Block Designs with SDP Parameters

It is traditional to call a quasi-symmetric design with certain parameters an SDP design if the symmetric difference of two different blocks is either a block or a block complement. In this note, we delete the requirements on the parameters and demand just that the symmetric difference of two blocks be a block, a block complement, or either the empty set or the whole point set. We obtain the parameters of such designs and use the result to prove Kantor’s theorem on the parameters of a symmetric SDP design. A spin-off of an exponential Diophantine equation considered by Ramanujan is at the core. 1 GP designs For a (v, k, λ) design D, let P be the point set, of size v, and let B be the block set, of size b = λv(v− 1)/(k(k− 1)). Each block has size k and each point is in r = λ(v− 1)/(k− 1) blocks. If D is a symmetric design, D is said to be an SDP (Symmetric Difference Property) design when the symmetric difference of any three blocks is either a block or a block complement. We identify subsets of P with their binary characteristic vectors, making the symmetric difference of subsets their (binary) sum. The complement of an SDP symmetric design is also one. W. M. Kantor proved that any symmetric SDP design (apart from the trivial (2, 1, 0) design) has parameters v = q, k = q 2 + δ √ q 2 , λ = q 4 + δ √ q 2 , (1) the electronic journal of combinatorics 19(3) (2012), #P11 1 where q is an even power of 2 and δ = ±1 [K, Theorem 3]. Both the derived and residual designs, with respect to a block, have the property that the sum of any two different blocks is either a block or a block complement (Lemma 6, ahead). The derived and residual designs have parameters derived: v = q 2 + δ √ q 2 , k = q 4 + δ √ q 2 , λ = q 4 + δ √ q 2 − 1 (2) residual: v = q 2 − δ √ q 2 , k = q 4 , λ = q 4 + δ √ q 2 , again with q an even power of 2. The designs of all of these parameter sets have orders r−λ = q/4. The complements of these designs also have the block-sum property. This collection of parameters also contains those of the complements (recall that the complement of a derived [residual] design is the residual [derived] design of the complement of the original). The block-sum property can be rephrased to say that the set {B,B′|B ∈ B}∪{∅,P}, where B′ is the complement P +B of B, is a group (an elementary Abelian 2-group). We make the following definition: Definition 1. A design is called a GP (group property) design if the set G = {B,B′|B ∈ B} ∪ {∅,P} is a group under symmetric difference. The complement of a GP design is also a GP design. It is traditional to call a nonsymmetric design an SDP design if it is a GP design and it (or its complement) has the parameters in (2) (see the summary by V. D. Tonchev in [CD, VII.1.9]). The purpose of this note is to show that any nonsymmetric GP design (or its complement) that is not a Hadamard 3-design does have the parameters in (2). From that we can infer Kantor’s theorem for symmetric SDP designs. Part of the point of doing this is that the proof of our result is an elementary number-theoretic argument, and obtaining Kantor’s theorem as a corollary avoids some of the complexities of that theorem’s proof. Moreover, the customary attendant specification of the parameters for GP designs can almost be omitted. Let D be a GP design. If B1 + B2 = B3, with Bi ∈ B, then as |B1 +B2| = 2(k − |B1 ∩B2|), |B1 ∩B2| = k/2. If B1 + B2 = B′ 3 instead, then |B1 ∩B2| = (3k − v)/2. Suppose first that these two intersection numbers are the same, which means that v = 2k. Then D cannot be symmetric, for if so, the second design equation r(k − 1) = λ(v − 1) (3) would entail the impossibility k(k−1) = λ(2k−1). The only other conceivable intersection number is 0, so D must be a quasi-symmetric design with intersection numbers x = 0 and y = k/2 (thus the complement of some block is also a block; that fact would imply that v = 2k). Then [SS, Proposition 3.17] applies: (r − 1)(y − 1) = (k − 1)(λ− 1). (4) the electronic journal of combinatorics 19(3) (2012), #P11 2 Solving this along with (3) gives r = 2k − 1 and λ = k − 1. As now b = 2v − 2, Theorem 5.8 of [CvL] implies that D is either a Hadamard 3-design or the unique (6, 3, 2) design [CD, II Example 1.18]. But this latter citation shows that the (6, 3, 2) design does not qualify as a GP design. The Hadamard matrices involved in the 3-designs must actually be of Sylvester type because of the group property. (A Sylvester type Hadamard matrix is one equivalent to a Kronecker power of [ 1 1 1 −1 ] , or (equivalently!) to the character table of an elementary Abelian 2-group; see [CvL, Example 1.31], for instance.) However, we can see all this identification rather more directly: if B is any block of D, then on counting the blocks other than B that meet B (a standard maneuver), we get k(r − 1)/y = 4k − 4 = b − 2. The missing two blocks can only be B and B′; so the complement of any block is also a block. If the group G for the GP property has size 2q, q a power of 2, then the design parameters are v = q, b = 2q − 2, r = q − 1, k = q/2, λ = q/2− 1. (5) Moreover, if B1 and B2 are different blocks with B2 ̸= B′ 1, then B1 +B2 must be a block. From this we infer that D is a Hadamard 3-design based on a Sylvester matrix. Another way to present the design is as the set of supports of the words of weight q/2 in the first order Reed-Muller code R(1,m), where q = 2 (see [AK] for these codes). Thus we have Proposition 2. If D is a GP design for which v = 2k, then D is a Hadamard 3-design corresponding to a Sylvester type Hadamard matrix. Now take v ̸= 2k. Then no block complement is also a block, and the two possible intersection numbers k/2 and (3k − v)/2 are different. Could D be symmetric? If so, then λ = k/2 or (3k − v)/2. Say that λ = k/2. Then k(k − 1) = k/2 × (v − 1) gives v = 2k − 1. Hence D is a Hadamard 2-design, and the group property again implies that the corresponding Hadamard matrix is of Sylvester type. If λ = (3k − v)/2, then v = 2k + 1 and λ = (k − 1)/2. This is also a Hadamard 2-design, with Sylvester matrix. Thus Proposition 3. If D is a symmetric GP design, then D is a Hadamard 2-design, and again the corresponding Hadamard matrix is of Sylvester type. The design with v = 2k − 1 can also be realized as the set of supports of the words of weight q/2 in the punctured first-order Reed-Muller code R(1,m)∗, q = 2. Incidentally, the sum of two different blocks is a block; but for k > 1, there are three blocks whose sum is ∅. Thus the design is not an SDP design–as it better well not be! There is one more special case: it could be that the intersection number (3k − v)/2 is 0, that is, that v = 3k. Now when we solve (4) and the design equations we get k = 3− 8/(r+ 3). The possibilities are r = 1 and 5, giving the trivial (3, 1, 0) design and the (6, 2, 1) design whose blocks are the 2-subsets of a 6-set. The parameters are those of (2) with q = 4 and δ = −1 in the residual set and with q = 16 and δ = −1 in the derived set. the electronic journal of combinatorics 19(3) (2012), #P11 3 Thus finally we may assume that D is a quasi-symmetric design with intersection numbers x = (3k − v)/2 and y = k/2, x and y different (D is proper) and both positive. We may also take k < v/2 by replacing D with its complement, if necessary. Let the order of G be 2q, q a power of 2 as before, so that now b = q − 1, since no block complement is a block. The equation generalizing (4) is k(r − 1)(x+ y − 1) + xy(1− b) = k(k − 1)(λ− 1) [SS, Lemma 3.23(i)]. This becomes (after a cancellation of k) (r − 1)(2k − v 2 − 1)− (q − 2) − v 4 = (k − 1)(λ− 1). Solving it with the design equations gives r = k(v − 1) v − (v − 2k)2 , λ = k(k − 1) v − (v − 2k)2 , q = 4k(v − k) v − (v − 2k)2 . Substituting v − (v − 2k) = 4k(v − k)/q from the third equation into the other two, we get λ = q(k − 1) 4(v − k) , r = q(v − 1) 4(v − k) . Now by [SS, Corollary 3.9], y − x divides both k − x and r − λ. Here y − x = v − 2k 2 , k − x = v − k 2 , r − λ = q 4 . Thus (v− 2k) divides q/2, so that v− 2k = q0, q0 also a power of 2, with q0 < q (we have assumed that k < v/2). Then v = q0 + 2k and the equation for q is q = 4k(q0 + k) q0 + 2k − q 0 . Thus 4k + (4q0 − 2q)k + qq 0 − qq0 = 0, making k = q 4 − q0 2 + δ q0 2 √( q 2q0 )2 − q + 1, (6) δ = ±1. So it must be that ( q 2q0 )2 − q + 1 = z (7) for some integer z. Equation (7) is a particular case of the exponential Diophantine equation