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I am looking for a simple degree conditon that ensures the existence of a k-factor in a graph. The k is supposed to be relatively high and I don't mind the condition being a bit strict. Ideally, something of the form $\delta(G) \geq f(k)$. Any suggestions? 10x!

To clarify a bit what I'm after: there is a theorem by Nishimura that ensures a k-factor for k not larger than n/4 or so. But I want a k-factor with k approaching n.


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Note that $\delta(G) \geq f(k)$ will not do since there are graphs of arbitrarily high minimum degree without 1-factors (e.g $K_{2n+1}$). –  Tony Huynh Apr 4 '12 at 4:16
You are right. What's the simplest I can get, then? –  Felix Goldberg Apr 4 '12 at 8:34
Hold on a sec: $K_{2n+1}$ does have a 2-factor, right? So maybe it's still possible? –  Felix Goldberg Apr 4 '12 at 8:57
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1 Answer

up vote 3 down vote accepted

A similar theorem is proved in "Relating minimum degree and the existence of a k-factor" by Hartke, Martin and Seacrest.

They show that a graph $G$ on $n$ vertices with minimum degree $\delta\geq \frac{n}{2}$ contains a $k$-factor if $kn$ is even and $$k< \frac{\delta+\sqrt{2\delta n-n^2+8}}{2}.$$

Moreover they show that this is optimal up to a small additive constant ($\le 1$). Notice that as $\delta\to n$ we have $k\to n$, as well.

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Thanks, this looks promising! –  Felix Goldberg Apr 4 '12 at 23:33
Following your lead, I also googled now this paper: arxiv.org/pdf/0708.0202.pdf –  Felix Goldberg Apr 5 '12 at 0:04
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