2
$\begingroup$

I am looking for ways to construct an infinite family of designs with parameters $2-(v,3,3)$ and apart from some doubling-type recursive constructions (such as in this paper) I haven't found anything in the literature.

So, are there "explicit" ways to construct a family of such designs?

$\endgroup$
4
  • $\begingroup$ When you say "design", do you mean a $2$-design? Usually designs have four parameters, and are then called $t$-$(v,k,\lambda)$ designs. $\endgroup$ Dec 10, 2013 at 9:59
  • $\begingroup$ @TomDeMedts Yes, $2$-designs. I made it clear now in the question. Thanks. $\endgroup$ Dec 10, 2013 at 10:00
  • 2
    $\begingroup$ Three copies of a Steiner triple system?? Or is that cheating too. $\endgroup$ Dec 10, 2013 at 13:02
  • $\begingroup$ @GordonRoyle Definitely not cheating but wouldn't work for my ulterior aims... $\endgroup$ Dec 10, 2013 at 14:54

3 Answers 3

7
$\begingroup$

An elementary counting argument shows that $2$-$(v,3,3)$ exists only if $v$ is odd (or, more precisely, for $\lambda \equiv 3 \pmod{6}$ a $2$-$(v,3,\lambda)$ exists only if $v \equiv 1 \pmod{2}$). This necessary condition is sufficient.

Arguably the simplest direct construction for the case $\lambda = 3$ that covers all odd $v$ is to use commutative quasigroups. Let $Q = (V, \otimes)$ be a commutative idempotent quasigroup of order $v \equiv 1 \pmod{2}$. A perpendicular array of strength $3$ is a ${{v+1}\choose{2}}\times 3$ array $L$ whose rows are the ordered triples $(x, y, x\otimes y)$ for $x, y \in V$ and $x \leq y$. Because $Q$ is idempotent, $L$ contains $v$ rows of the form $(x,x,x)$. Deleting such rows and ignoring the ordering on the rows, we obtain a set of triples. Every pair $\{a,b\}$ of distinct elements $a, b \in V$ occurs exactly three times in this set of triples, coming from the ordered triples $(a,b,a\otimes b)$, $(a,c,b)$ when $a \otimes c = c \otimes a = b$, and $(d,b,a)$ when $d\otimes b = b \otimes d =a$. Thus, it suffices to show that a commutative idempontet quasigroup exists for all odd $v = 2n+1$, which is true because we can define $\otimes$ as the binary operation $i\otimes j = (n+1)(i+j) \pmod{v}$ with $V = \{0,1,\dots,v-1\}$.

So, by defining $Q = (V, \otimes)$ as above, the construction is to take $(x,y,x\otimes y)$ for $x\leq y$, throw away $(x,x,x)$, and drop the ordering.

$\endgroup$
2
  • $\begingroup$ Thanks! I was actually going to post this myself, having spent a cozy afternoon with Colbourn & Rosa's Triple Systems. :) $\endgroup$ Dec 10, 2013 at 18:16
  • 1
    $\begingroup$ @Felix Ha ha. Looks like I beat you to it! This kind of construction is nicely explained in a very accessible way in "Design Theory" by Lindner and Rodger too if you're interested: books.google.com/books/about/… $\endgroup$ Dec 11, 2013 at 5:46
2
$\begingroup$

Yuichiro's is the best answer, but here's another "cheat" answer (which I nonetheless think is kind of neat):

Take a PBD$(v,\{3,5\})$, triple each block of size three, and replace each block of size five by the complete design $\binom{[5]}{3}$. Various PBD$(v,\{3,5\})$ can be found by cyclotomy.

$\endgroup$
1
$\begingroup$

{i,i+j,i+2j} for $i \in Z_{v}$ and $j=1,2,\ldots \frac{v-1}{2}$ should work.

$\endgroup$
1
  • 1
    $\begingroup$ This seems to be a simplification of Yuichiro Fujiwara's answer. $\endgroup$
    – S. Carnahan
    Jan 10, 2014 at 0:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.