OpenAI Generates A Recursive Function that Calculates the Fibonacci Sequence

That’s definitely faster than:

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

But it seems OpenAI needs to learn about functional programming and how to write idiomatic Elixir code.

According to the above blog post, the following implementation is the fastest!

defmodule Fibonacci do
  def fibonacci(number) do
   Enum.reverse(fibonacci_do(number))
  end

  def fibonacci_do(1), do: [0]
  def fibonacci_do(2), do: [1|fibonacci_do(1)]
  def fibonacci_do(number) when number > 2 do
    [x, y|_] = all = fibonacci_do(number-1)
    [x + y|all]
  end
end

It generated the list for fib(40): [0, 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, 10946, 17711, 28657, 46368, 75025, 121393, 196418, 317811, 514229, 832040, 1346269, 2178309, 3524578, 5702887, 9227465, 14930352, 24157817, 39088169, 63245986]
in an instant!

2nd best:

Rosetta code algorithm.

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

    def fib(_, prv, -1), do: prv
    def fib(prvprv, prv, n) do
        next = prv + prvprv
        fib(prv, next, n-1)
    end
end

IO.inspect Enum.map(0..10, fn i-> Fibonacci.fib(i) end)