Fibonacci sequence using reduce method

Repeatedly applying a function using reduce

This answer suggests writing up your own function repeated to repeatedly apply a function, rather than calling reduce with a dummy second argument.

We still use reduce, but in a more functional manner and with itertools.repeat.

from itertools import repeat
from functools import reduce

def repeated(func, n):
    def apply(x, f):
        return f(x)
    def ret(x):
        return reduce(apply, repeat(func, n), x)
    return ret

def fibonacci(n):
  get_next_pair = lambda p: (sum(p), p[0])
  first_pair = (1, 0)
  return repeated(get_next_pair, n)(first_pair)[1]

print(fibonacci(0), fibonacci(1), fibonacci(11))
# 0 1 89

Repeatedly apply a linear function using linear algebra

The function lambda a,b: b,a+b which you want to apply happens to be a linear function. It can be represented by a 2*2 matrix. Repeatedly applying the function to a two-element tuple is the same as repeatedly multiplying a two-element vector by the matrix.

This is cool, because taking the power of a matrix is much, much faster than repeatedly applying a function.

import numpy as np

def fibonacci(n):
  return np.linalg.matrix_power(np.array([[0, 1],[1,1]]), n).dot(np.array([0,1]))[0]

print(fibonacci(0), fibonacci(1), fibonacci(11))
# 0 1 89

If you don't like one-liners, here is the same function decomposed on a few lines with more explicit variable names:

import numpy as np

def fibonacci(n):
  next_pair_matrix = np.array([[0, 1],[1,1]])
  matrix_to_the_nth = np.linalg.matrix_power(next_pair_matrix, n)
  first_pair_vector = np.array([0,1])
  nth_pair_vector = matrix_to_the_nth.dot(first_pair_vector)
  return nth_pair_vector[0]

print(fibonacci(0), fibonacci(1), fibonacci(11))
# 0 1 89