Consider a bi-infinite chessboard (Z^2). What is the distance between (x1,y1) and (x2,y2) measured in number of knight-moves? [Incidentally, for a finite hXw chessboard, if min(h,w)>3 the bi-infinite and finite board exhibit the same distances in all cases except if the two squares are the corner of the board and diagonally adjacent to it, in which case the finite board has distance=4 while the bi-infinite board distance is 2.] After some fiddling, I came up with the following CONJECTURE: Define dx = |x1-x2|, dy = |y1-y2|, L1distance = dx + dy, KingDistance = max(dx, dy) F = ( max(2*L1distance, 3*KingDistance) + 4 ) / 6 Then KnightDistance = F rounded to the nearest eligible integer of the correct parity (correct parity: even if source and destination same-colored squares, otherwise odd) (eligible: must be non-negative, and if the two squares differ then must be positive) except for this list: (dx,dy) KnightDistance Formula (1,0) or (0,1) 3 1 (2,2) 4 2 (7,7) 4 6 in which my formula is off by 2. Are there any other exceptions? -- Warren D. Smith http://RangeVoting.org