方案：从嵌套列表中删除元素

Question

给定一个命题公式，即i。即((a and (b implies c) or (d and (e implies f))))， 我需要写一个计划函数删除连接词，and，implies，or。 函数的返回值包含公式中的所有变量。例如， 。

我不知道该如何开始，因为我不确定如何进入嵌套列表并删除某些变量和连接符。

Answer 1

我就开始了类似以下内容：

(define somelist 
    (list 'a 'and (list 'b 'implies 'c) 
     'or (list 'd 'and (list 'e 'implies 'f)))) 

(define (remove-vars xs ys) 
    (let ((xs-f (flatten xs))) 
    (filter-two xs-f ys))) 

(define (filter-two xs ys) 
    (foldr (lambda(y acc) 
      (filter (lambda(x) (not (eq? x y))) acc)) 
     xs 
     ys)) 

测试：

> (remove-vars somelist (list 'and 'or 'implies)) 
(a b c d e f) 
> (remove-vars somelist (list 'and 'or)) 
(a b implies c d e implies f) 

UPDATE：OK，@ karategeek6报道，说他没有flatten和filter在他Scheme解释，我不知道你做什么，所以让我们手动实现它们，因为在R 1 6RS要么没有filter和flatten：

(define (my-flatten xs) 
    (foldr 
    (lambda(x acc) 
    (if (list? x) 
     (append (my-flatten x) acc) 
     (cons x acc))) 
    (list) 
    xs)) 

(define (my-filter pred xs) 
    (let recur ((xs xs) 
       (acc (list))) 
    (if (empty? xs) 
     (reverse acc) 
     (if (pred (car xs)) 
      (recur (cdr xs) (cons (car xs) acc)) 
      (recur (cdr xs) acc))))) 

修改remove-vars和filter-two适当：

(define (remove-vars xs ys) 
    (let ((xs-f (my-flatten xs))) 
    (filter-two xs-f ys))) 

(define (filter-two xs ys) 
    (foldr (lambda(y acc) 
      (my-filter (lambda(x) (not (eq? x y))) acc)) 
     xs 
     ys)) 

你应该得到的输出与上面的程序以前的版本相同。