Getting recursively drunk with monoids

Disclaimer: I am not a mixologist. This is not professional cocktail advice!

Sam Horvath-Hunt blogged about modelling cocktails as monoids. This is a really cool example of FP modelling that I want to expand on. (Lennart Kolmodin once wrote that the dance steps of Tango form a monoid.)

First, I will demonstrate my ignorance by assuming that cocktail recipes are free commutative monoids over ingredients:

• The order in which you add ingredients does not matter,
• If you add two cocktails together, you get another cocktail,
• The identity cocktail is the empty cocktail with no ingredients in it.

Second, I want to up the ante with a recursive cocktail recipe.

It comes from a sketch in the computer science student revue at DIKU (University of Copenhagen): Superdrinks (2002); credits go to Uffe Christensen, Uffe Friis Lichtenberg, Jonas Ussing, Niels H. Christensen, Torben Æ. Mogensen, Jørgen Elgaard Larsen who either co-wrote or enacted the sketch.

A superdrinks consists of:

• 1 part gin,
• 2 parts lemon,
• 3 parts superdrinks.

Now, according to the sketch there are plenty of bad ways to materialize this drink; one such is through approximation: take 1 part gin, 2 parts lemon and 3 parts of whatever is your current best approximation of superdrinks. Iterate this process enough times and you will have a gradually finer superdrinks.

Recursively,

superdrinks(n) = 1 × gin + 2 × lemon + 3 × superdrinks (n-1) 

As for superdrinks(0), it could be water. It could be gin!

Experimenting a little,

superdrinks(1) = 1 × gin + 2 × lemon + 3 × superdrinks(0) superdrinks(2) = 1 × gin + 2 × lemon + 3 × superdrinks(1) = 1 × gin + 2 × lemon + 3 × (1 × gin + 2 × lemon + 3 × superdrinks(0)) = 4 × gin + 8 × lemon + 9 × superdrinks(0) 

The relationship between the number of parts of each ingredient can be expressed in closed form eliminating recursion:

superdrinks(n) = (3ⁿ - 1)/2 × gin + (3ⁿ - 1) × lemon + 3ⁿ × superdrinks(0) 

(You can find the closed form either by recognizing that the series 3 × 3 × ... with n occurrences is 3ⁿ, that there's always one less part lemon than superdrinks(0), and that there's always half the amount of gin of that; or you can solve their recurrence relation; or you can expand the three number series using a function,

> let superdrinks (gin, lemon, super) = (1 + 3*gin, 2 + 3*lemon, 3*super) > unzip3 $ take 6 $ iterate superdrinks (0,0,1) ([0,1,4,13,40,121],[0,2,8,26,80,242],[1,3,9,27,81,243]) 

and look them up.)

It is time to get schwifty.

The following ingredients are enough to make gin-tonic and superdrinks:

data Ingredient = Gin | Tonic | Lemon deriving (Eq, Ord, Show) 

A cocktail is any set of ingredients and their multiplicity:

newtype Cocktail = Cocktail (Map Ingredient Natural) deriving (Eq, Ord, Show) emptyCocktail :: Cocktail emptyCocktail = Cocktail Map.empty 

For convenience,

parts :: Natural -> Ingredient -> Cocktail parts n ingredient = Cocktail (Map.singleton ingredient n) combine :: Cocktail -> Cocktail -> Cocktail combine (Cocktail c1) (Cocktail c2) = Cocktail (Map.unionWith (+) c1 c2) 

One consequence of this modelling is:

> let gintonic = combine (1 `parts` Gin) (2 `parts` Tonic) > gintonic == combine gintonic gintonic False 

Since these are cocktail recipes, I'd like to normalize the quantities of each ingredient so that recipes don't eventually say "2 parts gin, 4 parts tonic" or "0 parts gin":

normalize :: Cocktail -> Cocktail normalize (Cocktail ingredients) = Cocktail . normalize' $ ingredients where scale = foldr1 gcd (Map.elems ingredients) normalize' = Map.map (`div` scale) . Map.filter (/= 0) 

(Note that while foldr1 is partial, because of Haskell's non-strict semantics, it is never evaluated when ingredients is empty because it is used within Map.map zero times.)

Demonstrating normalize:

> normalize emptyCocktail Cocktail (fromList []) > normalize (0 `parts` Gin) Cocktail (fromList []) > normalize $ combine (2 `parts` Gin) (4 `parts` Tonic) Cocktail (fromList [(Gin,1),(Tonic,2)]) 

It would be tempting to specialize the Eq Cocktail instance to use normalize so that c == combine c c for all c. But I don't like to do that because if you ever need to compare for structural equality, you can't, whereas equality under normalization can be achieved with:

> let (=~) = (==) `on` normalize > (1 `parts` Gin) =~ (2 `parts` Gin) True 

It would also be tempting to add normalization to combine so that the combination of two cocktails is a normalized cocktail. But since this blog post is about monoidal cocktails and combine is the best candidate for a composition operator, such choice actually breaks the law of associativity:

> let norm_combine c1 c2 = normalize (combine c1 c2) > gin1 `norm_combine` (gin1 `norm_combine` tonic1) Cocktail (fromList [(Gin,2),(Tonic,1)]) > (gin1 `norm_combine` gin1) `norm_combine` tonic1 Cocktail (fromList [(Gin,1),(Tonic,1)]) 

So while I like the notion of normalizing cocktail recipes, making it a part of the Semigroup Cocktail instance would probably be a bad idea, leaving the much simpler instances:

instance Semigroup Cocktail where (<>) = combine instance Monoid Cocktail where mempty = emptyCocktail 

This begs the question: Is the glass half full or half mempty?

As for superdrinks, the recipe can now be expressed as:

superdrinks :: Natural -> Cocktail -> Cocktail superdrinks n base = mconcat [ ((3^n - 1) `div` 2) `parts` Gin , (3^n - 1) `parts` Lemon , (3^n) `rounds` base ] rounds :: Natural -> Cocktail -> Cocktail rounds n = mconcat . genericReplicate n 

Whether the best approximation is using a base of mempty or, as the revue sketch suggests, n `parts` Gin, is highly subjective. For a 5th approximation of superdrinks using pure gin as 0th approximation,

> normalize $ superdrinks 5 (1 `parts` Gin) Cocktail (fromList [(Gin,182),(Lemon,121)]) 

Cheers!