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Problem 1.19
Friday, August 23, 2019; ago; Download .md

SICP Exercise 1.19

This was my favourite exercise so far. Calculating values of p’ and q’ was fairly easy. Applying the $$T_{pq}$$ transformation twice and factoring out a and b in each case did the trick.

$$ T_{pq} $$ is defined as:

$$ \begin{aligned} &a \leftarrow bq + aq + ap \ &b \leftarrow bp + aq \end{aligned} $$

We’re told applying $$T_{pq}$$ twice has the same effect as applying $$T_{p’q’}$$ once.

Applying $$T_{pq}$$ twice gives:

$$ a \leftarrow (bp+aq)q + (bq+aq+ap)q + (bq+aq+ap)p $$

Which can be simplified to:

$$ a \leftarrow b(2pq+q^2) + a(2pq+q^2) + a(q^2+p^2) $$

Similarly for b:

$$ \begin{aligned} &b \leftarrow (bp+aq)p + (bq+aq+ap)q \ &b \leftarrow b(p^2+q^2) + a(2pq+q^2) \end{aligned} $$

And by simple comparision, we get:

$$ \begin{aligned} &p’ \leftarrow p^2 + q^2 \ &q’ \leftarrow 2pq + q^2 \end{aligned} $$

Which can now simply be put into the fib-iter procedure.

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

(define (square x) (* x x))