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Let $H = ( V, E )$ be a $k$-uniform connected hypergraph, with $n = |V|$ vertices and $m = |E|$ hyperedges. Let $O_w$ be the number of edge induced subgraphs of $H$ having $w$ vertices and an odd number of hyperedges. Let $E_w$ be the number of edge induced subgraphs of $H$ having $w$ vertices and an even number of hyperedges. Let $\Delta_w = O_w - E_w$.

Let $b_w$ be the number of bits required to encode $\Delta_w$. Let $b = \displaystyle\max_{\substack{w}} b_w$.

I'm interested in how $b$ grows. I would like to determine the best possible upper bound for $b$ which is expressible as a function of only $n$, $m$ and $k$. More precisely, I would like to determine a function $f(n,m,k)$ having both the following properties:

  1. $b \leq f( n, m, k )$ for any $k$-uniform hypergraph $H$ having $n$ vertices and $m$ hyperedges.
  2. $f(n,m,k)$ grows slower than any other function which satisfies the 1st property.

In general both $O_w$ and $E_w$ are exponential in $m$, therefore I expect that their difference $\Delta_w$ is not exponential in $m$ and thus that $b \in o( m )$.

However for the moment I've no clue on how to try to prove this.

Questions

  • How does $b$ grow with respect to $n$, $m$ and $k$?
  • Are there any relevant results in the literature?
  • Any hint on how to try to prove $b \in o(m)$?

Update 13/09/2013
Here are some clarifications:

  • $O_w$ is the number of distinct edge-induced subgraphs in $H$. Repetitions are not allowed. The same holds for $E_w$ of course.
  • By "the number of bits required to encode $\Delta_w$" I mean $log_2 \ \Delta_w$.
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Can you explain what is meant by edge-induced? Especially, will the subgraphs be connected and k-uniform for large w? –  The Masked Avenger Sep 7 '13 at 15:49
    
If the subgraphs are not k-uniform, looking at k-complete graphs should get b greater than 2n - logn. –  The Masked Avenger Sep 7 '13 at 16:00
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Of course every edge induced subgraph is $k$-uniform. –  Giorgio Camerani Sep 7 '13 at 18:26
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Even so, complete hypergraphs should set the bar pretty high for b. –  The Masked Avenger Sep 7 '13 at 19:20

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