How do i do a recursive sum of 1253 to give me 1 2 5 3 which is 11 and adding to get 2 until my last-CodePudding

defmodule Kata do
    def digital_root(n) do
    n |> Integer.digits() |> Enum.reduce(0, &Kernel. /2)
    end
end
n = 1253

CodePudding user response：

You can use a multi clause function for this: in the first one, return n unchanged if it's a single digit. Otherwise, compute the sum of digits (like you've already figured out) and recursively call the function on it again. Also, your reduce can be replaced with Enum.sum/1.

defmodule Kata do
  def digital_root(n) when n < 10, do: n
  def digital_root(n) do
    n |> Integer.digits() |> Enum.sum() |> digital_root()
  end
end

Test:

iex(1)> Kata.digital_root(0)
0
iex(2)> Kata.digital_root(1)
1
iex(3)> Kata.digital_root(9)
9
iex(4)> Kata.digital_root(91)
1
iex(5)> Kata.digital_root(912)
3
iex(6)> Kata.digital_root(9123)
6
iex(7)> Kata.digital_root(1253)
2

CodePudding user response：

  def digit_sum(number) when number < 10, do: number
  def digit_sum(number) when is_integer(number) do
    digit_sum(
      for digit <- Integer.digits(number), reduce: 0 do
        acc -> acc   digit
      end
    )
  end

I like Dogbert's better.