Almost designs and their links with balanced incomplete block designs

Abstract

Almost designs (t-adesigns) were proposed and discussed by Ding as a certain generalization of combinatorial designs related to almost difference sets. Unlike t-designs, it is not clear whether t-adesigns need also be $(t-1)$-designs or $(t-1)$-adesigns. In this paper we discuss a particular class of 3-adesigns, i.e., 3-adesigns coming from certain strongly regular graphs and tournaments, and find that these are also 2-designs. We construct several classes of these, and discuss some of the restrictions on the parameters of such a class. We also construct several new classes of 2-adesigns, and discuss some of their properties as well.

Appendix

We will need some facts about cyclotomic classes and cyclotomic numbers. Let $q=ef+1$ be a prime power, and $\gamma$ a primitive element of the finite field $\mathbb {F}_{q}$ with q elements. The cyclotomic classes of order e are given by $D_{i}^{(e,q)}=\gamma ^{i}\langle \gamma ^{e} \rangle$ for $i=0,1,\ldots ,e-1$. The cyclotomic numbers of order e are given by $(i,j)_{e}=|D_{i}^{(e,q)}\cap (D_{j}^{(e,q)}+1)|$. It is obvious that there are at most $e^{2}$ different cyclotomic numbers of order e. When it is clear from the context, we will simply denote $(i,j)_{e}$ by (i, j).

We will need to use the cyclotomic numbers of order 2.

Lemma 7

[29] For a prime power q, if $q \equiv 1$ (mod 4), then the cyclotomic numbers of order two are given by

$$\begin{aligned} (0,0)= & {} \frac{q-5}{4}, \\ (0,1)= & {} (1,0) = (1,1) = \frac{q-1}{4}. \end{aligned}$$

If $q \equiv 3$ (mod 4) then the cyclotomic numbers of order two are given by

$$\begin{aligned} (0,1)= & {} \frac{q+1}{4}, \\ (0,0)= & {} (1,0) = (1,1) = \frac{q-3}{4}. \end{aligned}$$

Lemma 8

Let q be an odd prime power. Let $G=(\mathbb {F}_{q},+)\times (\mathbb {F}_{q},+)$, and define

$$\begin{aligned} D=\{(a,b)\in G\mid a\text { and }b\text { are both squares or both nonsquares}\}, \end{aligned}$$

and

$$\begin{aligned} \tilde{D}=\{(a,b)\in G\mid \text {one of }a,b\text { is square and the other is nonsquare}\}. \end{aligned}$$

Then both D and $\tilde{D}$ are $(q^{2},\frac{q^{2}-2q+1}{2},\frac{q^{2}-4q+7}{2},\frac{q^{2}-4q+3}{2})$ partial difference sets.

Proof

The case for D was shown in [32]. To show that for $\tilde{D}$, we count the number of solutions to the equation

$$\begin{aligned} (a,b)=(a_{1},b_{1})-(a_{2},b_{2}), \end{aligned}$$

(10)

where $(a_{1},b_{1}),(a_{2},b_{2})\in \tilde{D}$. We use a method similar to that used in [32].

Assume that a and b are both square. If $a_{1}$ and $a_{2}$ are square and $b_{1}$ and $b_{2}$ are nonsquare, then, using Lemma 7, the number of solutions to (10) is $(0,0)_{2}(1,1)_{2}$. There are three other cases depending on which of $a_{1},a_{2},b_{1}$ and $b_{2}$ are square and which are nonsquare, and the number of solutions to (10), as we run over these other possibilities, is one of $(0,1)_{2}(1,0)_{2},(1,0)_{2}(0,1)_{2}$ or $(1,1)_{2}(0,0)_{2}$. Summing over all four possibilities, the total number of solutions to (10) when a and b are both square is $\frac{q^{2}-4q+3}{4}$ (regardless of the residue of q modulo 4).

The other three cases where neither a nor b are zero can be argued similarly. When a and b are both nonsquare, the total number of solutions to (10) is $\frac{q^{2}-4q+3}{4}$, and when one of a and b is square and the other is nonsquare, the total number of solutions is $\frac{q^{2}-4q+7}{4}$. If $a\ne 0$ and $b=0$ then (10) becomes $(a_{2}a^{-1},b_{2})+(1,0)=(a_{1}a^{-1},b_{1})$ which, using Lemma 7 again, has $((0,0)_{2}+(1,1)_{2})\frac{q-1}{2}=\frac{q^{2}-4q+3}{4}$ solutions. A similar argument shows that when $a=0$ and $b\ne 0$ the number of solutions to (10) is again $\frac{q^{2}-4q+3}{4}$. Thus, if $x,y\in G$ are distinct, we have that each member of $\tilde{D}$ appears as a difference of two distinct members of $\tilde{D}$, $\frac{q^{2}-4q+7}{4}$ times, and each member of $G\setminus (\tilde{D}\cup \{(0,0)\})$ appears as a difference of two distinct members of $\tilde{D}$, $\frac{q^{2}-4q+3}{4}$ times. This completes the proof. $\square$