Roman number conversion challenge question

Hi guys!

I’m super new to Elixir, and my English can be a little bad, so I’m sorry about that.

Well, I’m here to get some advice on my code, it works well, but I don’t know if there’s anything I can do to make it more like the “Elixir way of things”. So if anyone can rate it and give me some tips :slight_smile:

Ps: The challenge is only for numbers up to 3000

Here’s the code

defmodule Converter do

  @one "I"
  @five "V"
  @ten "X"
  @fifty "L"
  @onehundread "C"
  @fivehundread "D"
  @onethousand "M"

  def to_roman_number(number,roman_number \\ "",n \\ 3)

  def to_roman_number(number,roman_number,_n) when number == 4,do: roman_number<>@one<>@five

  def to_roman_number(number,roman_number,_n) when div(number,10) == 4 do
    roman_conversion  = roman_number<>@ten<>@fifty
    rest = rem(number,10)
    to_roman_number(rest,roman_conversion,0)
  end
  def to_roman_number(number,roman_number,_n) when div(number,100) == 4 do
    roman_conversion  = roman_number<>@onehundread<>@fivehundread
    rest = rem(number,100)
    to_roman_number(rest,roman_conversion,1)
  end

  def to_roman_number(number,roman_number,_n) when number == 5, do: roman_number<>@five

  def to_roman_number(number,roman_number,_n) when div(number,10) == 5 do
    roman_conversion  = roman_number<>@fifty
    rest = rem(number,10)
    to_roman_number(rest,roman_conversion,0)
  end
  def to_roman_number(number,roman_number,_n) when div(number,100) == 5 do
    roman_conversion  = roman_number<>@fivehundread
    rest = rem(number,100)
    to_roman_number(rest,roman_conversion,1)
  end

  def to_roman_number(number,roman_number,_n) when number == 9,do: roman_number<>@one<>@ten

  def to_roman_number(number,roman_number,_n) when div(number,10) == 9 do
    roman_conversion  = roman_number<>@ten<>@onehundread
    rest = rem(number,10)
    to_roman_number(rest,roman_conversion,0)
  end
  def to_roman_number(number,roman_number,_n) when div(number,100) == 9 do
    roman_conversion  = roman_number<>@onehundread<>@onethousand
    rest = rem(number,100)
    to_roman_number(rest,roman_conversion,1)
  end


  def to_roman_number(number,roman_number,n) do
    base = trunc(:math.pow(10,n))
    quotient = div(number,base)
    rest = rem(number,base)
    cond do
      n == 3 ->
        roman_conversion = roman_number<>String.duplicate(@onethousand,quotient)
        to_roman_number(rest,roman_conversion,n-1)
      n == 2 ->
        roman_conversion = roman_number<>String.duplicate(@onehundread,quotient)
        to_roman_number(rest,roman_conversion,n-1)
      n == 1 ->
        roman_conversion = roman_number<>String.duplicate(@ten,quotient)
        to_roman_number(rest,roman_conversion,n-1)
      n == 0 ->
        roman_conversion = roman_number<>String.duplicate(@one,quotient)
        IO.puts "The converted number is "<>roman_conversion
    end
  end
end


It is repeating topic there, as I believe this is task in Exercism or other project like that. So check out if any other topic on that task helps you.

3 Likes

Since this is Exercism, I suggest requesting a mentor there to look at your code for some tips. That’s why they’re there :wink: Besides that, take a look at the most upvoted solutions for some inspiration. That really helped me think differently.

Edit: having said that, your code could definitely be more concise. Make use of the fact that you know all the possible mappings.

3 Likes

Sorry for the delay to answer hahahaha

I will check the other answers, thanks brother!

Sorry for the delay in responding too my brother

Thank you very much for the tip, the solution is a bit verbose, I’ll try to improve it!