Taras Bunyk

SICP 1.2.4 Exponentiation

Published: 2020-06-20T17:04:58.000Z

Today I learned that ancient mit-scheme REPL could be improved with history & tab completion. Thanks to this StackOverflow answer. sudo apt-get install rlwrap, and then run scheme as rlwrap scheme That answer is not very recent too, but here we are trying to learn really ancient magic.

And this section really starts to feel like magic. In the end, I learned that there is a way to compute n-th Fibonacci number with time complexity O(log(n)). And just few sections before, authors show how computing Fibonacci sequence using wrong approach could blow up exponencially.

Exercise 1.16: Fast exponentiation

(define (sqr x) (* x x))
(define (fast-exp x n) 
    (define (iter x n a) 
       ((= n 0) a)
       ((even? n) (iter (sqr x) (/ n 2) a))
       (else (iter x (- n 1) (* a x)))
    (iter x n 1)

Exercise 1.17: "Fast" multiplication

(define (double x) (+ x x))
(define (halve x) (/ x 2))

(define (fast-m a b) 
    (cond ((= b 1) a)
        ((even? b) (fast-m (double a) (halve b)))
        (else (+ a (fast-m a (- b 1)))

Exercise 1.18: Iterative multiplicaton

(define (fast-m a b) 
    (define (iter a b p) 
       ((= b 1) (+ a p))
       ((even? b) (iter (double a) (halve b) p))
       (else (iter a (- b 1) (+ p a)))
    (iter a b 0)

Exercise 1.19: Fast Fibonacci

With this exercise first, you discover that there is Fibonacci sequence inside Fibonacci formulas, and then, you figure out from where there appears exponential rise. Magical:

(define (fib n)
  (fib-iter 1 0 0 1 n))
(define (fib-iter a b p q count)
  (cond ((= count 0) b)
        ((even? count)
         (fib-iter a
                   (+ (* p p) (* q q))   ; compute p'
                   (+ (* q q) (* 2 p q)) ; compute q'
                   (/ count 2)))
        (else (fib-iter (+ (* b q) (* a q) (* a p))
                        (+ (* b p) (* a q))
                        (- count 1)))))

I not dediced yet on how to get latex in Hugo, so I'll not add here my calculations for p' and q'.