BishopMove

Problem:

TopCoder Problem Statement - BishopMove

Overview:

Count the number of moves it takes for a bishop to move from one square to another on a chess board.

Java Source:

01: /*
02: TopCoder
03: Single Round Match: 628
04: Division: 2
05: Level: 1
06: Points: 250
07: Description: http://community.topcoder.com/stat?c=problem_statement&pm=13280
08:  */
09: 
10: public class BishopMove {
11: 
12:  public int howManyMoves(int r1, int c1, int r2, int c2) {
13: 
14:         // Get the number of rows and columns that the bishop must move.
15:         int rowDiff = Math.abs(r2 - r1);
16:         int colDiff = Math.abs(c2 - c1);
17: 
18:         //  Check for no moves necessary
19:         if ((rowDiff == 0) && (colDiff == 0)) return 0;
20: 
21:         /*
22:          * If the number of rows that change is the same as the number of
23:          * columns that change, we can reach it in one move.
24:          */
25:         if (rowDiff == colDiff) return 1;
26: 
27:         /*
28:          * If the difference is not a multiple of 2 (i.e. up 1 row,
29:          * and over 2) then the square cannot be reached.
30:          */
31:         if (((rowDiff - colDiff) % 2) != 0) return -1;
32: 
33:         // In any other case, the square can be reached in 2.
34:         return 2;
35: 
36:  }
37: }

Notes:

It helps to think in terms of the basic properties of how a Bishop moves. The following three rules are all you need to solve this problem:

1. If the change in the number of rows from the starting square to the destination is the same as the change in the number of columns; then the Bishop can reach that square in 1 move.
2. From it's starting position, the Bishop can reach any square where the difference in rows minus the difference in columns is an even multiple of 2.
3. Any square that can be reached by the Bishop will take at most 2 moves.

This logic is implemented on lines 25, 31, and 34, and respectively.

Add to that a quick check at line 19 to see if any move is even necessary, and you're done.

Thank you for taking the time to read this solution. I welcome any feedback you may have.