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所以我需要做一个二十一点模拟器游戏，但似乎无法弄清楚洗牌有什么问题。它应该从牌堆中随机取出一张牌，然后将其放在牌堆的顶部。最后将其从其余部分中删除。所以 ：

(ace)(2)(3)(4)(5)...(k)
如果随机牌是 5
(5)(ace)(2)(3)(4)(5)...( k)
然后它删除第二个 5
(5)(ace)(2)(3)(4)(6)...(k)

这是代码：

    (define deck '((A . C) (2 . C) (3 . C) (4 . C) (5 . C) (6 . C) (7 . C) (8 . C) (9 . C) (10 . C) (V . C) (Q . C) (K . C)))

;auxilliary function for shuffle let you randomly select a card.
(define shuffAux
  (lambda (t)
    (define cardR
  (lambda (t)  (list-ref t (random 13))))
    (cardR t)))

;auxilliary function used to remove the card after the car to prevent
you from removing the randomly selected from the car(begining of the deck).
(define (removeDupC card deck)
      (delete card (cdr deck))
      )

(define shuffle2ndtry
  (lambda (deck seed)
    (define do-shuffle
      (lambda (deck seed)
        (if (> seed 0)(
        (cons (shuffAux deck) deck)
        (removeDupC (car deck)  deck)
        (- 1 seed))
        (write deck)   
        )
      )
      )
    (do-shuffle deck seed)))

(define (shuffle deck seed)
  (define cards (cons (shuffAux deck) deck))
  (write cards)
  (case (> seed 0)
   [(#t)
        (removeDupC (car cards) (cdr cards)) 
        (shuffle cards (- seed 1))]
   [(#f) (write cards)]))

(define random
 (let ((seed 0) (a 3141592653)
  (c 2718281829) (m (expt 2 35)))
  (lambda (limit)
   (cond 
   ((and (integer? limit))
    (set! seed (modulo (+ (* seed a) c) m))
    (quotient (* seed limit) m))
   (else
   (/ (* limit (random 34359738368))
   34359738368))))))


;function in which  you can delete an element from the list.
(define delete
  (lambda (item list)
    (cond
     ((equal? item (car list)) (cdr list))
     (else (cons (car list) (delete item (cdr list)))))))



(
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1 回答 1

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有一种更好的方法可以像这样洗牌。使用您的方法，可能会有一组卡片保持有序。

最好在牌组中循环，并为每张牌将其与随机位置交换。你会想为此使用一个向量。作为奖励，它会更快:)

于 2010-07-31T13:20:27.780 回答