Lunnon: It is a pleasure to report the defeat of the horrid spectre of topology, by dint of abandoning attempts to reason about attaching an extra step at the far end [x-2, y-1] in favour of considering (shorter) paths from neighbours [2, 1] etc. of the origin instead. Then straightforwardly the distance function d(x, y) satisfies the same recursion as the explicit expression f(x, y) , whence they are equal. A corrected and slightly expanded screed is posted at https://www.dropbox.com/s/nzmzjswtctju23f/knights_path.txt Reports of attempts to blow further holes in it would be welcome, even if unsuccessful. Particularly if unsuccessful! WDS: I don't agree that your lemma 2 is proven. (Same objection I gave before, I don't see that it was impacted in the slightest.) MInd you, I believe this general approach should work, just a lot more cases will be needed.