CODEWITHSUNDEEP
functional-programmingscheme
user2110714
user2110714

Reputation:

How can i use the removing for the following in scheme?

I am given two lists in scheme and check whether we can form the one of them from the elements of the other. In other words, i check whether this is an . To do this, i implemented a function called member?, which takes a symbol and a list, and if this symbol is in the list then this function removes that symbol from the list and returns the new list. The sample input-output for the function is like the following:

I am thinking about doing the following: Since we have two lists to check the , i take the first list and for each symbol in that list, by using member? function, i check whether that symbol occurs in the other list. Finally at the end, if it is an then we have an empty list. Here is some trials to do this:

(define member?
 (lambda (inSym inSeq)
(if (and (symbol? inSym) (sequence? inSeq)) ; can remove?       
   (remove-member inSym inSeq)             ; then remove!       
   'can-not-remove))) ; otherwise, present an error message

(define 
 (lambda (inSeq1 inSeq2)
    (if (and (sequence? inSeq1) (sequence? inSeq2)) ;if both are sequences
        (if(equal? '() (member? (car inSeq1) inSeq2))) ... ???
 )
)

I could not organize the recursion needed here. Can anyone help me with this?

Thank you.

Upvotes: 1

Views: 107

Answers (3)

Óscar López
Óscar López

Reputation: 235984

Just for the record: the simplest way to check if two sequences are anagrams is to sort them and see if they're equal:

(define symbol<?
  (lambda (s1 s2) ; compare two symbols
    (string<? (symbol->string s1)
              (symbol->string s2))))

(define anagram?
  (lambda (inSeq1 inSeq2)
    (and (sequence? inSeq1) ; a sequence is a list of symbols
         (sequence? inSeq2) ; a sequence is a list of symbols
         (equal? (sort inSeq1 symbol<?) (sort inSeq2 symbol<?)))))

Upvotes: 2

faisal
faisal

Reputation: 1339

Using sort is not efficient and for your implementation You do not need the member? function:

(define remove-member
(lambda (n lst)
(cond ((null? lst)
       '())
      ((equal? (car lst) n)
       (cdr lst))
      (else
       (cons (car lst)
             (remove-member n (cdr lst)))))))



(define anagram?
(lambda (lst1 lst2)
(cond ((and (null? lst1) (null? lst2))
       #t)
((or (null? lst1) (null? lst2))
       #f)
      (else
       (anagram? (cdr lst1) (remove-member (car lst1) lst2))
))))

Upvotes: 2

Jean-Bernard Pellerin
Jean-Bernard Pellerin

Reputation: 12670

(define canFind?
  (lambda (item sequence)
    (if (null? sequence)
      #f
      (if (equal? item (car sequence))
        #t
        (canFind? item (cdr sequence))
      )
    )
  )
)

(define remove
  (lambda (item sequence)
    (if (equal? item (car sequence))
      (cdr sequence)
      (append (list (car sequence)) (remove item (cdr sequence)))
    )
  )
)

(define isAnagram?
  (lambda (seq1 seq2)
    (if (and (null? seq1) (null? seq2))
      #t
      (if (or (null? seq1) (null? seq2))
        #f
        (if (canFind? (car seq1) seq2)
          (isAnagram? (cdr seq1) (remove (car seq1) seq2))
          #f
        )
      )
    )
  )
)

Upvotes: -1

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