April 98 Question 2 Revised

This a more efficient implementation of Question 2.

  boolean valid_position(boolean[][] board, int row, int column)
  {
   if (board[i][j])
     return false;
   else {
     int i,j;
     // check each square in the column
     j = column;
     for (i = 0; i < board.length; i++) {
       if (board[i][j])
         return false;
     }
     // check each square in the row
     i = row;
     for (j = 0; j < board[i].length; j++) {
       if (board[i][j])
         return false;
     }
     //check diagonals
     int incr_i, incr_j;
     return 
	(diagonal(-1,-1,board,row,column) && diagonal(1,-1,board,row,column)
	&& diagonal(1,1,board,row,column) && diagonal(-1,1,board,row,column));
    }	
  }


  boolean diagonal(int incr_i, int incr_j, boolean[][] board, int row,
	           int column)	    
  {      
     int i = row + incr_i; 
     int j = column + incr_j;
     while ( i < board.length && j < board[i].length && i>=0 && j>=0)
     {
       if (board[i][j]) return false;
       i += incr_i;
       j += incr_j;     
     }   
  }