FizzBuzz in recursion?

taro

I’m overthinking FizzBuzz in Elixir, refactoring everything possible Elixir way.

Now I’m trying to find a recursive way to put an IO list together so it will result in ["Fizz", ["Buzz"]]. But I’m quite stuck, recursion is difficult to grasp. So I’ve decided to ask smarter people.

defmodule FizzBuzz do
  ...

  def change(num) do
    # Recurse in with an integer in closure
    # Recurse out leaving string each step
  end

  defp name_multiple(num) when rem(num, 3) == 0, do: "Fizz"
  defp name_multiple(num) when rem(num, 5) == 0, do: "Buzz"
  defp name_multiple(_), do: ""
end

I’m not asking for the simpler solutions of FizzBuzz. I want to join FizzBuzz instead of returning them separately, so it’s more extensible. I also tried to use multi-clause functions and guards as much as I can.

My previous solution is

defmodule FizzBuzz do
  @moduledoc """
  Name multiples. 

    * Multiples of 3 => "Fizz"
    * Multiples of 5 => "Buzz"

  Concatenates when many are applicable.

    * Multiples of 3 and 5 => "FizzBuzz"

  Unapplicable numbers remain intact.
  """

  @spec list(integer) :: list
  def list(max) when is_integer(max), do: list(1..max)

  @spec list(Enumerable.t()) :: list
  def list(enum), do: Enum.map(enum, &change/1)

  @spec change(integer) :: integer | String.t()
  def change(num) do
    num
    |> prefer(fizz(num))
    |> prefer(buzz(num))
  end

  defp prefer(old, new) when not new, do: old
  defp prefer(old, new) when not is_binary(old), do: new
  defp prefer(old, new), do: old <> new

  defp fizz(num) when rem(num, 3) == 0, do: "Fizz"
  defp fizz(_), do: false

  defp buzz(num) when rem(num, 5) == 0, do: "Buzz"
  defp buzz(_), do: false
end

2 comments

#algorithm #quiz

0 668 2

2023-02-15 18:53:36 UTC

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Adzz

So the thing with recursion is you first want to ensure you know the conditions under which you want to stop recuring.

In this case that’s not clear, but I assume you want to halt once you have reached some maximum number of iterations. So let’s first call a recursive function, but implement the break condition:

defmodule FizBuzz do
  def change(max) do
     change(max, [], max - 1)
  end

  def change(_number, result, iterations) when iterations == 0 do
    result
  end 
end

If we call it like this FizzBuzz.change(0) we will be returned [].

Now we can add the other conditions

defmodule FizBuzz do
  def change(max) do
     change(max, [], max)
  end

  def change(number, result, iterations) when iterations <= 0 do
    result
  end

  def change(number, result, iterations) when rem(number, 15) == 0 do 
    change(number - 1, ["FizzBuzz" | result], iterations - 1)
  end

  def change(number, result, iterations) when rem(number, 5) == 0 do
    change(number - 1, ["Buzz" | result], iterations - 1)
  end

  def change(number, result, iterations) when rem(number, 3) == 0 do
    change(number - 1, ["Fizz" | result], iterations - 1)
  end

  def change(number, result, iterations) do
    change(number - 1, [number | result], iterations - 1)
  end
end

Post #1

100phlecs

I want to join FizzBuzz instead of returning them separately, so it’s more extensible

Here’s what I came up with.

defmodule FizzBuzz do
  def convert(num, lst \\ [])
  def convert(num, lst) when rem(num, 5) == 0, do: num |> extract(5) |> convert(["Buzz" | lst])
  def convert(num, lst) when rem(num, 3) == 0, do: num |> extract(3) |> convert(["Fizz" | lst])
  def convert(_num, lst) when length(lst) > 0, do: lst
  def convert(num, _lst), do: num

  defp extract(num, factor) when rem(num, factor) == 0, do: num |> div(factor) |> extract(factor)
  defp extract(num, _factor), do: num
end

maybe_join = fn
  elem when is_list(elem) -> Enum.join(elem)
  num when is_integer(num) -> num
end

1..100
|> Enum.map(&FizzBuzz.convert/1)
|> Enum.map(maybe_join)

Post #2

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