Bounds for equidistant codes and partial projective planes

Discrete Mathematics 17 (1977) 85-94. © North-Holland Publishing Company BOUNDS FOR EQUIDISTANT CODES AND PARTIAL PROJECTIVE PLANES J.I. HALL Technological University. Eindhoven, The Netherlands Received 16 December 1975 Revised 13 July 1976 o. Introduction Let V be a vector space of finite dimension n over GF(2) with some distinguished basis {£i: i = 1, n}. If x = ~iX£ we write x = (XI, X2"")' The Hamming distance between x and y is then d (x, y) : = /{i: Xi'll Yi}/. An equidistant m -code C is a subset of V such that, for all distinct x and y in C, d (x, y) = m. Any C such that / C / < 3 is an equidistant m -code for some m, so we shall always assume that / C / ~ 3. In particular we must have m = 2k, for some integer k. Suppose b = /C /. We shall frequently identify C with its row dictionary, a (0,1) matrix with b rows, these rows listing and exhausting the code words of C. C is non-trivial if in its row dictionary there is some column sum c such that 2 :s;; c :s;; b - 2. Otherwise C is trivial. Our main result is an extension of a result due to Deza [4] and Van Lint [9, 10]. e ~ ~ 4. There exists a non-trivial equidistant 2k-code with b words if and only if b :s;; k 2 + k + 2 and there exists a projective plane of order k. Theorem. Let b A partial projective plane P of order k is a pair ([j>, It) with [j> a finite set of elements (called points) and It a set of subsets of [j> (called lines) such that each line contains k + 1 points and each pair of distinct lines intersects in precisely one point. We assume each point to be on at least one line and call P trivial if there is some point which lies on every line of It. Otherwise P is non-trivial. The main theorem is a direct consequence of our Theorem 1 and Theorem 2. ~ Hk + 2)2 with k ~ 2. There exists a non-trivial equidistant 2k -code with b words if and only if there exists a non-trivial partial projective plane of order k with b - 1 lines. Theorem 1. Let b Theorem 2. Let b ~ e- 1 with k ~ 2. There exists a non-trivial partial projective 85 86 1.1. Hall plane of order k with b lines if and only if b .;;; k 2 + k + 1 and there exists a projective plane of order k. The relevant codes and partial planes with k < 2 are easily described and not of interest here. It is perhaps worthwhile to remark that the assumption of finite dimensionality for V and finite cardinality for fl/' does not enter into our proofs in any essential manner as long as we require that k be finite. A partial projective plane of order k may be thought of as the dual of a pairwise balanced design of index unity (see [1] or [12] for definitions) which possesses the additional property that all points occur in a fixed number, k + 1, of blocks. We then have, as a corollary to Theorem 2: Corollary. Let (X, d) be a pairwise balanced design of index unity such that all points of X occur in a fixed number, k + 1, of blocks of d. If 1XI;;;. k 2 - 1, then either d contains a block consisting of all points of X or (X, d) can be obtained by deleting k 2 + k + 1 -I X 1points from a projective plane of order k. Bose and Shrikhande [2] first proved a special (but important) case of the corollary. For k = 6, Janssen, Kolen and Van Lint (see [5]) and independently also McCarthy, Vanstone and DeWitte [3, 7, 11] proved the results given here. After having completed this paper the author learned of the work in [3, 7,8,11]. The case b ;;;. of Theorem 2 (and hence also of the corollary) has been proven by methods almost the same as those used here in [11]. Theorem 2 itself (and the corollary), except for the case b = k 2 - 1 and k = 6, has been (according to [3]) proven in Vanstone's 1974 Waterloo thesis. The last case is the topic of [3]. In [7] McCarthy and Vanstone prove a weak version of Theorem 1 and so, using the results of [3] and [11], derive independently several results somewhat weaker than our main theorem. e 1. Proof of Theorem I Given any partial projective plane P of order k with b lines, we form its blockpoint (0, 1) incidence matrix B. If J is a b by k - 1 matrix of 1's, then is the row dictionary of an equidistant 2k -code with b + 1 words. We refer to this as the standard embedding of P (or B) in an equidistant code. This construction immediately proves the "if" part of Theorem 1. Since Hamming distance is a translation invariant of V, the code C + x:= {c + x: c E C} is an equidistant 2k-code if and only if C is also. As we are working Bounds for equidistant codes and partial projective planes 87 over GF(2) this process just amounts to complementing certain columns of C. We shall call the two codes C + x and C equivalent. Clearly under this definition equivalence is indeed an equivalence relation. Theorem 1'. Let b;;:. ~(k + 2)2 with k;;:. 3 or b > !(k + 2)2 with k = 2. If C is a non-trivial equidistant 2k -code with b words and no constant columns then either C is equivalent to an incidence matrix for a partial projective plane of order k with b lines or C is equivalent to the standard embedding of an incidence matrix for such a plane with b - 1 lines. Since a projective plane of order 2 exists, Theorem 1 is a direct consequence of Theorem I' and the comment made above. We therefore need only prove Theorem I'. We now assume that C is a non-trivial equidistant 2k-code, for k ;;:. 2, with b words and no constant columns. The following important lemma is essentially due to Deza [4]. Lemma 1.1. Suppose a column of C has column sum c. Then c(b - c):e;; kb. If equality holds then c = ~b = 2k. Proof. We follow [10]. Let column 1 of C have exactly c ones, these be!ng in the first c rows. Let column i have ai ones in the first c rows and {3i ones in the last b - crows. Let ai = aJc and bi = (3J(b - c). We count distances between, respectively, pairs of rows from the top, pairs from the bottom, and pairs with one from each section, and find 2: ai (1 - ai) = k - k / c, (1) i>1 2: b;(1 - bi ) =k - k / (b - c), (2) i>1 2: (a;(l- b )+ b (1- ai» = 2k-1. i i (3) i>1 Combining, we have which gives the desired inequality. Equality only occurs when, for all i, ai = bi. Using (1) and (2) we then see that c = b - c, completing the proof of the lemma. Until the end of this section, we shall assume that C does not satisfy the conclusion to Theorem I'. Let column s have column sum c, and define t:= max, (min (c" b - c,». Then, in terms of k, the strongest bound on b given by Lemma 1.1 is 1.1. Hall 88 t(b - t):;;; kb. (4) We call a column light if c. :;;; t and heavy if c. ;;=!: b - t. Lemma 1.2. If t :;;; k + 1, then b :;;; ~(e + 3k + 4). In any case, b:;;; Hk + 2)2. Proof. Assume b > !(k 2 + 3k + 4) and t:;;; k + 1. Replacing C by an equivalent code, we may assume that the zero vector is a codeword. All non-zero codewords therefore have weight 2k (that is, 2k ones) and mutual inner products k. We claim that C has at most k + 1 heavy columns and that every non-zero row of C has at least k - 1 of its 1's in heavy columns. First, suppose we can select k + 2 heavy columns from C. Examining the non-zero codewords, we see that at most k + 2 rows can have k + 1 of their 1's in these columns. If some row has k + 2 of its 1's in these columns, all other rows have at most k. Hence, remembering that t:;;; k + 1, (k + 2)(k + 1) + (b - 1 - (k + 2»k ;;=!: (k + 2)(b - (k + 1». Isolating b we find b :;;; He + 3k + 4), against our assumptions. Thus C has at most k + 1 heavy columns. Now, if some row of C, x say, had at most k - 2 of its 1's in heavy columns, then the equivalent code C + x would contain the zero vector and have at least k + 2 heavy columns. This would contradict the above. Therefore all non-zero rows of C have at least k - 1 ones in heavy columns. C now has either k - 1, k, or k + 1 heavy columns. Suppose first that C has only k - 1 heavy columns. Then by the previous paragraph these columns are constantly equal to 1 off the zero vector. Thus C is the code of the standard embedding of a partial projective plane of order k with b - 1 lines, as in the conclusion to Theorem I'. Next we suppose that C has exactly k heavy columns. If these are all constantly equal to 1 off the zero vector, then the code is trivial. Hence some row, say x, has only k - 1 of its 1's in heavy columns. But then the equivalent code C + x has k + 2 heavy columns, contradicting the previous paragraph. Finally, assume that C has k + 1 heavy columns. If some row x has 1's in all these columns, then C + x has only k - 1 heavy columns and we are done as above. If there is an x with only k - 1 of its 1's in heavy columns, then C + x has k + 3 heavy columns, a contradiction. Therefore each non-zero row has exactly k of its 1's in the k + 1 heavy columns. Thus if x is the vector having 1's precisely in the positions of the heavy columns of C, then C + x is the incidence matrix of a partial projective plane of order k with b lines. Again C satisfies the conclusion to Theorem I', against our assumptions. We have thus completed the proof of the first sentence of the lemma. If we substitute for b in (4), we find that b > Hk + 2)2 implies t:;;; k + 1. As !(k + 2)2;;=!: He + 3k + 6) for all k > 1, the above shows that, for b > ~(k + 2)2, C would satisfy the conclusion to Theorem I'. Lemma 1.3. If k ;;=!: 3, then b < !(k + 2f. Bounds for equidistant codes and partial projective planes 89 Proof. In view of the previous lemma, we may assume that t;;:, k + 2. Now substituting b = !(k + 2)2 in (4) we find that t = k + 2 and that we have strict equality. Thus by Lemma 1.1, k + 2 = t =!b = l(k + 2)2 giving k = 2. This proves Lemma 1.3. Lemmas 1.2 and 1.3 imply Theorem I' and hence Theorem 1. 2. Proof of Theorem 2 It should be understood that when we refer to the partial plane P, we shall always mean that P is a partial projective plane of order k. We first give a well known result. Proposition 2.1. Suppose P is a non-trivial partial projective plane of order k. Then any point is on at most k + 1 lines, and if any point is on k + 1 lines then there are exactly k 2 + k + 1 points. P has at most k 2 + k + 1 lines, equality holding if and only if P is a projective plane of order k. Proof. If in any partial plane a point p lies on k + 2 lines, then any line not containing p can not meet (intersect) all the lines containing p. Hence all lines must contain p and the plane is trivial. A similar argument shows that if, in the non-trivial plane P, point p lies on k + 1 lines, then P has precisely P + k + 1 points. Now for any line I we can count the number of lines meeting I. Allowing for k intersections at each point of I, P has at most k 2 + k + 1 lines. In the case of equality, every point of P lies on k + 1 lines; and we must have a projective plane. One direction of Theorem 2 is clear. In view of Proposition 2.1, to prove Theorem 2 we need only show that the existence of a non-trivial partial plane with b lines satisfying k 2 - 1 ~ b < k 2 + k + 1 implies the existence of a non-trivial partial plane with P + k + 1 lines. We shall in fact prove a stronger result, for which the following definition is appropriate. Definition. A partial projective plane P of order k, P = (~, 9:), can be uniquely 9:), such extended to a projective plane, if there exists a projective plane, p) = (~t, that ~) ~ ~ and 9:) ~ 9: and further if, for any second projective plane, P2 = (~2' 9:2 ), with ~2 ~ p; and 9:2 ~ 9:, there is a design isomorphism of p) and P 2 which induces the identity on P; and If. Theorem 2'. For k ;;:, 2 a non-trivial partial projective plane of order k with at least k 2 - 1 lines can be uniquely extended to a projective plane. We remark without further comment that the corollary mentioned in the introduction is a direct consequence of Theorem 2'. 90 1.1. Hall We now let P, P = (gil, .;£), be a non-trivial partial projective plane of order k with v:= 1gil 1and b:= 1.;£ I. In our proof of Theorem 2' we shall study the (0,1) block-point incidence matrix A of P. Thus A is a b by v matrix with row sums k + 1. We take Cj to be the column sum of column i. Beginning with our matrix A with k 2 - 1 ~ b < k 2 + k + 1 we shall construct in an essentially unique manner the incidence matrix A' of a projective plane of order 2 k (that is, a partial plane with k + k + 1 lines), A' containing A in its upper left-hand corner (columns of zeroes may need to be added to A, corresponding to new points). We do this by showing that a new row, row b + 1, can always be added to A subject to (I) the new row has at most k + 1 ones and inner product 1 with all rows of A, and (II) any appropriate A' must contain the newly constructed row. The row is then the incidence vector of the intersection with P of an acceptable new line. Theorem 2' will follow by induction. We now assume that P satisfies k 2 - 1 ~ b < k 2 + k + 1 but does not satisfy the conclusion to Theorem 2', which is to say that no row b + 1 can be found satisfying (I) and (II). Lemma 2.2. For some i, Ci = k + 1. In particular v = k + k + 1. 2 Proof. In view of Proposition 2.1, we need only show that some i does not satisfy Cj ~ k. So we assume that Ci ~ k for each i. If we arrange A so that the 1's of row 1 are in columns 1 through k + 1, then ~:I Cj = b + k, and the average of these Cj is (b + k)/(k + 1). This is larger than k if b > k 2 , so we must have b ~ k 2 • If b = k 2 then all column sums are k, and P is the dual of an affine plane of order k. As such it can be uniquely extended to a projective plane (d., for example, [6, Theorem 12.3.3]). We now have b = k 2 - 1, and we may assume that CI = C2 = ... = Ck = k and CHI = k - 1. Thus every row has precisely one 1 in a column of weight k - 1, all other columns having weight k. If A has f columns of weight k and g of weight k - 1, we may count 1's and ordered row inner products to find fk + g (k - 1) = (e - 1)( k + 1) and fk (k - 1) + g(k - l)(k - 2) = (e - l)(e - 2). Solving, we find that f = k 2 - 1 and g = k + 1. Hence a new row b + 1 having 1's only in the columns which have weight k - 1 in A satisfies (I). The new matrix is again the dual of an affine plane. As v = f + g = k 2 + k, any suitable A' extending A must have a new column (point) added. With this in mind, it is easy to see that A' must contain the newly constructed row. Thus (II) is also satisfied, completing the proof of the lemma. 91 Bounds for equidistant codes and partial projective planes Lemma 2.3. For all i, Cj -,l k. Proof. Assume c, = k, the 1's of the column being in rows 1 through k. We may arrange the columns of A so that the remaining k ones of row i are in the columns (i - l)k + 2 through ik + 1 for 1 ~ i ~ k. Any row from k + 1 through b must have exactly one 1 in each of these intervals and a 0 in column 1, hence exactly one 1 in columns e + 2 through k 2 + k + 1. We may then add a row b + 1 with 1's in columns 1 and k 2 + 2 through k 2 + k + 1. This new row clearly satisfies (I) and (II). Lemma 2.4. i (Cj - kt = e+ k + 1 + (b - (k 2 -1»(b - (e+ k + 1». i-t Proof. In A we count the 1's and the ordered row inner products to find L Cj = b(k + 1), L As v = e+ k + 1, evaluating Lemma 2.5. b Cj(Cj -1) = b(b -1). j j =e ~, (Ci - - 1 and, for all i, k)2 gives the lemma. either Cj =k - 1 or Cj = k + 1. Proof. For b > k 2 - 1, the right-hand side of the equation of Lemma 2.4 is strictly less than e + k + 1. Considering the e + k + 1 terms of the left-hand side, there must be some i (indeed several) for which (Cj - k)2 = O. That is, Cj = k against Lemma 2.3. Hence b = 1. In this case Lemma 2.4 yields ~j (Cj - k)2 = + k + 1. By Lemma 2.3, (Cj - k)2 > 0 for all i, hence (Cj - k)2 = 1 in all cases. Thus, for any i, either Cj = k - 1 or Cj = k + 1. This gives the lemma. e- e We could now complete our proof of Theorem 2' by first noting that Janssen, Kalen and Van Lint (see [5]) and also DeWitte [3] have shown that for k = 6 the configuration of Lemma 2.5 does not exist and then using the embedding result of Bose and Shrikhande [2] to handle all knot 6. (This is, in fact, the method used by Vanstone and DeWitte [11, 3] in their proof of Theorem 2.) As the paper of Bose and Shrikhande appeals to a rather difficult theorem on the classification of certain strongly regular graphs, we prefer to complete our proof without reference to it. We first note that the configuration of Lemma 2.5 is trivial if k = 2, so we may assume k > 2. Suppose any specific row has d ones in columns of weight k - 1. Then b + k = e - 1 + k = d(k - 1) + (k + 1- d)(k + 1), giving d = !(k + 2). In particular, k is even. We say that two columns of A meet if there is a row which has a 1 in both columns (in which case the column inner product will be 1). If two columns do not meet, we say they are disjoint (column inner product 0). Note that a column of 92 1.1. Hall weight k + 1 meets all columns and that a column of weight k - 1 is disjoint from 2k columns, all of which must have weight k -1. We form a graph r with the columns of weight k - 1 as vertex set, two columns being joined if they are disjoint. Thus r is a regular graph of valency 2k on (k + I)O(k + 2» vertices. Suppose we choose the first column of A to have weight k - 1, its 1's in rows 1 through k - 1. Further we let the remaining 1's of row i, 1 Eo;; i Eo;; k - 1, be in positions (i - l)k + 2 through ik + 1. The last 2k columns of A are then those columns disjoint from column 1. Any row from k through b must have precisely one 1 in each of the intervals (i - l)k + 2 through ik + 1, hence two 1's in the last 2k columns. Now choosing column 2 to have weight k - 1, it meets 2(k - 2) of the last 2k columns. Thus we see that two columns of weight k -1 which meet are simultaneously disjoint from precisely 4 columns. In terms of our graph r, for any two nonadjacent points there are exactly 4 points joined to both. In a similar manner we find that any two adjacent points lie in k triangles (there are k points joined to both). This says precisely that r is a strongly regular graph with = k and prJ = 4 (cf. [2, p. 311]). To find a new row satisfying (I) we must find k + 1 columns of weight k - 1 in A, these columns pairwise disjoint. Then a new row having 1's in only these places has weight k + 1 and inner product 1 with all (k + 1)(k - 1) rows of A. Such a collection of columns of A corresponds in r to a clique of size k + 1. Bose and Shrikhande in [2] use the classification of all strongly regular graphs with the same parameters as r to construct k + 2 cliques of size k + 1, any two cliques intersecting in a point and each edge in a unique clique. These cliques then become the remaining k + 2 lines of a projective plane extending A. We have an advantage over this method in that if we furnish a single clique of size k + 1 then Lemma 2.5 and our assumptions guarantee the extension of A. We proceed to construct a (k + I)-clique in r. More precisely we construct a k-clique in the graph ro which r induces on the 2k points adjacent to some given point O. Adding 0 to the k -clique we find a (k + I)-clique which represents a line I satisfying (I) and containing O. We remark that any such line I satisfying (I) also must satisfy (II). For by Lemma 2.5 and our assumptions the existence of I ensures the extendibility of A to a projective plane. This in turn implies that ro has at least two disjoint k-cliques. But since in r every joined pair of points lies in k triangles ro has valency k. Therefore ro must be two k -cliques joined by a matching. Then as k > 2, ro has only two k -cliques. We see directly that I satisfies (II). We need now only find a k -clique in the graph ro. As mentioned above, ro has valency k and 2k points. Since any two nonadjacent points of r are both joined to 4 points, two nonadjacent points of ro are both joined to at most 3 points of From now on we no longer consider all of the graph r but only the graph ro. For k = 4 the appropriate 5 columns of A can be found directly by hand from A. For k = 6, Janssen, Kolen and Van Lint (see [5]) and also DeWitte [3] have shown the non-existence of the configuration of Lemma 2.5. As k is even and larger than 2, we may now assume that k ~ 8. p:. roo Bounds for equidistant codes and partial projective planes 93 In ro we choose a point, 1 say, and let T be the k points joined to 1 and T'those k - 1 points not joined to 1. Choosing point 2 in T' we see that 2 is joined to at most 3 points of T hence at least k - 3 points of T'. As T' has only k -1 points there is at most one point, 3 say, of T' not adjacent to 2. In that case the nonadjacent points 2 and 3 are both joined to the k - 3 other points of T'. As k - 3 > 3, this is a contradiction. Hence T' is a (k - I)-clique of roo That is, for any point of ro the points of ro not adjacent to it form a (k - I)-clique. The points of ro not adjacent to 2 form a (k -I)-clique S' disjoint from T'. There are two points not in S' or T' each of which is joined to at least !k points of either S' or T'. Without loss of generality we assume that one of these points, 4 say, is joined to at least !k points of T'. Suppose there is some point of T', 5 say, which is not joined to 4. Then there are at least !k points joined to each of the nonadjacent pair 4 and 5. As !k > 3 this is a contradiction. Therefore {4 U T'} is a k -clique in r o, as desired. This completes the proof of Theorem 2'. 3. The main theorem and remarks In view of the standard embedding, one direction of the main theorem is clear. Further, if k ;;;;. 5 then k 2;;;;. Hk + 2)2, so that Theorems 1 and 2 provide the converse. In fact, in this case Theorems l' and 2' guarantee that a code with b;;;;. k 2 can be "embedded" in a maximal code with k 2 + k + 2 words, a strong converse. For k = 2, 3, or 4 the projective planes exist. Thus we need only note that in those cases k 2 + k + 3 > Hk + 2)2, and again Theorems 1 and 2 imply that b ~ k 2 + k + 2 as required. Therefore the main theorem is proven. Both Theorem l' and Theorem 2' are in some small sense best possible. For k = 2 and 3 non-trivial equidistant 2k -codes exist which do not satisfy the conclusion to Theorem l' and have the largest number of words allowed by the theorem for such a code (8 for k = 2 and 12 for k = 3). However, these codes are related to symmetric designs with block size k + 2 and A = 2. This suggests that the best bound for Theorem l' would be b;;;;. He + 3k + 8). Lemma 1.2 would seem to support this conjecture. Many non-trivial equidistant 2k-codes with He + 3k + 6) words fail to satisfy the conclusion to Theorem 1', but the author knows of no such code with more words and this property. In a similar manner, there is a non-trivial partial projective plane of order 3 with 7 lines which cannot be extended, and thus the bound of Theorem 2' is sharp. However the reason would seem to be not that 7 is k 2 - 2 for k = 3, but that it is k 2 - k + 1. Indeed, if there exists a projective plane of order k - 1, then by adjoining to its incidence matrix an identity matrix we find a partial plane of order k with k 2 - k + 1 lines. This partial plane is not extendible as long as k ;;;;. 3. Again, we suspect that a non-trivial partial projective plane of order k with at least k 2 - k + 2 lines exists if and only if there is a projective plane of order k. Hall et al. [5], Mullin 1.1. Hall 94 and Vanstone [8] have proven this for k = 6, showing the non-existence of a non-trivial partial plane of order 6 with 32 words. We thank Professor J.H. van Lint for the argument which led to Lemma 1.3. References (1) (2) [3) [4) [5] [6] [7) [8] [9] [to] [11] [12] R.c. Bose and S.S. Shrikhande, On the composition of balanced incomplete block designs, Can. J. Math. 12 (1960) 177-187. R.C. Bose and S.S. Shrikhande, Embedding the completement of an oval in a projective plane of even order, Discrete Math. 6 (1973) 305-312. P. deWitte, The exceptional case in a theorem of Bose and Shrikhande, to appear. M. Deza, Une propriete extremale des plans projectifs finis dans une classe de codes equidistants, Discrete Math. 6 (1973) 343-352. J.I. Hall, A.J.E.M. Janssen, A.W.J. Kolen and J.H. van Lint, Equidistant codes with distance 12, Discrete Math. 17 (1976) 000-000 (this issue). M. Hall, Jr., Combinatorial Theory (Blaisdell, Waltham, MA, 1967). D. McCarthy and S.A. Vanstone, On (r, A)-designs and finite projective planes, to appear. R.c. Mullin and S.A. Vanstone, On regular pairwise balanced designs of order 6 and index 1, Utilitas Math. 8 (1975) 349--370. J.H. van Lint, A theorem on equidistant codes, Discrete Math. 6 (1973) 353-358. J.H. van Lint, Equidistant point sets, in: T.P. McDonough and V.C. Mavron (eds.), Combinatorics, Proc. Third British Combinatorial Conference, Aberystwyth, 1973, London Math. Soc. Notes 13 (Cambridge University Press, Cambridge, 1974). S.A. Vanstone, The extendibility of (r, I)-designs, in: Proc. Third Manitoba Conference on Numerical Mathematics, Winnipeg, 1973 (Utilitas Math., Winnipeg, 1974). R.M. Wilson, An existence theory for pairwise balanced designs I. Composition theorems and morphisms, J. Combin. Theory 13 (A) (1972) 220-245.