Wikipedia:Reference desk/Archives/Mathematics/2019 December 4

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December 4

number of corners + one offs?

For an N-cube, what is the maximum number of non overlapping sets where each set consists of a specific corner and the corners one away from it? So for N=2, the answer is 1, since that takes up three of the corners, for N=3, the answer is 2, and combined they cover everything. For N=4, only two can be put in, even though 3 would cover 15 of the 16 corners, they can't all be put in. Any idea how to calculate the number for any given N?Naraht (talk) 14:37, 4 December 2019 (UTC)

@Naraht: I do not have an answer, but each set contains N+1 vertices and there are 2^N vertices in total, so the upper bounding is ${\frac {2^{N}}{N+1}}$. --CiaPan (talk) 20:19, 4 December 2019 (UTC)

The relevant language to look for what's known here is coding theory, particularly error-correcting codes: you're asking for the maximum size of a binary code with minimum Hamming distance 3. The bound given by CiaPan is the Hamming bound (in the case q = 2, d = 3). --JBL (talk) 20:41, 4 December 2019 (UTC)

The answer to this particular question is A005864, whose known terms (beginning with N = 1) are 1, 1, 2, 2, 4, 8, 16, 20, 40, 72, 144, 256, 512, 1024, and 2048. (See chapter 9 of J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups" -- I was able to view it on google books but I don't know if that will be true for everyone.) In particular, the answer is not known for N > 15. --JBL (talk) 23:21, 4 December 2019 (UTC)

Joel B. Lewis Thanx. Can't see the book right now, that chapter isn't part of the preview for me. I'll look at the other oeis sources.Naraht (talk) 02:58, 7 December 2019 (UTC)

Fixed syntax of {{U}} template. --CiaPan (talk) 20:59, 9 December 2019 (UTC)