1.2.4. Exponentiation
Exercise 1.19

Fibonacci numbers through transformation

There is a clever algorithm for computing the Fibonacci numbers in a logarithmic number of steps. Recall the transformation of the state variables a and b in the fib-iter process of section 1.2.2: a ← a + b and b ← a. Call this transformation T, and observe that applying T over and over again n times, starting with 1 and 0, produces the pair Fib(n + 1) and Fib(n). In other words, the Fibonacci numbers are produced by applying Tⁿ, the nth power of the transformation T, starting with the pair (1,0). Now consider T to be the special case of p = 0 and q = 1 in a family of transformations Tpq, where Tpq transforms the pair (a,b) according to a ← bq + aq + ap and b ← bp + aq. Show that if we apply such a transformation Tpq twice, the effect is the same as using a single transformation Tp'q' of the same form, and compute p' and q' in terms of p and q. This gives us an explicit way to square these transformations, and thus we can compute Tⁿ using successive squaring, as in the fast-expt procedure. Put this all together to complete the following procedure, which runs in a logarithmic number of steps:

(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
                   <??>      ; compute p'
                   <??>      ; compute q'
                   (/ count 2)))
        (else (fib-iter (+ (* b q) (* a q) (* a p))
                        (+ (* b p) (* a q))
                        p
                        q
                        (- count 1)))))


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