Comments on: SICP sections 1.2.4 - 1.2.5

By: Samad

Samad — Thu, 13 Nov 2008 23:13:38 +0000

One thing I’d like to note that there is a very simple trick to Exercise 1.16, although not explicitly stated by Eli. From the hint, it’s obvious that a and n can change in each iteration. It is not quite so obvious that b can change too. It is this simple insight that produces a clean, elegant solution.

Also, why does the T_{p’q'} transformation produce a double-Fibonacci? While I could arrive at the solution for computing p’ and q’ without much difficulty, but it’s not so obvious why this works.

By: eliben

eliben — Fri, 07 Mar 2008 14:57:55 +0000

Brit,

Well, it looks exactly like the code I’ve posted, only translated to Scheme. The only difference would perhaps be in the function double. But that doesn’t affect the complexity of the code.

By: Brit Butler

Brit Butler — Tue, 04 Mar 2008 19:39:56 +0000

Eli,

Is this code for example 1.17 equivalent in complexity?

(define (* a b)
(define (double x)
(+ x x))
(define (halve x)
(/ x 2)
(cond ((= b 0) 0)
((= a 0) 0)
((= b 1) a)
((even? b) (* (double a) (halve b)))
(else (+ a (* a (- b 1))))))