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Zij gegeven:
:: BTree a = Tip a | Bin (BTree a) (BTree a)
map :: (a -> b) [a] -> [b]
map f [] = [] (1.)
map f [x:xs] = [f x : map f xs] (2.)
mapbtree :: (a -> b) (BTree a) -> BTree b
mapbtree f (Tip a) = Tip (f a) (3.)
mapbtree f (Bin t1 t2) = Bin (mapbtree f t1) (mapbtree f t2) (4.)
foldbtree :: (a a -> a) (BTree a) -> a
foldbtree f (Tip x) = x (5.)
foldbtree f (Bin t1 t2) = f (foldbtree f t1) (foldbtree f t2) (6.)
tips :: (BTree a) -> [a]
tips t = foldbtree (++) (mapbtree unit t) (7.)
unit :: a -> [a]
unit x = [x] (8.)
Te bewijzen:
voor alle functies f, voor alle eindige bomen t:
map f (tips t) = tips (mapbtree f t)
Bewijs:
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