# 15.2 Recursion

Here we go again...
Now that we've done a couple of nested for loops, let's take a look at our favorite problem: recursion.
Consider the recursive function f3 below:
public static int f3(int n) {
if (n <= 1)
return 1;
return f3(n-1) + f3(n-1);
}

#### What does this function do?

Let's think of an example of calling f3(4):
• The first call will return f3(4-1) + f3(4-1)
• Each f3(3-1) call will branch out to f3(2-1) + f3(2-1)
• Then for each f3(2-1) call, the condition if (n <= 1) will be true, which will return 1.
• What we observe at the end is that 1 will be returned 8 times, meaning we have f3(2-1) summed 8 times.
• Therefore,f3(4)will return 8.
We can visualize this as a tree, where each level represents a recursive call and each node value represents the argument to the function :
Visualization of f3's recursive calls
You can do a couple more examples, and see that this function returns
$2^N-1$
​ . Visualizing the recursive calls is extremely useful for getting a sense of what the function is doing, and we will discuss a few methods of determining runtime in recursive functions.

### Method 1: Intuition

Based on the visualization below, we can notice that every time we add one to n we double the amount of work that has to be done:
Then, adding one to n N times means doubling the amount of work N times, which results in the intuitive answer for runtime to be
$2^N$
.

### Method 2: Algebra

Another way to approach this problem is to count the number of calls to f3 involved. Utilizing the same tree visualization above, we can see that the number of calls to f3 at each recursive level is equivalent to the number of nodes at each recursive level of the tree. For instance, the number of calls we made to f3 at the top level is 1, at the second level is 2, at the third level is 4, etc.
The total number of calls to f3 then is the sum of the number of nodes at each recursive level, which can be expressed as the equation below:
$C(N)=1+2+4+...+2^{N-1}$
Applying the formula we saw earlier for the sum of the first powers of 2:
$1+2+4+8+...+Q=2Q−1$
Substituting
$Q$
with
$2^{N-1}$
, we get:
$C(N)=2Q−1=2(2^{N-1})-1=2^N-1 ​ ​​$
The work during each call is constant , so the overall runtime for f3 is
$\theta(2^N)$
.

### Method 3: Recurrence Relation (Out of Scope)

This method is not required reading and is outside of the course scope, but worth mentioning for interest's sake.
We can use a "recurrence relation" to count the number of calls, instead of an algebraic approach. This looks like:
$C(1)=1 C(N) = 2C(N-1) + 1$
Expanding this out with a method we will not go over but you can read about in the slides or online, we reach a similar sum to the one above. We can then again reduce it to
$2^N - 1$
, reaching the same result of
$\theta(2^N)$
.