Memoization fibonacci algorithm in python

Question

I have this memoization technique to reduce the number of calls getting a Fibonacci sequence number:

def fastFib(n, memo):
    global numCalls
    numCalls += 1
    print 'fib1 called with', n
    if not n in memo:
        memo[n] = fastFib(n-1, memo) + fastFib(n-2, memo)
        return memo[n]


def fib1(n):
    memo = {0:1, 1:1}
    return fastFib(n, memo)


numCalls = 0
n = 6
res = fib1(n)
print 'fib of', n,'=', res, 'numCalls = ', numCalls

But i am stuck at here: memo[n] = fastFib(n-1, memo) + fastFib(n-2, memo) and here memo = {0:1, 1:1}. How is it exactly reducing the number of calls each time i want to get fib of a number?

Answer 1

You should return memo[n] always, not only on unseccesful look up (last line of fastFib()):

def fastFib(n, memo):
    global numCalls
    numCalls += 1
    print 'fib1 called with', n
    if not n in memo:
        memo[n] = fastFib(n-1, memo) + fastFib(n-2, memo)
    #this should be outside of the if clause:
    return memo[n] #<<<<<< THIS

The number of calls is reduced this way, because for each value of n you actually compute and recurse from at most once, limitting the number of recursive calls to O(n) (upper bound of 2n invokations), instead of recomputing the same values over and over again, effectively making exponential number of recursive calls.

A small example for fib(5), where each line is a recursive invokation:

Naive approach:

f(5) = 
f(4) + f(3) = 
f(3) + f(2) + f(3) =
f(2) + f(1) + f(2) + f(3) =
f(1) + f(0) + f(1) + f(2) + f(3) = (base clauses) = 
1 + f(0) + f(1) + f(2) + f(3) = 
2 + f(1) + f(2) + f(3) =
3 + f(2) + f(3) = 
3 + f(1) + f(0) + f(3) = 
3 + 1 + f(0) + f(3) = 
5 + f(3) = 
5 + f(2) + f(1)  =
5 + f(1) + f(0) + f(1) =
5 + 1 + f(0) + f(1) =
5 + 2 + f(1) =
8

Now, if you use memoization, you don't need to recalculate a lot of things (like f(2), which was calculated 3 times) and you get:

f(5) = 
f(4) + f(3) = 
f(3) + f(2) + f(3) =
f(2) + f(1) + f(2) + f(3) =
f(1) + f(0) + f(1) + f(2) + f(3) = (base clauses) = 
1 + f(0) + f(1) + f(2) + f(3) = 
2 + f(1) + f(2) + f(3) =
3 + f(2) + f(3) =  {f(2) is already known}
3 + 2 + f(3) = {f(3) is already known}
5 + 3  = 
8

As you can see, the second is shorter than the first, and the bigger the number (n) becomes, the more significant this difference is.