This post contains my solutions to exercises from Section 1.2.4 “Exponentiation” of the book “Structure and Interpretation of Computer Programs” that I’m working through as part of my “Learning Computer Science” project. You can find an online version of the text at the companion website for the book.

## Exercise 1.16

(define (fast-expt b n)
(define (square n)
(* n n))
(define (iter a b n)
(cond ((= n 0) a)
((even? n) (iter a (square b) (/ n 2)))
(else (iter (* a b) (square b) (/ (- n 1) 2)))))
(iter 1 b n))


## Exercise 1.17

(define (double n)
(+ n n))
(define (halve n)
(/ n 2))
(define (fast-multiply a b)
(cond ((= b 0) 0)
((even? b) (double (fast-multiply a (halve b))))
(else (+ a (fast-multiply a (- b 1))))))


## Exercise 1.18

The invariant quantity I picked in this case is a+b*c. As defined in the hint of exercise 1.16, an invariant quantity remains constant throughout the iterative process. For the following procedure, this means that when the three arguments of iter are used to compute a+b*c, they will give the same value at any state of the generated iterative process (e.g. (iter 0 5 6), (iter 0 10 3), and (10 20 1), etc.).

(define (fast-multiply b c)
(define (double n)
(+ n n))
(define (halve n)
(/ n 2))
(define (iter a b c)
(cond ((= c 0) a)
((even? c) (iter a (double b) (halve c)))
(else (iter (+ a b) (double b) (halve (- c 1))))))
(iter 0 b c))


## Exercise 1.19

$T_{pq}=(bq+aq+ap,bp+aq)\\ T_{pq}^2=T_{pq}(bq+aq+ap,bp+aq)\\ \quad= ((bp+aq)q+(bq+aq+ap)q+(bq+aq+ap)p, (bp+aq)p+(bq+aq+ap)q)\\ \quad= (b(q^2+2pq)+a(q^2+2pq)+a(p^2+q^2), b(p^2+q^2)+a(q^2+2pq))\\ \quad= T_{p'q'}(a,b);\text{ where, } p'=p^2+q^2 \text{ and } q'=q^2+2pq$

Thus, the completed procedure is:

(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
b
(+ (* p p) (* q q))
(+ (* q q) (* 2 p q))
(/ count 2)))
(else (fib-iter (+ (* b q) (* a q) (* a p))
(+ (* b p) (* a q))
p
q
(- count 1)))))