Knight's Shortest Path on Chessboard

Question

I\'ve been practicing for an upcoming programming competition and I have stumbled across a question that I am just completely bewildered at. However, I feel as though it\'s

Answer 1

Yes, Dijkstra and BFS will get you the answer, but I think the chess context of this problem provides knowledge that can yield a solution that is much faster than a generic shortest-path algorithm, especially on an infinite chess board.

For simplicity, let's describe the chess board as the (x,y) plane. The goal is to find the shortest path from (x0,y0) to (x1,y1) using only the candidate steps (+-1, +-2), (+-2, +-1), and (+-2, +-2), as described in the question's P.S.

Here is the new observation: draw a square with corners (x-4,y-4), (x-4,y+4), (x+4,y-4), (x+4,y+4). This set (call it S4) contains 32 points. The shortest path from any of these 32 points to (x,y) requires exactly two moves.

The shortest path from any of the 24 points in the set S3 (defined similarly) to (x,y) requires at least two moves.

Therefore, if |x1-x0|>4 or |y1-y0|>4, the shortest path from (x0,y0) to (x1,y1) is exactly two moves greater than the shortest path from (x0,y0) to S4. And the latter problem can be solved quickly with straightforward iteration.

Let N = max(|x1-x0|,|y1-y0|). If N>=4, then the shortest path from (x0,y0) to (x1,y1) has ceil(N/2) steps.