Haskell's algebraic data types

Question

I'm trying to fully understand all of Haskell's concepts.

In what ways are algebraic data types similar to generic types, e.g., in C# and Java? And how are they different? What's so algebraic about them anyway?

I'm familiar with universal algebra and its rings and fields, but I only have a vague idea of how Haskell's types work.

Answer 1

Haskell's algebraic data types are named such since they correspond to an initial algebra in category theory, giving us some laws, some operations and some symbols to manipulate. We may even use algebraic notation for describing regular data structures, where:

• + represents sum types (disjoint unions, e.g. Either).
• • represents product types (e.g. structs or tuples)
• X for the singleton type (e.g. data X a = X a)
• 1 for the unit type ()
• and μ for the least fixed point (e.g. recursive types), usually implicit.

with some additional notation:

• X² for X•X

In fact, you might say (following Brent Yorgey) that a Haskell data type is regular if it can be expressed in terms of 1, X, +, •, and a least fixed point.

With this notation, we can concisely describe many regular data structures:

• Units: data () = ()

  1

• Options: data Maybe a = Nothing | Just a

  1 + X

• Lists: data [a] = [] | a : [a]

  L = 1+X•L

• Binary trees: data BTree a = Empty | Node a (BTree a) (BTree a)

  B = 1 + X•B²

Other operations hold (taken from Brent Yorgey's paper, listed in the references):

• Expansion: unfolding the fix point can be helpful for thinking about lists. L = 1 + X + X² + X³ + ... (that is, lists are either empty, or they have one element, or two elements, or three, or ...)

• Composition, ◦, given types F and G, the composition F ◦ G is a type which builds “F-structures made out of G-structures” (e.g. R = X • (L ◦ R) ,where L is lists, is a rose tree.

• Differentiation, the derivative of a data type D (given as D') is the type of D-structures with a single “hole”, that is, a distinguished location not containing any data. That amazingly satisfy the same rules as for differentiation in calculus:

  1′ = 0

  X′ = 1

  (F + G)′ = F' + G′

  (F • G)′ = F • G′ + F′ • G

  (F ◦ G)′ = (F′ ◦ G) • G′

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References:

• Species and Functors and Types, Oh My!, Brent A. Yorgey, Haskell’10, September 30, 2010, Baltimore, Maryland, USA
• Clowns to the left of me, jokers to the right (Dissecting Data Structures), Conor McBride POPL 2008