Calculating Scrabble scores with “reduce”

May 7, 2014 . By Reuven

This is the second installment of my series of blog posts on the “reduce” function/method.  For an introduction, see here.

I love to play Scrabble — or more commonly nowadays, I play Words with Friends on my phone. (I often say that the game should instead be called, “Words with people who used to be your friends,” given the competition that can result.)  The game, if you’re not familiar with it, is simple enough to understand: You choose seven English letters at a time from a bag or pool of letters.  Each letter has a different point value.  Your aim is to create, with your seven letters, the highest-scoring word that you can in each turn.

Fine, it’s a bit more complex than that: Your word needs to hook onto an existing word or letter, and you can get bonus points by completing more than one word, and there are squares which multiply your score, by doubling or tripling the value of an individual letter or the entire word.  I’m going to ignore all of these rules for now.

Let’s say that I want to write a Ruby program that will calculate the score of a given word in Scrabble. How can I do that?

Before I can begin, I’m going to need to need a list of the points that you get for each letter.  Because we’re dealing with a mapping between two values — a letter and its points — the easiest solution would be a hash:

points = {'a'=> 1, 'b'=> 3, 'c'=> 3, 'd'=> 2, 'e'=> 1, 'f'=> 4, 'g'=> 2, 'h'=>
 4, 'i'=> 1, 'j'=> 8, 'k'=> 5, 'l'=> 1, 'm'=> 3, 'n'=> 1, 'o'=> 1, 'p'=> 3,
 'q'=> 10, 'r'=> 1, 's'=> 1, 't'=> 1, 'u'=> 1, 'v'=> 4, 'w'=> 4, 'x'=> 8, 'y'=>
 4, 'z'=> 10}

Once we’ve defined our “points” hash, we can start calculating the number of points that we’ll get for a given word.  If you aren’t familiar with “reduce”, but have been programming in Ruby for at least a short while, you’ll probably realize that this is easily accomplished with a loop.  That is, we can iterate over each letter in the string, getting the point value for that letter, and adding it to the total.  For example, we could do something like this:

word = 'encyclopedia'
total = 0 
word.each_char do |letter|
  total += points[letter]
end

After the above loop, we’ll find that the total score is 22 points.  That’s fine, and it’ll work… but by using “reduce”, we can do the same thing in a single line:

word.split('').reduce(0) {|total, current| total + points[current]}

Now, how does the above work? First, it takes the string “word”, and turns it into an array, in which each letter is an element.  Then, we use “reduce” to calculate a score: Since it’s a numeric score, we’ll initialize the total with 0.  Then, for each letter in our word, we add the number of points associated with that word to our total.

To do the same thing in Python will require surprisingly similar code. First, we create a dictionary:

points = {'a': 1, 'b': 3, 'c': 3, 'd': 2, 'e': 1, 'f': 4, 'g': 2, 'h':
 4, 'i': 1, 'j': 8, 'k': 5, 'l': 1, 'm': 3, 'n': 1, 'o': 1, 'p': 3,
 'q': 10, 'r': 1, 's': 1, 't': 1, 'u': 1, 'v': 4, 'w': 4, 'x': 8, 'y':
 4, 'z': 10}

Then we run reduce in a similar way. Because strings in Python are sequences, we don’t need to turn our word into an an array before iterating over it; we can invoke “reduce” directly on the string:

>>> word = 'encyclopedia'
>>> reduce(lambda total, current: total + points[current], word, 0)
22

This is a simple demonstration of where “reduce” can be useful. I’m looking to apply the same operation on the elements of a sequence of data, adding the latest value to the total.  When I get to the end, I’d like to see what has accumulated so far.

In the next installment, we’ll see how we can use “reduce” to implement some other functions that we wouldn’t normally associate with functional programming, let alone with something called “reduce”.

 

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