Algebraic data type
m (Adding glossary category)
|(5 intermediate revisions by 4 users not shown)|
Latest revision as of 13:31, 27 June 2014
This is a type where we specify the shape of each of the elements. Wikipedia has a thorough discussion. "Algebraic" refers to the property that an Algebraic Data Type is created by "algebraic" operations. The "algebra" here is "sums" and "products":
- "sum" is alternation (, meaningA | BorAbut not both)B
- "product" is combination (, meaningA BandAtogether)B
- is a pair of numbers, andata Pair = P Int Doubleand aInttogether. The tagDoubleis used (in constructors and pattern matching) to combine the contained values into a single structure that can be assigned to a variable.P
- is just one number, either andata Pair = I Int | D Doubleor else aInt. In this case, the tagsDoubleandIare used (in constructors and pattern matching) to distinguish between the two alternatives.D
Sums and products can be repeatedly combined into an arbitrarily large structures.
Algebraic Data Type is not to be confused with *Abstract* Data Type, which (ironically) is its opposite, in some sense. The initialism "ADT" usually means *Abstract* Data Type, but GADT usually means Generalized *Algebraic* Data Type.
 1 Tree examples
Suppose we want to represent the following tree:
5 / \ 3 7 / \ 1 4
We may actually use a variety of Haskell data declarations that will handle this. The choice of algebraic data types determines its structural/shape properties.
 1.1 Binary search tree
In this example, values are stored at each node, with smaller values to the left, greater to the right.
data Stree a = Tip | Node (Stree a) a (Stree a)
and then our example tree would be:
etree = Node (Node (Node Tip 1 Tip) 3 (Node Tip 4 Tip)) 5 (Node Tip 7 Tip)
To maintain the order, such a tree structure is usually paired with a smart constructor.
 1.2 Rose tree
Alternatively, it may be represented in what appears to be a totally different stucture.
data Rose a = Rose a [Rose a]
In this case, the example tree would be:
retree = Rose 5 [Rose 3 [Rose 1 , Rose 4], Rose 7 ]