# fibonacci.py
"""
Calculates the Fibonacci sequence using iteration, recursion, memoization,
and a simplified form of Binet's formula
NOTE 1: the iterative, recursive, memoization functions are more accurate than
the Binet's formula function because the Binet formula function uses 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
RESULTS: (n = 20)
fib_iterative runtime: 0.0055 ms
fib_recursive runtime: 6.5627 ms
fib_memoization runtime: 0.0107 ms
fib_binet runtime: 0.0174 ms
"""
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_memoization(n: int) -> list[int]:
"""
Calculates the first n (0-indexed) Fibonacci numbers using memoization
>>> fib_memoization(0)
[0]
>>> fib_memoization(1)
[0, 1]
>>> fib_memoization(5)
[0, 1, 1, 2, 3, 5]
>>> fib_memoization(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")
# Cache must be outside recursuive function
# other it will reset every time it calls itself.
cache: dict[int, int] = {0: 0, 1: 1, 2: 1} # Prefilled cache
def rec_fn_memoized(num: int) -> int:
if num in cache:
return cache[num]
value = rec_fn_memoized(num - 1) + rec_fn_memoized(num - 2)
cache[num] = value
return value
return [rec_fn_memoized(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_memoization, num)
time_func(fib_binet, num)
In mathematics, the Fibonacci numbers commonly denoted F(n), form a sequence, called the Fibonacci sequence, such that each number is the sum of the two preceding ones, starting from 0 and 1. The Sequence looks like this:
[0, 1, 1, 2, 3, 5, 8, 13, 21, 34, ...]
Finding
N-th
member of this sequence would be useful in many Applications:
science including high energy physics, quantum mechanics, Cryptography and Coding.
Find 8-th
member of Fibonacci
| F(n+1) F(n) |
| F(n) F(n-1)|
Calculate matrix^1
| 1 1 |
| 1 0 |
Calculate matrix^2
| 2 1 |
| 1 1 |
Calculate matrix^4
| 5 3 |
| 3 2 |
Calculate matrix^8
| 34 21 |
| 21 13 |
F(8)=21