Difference between revisions of "Algebraic data type"
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===Rose tree=== |
===Rose tree=== |
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− | + | Alternatively, it may be represented in what appears to be a totally different stucture. |
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<haskell> |
<haskell> |
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data Rose a = Rose a [Rose a] |
data Rose a = Rose a [Rose a] |
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</haskell> |
</haskell> |
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− | In this case, the |
+ | In this case, the example tree would be: |
<haskell> |
<haskell> |
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retree = Rose 5 [Rose 3 [Rose 1 [], Rose 4[]], Rose 7 []] |
retree = Rose 5 [Rose 3 [Rose 1 [], Rose 4[]], Rose 7 []] |
Revision as of 23:26, 18 July 2009
This is a type where we specify the shape of each of the elements.
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.
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.
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 []]
The differences between the two are that the (empty) binary search tree Tip
is not representable as a Rose
tree, and a Rose tree can have an arbitrary and internally varying branching factor (0,1,2, or more).