Exercise 2.42
A solution to the eight-queens puzzle.
The “eight-queens puzzle” asks how to place eight queens on a chessboard so that no queen is in check from any other
(i.e., no two queens are in the same row, column, or diagonal). One possible solution is shown in figure 2.8.
One way to solve the puzzle is to work across the board, placing a queen in each column. Once we have placed k-1
queens, we must place the kth queen in a position where it does not check any of the queens already on the
board. We can formulate this approach recursively: Assume that we have already generated the sequence of all possible
ways to place k-1 queens in the first k-1 columns of the board. For each of these ways, generate an extended set
of positions by placing a queen in each row of the kth column. Now filter these, keeping only the positions
for which the queen in the kth column is safe with respect to the other queens. This produces the sequence
of all ways to place k queens in the first k columns. By continuing this process, we will produce not only one
solution, but all solutions to the puzzle.
We implement this solution as a procedure queens, which returns a sequence
of all solutions to the problem of placing n queens on an n×n chessboard. Queens has an internal procedure
queen-cols that returns the sequence of all ways to place queens in the first k columns of the board.
(define (queens board-size)
(define (queen-cols k)
(if (= k 0)
(list empty-board)
(filter
(lambda (positions) (safe? k positions))
(flatmap
(lambda (rest-of-queens)
(map (lambda (new-row)
(adjoin-position new-row k rest-of-queens))
(enumerate-interval 1 board-size)))
(queen-cols (- k 1))))))
(queen-cols board-size))
In this procedure rest-of-queens is a way to place k-1 queens in the first k-1 columns, and new-row is a
proposed row in which to place the queen for the kth column. Complete the program by implementing the
representation for sets of board positions, including the procedure adjoin-position, which adjoins a new row-column
position to a set of positions, and empty-board, which represents an empty set of positions. You must also write the
procedure safe?, which determines for a set of positions, whether the queen in the kth column is safe with
respect to the others. (Note that we need only check whether the new queen is safe – the other queens are already
guaranteed safe with respect to each other.)
(define (filter pred? seq)
(cond ((null? seq) '())
((pred? (car seq)) (cons (car seq) (filter pred? (cdr seq))))
(else
(filter pred? (cdr seq)))))
(define (flatmap pred seq)
(accumulate append '() (map pred seq)))
(define (accumulate op init seq)
(if (null? seq)
init
(op (car seq)
(accumulate op init (cdr seq)))))
(define (enumerate-interval low high)
(if (> low high)
'()
(cons low (enumerate-interval (+ 1 low) high))))
(define (queens board-size)
(define (queen-cols k)
(if (= k 0)
(list empty-board)
(filter
(lambda (positions) (safe? k positions))
(flatmap
(lambda (rest-of-queens)
(map (lambda (new-row)
(adjoin-position new-row k rest-of-queens))
(enumerate-interval 1 board-size)))
(queen-cols (- k 1))))))
(queen-cols board-size))
(define empty-board '())
(define (adjoin-position new-row k rest-of-queens)
(append rest-of-queens (list new-row) ))
(define (safe? k positions)
(let ((new-q-row (list-ref positions (- k 1))))
(safe-queen? new-q-row k positions 1)))
(define (safe-queen? new-q-row new-q-col positions col)
(cond ((null? positions) #t)
((and (= new-q-col col)
(= new-q-row (car positions))) #t) ;check for last col
((= new-q-row (car positions)) #f) ;rook moves
((=
(abs (- new-q-row (car positions))) ;bishop moves
(abs (- new-q-col col))) #f)
(else
(safe-queen? new-q-row new-q-col (cdr positions) (+ col 1)))))
> (length (queens 8))
92
> (length (queens 11))
2680
> (length (queens 10))
724
> (queens 4)
((2 4 1 3) (3 1 4 2))