SICP Exercise 2.59

(define (union-set set1 set2)
  (cond ((null? set1) set2)
        ((not (element-of-set? (car set1) set2))
         (union-set (cdr set1) (cons (car set1) set2)))
        (else 
         (union-set (cdr set1) set2))))

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This entry was posted on Friday, June 26th, 2009 at 10:38 pm and is filed under Scheme, SICP. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to SICP Exercise 2.59

Pawel says:

January 8, 2011 at 7:29 pm

I think there should be also condition:

((null? set2) set1)

The procedure would be faster for this case.

Reply
Shunu says:

July 13, 2011 at 3:24 pm

Yes, I agree with Pawel.

Here is my code:

(define (union-set set1 set2)
(cond ((and (null? set1) (null? set2)) ‘())
((null? set1) set2)
((null? set2) set1)
((not (element-of-set? (car set1) set2))
(cons (car set1) (union-set (cdr set1) set2)))
(else (union-set (cdr set1) set2))))

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Bart Gauquie says:

January 10, 2012 at 11:29 am

(define (union-set set1 set2)
(fold-right adjoin-set set2 set1))

Reply