12 Apr
2014
12 Apr
'14
1:57 p.m.
Correction: I should not have excluded the case L=0. --Dan On Apr 12, 2014, at 11:33 AM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
Let f(K,L) := the smallest number of knight moves {(+-2,+-1),(+-1,+-2)} it takes to get from square (0,0) to square (K,L) on an infinite chessboard.
WLOG assume K,L > 0.