嵌套映射

本小節代碼需要和”序列操作”那節的代碼在一起運行

;仿照了二重循環的方式
(define (make-pairs1 n)
  (define (i-iter i ans)
    (define (j-iter j ansi)
      (if (= j 0)
          ansi
          (j-iter (- j 1) (cons (list i j) ansi))))
    (if (= 0 i)
        ans
        (i-iter (- i 1) (append (j-iter (- i 1) nil) ans))))
  (i-iter n nil))
;
(define (flatmap proc seq)
  (accumulate append nil (map proc seq)))
(define (isprime? x)
  (define (smallest-divisor guess limit)
    (cond ((> guess limit) x)
          ((= (remainder x guess) 0) guess)
          (else (smallest-divisor (+ guess 1) limit))))
  (= (smallest-divisor 2 (sqrt x)) x))
(define (prime-sum? pair)
  (isprime? (+ (car pair) (cadr pair))))
(define (make-pair-sum pair)
  (list (car pair) (cadr pair) (+ (car pair) (cadr pair))))
(define (prime-sum-pairs n)
  (map make-pair-sum
       (filter prime-sum?
               (flatmap (lambda (i)
                          (map (lambda (j) (list i j))
                               (enumerate-interval 1 (- i 1))))
                        (enumerate-interval 1 n)))))
                                        ;
(define (remove item sequence)
  (filter (lambda (x) (not (= x item))) sequence))
(define (permutations s)
  (if (null? s)
      (list nil);attention!!
      (flatmap (lambda (x)
                 (map (lambda (sub-permutation)
                        (cons x sub-permutation))
                      (permutations (remove x s))))
               s)))
;2.40
(define (unique-pairs n)
  (flatmap (lambda (i)
             (map (lambda (j) (list i j));list only,because of 'prime-sum? and so on
                  (enumerate-interval 1 (- i 1))))
           (enumerate-interval 1 n)))
(define (prime-sum-pairs1 n)
  (map make-pair-sum
       (filter prime-sum?
               (unique-pairs n))))
;2.41
(define (3-pairs n)
  (define numbers (enumerate-interval 1 n))
  (flatmap (lambda (i)
             (flatmap (lambda (j)
                        (map (lambda (k) (list i j k))
                             (filter (lambda (x)
                                       (not (or (= x i) (= x j))))
                                     numbers)))
                      (filter (lambda (x) (not (= x i))) numbers)))
           numbers))

下面是八皇后問題，這裏方案的表示相當於將棋盤左右顛倒來看

;2.42
(define (queens board-size)
  (define rows (enumerate-interval 1 board-size))
  (define empty-board nil)
  (define (safe? k positions)
    (define new-row (car positions))
    (define (iter tmp-k tmp-positions)
      (if (null? tmp-positions)
          (= 1 1)
          (let ((tmp-row (car tmp-positions)))
            (if (or (= new-row (car tmp-positions))
                    (= (+ k new-row) (+ tmp-k tmp-row))
                    (= (- k new-row) (- tmp-k tmp-row)))
                (= 1 0)
                (iter (- tmp-k 1) (cdr tmp-positions))))))
    (iter (- k 1) (cdr positions)))
  (define (adjoin-position new-row k rest-of-queens)
    (cons new-row rest-of-queens))
  (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))
                         rows))
                  (queen-cols (- k 1))))))
  (queen-cols board-size))