递归笛卡尔积函数球拍

Question

我正在尝试实现一个递归函数来查找两个集合的笛卡尔积。我目前的代码如下：

    (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)))]))))

但是，我的逻辑有错误，我无法准确指出。感谢您的帮助！

Answer 1

用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))))

您在您的逻辑中发现了问题并试图在 b 中保存 set-2（您输入错误为 (set 2)），但此值将在每次递归时被覆盖称呼。如果您改为调用辅助函数，它循环遍历一组的所有元素以及另一组的第一个元素，您的问题就会消失。

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)))))

递归笛卡尔积函数球拍

Recursive Cartesian Product Function Racket

recursion

racket