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Apr 13, 2017 at 12:58 history edited CommunityBot
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Jan 16, 2012 at 20:45 comment added Seva @Ilya: right, this is what you get if you are not willing to sacrifice lower-order terms. My expression assumes a very marginal loss of accuracy, but a gain in clarity. (Just think of replacing $k-1$ with $k$ in the numerator of your fraction, and $n-k+1$ with $n$ in its denominator.)
Jan 16, 2012 at 13:20 answer added Ilya Bogdanov timeline score: 1
Jan 16, 2012 at 12:10 comment added Ilya Bogdanov It seems that a duoble-counting argument mentioned at the beginning leads to a bit different estimate, namely $\frac{Nn-2^{n-1}(k-1)}{n-k+1}$.
Dec 26, 2011 at 17:01 comment added Jyrki Lahtonen Andrew Klapper has worked on multi-covering problems (see cs.uky.edu/~klapper/multicov.html), but I don't know if the problem is too well-studied. What I was getting at with my other remark is that in coding theory balls are often called spheres. I know that this is bit against the standard of the rest of math :-)
Dec 23, 2011 at 8:15 comment added Seva @Jyrki: yes, the spheres are hollow; otherwise I would say Hamming balls. I am not sure of what exactly you mean by the problem of "bounding the size of multi-covering codes". It is quite possible that my question is related to some known coding-theory problems (and hence one of the tags); practically, I don't know of any such problem.
Dec 22, 2011 at 19:17 comment added Jyrki Lahtonen Are your spheres hollow? A bit non-standard, if you ask me. For example in the sphere packing bound of coding theory, the spheres consist of all the points at a distance at most the radius from the center. OTOH using the surface only may make the problem more interesting! I'm not an expert on variants of covering codes, but is your problem not a variant of bounding the size of multi-covering codes? Some people have worked on those.
Dec 16, 2011 at 15:47 history edited Seva CC BY-SA 3.0
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Dec 7, 2011 at 9:04 history edited Seva CC BY-SA 3.0
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Dec 7, 2011 at 8:45 history asked Seva CC BY-SA 3.0