giles
689438d12e
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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.
21 lines
446 B
OCaml
21 lines
446 B
OCaml
type m22 = { a : int; b : int; c : int; d : int }
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let mul x y =
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{ a = x.a * y.a + x.b * y.c;
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b = x.a * y.b + x.b * y.d;
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c = x.c * y.a + x.d * y.c;
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d = x.c * y.b + x.d * y.d }
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let rec mpow m n =
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if n = 0 then { a = 1; b = 0; c = 0; d = 1 }
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else if n mod 2 = 0 then
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let h = mpow m (n / 2) in mul h h
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else
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mul m (mpow m (n - 1))
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;;
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let fib_matrix = { a = 1; b = 1; c = 1; d = 0 } in
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let r = mpow fib_matrix 30 in
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r.b
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