CODEWITHSUNDEEP
lispschemeracketsicp
pvd
pvd

Reputation: 1343

SICP exercise 1.37: My iterative solution got the right answer but got wrong in 1.38

My iterative solution to SICP 1.37 is

(define (con-frac n d k)
  (define (iter i result)
    (if (= 1 i)
        result
        (iter (- i 1) (/ (n i) (+ (d i) result)))))
  (iter k (/ (n k) (d k))))

(con-frac (lambda (i) 1.0) (lambda (i) 1.0) 11)

(define (euler-d i)
  (if (= 2 (remainder i 3))
      (* (/ 2 3) (+ i 1))
      1))

(define (e)
  (+ 2 (con-frac (lambda (i) 1.0) euler-d 9)))

(e)

It returns:

Welcome to DrRacket, version 5.2.1 [3m].
Language: SICP (PLaneT 1.17); memory limit: 128 MB.
0.6180555555555556
2.39221140472879

It should be return:

Welcome to DrRacket, version 5.2.1 [3m].
Language: SICP (PLaneT 1.17); memory limit: 128 MB.
0.6180555555555556
2.718283582089552

I don't know what's wrong with my solution.

Upvotes: 2

Views: 428

Answers (2)

Óscar López
Óscar López

Reputation: 235984

As has been pointed in @soegaard's answer, there's a simple off-by-one error in your code. Just replace this line:

(if (= 1 i)

With this one - it's a bit more idiomatic than asking if (= 0 i):

(if (zero? i)

... And that's all!

Upvotes: 2

soegaard
soegaard

Reputation: 31147

You are off by one in the iter.

In 

  (define (iter i result)
    (if (= 1 i)
        result
        (iter (- i 1) (/ (n i) (+ (d i) result)))))

change (= 1 i) to (= 0 i.

  (define (iter i result)
    (if (= 0 i)
        result
        (iter (- i 1) (/ (n i) (+ (d i) result)))))

The test using phi doesn't catch this since all numerators and denominators are equal.

Upvotes: 6

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