New nonbinary code bounds based on divisibility arguments

For \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$q,n,d \in \mathbb {N}$$\end{document}q,n,d∈N, let \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_q(n,d)$$\end{document}Aq(n,d) be the maximum size of a code \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C \subseteq [q]^n$$\end{document}C⊆[q]n with minimum distance at least d. We give a divisibility argument resulting in the new upper bounds \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_5(8,6) \le 65$$\end{document}A5(8,6)≤65, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_4(11,8)\le 60$$\end{document}A4(11,8)≤60 and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_3(16,11) \le 29$$\end{document}A3(16,11)≤29. These in turn imply the new upper bounds \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_5(9,6) \le 325$$\end{document}A5(9,6)≤325, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_5(10,6) \le 1625$$\end{document}A5(10,6)≤1625, \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_5(11,6) \le 8125$$\end{document}A5(11,6)≤8125 and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_4(12,8) \le 240$$\end{document}A4(12,8)≤240. Furthermore, we prove that for \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu ,q \in \mathbb {N}$$\end{document}μ,q∈N, there is a 1–1-correspondence between symmetric \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(\mu ,q)$$\end{document}(μ,q)-nets (which are certain designs) and codes \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$C \subseteq [q]^{\mu q}$$\end{document}C⊆[q]μq of size \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu q^2$$\end{document}μq2 with minimum distance at least \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mu q - \mu $$\end{document}μq-μ. We derive the new upper bounds \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_4(9,6) \le 120$$\end{document}A4(9,6)≤120 and \documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$A_4(10,6) \le 480$$\end{document}A4(10,6)≤480 from these ‘symmetric net’ codes.


Introduction
For any m ∈ N, we write [m] := {1, . . ., m}.Fix n, q ∈ N. A word is an element v ∈ [q] n .So [q] serves as the alphabet.(If you prefer {0, 1, . . ., q − 1} as alphabet, take the letters mod q.)For two words u, v ∈ [q] n , their (Hamming) distance d H (u, v) is the number of indices i with u i = v i .A code is a subset of [q] n .For any code C ⊆ [q] n , the minimum distance d min (C) of C is the minimum distance between any two distinct code words in C. For d ∈ N, an (n, d) q -code is a set C ⊆ [q] n that satisfies d min (C) ≥ d.Define A q (n, d) := max{|C| | C is an (n, d) q -code}. (1) Computing A q (n, d) and finding upper and lower bounds for it is a long-standing research interest in combinatorial coding theory (cf.MacWilliams and Sloane [12]).In this paper we find new upper bounds on A q (n, d) (for some q, n, d), based on a divisibility-argument.In some cases, it will sharpen a combination of the following two well-known upper bounds on A q (n, d).
Fix q, n, d ∈ N. Then qd > (q − 1)n =⇒ A q (n, d) ≤ qd qd − n(q − 1) . ( This is the q-ary Plotkin bound.Moreover, A proof of these statements can be found in [12].Plotkin's bound can be proved by comparing the leftmost and rightmost terms in (4) below.The second bound follows from the observation that in a (n, d) q -code any symbol can occur at most A q (n − 1, d) times at the first position.
We view an (n, d) q -code C of size M as an M × n matrix with the words as rows.Two codes C, D ⊆ [q] n are equivalent (or isomorphic) if D can be obtained from C by first permuting the n columns of C and subsequently applying to each column a permutation of the q symbols in [q] (we will write 'renumbering a column' instead of 'applying a permutation to the symbols in a column').
If an (n, d) q -code C is given, then for j = 1, . . ., n, let c α,j denote the number of times symbol α ∈ [q] appears in column j of C. For any two words u, v ∈ [q] n , we define g(u, v) := n − d H (u, v).In our divisibility arguments, we will use the following observations (which are well known and often used in coding theory and combinatorics).
where m := M/q and r := qm − M , so that M = qm − r and 0 ≤ r < q.Moreover, writing L and R for the leftmost term and the rightmost term in (4), respectively, we have i.e., the number of pairs of distinct words {u, v} ⊆ C with distance unequal to d is at most the leftmost term minus the rightmost term in (4).
Proof.The first inequality in (4) holds because n − d ≥ g(u, v) for all u, v ∈ C. The equality is obtained by counting the number of equal pairs of entries in the same columns of C in two ways.The second inequality follows from the (strict) convexity of the binomial coefficient F (x) := x(x − 1)/2.Fixing a column j, the quantity α∈[q] F (c α,j ), under the condition that α∈[q] c α,j = M , is minimal if the c α,j are as equally divided as possible, i.e., if c α,j ∈ { M/q , M/q } for all α ∈ [q].The desired inequality follows.
To prove the second assertion, note that it follows from (4) that {u,v}⊆C, In the next sections we will use (i), (ii) and the bound in (5) to give (for some q, n, d) new upper bounds on A q (n, d), based on divisibility arguments.Furthermore, in Section 5, we will prove that, for µ, q ∈ N, there is a 1-1-correspondence between symmetric (µ, q)-nets (which are certain designs) and (n, d) q = (µq, µq − µ) q -codes C with |C| = µq 2 .We derive some new upper bounds from these 'symmetric net' codes.

The divisibility argument
In this section, we describe the divisibility argument and illustrate it by an example.Next, we show how the divisibility argument can be applied to obtain upper bounds on A q (n, d) for certain q, n, d.In subsequent sections, we will see how we can improve upon these bounds for certain fixed q, n, d.We will use the following notation.
Definition 2.1 (k-block).Let C be an (n, d) q -code in which a symbol α ∈ [q] is contained exactly k times in column j.The k × n matrix B formed by the k rows of C that have symbol α in column j is called a (k-)block (for column j).In that case, columns [n] \ {j} of B form an (n − 1, d) q -code of size k.
At the heart of the divisibility arguments that will be used throughout this paper lies the following observation.
Proposition 2.1 (Divisibility argument).Suppose that C is an (n, d) q -code and that B is a block in C (for some column j) containing every symbol exactly m times in every column except for column j.  ∈ {6, 8}.However, since all (7, 6) 5 -codes of size 15 are equidistant, all distances in C belong to {6, 8}: either two words are contained together in some 15-block (hence their distance is 6) or there is no column for which the two words are contained in a 15-block (hence their distance is 8).This implies that an (8, 6) 5 -code C of size 75 cannot exist.Hence A 5 (8,6) ≤ 74.Theorem 2.2 and Corollary 2.3 below will imply that A 5 (8,6) ≤ 70 and in Section 3 we will show that, with some computer assistance, the bound can be pushed down to A 5 (8,6) ≤ 65.
To exploit the idea of Proposition 2.1, we will count the number of so-called irregular pairs of words occuring in a code.Definition 2.2 (Irregular pair).Let C be an (n, d) q -code and u, v ∈ {d, n}, we call {u, v} an irregular pair.
For any code C ⊆ [q] n , we write Using Proposition 2.1, we can for some cases derive a lower bound on |X|.If we can also compute an upper bound on |X| that is smaller than the lower bound, we derive that the code C cannot exist.The proof of the next theorem uses this idea.For fixed q, n, d, m ∈ N with q ≥ 2, define the following quadratic polynomial in r: Proof.By Plotkin's bound (2) we have Let D be an (n − 1, d) q -code of size mq − t with t < q.Note that d = m(n − 1)(q − 1)/(mq − 1).Then the right-hand side in (5) (taking Hence Therefore, all (n − 1, d) q -codes D of size mq are equidistant (then t = 0) and each symbol occurs m times in every column of D. Now let C be an (n, d) q -code of size M := mq 2 − r with r ∈ {1, . . ., q − 1}.Consider an mq-block B for some column of C. As n − d does not divide m(n − 1), by Proposition 2.1 we know Let B 1 , . . ., B s be mq-blocks in C for some fixed column.Since |C| = mq 2 − r, the number of mq-blocks for any fixed column is at least q − r (so we can take s = q − r).Then, with (11), one obtains a lower bound on the number |X| of irregular pairs in C. Every pair {B i , B k } of mq-blocks gives rise to mq irregular pairs: for each word u ∈ B i , there is a word v ∈ B k such that {u, v} ∈ X.This implies that in ∪ s i=1 B i ⊆ C there are at least s 2 mq irregular pairs.Moreover, for each word u in C \ ∪ s i=1 B i (there are M − mq • s of such words) there is, for each i = 1, . . ., s, a word v i ∈ B i with {u, v i } ∈ X.This gives an additional number of at least (M − mqs)s irregular pairs in C. Hence: On the other hand, note that the ith block for the jth column has size mq − r i,j for some integer r i,j ≥ 0 by (9), where q i=1 r i,j = r ≤ q − 1 (hence each r i,j < q).So by (10), the number of irregular pairs in C that have the same entry in column j is at most As each irregular pair {u, v} has u j = v j for at least one column j, we conclude Here the last inequality follows by convexity of the binomial function, since (for fixed j) the sum q i=1 r i,j 2 under the condition that q i=1 r i,j = r is maximal if one of the r i,j is equal to r and the others are equal to 0.
If each r i,j ∈ {0, 1}, then |X| = 0 by (14).As q − r ≥ 1, there is at least one mq-block for any fixed column, so |X| ≥ 1 by (11), which is not possible.Hence we can assume that r i,j ≥ 2 for some i, j (this also implies A q (n, d) ≤ mq 2 −2).Then the number s of mq-blocks for column j satisfies s ≥ q − r + 1.This gives by ( 12) and ( 14) that Subtracting the left hand side from the right hand side in (15) yields φ(r)/2 ≥ 0, i.e., φ(r) ≥ 0. So if φ(r) < 0, then A q (n, d) < mq 2 − r, as was needed to prove.
We give two interesting applications of Theorem 2.2.
Theorem 2.2 also gives an upper bound on A q (n, d) = A q (kq + k + q, kq), where q ≥ 2 and k does not divide q(q + 1) (which is useful for k < q − 1; for k ≥ q + 1 the Plotkin bound gives a better bound).One new upper bound for such q, n, d is obtained: This implies the following bound, which is also new: 3 Kirkman triple systems and A 5 (8,6).
As in the proof of Theorem 2.2, we will compare upper and lower bounds on |X|.But since an (8,6) 5 -code C of size at most 70 does not necessarily contain a 15-block (as 70 = 5 • 14), we need information about 14-blocks.To this end we show, using an analogous approach as in [6] (based on occurrences of symbols in columns of an equidistant code): Proposition 3.1.Any (7, 6) 5 -code C of size 14 can be extended to a (7, 6) 5 -code of size 15.
To establish the claim we must prove that d H (u, w) ≥ 6 for all w ∈ C. Suppose that there is a word w ∈ C with d H (u, w) < 6.We can renumber the symbols in each column of C such that w = 1.Since C is equidistant, each word in C \ {w} contains precisely one 1.On the other hand, there are two column indices j 1 and j 2 with u j 1 = 1 and u j 2 = 1.Then C \ {w} contains at most 1 + 1 + 5 • 2 = 12 occurrences of the symbol 1 (since in columns j 1 and j 2 there is precisely one 1 in C \ {w}).But in that case, since |C \ {w}| = 13 > 12, there is a row in C that contains zero occurrences of the symbol 1, contradicting the fact that C is equidistant.
Note that a code of size more than 65 must have at least one 15-or 14-block, and therefore it must have a subcode of size 65 containing at least one 15-or 14-block.We shall now prove that this is impossible because each (8,6)  Proof.First consider a (7, 6) 5 -code D of size 15 or size 14 and define For any u ∈ S, define This can be checked efficiently with a computer 4 by checking all possible (7, 6) 5 -codes of size 15 and 14 up to equivalence.Here we note that a (7, 6) 5 -code D (which must be equidistant, see Example 2.1) of size 15 corresponds to a solution to Kirkman's school girl problem [15]. 5So to establish (21), it suffices to check 6 all (7, 6) 5 -codes of size 15, that is, Kirkman systems (there are 7 nonisomorphic Kirkman systems [8]), and all (7, 6) 5 -codes of size 14, of which there are at ) is given by (5).The resulting values h(k) are given in Table 2. Proof.Let a k be the number of symbols that appear exactly k times in column j of C. Then the number of irregular pairs that have the same entry in column j is at most 15 k=5 a 4 All computer tests in this paper are small and can be executed within a minute on modern personal computers. 5Kirkman's school girl problem asks to arrange 15 girls 7 days in a row in groups of 3 such that no two girls appear in the same group twice.The 1-1-correspondence between (n, d)q = (7, 6)5-codes D of size 15 and solutions to Kirkman's school girl problem is given by the rule: girls i1 and i2 walk in the same triple on day j ⇐⇒ Di 1 ,j = Di 2 ,j .
6 By 'check' we mean that given a (7, 6)5-code D of size 14 or 15, we first compute S, then α(u) for all u ∈ S, and subsequently verify (21).
(There are 30 a ∈ Z 15 ≥0 with k a k k = 65 and k a k = 5.So there are 900 possible pairs a, b.A computer now quickly verifies (24).) By permuting the columns of C we may assume that max j f (a 15 , a ). Hence if f (a 14 ) > 0, then < f (a (where we used Proposition 3.2 in the last inequality), contradicting (23).So f (a 14 ) = 0 for all j, which implies (for a (j) ∈ Z 15 ≥0 with k a (j) 14 = 0 for all j, hence each symbol appears exactly 13 times in each column of C. 4 Improved bound on A 3 (16,11).
We show that A 3 (16,11) ≤ 29 using a surprisingly simple argument.Proof.Suppose that C is an (n, d) q = (16, 11) 3 -code of size 30.We can assume that 1 ∈ C. It is known that A 3 (15, 11) = 10, so the symbol 1 is contained at most 10 times in every column of C. Since |C| = 30, the symbol 1 appears exactly 10 times in every column of C, so the number of 1's in C is divisible by 5. On the other hand it is easy to check that a (15,11) 3 -code of size 10 is equidistant (using (5), as L = R).This implies that all distances in a (16,11) 3 -code of size 30 belong to {11, 16}.So the number of 1's in any code word = 1 is 0 or 5.As 1 contains 16 1's, it follows that the total number of 1's is not divisible by 5, a contradiction.

Codes from symmetric nets
In this section we will show that there is a 1-1-correspondence between symmetric (µ, q)-nets and (n, d) q = (µq, µq − µ) q -codes of size µq 2 .From this, we derive in Section 6 the new upper bound A 4 (9, 6) ≤ 120, implying A 4 (10, 6) ≤ 480.Definition 5.1 (Symmetric net).Let µ, q ∈ N. A symmetric (µ, q)-net (also called symmetric transversal design [2]) is a set X of µq 2 elements, called points, together with a collection B of subsets of X of size µq, called blocks, such that: (s1) B can be partitioned into µq partitions (block parallel classes) of X. (s2) Any two blocks that belong to different parallel classes intersect in exactly µ points.
(s3) X can be partitioned into µq sets of q points (point parallel classes), such that any two points from different classes occur together in exactly µ blocks, while any two points from the same class do not occur together in any block. 7emark 5.1.From the 1-1-correspondence between symmetric (µ, q)-nets and (n, d) q = (µq, µq− µ) q -codes C of size µq 2 in Theorem 5.1 below it follows that (s2) and (s3) can be replaced by the single condition: (s') Each pair of points is contained in at most µ blocks, since the only condition posed on such a code is that g(u, v) ≤ µ for all distinct u, v ∈ C. By labeling the points as x 1 , . . ., x µq 2 and the blocks as B 1 , . . ., B µq 2 , the µq 2 × µq 2 -incidence matrix N of a symmetric (µ, q)-net is defined by An isomorphism of symmetric nets is a bijection from one symmetric net to another symmetric net that maps the blocks of the first net into the blocks of the second net.That is, two symmetric nets are isomorphic if and only if their incidence matrices are the same up to row and column permutations.Symmetric nets are, in some sense, a generalization of generalized Hadamard matrices.Each generalized Hadamard matrix GH(n, G) gives rise to a symmetric (n/|G|, |G|)-net: by replacing G by a set of |G| × |G|-permutation matrices isomorphic to G (as a group), one obtains the incidence matrix of a symmetric net.Not every symmetric (n/q, q)-net gives rise to a generalized Hadamard matrix GH(n, q), see [13].But if the group of automorphisms (bitranslations) of a symmetric (n/q, q)-net has order q, then one can construct a generalized Hadamard matrix GH(n, q) from it.See [2] for details.occurs exactly n/q times at each position, since the leftmost term equals the rightmost term in (4) for (n − 1, d) q -codes of size n.)We see that also M T M = A. Hence, M is the incidence matrix of a symmetric (µ, q)-net (see [2], Proposition I.7.6 for the net and its dual).
Note that one can do the reverse construction as well: given a symmetric (µ, q)-net, the incidence matrix of M can be written (after possible row and column permutations) as a matrix of permutation matrices such that M M T = M T M = A, with A as in (28).From M we obtain a code C of size µq 2 of the required minimum distance by mapping the rows (i, w i ) ∈ [µq] × [q] to w ∈ [q] µq .Observe that equivalent codes yield isomorphic incidence matrices M and vice versa.
As A 4 (8, 6) = 32, any (9, 6) 4 -code of size more than 120 must contain at least one 31-or 32block, and therefore it contains a subcode of size 120 containing at least one 31-or 32-block.We will show (using a small computer check) that this is impossible because a (9, 6) 4 -code of size 120 does not contain any 31-or 32-blocks.Therefore A 4 (9, 6) ≤ 120.In order to do prove this, we need information about (8,6) 4 -codes of size 31.Proposition 6.1.Let q, n, d ∈ N satisfy qd = (q − 1)n.Any (n, d) q -code C of size qn − 1 can be extended to an (n, d) q -code of size qn.
Proof.Let C be an (n, d) q -code of size qn − 1.By Plotkin's bound, A q (n − 1, d) ≤ n, so each symbol occurs at most n times in each column of C, hence there exists for each j ∈ [n] a unique β j ∈ [q] with c β j ,j = n − 1 and c α,j = n for all α ∈ [q] \ {β j }.We can define a qn-th codeword u by putting u j := β j for all j = 1, . . ., n.We claim that C ∪ {u} is an (n, d) q -code of size qn.
To establish the claim we must prove that d H (u, w) ≥ d for all w ∈ C. Let w ∈ C with d H (u, w) < n.We can renumber the symbols in each column of C such that w = 1.Then w is contained in an (n − 1)-block B for some column in C (otherwise d H (u, w) = n).The number of 1's in B is n + (n − 2)n/q (since any (q, n − 1, d)-code of size n − 1 is equidistant, as L − R = 0 in (5) for (n − 1, d) q -codes of size n − 1) and the number of 1's in C \ B is (q − 1)(n − 1)n/q (since in any (n − 1, d) q -code of size n, each symbol appears exactly n/q times in each column, as the leftmost term equals the rightmost term in (4) for (n − 1, d) qcodes of size n).Adding these two numbers we see that the number of 1's in C is n 2 − n/q.Since C ∪ {u} contains each symbol n 2 times by construction, u contains symbol 1 exactly n/q times, hence d H (u, w) = n − n/q = d, which gives the desired result.Proposition 6.2.A 4 (9, 6) ≤ 120.

( 8 , 6 ) ≤ 5 •
15 = 75.Since, for (n, d) q = (7, 6) 5 and M = 15, the left hand side equals the right hand side in (4), any (7, 6) 5 -code D of size 15 is equidistant and each symbol appears exactly m = 3 times in every column of D. Note 2 = n−d m(n−1) = 21.Suppose there exists a (8, 6) 5 -code C of size 75.As A 5 (7, 6) ≤ 15, for each column, C is divided into five 15-blocks.Let B be a 15-block for the jth column and let u ∈ C \ B. Note that the other columns of B contain each symbol 3 times, and 3(n − 1) = 3 • 7 = 21 is not divisible by n − d = 2.So by Proposition 2.1, there must be a word v ∈ B with d H (u, v) /

most 7 • 2
15 by Proposition 3.1.Let G = (C, X) be the graph with vertex set V (G) := C and edge set E(G) := X.Consider a 15-block B determined by column j.By (21), each u ∈ C \ B has ≥ 1 neighbour in B. We observed this also in Example 2.1: for any u ∈ C \ B there exists at least one v ∈ B such that d H (u, v) / ∈ {6, 8}, so d H (u, v) = 7 and {u, v} ∈ X.In (21) this is represented as: if |D| = 15 then |{u ∈ S | α(u) = 0}| = 0, i.e., for any word u of length 7 that has distance ≥ 5 to all words in a (7, 6) 5 -code D of size 15, there is at least one v ∈ D such that d H (u , v ) = 6.Furthermore, (21) gives that all but ≤ 21 elements u ∈ C \ B have ≥ 3 neighbours in B. So by adding ≤ 2 • 21 new edges, we obtain that each u ∈ C \ B has ≥ 3 neighbours in B. Similarly, for any 14-block B determined by column j, by adding ≤ 8 new edges we achieve that each u ∈ C \ B has ≥ 1 neighbour in B. Hence, by adding ≤ (2 • 21 • x + 8 • y) edges to G, we obtain a graph G with |E(G )| ≥ (3x + y)(65 − 15x − 14y) + 3 • 15 x the required bound, except for the term with the indicator function.That term can be added because |{u ∈ S | α(u) ≤ 1}| ≤ 39 if |D| = 14, by (21).It is also possible to give an upper bound on |X|.If D is a (7, 6) 5 -code of size k, an upper bound h(k) = L − R on the number of pairs {u, v} ⊆ D with u = v and d H