25 May
2014
25 May
'14
4:37 p.m.
Q2. Prove(!) that there is just a single shortest knight path from (0,0) to (2r,r) taking distance r steps.
Consider the `energy' E = 2x + y as a function of time. This can increase by at most 5 per move, and clearly the only move that increases it by 5 is (+2, +1). Quod erat demonstrandum. Sincerely, Adam P. Goucher