# fibonacci.py """ Calculates the Fibonacci sequence using iteration, recursion, and a simplified form of Binet's formula NOTE 1: the iterative and recursive functions are more accurate than the Binet's formula function because the iterative function doesn't use floats NOTE 2: the Binet's formula function is much more limited in the size of inputs that it can handle due to the size limitations of Python floats """ from math import sqrt from time import time def time_func(func, *args, **kwargs): """ Times the execution of a function with parameters """ start = time() output = func(*args, **kwargs) end = time() if int(end - start) > 0: print(f"{func.__name__} runtime: {(end - start):0.4f} s") else: print(f"{func.__name__} runtime: {(end - start) * 1000:0.4f} ms") return output def fib_iterative(n: int) -> list[int]: """ Calculates the first n (0-indexed) Fibonacci numbers using iteration >>> fib_iterative(0) [0] >>> fib_iterative(1) [0, 1] >>> fib_iterative(5) [0, 1, 1, 2, 3, 5] >>> fib_iterative(10) [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55] >>> fib_iterative(-1) Traceback (most recent call last): ... Exception: n is negative """ if n < 0: raise Exception("n is negative") if n == 0: return [0] fib = [0, 1] for _ in range(n - 1): fib.append(fib[-1] + fib[-2]) return fib def fib_recursive(n: int) -> list[int]: """ Calculates the first n (0-indexed) Fibonacci numbers using recursion >>> fib_iterative(0) [0] >>> fib_iterative(1) [0, 1] >>> fib_iterative(5) [0, 1, 1, 2, 3, 5] >>> fib_iterative(10) [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55] >>> fib_iterative(-1) Traceback (most recent call last): ... Exception: n is negative """ def fib_recursive_term(i: int) -> int: """ Calculates the i-th (0-indexed) Fibonacci number using recursion """ if i < 0: raise Exception("n is negative") if i < 2: return i return fib_recursive_term(i - 1) + fib_recursive_term(i - 2) if n < 0: raise Exception("n is negative") return [fib_recursive_term(i) for i in range(n + 1)] def fib_binet(n: int) -> list[int]: """ Calculates the first n (0-indexed) Fibonacci numbers using a simplified form of Binet's formula: https://en.m.wikipedia.org/wiki/Fibonacci_number#Computation_by_rounding NOTE 1: this function diverges from fib_iterative at around n = 71, likely due to compounding floating-point arithmetic errors NOTE 2: this function doesn't accept n >= 1475 because it overflows thereafter due to the size limitations of Python floats >>> fib_binet(0) [0] >>> fib_binet(1) [0, 1] >>> fib_binet(5) [0, 1, 1, 2, 3, 5] >>> fib_binet(10) [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55] >>> fib_binet(-1) Traceback (most recent call last): ... Exception: n is negative >>> fib_binet(1475) Traceback (most recent call last): ... Exception: n is too large """ if n < 0: raise Exception("n is negative") if n >= 1475: raise Exception("n is too large") sqrt_5 = sqrt(5) phi = (1 + sqrt_5) / 2 return [round(phi ** i / sqrt_5) for i in range(n + 1)] if __name__ == "__main__": num = 20 time_func(fib_iterative, num) time_func(fib_recursive, num) time_func(fib_binet, num)