z) is an element of the code in Example 22.1, which has d 2 12. This is a contradiction and so d 2 10. Example 22.5: 71 = 22, R = {4.7.11,12,13.14,16.17.19. Theorem 3 (Orthogonal Suhcode Representation): Suppose 20.21}. The conditions in Corrollary 1 are satisfied (see C and COare ternary cyclic codes of length 71 generated by Example 22.2) and ~ ~ H I F=T6. Using Corollary 1 b) and the fact that the code in Example 22.2 has d 2 9, we get d 2 7. Example 22.6: 71, = 22, R = {2.6,7,8.10.11.13: 17,18, 19%21}.The conditions in Corollary 1 are satisfied (see respectively, where t > s . Suppose that if y is a zero of Example 22.3) and ~ ~ H I F=T5 . Using Corollary 1 a) we get go(x)/g(x),then 7-l is a zero of go(:r). Then every c(x) E C d 2 6. Example 22.7: 7) = 22, R = (4.7.12. 13,14,16, 17,19,20, can be written as C ( X ) = c o ( : E ) c1(x), where cg(:~;) E CO 21). Here &HIFT = 6. We wish to apply Theorem 3 to and c ~ ( : eis) an element of the code C1, that is generated by prove that (1 2 7. Let the code C with defining set R :Cn - 1 be generated by g(:7:). Let CO be generated by g(z)(z l ) ( . r 1). Then the condition in Theorem 3 is satisfied. CO is the code of Example 22.2 and is selforthogonal and has d 2 9. Furthermore, C1 is generated by ( x ' ~- 1)/(x2 - 1) Moreover, wt(c(x)) = wt(cO(:i:)) w t ( q ( . c ) ) ~rioc-l3 . and q ( : r ) E C1 can only have a weight divisible by 3 if Proof: It is easy to see that gcd(go(.c),gl(.c)) = ,q(x). cl(.r) e 0. Now suppose c ( r ) E G has weight 6. Write So g ( z ) = o,(:c)gO(x) b ( x ) g l ( : c ) for certain polynomials C ( T ) = t ~ ~ ] ( . t : ) q ( z ) ,where q,(:r) E Go and q ( z ) E C1. Then by the congruence relation wt(c(x)) wt(co(x)) o,(:c), b ( x ) E GF(3)[.c]/(.c'L- 1). So also C ( J ) = c ~ ( : c ) w t ( q ( . I . ) ) iiiotl 3, and from the observations previously made q ( x ) for certain C O ( : I ; ) E CO and r1(.1;) E C1. Further0. So .(:I;) E CO,which is a contradiction, more ,qn(x)gl(:Y1) = 0 mod (:eT' - l ) , because if y is not we have (:I(:/-) a zero of go(:.), then by assumption it is not a zero of since COhas no codewords of weight 6. So d 2 7. Example 23.2: 71 = 23, R = {5,7,10,11,14,15,17,19,20. go(:~;-')/g(x-') (i.e., y is a zero of , q ~ ( : r ; - ' ) ) . But now we have c " ( : L ) ~ ( : I ; - ' ) = 0 rnod (:r:" - l ) , and this means that 21-22}. The conditions of Corrolary 1 are satisfied (see Example 23.1) and SHIFT = 6. Suppose a codeword of (co(.T:),cl(.c)) = 0 mod 3 . SO weight 6 occurs, then by Corollary 1 b) it is in the code of wt(c(:r)) ( . ( . E ) . C ( . X ) ) Example 23.1, which has (1 2 9. A contradiction, so d 2 7. Applying Corollary 1 a), we get d 2 8 (this is also shown (co(:c) q ( z ) eo(..) , q(x)) in [SI). (c(l(:c).co(:r)) 2(c0(z). q ( : r ) ) ( q ( x ) ,e1(:1;)) Example 26.4: 'rt, = 26, R = { 0 , 1 . 2 , 3 , 6 , 9 ,IS}. The (cg(z). Co(Z)) (q(:1;). q ( : r ) ) conditions of Corollary 1 are satisfied (see Example 26.1). wt(co(x)) w t ( c l ( x ) ) iriotl 3 . 0 Here SHIFT = 12, but by Corollary 1 b) we must have d 2 13 (see Example 26.1). And by Corollary 1 a) we get d 2 14. Corollary I : Suppose C is a ternary cyclic code of length Example 26.5: 7) = 26, R = {7.8.11.13,14,16,17,20,21, 11, with defining set R not containing 0. Also suppose that the 22,23,24.25}. Here ~ ~ H I F=T 7, and by Corollary 1 a) we code COcorresponding to R U (0) is selforthogonal. Then, the have r l 2 8 (see Example 26.2). following holds for all c ( x ) E C: Since our codes are ternary, 3 is not a divisor of n. So if 71, a) wt(c(z)) mod 3 or wt(c(z)) 71, mod 3, is not too big (say n 5 SO), then 2 is a divisor of n, unless b) if wt(c(z)) c 0 mod 3, then c(x) E Co. TI, is a prime or n = 25, 35, or 49. But if 2 divides n, then
=
+
+
+
+
+
+
+ +
+
+
+
+
+
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
412
we can use knowledge of the cyclic codes of length n/2 to find better lower bounds. Definition 2: Suppose C is a ternary cyclic code of length m . Then the code of length 2m defined by:
c2:= {c(z2) + zc’(z2) Ic(x), c’(x) E c }
Theorem 4: A ternary cyclic code D of length n = 2m is a square, if and only if R = m R, where R is the defining set of D and m R:= {m T mod n)T E R}.
+
+
Proof: j: Suppose D = C 2 , where C is a cyclic code of length m . Suppose a is a primitive nth root of unity. Then obviously am = -1 and so am+i = -ai for all i. Let i E R. Because D is a square of C , we know that c(a22) aicJ(azi)= 0, for all c ( x ) , c’(x) E C. But then also - ( ~ ~ c ’ ( a ’= ~ ) 0, for all c ( x ) , c’(x) E C , since c’(z) E C implies -c’(x) E C. So we also have c(( -ai)2> (-aZ)c’( ( -ai)2) = 0, for all c(x), c’(x) E C, and so -a2= a m f i is a zero of all elements of D. This means that also m i E R. e:Again a is a primitive nth root of unity. We write p:= a2 and /3 is a primitive mth root of unity. We shall prove that if R = m R, then D = C 2 ,where C has defining set R’ = R mod m. Take d(x) E D and write d(z) d ( - x ) = a ( z 2 ) ,d ( z ) - d ( - x ) = xb(z2),where u ( z ) , b(x) E GF(3)[z]/(xm - 1). We have d(a2) = d(-ai) = 0 for all i E R. So also .(a2;) = b(aZi)= 0 for all i E R. But this means that a ( @ ) = b(p2) = 0 for all z E €2’ d(-2)) so a ( z ) , b(z) E C and so d(x) = 2((d(z) (d(z) - d ( - 2 ) ) ) = 2(u(x2) z b ( x 2 ) )E C2. On the other hand, if c(@) = c’(T) = 0 for all i E R’ (i.e., c(z), c‘(a) E C), then c(aZi) f aic’(aZi) = 0 for all i E R’. But then c ( x 2 ) xc’(z2) has zeros aiand -ai = am+i for 0 all i E RI. So c(x2) xc’(x2) E D.
+
,+ +
+
+
+
+
+
e
0
is called the square of C. It is easy to see that C2 is a ternary cyclic code again, which has the same minimum distance as C. Moreover the following holds.
+
are squares. Of course CL 5 C 5 Cu. Suppose c(x) E C and ~ ( z ) CL. Then c(z) is nonzero both on the even positions and on the odd positions, since otherwise .(ai) = 0 would imply c(a-’) = c(am+2)= 0, and c(z) would be an element of CL.But c(x) is also in the square Cu. So wt(c(x)) 2 2du.
+
Corollary2: If dL 2 2du, then d = dL. Proof: From Theorem 5 we have: d 2 dl;. But CL 5 C,
so we also have: d 5
Example 22.9: n = 22, R = {2,4,6,7,8,10,12,13,14,16, 17,18,19,20,21}. Cu is the code of Example 22.8 and so dU 2 5. CL has defining set {0,11} and so d L = 11. Using Theorem 5, we get: d 2 10. Theorem 6: Suppose C is a ternary cyclic code of length n = 2m with defining set R. Let CE be the cyclic code of length n with defining set R U {jlj 0 mod 2) and CO the cyclic code of length n with defining set R U { j [ j G 1 mod 2). Then every element of C can be written as (a b, -a b), where (a,-a) E C E , and ( b , b ) E CO (a and b are of length m), and a) w t ( ( a + b , - a + b ) ) = 3(m-loo)-wt(a)-wt(b), where 100 := I{ilai = bi = O}l, b) w t ( ( a + b , - a + b ) ) 2 2max{wt(a), wt(b)} -min{wt. (a),wt(b)}, c) wt((a b,-a b ) ) 2 min{dE, d o , max{d~/2,d0/2}}, where d E and d o are the minimum distances of CE and CO, respectively, and (a b, -a b) # 0. Proof: By Theorem 3, we can write every codeword of C as the sum of a codeword of CE and a codeword of CO. But every codeword of CE is of the form (a,-a) and every codeword of CO is of the form (b,b) (since xm - 1 and xm 1 are divisors of their respective generators). Now, define for given a and b in GF(3)” : l,, := I{ilut = T and bi = s}I(r,s = 0,1,2). Then, we have
Theorem 5 (“Paris by Night”): Suppose C is a ternary cyclic code of length n = 2m and defining set R. Let CL be the cyclic code of length n with defining set R U (m R) (the so called lower square of C ) and Cu the cyclic code of length n with defining set R n ( m R) (the upper square of C). Call d L and dU the minimum distances of CL and Cu, respectively. Then, for the minimum distance d of C, we have d 2 min{d~,2du}.
+
+
Proof: Notice that CL and C” are ternary cyclic codes. Moreover, using Theorem 4, it is easy to see that CL and Cu
+
+
+
+
+
+
+
+
Notice that the code of Example 22.3 is the square of the code of Example 11.1 Example 22.8: n = 22, R = {2,6,7,8,10,13,17,18,19, 21). By Theorem 4, this code is the square of the code of Example 11.2. So d 2 5. This also explains why the code has no words of weight 7 or 21. In the explanation of Table 111, the following theorem will be quite useful. We dedicate it to Paris, since it was proved (at night) in a hbtel room in that city.
0
dL.
wt(a)
=
+ wt(b) + wt(a + b) + wt(-a + b)
Erf0 E, +
=
+
j,,
E,c,,,
E,
&,
+
E,E,+,
Er#o .
(E, lr,
+
1,s
Esf0 = Crfo (E, +3
+
E,#-,
+
c,,,
IT,)
10s
1,s
+ 2 E, 1rs) + 3
= 3 ( x . # 8 ~ ’ . .+xs#olos =3(Cr
1,s
E, -
= 3(m - loo).
1,s
100)
10s
__
VAN EUPEN AND VAN LINT: ON THE MINIMUM DISTANCE OF TERNARY CYCLIC CODES
And so wt((a
+ b, -a + b ) ) = wt(a + b)
+ wt(-a + b)
-
loo) - wt(a) - wt(b)
3("
-
L 3 max{wt(a). wt(b)) - wt(a) - wt(b) - 2max{wt(a), wt(b)} -
min{wt(a), wt(b))
, ( b )) . L max{ wt ( a ) wt
413
so wt(c) 2 13) or wt(c') = 0. So, if wt(c) < 13, then wt(c) is even. This is also obvious if we use Theorem 6 and observe that CO has defining set (0). Suppose C is a ternary cyclic code of length n = 2m and c = ( a b, -a b ) , where ( a ,- a ) E CE and ( b , b ) E CO, is a codeword of C. Then obviously c' = 2b (see Lemma 1) and so wt(c') = wt(b). So actually d o / 2 equals the minimum distance of C'. One could also define C": = { ( C O - e,, -c1+ cm+l. . . *rm-l c ~ ~ - ~ ) I c E C} and derive something similar for dE/2, but unfortunately C" need not be cyclic. We shall not go into detail about this. The reader can verify easily that in the ternary case Theorem 6 is always at least as powerful as contraction to length n/2. A ternary code of length n = 2m is called double if { , j I , j 1 mod 2) c R, and it is easy to see that the minimum distance of a double code is twice the minimum distance of its contraction.
+
+
So a) and b) have been proved. To prove c), we observe that there are three possibilities: 1) wt(a) = 0; then w t ( ( a + b . - a + b ) ) 2 d o ; Special Cases 2) wt(b) = 0; then wt((a b, -a b ) ) 2 d ~ ; 3) wt(a) # 0, wt(b) # 0; then max{wt(a).wt(b)} 2 Example 26.6: n = 26, R = { 1,2,3,6.8,9,17,18,20,23, max{dE/2, d0/2}. 0 24.25). ~ ~ H I F=T6. We wish to prove that d 2 7. Suppose Example 22.10: n = 22, R = {0.1.3.5,9,15). CE is c(.x) is a codeword of weight 6. Obviously we have equivalent to the code of Example 22.1 and so d E 2 12. CO (e(.) c(-.z)j2 - (.(:Ti) = c(x)c(-x). (1) has defining set (0) and so d o = 22. Applying Theorem 6c) we get d 2 11. But a codeword of weight 11 cannot occur by Notice that c ( x ) c - x ) is in the code with defining set {0,13}. Corollary 1 a). So d 2 12. Moreover, gcd(c(z).z2 - 1) = gcd(c(-z),x2 - 1) = 1 (since Theorem 5 and Theorem 6 will be used more optimally in the codes corresponding to R U (1) and R U {-I) have the next section. We shall now discuss the relationship between d 2 8), and so c(.c)c(-x) is equal to (.zz6- 1)/(z2- I), up to Theorem 6 and contraction. The method of contraction was cyclic shift or multiplication by 2. We wish to prove that this is introduced in [l].For the parameters that we shall treat in this not possible for any W O , 1111, where 7110: = wt(c(z) c ( - x ) ) paper, we could actually do without the method. However, one and 7/11: = wt,(c(:r) - c(-x)) = fi - 7110. First, we have to special case is so often quite useful that we mention it here. make some preparations. We apply Theorem 3 to get e(.) = Lemma 1: Let C be a cyclic code of length 2m with c g ( z ) c ~ ( x )where , C " ( : I : ) is in the code with defining set defining set R, containing as even integers the set 2R' (where RU (0, 13) and c1(.c) is in the code with defining set {0,13}. R' c {0,1;...m - 1)). Define the contraction C' as We do this because looking at the zeros of co(z) and ~ ( zwe ) 0 and c o ( ~ ; ) c ~ ( - x - ' E ) 0, and so have co(:e)co(--~c-~) (co(:I;),c"(-.c)) = (co(.c).c1(-2)) = ( c " ( - x ) , c l ( z ) ) = 0. s o wg = (e(.) C ( - 3 ; ) . e(.) e(-.)) = Z ( c ( x ) ,C(-.)) = and write e::= e, e,+, (0 5 i < m). Then C' is a cyclic Z(co(:c) c ~ ( x ) , c ~ ( - x + C ( - I C ) ) ~and ( C ( Z ) - C ( - I C ) ) ~ , of their sum must be 13. But also by (l),we must then have that all nonzero coefficients in ( c ( x ) C ( - X ) ) ~ have the same sign. This is a contradiction, since the nonzero coefficients in C(Z) +e( - E ) have the same sign. So c ( x ) cannot have weight 6 and so d 2 7. Example 26.7: n = 26, R = (0,13,14,16,17,22,23,25}. Here, d s H I F T = 5 and the Roos bound equals 6 (which is rather remarkable). To prove this, observe that the consecutive sets (13 3 j , 14 3 j ) are contained in R if j E (0, 1 , 3 , 4 } . By Roos [6],this means that d 2 3 4 - 1 = 6.
+
+
+
+
Explanation of Table I By computer we calculated the minimum distance of all ternary cyclic codes with length < 40. In many cases this minimum distance is equal to the BCH bound. In Table I we give the minimum distance of all ternary cyclic codes of length < 40, and minimum distance not equal to the BCH bound. So the minimum distances of all other ternary cyclic codes of length < 40 can easily be found by computing the BCH bound. Also by computer we calculated the SHIFT (see Definition 1) of the codes listed in Table 1. We see from Table I that in many cases the minimum distance equals the &HIFT. In the other cases, some minimum distances are equal to bounds given by other theorems in this section (or belong to the special cases). In these cases we refer to an example in this section. Unfortunately some minimum distances are left, that could not be derived theoretically by us. These are indicated by a question mark. By the shifting certificate we mean the sequence r l , rz, . . . ,rdsHrFT-l, which has been put in the order of Definition 1. Of course we did not mention codes that were equivalent to a code that was already in the list. We saw that we could call two codes equivalent, if their defining sets were the same up to multiplication with an integer coprime to n. But if n = 2m, we can also call two codes equivalent, if their defining sets are the same up to a shift over m (this corresponds to a substitution of IC by --z in the codewords c(z), and so the minimum distance stays invariant). So we call two codes equivalent if their defining sets can be obtained from each other by some combination of multiplying with an integer coprime to n and shifting over m (where n = 2m). To find the defining sets of the codes in Table I and to find equivalent codes, we refer to the Appendix.
m.
LONGER CODES
In this section, ternary cyclic codes will be studied with 40 5 n 5 50. Computing exact minimum distances for these
codes is very time consuming, so we only computed lower bounds. Of course, first of all we used shifting, since this method seems to be very powerful as we saw in Section 11. But, for some codes, shifting also took too much time and so we had to compute a lower bound using one of the other theorems in Section 11. Of course, we also tried to improve the dsHIFT using one of the other theorems.
n=40 There are too many ternary cyclic codes of length 40 to give a list of all of them. So for each dimension k we computed the best of the lower bounds for the minimum distance of all codes of dimension IC, and only listed the corresponding codes (Table 11). For some codes in Table I1 we were able to find an upper bound for the minimum distance (just by computing some codewords), that equaled the lower bound for its minimum distance. This is indicated by a boldface entry in the column “d,, 2.” In the last column of Table 11, an explanation is given for the lower bound. If this lower bound is not the BCH bound and is not explained in some example, then the bound equals the d s H I F T and the sequence rl,rz,..‘,rdSHIFT-lis given. Example 40.1: n = 40, fi = (1,3,9,27}. We will use Theorem 5 to prove that d 2 24. Notice that CL has as defining set all integers modulo 40, and so we may take d L infinite. Cu is the square of the code of length 20 with defining set (0,2,4,5,6,8,10,11,12,13,14,15,16,17,18,19}, which has minimum distance at least 12 by the BCH bound. So by Theorem 5 we have d 2 24. Example 40.2: n = 40, R = (0,1,3,9,27}. Again we will use “Paris by night” to prove that d 2 22. Also again d L is infinite. Cu is the square of the code of length 20 with defining set {0,1,3,7,9}, which has minimum distance at least 11 by the BCH bound. So by Theorem 5, we have d 2 22.
40<ns50 In Table 111, we give a complete list of all ternary cyclic codes of length 40 < n 5 50, having a lower bound for the minimum distance more than the BCH bound. Again we also computed upper bounds and if the lower bound equals the upper bound, we indicate this with a boldface entry in the column “d 2.”In the last column one can find an explanation for the lower bound for d. If shifting gives the best bound, then a shifting certificate is given. Otherwise we either refer to a theorem (“Paris” means Theorem 5) or to an example. Sometimes we also give an explanation for the upper bound, by giving a subcode with a known minimum distance contained in our code (e.g., “>nr.13”). The “-”-sign denotes “a code equivalent to.” If the code is double or a square, then the minimum distance can be calculated from Table I and there is nothing else to explain. Notice that not for all codes the shifting bound is given in Table 111. Here the computer spent too much time to compute the shifting bound. Because we need lower bounds for the minimum distance of all proper subcodes of a code to compute the shifting bound, we have put other lower bounds in the computer where shifting bound was not known. Notice that this means that the entries in the column
415
VAN EUPEN AND VAN L I N T ON THE MINIMUM DISTANCE OF TERNARY CYCLIC CODES
I, r 1 1
13
- ir , > > , >
13 16
> I
> 3 >
, 3 ,
3
1 > 1
$
I1 I1
3 4 5 6
TABLE I (PART 1)
k 1
L6
1 > 1 >
8
16
9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
16
>
16
3
1 > 1
20 20 20 20 20
5
3
5
7
5
8
8
3 10 12 12 13 13 13 14 14 I5 5 6 6 7 10 10 11 11 12 12 15 16 11 12 6
5
9 B 5
41
42 43 44 45 46 47 48 49 50
51 52 53 54 55
56 57 58 59
60 61 62 63 64 65 66
67 68 69
TO 71
72 i3
74 i5 i6 77
20 20 20
20 20 20 20 20 22 22 22 22 22 22 22 22 22 22 22 22 23 23 26 26
6
26 26 26 26 26 26 26
6 7 7 7 8 8 9
26
9 9 9 10 10 10 10
26 26 26 26 26 26 26 26 26 26 26 26 26 26 26
IO 10 10 11
11 11 11
12 12
2€
12
2z 2f 2f 2f 2f 2f 2f 2t 2t 2f 2t 2f 2t
12 12 12 12
2t
- -
13
13 13 13 13
13 13 13 13 13
5 4 4 4 4
I 4
4 4 12 12 10 10 9 6 7 6 7 5 4 4 9 8
15 12 12 14 12 12
11 10 9 9 9 11 9 9 8
9 10 10 3 8 6 7
I 4 4 4 4 4 4
10 10 8 8 7 5 6 5 6 4 4 4 7 6 13 12 12
12 12 12 11
10 9 9 9 10 9 9
8 9 10 10 9 8
6 7
2
1,4,7 1~2.7
5 6 5 3 4 3
5,10,11
3 3 3 3 3
4.5,lO 2,5,10 Q,10.11 4.5 10,11 0.1,4,7.11 2.4,7.11 1,4,7,11 Z34,7
,
8
7 5 4 5 4 4 4
3 3 6
5
,
i 6
8 9 7 7 5 6
6 6 6 8 8 6 8 7 6 8
6 6 6
6 7 6 5 5
8
4.5,7,8,13.14,17 2,4,7,8,13,14,17
8 7 9 7
8 8
8
11.12.13,22.10,19,20,21,14,15,4,0 19,21,13,23,20,22,14,24,7,0.1
6 8
8
6
4,7,11 ’2.7,ll 4!7 2.7 0,7,11 4,11 0,5 5
10 10
5
- -
10,16,11,1i,12,13,0 10,11,16,12,17,13,2 11,4,16,12,15.10.0 11.15.10,4.5 11.12.19.10.0 10,19.0 4,15,0 11,17,5 4.15,5 14.5,2 10,19.0 4,15.5 11,17,0 13.19.14,20.11,12,3.0,1 10.17,12,13,11,6,16,7,2 11,12,13,14,3,4,1 12.6,17.16,18,7,2 11,12,20,19,13.0 7,10,6,0 11,12.19,13,4 6.17.7,2 16,13,12,19.4 6.7,2 19.1 1,o 12,14,4 14,10,21.19,20,0 14,10.21.19,5
11
9 10 8 8
5 6 5 5
6 4 7 5 5 5
Expl.
Ex 11.1 Ex 11.2
k,i,5
3
i
6
L.5,8,10
3,4.5,10,11 4,5,10,11 3,4,10,11 3.5,10.11 3,4.5,10
8
..
7,10.6.0 5.7,~ 10.11,7,12.0 5,0,11,5,2 7.13.8.14,3,0,1 4.i.12.13.5 4,7,0 7,14,13,5
1,1,5.8.10
5 3 3
8
7 8 6 7 7 6 8 6 6
t.i.0
5
9 9 8 9
8
Shifting certificate
1,4,5 ),2 4
6 6 5 5
6 4
c
2,i,8,13,14,17 1.i,8.13.14.17
4~5,1,l3~14~17 1.4.7.13.14.17 2,i,8,14,17 2.4.8.14.li 1,4,8,14,17 1,2.8,14.17 O,i.8.13,14.17 0,5.8.13,14.17
0,1,7,13,14,17 0,1,4,13,14,17 0,4,7,8.13,17 7,8,13,14,17 5.8,13,14.17 4,8.13.14.17 1.8.13.14.17 4.i.13.14.17 1,7,13,14,1i 4,5,13,14,17 2,4.13.14,17 1.4,13.14.17 1.2.13.14.17
SHIFT" need not equal the ~ ~ H I FofTDefinition 1 (but it is easy to see that the d s H l F T from Table I11 is still at most d). Example44.1: n = 44, R = {2,6,10,11,18,30,33}. We shall use Theorem 5 to prove that d 2 22. CL has defining set {11,33} and so d L = 22 CLI is the square of the code of length 22 with defining set {0,1,3,4,5,7,9,12,13,14,15,16,17,19,20,21},which has minimum distance 12, since it is equivalent to the code of
10,11,12,13,21,20,19.14,15,4.5 16,6,12,20.10.21,22,11,23,7,2 1,9,3,13.17,24,23,12,8,2,4 19,22,10,23,20.14,15,21,4,5
E x . 22.1 E x . 22.10
Ex.
22.4
E x . 22.9 E x . 22.2
Ex. Ex. Ex. Ex. Ex.
22.3 22.5 22.6 22.7 22.8
Ex. 23.1 E x 23.2 E x . 26.1
E x . 26.4
1,9.17,22.3,24.23,2,4 10,22,21,20,11,12.7,0 20,21,22,16,24.6,7,0 22,12,8,25,23,24,0,1 22,24.25,0,16.1,23,2.3 10.11.22,21,20,12,7,4 20,16.6,22,21.23,7,2 7.23.21.20,22,8.1 23,16,9,13,12,8,3,1 2 2 , 1i,8,23.1.13,24,2.3 12,13,21,15,11,10,14,4,5
10.3,16,22,12.11.9.1 20,16,6,23.21,7,2 22.16,23,17,2 22,9.24,3,8.1 22.23,1,16.2.3 20.21,22,23.24,7,0 13,22,14.23.15.16.0 23.24,13.0.22,8,1 11,12,16,25,21,10,0 22,13.0.16,1 9,0.12,3,1 10,11.7,12.0 20.21,22,23.i,8 13.14.22.15,23,16.5 22.12.23.13.4 23,16,13,24,8.1 12,21.10,11.13.4 22,16,21,13.1 13.14,22.16,12,4.5 12.13,17.16.2 9.13,22,3,1 1.17.13,2.16.22.3
E x . 26.2 E x . 26.3 7
Ex. 26.5
Example 22.10. So also d~ = 12. By Theorem 5, we have d 2 22. By Corollary 2, we have d = 22. Example 44.2: n = 44,R = {0,2,6,10,11,18,30,33}. We wish to apply Theorem 6 to prove that 21 is the best lower bound for the minimum distance that we can find (notice that “Paris by Night” gives d 2 20). Suppose c = (a b, -a b) is a nonzero element of the code of weight less than 21, where (a,-a) E CE and ( b , b ) E Co. If c E CE or c E CO, then
+
+
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
416
TABLE I (PART2)
- - - nr - - 6 78 3 L
79 80 81 82
83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107
108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 I4e 141 14f 14: 15C 151 152 15: 154 15: 15f
-
6 6 6 6 6 6 6 6 6
6 6 6 6 6 6 ,6 26 26 76 26 26 26 26 26 26 26 26 26 26 28 28 28 28 28 28 28 28 32
3 4 4 4 4 4 4 4 4 5 5 5 5 6 L6 16 I6 16 16 16 16 16 17
I7 18 18 19 19 20 6 7 8 10
18
32 32 32 32 32
19 19 20 4 7 8 9 9 13 14 14 15 15 16 16 16 17 17 17 17 18 18
32
18
32 32 32 32
18 18 19 19 19 19 20 2c 21 21 1: 1:
32 32 32 32 32
32 32 32 32
32 32
32 32
32
32 32 32 32
32 3: 3:
3: 35
1f
3: 3:
1f 2: 2:
3! 3: 3: 3' 3;
-
2:
11 1: -
ELK
b
>
>
,
> >
i
>
i >
>
, > , 4 5 4 4 12 12 12 7 4 4 4 4 16 12
8 10 8
6 6 6 6 6 4 6 6 4 6 6 6 4 4 4 5 5 4 4 4 5 4 4 4 4 li 12 8
li
2A
>
, I > 4 4 12 I2 12 7 4 4
I
I
I I
I
IO IO i
, 3 3
6 4
3
16 12 B 10 B 6
12
6
5 5 5 5
6 6 6 4 6 6 4 6 6 6
4 4 4 5 5 4 4 4 5 4 4 4 4 12 11 8 7 7 6
5 4 4 8 8
3
IO 3
3 5 3
3 5 5
3 5 5 3 3 3 3 4 1 3 3 3 4 3 3 3 3 10 10 6 6 5 4 4 3 3 5 5
ihifting certificate .i,8,13,17 .7.8,13,17 .8,14.17 ?8,14,17 ',8,14,17 -8*14,17 1?7,14,17 ,7,14,17 ,7,8,17 ,2,8,li l,7,13,14,17 1,4,13,14,17 1,7,8,13.17 l,4,8,13,17 ',13,14,17 l,13,14,17 2,13,14,17 l.13,14.17 r,8.13.17 4,8,13,17 l,8,13,17 4.7,13,17 2.4,13.17 4.14.17 2,8,1? ),13,14,17 ).7,13,17 13,14,17 ),7,17 14,17 D,2,4,5,7,14 274v5,7>14 2.4A7 2,4,5 D,4,7,14 4,7>14 2,i,14 4>7 0,1,2,5,10,16,2b 1,5,8,10,16,20 0,1.5.8,10,16 1.5,10,16,20 1,5,8,10,16 2,5,10,16,20 0,1,4,10,16,20 4.5.8.10.20 5,8,10,16,20 1,4,10,16.20 0,4.5.8,16,20 0,1,10,16.20 5,8,10,20 4,5.8,16,20 5.10.16,20 1.10.16,20 2,5,16,20 0,5.8.16,20 0.4,5,16.20 4,5,8,20 0,1,10,16 5,8.10 5.8.16,20 1.8.16,20 4,5,16,20 1.10.16 0,5.16.20 5,8,20 5,16,20 1,16,20 0.2,5,7 2.5,7 0.2.5 2,s 0.2.7 0,2 2 0.5.7 5.7 0.2 2
wt(c) 2 22 (C, has defining set { 11,33}and CO is the code number 7 in Table 111). If wt(a) = 22, then by Theorem 6b) we also have wt(c) 2 22. So we can assume that wt(a) = 11. Now we can say immediately that wt(c) e 2wt(a)+2wt(b) e 1 2wt(b) $ 2 mod 3 (since b is an element of the code of Example 22.10 and so wt(b) f 2 mod 3) and so wt(c) # 20 so d 2 21. We shall show that we cannot improve this bound
+
0,23.11,7,4 !0.8,24.7,1 4.15,23.22,16,5 6,24,22.23.4 !2,16,23,17,2 !3,16,8,24,22.1 .1.16,10,21,4 !1,23,7,1,3 !0,24,8,7,1 !3,24,2,1,3 !3,22,21,16,0 !2,13,12,16,0 23.24,7.20,0 12,25.23,24,0 22,16,21.13,7 12.10,23,14,4 22,17,16.13,2 14.1,25,23,3 7.24.23.20,8 3,20.4.13,10 23.13.9,20,1 10,23.21.11,4 10,4,2,23,6 12,10.14.4 17,24.23,2 13,22,16,0 23.11,O 13,22.16,14 23,21,0 22,16,14 4,14,5,15,10,11,12,13.6,7.0 4,14,10,12,20,11,13,5,6,7,2 10,5,11,6.16,15,4,20,21,7.2 11,4.22.10,5,2 20,7,0 20,7.4 6.21,2 20,7,4 13,25.14,26,15,27,16,28,17,18,19,0,1,2,3 7.13,23,25,27.24.26.8,9,10,1 7,13,8,14,9,0,1 13,27.25,14,28.26,9,10,1 7,23,13,8,9,10,1 13,20,5.6,2 9,11,0,3,1 4,7,12.13,5 13,28.14,7,5 9.25,10,3,1 4.7,O 25,9,26.0,1 13,29,28.7.5 4.7.5 13,28.14,15,5 25.9,26,10,1 21,5,20,6,2 20.7,O 4.15,O 4,7,5 9,10,0.1 i,14.13.5 20,7,5 8.19,l 4.15,5 9,25,10,1 20.15,O 20,7,5 20.15,5 16,19,1 18,21.30,24,22,6,19,20,5,7,0 18,19,20.21,23.5.22,6,7.2 18,24,23.30.22.5,0 18,30,22.23,5,2 18.23,21.31.6,0 34.23.31.18.0 31,23,18,2 20.14,O 20.14.5 13.18.14.19,31.5.0 13.18.14,19.31,5,2
:xpl
Ex. 26.6
I
Ex. 26.7
, ?
, 7
by using Theorem 6. Let b(z)be the polynomial corresponding to b. Notice that b(z) dz b(-x) is an element of the code of Example 22.8 (minimum distance 5 ) and cannot be the zero word (since otherwise wt(c) 2 22f10 = 32 or c E CL,which has minimum distance 22). Since a is either zero on the even or zero on the odd positions, we have loo 5 11 - 5 = 6. So 22 - loo 2 22 - 6 = 16 and, by Theorem 6a), we have
VAN EUPEN AND VAN LINT ON THE MINIMUM DISTANCE OF TERNARY CYCLIC CODES
417
TABLE I1
k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
18 19 20 21 22 23 24
25 26 27 28 29 30 31 32 33 34 35
-
dmaz 2
40 30 25 24 22 24 22 20 18 18
16 16 14 14 14 12 12 11 11 10 9 8 8 8 8 7 6 6 6 5 5
4 4 4 3
G
Explanation
1,?.4,5,7,8,10.11,13,~0,22,?5 0,1,2,4,7,8,10,11,13,20,22,25 1,2.4.7,8,10,11,13.20,22,25 0,2,4,5,7,8,10.11,13.20,22,25 2,4,5,7,8,10,11.13,20,2?,25
BCH BCH BCH
0,1,4,7,8,10,11,13,?0,22,25
BCH BCH BCH
1,4,7,8,10,11,13,20,22,25 1,4,7,8,10,11,13,22,25 4,5,7,8,10,11,13,20,22,25 0,4,7,8,10,11,13,20,22,25 2.7,8,10,11,13,20,22,25 0,4,7,8,11,13,20,22,25 1,4,5,8,10,13,20,?2,25 0,7,8,10,11,13,20,22.25 1,2,5,8,13,20,22,25 0,5.7,8,10,13.20.22,25 5,7,8,10,13,20,22,25 0,7,8,10,13,20,22,25 7,8,10.11,20,22,25 0,7,8,13,20,22,25 8,11,13,20,22,25 0,10,11,13,20,22,25 10,11,13,20,22,25 0,1,10,13,20,22 7,11,20,22,25 4,13,22,25 5,7,20,22,25 0,13,20,?2,25 7,20,22,25 0,13.20,22 7,20,?? 5,22,25 20.22,25 22,25 13,20
wt(c) 2 3 * 16 - wt(a) - wt(6) = 37 - wt(6). We have three cases. 1) wt(b) 5 15: wt(C) 2 37 - 15 = 22. 2) wt(b) = 16: wt(c) 2 37 - 16 = 21, and also by Theorem 6b): wt(c) 2 2 * 16 - 11 = 21. 3) wt(b) > 16: wt(c) > 21 by Theorem 6b). Example 44.3: n = 44, R = {2,4,G. 10.11.12,16>18.20. 30,33,36}. We wish to apply Theorem 6 to prove that d 2 16. Suppose c = ( a b, -a b ) is a nonzero codeword of weight less than 16, where ( a ,- a ) E CE (having defining set {11,33}) and ( b , b ) E CO (code number 14). As in Example 44.2, we may assume that w t ( a ) = 11. If b(.r) is the polynomial corresponding to b, then b(.r) f b(-.r) is an element of the code of length 22 with defining set {0. ll} and we may assume that b ( r )fb( -.r) $ 0 (since otherwise wt ( b ) = 0 and so wt(c) = 22). So lo0 5 11-2 = 9 and 22-10" 2 13. So by Theorem 6a), we have wt(c) 2 3 * 13 - wt ( a )- wt(b) = 28 - wt(b). But wt(b) E 0 mod 3 (by Theorem 2 ) and we must be in one of the following two cases. 1) wt(b) 5 12: wt(c) 2 28 - 12 = 16. 2) wt(b) 2 15: wt(c) 2 2 * 15 - 11 = 19 by Theorem 6b). So d 2 16. Example 44.4: n = 44, R = { 1 , 2 , 3 , 4 , 5 . 6 . 9 , 1 0 . 1 2 , 1 5 , 16,18,20,23,25,27.30.31,36,37}. This code satisfies the condition of Theorem 2, and so d 5 0 mod 3. ~ ~ H I F=T 10 and so d 2 12. Moreover d = 12, since this code contains number 13 in Table 111. Example 44.5: n = 44, R = {0,1,2,3,4.5,6,9,10,12.15. 16,18,20,23,25,27,30,31,36,37}. This code satisfies the conditions of Corollary 1 a) (see Example 44.4). Here
+
+
Example 40.1 Example 40.2
19,20,33,32,~9,30,21,34.28,31,24,15,11,22,10,4,5 28,31.32,19,33,29,20,34,30,21,35.22,36,10,11,7,0 30,6,24.32.33,29.?1,19,37,18,17,16,22,7,2 31,32,19,35,36,23,37,34,11,24,33,20,21,7,0 8,27,4,12,37,31,35,30.26.?4,9,3,1 29,?0,23,24,38,30,37,33,32,21,19,7,0
BCH 20,21,22,30,38,29,37,24,23,7,0 23,24,15,31,7,29,21,30,22,20,5 37,24,38,30,21,22,20,23,7,0 19,20,23,32,22,33,?4,21,7,8 22,23,31,7,39,20,25,?1,0 34,16,37,38,19,32,24,0 19,10,37,34,30,25,0 30,10,37,33,19.25,11 26,30,9,39,37,0,1 21,25,34,22,19.?0,7
BCH 21,34,22,20,5 37.38,25,39,0 ?1.34,21,?0,7
BCH BCH 34,25,5 34,25,20 34,25,22
BCH
SHIFT
= 10, so we have by Corollary 1 c) that d 2 11. Example 44.6: rt = 44, R = {0.7,8,13,14,17,19,21,22. 24,26,28,29.32.34.35.38,39,40,41,42,43}. We wish to apply Theorem 6 to prove that d 2 10. Suppose c = (a b, -a+b) is a nonzero codeword of weight less than 10, where ( a ,- a ) E CE (number 19) and ( b , b ) E CO (number 14). Notice that CE is a square with minimum distance 10, i.e., if wt(a, - a ) < 20, then ( a ,- a ) (and also a) is either zero on the even or on the odd positions (and if it is zero both on the even and on the odd positions, then wt(c) 2 18). Let b ( r ) be the polynomial corresponding to b, then b ( s ) f b ( - s ) has zeros 1 and -1 and so b must have weight at least 2 on both the even and the odd positions (if b is zero on either the even or on the odd positions, then b = 0 and wt(c) 2 lo). We have three cases. 1) w t ( a ) > 9 or wt(b) > 9. Then, wt(c) 2 10 by Theorem 6b). 2 ) wt(a) < 9 and wt(b) = 9. Then, wt(c) 2 10 by Theorem 6b). 3) w t ( a ) = 9 and wt(b) = 9. Then by the observations just made loo 5 2 (11 - 2) = 11 and so by Theorem 6a) we have: wt(c) 5 3 * 11 - 9 - 9 = 15. So d 2 10, and since code nr.19 in Table 111 is contained in this code, we have d = 10. Example 44.7: n = 44, R = { 1.3,5,8,9,11,15,22.23,24, 25.27.28,31,32,33,37,40}. We wish to apply Theorem 6 to prove that d 2 8. Suppose c = ( a b. -a b) is a nonzero codeword of weight < 8, where ( a .- a ) E C:E(equivalent to number 13) and ( b . 6 ) E CO (number 38). CE is selforthogonal and has minimum distance 12, so wt(c) can only be
+
+
+
+
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
418
n
-
ii-
P
2
41
3 4 5
41 41 41 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44 44
3 17 14
nr. 1
6 7 8 9 10 11
12 13 14 15 16 17 18 19
20 21
22 23 24 25 28
27
28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58
59
60 61
62 63 64 65 66 67 68 69 70
71
72 73 74 75 76 77
-
k
-
!L -rpr 0 8
8
2
2
5 6
6 7 7 8
8 10 10 10 11 11 11 12 12 12 12 12
12 13 13 13 14 14 14 15 15 15
--
F
I
25
I2
IO
L 3
6 4
6 3 4
12 L8
La 14 L2 LO 16
La La 14 LO LO L4
La LO
I1 LO
L2 12 B
1 IO
LO LO
I
I IO
LO
I
3
I LO
LO LO
r
5
3
14 10 8 11 10
LO
10
10 8
P 5 6
7 3 3
7 7 S T
16
12
10
7
16 16 16 16 17 17 17 17 17 18
11
10 10 10 8 8 8 8
7 5
10
7 7
12
ia 8
10
10 B
18
12 11 10 10
18 18
10
18 19 I9
20 20 20 20 21 21 21 21
22 22 22 22 22
22 22 23 23 23 23 24 24 24 25 25 26
28 21
-
I)
6 8 8 6
7 8
10 8 8 8 8 8 8 8 10 7 8 5 10 7 5 8 8 6 8
4
8
5 8
6 5 8 6
5 4
10 8
10 10
ia
8 8
e 11 Q
e 8
10
e 7
e. 8
I e. 8
8 7 K 8
e e.
8
7 -
Explanation
L2.4.7.8
~8,34,30,36,17,5,23,19,~,25,15,10,12,18,20,16,4,6,0
44.7.8 L.73
L,33,21,5,34,22.23.17,16,15,20.10,18,19.6.7,2 r,21 p , i i,20,15,25.29.4.i0,5 1.33,31,39,8,6,2 r,21.16,30,19,10,8 iouble iouble iouble Example 44.1 double Example 44.2 ‘Paris‘,3 n r . 8 iquare double double ‘Paris’,>nr.l3 double double square Example 44.3 square square double double 1Paris’,3nr.19 21,22,24,25,27,13,41.3t,39,23,7,8,l.>nr.17 ‘Paris’, 30,121 rquare 23,37,24,38.25,39,26,7,8~ 1 square ‘Paris’, 311.13 ‘Paris’, 3 n r . 1 3 double ‘Parid’, 311.13 ‘Paris’,3 n r . 1 3 ‘Paris’, 3 ? m r . 1 3 ‘ParL’,>~nr.l3 double ‘Paris’,Dnr.lS ‘Paris’83 n r . 1 9
42.8
I 14 I4 IO I2 IO ’1
La
TABLE I11 (PART 1)
5
8 8 6 4 7 5 5 7 8
7
7 8 8
6 7 7 8 3 5
4
7 5
4 5
8 5 5 7
7
5 4 5 7 4 4 5 4 4 5 5
r,a
1,1,4,7,8,11.14,22 1,4,7,8,1l.I4,22 L.2.7.8.11,14.22 ).1,4,7,8,14,22 1,4,7,8,11,14 L.4,7,8.14,22 1,2,7,0,14.22 1,2,4,7,8,11,14,22 1,1,7,8,11,14,22 1,1,4,7,11,14,22 2,4,7,8,11,14,22 1,7,8,11,14,22 1,4,7,11,14,22 D.2.4.7,8,14p22 D,1,7,8,14,22 0,1,4,7,14,22 2,4,7,8,11,14 1,7,0,11,14 1,4,,7,11,14 2,4,7,8,14,22 1,7,8,14,22 1.4,7B14.22 2,4.7,8,14 1,7,8,14 1,4,7,14 0,4,7,8,11,14,22 o,a,7,8,ii,i4,22 0.1.7,11,14,22 4,7,8,11,14,22 2,7,8,11.14,22 1,4,8,11,14,22 1,2,8.11.14.22 1,7,8,11,22 0,4,7,8,14,22 0,2.7,8,14,22 0,1,7,14,22 4,7,8,11,14 2,7,8,11,14 4.7,8,14,22 2.7,0,14,22 1,4,8,14,22 1,2,8,14,22 1,7,8,22 4.7.8,14 2,7,8,14 0,7,8,11,14,22 0,4,7,11,14,22 0,2,7.11,14,22 0,1,4,11,14,22 7,8,11,14,22 4,7,11,14,22 1,4,11,14,22 4.7,8,11,22 0,7.8,14,22 0#4,7,14,22 0,2,7,14,22 0,1,4,14,22 7.8,11,14 4.7,11,14 1,4,11,14 7,8.14.22 4.7,14,22 1,4.14.22 4,7,8.22 7.8,14 4,7,14 1,4,14 0,7,11,14,22 0,1,11,14,22 7,8.11.22 1,8,11,22 0.7,14.22
37,21,13,25,42.0,1,3nr.33 ‘Paris’,3111.22 <Paris’,3 n r . 2 2 ’Paris’.>nr. 19 ‘Paris’,3 n r . 1 9 ‘Paris’,>zznr.lS ‘Paris’,3 z n r . 19 21,39,23,27,7,8,1,>nr.38 ‘Paris’,3 n r . 2 8 ‘Paris,, 3 n r . 2 8 E x a m p l e 44.4 square 10,18.38,43,41.8,0 square E x a m p l e 44.5 ‘Paris’,3nr.52 ‘Paris’,3 n r . 5 4 32,21,40,16,19,7,4 E x a m p l e 44.6 square 38,13,34.17.6,21,0,3nr.53 square 38.39,32,40.33,7,8 square square 38,39,13,28,34.7,8 ’Paris’, 3nr.60 ‘Paris’. 3111.62 16,35,12.20,39,7,4, Jnr.58 38,39,13,28,34.7.8 square square ‘Paris’, 311.33 ‘Paris’, 3111.54 Theorem 6 ( i i i ) , 3 z n r . 5 4 E x a m p l e 44.7 ‘Paris’, 3111.60
2) m,- loa 2 7. Then, by Theorem 6a), we have: wt(c) 2 smaller than 8 if wt(a) = 6 (if wt(a) = 0, then c E CO, 21-6-6=9. which has minimum distance 8). But if wt(a) = 6, then a is zero on either the even or the odd positions. By Theorem 6b) So d 2 8, and since code number 38 is contained in it, we wt(c) can only be smaller than 8 if wt(b) 5 6. So we have have d = 8. two cases. Example 44.8: n = 44, R = {7,8,13,17,19,21,22,24,28, 1) m - ZOO = 6. Then c is in the lower square CL by the 29,32,35,39,40,41,43}. We wish to apply Theorem 6 to b) is a nonzero observations just made (c is zero on either the even or prove that d 2 6. Suppose c = ( a b, -a codeword of weight < 6, where ( a , - a ) E CE (number 19) the odd positions). But d L = 9, so this is impossible.
+
+
419
VAN EUPEN AND VAN LINT: ON THE MINIMUM DISTANCE OF TERNARY CYCLIC CODES
TABLE 111 (PART2)
- - -
-
nr n k 78 44 27 79 80 81
44 44 44 44 44 44 44 44 44 44 44 44 44 44 14 44 46 46 46 46 46 46 46 46 46 46 46 46 47 47
82
83 84 85 86 87 88
89 90 91 92 93 94 95 96 97 98 99 100 101
102 103 104 105 106 107 108
-
G 0,1.14.22 7.8.22 1.8.22 0 , 4 , 1 1, l 4 . 2 2 0.2,11.14.22
d >
0
28
28 30 30 30 31 31 31 31 32 32 32 33 35 36 37 11 12 12 13
5
6
5
7
6 4 4 4 4 4 4 4 4 4 4 4 4 4 4
4 4 4 4 4 4 4 4 4 4 4 4 4 4 18
18 10
4 5
3 3 3 3 3 3
3 3 3 3 3
18
le 9 9 9 9
22 22
23 23 24 24
8
8 4 4
33
34 23 24
-
3 3 3 12 12 10 10
7 9 7
6 6
8 6 8
6
9
4 4 9
8
8
0,7,11.22
4,11,14,22 2,11,14,22 4,8,11,22 7,11,22 0,7,22 4,11,14 2,11,14 7,22 0,11,14,22 8,11,22 11,14 0,2.5.10,23 2.5,10.23 1.5,10.23 2.5,lO 0,5.10.23 0.1,10.23 5,10.23 1.10,23 5,lO 1.10 0,10,23 10,23 0.5 5
6
5 6 3 3 6 5
Explanation ' P a r i s ' , > n r 62 Example 4 4 8 Example 44 9 square 10.33.0 square 11.33.4 10.33.2 32.11.4 19.11,7 square square 10,33,2 19,17,7 38.22.0 28,24,8 34,38.11 -.double 16,10.6,18,14.4,20,22,32,36,42,5,17,15,19,7,2,~nr.95
double 'Paris'. 3 r r n r . 9 7 square 38,27,39,22.41.0.9,1 'Paris', 3 n r . 9 9 E x a m p l e 46.1 square 38,27,28,39,29.9.1 20,28,0
20.28.10 38,44.30,19.43,10,22,0 43.29.44.38.39.10.5
TABLE IV (PART1 ) n
4
5 i
8
IO
m i n i m a l polvnomialr = + 2 1 2 2 + 1 3 r + 1 trivial trivial 1 r + 2
zeros e x p o n e n t s 0 1,3
2 3 4 5
r 2 + r + ?
1.3
2 + 1 r + l
2.6
13
1
X + 2
2
r 3 + r 2 t r + 2 r 3 t r 2 + 2 r 3 t 21' 21 2 r 3 + 2 r t 2
1 2
3
14
4 5 1 2
3 4 16
0
3
3 4 11
2
r2+2r+2 r + 2 r4 t 2r3 t r 2 + 2 z + 1 r 4 + r 3 + x 2 t r t 1 r+1 r t 2 r5+2r3+r2+2rt2 r5 r4 zr3 t r2 t 2
1 2
1 2
3
-+
+
+
14-2
+
+
r6 + 2 r 5 r4 + 2 r 3 + r2 + 2 1 + 1 s6 r5 t r 4 s3 r2 t r t 1
+
+
+
r+1
X t 2 r 4 + r 2 + 2 r2+r+2 ,=+I s4 2r2 + 2
4 5 r+1 6 7 r2+2r+2 trivial trivial 1 r + 2 2 r1+r3+2r+1 zr3 r2 2r t 1 3 4 s 4 + r 3 t 2 t r + 1 s ,'+I 6 r+1
+
17 19 20
+
7
+
permutations 1 2 3 3 2 1
+
r 4 + 2 2 + r + l
4 5*7 0 1.3,7,9 2,4,6,8 5 0 1,3,4.5,9 2,6,7,8,10 0 1,3,9 2.5.6 4,10,12 7.8.11 0 1,3,5,9,11,13 2.4.6.8,10.12 7 0 1,3,9,11 2,6 4,12 5,7,13,15
1 1 4
2 5 5
3 3 3
4 1 1
5 2 2
1 4
2 3
3 2
4 1
1 1
2 3
3 2
1 1 1 1
2 3 4 5
3 1 5 2
4 5 2 3
1 4
2
3
3
2
4 1
1 1 6
2 5 2
3 7 7
4 4 4
5 2 5
6 6 1
7 3 3
I 1 6
2 7 7
3 3 4
4 4 3
5 5 5
6 6 1
7 2 2
5 2 3 4
8
10.14
0 1,3,7,9 2,6,14,18 4,8.12,16 5,15 10
11.13,17,19
and (b, b) E CO (number 38). The only way to get wt(c) < 6, is to take wt(a) = 5 and wt(b) = 4 or 5. But CE is a square and so wt(a) = 5 implies that a is zero on either the even or
the odd positions. We have two cases. 1) m,- loo = 5. Then c is an element of the lower square, which has minimum distance 6,
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
420
TABLE IV (PART2) minimal polynomials 1 s+z z 2 zt4 zS3 2 2 I 3 2 z t 3 2 zS 2 4 .5 t4 2s3 t2 2 s ,5 2r3 2 2 t 2t 1 t + l 6 1 s+Z 2 2 1 2 2 2x6 s4 s3+ 21 2 +2s2 3 tll-+tlo+ t 9+ 2 s 8 + 2 s 7 +
n
22
zeros exponents 0 1,3,5.9.15 2,6,8,10,18 4.12,14.16,20 7.13.17.19,21 11 0
+ + + + + + + + + + + + + + +
23
+
25 I
28
+
1
2 3 4 5 6 7 8 9 10
I
1
2
+
+ + +
+
+
+
+
+
+
4 5 zS5 zS3 Z+ 6 t 2 + 1 7 s+1 trivial trivial 1 s+2 2 4 + s 4 + 2 3 s 4 + 2 + 2 4 s2+.+2 5 2 +zS4 + Z 6 s2+l 7 s4 2x2 + 2 8 S+l 9 z2+21+2 1 s + 2 2 ,I6 2J5 s14 ,I5 zl‘ 3 t16 4 t+i 1 s + 2
+
+ +I
+ +1
+
34
35
+
I
+
2 3
37
4 5 1 2
3
+
1
2 3 4
+ 2, + 1 + I
+1
4,8,12.16.20,24 5,11,13,15,17,23 7,21 14
0 1.3.9.11,17.19,25.27 2.6,18,22 4,lZ 5,7.13,15,21,23,23,31 8,24 10.14.26,30 I6 20.28 0 1.3,5,7. ,33 2.4.6.8 32 17 0
+=7
33
t2
S4+S3+.2+S+l s + 2 2 8 2,16 zt14 2 P + +P ,7 zS5 zS4+ 1 + 2 2 S18+s17+2z16+2s15 +s14+ 2s” 2s” Zs9+ +22 2 2 zS5 t4 zS3+ s 1 +22 t + Z Zt17 tl‘ . . . +It I *I7 SI5 . . .s 1 SI8
+
+ +
+ +
+d3 +
+
+
t+l
+
+
+
+
+
+
+
+
+
+
+
+
6 6 1
3 2
1 1 1 i 8
2 . 6 1 9
3 4 5 6 7 8 9 1 5 4 7 1 0 2 9 8 3 9 3 ’ 2 1 0 4 8 7 0 7 9 6 5 3 8 4 5 1 0 3 7 6 1 2
I 1 7
2 5 5
3 3 4
4 4 3
5 2 2
1 1 8
2 5 2
3 7 3
1 9 9
5 2 5
1 4
2 3
3 2
4 1
6 6 6
0 6 5 2 1
7 7 1
1
1.3,4,9,11.12.13.16.17,27,29.
+ + 2 2 + +Z + 1 s 1 2 + 210 + 2 2 + s 7 + s 5 + +zr4 + ,3 + 2 2 + zS + 1 z6 + z5 + z4 + s3 + +s+1
5 2 3
I none
s12_+2t~’1+2r10+_t9+2s8+
+ + + +
38
+ +
+
2 3
4 3 2
1
(1,2,3,4) (0,5,1O,I5,20} 5,10,15.20 0 1.3.9 2.6,18 4,10,12 5.15.19 7,11,21 8.20.24 13 14,16,22 17,23.25 0 1,3,9,19,25,27 2,6.10.18,22,26
+
+
5.7.10.11,14,15,17.19.20,21,
22
1 0
t 4 + . 3 + 2 + s + 1 t + Z z3+21+1 ,3+,2+.+2 s 3 + 2 + 2 S 3 + 2 2 + S + 1 z3+s2+2z+1 13 212 2s 2 S+l t3+21+2 t 3 t 2x2 1 s+2 S 6 + S ~ + S 3 + S + l zS5 ,4 zt3 2+ + I = +1 t6 2’ z4 s3 t2 s
+
1 1
1.2,3,4.6.8,3.12,13,16,18
+ 23 + 2 220 + S I 5 + 2 0 + 2 5 + 1
3
29 31 32
+
+t5 r + 2
1 2 3
_I
+
+ +
permutations 1 2 3 1 5 4 6 4 5
1
2
3
4
5
- 1
3
2
4
5
1 1
2 3
3 2
6 6 6
7 3 7
8 8 1
9 4 4
2,6,8,18,19,22,23,24,26,31,
32,34 5,10,15.20.25.30 7,14,21,28 0 1,3.4.~.9,10.11.12.16.21.
25,26.27.28.30.33,34,36 2.5.6.8.13.14,15.17,18,19,
20,22.23,24,29.31.32,35
0 1 . 3 , 5 , 7 , . . . .37 2,4,6,8,. . . .36 19
2) m - loo >_ 6. Then, by Theorem 6a), we have wt(c) 2 1 8 - 5 - 5 = 8. So d 2 6, and since the lower square has minimum distance 6, we have d = 6. Example 44.9: n = 44, R = {1,3,5,8,9,15,22,23,24,25, 27,28,31,32,37,40}. We wish to apply Theorem 6 to prove that d 2 7. Suppose c = (a+b, -a+b) is a nonzero codeword of weight < 7, where (a,-a) E CE (equivalent to number 19)
1234 4
3
2
1
and (b, b) E CO (number 38). The only way to get wt(c) < 7 is by taking wt(a) = 5 and wt(b) = 4 or 5 or wt(a) = 6 and wt(b) = 6. But CE is a square and so wt(a) < 10 implies that a is zero on either the even or the odd positions. We have two cases.
1) m - loo = wt(a). Then, c is an element of the lower square, which has minimum distance 7. 2) m - loo 2 wt(a) 1. Then, by Theorem 6a), we have
+
VAN
EUPEN
AND VAN LINT:
ON THE
42 1
MINIMUM DISTANCE OF TERNARY CYCLIC CODES
TABLE IV (PART3 )
n 40
~~~~
~~~~
minimal polynomials s+2
1
2 3 4 5 6 7 8 9 IO
+
+
+
11
41
s 4 + s 3 + r 2 t 1 s4 + r 3 +IS+ 1 2r3 s2 2s I r4 s 2 + s + 2 r4 2 s 3 r2 1 z 4 + r 3 + r 2 + s + ~ ,'+I r 4 + 2 + 2 1 + 1 r 4 + s 2 + r + 1 r + 1 r4+2z3+s+1 s2 2s+ 2 s + 2 28 216 s5 r3+ 1 +2s2 2 s7 zS4 s2+ +s 1 s8 s7 216 2 2 +
12 13 1
2
+
+
+
+
4
+
+
+
+ + +
+
+
+
3
+
+
+
+
43 44
46
aero$ exponents 0 1,3,9.27 2,6,14,18 4,12,28,36 5,151 7,21,23.29 8.16,24,32 10,30 11,17,19,33 13,31,37.39 20 22,26,34.38 25.35 0 1.3.9.14.27,32.38,40
permutations 1 2 3 1 6 3 4 1 9 1 2 0
12
11
6
12
4 1 4 4 7
1 1 1 1 1
2 3 4 5 6
3 4 6 2 5
4 6 5 3 2
5 2 3 6 4
6 5 2 4 3
0 1,3,5,9,15,23,25,27, 31,37 2,6,10,18,30 4.12,16,20,36 7,13,17.19,21,29,35, 39,41,43 8,24.28.32,40 11.33 14,26,34,38,42 22
1 1 9
2 5 2
3 8 6
4 6 8
5 2 5
6 4 3
0 1,3,9,13,25,27,29, 31,35,39,41 2,4.6.8,12,16,18,24, 26,32,36 5,7,11,15,19,21,33, 37,43,45 10,14,20,22,28,30, 34.38,40,42.44 23 0
1 1 6
2 4 3
3 5 2
4 2 5
5 3 4
6 6 1
1 1
2 3
3 2
2 3
3 2
4 5
5 4
1
1
2.6.13.18.23,28.35, 39 4,5,12,15,26,29,36. 37 7,16.19,20,21,22,25, 34 8,10,11,17,24,30,31,
1
5 3 2 5 1 13 13
6 7 0 9 2
7 8 7 7 4
8 1 8
9 0 2
8
6 10
8 3 4
9 9 1
8
10 9 6
1 1 2 9
11 12 1 1 2 1 3 1 11 3 1 3
13 5 3 5
5
33
9 1 2
3
r+l r + 2 S I 1 + 2 2 0 + r 9 + sa+ +zS7 ,5 ,3 I SI1 + r 1 0 + s 9 + 2 s 8 + +zS7 r 5 s 3 2
+ +
+ +
+ +
47
1,2.3.4,6,7,8,9,12, 14.16.17,18,21,24. 25,27,28,32,34,36, 37,42 5,10,11,13,15,19,20, 22,23,26,29,30,31, 33,35,38,39,40,41, 43,44.45.46 0 [1.2.3.4.5,6} IO, 7.14.21,28,35,42} 7,14,21.28,35,42
49 50
7 7 7
none
0
1
(1,3,7,9}*(0,10.20,30,40}~6
6
( 2 , 4 , 6 . 8 } I [O, 1 0 . 2 0 , 3 0 , 4 0 ) 5.15.35.45 10,20,30,40 25
wt(c) wt(b)
2
S(wt(a) 2 8.
+3
+ 1)
-
wt(a)
-
wt(b) = 2wt(a)
-
So d 2 7, and since the lower cquare has minimum distance 7, we have (1! = 7 . Example 46.1: 71 = 46, R = { 1 . 3 . 9 . 10.13.13.20.22.23. 25.27.28.29.30.31.34.35.38.39.40.41. 42.44). ~ ~ H I F=T 8. The conditions in Corollary 1 are satisfied. By Corollary 1 a) weights are congruent to 0 or 1 mot1 3. So d 2 9.
APPENDIX
In Table IV, we give a complete list of all irreducible factors of .r" - 1, 71 5 50, over GF(3) (a list for n 5 100 can be found in [9]). We also give the exponents of the zeros of these polynomials for some chosen nth root of unity. As we saw, a cyclic code of length 71 can be represented by a set of irreducible factors of .I.'~ - 1. In the last column, we give the permutations on the minimal polynomials, that permute
IEEE TRANSACTIONS ON INFORMATION THEORY, VOL. 39, NO. 2, MARCH 1993
422
+
+
+
codes into equivalent codes. If xn-’ xn-2 . . . + x 1 is an irreducible factor of xn - 1, this will be denoted by “trivial.”
ACKNOWLEDGMENT The authors are very grateful to F. Bussemaker, Department of Mathematics, Eindhoven University of Technology, for writing the programs.
REFERENCES [l] J.H. van Lint and R.M. Wilson, “On the minimum distance of cyclic codes,” IEEE Trans. Inform. Theory, vol. IT-32, pp. 23-40; Jan. 1986.
M. van Eupen and C. van Eijl, “Minimum afstanden van temaire cyclische codes,” rep., Eindhoven Univ. of Technol., 1989. J.H. van Lint, Introduction to Coding Theory. New York: SpringerVerlag, 1982. F. J. MacWilliams and N. J. A. Sloane, The Theory of Error-Correcting Codes. Amsterdam: North Holland, 1977. C. R. P. Hartmann and K. K. Tzeng, “Generalizations of the BCH bound,” Inform. Contr., vol. 20, pp. 489-498, 1972. C. Roos, “A new lower bound for the minimum distance of a cyclic code,” IEEE Trans. Inform. Theory, vol. IT-29, pp. 330-332, May 1983. R. J. McEliece, “Weight congruences for p-ary cyclic codes,” Discrete Math., vol. 3, pp. 117-192, 1972. E. F. Assmus, Jr. and H. F. Mattson, Jr., “On weights in quadratic residue codes,” Discrete Math., vol. 3, pp. 1-20, 1972. R. J. McEliece, “Factorization of polynomials over finite fields,” Math. Computat., vol. 23, pp. 861-867, 1969.
