A Linear Algebra Trick for Computing Fibonacci Numbers Fast - eviltoast
    • IAm_A_Complete_Idiot@sh.itjust.works
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      1 year ago

      According to the benchmark in the article it’s already way faster at n = 1000. I think you’re overestimating the cost of multiplication relative to just cutting down n logarithmically.

      log_2(1000) = roughly a growth factor of 10. 2000 would be 11, and 4000 would be 12. Logs are crazy.

      • cbarrick@lemmy.world
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        1 year ago

        The article is comparing to the dynamic programming algorithm, which requires reading and writing to an array or hash table (the article uses a hash table, which is slower).

        The naive algorithm is way faster than the DP algorithm.

        • t_veor@sopuli.xyz
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          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