So far our recursive functions have all included a **single** recursive call within the function definition.

Let’s explore a problem which pushes us to use multiple recursive calls within the function definition.

Fibonacci numbers are integers which follow a specific sequence: the next Fibonacci number is the **sum of the previous two Fibonacci numbers.**

We have a self-referential definition which means this problem is a great candidate for a recursive solution!

We’ll start by considering the base case. The Fibonacci Sequence starts with `0`

and `1`

respectively. If our function receives an input in that range, we don’t need to do any work.

If we receive an input greater than `1`

, things get a bit trickier. This recursive step requires **two previous** Fibonacci numbers to calculate the **current** Fibonacci number.

That means we need **two recursive calls** in our recursive step. Expressed in code:

fibonacci(3) == fibonacci(1) + fibonacci(2)

Let’s walk through how the recursive calls will accumulate in the call stack:

call_stack = [] fibonacci(3) call_stack = [fibonacci(3)]

To calculate the 3rd Fibonacci number we need the previous two Fibonacci numbers. We start with the previous Fibonacci number.

fibbonacci(2) call_stack = [fibbonacci(3), fibbonacci(2)]

`fibonacci(2)`

is a base case, the value of `1`

is returned…

call_stack = [fibbonacci(3)]

The return value of `fibonacci(2)`

is stored within the execution context of `fibonacci(3)`

while ANOTHER recursive call is made to retrieve the **second most previous** Fibonacci number…

fibonacci(1) call_stack = [fibonacci(3), fibonacci(1)]

Finally, `fibonacci(1)`

resolves because it meets the base case and the value of `1`

is returned.

call_stack = [fibonacci(3)]

The return values of `fibonacci(2)`

and `fibonacci(1)`

are contained within the execution context of `fibonacci(3)`

, which can now return the sum of the previous two Fibonacci numbers.

As you can see, those recursive calls add up fast when we have multiple recursive invocations within a function definition!

Can you reason out the big O runtime of this Fibonacci function?

### Instructions

**1.**

Define our `fibonacci()`

function that takes `n`

as an argument.

Let’s address our base cases:

- if the input is 1, we return 1
- if the input is 0, we return 0

**2.**

Now take care of the recursive step.

This step involves summing two recursive calls to `fibonacci()`

.

We need to retrieve the second to last and last Fibonacci values and return their sum.

We can get the second to last Fibonacci by decrementing the input by 2 and the last by decrementing the input by 1.

**3.**

Add print statements within `fibonacci()`

to explore the different recursive calls.

Set `fibonacci_runtime`

to the appropriate big O runtime.

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