t_veor
@t_veor@sopuli.xyz
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- Comment on A Linear Algebra Trick for Computing Fibonacci Numbers Fast 1 year ago:
It’s not that hard to check yourself. Running the following code on my machine, I get that the linear algebra algorithm is already faster than the naive algorithm at around n = 100 or so. I’ve written a more optimised version of the naive algorithm, which is beaten somewhere between n = 200 and n = 500.
Try running this Python code on your machine and see what you get:
import timeit def fib_naive(n): a = 0 b = 1 while 0 < n: b = a + b a = b - a n = n - 1 return a def fib_naive_opt(n): a, b = 0, 1 for _ in range(n): a, b = b + a, b return a def matmul(a, b): return ( (a[0][0] * b[0][0] + a[0][1] * b[1][0], a[0][0] * b[0][1] + a[0][1] * b[1][1]), (a[1][0] * b[0][0] + a[1][1] * b[1][0], a[1][0] * b[0][1] + a[1][1] * b[1][1]), ) def fib_linear_alg(n): z = ((1, 1), (1, 0)) y = ((1, 0), (0, 1)) while n > 0: if n % 2 == 1: y = matmul(y, z) z = matmul(z, z) n //= 2 return y[0][0] def time(func, n): times = timeit.Timer(lambda: func(n)).repeat(repeat=5, number=10000) return min(times) for n in (50, 100, 200, 500, 1000): print("========") print(f"n = {n}") print(f"fib_naive:\t{time(fib_naive, n):.3g}") print(f"fib_naive_opt:\t{time(fib_naive_opt, n):.3g}") print(f"fib_linear_alg:\t{time(fib_linear_alg, n):.3g}")
Here’s what it prints on my machine:
======== n = 50 fib_naive: 0.0296 fib_naive_opt: 0.0145 fib_linear_alg: 0.0701 ======== n = 100 fib_naive: 0.0652 fib_naive_opt: 0.0263 fib_linear_alg: 0.0609 ======== n = 200 fib_naive: 0.135 fib_naive_opt: 0.0507 fib_linear_alg: 0.0734 ======== n = 500 fib_naive: 0.384 fib_naive_opt: 0.156 fib_linear_alg: 0.112 ======== n = 1000 fib_naive: 0.9 fib_naive_opt: 0.347 fib_linear_alg: 0.152