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I am interested in shortest paths in the Cayley graph of the alternating group $A_{12}$ acting on the vertices of the icosahedron, where the generators are given by 5-cycles on the neighbors of any particular vertex.

Is there a decent algorithm for computing shortest paths in such a highly symmetric graph, given an explicit list of the generators? Brute force is doable, since there are only $12!/2$ different elements, but it would be nice to have a faster algorithm if one is available.

Background: If you place 12 unit spheres around a central unit sphere in 3D in an icosahedral configuration, each such generator can be realized without intersections or loss of contact by moving the 5 neighbors of an outer sphere P towards P inwards and spinning them around. http://en.wikipedia.org/wiki/Kissing_number#cite_note-1

Geoffrey Irving
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    I find this a strange question, because the generators of $A_{12}$ that you describe are not isometries of the icosahedron (its isometry group is isomorphic to $A_5 \times C_2$), so why are you interested in this particular generating set? In general, finding shortest paths in Cayley graphs is computationally difficult, and the algorithms used are not much cleverer than brute force. Just consider how hard this is for the Rubik cube group! – Derek Holt Mar 26 '13 at 10:03
  • I added the motivation for this particular set of generators. By "not much cleverer than brute force", do you mean the same order exponential? I would be quite happy with a birthday paradox-ish algorithm that achieved $\sqrt{|A_{12}|}$ time or something similar, for example. – Geoffrey Irving Mar 26 '13 at 16:22
  • This is a quite related question: http://mathoverflow.net/questions/90923/shortest-path-in-cayley-graphs. Unfortunately, frontier search doesn't help much here because the layer sizes are exponential in the distance. – Geoffrey Irving Mar 26 '13 at 16:50

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Let $G$ be a finite group and $A \subset G$ a set of generators closed under inversion. Given $g \in G$, we seek a minimum length sequence $$g = a_1 \cdots a_k.$$ A sequence of length $k$ exists iff $\exists a_1 \cdots a_k = g$ iff $\exists a_1 \cdots a_{\lfloor k/2 \rfloor} = g a_1 \cdots a_{\lceil k/2 \rceil}$ iff $\exists a_L \in A^{\lfloor k/2 \rfloor}, a_R \in A^{\lceil k/2 \rceil}$ s.t. $a_L = g a_R$ iff $A^{\lfloor k/2 \rfloor} \cap g A^{\lceil k/2 \rceil} \ne \emptyset$.

If $|A^k|$ is exponential in $k$ up to near the diameter of $G$, this is a substantial savings over full brute force (something like $O(\sqrt{|G|})$ depending on the structure of $G$).

Geoffrey Irving
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  • This could be reduced to cube root time given http://cstheory.stackexchange.com/questions/17039/checking-whether-ab-intersects-c-in-a-group (which may be unlikely). – Geoffrey Irving Mar 27 '13 at 00:41