## 1. Overview

In this article, we’ll implement two common algorithms that evaluate the *n*^{th} number in a Fibonacci Sequence. We’ll then step through the process of analyzing time complexity for each algorithm.

Let’s start with a quick definition of our problem.

## 2. Fibonacci Sequence

**The Fibonacci Sequence is an infinite sequence of positive integers, starting at 0 and 1, where each succeeding element is equal to the sum of its two preceding elements.**

If we denote the number at position *n* as *F _{n}*, we can formally define the Fibonacci Sequence as:

*F _{n}* = o for

*n*= 0

*F*= 1 for

_{n}*n*= 1

*F*

_{n}= F_{n-1}+

*F*

_{n-2}for

*n*> 1

Therefore, the start of the sequence is:

0, 1, 1, 2, 3, 5, 8, 13, …

So, how can we design an algorithm that returns the *n*^{th} number in this sequence?

## 3. Recursive Algorithm

Our first solution will implement recursion. This is probably the most intuitive approach, since the Fibonacci Sequence is, by definition, a recursive relation.

### 3.1. Method

Let’s start by defining *F*(*n*) as the function that returns the value of *F _{n}*.

**To evaluate F(n) for n > 1, we can reduce our problem into two smaller problems of the same kind: F(n-1) and F(n-2).** We can further reduce

*F*(

*n*-1) and

*F*(

*n*-2) to

*F*((

*n*-1)-1) and

*F*((

*n*-1)-2); and

*F*((

*n*-2)-1) and

*F*((

*n*-2)-2), respectively.

If we repeat this reduction, we’ll eventually reach our known base cases and, thereby, obtain a solution to *F*(*n*).

Employing this logic, our algorithm for *F*(*n*) will have two steps:

- Check if
*n*≤ 1. If so, return*n*. - Check if
*n*> 1. If so, call our function*F*with inputs*n*-1 and*n*-2, and return the sum of the two results.

Here’s a visual representation of this algorithm:

### 3.2. Pseudocode

Now that we understand how this algorithm works, let’s implement some pseudocode:

## 4. Analysis of Time Complexity

**We can analyze the time complexity of F(n) by counting the number of times its most expensive operation will execute for n number of inputs.** For this algorithm, the operation contributing the greatest runtime cost is addition.

### 4.1. Finding an Equation for Time Complexity

Let’s use *T*(*n*) to denote the time complexity of *F*(*n*).

The number of additions required to compute *F*(*n*-1) and *F*(*n*-2) will then be *T*(*n*-1) and *T*(*n*-2), respectively. We have one more addition to sum our results. Therefore, for *n* > 1:

*T*(*n*) = *T*(*n*-1) + *T*(*n*-2) + 1

When *n* = 0 and *n* = 1, no additions occur. This implies that:

*T*(0) = *T*(1) = 0

Now that we have our equation, we need to solve for *T*(*n*).

There are several ways we could do this. We’ll implement a fairly straightforward approach, where we slightly simplify *T*(*n*) and find its solution using backward substitution. The result will give us an upper bound on the time complexity of *T*(*n*).

### 4.2. Simplifying T*(n)*

Let’s start by assuming that ** T(n-2) ≈ T(n-1)**. Don’t worry about

*why*just yet – this will become apparent shortly.

Substituting the value of *T*(*n*-1) = *T*(*n*-2) into our relation *T*(*n*), we get:

*T*(*n*) = *T*(*n*-1) + *T*(*n*-1) + 1 = 2**T*(*n*-1) + 1

By doing this, **we have reduced T(n) into a much simpler recurrence.** As a result,

**we can now solve for**

*T*(*n*) using backward substitution.### 4.3. Solving T(*n*) Using Backward Substitution

To do this, we first substitute *T*(*n*-1) into the right-hand side of our recurrence. Since *T*(*n*-1) = 2**T*(*n*-2) + 1, we get:

*T*(*n*) = 2*[2**T*(*n*-2) + 1] + 1 = 4**T*(*n*-2) + 3

Next, we can substitute in *T*(*n*-2) = 2**T*(*n*-3) + 1:

*T*(*n*) = 2*[2*[2*T(n-3) + 1] + 1] + 1 = 8*T(n-3) + 7

And once more for *T*(*n*-3) = 2**T*(*n*-4) + 1:

*T*(*n*) = 2*[2*[2*[2**T*(*n*-4) + 1]+ 1] + 1] + 1 = 16**T*(*n*-4) + 15

We can see a pattern starting to emerge here, so let’s attempt to form a general solution for *T*(*n*). It appears to stand that:

*T*(*n*) = 2* ^{k}**

*T*(

*n*–

*k*) + (2

*-1)*

^{k}For any positive integer *k*. We can prove this equation holds through simple induction – for brevity’s sake, we’ll skip this process.

Finally, we can find *k* and, thereby, solve *T*(*n*), by substituting in *T*(0) = 1.

For *T*(0), we can see that *n* – *k* = 0. Rearranging, we get *k* = *n*. Now, substituting in our values for *T*(0) and *k*, we get:

*T*(n) = 2* ^{n}**

*T*(0) + (2

*-1) = 2*

^{n}*+ 2*

^{n}*– 1 = O(2*

^{n}*)*

^{n}### 4.4. Analyzing Our Solution

Recall the assumption we made earlier that *T*(*n*-2) ≈ *T*(*n*-1). Since *T*(*n*-2) ≤ *T*(*n*-1) will always hold, **our solution O(2^{n}) is an upper bound for the time complexity of F(n).**

It does not, however, give us the tightest upper bound. Our initial assumption removed a bit of precision. The tightest upper bound of *F*(*n*) works out to be:

*T*(*n*) = O(Φ* ^{n}*)

Where Φ = (1+√5) / 2.

Both of these solutions reveal that **the run time of our algorithm will grow exponentially in n.** This observation is quite intuitive if we consider that each time we call

*F*(

*n*), it will potentially make two more calls to

*F*, thus doubling the cost of

*F*(

*n*). This is some very undesirable behavior!

## 5. Iterative Algorithm

Let’s move on to a much more efficient way of calculating the Fibonacci Sequence.

For this algorithm, we’ll start at our known base cases and then evaluate each succeeding value until we finally reach the *n*^{th} number. We’ll store our sequence in an array *F*[].

### 5.1. Method

First, we’ll store 0 and 1 in *F*[0] and *F*[1], respectively.

Next, we’ll iterate through array positions 2 to *n-1*. At each position *i*, we store the sum of the two preceding array values in *F*[*i*].

Finally, we return the value of *F*[*n*-1], giving us the number at position *n* in the sequence.

Here’s a visual representation of this process:

### 5.2. Pseudocode

Let’s take a look at the pseudocode for this approach:

### 5.3. Time Complexity

Analyzing the time complexity for our iterative algorithm is a lot more straightforward than its recursive counterpart.

In this case, our most costly operation is assignment. Firstly, our assignments of *F*[0] and *F*[1] cost O(1) each. Secondly, our loop performs one assignment per iteration and executes (*n*-1)-2 times, costing a total of O(*n*-3) = O(*n*).

**Therefore, our iterative algorithm has a time complexity of O( n) + O(1) + O(1) = O(n).**

This is a marked improvement from our recursive algorithm!

## 6. Conclusion

In this article, we analyzed the time complexity of two different algorithms that find the *n*^{th} value in the Fibonacci Sequence. First, we implemented a recursive algorithm and discovered that its time complexity grew exponentially in *n*.

Next, we took an iterative approach that achieved a much better time complexity of O(*n*).