Sunday 31 March 2024

Python Coding challenge - Day 161 | What is the output of the following Python Code?



def fibonacci(n):

    if n <= 1:

        return n


        return fibonacci(n-1) + fibonacci(n-2)


Solution and Explanation:

The provided code defines a Python function fibonacci(n) that computes the nth Fibonacci number recursively, and then prints the 6th Fibonacci number.

Here's how it works:

The fibonacci function takes an integer n as input.

It starts with a base case: if n is 0 or 1, it returns n itself. This is because the Fibonacci sequence starts with 0 and 1, and from there each subsequent number is the sum of the two preceding ones.

If n is greater than 1, it recursively calls the fibonacci function for n-1 and n-2, and returns the sum of these two calls.

The function keeps recursively calling itself until it reaches one of the base cases, where it directly returns a value.

Finally, it prints the 6th Fibonacci number by calling fibonacci(6) and prints the result.

Let's walk through fibonacci(6):

fibonacci(6) calls fibonacci(5) and fibonacci(4).

fibonacci(5) further calls fibonacci(4) and fibonacci(3).

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

This process continues until it reaches the base cases (n <= 1).

Eventually, it computes the Fibonacci numbers for n=2, n=3, n=4, n=5, and n=6 and sums them up to get the 6th Fibonacci number.

So, when you run print(fibonacci(6)), it will output the 6th Fibonacci number, which is 8.


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