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Came here for this - I knew there must be an analytical solution involving Markov chains.
by gbrown 6y ago
Came here for this - I knew there must be an analytical solution involving Markov chains.
- tetha 6y agoI'm pretty sure this can be modeled with markov chains. Your vertices would be (x, y, f), f being facing. Edges are along the lines of (x, y, 0) - (x+1, y, {0,1,2,3}\1); (x, y, 1) - (x-1, y, {0,1,2,3}\0); (x, y, 2) - (x, y+1, {0,1,2,3}\3); (x, y, 3) - (x, y-1, {0,1,2,3}\2). Probabilities are modeled in the video and need to be rotated 4 times for the four facings. After that, I just don't recall my eigenvector calculation anymore. It feels like a good markov exercise, though.