Python Coding challenge - Day 438| What is the output of the following Python Code?

 

Code Explanation:

1. Importing lru_cache from functools

from functools import lru_cache

lru_cache is a decorator provided by the functools module. It stands for "Least Recently Used Cache." The decorator caches the results of function calls so that subsequent calls with the same arguments can return the cached result, rather than recomputing it.

Why use lru_cache?

Memoization is an optimization technique to improve performance by storing the results of expensive function calls and reusing them when the same inputs occur again. This avoids redundant computations in recursive functions, which is especially useful in problems that involve overlapping subproblems, like the Fibonacci sequence.

2. The Function f(x)

@lru_cache()

def f(x):

    if x < 3: 

        return x

    return f(x - 1) + f(x - 2)

@lru_cache(): This decorator applies caching to the f(x) function. It stores previously computed values of f(x) so that if f(x) is called again with the same x, the function doesn't need to recompute the result.

Base Case (if x < 3):

The function returns x directly when x is less than 3. This is the base case of the recursion. For x = 0, 1, 2, the function just returns x.

Recursive Case:

For x >= 3, the function calls itself recursively:

return f(x - 1) + f(x - 2)

This means that f(x) is the sum of the previous two values: f(x - 1) and f(x - 2).

This recursive structure is similar to the Fibonacci sequence, where each term is the sum of the previous two terms.

3. Calling f(5)

print(f(5))

Now, let's walk through the computation of f(5) step-by-step:

f(5):f(5) calls f(4) and f(3).

f(4):f(4) calls f(3) and f(2).

f(3):f(3) calls f(2) and f(1).

Base Cases:

f(2) and f(1) return 2 and 1, respectively (as per the base case).

Now we have:

f(2) returns 2.

f(1) returns 1.

f(3) becomes f(2) + f(1) = 2 + 1 = 3.

Continuing to f(4):

Now, we can compute f(4):

f(4) calls f(3) and f(2).

f(3) is already computed as 3 (due to caching from the earlier calculation).

f(2) is also cached as 2.

Therefore, f(4) = f(3) + f(2) = 3 + 2 = 5.

Finally, computing f(5):

Now we can compute f(5):

f(5) calls f(4) and f(3).

f(4) is already cached as 5.

f(3) is already cached as 3.

Therefore, f(5) = f(4) + f(3) = 5 + 3 = 8.

4. Memoization and Caching

The memoization (enabled by @lru_cache()) ensures that intermediate results like f(4), f(3), f(2), and f(1) are computed only once. After the first computation of each value, the results are stored in the cache, and subsequent calls with the same argument will simply return the cached result, making the function much faster.

For example, when computing f(5), f(3) is computed only once, and its result is reused when computing f(4) and f(5).

5. Output

print(f(5))

Finally, f(5) evaluates to 8, so the output of the program will be:

8