Fibonacci via repeated-squaring matrix exponentiation:
[[1, 1], [1, 0]] ^ n = [[F(n+1), F(n)], [F(n), F(n-1)]]
Recursive O(log n) power:
let rec mpow m n =
if n = 0 then identity
else if n mod 2 = 0 then let h = mpow m (n / 2) in mul h h
else mul m (mpow m (n - 1))
Returns the .b cell after raising to the 30th power -> 832040 = F(30).
Tests record literal construction inside recursive function returns,
record field access (x.a etc), and pure integer arithmetic in the
matrix multiply.
165 baseline programs total.
giles
689438d12e