I was surprised none of the solutions exploited the symmetry of the graph. There are 4 sets of numbers you can be on in a path of length 2+, and it only matters

I was surprised none of the solutions exploited the symmetry of the graph. There are 4 sets of numbers you can be on in a path of length 2+, and it only matters which set you are in: {0}, {1, 3, 7, 9}, {2, 8} and {4, 6}. 5 should be special cased since only paths of length 1 are possible. That doesn't give you a big-O speedup, but it should be a 2x performance improvement for the algorithm that is linear in the number of hops.