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This question caught my atention recently. Most (if not all) of the answers aproached the problem via a brute force attack. Surely there is a more elegant way to deal with this, given the inherent symmetry.

My question is can this be generalised to an $n$ dimensional lattice? Clearly the placement of the nearest neighbour is critical in this problem (ie, it is unclear where $n=5$ in $2$ dimensions would be places in terms of the classical Descatrian coordinate system), but it could also be generalised to $n$ in any dimension. Is this a Hamiltonian path (ie NP hard problem), or can the symmetry be utilised to present a more analytic solution?

Note

If standard MSE policy of questions being self-contained is desired, please note, and I will alter.

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  • $\begingroup$ What exactly do you mean by symmetry? $\endgroup$ – Matt Samuel Dec 2 '15 at 23:50
  • $\begingroup$ @MattSamuel Clearly one vertec needs to be consisered and mutliplied by 4. The same goes of the mid- sides, and the centre is the exception. The problem lies of course in the topological issue of not retracing your steps. $\endgroup$ – martin Dec 2 '15 at 23:52
  • $\begingroup$ Surely there is a more elegant way to deal with this, given the inherent symmetry. - Surely there is a God, given the inherent order present in the Universe. $\endgroup$ – Lucian Dec 3 '15 at 2:18
  • $\begingroup$ @Lucien oh, the irony ;) $\endgroup$ – martin Dec 3 '15 at 2:19
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In general your "dots" form the vertices of a graph, and I think you're asking for the number of self-avoiding walks of a given length on this graph. There is a huge literature on self-avoiding walks. In general, counting self-avoiding walks is #P-complete, and thus very difficult: see e.g. this paper.

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  • $\begingroup$ this is an interesting comment. I suppose the question of how to apply it will rely on my own research - made easier by the link you provided - many thanks :) $\endgroup$ – martin Dec 2 '15 at 23:54

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