How to write a scheme function that takes two lists and returns four lists

前端 未结 4 1615
后悔当初
后悔当初 2020-12-18 11:32

I have 2 lists of elements \'(a b c) \'(d b f) and want to find differences, union, and intersection in one result. Is that possible? How?

I wrote a member function

相关标签:
4条回答
  • 2020-12-18 11:47

    Sure it is possible. Assuming that you have function to compute the differences, union intersection etc:

     (define (checkResult lis1 list2)
       (list (difference lis1 lis2)
            (union ...
    
    0 讨论(0)
  • 2020-12-18 11:49

    Sure it's possible. Here are a couple hints:

    1. what's the result of combining a list and an empty list?
    2. You don't have to do it all at once. Take a piece at a time.
    0 讨论(0)
  • 2020-12-18 12:12

    Like others have said, all you need to do is create separate functions to compute the intersection, union, and subtraction of the two sets, and call them from checkresult:

    (define (checkresult a b)
      (list (subtract a b)
            (subtract b a)
            (union a b)
            (intersect a b)))
    

    Here are some example union, intersection, and subtraction functions:

    (define (element? x lst)
      (cond ((null? lst) #f)
            ((eq? x (car lst)) #t)
            (#t (element? x (cdr lst)))))
    
    (define (union a b)
      (cond ((null? b) a)
            ((element? (car b) a)
             (union a (cdr b)))
            (#t (union (cons (car b) a) (cdr b)))))
    
    (define (intersect a b)
      (if (null? a) '()
          (let ((included (element? (car a) b)))
            (if (null? (cdr a))
                (if included a '())
                (if included
                    (cons (car a) (intersect (cdr a) b))
                    (intersect (cdr a) b))))))
    
    (define (subtract a b)
      (cond ((null? a) '())
            ((element? (car a) b)
             (subtract (cdr a) b))
            (#t (cons (car a) (subtract (cdr a) b)))))
    

    Note: since these are sets and order doesn't matter, the results are not sorted. Also, the functions assume that the inputs are sets, and therefore don't do any duplicate checking beyond what's required for union.

    0 讨论(0)
  • 2020-12-18 12:14

    On top of Charlie Martin's and tomjen's answers, I have come up with this source:

    Union Intersection and Difference

    Implementation of the distinct functions can be found with nice explanations.

    0 讨论(0)
提交回复
热议问题