On the nonexistence of some quaternary linear codes meeting the Griesmer bound

Journal of Statistical Planning and Inference 72 (1998) 303–322

On the nonexistence of some quaternary linear codes meeting the Griesmer bound Noboru Hamada ∗ Department of Applied Mathematics, Osaka Women’s University, Daisen-cho, Sakai, Osaka 590, Japan

Abstract It is unknown whether or not there exists a quaternary linear code with parameters [293, 5, 219], [289, 5, 216] or [277, 5, 207]. The purpose of this paper is to prove the nonexistence of quaternary linear codes with parameters [293, 5, 219], [289, 5, 216] or [277, 5, 207] using the nonexistence of {48; 11; 4; 4}-minihypers, {52; 12; 4; 4}-minihypers and {64; 15; 4; 4}-minihypers, respectively. Since there exists a quaternary linear code with parameters [294, 5, 219], [290, 5, 216] or [278, 5, 207], this implies that n4 (5; 219) = 294, n4 (5; 216) = 290 and n4 (5; 207) = 278. c 1998 Elsevier Science B.V. All rights reserved.

AMS classi cation: 05B25; 62K15 Keywords: Linear code; Griesmer bound; Minihyper

1. Introduction Let V (n; q) be an n-dimensional vector space consisting of row vectors over the Galois eld GF(q) of order q, where n¿3 and q is a prime power. A k-dimensional subspace C of V (n; q) is called an [n; k; d; q]-code (or a q-ary linear code with length n, dimension k, and minimum distance d) if the minimum Hamming distance of the code C is equal to d (cf. MacWilliams and Sloane, 1977). A central problem in coding theory is to optimize one of the parameters n; k; d for given values of the other two. In this paper, we shall consider the following two problems. Problem 1.1. (1) Find dq (n; k), the largest value of d for which there exists an [n; k; d; q]-code for given values of n, k, and q. (2) Construct an [n; k; d; q]-code with d = dq (n; k) for given values of n; k; and q. ∗

E-mail: [email protected]

c 1998 Elsevier Science B.V. All rights reserved. 0378-3758/98/$ – see front matter PII: S 0 3 7 8 - 3 7 5 8 ( 9 8 ) 0 0 0 3 9 - 1

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Problem 1.2. (1) Find nq (k; d), the smallest value of n for which there exists an [n; k; d; q]-code for given values of k, d, and q. (2) Construct an [n; k; d; q]-code with n = nq (k; d) for given values of k; d; and q. Problem 1.1 is known as the “packing problem” and has been investigated by many researchers of statistics and combinatorics. It is known that (i) linear codes and symmetrical factorial designs are closely related (cf. Bose, 1961; Bose and Srivastava, 1964) and (ii) in order to solve Problem 1.1, it is sucient to solve Problem 1.2 for all values of k; d; and q and (iii) if there exists an [n; k; d; q]-code for given values of k; d; and q, then   k−1 P d ; (1.1) n¿gq (k; d) := i i=0 q where dxe denotes the smallest integer ¿x. The bound (1:1) is called Griesmer bound (cf. Griesmer, 1960; Solomon and Stier, 1965). Hence in order to solve Problem 1.2, it is sucient to solve the following problem when there exists a [gq (k; q); k; d; q]-code or a [gq (k; d) + 1; k; d; q]-code with B2 = 0, where B2 denotes the number of codewords of weight 2 in its dual code. Problem 1.3. (1) Find a necessary and sucient condition for integers k; d; and q such that there exists a [gq (k; d); k; d; q]-code. (2) Characterize all [gq (k; d); k; d; q]-codes for given values of k; d; and q when such codes exist. (3) Characterize all [gq (k; d) + 1; k; d; q]-codes with B2 = 0 for given values of k; d; and q when there is no [gq (k; d); k; d; q]-code. In the case q = 2 and 16d62k−1 , (1) and (2) in Problem 1.3 were solved completely by Helleseth (1981). Hence we restrict ourselves to the case q¿3; k¿3 and 16d¡q k−1 in what follows. In this case, d can be expressed uniquely as follows: d = q k−1 −

k−2 P i=0

i q i

(1.2)

using some ordered set (0 ; 1 ; : : : ; k−2 ) in E(k − 1; q) and the Griesmer bound (1:1) can be expressed as follows: n¿gq (k; d) = vk −

k−2 P i=0

i vi+1 ;

(1.3)

where vi = (q i − 1)=(q − 1) for any integer i¿0 and E(k − 1; q) denotes the set of all ordered sets (0 ; 1 ; : : : ; k−2 ) of integers i such that 06i 6q−1 and (0 ; 1 ; : : : ; k−2 ) 6= (0; 0; : : : ; 0). De nition 1.1. Let F be a set of f points in a nite projective geometry PG(t; q), where f¿1 and t¿2. If |F ∩ H |¿m for every hyperplane (i.e., (t − 1)- at) H in

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PG(t; q) and |F ∩ H | = m for some hyperplane H in PG(t; q), then F is called an {f; m; t; q}-minihyper, where m¿0 and |A| denotes the number of elements in the set A. The concept of a minihyper was introduced by Hamada and Tamari (1978). It was shown by Hamada (1993a) (cf. Corollaries A.1 and A.2 in Appendix A) that in order to solve Problem 1.3, it is sucient to solve the following problem. In what follows, let t = k − 1. Problem 1.4. (1) Find a necessary and sucient condition for integers t; q; 0 ; 1 ; Pt−1 Pt−1 : : : ; t−2 and t−1 such that there exists a { i=0 i vi+1 ; i=0 i vi ; t; q}-minihyper for given values of t; q; 0 ; 1 ; : : : ; t−2 and t−1 , where t¿2, q¿3, (0 ; 1 ; : : : ; t−1 ) ∈ E(t; q) and E(t; q) denotes the set de ned in De nition A.2. Pt−1 Pt−1 (2) Characterize all { i=0 i vi+1 ; i=0 i vi ; t; q}-minihyper for given values of t; q; 0 ; 1 ; : : : ; t−2 and t−1 , when there exist such minihypers. Recently Problems 1.1–1.4 were solved for many cases. In the case q = 3, the value of n3 (k; d) is known for k65 and all d and a table of the bounds for n3 (6; d), 16d6243, was given by Hamada (1993b). Recently it has been updated by Hamada and Watamori (1996). In the case q = 4, the value of n4 (k; d) is known for k64 and all d. But in the case q = 4 and k = 5, the value of n4 (5; d) is unknown for many integers d and a table of the bounds for n4 (5; d); 16d6256, has been given by Hamada (1996) using recent results. It is unknown whether or not there exists a quaternary linear code with parameters [293; 5; 219], [289; 5; 216] or [277; 5; 207]. The purpose of this paper is to prove the nonexistence of quaternary linear codes with parameters [293; 5; 219], [289; 5; 216] and [277; 5; 207] using the nonexistence of {v1 + v2 + 2v3 ; v0 + v1 + 2v2 ; 4; 4}-minihypers, {2v2 + 2v3 ; 2v1 + 2v2 ; 4; 4}-minihypers and {v1 + 3v3 ; v0 + 3v2 ; 4; 4}-minihypers, respectively. In what follows, let q = 4 and vi = (4i − 1)=(4 − 1) for any integer i¿0. That is, v0 = 0, v1 = 1, v2 = 5, v3 = 21, v4 = 85 and v5 = 341. Theorem 1.1. There is no {v1 + v2 + 2v3 ; v0 + v1 + 2v2 ; 4; 4}-minihyper. Theorem 1.2. There is no {2v2 + 2v3 ; 2v1 + 2v2 ; 4; 4}-minihyper. Theorem 1.3. There is no {v1 + 3v3 ; v0 + 3v2 ; 4; 4}-minihyper. It is known that (1) there exists a quaternary [256; 5; 192] code which has a codeword of weight 256 and (2) there exists a quaternary linear code with parameters [38; 4; 27], [34; 4; 24] and [22; 4; 15] (cf. Table 1 in Appendix B). Hence it follows from Lemma 2.11 in Hill and Newton (1992) that there exists a quaternary linear code with parameters [294; 5; 219], [290; 5; 216] and [278; 5; 207]. From Corollary A.1 in Appendix A and Theorems 1.1–1.3, we have the following Corollaries 1.1–1.3, respectively.

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Corollary 1.1. There is no [293; 5; 219; 4]-code meeting the Griesmer bound and n4 (5; 219) = 294: Corollary 1.2. There is no [289; 5; 216; 4]-code meeting the Griesmer bound and n4 (5; 216) = 290: Corollary 1.3. There is no [277; 5; 207; 4]-code meeting the Griesmer bound and n4 (5; 207) = 278: In order to prove Theorems 1.2 and 1.3, we use the following two theorems whose proofs will be given in Sections 5 and 6, respectively. Theorem 1.4. (1) K is a {3v1 + v3 ; 3v0 + v2 ; 3; 4}-minihyper if and only if K = V ∪ {P1 ; P2 ; P3 } for some 2- at V and some three points P1 ; P2 and P3 in PG(3; 4). (2) If K is a {3v1 + v3 ; 3v0 + v2 ; 3; 4}-minihyper; then |K ∩ G| = 5; 6; 7; 8 or 21 for any 2- at G in PG(3; 4) and (m5 ; m6 ; m7 ; m8 ; m21 ) = (31; 48; 0; 5; 1) or (35; 36; 12; 1; 1) according as the three points P1 ; P2 and P3 are collinear or not; where mi denotes the number of 2- ats G in PG(3; 4) such that |K ∩ G| = i. Theorem 1.5. If K is a {4v2 ; 4v1 ; 3; 4}-minihyper; then |K ∩ G| = 4; 5; 6; 8; 16 or 20 for any 2- at G in PG(3; 4) and (m4 ; m5 ; m6 ; m8 ; m16 ; m20 ) = (20; 64; 0; 0; 0; 1); (55; 0; 24; 5; 1; 0) or (65; 0; 0; 20; 0; 0).

2. The proof of Theorem 1.1 It follows from Corollary A.1 and Table 1 (d = 15; 16; 24; 27; 28; 30; 31; 32; 39; 40; 42; 43; 44) that the following lemma holds. Lemma 2.1. There is no {0 v1 + 1 v2 + 2 v3 ; 0 v0 + 1 v1 + 2 v2 ; 3; 4}-minihyper for any ordered set (0 ; 1 ; 2 ) in M (3; 4); where M (3; 4) = {(1; 0; 3); (0; 0; 3); (0; 2; 2); (1; 1; 2); (0; 1; 2); (2; 0; 2); (1; 0; 2); (0; 0; 2); (1; 2; 1); (0; 2; 1); (2; 1; 1); (1; 1; 1); (0; 1; 1)}. To prove Theorems 1.1–1.5, we use repeatedly the following two known lemmas. Lemma 2.2. Let F be an {f; a; 4; 4}-minihyper and let H and G be a 3- at in PG(4; 4) and a 2- at in H; respectively. Then 4 P i=1

|F ∩ Hi | = |F| − |F ∩ H | + 4|F ∩ G|;

(2.1)

where H1 ; H2 ; H3 and H4 denote four 3- ats in PG(4; 4); except for H; which contain G.

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(2) Let K be a {k; b; 3; 4}-minihyper and let G and L be a 2- at in PG(3; 4); and a 1- at in G; respectively. Then 4 P i=1

|K ∩ Gi | = |K| − |K ∩ G| + 4|K ∩ L|;

(2.2)

where G1 ; G2 ; G3 and G4 denote four 2- ats in PG(3; 4); except for G; which contain L. Lemma 2.3. (1) If there exists an {f; a; 4; 4}-minihyper F; then     85 85 85 P P P f i ni = v5 ; ini = fv4 and v3 ; ni = 2 2 i=a i=a i=a where ni denotes the number of 3- ats H in PG(4; 4) such that |F ∩ H | = i. (2) If there exists a {k; b; 3; 4}-minihyper K; then     21 21 21 P P P k i mi = v4 ; imi = kv3 and v2 ; mi = 2 2 i=b i=b i=b

(2.3)

(2.4)

where mi denotes the number of 2- ats G in PG(3; 4) such that |K ∩ G| = i. Lemma 2.4. If there exists a {v1 + v2 + 2v3 ; v0 + v1 + 2v2 ; 4; 4}-minihyper F; then |F ∩ H | = 11; 12; 15; 16; 20; 21; 22; 23; 24 or 36 for any 3- at H in PG(4; 4). Proof. Suppose there exists a {v1 + v2 + 2v3 ; v0 + v1 + 2v2 ; 4; 4}-minihyper F. Let H be any 3- at in PG(4; 4). Since v1 + 2v2 6|F ∩ H |6v1 + v2 + 2v3 , it follows from Theorem A.2 ( = 4) that |F ∩ H | = 0 v1 +1 v2 +2 v3 for some ordered set (0 ; 1 ; 2 ) in N1 (3; 4), where N1 (3; 4) = {(1; 2; 0); (2; 2; 0); (0; 3; 0); (1; 3; 0); (0; 4; 0); (0; 0; 1); (1; 0; 1); (2; 0; 1); (3; 0; 1); (0; 1; 1); (1; 1; 1); (2; 1; 1); (0; 2; 1); (1; 2; 1); (0; 3; 1); (0; 0; 2); (1; 0; 2); (2; 0; 2); (0; 1; 2); (1; 1; 2)}. Suppose there exists a 3- at H in PG(4; 4) such that |F ∩ H | = 0 v1 + 1 v2 + 2 v3 for some ordered set (0 ; 1 ; 2 ) in M (3; 4) ∩ N1 (3; 4). It follows from Theorem A.2 ( = 4; q = 4; = 0) that F ∩ H is a {0 v1 + 1 v2 + 2 v3 ; 0 v0 + 1 v1 + 2 v2 ; 4; 4}minihyper in the 3- at H . Since F ∩ H ⊂ H , it follows from Remark A.1 that there exists a {0 v1 + 1 v2 + 2 v3 ; 0 v0 + 1 v1 + 2 v2 ; 3; 4}-minihyper. This is contradictory to Lemma 2.1. Hence |F ∩ H | = v1 + 2v2 ; 2v1 + 2v2 ; 3v2 ; v1 + 3v2 ; 4v2 ; v3 ; v1 + v3 ; 2v1 + v3 ; 3v1 + v3 or 3v2 + v3 . Since v1 = 1; v2 = 5 and v3 = 21; this completes the proof. Lemma 2.5. If K is a {3v2 + v3 ; 3v1 + v2 ; 3; 4}-minihyper; then |K ∩ G| = 8; 12; 16 or 20 for any 2- at G in PG(3; 4) and (m8 ; m12 ; m16 ; m20 ) = (68; 16; 0; 1) or (69; 13; 3; 0). Proof. Let K be a {3v2 + v3 ; 3v1 + v2 ; 3; 4}-minihyper. It follows from Theorem A.2 ( = 4) that there is no 2- at G in PG(3; 4) such that |K ∩ G| = 4v1 + v2 = 9: Hence it follows from Theorem A.2 ( = 4; q = 4; s = 1) that there is no 2- at G such that |K ∩ G| = v1 + 2v2 = 11. Suppose there exists a 2- at G in PG(3; 4) such that |K ∩ G| = 2v2 = 10. It follows from Theorem A.2 ( = 4; q = 4; = 0) that K ∩ G is a {2v2 ; 2v1 ; 3; 4}-minihyper

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in the 2- at G. Since K ∩ G ⊂ G, it follows from Remark A.1 that there exists a {2v2 ; 2v1 ; 2; 4}-minihyper. This is contradictory to Theorem 4.1 (t = 2;  = 1; q = 4) in Hamada (1993a). Hence there is no 2- at G in PG(3; 4) such that 3v1 +v2 ¡|K ∩ G|63 + (3v1 + v2 ). Since 3v1 + v2 6|K ∩ G|6v3 , it follows from Theorem A.2 ( = 4; q = 4; s = 3) that |K ∩ G| = 3v1 + v2 ; 2v1 + 2v2 ; v1 + 3v2 or 4v2 (i.e., |K ∩ G| = 8; 12; 16 or 20) for any 2- at G in PG(3; 4). (A) In the case |K ∩ G| = 4v2 = 20; it follows from |G| = v3 = 21 and Theorem A.2 ( = 4; q = 4; = 0) that K ∩ G = G\{Q} for some point Q in G and K ∩ G is a {4v2 ; 4v1 ; 3; 4}-minihyper in the 2- at G. Hence it follows from |L| = v2 = 5 that |K ∩ L| = |(K ∩ G) ∩ L| = 4 or 5 for any 1- at L in G. Since |(K ∩ G) ∩ L| = 4 if and only if Q ∈ L, there are v2 1- ats L in G such that |K ∩ L| = 4 and there are (v3 −v2 ) 1- ats L in G such that |K ∩ L| = 5. Let Li (16i65) and Lj (66j621) be 1- ats in G such that |K ∩ Li | = 4 and |K ∩ Lj | = 5, respectively. Let Gi1 ; Gi2 ; Gi3 and Gi4 be four 2- ats in PG(3; 4), except for G, which contain Li for i = 1; 2; : : : ; 21, where |K ∩ Gi1 |6|K ∩ Gi2 |6|K ∩ Gi3 |6|K ∩ Gi4 |. Since |K| = 36; |K ∩ G| = 20 and |K ∩ Li | = 4 or 5 according as 16i65 or 66i621, it follows from (2:2) that 4 P j=1

|K ∩ Gij | = |K| − |K ∩ G| + 4|K ∩ Li | = 32 or 36

(2.5)

according as 16i65 or 66i621. Since |K ∩ Gij | = 8; 12; 16 or 20, it follows that (|K ∩ Gi1 |; |K ∩ Gi2 |; |K ∩ Gi3 |; |K ∩ Gi4 |) = (8; 8; 8; 8) or (8; 8; 8; 12) according as 16i 65 or 66i621. Since |K ∩ G| = 20, this implies that m8 = 68, m12 = 16, m16 = 0 and m20 = 1. (B) In the case |K ∩ G| = 8; 12 or 16 for any 2- at PG(3; 4), it follows from Eq. (2.4) that m8 + m12 + m16 = 85; 8m8 + 12m12 + 16m16 = 756 and ( 82 )m8 + ( 12 2 )m12 + ( 16 )m = 3150. Hence m = 69, m = 13, m = 3 and m = 0. This completes the 16 8 12 16 20 2 proof. From Eq. (2.4), Theorem 3.1 (t = 3;  = 1; 1 = 2; t = 3;  = 2; 1 = 0; 2 = 2) and Theorem 5.5 (t = 3; 1 = 2 = 0; 3 = 2) in Hamada (1993a), we have: Lemma 2.6. If K is an {v1 +v3 ; v0 +v2 ; 3; 4}-minihyper for some integer  in {0; 1; 2}; then: (1) K consists of one 2- at and  points in PG(3; 4). (2) In the case  = 0; |K ∩ G| = 5 or 21 for any 2- at G in PG(3; 4) and (m5 ; m21 ) = (84; 1). (3) In the case  = 1; |K ∩ G| = 5; 6 or 21 for any 2- at G in PG(3; 4) and (m5 ; m6 ; m21 ) = (63; 21; 1). (4) In the case  = 2; |K ∩ G| = 5; 6; 7 or 21 for any 2- at G in PG(3; 4) and (m5 ; m6 ; m7 ; m21 ) = (47; 32; 5; 1).

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Lemma 2.7. There is no {v1 + v2 + 2v3 ; v0 + v1 + 2v2 ; 4; 4}-minihyper F such that |F ∩ H | = 21; 22 or 23 for some 3- at H in PG(4; 4). Proof. Suppose there exists a {48; 11; 4; 4}-minihyper F such that |F ∩ H | = 21; 22 or 23 for some 3- at H in PG(4; 4). Then there exists an integer  in {0; 1; 2} such that |F ∩ H | = v3 + . Hence it follows from Theorem A.2 ( = 4, q = 4, = 0) and Lemma 2.6 that F ∩ H is an {v1 + v3 ; v0 + v2 ; 4; 4}-minihyper in H and there exists a 2- at G in H such that G ⊂ F ∩ H , i.e., G ⊂ F. Let H1 ; H2 ; H3 and H4 be four 3- ats in PG(4; 4), except for H , which contain G, where |F ∩ H1 |6|F ∩ H2 |6|F ∩ H3 |6|F ∩ H4 |. Since |F| = 48; |F ∩ G| = 21 and |F ∩ H | = 21; 22 or 23 according as  = 0; 1 or 2, it follows from (2:1) and Lemma 2.4 that |F ∩ H4 | = 3v2 + v3 = 36. Hence it follows from Theorem A.2 ( = 4, q = 4, = 0) and Lemma 2.5 that F ∩ H4 is a {3v2 + v3 ; 3v1 + v2 ; 4; 4}-minihyper in H4 and |F ∩
| 6 21) for any 2- at
in H4 , respec= |(F ∩ H4 ) ∩
| = 8; 12; 16 or 20 (i.e., |F ∩
| = tively. On the other hand, it follows from G ⊂ F ∩ H4 that G is a 2- at in H4 such that |F ∩ G| = |(F ∩ H4 ) ∩ G| = |G| = v3 = 21: This is a contradiction. Lemma 2.8. If K is a {v1 + 2v2 ; v0 + 2v1 ; 3; 4}-minihyper; then |K ∩ G| = 2; 3; 6 or 7 for any 2- at G in PG(3; 4) and (m2 ; m3 ; m6 ; m7 ) = (56; 19; 8; 2). Proof. Let K be a {v1 + 2v2 ; v0 + 2v1 ; 3; 4}-minihyper. It follows from Theorem 5.5 (t = 3, q = 4, 1 = 0, 2 = 3 = 1) in Hamada (1993a) that K = L1 ∪ L2 ∪ {P} for some disjoint 1- ats L1 ; L2 and some point P in PG(3; 4). Since |G ∩ Li | = 1 or 5 for any 2- at G in PG(3; 4) and there is no 2- at G in PG(3; 4) such that L1 ∪ L2 ⊂ G, it follows that |K ∩ G| = |(L1 ∪ L2 ∪ {P}) ∩ G| = 2; 3; 6 or 7 for any 2- at G in PG(3; 4). Since |K ∩ G| = 7 if and only if P ∈ G and Li ⊂ G for i = 1 or 2, we have m7 = 2: Hence it follows from Eq. (2.4) that m2 = 56; m3 = 19 and m6 = 8. Lemma 2.9. If K is a {3v2 ; 3v1 ; 3; 4}-minihyper; then |K ∩ G| = 3; 5; 7 or 15 for any 2- at G in PG(3; 4) and (m3 ; m5 ; m7 ; m15 ) = (60; 24; 0; 1) or (70; 0; 15; 0): Proof. Let K be a {3v2 ; 3v1 ; 3; 4}-minihyper. It follows from Theorem 5.2 (t = 3, q = 4) in Hamada (1993a) that either () K = V \S for some 2- at V in PG(3; 4) and some 6-arc S in V , or ( ) K = L1 ∪ L2 ∪ L3 for some disjoint three 1- ats L1 ; L2 and L3 in PG(3; 4), or ( ) K = {(w0 ); (w1 ); (w2 ); (w3 ); (w0 + w1 ); (w0 + w2 ); (w0 + w3 ); (w1 + w2 ); (w1 + w3 ); (w2 + w3 ); (w0 + w1 + w2 ); (w0 + w1 + w3 ); (w0 + w2 + w3 ); (w1 + w2 + w3 ); (w0 + w1 + w2 + w3 )} for some four linearly independent points (w0 ); (w1 ); (w2 ) and (w3 ) in PG(3; 4). In the case (), it can be shown that |K ∩ G| = 3; 5 or 15 for any 2- at G in PG(3; 4) and (m3 ; m5 ; m7 ; m15 ) = (60; 24; 0; 1). In the case ( ) or ( ), it can be shown that |K ∩ G| = 3 or 7 for any 2- at G in PG(3; 4) and (m3 ; m5 ; m7 ; m15 ) = (70; 0; 15; 0). This completes the proof.

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Proof of Theorem 1.1. Suppose there exists a {v1 +v2 +2v3 ; v0 +v1 +2v2 ; 4; 4}-minihyper F. It follows from the de nition of a minihyper that there exists a 3- at H in PG(4; 4) such that |F ∩ H | = v1 + 2v2 = 11. Hence it follows from Theorem A.2 ( = 4, q = 4, = 0) and Lemma 2.8 that F ∩ H is a {v1 + 2v2 ; v0 + 2v1 ; 4; 4}-minihyper in H and there exists a 2- at G in H such that |F ∩ G| = |(F ∩ H ) ∩ G| = 6. Let H1 ; H2 ; H3 and H4 be four 3- ats in PG(4; 4), except for H , which contain G; where |F ∩ H1 |6|F ∩ H2 |6|F ∩ H3 |6|F ∩ H4 |. Since |F| = 48; |F ∩ H | = 11 and |F ∩ G| = 6; it follows from Lemmas 2.4, 2.7 and (2:1) that (|F ∩ H1 |; |F ∩ H2 |; |F ∩ H3 |; |F ∩ H4 |) = (11; 11; 15; 24); (11; 15; 15; 20) or (15; 15; 15; 16). This implies that |F ∩ H3 | = 3v2 = 15 and |(F ∩ H3 ) ∩ G| = |F ∩ G| = 6. Since F ∩ H3 is a {3v2 ; 3v1 ; 4; 4}minihyper in H3 , it follows from Remark A.1 and Lemma 2.9 that there is no 2- at G in H3 such that |(F ∩ H3 ) ∩ G| = 6. This is a contradiction.

3. The proof of Theorem 1.2 In order to prove Theorem 1.2, we shall use the following three lemmas. Lemma 3.1. If there exists a {2v2 +2v3 ; 2v1 +2v2 ; 4; 4}-minihyper F; then |F ∩ H | = 12; 16; 20; 24 or 36 for any 3- at H in PG(4; 4). Proof. Let F be a {2v2 + 2v3 ; 2v1 + 2v2 ; 4; 4}-minihyper. Since 2v1 + 2v2 = 12; 3v1 + 2v2 = 13; 4v1 + 2v2 = 14 and 3v2 = 15; it follows from (1) and (3) in Theorem A.2 ( = 4, q = 4, s = 2) that there is no 3- at H in PG(4; 4) such that 2v1 +2v2 ¡|F ∩ H |6 3 + (2v1 + 2v2 ). Hence it follows from Theorem A.2 ( = 4, q = 4, s = 3) and 2v1 + 2v2 6|F ∩ H |62v2 +2v3 that |F ∩ H | = 0 v1 +1 v2 +2 v3 for some ordered set (0 ; 1 ; 2 ) in N2 (3; 4), where N2 (3; 4) = {(2; 2; 0); (1; 3; 0); (0; 4; 0); (3; 0; 1); (2; 1; 1); (1; 2; 1); (0; 3; 1); (2; 0; 2); (1; 1; 2); (0; 2; 2)}. Using a method similar to the proof of Lemma 2.4, it can be shown that |F ∩ H | = 2v1 + 2v2 ; v1 + 3v2 ; 4v2 ; 3v1 + v3 or 3v2 + v3 for any 3- at H in PG(4; 4). This completes the proof. Lemma 3.2. There is no {2v2 +2v3 ; 2v1 +2v2 ; 4; 4}-minihyper F such that |F ∩ H | = 36 for some 3- at H in PG(4; 4). Proof. Suppose there exists a {2v2 +2v3 ; 2v1 +2v2 ; 4; 4}-minihyper F such that |F ∩ H | = 3v2 +v3 = 36 for some 3- at H in PG(4; 4). It follows from Theorem A.2 and Lemma 2.5 that F ∩ H is a {3v2 + v3 ; 3v1 + v2 ; 4; 4}-minihyper in H and |F ∩ G| = |(F ∩ H ) ∩ G| = 8; 12; 16 or 20 for any 2- at G in H and (m8 ; m12 ; m16 ; m20 ) = (68; 16; 0; 1) or (69,13, 3,0), where mi denotes the number of 2- ats G in H such that |F ∩ G| = i. Let G be a 2- at in H and let H1 ; H2 ; H3 and H4 be four 3- ats in PG(4; 4), except for H , which contain G; where |F ∩ H1 |6|F ∩ H2 |6|F ∩ H3 |6|F ∩ H4 |. Since 3v1 + v3 = 24 and 4v2 = 20, it follows from Theorems A.2, 1.4 and 1.5 that there is no 2- at G in Hi such that (|F ∩ G|; |F ∩ Hi |) ∈ {(12; 20); (12; 24); (16; 24); (20; 24)}. Hence it

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follows from Lemma 3.1 and (2:1) that (|F ∩ G|; |F ∩ H1 |; |F ∩ H2 |; |F ∩ H3 |; |F ∩ H4 |) = (8; 12; 12; 12; 12); (12; 16; 16; 16; 16); (16; 20; 20; 20; 20) or (20; 20; 20; 20; 36). This implies that n12 = 4m8 ; n16 = 4m12 ; n20 = 4m16 + 3m20 and n36 = m20 + 1. (A) In the case (m8 ; m12 ; m16 ; m20 ) = (68; 16; 0; 1), it follows that n12 = 272; n16 = 64, n20 = 3 and n36 = 2. Since           16 20 36 52 12 n16 + n20 + n36 − v3 = −384; (3.1) n12 + 2 2 2 2 2 this is contradictory to the third equation of (2:3). (B) In the case (m8 ; m12 ; m16 ; m20 ) = (69; 13; 3; 0), it follows that n12 = 276; n16 = 52, n20 = 12 and n36 = 1. Since           16 20 36 52 12 n16 + n20 + n36 − v3 = −480; (3.2) n12 + 2 2 2 2 2 this is a contradiction. Lemma 3.3. There is no {2v2 +2v3 ; 2v1 +2v2 ; 4; 4}-minihyper F such that |F ∩ H | = 24 for some 3- at H in PG(4; 4). Proof. Suppose there exists a {2v2 +2v3 ; 2v1 +2v2 ; 4; 4}-minihyper F such that |F ∩ H | = 3v1 + v3 = 24 for some 3- at H in PG(4; 4). Since F ∩ H is a {3v1 + v3 ; 3v0 + v2 ; 4; 4}minihyper in H , it follows from Theorem 1.4 that there exists a 2- at G in H such that |F ∩ G| = |(F ∩ H ) ∩ G| = 21. Let H1 ; H2 ; H3 and H4 be four 3- ats in PG(4; 4), except for H , which contain G. Since |F| = 52; |F ∩ H | = 24 and |F ∩ G| = 21; it follows from (2:1) that 4 P i=1

|F ∩ Hi | = |F| − |F ∩ H | + 4|F ∩ G| = 112:

(3.3)

Since |F ∩ Hi |624 for i = 1; 2; 3; 4; this is a contradiction. Proof of Theorem 1.2. Suppose there exists a {2v2 + 2v3 ; 2v1 + 2v2 ; 4; 4}-minihyper F. It follows from Lemmas 3.1–3.3 that |F ∩ H | = 12; 16 or 20 for any 3- at H in PG(4; 4). Hence it follows from (2:3) that n16 = −32¡0. This is a contradiction. 4. The proof of Theorem 1.3 Lemma 4.1. If there exists a {v1 +3v3 ; v0 +3v2 ; 4; 4}-minihyper F, then |F ∩H | = 15; 16; 20; 21; 22; 23; 24 or 36 for any 3- at H in PG(4; 4): Proof. Suppose there exists a {v1 +3v3 ; v0 + 3v2 ; 4; 4}-minihyper F. Let H be any 3 at in PG(4; 4). Since 3v2 6|F ∩ H |6v1 +3v3 , it follows from Theorem A.2 ( = 4) that |F ∩H | = 0 v1 +1 v2 +2 v3 for some ordered set (0 ; 1 ; 2 ) in N3 (3; 4); where

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N3 (3; 4) = {(0; 3; 0); (1; 3; 0); (0; 4; 0); (0; 0; 1); (1; 0; 1); (2; 0; 1); (3; 0; 1); (0; 1; 1); (1; 1; 1); (2; 1; 1); (0; 2; 1); (1; 2; 1); (0; 3; 1); (0; 0; 2); (1; 0; 2); (2; 0; 2); (0; 1; 2); (1; 1; 2); (0; 2; 2); (0; 0; 3); (1; 0; 3)}. Hence it follows from Theorem A.2 and Lemma 2.1 that Lemma 4.1 holds. From Theorem A.2 and Lemma 2.6 ( = 0; 1); we have: Lemma 4.2. For any {v1 +3v3 ; v0 +3v2 ; 4; 4}-minihyper F, there is no 3- at H in PG(4; 4) such that (|F ∩ G|; |F ∩H |) = (7; 21) or (7; 22) for some 2- at G in H. Proof of Theorem 1.3. Suppose there exists a {v1 + 3v3 ; v0 + 3v2 ; 4; 4}-minihyper F. It follows from the de nition of a minihyper that there exists a 3- at H in PG(4; 4) such that |F ∩ H | = 3v2 = 15: Hence it follows from Theorem A.2 ( = 4; q = 4; = 0) and Lemma 2.9 that F ∩ H is a {3v2 ; 3v1 ; 4; 4}-minihyper in H and |F ∩G| = |(F ∩H ) ∩G| = 3; 5; 7 or 15 for any 2- at G in H and (m3 ; m5 ; m7 ; m15 ) = (60; 24; 0; 1) or (70; 0; 15; 0): (A) In the case (m3 ; m5 ; m7 ; m15 ) = (60; 24; 0; 1); there exists a 2- at G in H such that |F ∩ G| = 15: Let H1 ; H2 ; H3 and H4 be four 3- ats in PG(4; 4); except for H, which contain G, where |F ∩H1 |6|F ∩ H2 |6|F ∩ H3 |6|F ∩H4 |: Since |F| = 64; |F ∩H | = 15 and |F ∩G| = 15; it follows from (2.1) and Lemma 4.1 that (|F ∩ H1 |; |F ∩ H2 |; |F ∩ H3 |; |F ∩ H4 |) = (15; 22; 36; 36) or (16; 21; 36; 36): Hence |F ∩ H4 | = 3v2 + v3 = 36 and G is a 2- at in H4 such that |(F ∩H4 )∩G| = |F ∩G| = 15. Since F ∩H4 is a {3v2 + v3 ; 3v1 + v2 ; 4; 4}-minihyper in H4 , it follows from Lemma 2.5 that there is no 2- at G in H4 such that |(F ∩H4 )∩G| = 15: This is a contradiction. (B) In the case (m3 ; m5 ; m7 ; m15 )=(70; 0; 15; 0); let Gi (16i670) and Gj (716j6 85) be 2- ats in H such that |F ∩Gi | = 3 and |F ∩ Gj | = 7, respectively. Let Hi1 ; Hi2 ; Hi3 and Hi4 be four 3- ats in PG(4; 4); except for H; which contain Gi for i = 1; 2; : : : ; 85; where |F ∩ Hi1 |6|F ∩ Hi2 |6|F ∩ Hi3 |6|F ∩ Hi4 |: Since |F| = 64 and |F ∩H |= 15; it follows from (2.1) and Lemmas 4.1 and 4.2 that (|F ∩Hi1 |; |F ∩ Hi2 |; |F ∩ Hi3 |; |F ∩ Hi4 |) = (15; 15; 15; 16) for i = 1; 2; : : : ; 70 and (|F ∩Hi1 |; |F ∩ Hi2 |; |F ∩ Hi3 |; |F ∩ Hi4 |) = (15; 15; 23; 24) or (15; 16; 23; 23) for i = 71; 72; : : : ; 85: Since |F ∩H | = 15; this implies that n15 = x + 226; n16 = 85 − x, n23 = 30 − x and n24 = x; where x denotes the number of integers i in {71; 72; : : : ; 85} such that (|F ∩ Hi1 |; |F ∩ Hi2 |; |F ∩ Hi3 |; |F ∩ Hi4 |) = (15; 15; 23; 24): Hence it follows from the third equation of (2.3) that x = 102: Since x615; this is a contradiction. This completes the proof.

5. The proof of Theorem 1.4 From Theorem 4.1 (t =2; q =4;  =1) and Theorem 5.5 (t = 2; q = 4; 1 = 0; 2 = 3 = 1) in Hamada (1993a), we have: Lemma 5.1. There is no {0 v1 +1 v2 ; 0 v0 + 1 v1 ; 2; 4}-minihyper for any ordered set (0 ; 1 ) in {(0; 2); (1; 2)}:

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Lemma 5.2. If K is a {3v1 +v3 ; 3v0 +v2 ; 3; 4}-minihyper, then |K ∩G| = 5; 6; 7; 8; 12; 15; 16; 20 or 21 for any 2- at G in PG(3; 4). Proof. Let K be a {3v1 +v3 ; 3v0 +v2 ; 3; 4}-minihyper and let G be any 2- at in PG(3; 4): Since v2 6|K ∩G|6v3 ; it follows from Theorem A.2 ( = 4) that |K ∩G| = v3 or 0 v1 + 1 v2 for some ordered set (0 ; 1 ) in M1 (2; 4); where M1 (2; 4) = {(0; 1); (1; 1); (2; 1); (3; 1); (0; 2); (1; 2); (2; 2); (0; 3); (1; 3); (0; 4)}: Hence it follows from Theorem A.2 and Lemma 5.1 that Lemma 5.2 holds. Lemma 5.3. If K is a {3v1 +v3 ; 3v0 +v2 ; 3; 4}-minihyper such that |K ∩V | = 21 for some 2- at V in PG(3; 4); then: (1) K = V ∪{P1 ; P2 ; P3 } for some three points P1 ; P2 and P3 in PG(3; 4): (2) |K ∩G| = 5; 6; 7; 8 or 21 for any 2- at G in PG(3; 4). (3) (m5 ; m6 ; m7 ; m8 ; m21 ) = (31; 48; 0; 5; 1) or (35; 36; 12; 1; 1) according as the three points P1 ; P2 and P3 are collinear or not. Proof. Let K be a {3v1 +v3 ; 3v0 +v2 ; 3; 4}-minihyper such that |K ∩V | = 21 for some 2- at V in PG(3; 4). (1) Since |V | = v3 = 21; it follows from |K| = 24 and |K ∩V | = 21 that (1) holds. (2) Since |V ∩ G| = 5 or 21 for any 2- at G in PG(3; 4), it follows from (1) and |G| = 21 that (2) holds. (3) Since m21 = 1 and m8 = 5 or 1 according as the three points P1 ; P2 and P3 are collinear or not, it follows from (2.4) that (3) holds. This completes the proof. Lemma 5.4. There is no {3v1 +v3 ; 3v0 +v2 ; 3; 4}-minihyper K such that |K ∩G| = 20 for some 2- at G in PG(3; 4). Proof. Suppose there exists a {24; 5; 3; 4}-minihyper K such that |K ∩G| = 20 for some 2- at G in PG(3; 4). Since |K ∩ G| = 20; |G| = 21 and |K| = 24; it follows that K ∩ G = G\{Q} for some point Q in G and K = (G\{Q})∪{P1 ; P2 ; P3 ; P4 } for some four points P1 ; P2 ; P3 and P4 in G c : Case 1: P1 ; P2 ; P3 and P4 are collinear. Let L denote the 1- at in PG(3; 4) passing through two points P1 and P2 . Since |G ∩ L| = 1; there exists a point R in G such that G ∩L = {R}; i.e., L = {R; P1 ; P2 ; P3 ; P4 }: (A) In the case R = Q; let M be a 1- at in G passing through Q and let G1 ; G2 ; G3 and G4 be four 2- ats in PG(3; 4); except for G, which contain M, where |K ∩G1 |6|K ∩ G2 |6|K ∩ G3 |6|K ∩G4 |: Since K =(G\{Q}) ∪L and G ∩L= {Q}; it follows that L ⊂ G4 and K ∩ G1 =K ∩M =M \{Q}: This implies that G1 is a 2- at in PG(3; 4) such that |K ∩G1 | = |M \{Q}|=4¡5: Since K is a {24; 5; 3; 4}-minihyper, this is a contradiction. (B) In the case R 6= Q; let M be the 1- at in G passing through Q and R. Let G1 ; G2 ; G3 and G4 be four 2- ats in PG(3; 4), except for G, which contain M, where |K ∩ G1 |6|K ∩ G2 |6|K ∩ G3 |6|K ∩ G4 |: Since L ⊂G4 ; it follows that |K ∩G1 |= |M \{Q}|=4¡5: This is a contradiction.

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Case 2: P1 ; P2 ; P3 and P4 are noncollinear. Without loss of generality, we can assume that P1 ; P2 and P3 are noncollinear. Let
be the 2- at in PG(3; 4) which contain the three points P1 ; P2 and P3 . Since G ∩
is a 1- at (denoted by N ) in G, there are three points Q1 ; Q2 and Q3 in N such that N ∩(P1 ⊕ P2 ) = {Q3 }, N ∩(P1 ⊕ P3 ) = {Q2 } and N ∩(P2 ⊕ P3 ) = {Q1 }, where R1 ⊕R2 denotes the 1- at passing through the two points R1 and R2 : (A) In the case Q ∈ N; let G1 ; G2 and G3 be three 2- ats in PG(3; 4), except for G and
, which contain N, where |K ∩G1 |6|K ∩G2 |6|K ∩G3 |: Since {P1 ; P2 ; P3 } ⊂
and either P4 ∈
or P4 ∈ G3 ; it follows that G1 ∩ {P1 ; P2 ; P3 ; P4 } = ∅: Hence |K ∩ G1 |= |N \{Q}| = 4¡5; a contradiction. (B) In the case Q ∈N; = let G1 ; G2 ; G3 and G4 be four 2- ats in PG(3; 4), except for G, which contain Q ⊕ Q3 , where |K ∩ G1 |6|K ∩G2 |6|K ∩G3 |6|K ∩G4 |: Without loss of generality, we can assume that P3 ∈ G3 and {P1 ; P2 } ⊂ G4 : Since P4 ∈Gi for i = 2; 3 or 4, it follows that G1 ∩ {P1 ; P2 ; P3 ; P4 } = ∅: Hence |K ∩G1 | =4¡5; a contradiction. This completes the proof. Lemma 5.5. There is no {3v1 +v3 ; 3v0 +v2 ; 3; 4}-minihyper K such that |K ∩G| = 12; 15 or 16 for some 2- at G in PG(3; 4): Proof. Suppose there exists a {3v1 +v3 ; 3v0 + v2 ; 3; 4}-minihyper K such that |K ∩ G| = 12; 15 or 16 for some 2- at G in PG(3; 4): (A) In the case |K ∩G| = v1 +3v2 = 16; it follows from Theorem A.2 that K ∩G is a {v1 +3v2 ; v0 +3v1 ; 3; 4}-minihyper in G. Hence there exists a 1- at L in G such that |K ∩ L| = |(K ∩G)∩ L| = 3: Let G1 ; G2 ; G3 and G4 be four 2- ats in PG(3; 4); except for G, which contain L. Since |K| = 24; |K ∩G| = 16 and |K ∩L| = 3; it follows from (2.2) and Lemma 5.2 that (|K ∩ G1 |; |K ∩G2 |; |K ∩G3 |; |K ∩ G4 |) = (5; 5; 5; 5) and |(K ∩ G1 ) ∩ L| = |K ∩ L| = 3: Since |K ∩G1 | = v2 = 5; it follows from Theorem A.2 and (1) of Theorem 3.1 in Hamada (1993a) that K ∩ G1 is a {v2 ; v1 ; 3; 4}-minihyper in G1 and K ∩ G1 is a 1- at in G1 : Hence |(K ∩ G1 )∩ L| = 1 or 5. This is a contradiction. (B) In the case |K ∩G| = 3v2 = 15; it follows from Theorem A.2 that K ∩ G is a {3v2 ; 3v1 ; 3; 4}-minihyper in G. Hence there exists a 1- at L in G such that |K ∩L| = |(K ∩ G) ∩ L| = 3: Using a method similar to (A), we have a contradiction. (C) In the case |K ∩ G|=2v1 +2v2 =12; it follows that K ∩G is a {2v1 + 2v2 ; 2v0 + 2v1 ; 3; 4}-minihyper in G and there exists a 1- at L in G such that |K ∩ L| = |(K ∩G) ∩ L| = 2: Using a method similar to (A), we have a contradiction. Lemma 5.6. There is no {3v1 +v3 ; 3v0 +v2 ; 3; 4}-minihyper K such that |K ∩G| = 5; 6; 7 or 8 for any 2- at G in PG(3; 4): Proof. Suppose there exists a {3v1 +v3 ; 3v0 +v2 ; 3; 4}-minihyper K such that |K ∩G| = 5; 6; 7 or 8 for any 2- at G in PG(3; 4). It follows from the de nition of a minihyper that there exists a 2- at
in PG(3; 4) such that |K ∩
| = v2 = 5: Since K ∩
is a {v2 ; v1 ; 3; 4}-minihyper in
; K ∩
is a 1- at (denoted by L) in
: Let G1 ; G2 ; G3

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and G4 be four 2- ats in PG(3; 4), except for
; which contain L. Since |K| = 24; P4 |K ∩
| = 5 and |K ∩ L| = |L| = 5, it follows from (2.2) that i=1 |K ∩ Gi | = 39: Since P4 |K ∩Gi |68 for i = 1; 2; 3; 4; it follows that i=1 |K ∩ Gi |632: This is a contradiction. Proof of Theorem 1.4. If K = V ∪{P1 ; P2 ; P3 } for some 2- at V and some three points P1 ; P2 and P3 in PG(3; 4), then |K| = v3 +3v1 and |K ∩G|¿|V ∩G|¿v2 for any 2- at G in PG(3; 4) and |K ∩G| = v2 for some 2- at G in PG(3; 4). This implies that K is a {3v1 +v3 ; 3v0 +v2 ; 3; 4}-minihyper. Hence it follows from Lemmas 5.2–5.5 that Theorem 1.4 holds. This completes the proof.

6. The proof of Theorem 1.5 Using a method similar to the proof of Lemma 5.2, we have: Lemma 6.1. If K is a {4v2 ; 4v1 ; 3; 4}-minihyper, then |K ∩G| = 4; 5; 6; 7; 8; 12; 15; 16 or 20 for any 2- at G in PG(3; 4): Lemma 6.2. If K is a {4v2 ; 4v1 ; 3; 4}-minihyper such that |K ∩V | = 20 for some 2- at V in PG(3; 4); then: (1) K = V \{Q} for some point Q in V. (2) |K ∩G| = 4; 5 or 20 for any 2- at G in PG(3; 4): (3) (m4 ; m5 ; m20 ) = (20; 64; 1): Proof. Let K be a {4v2 ; 4v1 ; 3; 4}-minihyper such that |K ∩V | = 20 for some 2- at V in PG(3; 4): (1) Since |K| = 20; |K ∩V | = 20 and |V | = 21; there exists a point Q in V such that K = V \{Q}: (2) Since V ∩G is a 1- at in V for any 2- at G in PG(3; 4) such that G 6= V; it follows that |K ∩G| = 4 or 5 according as Q ∈ G or not in the case G 6= V: Hence |K ∩G| = 4; 5 or 20 for any 2- at G in PG(3; 4). (3) It follows from (2) and (2.4) that (m4 ; m5 ; m20 ) = (20; 64; 1): Lemma 6.3. There is no {4v2 ; 4v1 ; 3; 4}-minihyper K such that |K ∩ G| = 15 for some 2- at G in PG(3; 4): Proof. Suppose there exists a {4v2 ; 4v1 ; 3; 4}-minihyper K such that |K ∩G| = 3v2 = 15 for some 2- at G in PG(3; 4): Since K ∩ G is a {3v2 ; 3v1 ; 3; 4}-minihyper in G, there exists a 1- at L in G such that |K ∩ L| = |(K ∩G) ∩L| = 3: Let G1 ; G2 ; G3 and G4 be four 2- ats in PG(3; 4), except for G, which contain L, where |K ∩G1 |6|K ∩G2 |6|K ∩G3 | 6|K ∩G4 |: Since |K| = 20; |K ∩ G| = 15 and |K ∩L| = 3; it follows from (2.2) and Lemma 6.1 that (|K ∩G1 |; |K ∩ G2 |; |K ∩ G3 |; |K ∩G4 |) = (4; 4; 4; 5): Since K ∩G4 is a {v2 ; v1 ; 3; 4}-minihyper in G4 ; K ∩ G4 is a 1- at in G4 : Hence |K ∩L| = |(K ∩G4 )∩ L| = 1 or 5. This is contradictory to |K ∩ L| = 3:

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From Theorem 5.4 (t = 2; q = 4;  = 0) in Hamada (1993a), we have: Lemma 6.4. If J is a {2v1 +v2 ; 2v0 +v1 ; 2; 4}-minihyper, then (1) either () J = M ∪ {P1 ; P2 } for some 1- at M and some points P1 and P2 in PG(2; 4) or ( ) J = {(w0 ); (w1 ); (w2 ); (w0 + w1 ); (w0 + w2 ); (w1 + w2 ); (w0 + w1 + w2 )} for some three linearly independent points (w0 ); (w1 ) and (w2 ) in PG(2; 4): (2) in the case (); |J ∩ L| = 1; 2; 3 or 5 for any 1- at L in PG(2,4) and (l1 ; l2 ; l3 ; l5 ) = (11; 8; 1; 1): (3) in the case ( ); |J ∩L| = 1 or 3 for any 1- at L in PG(2; 4) and (l1 ; l3 ) = (14; 7): Lemma 6.5. For any {4v2 ; 4v1 ; 3; 4}-minihyper K, there is no 2- at G in PG(3; 4) such that (|K ∩ L|; |K ∩ G|) = (2; 5); (3; 5); (4; 5); (3; 6); (4; 6) or (4,7) for some 1- at L in G. Proof. Suppose there exists a 2- at G in PG(3; 4) which satis es the condition in Lemma 6.5. (A) In the case |K ∩ G| = v2 = 5, it follows that K ∩ G is a 1- at (denoted by M ) in G. Since |M ∩ L| = 1 or 5, it follows from L ⊂ G that |K ∩ L| = |(K ∩ G) ∩ L| = |M ∩ L| = 1 or 5. This is contradictory to |K ∩ L| = 2; 3 or 4. (B) In the case |K ∩ G| = v1 +v2 = 6, it follows that K ∩ G = M ∪ {P} for some 1- at M and some point P in G. Hence |K ∩ L| = |(K ∩ G) ∩ L| = 1; 2 or 5, a contradiction. (C) In the case |K ∩ G| = 2v1 + v2 = 7, it follows from Theorem A.2 and Lemma 6.4 that |K ∩ L| = |(K ∩ G) ∩ L| = 1; 2; 3 or 5, a contradiction. This completes the proof. Lemma 6.6. If K is a {4v2 ; 4v1 ; 3; 4}-minihyper such that |K ∩ V | = 16 for some 2 at V in PG(3; 4); then |K ∩ G| = 4; 6; 8 or 16 for any 2- at G in PG(3; 4) and (m4 ; m6 ; m8 ; m16 ) = (55; 24; 5; 1). Proof. Let K be a {4v2 ; 4v1 ; 3; 4}-minihyper such that |K ∩ V | = v1 +3v2 = 16 for some 2- at V in PG(3; 4). Since K ∩ V is a {v1 +3v2 ; v0 +3v1 ; 3; 4}-minihyper in V , it follows that |K ∩ L| = |(K ∩ V ) ∩ L| = 3; 4 or 5 for any 1- at L in V and (l3 ; l4 ; l5 ) = (10; 5; 6). Let L be a 1- at in V and let G1 ; G2 ; G3 and G4 be four 2- ats in PG(3; 4), except for V , which contain L, where |K ∩ G1 | 6 |K ∩ G2 | 6 |K ∩ G3 | 6 |K ∩ G4 |. Since 4 X

|K ∩ Gi | = |K| − |K ∩ V | + 4|K ∩ L| = 16; 20 or 24

(6.1)

i=1

according as |K ∩ L| = 3; 4 or 5, it follows from Lemmas 6.1 and 6.5 that (|K ∩ L|; |K ∩ G1 |; |K ∩ G2 |; |K ∩ G3 |; |K ∩ G4 |) = (3; 4; 4; 4; 4), (4; 4; 4; 4; 8), (5; 5; 5; 6; 8), (5; 5; 5; 7; 7), (5; 5; 6; 6; 7) or (5; 6; 6; 6; 6). Since |K ∩ V | = 16, it follows that m4 = 4l3 +3l4 =55, m5 = 2x1 +2x2 +x3 , m6 = x1 +2x3 +4x4 , m7 = 2x2 +x3 , m8 = l4 +x1 , m16 = 1 and x1 +x2 + x3 + x4 = l5 = 6, where x1 ; x2 ; x3 and x4 denote the number of 1- ats L in V such that |K ∩ L| = 5 and (|K ∩ G1 |; |K ∩ G2 |; |K ∩ G3 |; |K ∩ G4 |) = (5; 5; 6; 8), (5; 5; 7; 7; ), (5; 6; 6; 7) and (6; 6; 6; 6), respectively. Hence it follows from the third equation of (2:4) and

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317

x1 + x2 + x3 + x4 = 6 that 590 + 63x1 + 62x2 + 61x3 + 60x4 = 950 and 3x1 + 2x2 + x3 = 0. Since xi ¿0 for i = 1; 2; 3, this implies that x1 = x2 = x3 = 0, x4 = 6, m4 = 55, m5 = 0, m6 = 24, m7 = 0, m8 = 5 and m16 = 1. This completes the proof. Remark 6.1. Let V be the 2- at in PG(3; 4) generated by three points (1; 0; 0; 0), (0; 1; 0; 0) and (0; 0; 1; 0) in S(4; 4). Let S = {(0; 1; 0; 0), (0; 0; 1; 0), (1; 1; 1; 0), (1; ;  2 ; 0), (1;  2 ; ; 0)} and let K = (V \S) ∪ L for a 1- at L in PG(3; 4) such that V ∩ L = {(1; 0; 0; 0)}, where  is a primitive element in GF(2 2 ) such that  2 +  + 1 = 0. Then K is a {4v2 ; 4v1 ; 3; 4}-minihyper such that |K ∩ V | = 16 for the 2- at V in PG(3; 4). Lemma 6.7. If K is a {4v2 ; 4v1 ; 3; 4}-minihyper such that |K ∩ G| = 4; 5; 6; 7 or 8 for any 2- at G in PG(3; 4), then |K ∩
| = 4 or 8 for any 2- at
in PG(3; 4) and (m4 ; m8 ) = (65; 20). Proof. Let K be a {4v2 ; 4v1 ; 3; 4}-minihyper which satis es the condition in Lemma 6.9. Suppose there exists a 2- at
in PG(3; 4) such that |K ∩
| = 5 or 6. It follows that (i) in the case |K ∩
| = 5, K ∩
= L for some 1- at L in
and (ii) in the case |K ∩
| = 6, K ∩
= L ∪ {P} for some 1- at L and some point P in
. Let G1 ; G2 ; G3 and G4 be four 2- ats in PG(3; 4), except for
, which contain L. Then 4 P i=1

|K ∩ Gi | = |K| − |K ∩
| + 4|L| = 35 or 34

(6.2)

according as |K ∩
| = 5 or 6. Since |K ∩ Gi |68 for i = 1; 2; 3; 4, this is a contradiction. Hence |K ∩
| = 4; 7 or 8 for any 2- at
in PG(3; 4). Since m4 +m7 +m8 = v4 , 4m4 +7m7 +8m8 = 20v3 and ( 42 )m4 +( 72 )m7 +( 82 )m8 = ( 20 2 )v2 , it follows that m4 = 65, m7 = 0 and m8 = 20. This completes the proof. Remark 6.2. Let L1 , L2 , L3 and L4 be four disjoint 1- ats in PG(3; 4) and let K = L1 ∪ L2 ∪ L3 ∪ L4 . Then K is a {4v2 ; 4v1 ; 3; 4}-minihyper such that |K ∩ G| = 4 or 8 for any 2- at G in PG(3; 4). Lemma 6.8. If J is a {2v1 + 2v2 ; 2v0 + 2v1 ; 2; 4}-minihyper, then |J ∩ L| = 2; 3; 4 or 5 for any 1- at L in PG(2; 4) and (l2 ; l3 ; l4 ; l5 ) = (12; 0; 9; 0), (11; 3; 6; 1), (10; 6; 3; 2) or (9; 9; 0; 3), where li denotes the number of 1- ats in PG(2; 4) such that |J ∩ L| = i. Proof. Let J be a {2v1 +2v2 ; 2v0 +2v1 ; 2; 4}-minihyper. It is obvious that |J ∩ L| = 2; 3; 4 or 5 for any 1- at L in PG(2; 4). Since l2 + l3 + l4 + l5 = v3 , 2l2 + 3l3 + 4l4 + 5l5 = 12v2 and ( 22 )l2 + ( 32 )l3 + ( 42 )l4 + ( 52 )l5 = ( 12 2 )v1 , it follows that l2 = 12 − l5 , l3 = 3l5 and l4 = 3(3 − l5 ). This implies that Lemma 6.8 holds. Lemma 6.9. There is no {4v2 ; 4v1 ; 3; 4}-minihyper K such that |K ∩ G| = 12 for some 2- at G in PG(3; 4).

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Proof. Suppose there exists a {4v2 ; 4v1 ; 3; 4}-minihyper K such that |K ∩ G| = 2v1 + 2v2 = 12 for some 2- at G in PG(3; 4). Since K ∩ G is a {2v1 + 2v2 ; 2v0 + 2v1 ; 3; 4}minihyper in G, it follows from Remark A.1 and Lemma 6.8 that |K ∩ L| = |(K ∩ G) ∩ L| = 2; 3; 4 or 5 for any 1- at L in G and (l2 ; l3 ; l4 ; l5 ) = (12; 0; 9; 0), (11; 3; 6; 1), (10; 6; 3; 2) or (9; 9; 0; 3). Let L be a 1- at in G and let G1 , G2 , G3 and G4 be four 2- ats in PG(3; 4), except for G, which contain L, where |K ∩ G1 |6|K ∩ G2 |6|K ∩ G3 |6|K ∩ G4 |. Since |K| = 20 and |K ∩ G| = 12, it follows from (2:2) and Lemmas 6.1 and 6.5 that (|K ∩ L|; |K ∩ G1 |; |K ∩ G2 |; |K ∩ G3 |; |K ∩ G4 |) = (2; 4; 4; 4; 4), (3; 4; 4; 4; 8), (4; 4; 4; 4; 12), (4; 4; 4; 8; 8), (5; 5; 5; 6; 12), (5; 5; 7; 8; 8), (5; 6; 6; 8; 8), (5; 6; 7; 7; 8) or (5; 7; 7; 7; 7). Note that (|K ∩ G1 |; |K ∩ G2 |; |K ∩ G3 |; |K ∩ G4 |) = (4; 4; 4; 4)

(6.3)

for any 1- at L in G such that |K ∩ L| = 2 and (|K ∩ G1 |; |K ∩ G2 |; |K ∩ G3 |; |K ∩ G4 |) = (4; 4; 4; 8)

(6.4)

for any 1- at L in G such that |K ∩ L| = 3. Let y denote the number of 1- ats L in G such that |K ∩ L| = 4 and (|K ∩ G1 |; |K ∩ G2 |; |K ∩ G3 |; |K ∩ G4 |) = (4; 4; 4; 12). Let x1 , x2 , x3 and x4 denote the number of 1- ats L in G such that |K ∩ L| = 5 and (|K ∩ G1 |; |K ∩ G2 |; |K ∩ G3 |; |K ∩ G4 |) = (5; 5; 6; 12), (5; 7; 8; 8), (6; 6; 8; 8), (6; 7; 7; 8) and (7; 7; 7; 7), respectively. Since |K ∩ G| = 12, it follows that m4 = 4l2 +3l3 +2l4 +y, m5 = 2x1 + x2 , m6 = x1 + 2x3 + x4 , m7 = x2 + 2x4 + 4x5 , m8 = l3 + 2l4 + 2x2 + 2x3 + x4 − 2y, m12 = y + x1 + 1 and x1 + x2 + x3 + x4 + x5 = l5 :

(6.5)

Hence it follows from the third equation of (2:4) that 24l2 + 46l3 + 68l4 + 16y + 101x1 + 87x2 + 86x3 + 85x4 + 84x5 = 884:

(6.6)

Case 1: (l2 ; l3 ; l4 ; l5 ) = (12; 0; 9; 0). It follows from Eqs. (6.5) and (6.6) that x1 = x2 = x3 = x4 = x5 = 0 and y = −1. Since y¿0, this is a contradiction. Case 2: (l2 ; l3 ; l4 ; l5 ) = (11; 3; 6; 1). It follows from (6:6)−(6:5)×84 that 16y+17x1 + 3x2 + 2x3 + x4 = −10. Since y¿0 and xi ¿0 for i = 1; 2; 3; 4, this is a contradiction. Case 3: (l2 ; l3 ; l4 ; l5 ) = (10; 6; 3; 2). It follows from (6:6) − (6:5) × 84 that 16y + 17x1 + 3x2 + 2x3 + x4 = −4¡0, a contradiction. Case 4: (l2 ; l3 ; l4 ; l5 ) = (9; 9; 0; 3). It follows from (6:6)−(6:5)×84 that 16y +17x1 + 3x2 + 2x3 + x4 = 2. Hence y = x1 = x2 = 0 and (x3 ; x4 ; x5 ) = (1; 0; 2) or (0; 2; 1). Since x5 ¿1, there exists a 1- at L in G such that |K ∩ L| = 5 and (|K ∩ G1 |; |K ∩ G2 |; |K ∩ G3 |; |K ∩ G4 |) = (7; 7; 7; 7). Hence there exist two points Pi1 and Pi2 in Gi \L such that K ∩ Gi = L ∪ {Pi1 ; Pi2 } for i = 1; 2; 3; 4. Since |L ∩ (Pi1 ⊕ Pi2 )| = 1 for i = 1; 2; 3; 4, there exists an integer l in {1; 2; 3; 4} such that L ∩ (Pi1 ⊕ Pi2 ) = {Rl } for each integer i, where L = {R1 ; R2 ; R3 ; R4 ; R5 }. Let Lj1 , Lj2 , Lj3 and Lj4 be four 1- ats in G, except for L, which contain Rj for j = 1; 2; 3; 4; 5. Since (l2 ; l3 ; l4 ; l5 ) = (9; 9; 0; 3) and |K ∩ L| = 5, we can assume without

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319

loss of generality that (|K ∩ Lj1 |; |K ∩ Lj2 |; |K ∩ Lj3 |; |K ∩ Lj4 |) = (2; 2; 2; 5) for j = 1; 2 and (|K ∩ Lj1 |; |K ∩ Lj2 |; |K ∩ Lj3 |; |K ∩ Lj4 |) = (2; 3; 3; 3) for j = 3; 4; 5. Note that K = (K ∩ G) ∪ {P11 ; P12 ; P21 ; P22 ; P31 ; P32 ; P41 ; P42 }. (A) In the case L ∩ (Pi1 ⊕ Pi2 ) = {R1 } for some integer i in {1; 2; 3; 4}, we can assume without loss of generality that L ∩ (P11 ⊕ P12 ) = {R1 }. Let
k be the 2- at in PG(3; 4) which contain two 1- ats L1k and P11 ⊕ P12 for k = 1; 2; 3; 4. Since (|K ∩ L11 |; |K ∩ L12 |; |K ∩ L13 |; |K ∩ L14 |) = (2; 2; 2; 5) and {P11 ; P12 } ⊂
k for k = 1; 2; 3; 4, it follows that (|K ∩
1 |; |K ∩
2 |; |K ∩
3 |; |K ∩
4 |) = (4+b1 ; 4+b2 ; 4+b3 ; 7+b4 ), where bk denotes the number of points in {P21 ; P22 ; P31 ; P32 ; P41 ; P42 } which are contained in
k . Since L11 , L12 and L13 are 1- ats in G such that |K ∩ L11 | = |K ∩ L12 | = |K ∩ L13 | = 2, it follows from (6:3) that b1 = b2 = b3 = 0. Since b1 + b2 + b3 + b4 = 6, this implies that b4 = 6 and |K ∩
4 | = 13, a contradiction. (B) In the case L ∩ (Pi1 ⊕ Pi2 ) = {R2 } for some integer i in {1; 2; 3; 4}, we have a contradiction using a method similar to (A). (C) In the case L ∩ (Pi1 ⊕ Pi2 ) = {R3 } for some integer i in {1; 2; 3; 4}, we can assume without loss of generality that L ∩ (P11 ⊕ P12 ) = {R3 }. Let
k be the 2- at in PG(3; 4) which contain two 1- ats L3k and P11 ⊕ P12 for k = 1; 2; 3; 4. Since (|K ∩ L31 |; |K ∩ L32 |; |K ∩ L33 |; |K ∩ L34 |) = (2; 3; 3; 3) and {P11 ; P12 } ⊂
k , it follows that (|K ∩
1 |; |K ∩
2 |; |K ∩
3 |; |K ∩
4 |) = (4 + b1 ; 5 + b2 ; 5 + b3 ; 5 + b4 ). Hence it follows from (6:3) and (6:4) that b1 = 0 and b2 = b3 = b4 = 3. Since b1 + b2 + b3 + b4 = 6, this is a contradiction. (D) In the case L ∩ (Pi1 ⊕ Pi2 ) = {R4 } or {R5 } for some integer i in {1; 2; 3; 4}, we have a contradiction using a method similar to (C). It follows from (A)–(D) that there is no {4v2 ; 4v1 ; 3; 4}-minihyper K such that |K ∩ G| = 12 for some 2- at G in PG(3; 4). This completes the proof. Proof of Theorem 1.5. It follows from Lemmas 6.1, 6.2, 6.3, 6.6, 6.7 and 6.9 that Theorem 1.5 holds. This completes the proof.

Acknowledgements This research was partially supported by Grant-in-Aid for Scienti c Research of the Ministry of Education, Science and Culture under Contract Numbers 304-350807640326 and 304-5009-09640286.

Appendix A. Connections between minihypers and codes Let S(k; q) be the set of column vectors c, cT = (c1 ; c2 ; : : : ; ck ), in W (k; q) such that either c1 = 1 or c1 = c2 = · · · = ci−1 = 0, ci = 1 for some integer i in {2; 3; : : : ; k}, where k¿3 and W (k; q) denotes a k-dimensional vector space consisting of column vectors

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over GF(q). Then S(k; q) consists of all (q k − 1)=(q − 1) projectively distinct nonzero vectors in W (k; q) which may be regarded as (q k − 1)=(q − 1) points in PG(k − 1; q). Theorem A.1 (Hamada, 1987, 1993a). Let F be a set of f vectors in S(k; q) and let C be the subspace of V (n; q) generated by a k × n matrix (denoted by G) whose column vectors are all the vectors in S(k; q)\F, where n = vk − f, 16f¡vk − 1 and vi = (q i − 1)=(q − 1) for any integer i¿0. (1) Let Hz = {y ∈ S(k; q) | z · y = 0 over GF(q)} for a nonzero vector z in W (k; q). Then Hz is a hyperplane in PG(k − 1; q) and the weight of the code vector zT G in C is equal to |F ∩ Hz | − vk−1 + n, where zT denotes the transpose of the vector z. (2) In the case k¿3 and 16d¡q k−1 , C is an [n; k; d; q]-code meeting the Griesmer bound if and only if F is a {vk − n; vk−1 − n + d; k − 1; q}-minihyper. De nition A.1. Two [n; k; d; q]-codes C1 and C2 are side to be equivalent if there exists a k × n generator matrix G2 of the code C2 such that G2 = G1 PD (or G2 = G1 DP) for some permutation matrix P and some nonsingular diagonal matrix D with entries from GF(q), where G1 is a k × n generator matrix of C1 . De nition A.2. (1) Let E(t; q) denote the set of all ordered sets (0 ; 1 ; : : : ; t−1 ) of integers i such that 06i 6q − 1 and (0 ; 1 ; : : : ; t−1 ) 6= (0; 0; : : : ; 0).  q) denote the set of all ordered sets (0 ; 1 ; : : : ; t−1 ) of integers i such (2) Let E(t; that (a) (0 ; 1 ; : : : ; t−1 ) ∈ E(t; q) or (b) 0 = q, 061 6q − 1; : : : ; 06t−1 6q − 1, or (c) 0 = 1 = · · · = −1 = 0,  = q, 06+1 6q−1; : : : ; 06t−1 6q−1 for some integer  in {1; 2; : : : ; t − 1}. P k−2 P k−2 Corollary A.1. In the case d = q k−1 − i=0 i qi and n = vk − i=0 i vi+1 for some ordered set (0 ; 1 ; : : : ; k−2 ) in E(k − 1; q); there is a one-to-one correspondence between the set of all nonequivalent [n; k; d; q]-codes meeting the Griesmer bound and P k−2 P k−2 the set of all { i=0 i vi+1 ; i=0 i vi ; k − 1; q}-minihypers. P k−2 P k−2 Corollary A.2. In the case d = q k−1 − i=0 i qi and n = 1 + vk − i=0 i vi+1 for some ordered set (0 ; 1 ; : : : ; k−2 ) in E(k − 1; q) such that 0 = 1 = · · · =  = 0 and +1 ¿1 for some integer  in {0; 1; : : : ; k − 3}; there is a one-to-one correspondence between the set of all nonequivalent [n; k; d; q]-codes with B2 = 0 and the set of all P k−2 P k−2 { i=0 i vi+1 ; i=0 i vi ; k − 1; q}-minihypers; where 0 = 1 = · · · = −1 = 0;  = q; +1 = +1 − 1; i = i for i =  + 2;  + 3; : : : ; k − 2 and B2 denotes the number of codewords of weight 2 in its dual code. The following theorem is due to Hamada and Helleseth (1994, 1996). Theorem A.2 plays an important role in characterizing minihypers. Pt−1 Pt−1 Theorem A.2. If there exists a { i=0 i vi+1 ; i=0 i vi ; t; q}-minihyper F for some ordered set (0 ; 1 ; : : : ; t−1 ) in E(t; q) and H is a (t − 1)- at in PG(t; q) such that

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321

Pt−1 |F ∩ H | = i=0 i vi+1 for some ordered set (0 ; 1 ; : : : ; t−1 ) in E(t; q); then: Pt−1 Pt−1 (1) i=0 i 6; where  = i=0 i . Pt−1 (2) In the case  − ( + 1)q¡ i=0 i 6 − q for some integer ¿0; F ∩ H is a Pt−1 Pt−1 { i=0 i vi+1 ; i=0 i vi − ; t; q}-minihyper in H for some integer in {0; 1; : : : ; }. Pt−1  v ¡|F ∩ |6s + (3) If there is no (t − 1)- at  in PG(t; q) such that i=1 Pt−1 Pt−1 Pt−1i i  v for some positive integer s¡q; then  =  or i i i i=1 i=0 i=0 i ¡ − s: Remark A.1. There exists an {f; m; t; q}-minihyper F such that F ⊂ H for some (t−1) at H in PG(t; q) if and only if there exists an {f; m; t − 1; q}-minihyper, where 06m¡f¡vt . Remark A.2. Using minihypers, Problem 1.2 has been solved for many cases by Hamada (1987, 1993a, 1993b, 1996), Hamada and Deza (1991), Hamada and Helleseth (1990, 1992, 1994, 1996), and Hamada et al. (1992, 1993). Appendix B. Tables of the values of g4 (4; d) and n4 (4; d) for 16d643 P2 P2 Let d = 43 − i=0 i 4i ; g = g4 (4; d) = v4 − i=0 i vi+1 and  = (0 ; 1 ; 2 ), where i = 0; 1; 2 or 3 for i = 0; 1; 2. Let n = n4 (4; d) denote the smallest length of codes of dimension 4 and minimum distance d over the Galois eld GF(4). Recently it has been shown by Landgev et al. (1996) that n4 (4; 37) = 52. Hence we have the following table from Hamada (1996). Table 1 The values of g4 (4; d) and n4 (4; d) for 16d664 d



g

n

d



g

n

d



g

n

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22

333 233 133 033 323 223 123 023 313 213 113 013 303 203 103 003 332 232 132 032 322 222

4 5 6 7 9 10 11 12 14 15 16 17 19 20 21 22 25 26 27 28 30 31

4 5 7 8 9 10 12 13 14 15 16 17 20 21 22 23 25 26 27 28 30 31

23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44

122 022 312 212 112 012 302 202 102 002 331 231 131 031 321 221 121 021 311 211 111 011

32 33 35 36 37 38 40 41 42 43 46 47 48 49 51 52 53 54 56 57 58 59

33 34 36 37 38 39 41 42 43 44 46 47 48 49 52 53 54 55 57 58 59 60

45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64

301 201 101 001 330 230 130 030 320 220 120 020 310 210 110 010 300 200 100 000

61 62 63 64 67 68 69 70 72 73 74 75 77 78 79 80 82 83 84 85

61 62 63 64 67 68 69 70 72 73 74 75 77 78 79 80 82 83 84 85

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