Dan's Thoughts Thinking somewhat carefully

Exercise 1.11 of SICP

Exercise 1.11: A function f is defined by the rule that f(n) = n if n<3 and f(n) = f(n - 1) + 2f(n - 2) + 3f(n - 3) if n> 3. Write a procedure that computes f by means of a recursive process. Write a procedure that computes f by means of an iterative process.

The idea is to use a triple of integers a,b,c initialized to f(2) = 2 f(1) = 1 and f(0) = 0, and to repeatedly apply the simultaneous transformations

a is now f(n)

b is now f(n-1)

c is now f(n-2)

This is nothing but a slightly more complex Fibonacci iteration example from earlier.
F(n) = F(n-1) + F(n-2) where F(n) = n if n<2


a is now F(n)

b is now F(n-1)

Recursive implementation of f(n) is straightforward:

(define (f-rec n)
  (if (< n 3)
    n
    (+ (f-rec (- n 1)) (* 2 (f-rec (- n 2))) (* 3 (f-rec (- n 3))))))

> (f-rec 1)
1
> (f-rec 3)
4
> (f-rec 6)
59
> (f-rec 8)
335

Iterative implementation of f(n)

(define (f n)
  (f-iter 2 1 0 n))
(define (f-iter a b c count)
    (cond ((< count 2) count)
          ((= count 2) a)
          (else
            (f-iter (+ a (* 2 b) (* 3 c))
                    a
                    b
                    (- count 1)))))

> (f 1)
1
> (f 3)
4
> (f 6)
59
> (f 8)
335