1
$\begingroup$

Let S = {1, .. ,n}.
Let H = (S, E) be the m-uniform hypergraph with r edges. Let F(H, k) = #{B | |B| = k, $\exists R \in E, B \cap R = \emptyset$ } - a number of k-subsets, that doesn't intersect with edges of H.

I am interested in following two problems:

Let r, m, k be fixed parameters.

  1. For which hypergraph H, F(H, k) be minimal?
  2. For which H, F(H, k) be maximal?

I have a conjecture for the first problem, but can't prove it.

Introduce a total order $\preceq$ on m-element subsets of S: $a \preceq b$ iff $\exists s$, such that $s \in a, s \notin b$ and $\forall i < s, i \in a \cap b \mbox{ or } i \notin a \cup b$.
Example: 3-subsets from 5 element set in ascending order according to $\preceq$.
11100
11010
11001
10110
10101
10011
01110
01101
01011
01111

Conjecture For all k minimal value of F(H, k) achieved on hypergraph, whose edges are r first sets from ${S \choose m}$ according to $\preceq$ order.

$\endgroup$

1 Answer 1

2
$\begingroup$

The answer to your first question should follow from a more or less straightforward application of Kruskal-Katona.

Namely, if you let $H^c=(S,E')$ be the family of complements, i.e. $X \in E'$ iff $S \setminus X \in E$, then $F(H,k)$ is simply the number of $k$-element subsets contained in edges of $H^c$, which is simply the $l$-th shadow of $H^c$, where $l=(n-m) - k$. You should get that the minimal value of $F(H,k)$ corresponds to the initial segment under the colex ordering for $H^c$; taking the complements of that family should resolve your conjecture.

$\endgroup$
1
  • $\begingroup$ Thank you! I did the same reformulation of problem as you, but i didn't know about Kruskal-Katona theorem. $\endgroup$
    – ailyr
    Mar 19, 2010 at 20:12

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.