scheme word lists eq?

Question

i\'ve got a problem: I need to find if list equal to the second one, for example:

(set%eq? \'(1 2 3) \'(1 2 3))     ===> #t

(set%eq? \'(1 2 3) \'(2 3 4))             


        

Answer 1

Base on the comments from OP it's clear that these are set-eq?

(set-eq? '(a b c) '(c b a)) ; ==> #t
(set-eq? '(a b c) '(b c a)) ; ==> #t
(set-eq? '() '())           ; ==> #t
(set-eq? '(a b) '(a b c))   ; ==> #f
(set-eq? '(a b c) '(a c))   ; ==> #f

If the lists are without duplicates one could iterate the first list and try to find it in the second. If found we recurse with the two lists without the match.

#!r6rs
(import (rnrs)
        (rnrs lists))

(define (set-eq? xs ys)
  (if (null? xs) 
      (null? ys) ; #t if both sets are empty, otherwise #f
      (let ((cur (car xs)))        
        (and (memq cur ys) ; first element is found 
             (set-eq?  (remq  )))))) ; recurse without first in both lists

There are ways to get this faster. E.q. Hash the first list and iterate the second. If all matches and hashtable-size is the same as the number of iterations it's #t.