- Email: [email protected]
Weight hierarchies of linear codes of dimension 3
Journal of Statistical Planning and Inference 94 (2001) 167–179
www.elsevier.com/locate/jspi
Weight hierarchies of linear codes of dimension 3 a Institute
Wende Chena , Torleiv KlHveb;∗ of Systems Science, Academia Sinica and State Laboratory of Information Security, Graduate School of Academia Sinica, Beijing 100080, People’s Republic of China, China b Department of Informatics, University of Bergen HIB, N-5020 Bergen, Norway
Abstract The weight hierarchy of a linear [n; k; q] code C over GF(q) is the sequence (d1 ; d2 ; : : : ; dk ); where dr is the smallest support of an r-dimensional subcode of C. The weight hierarchies of [n; 3; q] codes are studied. In particular, for q65 the possible weight hierarchies of [n; 3; q] codes c 2001 Elsevier Science B.V. All rights reserved. are determined. MSC: primary 94B75; secondary 94B05; 05B25 Keywords: Weight hierarchy; Support weight; q-ary code; Chain condition; Di;erence sequence
1. Introduction The weight hierarchy of linear codes has been studied by a number of researchers. For a code of dimension k, it is a sequence of parameters (d1 ; d2 ; : : : ; dk ). In particular, d1 is the minimum distance of the code. The parameters were
Supported by The Norwegian Research Council. Corresponding author. E-mail address: [email protected] (T. KlFve).
∗
c 2001 Elsevier Science B.V. All rights reserved. 0378-3758/01/$ - see front matter PII: S 0 3 7 8 - 3 7 5 8 ( 0 0 ) 0 0 2 5 0 - 0
168
W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
(1995,1996) we studied the possible weight hierarchies of linear codes of dimension 4 or less over arbitrary
2. Notations and problem formulation Throughout this paper, unless otherwise stated, C will be an [n; 3; q] code, that is, a linear code of length n and dimension 3 over GF(q). For any subcode D of C, we de
W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
169
If y is a column in G, and x = y for some non-zero ∈ GF(q), then we may replace y by x without changing the support weight of any subcode. Therefore, we look at the vectors as projective points. Let V2 be the projective plane PG(2; q). A value assignment is a function m : V2 → N = {0; 1; 2; : : :}. For p ∈ V2 we call m(p) the value of p. A value assignment de
The di?erence sequence (DS) (i0 ; i1 ; i2 ) of a [d3 ; 3; q] code is de
i1 = d2 − d1 ;
i2 = d1 :
The di;erence sequence can easily be computed from the weight hierarchy and vice versa. In the geometric formulation of the problem it is more convenient to use the DS. We can reformulate the problem as follows. Problem 2. Given a triple (i0 ; i1 ; i2 ); is this a DS; that is; is there a value assignment m such that max{m(p) | p ∈ V2 } = i0 ; max{m(U ) | U is a line in V2 } = i0 + i1 ; m(V2 ) = i0 + i1 + i2 : In the terminology of Wei and Yang (1993), we say that C satis
and
m(V1 ) = i0 + i1 :
The corresponding di;erence sequence will be called a chain di?erence sequence (CDS). If C does not satisfy the chain condition, this can be formulated as follows: For any point p on any line L in V2 we have m(p) ¡ i0
or
m(L) ¡ i0 + i1 :
The corresponding di;erence sequence will be called a non-chain di?erence sequence (NDS).
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W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
Remark. A DS may be both q = 2, the codes generated by 111000000 000111000 and 000000111
CDS and NDS (for di;erent codes). For example, for the matrices 111000000 000111100 000001111
both have weight hierarchy (3; 6; 9). The
W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
171
Fig. 1. Representation of V2 = PG(2; 3).
Proof. Let V1 be a line and V0 a point on V1 such that m(V0 ) = i0 and m(V1 ) = i0 + i1 . Since m(p)6i0 for all points on V1 we have i0 + i1 = m(V1 )6(q + 1)i0 : This proves (i). Next, each line through V0 has value at most i0 + i1 . Since each point in V2 \{V0 } is on exactly one line through V0 we get i0 + i1 + i2 = m(V2 )6i0 + (q + 1)i1 and so (ii) follows. For q = 2, Lemma 1(i) and (ii) are special cases of Theorems 1 and 9, respectively, in Helleseth et al. (1992). Lemma 2. (i) If (i0 ; i1 ; i2 ) is a CDS and i2 ¿ 1; then (i0 ; i1 ; i2 − 1) is a CDS. (ii) If (i0 ; i1 ; i2 ) is a CDS; then (i0 + 1; i1 ; i2 ) is a CDS. (iii) If (i0 ; i1 ; i2 ) is a CDS; then (i0 + 1; i1 + q; i2 + q2 ) is a CDS. Proof. We use the same notations as in the previous proof. We have m(V2 \V1 ) = i2 . If i2 ¿ 1, let p ∈ V2 \V1 such that m(p) ¿ 0. If we decrease the value of m(p) by 1, we get a value assignment for the sequence (i0 ; i1 ; i2 − 1). This proves (i). If we increase m(V0 ) by 1, we get a value assignment for the sequence (i0 + 1; i1 ; i2 ). This proves (ii). If we increase the value of all points in V2 by 1 we get a value assignment for the sequence (i0 + 1; i1 + q; i2 + q2 ). This proves (iii).
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To describe the constructions, we let V0 = C and V1 = L(0). Construction 1. Suppose (i0 ; i1 ; i2 ) is a triple of positive integers and non-negative integers such that
and ! are
!6 ; i0 = ; i1 = (q − 1) + !; i2 = qi1 : Let
m(p) =
!
if p ∈ L(q); otherwise:
The corresponding DS is (i0 ; i1 ; i2 ) and it is an CDS. Proof. Clearly m(V0 ) = ¿m(p)
for all p ∈ V2 :
Further m(V2 ) = (q + 1)! + q2 = + { (q − 1) + !} + q{ (q − 1) + !} = i 0 + i1 + i2 : Finally m(L(q)) = (q + 1)!6 + (q − 1) + ! = i0 + i1 and since any line L = L(q) has exactly one point in common with L(q), we have m(L) = ! + q = i0 + i1 = m(V1 ): From this construction and Lemma 2 we get the following result: Corollary 1. Suppose i1 = (q − 1) + !; where 06!6 . Then (i0 ; i1 ; i2 ) is CDS for all i0 ¿ and 16i2 6qi1 . Construction 2. Suppose (i0 ; i1 ; i2 ) is a triple of positive integers and non-negative integers such that ¡ !6q − 2; %6q − ! − 1; i0 = q − ! + − %; i1 = (q − 1) + !; i2 = qi1 − %(! − )¿1:
; !; % are
W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
Let
173
0 i 0
for p = A; for p = C; m(p) = + 1 for p = P(i; 0) where 16i6!; + 1 for p = P(i; j) where 16i6q − %; 16j6! − ; otherwise: The corresponding DS is (i0 ; i1 ; i2 ) and it is a CDS. Proof. First, qi0 ¿i1 = q + ! − ¿ q , and so m(V0 ) = i0 ¿ + 1¿m(p) for all p = V0 . Next m(V2 ) = i0 + (q2 + q − 1) + ! + (! − )(q − %) = i 0 + i1 + i2 : We have m(V1 ) = i0 + (q − 1) + ! = i0 + i1 : For the other lines L through A we have m(L)6q + (q − %) = (q − 1) + ! + (q − ! + − %) = i0 + i1 : Any line L not through A meets all the lines L(i) for 06i6q and so m(L) 6 m(L ∩ L(0)) + q + (! − ) 6 i0 + (q − 1) + ! = i0 + i1 : Corollary 2. Suppose i1 = (q − 1) + !; where 06 ¡ !6q − 2. Let 06%6q − ! − 1. Then (i0 ; i1 ; i2 ) is CDS for all i0 ¿q − ! + − % and 16i2 6qi1 − %(! − ). Example. Consider q = 4. Let
= i1 =4 . Then
i1 = 3 + !; where ! ∈ { ; − 1; − 2; − 3}. Here !¿0 except in three cases, i1 = 1; 2; 5. By Corollary 1, all triples (i0 ; i1 ; i2 ) of positive integers such that i1 64i0 , i2 64i1 , i1 = 1; 2; 5 are CDS. Consider i1 = 1; 2; 5. By Corollary 2 we get (i0 ; 1; 4) (i0 ; 1; 3) (i0 ; 1; 2) (i0 ; 2; 8) (i0 ; 2; 6) (i0 ; 5; 20) (i0 ; 5; 19)
is is is is is is is
a a a a a a a
CDS CDS CDS CDS CDS CDS CDS
for for for for for for for
i0 ¿3 i0 ¿2 i0 ¿1 i0 ¿2 i0 ¿1 i0 ¿3 i0 ¿2
(use (use (use (use (use (use (use
= 0; = 0; = 0; = 0; = 0; = 1; = 1;
! = 1; ! = 1; ! = 1; ! = 2; ! = 2; ! = 2; ! = 2;
% = 0); % = 1); % = 2); % = 0); % = 1); % = 0); % = 1):
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Hence our constructions show that if (i0 ; i1 ; i2 ) is a triple of positive integers such that i1 64i0 and i2 64i1 , then (i0 ; i1 ; i2 ) is a CDS except possibly in the following six cases: (2; 1; 4); (1; 1; 3); (1; 1; 4); (1; 2; 7); (1; 2; 8); (2; 5; 20): However, a [6; 3; 4; 4] code is known to exist (see Brouwer, 1996). The corresponding DS is (1; 1; 4). This is not an NDS (see e.g. Corollary 4 in the next section) and so it has to be a CDSs. From Lemma 2(i), (ii), and (iii), respectively, we can conclude that (1; 1; 3), (2; 1; 4), and (2; 5; 20) are also CDS. On the other hand, no [10; 3; 7; 4] code exists (see Brouwer, 1996), and so (1; 2; 7) is not a DS. By Lemma 2(i), (1; 2; 8) is not a CDS either. For q65 we have done a similar analysis. The result is summarized in the following theorem: Theorem 1. For q65; all triples (i0 ; i1 ; i2 ) of positive integers such that i1 6qi0 and i2 6qi1 are CDS; except the following: q
Exceptions
2 3 4 5
None (1,1,3) (1,2,7), (1,2,8) (1,1,5), (1,2,9), (1,2,10), (1,3,13), (1,3,14), (1,3,15) For general q, we have the following corollary:
Corollary 3. If 16i1 6qi0 ; 16i2 6qi1 ; and i0 ¿q − 1; then (i0 ; i1 ; i2 ) is a CDS. Proof. Let = i1 =q :
(1)
Then i1 = (q − 1) + !;
where
− (q − 1)6!6 :
Since qi0 ¿i1 , (1) implies that i0 ¿ . Hence, if !¿0, then Corollary 1 implies that (i0 ; i1 ; i2 ) is a CDS. On the other hand, if − (q − 1)6!6 − 1, let
= −1
and
! = ! + q − 1:
Then i1 =
(q − 1) + !
and
¡ ! 6q − 2:
From Corollary 2 (with %=0), we conclude that (i0 ; i1 ; i2 ) is a CDS for i0 ¿q−(! − ), and in particular for i0 ¿q − 1.
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175
4. Weight hierarchies of codes not satisfying the chain condition Lemma 3. Let (i0 ; i1 ; i2 ) be an NDS. Then (i) i1 6qi0 − (q + 1); (ii) i2 6qi1 − (q + 1); (iii) i2 ¿i0 . Proof. By de
or
m(U ) ¡ i0 + i1 :
Let V0 be a point such that m(V0 ) = i0 and V1 a line such that m(V1 ) = i0 + i1 . Since m(p)6i0 − 1 for all points on V1 , we have i0 + i1 = m(V1 )6(q + 1)(i0 − 1): This proves (i). Next, each line through V0 has value at most i0 + i1 − 1. Since each point in V2\{V0 } is on exactly one line through V0 , we get i0 + i1 + i2 = m(V2 )6i0 + (q + 1)(i1 − 1) and so (ii) follows. Finally, i0 + i1 + i2 = m(V2 )¿m(V0 ) + m(V1 ) = i0 + i0 + i1 and (iii) follows. Corollary 4. Let (i0 ; i1 ; i2 ) be an NDS. Then i0 ¿2; i1 ¿2; and i2 ¿2. Remark. Corollary 4 shows in particular that the exceptional triples listed in Theorem 1 are not NDS. Hence all DS are CDS for q65. Whether this is true for all q is an open question. Lemma 4. If (i0 ; i1 ; i2 ) is an NDS and i2 ¿ i0 ; then (i0 ; i1 ; i2 − 1) is an NDS. Proof. We use the same notations as in the previous proof. We have m(V2 \V1 ) = i2 . If i2 ¿ i0 , let p ∈ V2 \(V1 ∪ V0 ) such that m(p) ¿ 0. If we decrease the value of m(p) by 1, we get a value assignment for the sequence (i0 ; i1 ; i2 − 1). The lower bound on i0 in Corollary 4 can be sharpened. To this end, and also to be able to describe our constructions, we introduce some further notations. De
(2)
where 06'6q. By Lemma 3(i), ¿1. Next, de
(3)
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W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
where 06)6q − 2. From (2) and (3) we get +'+1 (=+ : q−1 Finally, if ' ¿ 0, de
(4)
where 06+6). Lemma 5. We have i0 ¿(: Proof. Let P be a point on V1 with maximal value. By (2), m(P)¿i0 − . Let , be the average value of the points on PV0 \{P; V0 }. Then i0 + m(P) + (q − 1), = m(PV0 )6i0 + i1 − 1 and so (q − 1),6i1 − 1 − m(P)6i1 − 1 − (i0 − ) = (q − 1)(i0 − () + ): Therefore, there exists a point p on PV0 such that 06m(p)6,6i0 − (: To describe our constructions, we let V0 = B and V1 = L(0). Construction 3. Using notations (2)–(3); let (i0 ; i1 ; i2 ) be a triple such that (i) i1 6qi0 − (q + 1); (ii) i2 = qi1 − (q + 1); (iii) i2 ¿i0 ; (iv) i0 ¿(; (v) ' = 0: Let i0 for p = V0 ; i − for p ∈ V1 ; 0 m(p) = i0 − ( for p = P(i; q) where ) + 16i6q − 1; i − ( for p = P(i; j) where 06i6q − 1; ) + 16j6q − 1; 0 i0 − ( + 1 otherwise: The corresponding DS is (i0 ; i1 ; i2 ) and it is an NDS. Construction 4. Using notations (2) – (4); let (i0 ; i1 ; i2 ) be a triple such that (i) i1 6qi0 − (q + 1); (ii) i2 = qi1 − (q + 1); (iii) i2 ¿i0 ;
W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
177
(iv) i0 ¿(; (v) ' ¿ 0; (vi) (+ = 0 and ( + ) + *¿q − 1) or (+ ¿ 0 and ( + ) + *¿q): Let i0 for p = V0 ; i − for p = P(i; 0) where 06i6q − '; 0 i0 − − 1 for p = P(i; 0) where q − ' + 16i6q; i0 − ( for p = P(i; q) where ) + 26i6q − 1; i − ( for p = P(i; j) where 16j6+; 0 m(p) = and (j − 1)(* − 1)6i6j(* − 1) − 1; i0 − ( for p = P(i; j) where + + 16j6) + 1; and (j − 1)* − +6i6j* − + − 1; i − ( for p = P(i; j) where ) + 26j6q − 1; 0 and 06i6q − 1; i0 − ( + 1 otherwise: The corresponding DS is (i0 ; i1 ; i2 ) and it is an NDS. Proof. We prove that Construction 4 has the stated property, the proof for Construction 3 is similar and simpler and is omitted. It is clear that m(p)6i0 =m(V0 ) for all p ∈ V2 . Let us consider the lines. We have m(V1 ) = (q + 1)(i0 − ) − ' = i0 + i1 ; m(L(q)) = i0 + (i0 − − 1) + (q − 1)(i0 − () + ) + 1 = i0 + i1 − 1: For 16j6q − 1 we have m(L(j)) 6 (i0 − − 1) + q(i0 − ( + 1) − (* − 1) = i0 + i1 + q − (( + ) + *) − 1 ¡ i0 + i1 if + ¿ 0 and similarly if + = 0. For 06i6q − ' we have m(L∗ (i)) = i0 + (i0 − ) + (q − 1)(i0 − () + ) = i0 + i1 − 1; and similarly for q − ' + 16i6q − 1. If L is a line not containing A or B, then m(L)6(i0 − ) + q(i0 − () + ) + 2 = i0 + i1 + 1 − (6i0 + i1 : Finally, m(V2 ) = m(V0 ) +
q i=0
m(L∗ (q)\V0 ) = i0 + (q + 1)(i1 − 1) = i0 + i1 + i2 :
Combining the constructions with Lemma 4 we get NDS for i0 6i2 6qi1 − (q + 1). Example. Let q = 3. To use the constructions, we consider 3i0 − i1 modulo 8. We list the corresponding , ', etc. in the following table.
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Fig. 2. An example of value assignment.
3i0 − i1
'
(
)
8a 8a + 1 8a + 2 8a + 3 8a + 4 8a + 5 8a + 6 8a + 7
2a 2a 2a 2a 2a + 1 2a + 1 2a + 1 2a + 1
0 1 2 3 0 1 2 3
3a + 1 3a + 1 3a + 2 3a + 2 3a + 2 3a + 3 3a + 3 3a + 4
1 0 1 0 0 1 0 1
*
+
(+)+*
3 1 1
0 0 0
3a + 4 3a + 4 3a + 3
2 2 1
1 0 1
3a + 6 3a + 5 3a + 6
From the table we see that condition (vi) in Construction 4 is satis
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179
m(L(j)) is reduced by (+)+*−q+1 for 16j6+ and (+)+*−q for ++16j6)+1, then the construction works for i2 6qi2 − (q + 1) − () + 1)(( + ) + * − q) − +. References Barbero, A.I., Tena, J.G., 1995. Weight hierarchy of a product code. IEEE Trans. Inform. Theory 41, 1475–1479. Brouwer, A.E., 1996. Net accessible table of parameters of [n; k; d; q] codes. Web address: html:==www.win.tue.nl=win=math=dw=voorlincod.html Chen, W., KlHve, T., 1995. The weight hierarchies of q-ary codes of dimension 2 and 3. Unpublished manuscript. Chen, W., KlHve, T., 1996. The weight hierarchies of q-ary codes of dimension 4. IEEE Trans. Inform. Theory 42, 2265–2272. Chen, W., KlHve, T., 2001. Weight hierarchies of binary linear codes of dimension 4. Discrete Math. (to appear). Encheva, S., KlHve, T., 1994. Codes satisfying the chain condition. IEEE Trans. Inform. Theory 40, 175–180. Forney, G.D., 1994. Dimension=length pro
www.elsevier.com/locate/jspi
Weight hierarchies of linear codes of dimension 3 a Institute
Wende Chena , Torleiv KlHveb;∗ of Systems Science, Academia Sinica and State Laboratory of Information Security, Graduate School of Academia Sinica, Beijing 100080, People’s Republic of China, China b Department of Informatics, University of Bergen HIB, N-5020 Bergen, Norway
Abstract The weight hierarchy of a linear [n; k; q] code C over GF(q) is the sequence (d1 ; d2 ; : : : ; dk ); where dr is the smallest support of an r-dimensional subcode of C. The weight hierarchies of [n; 3; q] codes are studied. In particular, for q65 the possible weight hierarchies of [n; 3; q] codes c 2001 Elsevier Science B.V. All rights reserved. are determined. MSC: primary 94B75; secondary 94B05; 05B25 Keywords: Weight hierarchy; Support weight; q-ary code; Chain condition; Di;erence sequence
1. Introduction The weight hierarchy of linear codes has been studied by a number of researchers. For a code of dimension k, it is a sequence of parameters (d1 ; d2 ; : : : ; dk ). In particular, d1 is the minimum distance of the code. The parameters were
Supported by The Norwegian Research Council. Corresponding author. E-mail address: [email protected] (T. KlFve).
∗
c 2001 Elsevier Science B.V. All rights reserved. 0378-3758/01/$ - see front matter PII: S 0 3 7 8 - 3 7 5 8 ( 0 0 ) 0 0 2 5 0 - 0
168
W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
(1995,1996) we studied the possible weight hierarchies of linear codes of dimension 4 or less over arbitrary
2. Notations and problem formulation Throughout this paper, unless otherwise stated, C will be an [n; 3; q] code, that is, a linear code of length n and dimension 3 over GF(q). For any subcode D of C, we de
W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
169
If y is a column in G, and x = y for some non-zero ∈ GF(q), then we may replace y by x without changing the support weight of any subcode. Therefore, we look at the vectors as projective points. Let V2 be the projective plane PG(2; q). A value assignment is a function m : V2 → N = {0; 1; 2; : : :}. For p ∈ V2 we call m(p) the value of p. A value assignment de
The di?erence sequence (DS) (i0 ; i1 ; i2 ) of a [d3 ; 3; q] code is de
i1 = d2 − d1 ;
i2 = d1 :
The di;erence sequence can easily be computed from the weight hierarchy and vice versa. In the geometric formulation of the problem it is more convenient to use the DS. We can reformulate the problem as follows. Problem 2. Given a triple (i0 ; i1 ; i2 ); is this a DS; that is; is there a value assignment m such that max{m(p) | p ∈ V2 } = i0 ; max{m(U ) | U is a line in V2 } = i0 + i1 ; m(V2 ) = i0 + i1 + i2 : In the terminology of Wei and Yang (1993), we say that C satis
and
m(V1 ) = i0 + i1 :
The corresponding di;erence sequence will be called a chain di?erence sequence (CDS). If C does not satisfy the chain condition, this can be formulated as follows: For any point p on any line L in V2 we have m(p) ¡ i0
or
m(L) ¡ i0 + i1 :
The corresponding di;erence sequence will be called a non-chain di?erence sequence (NDS).
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W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
Remark. A DS may be both q = 2, the codes generated by 111000000 000111000 and 000000111
CDS and NDS (for di;erent codes). For example, for the matrices 111000000 000111100 000001111
both have weight hierarchy (3; 6; 9). The
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171
Fig. 1. Representation of V2 = PG(2; 3).
Proof. Let V1 be a line and V0 a point on V1 such that m(V0 ) = i0 and m(V1 ) = i0 + i1 . Since m(p)6i0 for all points on V1 we have i0 + i1 = m(V1 )6(q + 1)i0 : This proves (i). Next, each line through V0 has value at most i0 + i1 . Since each point in V2 \{V0 } is on exactly one line through V0 we get i0 + i1 + i2 = m(V2 )6i0 + (q + 1)i1 and so (ii) follows. For q = 2, Lemma 1(i) and (ii) are special cases of Theorems 1 and 9, respectively, in Helleseth et al. (1992). Lemma 2. (i) If (i0 ; i1 ; i2 ) is a CDS and i2 ¿ 1; then (i0 ; i1 ; i2 − 1) is a CDS. (ii) If (i0 ; i1 ; i2 ) is a CDS; then (i0 + 1; i1 ; i2 ) is a CDS. (iii) If (i0 ; i1 ; i2 ) is a CDS; then (i0 + 1; i1 + q; i2 + q2 ) is a CDS. Proof. We use the same notations as in the previous proof. We have m(V2 \V1 ) = i2 . If i2 ¿ 1, let p ∈ V2 \V1 such that m(p) ¿ 0. If we decrease the value of m(p) by 1, we get a value assignment for the sequence (i0 ; i1 ; i2 − 1). This proves (i). If we increase m(V0 ) by 1, we get a value assignment for the sequence (i0 + 1; i1 ; i2 ). This proves (ii). If we increase the value of all points in V2 by 1 we get a value assignment for the sequence (i0 + 1; i1 + q; i2 + q2 ). This proves (iii).
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To describe the constructions, we let V0 = C and V1 = L(0). Construction 1. Suppose (i0 ; i1 ; i2 ) is a triple of positive integers and non-negative integers such that
and ! are
!6 ; i0 = ; i1 = (q − 1) + !; i2 = qi1 : Let
m(p) =
!
if p ∈ L(q); otherwise:
The corresponding DS is (i0 ; i1 ; i2 ) and it is an CDS. Proof. Clearly m(V0 ) = ¿m(p)
for all p ∈ V2 :
Further m(V2 ) = (q + 1)! + q2 = + { (q − 1) + !} + q{ (q − 1) + !} = i 0 + i1 + i2 : Finally m(L(q)) = (q + 1)!6 + (q − 1) + ! = i0 + i1 and since any line L = L(q) has exactly one point in common with L(q), we have m(L) = ! + q = i0 + i1 = m(V1 ): From this construction and Lemma 2 we get the following result: Corollary 1. Suppose i1 = (q − 1) + !; where 06!6 . Then (i0 ; i1 ; i2 ) is CDS for all i0 ¿ and 16i2 6qi1 . Construction 2. Suppose (i0 ; i1 ; i2 ) is a triple of positive integers and non-negative integers such that ¡ !6q − 2; %6q − ! − 1; i0 = q − ! + − %; i1 = (q − 1) + !; i2 = qi1 − %(! − )¿1:
; !; % are
W. Chen, T. Kl3ve / Journal of Statistical Planning and Inference 94 (2001) 167–179
Let
173
0 i 0
for p = A; for p = C; m(p) = + 1 for p = P(i; 0) where 16i6!; + 1 for p = P(i; j) where 16i6q − %; 16j6! − ; otherwise: The corresponding DS is (i0 ; i1 ; i2 ) and it is a CDS. Proof. First, qi0 ¿i1 = q + ! − ¿ q , and so m(V0 ) = i0 ¿ + 1¿m(p) for all p = V0 . Next m(V2 ) = i0 + (q2 + q − 1) + ! + (! − )(q − %) = i 0 + i1 + i2 : We have m(V1 ) = i0 + (q − 1) + ! = i0 + i1 : For the other lines L through A we have m(L)6q + (q − %) = (q − 1) + ! + (q − ! + − %) = i0 + i1 : Any line L not through A meets all the lines L(i) for 06i6q and so m(L) 6 m(L ∩ L(0)) + q + (! − ) 6 i0 + (q − 1) + ! = i0 + i1 : Corollary 2. Suppose i1 = (q − 1) + !; where 06 ¡ !6q − 2. Let 06%6q − ! − 1. Then (i0 ; i1 ; i2 ) is CDS for all i0 ¿q − ! + − % and 16i2 6qi1 − %(! − ). Example. Consider q = 4. Let
= i1 =4 . Then
i1 = 3 + !; where ! ∈ { ; − 1; − 2; − 3}. Here !¿0 except in three cases, i1 = 1; 2; 5. By Corollary 1, all triples (i0 ; i1 ; i2 ) of positive integers such that i1 64i0 , i2 64i1 , i1 = 1; 2; 5 are CDS. Consider i1 = 1; 2; 5. By Corollary 2 we get (i0 ; 1; 4) (i0 ; 1; 3) (i0 ; 1; 2) (i0 ; 2; 8) (i0 ; 2; 6) (i0 ; 5; 20) (i0 ; 5; 19)
is is is is is is is
a a a a a a a
CDS CDS CDS CDS CDS CDS CDS
for for for for for for for
i0 ¿3 i0 ¿2 i0 ¿1 i0 ¿2 i0 ¿1 i0 ¿3 i0 ¿2
(use (use (use (use (use (use (use
= 0; = 0; = 0; = 0; = 0; = 1; = 1;
! = 1; ! = 1; ! = 1; ! = 2; ! = 2; ! = 2; ! = 2;
% = 0); % = 1); % = 2); % = 0); % = 1); % = 0); % = 1):
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Hence our constructions show that if (i0 ; i1 ; i2 ) is a triple of positive integers such that i1 64i0 and i2 64i1 , then (i0 ; i1 ; i2 ) is a CDS except possibly in the following six cases: (2; 1; 4); (1; 1; 3); (1; 1; 4); (1; 2; 7); (1; 2; 8); (2; 5; 20): However, a [6; 3; 4; 4] code is known to exist (see Brouwer, 1996). The corresponding DS is (1; 1; 4). This is not an NDS (see e.g. Corollary 4 in the next section) and so it has to be a CDSs. From Lemma 2(i), (ii), and (iii), respectively, we can conclude that (1; 1; 3), (2; 1; 4), and (2; 5; 20) are also CDS. On the other hand, no [10; 3; 7; 4] code exists (see Brouwer, 1996), and so (1; 2; 7) is not a DS. By Lemma 2(i), (1; 2; 8) is not a CDS either. For q65 we have done a similar analysis. The result is summarized in the following theorem: Theorem 1. For q65; all triples (i0 ; i1 ; i2 ) of positive integers such that i1 6qi0 and i2 6qi1 are CDS; except the following: q
Exceptions
2 3 4 5
None (1,1,3) (1,2,7), (1,2,8) (1,1,5), (1,2,9), (1,2,10), (1,3,13), (1,3,14), (1,3,15) For general q, we have the following corollary:
Corollary 3. If 16i1 6qi0 ; 16i2 6qi1 ; and i0 ¿q − 1; then (i0 ; i1 ; i2 ) is a CDS. Proof. Let = i1 =q :
(1)
Then i1 = (q − 1) + !;
where
− (q − 1)6!6 :
Since qi0 ¿i1 , (1) implies that i0 ¿ . Hence, if !¿0, then Corollary 1 implies that (i0 ; i1 ; i2 ) is a CDS. On the other hand, if − (q − 1)6!6 − 1, let
= −1
and
! = ! + q − 1:
Then i1 =
(q − 1) + !
and
¡ ! 6q − 2:
From Corollary 2 (with %=0), we conclude that (i0 ; i1 ; i2 ) is a CDS for i0 ¿q−(! − ), and in particular for i0 ¿q − 1.
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4. Weight hierarchies of codes not satisfying the chain condition Lemma 3. Let (i0 ; i1 ; i2 ) be an NDS. Then (i) i1 6qi0 − (q + 1); (ii) i2 6qi1 − (q + 1); (iii) i2 ¿i0 . Proof. By de
or
m(U ) ¡ i0 + i1 :
Let V0 be a point such that m(V0 ) = i0 and V1 a line such that m(V1 ) = i0 + i1 . Since m(p)6i0 − 1 for all points on V1 , we have i0 + i1 = m(V1 )6(q + 1)(i0 − 1): This proves (i). Next, each line through V0 has value at most i0 + i1 − 1. Since each point in V2\{V0 } is on exactly one line through V0 , we get i0 + i1 + i2 = m(V2 )6i0 + (q + 1)(i1 − 1) and so (ii) follows. Finally, i0 + i1 + i2 = m(V2 )¿m(V0 ) + m(V1 ) = i0 + i0 + i1 and (iii) follows. Corollary 4. Let (i0 ; i1 ; i2 ) be an NDS. Then i0 ¿2; i1 ¿2; and i2 ¿2. Remark. Corollary 4 shows in particular that the exceptional triples listed in Theorem 1 are not NDS. Hence all DS are CDS for q65. Whether this is true for all q is an open question. Lemma 4. If (i0 ; i1 ; i2 ) is an NDS and i2 ¿ i0 ; then (i0 ; i1 ; i2 − 1) is an NDS. Proof. We use the same notations as in the previous proof. We have m(V2 \V1 ) = i2 . If i2 ¿ i0 , let p ∈ V2 \(V1 ∪ V0 ) such that m(p) ¿ 0. If we decrease the value of m(p) by 1, we get a value assignment for the sequence (i0 ; i1 ; i2 − 1). The lower bound on i0 in Corollary 4 can be sharpened. To this end, and also to be able to describe our constructions, we introduce some further notations. De
(2)
where 06'6q. By Lemma 3(i), ¿1. Next, de
(3)
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where 06)6q − 2. From (2) and (3) we get +'+1 (=+ : q−1 Finally, if ' ¿ 0, de
(4)
where 06+6). Lemma 5. We have i0 ¿(: Proof. Let P be a point on V1 with maximal value. By (2), m(P)¿i0 − . Let , be the average value of the points on PV0 \{P; V0 }. Then i0 + m(P) + (q − 1), = m(PV0 )6i0 + i1 − 1 and so (q − 1),6i1 − 1 − m(P)6i1 − 1 − (i0 − ) = (q − 1)(i0 − () + ): Therefore, there exists a point p on PV0 such that 06m(p)6,6i0 − (: To describe our constructions, we let V0 = B and V1 = L(0). Construction 3. Using notations (2)–(3); let (i0 ; i1 ; i2 ) be a triple such that (i) i1 6qi0 − (q + 1); (ii) i2 = qi1 − (q + 1); (iii) i2 ¿i0 ; (iv) i0 ¿(; (v) ' = 0: Let i0 for p = V0 ; i − for p ∈ V1 ; 0 m(p) = i0 − ( for p = P(i; q) where ) + 16i6q − 1; i − ( for p = P(i; j) where 06i6q − 1; ) + 16j6q − 1; 0 i0 − ( + 1 otherwise: The corresponding DS is (i0 ; i1 ; i2 ) and it is an NDS. Construction 4. Using notations (2) – (4); let (i0 ; i1 ; i2 ) be a triple such that (i) i1 6qi0 − (q + 1); (ii) i2 = qi1 − (q + 1); (iii) i2 ¿i0 ;
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177
(iv) i0 ¿(; (v) ' ¿ 0; (vi) (+ = 0 and ( + ) + *¿q − 1) or (+ ¿ 0 and ( + ) + *¿q): Let i0 for p = V0 ; i − for p = P(i; 0) where 06i6q − '; 0 i0 − − 1 for p = P(i; 0) where q − ' + 16i6q; i0 − ( for p = P(i; q) where ) + 26i6q − 1; i − ( for p = P(i; j) where 16j6+; 0 m(p) = and (j − 1)(* − 1)6i6j(* − 1) − 1; i0 − ( for p = P(i; j) where + + 16j6) + 1; and (j − 1)* − +6i6j* − + − 1; i − ( for p = P(i; j) where ) + 26j6q − 1; 0 and 06i6q − 1; i0 − ( + 1 otherwise: The corresponding DS is (i0 ; i1 ; i2 ) and it is an NDS. Proof. We prove that Construction 4 has the stated property, the proof for Construction 3 is similar and simpler and is omitted. It is clear that m(p)6i0 =m(V0 ) for all p ∈ V2 . Let us consider the lines. We have m(V1 ) = (q + 1)(i0 − ) − ' = i0 + i1 ; m(L(q)) = i0 + (i0 − − 1) + (q − 1)(i0 − () + ) + 1 = i0 + i1 − 1: For 16j6q − 1 we have m(L(j)) 6 (i0 − − 1) + q(i0 − ( + 1) − (* − 1) = i0 + i1 + q − (( + ) + *) − 1 ¡ i0 + i1 if + ¿ 0 and similarly if + = 0. For 06i6q − ' we have m(L∗ (i)) = i0 + (i0 − ) + (q − 1)(i0 − () + ) = i0 + i1 − 1; and similarly for q − ' + 16i6q − 1. If L is a line not containing A or B, then m(L)6(i0 − ) + q(i0 − () + ) + 2 = i0 + i1 + 1 − (6i0 + i1 : Finally, m(V2 ) = m(V0 ) +
q i=0
m(L∗ (q)\V0 ) = i0 + (q + 1)(i1 − 1) = i0 + i1 + i2 :
Combining the constructions with Lemma 4 we get NDS for i0 6i2 6qi1 − (q + 1). Example. Let q = 3. To use the constructions, we consider 3i0 − i1 modulo 8. We list the corresponding , ', etc. in the following table.
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Fig. 2. An example of value assignment.
3i0 − i1
'
(
)
8a 8a + 1 8a + 2 8a + 3 8a + 4 8a + 5 8a + 6 8a + 7
2a 2a 2a 2a 2a + 1 2a + 1 2a + 1 2a + 1
0 1 2 3 0 1 2 3
3a + 1 3a + 1 3a + 2 3a + 2 3a + 2 3a + 3 3a + 3 3a + 4
1 0 1 0 0 1 0 1
*
+
(+)+*
3 1 1
0 0 0
3a + 4 3a + 4 3a + 3
2 2 1
1 0 1
3a + 6 3a + 5 3a + 6
From the table we see that condition (vi) in Construction 4 is satis
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m(L(j)) is reduced by (+)+*−q+1 for 16j6+ and (+)+*−q for ++16j6)+1, then the construction works for i2 6qi2 − (q + 1) − () + 1)(( + ) + * − q) − +. References Barbero, A.I., Tena, J.G., 1995. Weight hierarchy of a product code. IEEE Trans. Inform. Theory 41, 1475–1479. Brouwer, A.E., 1996. Net accessible table of parameters of [n; k; d; q] codes. Web address: html:==www.win.tue.nl=win=math=dw=voorlincod.html Chen, W., KlHve, T., 1995. The weight hierarchies of q-ary codes of dimension 2 and 3. Unpublished manuscript. Chen, W., KlHve, T., 1996. The weight hierarchies of q-ary codes of dimension 4. IEEE Trans. Inform. Theory 42, 2265–2272. Chen, W., KlHve, T., 2001. Weight hierarchies of binary linear codes of dimension 4. Discrete Math. (to appear). Encheva, S., KlHve, T., 1994. Codes satisfying the chain condition. IEEE Trans. Inform. Theory 40, 175–180. Forney, G.D., 1994. Dimension=length pro