CODEWITHSUNDEEP
schemeracket
Robbie
Robbie

Reputation: 67

Scheme: Number of occurrences in a list

I'm trying to write a Scheme function called "p" with one parameter X which is a list of letters. the function should return true if the number of a's is one less than the number of b's. This is what I have but can't get around argument errors. Any help is greatly appreciated.

#lang scheme

(define p
(lambda (X)
(let ((countA 0))
(let ((countB 0))
(count(countA countB X)
(if (= countA (- countB 1))
#t
#f))))))

(define count
(lambda (A B X)
(if (null? (cdr X))
(car X)
((if (string=? "a" (car X))
((+ A 1) (p(cdr X)))
((if (string=? "b" (car X))
((+ B 1) (p(cdr X)))
(p(cdr X)))))))))

Upvotes: 1

Views: 3816

Answers (1)

Óscar López
Óscar López

Reputation: 236004

I'm betting this is a homework. If it weren't, this is the idiomatic way to solve it:

(define (p lst)
  (= (count-letters "a" lst)
     (- (count-letters "b" lst) 1)))

(define (count-letters letter lst)
  (count (lambda (e) (string=? e letter))
         lst))

... But of course, you're expected to solve it by hand. Some feedback on your code:

  • There are multiple misplaced parenthesis
  • In particular, it's incorrect to put double parenthesis around an if
  • The way you're using let for this problem is incorrect, it might work if you were mutating the value of the variables defined in the lets, but I really doubt that's the best approach to write the solution in this case
  • You're incorrectly assuming that the parameters passed to the count procedure will somehow store the result after count returns - that's not right, it won't work like that

Take ideas from the above code to write your own solution - first, you need to define a count-letters function that doesn't use count and that, for a given letter, counts the number of times it occurs in the list. With that procedure in hand, it's pretty easy to write the p procedure (in fact, you can just snarf my implementation above!). Don't forget to test it:

(p '("a" "b" "b"))
=> #t
(p '("a" "a" "b" "b"))
=> #f

Upvotes: 1

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