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Tue, 13 Nov 2018
Counting paths through polyhedra
A while back someone asked on math stack exchange how many paths there were of length !!N!! from one vertex of a dodecahedron to the opposite vertex. The vertices are distance 5 apart, so for !!N<5!! the answer is zero, but the paths need not be simple, so the number grows rapidly with !!N!!; there are 58 million paths of length 19. This is the kind of thing that the computer answers easily, so that's where I started, and unfortunately that's also where I finished, saying:
Another user reminded me of this and I took another whack at it. I couldn't remember what my idea had been last year, but my idea this time around was to think of the dodecahedron as the Cayley graph for a group, and then the paths are expressions that multiply out to a particular group element. I started by looking at a tetrahedron instead of at a dodecahedron, to see how it would work out. Here's a tetrahedron. Let's say we're counting paths from the center vertex to one of the others, say the one at the top. (Tetrahedra don't have opposite vertices, but that's not an important part of the problem.) A path is just a list of edges, and each edge is labeled with a letter !!a!!, !!b!!, or !!c!!. Since each vertex has exactly one edge with each label, every sequence of !!a!!'s, !!b!!'s, and !!c!!'s represents a distinct path from the center to somewhere else, although not necessarily to the place we want to go. Which of these paths end at the bottom vertex? The edge labeling I chose here lets us play a wonderful trick. First, since any edge has the same label at both ends, the path !!x!! always ends at the same place as !!xaa!!, because the first !!a!! goes somewhere else and then the second !!a!! comes back again, and similarly !!xbb!! and !!xcc!! also go to the same place. So if we have a path that ends where we want, we can insert any number of pairs !!aa, bb, !! or !!cc!! and the new path will end at the same place. But there's an even better trick available. For any starting point, and any letters !!x!! and !!y!!, the path !!xy!! always ends at the same place as !!yx!!. For example, if we start at the middle and follow edge !!b!!, then !!c!!, we end at the lower left; similarly if we follow edge !!c!! and then !!b!! we end at the same place, although by a different path. Now suppose we want to find all the paths of length 7 from the middle to the top. Such a path is a sequence of a's, b's, and c's of length 7. Every such sequence specifies a different path out of the middle vertex, but how can we recognize which sequences end at the top vertex? Since !!xy!! always goes to the same place as !!yx!!, the order of the
seven letters doesn't matter.
A complicated-seeming path like Since the paths we want are those that go to the same place as the trivial path !!c!!, we want paths that have an even number of !!a!!s and !!b!!s and an odd number of !!c!!s. Any path fitting that description will go to same place as !!c!!, which is the top vertex. It's easy to enumerate such paths:
Here something like “cccbbbb” stands for all the paths that have three c's and four b's, in some order; there are !!\frac{7!}{4!3!} = 35!! possible orders, so 35 paths of this type. If we wanted to consider paths of arbitrary length, we could use Burnside's lemma, but I consider the tetrahedron to have been sufficiently well solved by the observations above (we counted 547 paths by hand in under 60 seconds) and I don't want to belabor the point. Okay! Easy-peasy! Now let's try cubes: Here we'll consider paths between two antipodal vertices in the upper left and the lower right, which I've colored much darker gray than the other six vertices. The same magic happens as in the tetrahedron. No matter where we
start, and no matter what !!x!! and !!y!! are, the path !!xy!! always
gets us to the same place as !!yx!!. So again, if some complicated
path gets us where we want to go, we can permute its components into
any order and get a different path of the same length to the same
place. For example, starting from the upper left, And again, because !!xx!! always make a trip along one edge and then
back along the same edge, it never goes anywhere. So the three paths
in the previous paragraph also go to the same place as We want to count paths from one dark vertex to the other. Obviously
To get paths of length 5, we must insert a pair of matching letters
into one of the paths of length 3. Without loss of generality we can
assume that we are inserting To get paths of length 7, we must insert two pairs. If the two pairs are the same, there are !!\frac{7!}{5!} = 42!! possible orders and 3 choices about which letters to insert, for a total of 126. If the two pairs are different, there are !!\frac{7!}{3!3!} = 140!! possible orders and again 3 choices about which pairs to insert, for a total of 420, and a grand total of !!420+126 = 546!! paths of length 7. Counting the paths of length 9 is almost as easy. For the general case, again we could use Burnside's lemma, or at this point we could look up the unusual sequence !!6, 60, 546!! in OEIS and find that the number of paths of length !!2n+1!! is already known to be !!\frac34(9^n-1)!!. So far this technique has worked undeservedly well. The original problem wanted to use it to study paths on a dodecahedron. Where, unfortunately, the magic property !!xy=yx!! doesn't hold. It is possible to label the edges of the dodecahedron so that every sequence of labels determines a unique path: but there's nothing like !!xy=yx!!. Well, nothing exactly like it. !!xy=yx!! is equivalent to !!(xy)^2=1!!, and here instead we have !!(xy)^{10}=1!!. I'm not sure that helps. I will probably need another idea. The method fails similarly for the octahedron — which is good, because I can use the octahedron as a test platform to try to figure out a new idea. On an octahedron we need to use four kinds of labels because each vertex has four edges emerging from it: Here again we don't have !!(xy)^2=1!! but we do have !!(xy)^3 = 1!!. So it's possible that if I figure out a good way to enumerate paths on the octahedron I may be able to adapt the technique to the dodecahedron. But the octahedron will be !!\frac{10}3!! times easier. Viewed as groups, by the way, these path groups are all examples of Coxeter groups. I'm not sure this is actually a useful observation, but I've been wanting to learn about Coxeter groups for a long time and this might be a good enough excuse. [Other articles in category /math] permanent link |