Exercise 1.19: 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 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 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 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)))))
T is the transformation of applying
a <- a + b b <- a
a <- bq + aq + ap b <- bp + aq
and T is T10 actually.
Applying Tpq twice, it transforms (a, b) into (a2, b2)
a1 = bq + aq + ap b1 = bp + aq a2 = b1q + a1q + a1p = (bp + aq)q + (bq + aq + ap)q + (bq + aq + ap)p = bpq + aqq + bqq + aqq + apq + bqp + aqp + app = b(pq + qq + qp) + a(qq + qq + pq + qp + pp) = b(pq + qq + qp) + a(pq + qq + qp) + a(qq + pp) b2 = b1p + a1q = (bp + aq)p + (bq + aq + ap)q = bpp + aqp + bqq + aqq + apq = b(pp + qq) + a(qp + qq + pq)
let p′ = pp + qq, and q′ = 2pq + qq, we can simplify the above to
a2 = bq′ + aq′ + ap′ b2 = bp′ + aq′
That is to say: .
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)) ;compute p' (+ (* 2 p 1) (* q q)) ;compute q' (/ count 2))) (else (fib-iter (+ (* b q) (* a q) (* a p)) (+ (* b p) (* a q)) p q (- count 1))))) (fib 0)