From a previous post on recursion we see that sometimes the sub problems of a recursive solution are related or overlapping in certain ways, the Fibonacci number sequence has sub-problems which overlap. From the recursion tree we notice that most calculations repeat, which causes inefficiency in the algorithm.

**Recursion Tree:**

The solution for Fibonacci 4, 3, 2, 1 and 0 is repeated on the left and right of the recursion tree, which makes the algorithm very inefficiently. What if there was a way we could store values for previously calculated Fibonacci numbers so that in the event that we need those solutions we could find them O(1) without doing any calculations.

**Memoization**

In computing, **memoization** is an optimization technique used primarily to speed up computer programs by having function calls avoid repeating the calculation of results for previously processed inputs. Memoization has also been used in other contexts (and for purposes other than speed gains), such as in simple mutually recursive descent parsing^{[1]} in a general top-down parsingalgorithm^{[2]}^{[3]} that accommodates ambiguity and left recursion in polynomial time and space. Although related to caching, memoization refers to a specific case of this optimization, distinguishing it from forms of caching such as buffering or page replacement. In the context of some logic programming languages, memoization is also known as tabling;^{[4]} see also lookup table more…

Original Algorithm:

def fib(num):
if num < 2:return num if num > 0 :return fib(num-1) + fib(num-2)

Memoized Version:

cache = {}
def fib(num):
if num in cache:
return cache[num]
else:
cache[num] = num if num < 2 else fib(num-1) + fib(num-2)
return cache[num]

**Pruned Recursion Tree:**

The red lines on the above diagram shows the pruning on the recursion tree caused by the memoization of the algorithm. The algorithm is now much more efficient because it no longer needs to traverse the recursion tree for a value which has already been calculated. Until next time have fun :-)

Refrences

Wikipedia.com

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Awesome Algorithm

Made my day :)

The red lines on the above diagram shows the ***proning*** on the recursion tree … maybe you meant “pruning”

Thank for the correction Security Dude.

Good writing.

The return statement can be taken out from the if/else blocks.

cache = {}

def fib(num):

if num not in cache:

cache[num] = num if num < 2 else fib(num-1) + fib(num-2)

return cache[num]