Question

我正在尝试实现递归函数来查找两组的笛卡尔积。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)))]))))

然而，我的逻辑存在错误，我无法精确定位。任何帮助表示赞赏！

Answer 1

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))))

您在逻辑中发现了该问题，并尝试在set-2中保存(set 2)（您在b中输入错误标记为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)) '()），但此值将在每次递归时被覆盖呼叫。如果你改为调用辅助函数，它循环遍历一组中的所有元素以及另一组的第一个元素，那么你的问题就会消失。

(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)))))

或者，类似于球拍的东西：

{{1}}

递归笛卡尔积函数球拍

1 个答案: