human
Can you provide a Python code to compute the return value recursively, but instead return the existing one that has already been constructed given the following:
I am working on the file "CS/Part_1_Complexity_RunTimeAnalysis.ipynb".
The first task is:
3. Can we do better?
A polynomial algorithm for $fib$
Let’s try to understand why $fib$ is so slow. fib.call_count shows the count of recursive invocations triggered by a single call to $fib(5)$, which is 15. If you sketched it out, you will notice that many computations are repeated!
A more sensible scheme would store the intermediate results—the values $F_0 , F_1 , . . . , F_{n−1}$ as soon as they become known. 
Lets do exactly that through memoization. Note that you can also do this by writing a polynomial algorithm.
Memoization
Tree-recursive computational processes can often be made more efficient through memoization, a powerful technique for increasing the efficiency of recursive functions that repeat computation. A memoized function will store the return value for any arguments it has previously received. A second call to fib(30) would not re-compute the return value recursively, but instead return the existing one that has already been constructed.
Memoization can be expressed naturally as a higher-order function, which can also be used as a decorator. The definition below creates a cache of previously computed results, indexed by the arguments from which they were computed. The use of a dictionary requires that the argument to the memoized function be immutable.
Can you write Python code for it?
assistant
#Here is Python code to compute the return value recursively, but instead return the existing one that has already been constructed.

def memo(f):
    cache = {}
    def memoized(n):
        if n not in cache:
            cache[n] = f(n) # Make a mapping between the key "n" and the return value of f(n)
        return cache[n]
    return memoized

fib = memo(fib)

t0 = time()

n = 400
fib(n)

print ('This memoized implementation of fib(', n, ') took', round(time() - t0, 4), 'secs')

t0 = time()

n = 300
fib(n)

print ('This memoized implementation of fib(', n, ') took', round(time() - t0, 4), 'secs')

# Here is the polynomial algorithm for fibonacci sequence
def fib2(n):
    if n == 0:
        return 0
    
    f = [0] * (n+1) # create an array f[0 . . . n]
    f[0], f[1] = 0, 1
    
    for i in range(2, n+1):
        f[i] = f[i-1] + f[i-2]
        
    return f[n] 

fib2 = count(fib2)

t0 = time()

n = 3000
fib2(n)

print ('This polynomial implementation of fib2(', n, ') took', round(time() - t0, 4), 'secs')

fib2.call_count