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Solving for the nth Fibonacci number

theisomorphic profile image SATOSHI 💬 Originally published at ・4 min read

Cover Photo Credit: Chloe Evans on Unsplash

A classic interview question: “Please write a function fibonacci that takes an integer n and returns the nth Fibonacci number.” The Fibonacci sequence follows the following pattern:

0, 1, 1, 2, 3, 5, 8, 13…

The pattern continues by adding the previous two Fibonacci numbers together and therefore, the next value above would be 21. Now let’s write a function to get the nth Fibonacci value so that,

// base Fibonacci numbers
fibonacci(0) // returns 0
fibonacci(1) // returns 1
// generated Fibonacci numbers
fibonacci(2) // returns 1
fibonacci(3) // returns 2
fibonacci(4) // returns 3
fibonacci(5) // returns 5
fibonacci(6) // returns 8
// ...

Solution 1: Recursion

This is the most popular way of solving this problem because it is easier to reason about since,

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

Let’s write this as a function:

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

This is great but, this has no stopping condition so it will go on forever. We need to specify that if n is 0 or 1 (our base Fibonacci numbers) we return 0 and 1, respectively.

function fibonacci(n) {
  if (n === 0) return 0
  else if (n === 1) return 1
  else return fibonacci(n - 1) + fibonacci(n - 2)

Great! Try the function out for n = 1, n = 5, and n = 50.

  • fibonacci(1) should return 1.
  • fibonacci(5) should return 5.
  • fibonacci(50) should return 12586269025.

You might have noticed that fibonacci(50) hangs in the console for some time. In fact, it took my console around eight minutes to execute!

console showing an 8 minute execution of f(50)

This is the downside to this solution. For large n , the computation time takes way too long. The second solution fixes this problem.

Solution 2: Using a Generator Function

So the previous solution worked but, is super slow for large values of n.
Why is this the case? Well, let’s calculate fibonacci(10) as an example by hand (I will denote fibonacci as f for sake of simplicity.)

long recursive tree for f(10)

We are having to dive into a bunch of the same rabbit holes over and over again to calculate fibonacci(10). Why do we have to do this if all we need is the previous two Fibonacci numbers? Is there a way we can just remember the previous two Fibonacci numbers and generate the next Fibonacci number in the sequence? Yes! We can use generators to create an infinite sequence of Fibonacci numbers. Generators are interesting. They are like regular functions but with super powers. They are able to return values without completely ending the execution of a function. It does this by making use of the special yield statement. Let’s look at a trivial example of a generator function.

function* x() {
  // the "*" denotes that function x is a generator
  yield 'One taught me love'
  yield 'One taught me patience'
  yield 'And one taught me pain'

Great! Let’s invoke this function to see how it works:

const thanku = x() // we invoke the generator

// invoke the `next` method on the generator prototype // returns {value: "One taught me love", done: false} // {value: "One taught me patience", done: false} // {value: "And one taught me pain", done: false} // {value: undefined, done: true}

// Now aren't you grateful for your x?

Ariana Grande Wink - Thank u next.

For every call to the next method on the generator prototype, we get an object with two properties: value and done which is the value you are yielding from the generator and whether or not your generator is done generating values, respectively. Let’s look at a more interesting example. Let’s generate an infinite sequence of even numbers:

function* even() {
  let start = 0
  yield start // yield 0 as our first even number
  while (true) {
    // the start of our infinite sequence!
    start += 2 // add 2 to start
    yield start

function helper() {
  const value =
  console.log(`NEXT: ${JSON.stringify(value)}`)

const evenNumbers = even()

setTimeout(helper, 1000)
setTimeout(helper, 2000)

Let’s go though the execution of the code above together:

  1. We first Initialize the variable evenNumbers with the invocation of even generator.
  2. We then wait 1000 milliseconds for the first invocation of helper.
  3. 1000 milliseconds pass, and helper is invoked
    1. We initialize value with the invocation of
      1. We initialize start with 0
      2. Then we yield start and pause the generator.
    2. Now we console.log the value
  4. Wait another 1000 milliseconds for the second invocation of helper
    1. We enter the while loop
      1. Increment start by 2.
      2. yield start and pause the generator.
    2. Now we console.log the value.

Great! So how do we use the generator function to get the nth Fibonacci number? What we want to do is

  1. Create an infinite sequence of Fibonacci numbers using a generator.
  2. Keep invoking until we get the nth Fibonacci number.

1. Create an infinite sequence of Fibonacci numbers using a generator

function* Fibonacci() {
  let a = 0,
    b = 1 // base Fibonacci numbers
  while (true) {
    const c = a + b // next Fibonacci number
    yield c
    a = b // new a will be what used to be b
    b = c // new b will be what used to be c

2. Keep invoking until we have the nth number

We can do this by using a standard for loop:

function fibonacci(n) {
  if (n === 0) return 0
  else if (n === 1) return 1
  else {
    const Fib = Fibonacci()
    let value
    for (let i = 0; i < n - 1; i++) {
      value =
    return value

And there you have it: a faster function to find the nth Fibonacci number. Look at the difference in speed! ~8 minutes vs ~0.029 milliseconds!

console showing a 0.03 millisecond execution of f(50)

Discussion (2)

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michaeltharrington profile image
Michael Tharrington (he/him)

Oh wow, didn't expect another Ariana Grande reference on DEV!

In case you haven't seen it yet, check it out:

theisomorphic profile image
SATOSHI 💬 Author

Whoa thanks for sharing haha!