有很多方法可以解决这个问题.对于初学者，我建议使用更简单的电路板表示法，我选择使用数字列表.列表中的索引从1开始，表示皇后的列及其行的值(坐标原点在左上角，新位置在列表末尾相邻)；所有其他位置都假定为空.例如，以下板:

There are many ways to solve this problem. For starters, I'd suggest a simpler representation for the board, I chose to use a list of numbers. The indexes in the list start from one and indicate the queen's column and the value its row (origin of coordinates is on the upper-left corner, new positions are adjoined at the end of the list); all the other positions are assumed to be empty. For instance, the following board:

(. Q)
(Q .)

将由列表 '(2 1) 表示.在我的表示中，safe? 过程看起来像这样 - 请注意 diagonals? 检查实现起来有点棘手:

Would be represented by the list '(2 1). With my representation, the safe? procedure looks like this - and notice that the diagonals? check is a bit trickier to implement:

; a new queen is safe iff there are no other queens in the same
; row nor in any of the diagonals preceding its current position
; we don't need to check the column, this is the only queen on it

(define (safe? col board)
  (let ((row (list-ref board (- col 1))))
    (and (<= (number-occurrences row board) 1)
         (diagonals? row board))))

; counts how many times an element appears on a list

(define (number-occurrences e lst)
  (count (curry equal? e) lst))

; traverses the board looking for other queens
; located in one of the diagonals, going backwards
; starting from the location of the newest queen

(define (diagonals? row board)
  (let loop ((lst   (cdr (reverse board)))
             (upper (sub1 row))
             (lower (add1 row)))
    (or (null? lst)
        (and (not (= (car lst) upper))
             (not (= (car lst) lower))
             (loop (cdr lst) (sub1 upper) (add1 lower))))))

结果如预期:

(safe? 4 '(2 4 1 3))
=> #t

如果您愿意，您可以修改上述代码以使用不同的坐标原点，或者使用坐标对来表示皇后.

You can adapt the above code to use a different origin of coordinates if you wish so, or to use pairs of coordinates to represent the queens.