On 26/05/2014 17:50, Fred Lunnon wrote:
Anyway, if Gareth cares to show me (or all of us) his A3, I'll show him mine ...
OK. Note that the following is full of handwaving that I'm not strongly motivated to make more rigorous, especially as it may well just be wrong. So, we're interested in the number of shortest paths - from (0,0) to A := (4N, 3), and - from (0,0) to B := (3N, 3N-2). I think the shortest paths from O to A are made up of N-1 NEE, N SEE, 2 NNE -- (2N+1 choose N-1;N;2) or else N+1 NEE, N-2 SEE, 1 NNE, 1 SSE -- (2N+1 choose N+1;N-2;1;1) and the shortest paths from 0 to B are made up of N NNE, N-1 NEE, 1 SEE -- (2N choose N;N-1;1) If so, then the paths to A greatly outnumber the paths to B for large N. I should repeat at this point that I am by no means confident that any of this is correct. -- g