递归笛卡尔积函数球拍笛卡尔、递归、球拍、函数

2023-09-03 08:49:06 作者:愿化蝶纷飞

我正在尝试实现一个递归函数来查找两个集合的笛卡尔乘积。我目前拥有的代码如下:

    (define (cartesian-product set-1 set-2)
        (let (b (set 2))
             (cond [(empty? set-1) '()]
                   [(empty? set-2)  (cartesian-product (rest set-1) b)] 
                   [else (append (list (list (first set-1) (first set-2))) (cartesian product set-1 (rest set-2)))]))))

然而,我的逻辑中有一些错误,我还不能准确地指出。如有任何帮助,我们将不胜感激!

推荐答案

笛卡尔心形函数表达式 说笛卡尔心形图,是数学史上最美公式我不服,那是你不记得这个了...

有两个循环而不是一个循环怎么样?

(define (cartesian-product set-1 set-2)
  (define (cartesian-product-helper element set)
    (if (empty? set)
        set
        (cons (list element (first set))
              (cartesian-product-helper element (rest set)))))
  (if (or (empty? set-1)
          (empty? set-2))
      empty
      (cons (cartesian-product-helper (first set-1) set-2)
            (cartesian-product (rest set-1) set-2))))
您发现了逻辑中的问题,并尝试将set-2(您键入为(set 2))保存在b中,但此值将在每次递归调用时被覆盖。如果改为调用helper函数,该函数将循环访问一个集合的所有元素以及另一个集合的第一个元素,您的问题将不复存在。

Welcome to DrRacket, version 6.1.1 [3m].
Language: racket; memory limit: 128 MB.
> (cartesian-product '(1 2 3) '(x y z))
'(((1 x) (1 y) (1 z))
  ((2 x) (2 y) (2 z))
  ((3 x) (3 y) (3 z)))
> (cartesian-product '(1 2 3) '())
'()
> (cartesian-product '() '(x y z))
'()

或者,更像球拍的东西:

(define (cartesian-product set-1 set-2)
  (if (or (empty? set-1)
          (empty? set-2))
      empty
      (for/list ([i set-1])
        (for/list ([j set-2])
          (list i j)))))