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! Fred Lunnon On 4/3/14, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Warren's observation that my lemma proof failed to deal with both boundary edges of the first octant region is well spotted, though straightforward (if tedious) to repair.
The second problem he raises was obscured by my failure to state the inductive hypothesis explicitly. But the actual gremlin is unjustified assumption of the `obvious' d(x-2, y-1) < d(x, y) for [x, y] sufficiently far into first octant: so I ought not to rely on d(x-2, y-1) < f .
It begins to look as though the geometric material at the end of that screed may constitute a topological component essential to a successful proof.
Back to the drawing board! WFL