What Does Fibonacci Look Like in Elixir?

I recently had a conversation about Elixir since I have been using it more and it was not a language this person was familiar with. As curious engineers, one basic question that arose was, "what does Fibonacci look like?". I was happy to comply and provide some code.

As a quick refresher, we want to write a function that takes in an integer that represents the nth number in the Fibonacci sequence. For this implementation, we're assuming that the input is a non-negative integer. Okay, let's go!


So the typical recursive solution has a function fib(n) and we return an integer of 1 if n is equal to 0 or 1 and otherwise, we want to recursively call fib with the last two previous indexes.

def fib(0), do: 1
def fib(1), do: 1
def fib(n), do: fib(n-1) + fib(n-2)

Happy days! We just need three lines to implement this! Now, if you recall my previous post on Elixir function conditionals, that is the same thing we're doing here. We're returning 1 if we pass in a 0 or 1 index, otherwise we'll just do the recursive logic that we want. That's it!


"So what about an iterative solution?" you ask? Yes, that's actually what was discussed next anyhow. So, the typical solution for an iterative solution is to have a loop and have two variables track the previous and current values. So, that's fine, but something just felt wrong with that not being quite Elixir-y. So, I thought about it for a bit and came up with the following solution.

  def iter_fib(0), do: 1
  def iter_fib(1), do: 1
  def iter_fib(index) do
    Enum.reduce(2..index, [1, 1], fn(_i, acc) ->
      # Calculate the Fibonacci value
      fib_val =
        # Take the accumulator
        # Flatten the list since we're appending fib_val by a new list
        |> List.flatten()
        # Sum those values
        |> Enum.sum()
      [Enum.take(acc, -1), fib_val]
    # This is now the Fibonacci value for the index and its previous value, so just take the last value
    |> List.last()
So, what I've done here is create a list that represents the Fibonacci sequence. We have the same overloaded function signatures for index values of 0 and 1, and otherwise, the bulk of our code goes into our Enum.reduce/3. What we are doing is constantly keeping the list length at 2 so we can easily just sum the values and compute the next Fibonacci value. My first implementation actually was a bit memory hungry because I was just appending my list continuously and then taking a sum of the last two values. Why keep the list long if you only want to compute that last value right?

And Bob's Your Uncle

That's it! Plain. Simple. Memory smart. Anyhow, you can checkout the full source here in this gist. Cheers!

By Adrian Cruz | Published Aug. 19, 2017, 10:08 p.m. | Permalink | tags: elixir

Elixir Conditionals with Function Signatures

On Learning Elixir

So, I've been learning and doing a bit more Elixir lately. I'm only a couple months in, but Elixir has been the primary language I have been coding in at work. Happy days!

This is not an introductory write-up nor a tutorial, but rather a quick look at how conditionals and guards are done in Elixir. Onward.

Let's Take a Look at FizzBuzz

So, yes, FizzBuzz; the classic programming problem and still a favourite interview question. As a quick refresher, this game is played by counting from one through n, but for multiples of three, we'll have "Fizz", multiples of five, "Buzz", and multiples of both, "FizzBuzz". I think that is the simplest I can word it.

Let's have a look at an example in Python first.

def fizz_buzz(n):
    if 0 == (n % 3) and 0 == (n % 5):
        return "FizzBuzz"
    elif 0 == (n % 3):
        return "Fizz"
    elif 0 == (n % 5):
        return "Buzz"
        return str(n)
There is nothing crazy going on here, the logic is simple and it does exactly what we need it to. It uses the usual if/else logic that you would expect.

So, now let's look at a solution in Elixir.

def fizz_buzz(n) when 0 === rem(n, 3) and 0 === rem(n, 5) do
def fizz_buzz(n) when 0 === rem(n, 3), do: "Fizz"
def fizz_buzz(n) when 0 === rem(n, 5), do: "Buzz"
def fizz_buzz(n), do: n
If you've never done any Elixir before, I'm sure you may be a bit confused.

Function Signatures as Conditionals

So, let me first say that, yes, if/else does exist in Elixir. "Then, why isn't it used here at all?", you ask? Well, Elixir has a good foundation of having their functions be explicit in what they do and being able to pipe your code is very useful. I won't write much about the pipe operator here, but if you don't know much about it in Elixir, I highly recommend learning about it.


So what we see in the Elixir FizzBuzz is an overloaded fizz_buzz/1 function. As I'm sure you've looked at it and studied it a bit by now, yes, that is being done in lieu of if/else. The important piece is the when part of the function signature which is called a guard. Using guards, we now have logic for which function we want to match on when we pass in the variable n. So now we see that each function signature matches up exactly to what we have done in Python with if/elif/else logic. Pretty neat eh?

By Adrian Cruz | Published March 14, 2017, 7:06 p.m. | Permalink | tags: elixir