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In SICP exercise 1.37

Section 1.3.3 in SICP Scroll Down to the end of section (just before 1.3.4) to find the exercise [3rd exercise in the section].

According to the problem, I defined cont-frac as 

(define (cont-frac n d k)
    (if (= k 0)
     0 
     (/ n (+ d (cont-frac n d (- k 1))))
    )
)

Link to solution for the exercise

According to the solution link the above code seems to be consistent. The problem arise in the second part of the solution when n and d are substituted as (lamda (i) 1.0), in part (a) of the solution, which is a procedure.

I can't understand how this will work when substituted in the procedure of cont-frac. When I tried, there is the error which says Wrong type of argument

Edit 1

I have added my whole solution. It solves the problem but doesn't capture the essence of the section. This is the solution to exercises 1.37, 1.38 and 1.39. The program does not use Procedure as General Method which solutions of the below links do Solution to 1.37, Solution to 1.38 and Solution to 1.39

In the below program
in the procedures phi and e-2-val, k is no of steps in continued fraction
in the procedure tan, k is angle in radians (No of steps is 1000 for accurate value)

#!/usr/local/bin/guile \
-e main -s
!#
(define (finite-cont-frac n d k)
    (if (= k 0)
        0 
        (/ n (+ d (finite-cont-frac n d (- k 1))))))

(define (e-2 n d k1 c k)
    (define (d-val) 
        (if (= (modulo k1 3) 1)
            (+ c 2)
            1))
    (define (c-val)
        (if (= (d-val) 1) c (d-val)))
    (if (= k 0)
        0 
        (/ n (+ (d-val) (e-2 n (d-val) (+ k1 1) (c-val) (- k 1))))))

(define (tan-cf n k d k1)
    (define (d-val)
        (if (= k1 0) 1 (+ d 2)))
    (if (= k 0) 
        0
        (/ n (+ (d-val) (tan-cf n (- k 1) (d-val) (+ k1 1))))))

(define (tan-man x kk)
    (let ((a (- (* x x))))
        (tan-cf a kk 1 0)))
(define rrr 80.0)
(define (main args)
    (let* ((k (string->number (list-ref args 1)))
           (phi (/ 1.0 (finite-cont-frac 1.0 1.0 k)))
           (e-2-val (e-2 1.0 1 0.0 0 k))
           (tt (/ (tan-man k 1000) (- 0.0 k))))
        (display tt)
        (newline)))
Tarun Maganti
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  • Note that closing parentheses shouldn't be on their own line like C programming curlies. They should be on the line above next to the last closing parenthesis. – Sylwester Jul 20 '16 at 21:48
  • Is it like an idiomatic rule? Because this doesn't give any error. – Tarun Maganti Jul 21 '16 at 02:53
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    *TL;DR It's a consensus.* It's true that it doesn't affect the interpretations. You could have had it one symbol or parenthesis on each line making your example several pages long line of unreadable, but perfectly runnable, code. Lisp has very little structure in its code so how you style the code is how you make it readable.There are [style guides](http://mumble.net/~campbell/scheme/style.txt) that lispers agree on and questions about it on [PSE](http://programmers.stackexchange.com/questions/99229/why-does-the-lisp-community-prefer-to-accumulate-all-the-parentheses-at-the-end) – Sylwester Jul 21 '16 at 11:25
  • Okay, I understand. Thank you for the style guide. – Tarun Maganti Jul 21 '16 at 17:54

1 Answers1

1

The linked answer looks wrong, you're supposed do pass procedures, not numbers as actual parameters. Using a helper procedure called accumulate:

(define (accumulate combiner null-value term1 term2 a next b)
  (if (> a b)
      null-value
      (combiner (term1 a)
                (term2 a)
                (accumulate combiner
                            null-value
                            term1
                            term2
                            (next a)
                            next
                            b))))

(define (cont-frac n d k)
  (accumulate (λ (x y rec) (/ x (+ y rec)))
              0 n d 1 add1 k))

Now we can call the procedure as expected:

(cont-frac (lambda (i) 1.0)
           (lambda (i) 1.0)
           10)
=> 0.6179775280898876
Óscar López
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  • I just realized what blunder I made and returned to stackoverflow to delete the question. Thank you for the solution though. – Tarun Maganti Jul 13 '16 at 21:14
  • The linked answer is right. My solution is not appropriate for the section. My solution isn't wrong in a sense but the solution in the link uses n and d as procedures passing k as parameter. The solution in the link is much generalized, I had to write a lot of other procedures to define n and d in exercises 1.38 and 1.39 but using the generalized solution of the link, passing λ functions was enough. – Tarun Maganti Jul 13 '16 at 21:23