常用Lisp:使用(let)计算递归函数

Question

所以我写了一些东西，返回一个最大子集和，给出一个正整数列表。但是，我想使用(let)，以便使代码更高效。我想知道这是否可能或如何可能。

(defun subset-sum1 (numbers capacity)
  (subset-sum numbers capacity 0))

(defun subset-sum (numbers capacity counter) 
  (cond ((null numbers) counter)
        ((= counter capacity) counter) ;; return if counter 'hits' the capacity
        ((< capacity (+ (car numbers) counter))
         (subset-sum (cdr numbers) capacity counter)) ;; cdr if car branch exceeds capacity
        ((<= (subset-sum (cdr numbers) capacity counter)
             (subset-sum (cdr numbers) capacity (+ (car numbers) counter)))
         (subset-sum (cdr numbers) capacity (+ (car numbers) counter))) ;; choose car
        (t (subset-sum (cdr numbers) capacity counter)))) ;; choose cdr

上面的代码在普通lisp中工作得很好。但是，我想做下面这样的事情，因为我觉得使用let会使代码更好。但是我写的是一个无限循环:(这是一个人工智能入门类的作业.帮帮新手吧！

(defun subset-sum (numbers capacity counter)
  (let ((exclude (subset-sum (cdr numbers) capacity counter))
   (include (subset-sum (cdr numbers) capacity (+ (car numbers) counter))))
     (cond ((null numbers) counter)
       ((= counter capacity) counter)
       ((< capacity (+ (car numbers) counter)) exclude)
       ((<= exclude include) include)
       (t (exclude)))))

Answer 1

由于这些行，您得到了一个无限循环：

(defun subset-sum (numbers capacity counter)
  (let ((exclude (subset-sum (cdr numbers) capacity counter))

您一直递归地调用子集和，并且它永远不会终止。即使到了列表的末尾，而且numbers是()，也可以继续下去，因为(cdr '())是'()。您在原始版本中通过检查(空号)来处理这个问题(尽管使用(endp数字)可能更符合惯例)。现在你可以做这样的事情：

(defun subset-sum (numbers capacity counter)
  (if (endp numbers)
    counter
    (let ((exclude (subset-sum (cdr numbers) capacity counter))
          ;...
         )
      (cond ; ...
          ))))