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A $t$-design on $v$ points with block size and index $\lambda$ is a collection $\mathcal{B}$ of subsets of a set $V$ with $v$ elements satisfying the following properties:

(a) every $B\in\mathcal{B}$ has $k$ elements,

(b) for every subset $T$ of $V$ with $t$ elements, there are exactly $\lambda$ sets $B\in \mathcal{B}$ such that $T\subset B$.

The Ray-Chaudhuri-Wilson inequality states that, for any $s\leq \min(t/2,v-k)$, $$b\geq \binom{v}{s},$$ where $b=|\mathcal{B}|$, the number of elements of $\mathcal{B}$.

Question: what happens if you drop the assumption that all sets in $\mathcal{B}$ have the same number of elements? Can you still give a lower bound on $b$ similar to the above?

Feel free to assume that every $B\in\mathcal{B}$ satisfies $|B|\leq 2 |V|/3$, say.

(Notes: 1) the case $t=2$ is known ("nonuniform Fisher's inequality"); 2) there are papers by Frankl-Wilson and Babai on "the nonuniform Ray-Chaudhuri-Wilson inequality", but they generalize something else from the same paper.)

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  • $\begingroup$ You mean $b=\vert \mathcal{B}\vert$? Also you've two different fonts for the collection $\mathcal{B}$. $\endgroup$ – T. Amdeberhan Dec 30 '16 at 23:10
  • $\begingroup$ We have $\lambda\binom{v}t=b\binom{k}t$ by counting pairs $T\subset B$, $|T|=t$, $B\in \mathcal{B}$, yes? So, $b$ may be eleminated from this inequality? $\endgroup$ – Fedor Petrov Dec 31 '16 at 12:20
  • $\begingroup$ If you insist... $\endgroup$ – H A Helfgott Dec 31 '16 at 14:19

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