# Fibonacci sequence with Python recursion and memoization

Kinyanjui Wangonya Originally published at wangonya.com γ»3 min read

The Fibonacci sequence is a sequence of numbers such that any number, except for the first and second, is the sum of the previous two. For example:

0, 1, 1, 2, 3, 5, 8, 13, 21...


Fun fact: November 23 is celebrated as Fibonacci day because when the date is written in the mm/dd format (11/23), the digits in the date form a Fibonacci sequence: 1,1,2,3.

We can represent this in the formula:

fib(n) = fib(n-1)+fib(n-2)


With this formula, we can write a simple recursive function to solve for fib(n).

Note: Only use this to test for small numbers, preferably n < 10. I'll explain later

def fib(n):
if n < 2:  # base case
return n
return fib(n-1) + fib(n-2)

if __name__ == "__main__":
print(fib(7))


### Why do we need a base case?

It's easy enough to convert the formula directly into

def fib(n):
return fib(n-1) + fib(n-2)


The problem with this though is that when you run it, it throws a RecursionError. This simply means that there's possibly an infinite loop in the program. A base case in a recursive function tells the function when to stop (to avoid going into an infinite loop) and is usually something that is already known or that can be solved for easily without needing the function. In our case here, we know from the definition that any number in the sequence, except for the first and second, is the sum of the previous two. We use that to form our base case if n < 2: return n.

## Making it more efficient

Remember when I told you to only test the program with small values of n? Here's why.

As it stands, every call to fib() results in two more calls to fib() in the return statement. The call tree grows exponentially. 15 calls are required to compute fib(5), 177 calls for fib(10), 21,891 for fib(20)... you get the point. To solve this problem, we can use memoization.

Memoization helps avoid unnecessary calculation of the same values if they were already previously calculated. It works just like memorization for humans. You already have 2 x 2 memorized and can give the answer immediately without having to use a calculator. 799 x 377? You probably need to use a calculator for that. If, for some reason, you find that you get asked 799 x 377 a lot, it would be nice to have it memorized so you don't have to calculate it every other time. The value of 799 x 377 will always remain the same, so all you have to do is calculate it once, save the value in your "cache" (memory), and retrieve it every time you need it.

Luckily, python has a built-in decorator that does just that.

from functools import lru_cache

@lru_cache(maxsize=None)
def fib(n):
if n < 2:  # base case
return n
return fib(n-1) + fib(n-2)

if __name__ == "__main__":
print(fib(50))


Now you can test with large numbers safely.

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### Discussion

Good post! ππ

func fib(n: int): int =
if n < 2: n else: fib(n - 1) + fib(n - 2)

echo fib(9)


Nim lang version.