The Fine Intersection Problem for Steiner Triple Systems (original) (raw)
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Steiner Triple Systems Intersecting in Pairwise Disjoint Blocks
The Electronic Journal of Combinatorics, 2004
Two Steiner triple systems (X,calA)(X,{\cal A})(X,calA) and (X,calB)(X,{\cal B})(X,calB) are said to intersect in mmm pairwise disjoint blocks if ∣calAcapcalB∣=m|{\cal A}\cap{\cal B}|=m∣calAcapcalB∣=m and all blocks in calAcapcalB{\cal A}\cap{\cal B}calAcapcalB are pairwise disjoint. For each vvv, we completely determine the possible values of mmm such that there exist two Steiner triple systems of order vvv intersecting in mmm pairwise disjoint blocks.
Journal of Combinatorial Designs, 2005
In this paper, we present a conjecture that is a common generalization of the Doyen-Wilson Theorem and Lindner and Rosa's intersection theorem for Steiner triple systems. Given u, v v 1, 3 ðmod 6Þ, u < v v < 2u þ 1, we ask for the minimum r such that there exists a Steiner triple system ðU, BÞ, jUj ¼ u such that some partial system ðU, Bnq qÞ can be completed to an STSðv vÞ, ðV, B 0 Þ, where jq qj ¼ r. In other words, in order to ''quasi-embed'' an STSðuÞ into an STSðv vÞ, we must remove r blocks from the small system, and this r is the least such with this property. One can also view the quantity ðuðu À 1Þ=6Þ À r as the maximum intersection of an STSðuÞ and an STSðv vÞ with u < v v. We conjecture that the necessary minimum r ¼ ðv v À uÞð2u þ 1 À v vÞ=6 can be achieved, except when u ¼ 6t þ 1 and v v ¼ 6t þ 3, in which case it is r ¼ 3t for t 6 ¼ 2, or r ¼ 7 when t ¼ 2. Using small examples and recursion, we solve the cases v v À u ¼ 2 and 4, asymptotically solve the cases v v À u ¼ 6, 8, and 10, and further show for given v v À u > 2 that an asymptotic solution exists if solutions exist for a run of consecutive values of u (whose required length is no more than v v À u). Some results are obtained for v v close to 2u þ 1 as well. The cases where v v % 3u=2 seem to be the hardest.
Sets of three pairwise orthogonal Steiner triple systems
Journal of Combinatorial Theory, Series A, 2003
Two Steiner triple systems (STS) are orthogonal if their sets of triples are disjoint, and two disjoint pairs of points defining intersecting triples in one system fail to do so in the other. In 1994, it was shown (Canad. J. Math. 46(2) (1994) 239-252) that there exist a pair of orthogonal Steiner triple systems of order v for all v 1; 3 (mod 6), with vX7; va9: In this paper we show that there exist three pairwise orthogonal Steiner triple systems of order v for all v 1 ðmod 6Þ; with vX19 and for all v 3 ðmod 6Þ; with vX27 with only 24 possible exceptions. r
The flower intersection problem for Kirkman triple systems
Journal of Statistical Planning and Inference, 2003
The ower at a point x in a Steiner triple system (X; B) is the set of all triples containing x. Denote by I * R [r] the set of all integers k such that there exists a pair of KTS(2r + 1) having k + r triples in common, r of them being the triples of a common ower. In this article we determine the set I * R [r] for any positive integer r ≡ 1 (mod 3) (only nine cases are left undecided for r = 7; 13; 16; 19), and establish that I * R [r] = J [r] for r ≡ 1 (mod 3) and r ¿ 22 where J [r] = {0; 1; : : : ; 2r(r − 1)=3 − 6; 2r(r − 1)=3 − 4; 2r(r − 1)=3}.
Independent sets in Steiner triple systems
Ars Combinatoria, 2004
A set of points in a Steiner triple system (STS(v)) is said to be independent if no three of these points occur in the same block. In this paper we derive for each k 8 a closed formula for the number of independent sets of cardinality k in an STS(v). We use the formula to prove that every STS(21) has
Intersection Numbers of Kirkman Triple Systems
Journal of Combinatorial Theory, Series A, 1999
Let J R (v) denote the set of all integers k such that there exists a pair of KTS(v) with precisely k triples in common. In this article we determine the set J R (v) for v#3 (mod 6) (only 10 cases are left undecided for v=15, 21, 27, 33, 39) and establish that J R (v)=I(v) for v#3 (mod 6) and v 45, where I(v)=[0, 1, ..., t v &6, t v &4, t v ] and t v = 1 6 v(v&1).