Added my solutions so far

ex-1-19.rkt

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+#lang sicp
+
+;; Yeah okay, this one was extremely satisfying.
+;; I just did the calculations on-paper.
+;; it's some simple algebra anyway, but I'll type it here:
+;; assuming Tpq on (a0, b0):
+;; a1 = b0q + a0 * ( p + q)
+;; b1 = b0p + a0q
+;; a2 = (b0p + a0q) * p + (b0q + a0 * ( p + q)) * (p + q)
+;; b2 = (b0p + a0q) * p + (b0q + a0 * ( p + q)) * q
+;;
+;; rearrange a2 and b2 into a similar form to the definition of a1 and b1:
+;; (i.e. define a2 and b2 in terms of a0 and b0 and p and q)
+;; a2 = b0 * (q^2 + 2*p*q) + a0 * (2*q^2 + 2*p*q + p^2)
+;; b2 = b0 * (p^2 + q^2) + a0 * (q^2 + 2*p*q)
+;;
+;; from here we can see that p'= p^2 + q^2, and q'= q^2 + 2 * p * q
+;; and as a result, we have logarithmic fibonacci!
+
+;; printing the resulting number takes longer than calculating it lmao.
+
+(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′
+                   (+ (* q q) (* 2 (* p q))) ; compute q′
+                   (/ count 2)))
+        (else (fib-iter (+ (* b q) (* a q) (* a p))
+                        (+ (* b p) (* a q))
+                        p
+                        q
+                        (- count 1)))))