Scoring Scrabble Words with Elm

The Problem

In approximately 6th century B.C., Cleobulus of Lindos, one of the Seven Sages of Greece, presented to the world a word-based mind game¹. He called it "Scrabble". When outlining his game, he omitted an important part, perhaps on purpose -- how to compute the score of a word. A universal solution has yet to arise.

Assumptions

In order to provide a reasonable solution that will hold in this dimension, we must first make some assumptions.

1) The letters A, E, I, O, U, L, N, R, S, and T have the value 1.
2) The letters D and G have the value 2.  
3) The letters B, C, M, and P have the value 3.  
4) The letters F, H, V, W, and Y have the value 4.  
5) The letter K has the value 5.  
6) The letters J and X have the value 8.  
7) The letters Q and Z have the value 10.  

A Solution

Let us begin.

Breaking the problem down and outlining a solution in pseudocode is good for simple scripting problems like this one, so let's start there.

-- given a word
-- split the word up into letters
-- get the points value for each letter
-- store / record the point values for each letter
-- add up all the point values
-- return the sum

I find it helpful to start with the type signature when constructing functions in Elm. The function we want to build will take a word and return a score. In terms the compiler can understand, it will take a String and return an Int.

scoreWord: String -> Int

Next, we can start building our function.

scoreWord: String -> Int
scoreWord word = 
   word

Okay, so splitting the word up into letters is easy:

scoreWord: String -> Int
scoreWord word = 
  String.split "" word

Now we need to get the points value for each letter. This introduces a bit of complexity into our problem -- where do the point values come from? Elm doesn't know what a letter is worth. One way to tell it is to make a record, a series of key-value pairs.

pointsCategories =
    { one = [ "A", "E", "I", "O", "U", "L", "N", "R", "S", "T" ]
    , two = [ "D", "G" ]
    , three = [ "B", "C", "M", "P" ]
    , four = [ "F", "H", "V", "W", "Y" ]
    , five = [ "K" ]
    , eight = [ "J", "X" ]
    , ten = [ "Q", "Z" ]
    }

Each key of our record is the word version of the point value. Each value is a list containing every letter that has that point value. Records in Elm cannot have number literals for keys, so this is the closest we can get.

Now, we need a function that will score a letter by looking up it's point value in the pointsCategories record and converting that to a number.

getPointsForLetter : String -> Int
getPointsForLetter letter =
    if List.member letter pointsCategories.one then
        1
    else if List.member letter pointsCategories.two then
        2
    else if List.member letter pointsCategories.three then
        3
    else if List.member letter pointsCategories.four then
        4
    else if List.member letter pointsCategories.five then
        5
    else if List.member letter pointsCategories.eight then
        8
    else if List.member letter pointsCategories.ten then
        10
    else
        0

Okay, this function is a little (very) redundant. I'm starting to get the feeling that this program is a wee bit smelly and could use some tidying up, but let's get this solution working before we refactor.

This would be a good time to note that all the letters in our pointsCategories record are uppercase, but we do not have that guarantee for a given word, unless you're playing a very loud version of Cleobulus's game.

In order to address this, we might as well make all of our letters uppercase, since that's how they appear in the pointsCategories record -- WITH ENERGY!

scoreWord : String -> Int
scoreWord word =
  List.map String.toUpper (String.split "" word)

Now we have a list of uppercase letters to work with so we can use our getPointsValue function to convert letters to points.

scoreWord : String -> Int
scoreWord word = 
  List.map getPointsForLetter (List.map String.toUpper (String.split "" word))

These parentheses are getting out of hand -- time to take advantage of the Elm's brilliant pipe operator to make our solution readable.

The pipe operator is nice because it turns

f(g(h(i(x))))

into

x 
  |> i 
  |> h 
  |> g 
  |> f

where x is some value and i, h, g, and f are functions.

Applying this nifty tool to our solution, we get:

scoreWord : String -> Int
scoreWord word =
  word 
    |> String.split ""
    |> List.map String.toUpper
    |> List.map getPointsForLetter

The astute reader will notice we are not yet returning an Int, but rather a list of Int's (or List Int using Elm syntax). Let's fix that.

scoreWord : String -> Int
scoreWord word = 
  word 
    |> String.split ""
    |> List.map String.toUpper
    |> List.map getPointsForLetter
    |> List.sum

This solution works, but I detest the repetitive chain of if else if's in the getPointsForLetter function. Unfortunately, this is the crux of the solution, so this refactor is going to be a big one!

All Aboard the Refactor Tractor

The main issue with the above solution is the fact that we can't use a number literal as the key in our pointsCategories record. This means we are separating looking up the value and converting it to an Int. We will need to reach into our toolbox and find a new data structure to use in order to solve this. I propose we use a dictionary (Dict) to tie a number to a character. We need to import Dict to use it.

import Dict exposing (..)

pointValues : Dict Char Int
pointValues =
    Dict.fromList
        [ ( 'A', 1 )
        , ( 'E', 1 )
        , ( 'I', 1 )
        , ( 'O', 1 )
        , ( 'U', 1 )
        , ( 'L', 1 )
        , ( 'N', 1 )
        , ( 'R', 1 )
        , ( 'S', 1 )
        , ( 'T', 1 )
        , ( 'D', 2 )
        , ( 'G', 2 )
        , ( 'B', 3 )
        , ( 'C', 3 )
        , ( 'M', 3 )
        , ( 'P', 3 )
        , ( 'F', 4 )
        , ( 'H', 4 )
        , ( 'V', 4 )
        , ( 'M', 4 )
        , ( 'Y', 4 )
        , ( 'K', 5 )
        , ( 'J', 8 )
        , ( 'X', 8 )
        , ( 'Q', 10 )
        , ( 'Z', 10 )
        ]

Now we can access the point value for a letter using the Dict.get method. This change of data structure is crucial because we have moved the redundancy out of our program and into the data itself. Redundancy in data is gravy, redundancy in code is gunk.

Let's write a function that takes a character and returns it's point value.

getPointValue : Char -> Int
getPointValue letter = 
    let
        letterScore =
            Dict.get letter pointValues
    in
    case letterScore of
        Just score ->
            score

        Nothing ->
            0

Dict.get is not guaranteed to return something since it can be called with a value that may not correspond to a key in the dictionary, like if we tried to get the point value of a Chinese character. In terms that Elm's compiler can understand, Dict.get returns a Maybe type, so we need to tell the program how to handle both cases of the Maybe type -- Just (we got something) and Nothing (no luck).

If we get a score back, then return that score. If not, default to 0.

Time to rewrite scoreWord so it splits our word into a list of Char's (String.toList) and uses our new getPointValue function to get the value of each character in the list.

scoreWord : String -> Int
scoreWord word =
    word
        |> String.toUpper
        |> String.toList
        |> List.map getPointValue
        |> List.sum

This works, but I still think getPointValue is a bit verbose. Let's use a couple tricks to make it easier to understand.

flip is a neat tool that lets you flip the order of arguments to a function. Maybe.withDefault is a shorthand way to do what we did in our case statement. Taking advantage of these two handy mechanisms and the pipe operator, we can write getPointValue like this:

getPointValue : Char -> Int
getPointValue letter =
  letter 
    |> flip Dict.get pointValues
    |> Maybe.withDefault 0

Ah, much better.

To cap off the post, here's the entire script:

module ScrabbleScore exposing (..)

import Dict exposing (..)


pointValues : Dict Char Int
pointValues =
    Dict.fromList
        [ ( 'A', 1 )
        , ( 'E', 1 )
        , ( 'I', 1 )
        , ( 'O', 1 )
        , ( 'U', 1 )
        , ( 'L', 1 )
        , ( 'N', 1 )
        , ( 'R', 1 )
        , ( 'S', 1 )
        , ( 'T', 1 )
        , ( 'D', 2 )
        , ( 'G', 2 )
        , ( 'B', 3 )
        , ( 'C', 3 )
        , ( 'M', 3 )
        , ( 'P', 3 )
        , ( 'F', 4 )
        , ( 'H', 4 )
        , ( 'V', 4 )
        , ( 'M', 4 )
        , ( 'Y', 4 )
        , ( 'K', 5 )
        , ( 'J', 8 )
        , ( 'X', 8 )
        , ( 'Q', 10 )
        , ( 'Z', 10 )
        ]


getPointValue : Char -> Int
getPointValue letter =
    letter
        |> flip Dict.get pointValues
        |> Maybe.withDefault 0


scoreWord : String -> Int
scoreWord word =
    word
        |> String.toUpper
        |> String.toList
        |> List.map getPointValue
        |> List.sum

While we can never know exactly how Cleobulus intended for a word to be scored, we can give our best guess and this is mine.

If you like my style, check out my blog or follow me on twitter at @_vincecampanale for occasional web development goodies in between memes and retweets :)

¹ I made this up.