I am trying to use an unfold function to build trees.
Tree t = Leaf | Node (Tree t) t (Tree t)
unfoldT :: (b -> Maybe (b, a, b)) -> b -> Tree a
unfoldT f b =
case f b of
Nothing -> Leaf
Just (lt, x, rt) -> Node (unfoldT f lt) x (unfoldT f rt)
The build function needs to create a tree that has a height equal to the number provided, as well as be numbered in an in-order fashion. The base case being build 0 = Leaf and the next being build 1 = Node (Leaf 0 Leaf).
build :: Integer -> Tree Integer
My attempt at solving it:
build n = unfoldT (\x -> Just x) [0..2^n-2]
I am not entirely sure how to go about constructing the tree here.
Would love it if somebody could point me in the right direction.
Edit 1:
If I was to use a 2-tuple, what would I combine? I need to be able to refer to the current node, its left subtree and its right subtree somehow right?
If I was to use a 2-tuple, what would I combine?
I would recommend to pass the remaining depth as well as the offset from the left:
build = unfoldT level . (0,)
where
level (_, 0) = Nothing
level (o, n) = let mid = 2^(n-1)
in ((o, n-1), o+mid-1, (o+mid, n-1))
If I was to use a 2-tuple, what would I combine?
That's the key question behind the state-passing paradigm in functional programming, expressed also with the State Monad. We won't be dealing with the latter here, but maybe use the former.
But before that, do we really need to generate all the numbers in a list, and then work off that list? Don't we know in advance what are the numbers we'll be working with?
Of course we do, because the tree we're building is totally balanced and fully populated.
So if we have a function like
-- build2 (depth, startNum)
build2 :: (Int, Int) -> Tree Int
we can use it just the same to construct both halves of e.g. the build [0..14] tree:
build [0..14] == build2 (4,0) == Node (build2 (3,0)) 7 (build2 (3,8))
Right?
But if we didn't want to mess with the direct calculations of all the numbers involved, we could arrange for the aforementioned state-passing, with the twist to build2's interface:
-- depth, startNum tree, nextNum
build3 :: (Int, Int) -> (Tree Int, Int)
and use it like
build :: Int -> Tree Int -- correct!
build depth = build3 (depth, 0) -- intentionally incorrect
build3 :: (Int, Int) -> (Tree Int, Int) -- correct!
build3 (depth, start) = Node lt n rt -- intentionally incorrect
where
(lt, n) = build3 (depth-1, start) -- n is returned
(rt, m) = build3 (depth-1, n+1) -- and used, next
You will need to tweak the above to make all the pieces fit together (follow the types!), implementing the missing pieces of course and taking care of the corner / base cases.
Formulating this as an unfold would be the next step.
Related
I try to create random trees from this structure:
data Tree a = Leaf a
| Funktion (Tree a) (Tree a)
| Lambda (Tree a)
deriving (Show)
The problem I have is that I don't even know how to generate a tree with the depth of (for example) 2 which only has "Lambda" as nodes. If someone could help me with generating this simple tree (depth 2) I could generate them randomly.
If I implement a function like this:
build (Tree a) 0 = Leaf "A"
build (Tree a) n = build (Lambda a) (n-1)
It won't work since the function build itself expects a Tree as input. Actually I need trees which have the nodes Lambda or Funktion, but first of all I need to understand how to generate a simple version of this structure.
It sounds like you want something like
build :: Natural -- ^ Number, n, of Lambda nodes
-> a -- ^ Value, v, to store in the Leaf
-> Tree a -- ^ Lambda^n (Leaf v)
build 0 a = Leaf a
build n a = Lambda (build (n - 1) a)
So that build 4 "A" will produce
Lambda (Lambda (Lambda (Lambda (Leaf "A"))))
Unfortunately, the rest of your question (about generating random trees) really requires substantially more context to answer.
You're close - your build function at the moment will always return a Leaf, as the base case of the recursion doesn't do anything with its argument. You don't actually need the argument:
build :: Integer -> Tree String
build 0 = Leaf "A"
build n = Lambda $ build (n-1)
This would produce what your build function seems to be intending, that is a simple "tree" of given depth n composed of a Leaf node and Lambda nodes. Eg:
λ> build 2
Lambda (Lambda (Leaf "A"))
To get your randomly generated tree, you need to look at the System.Random module.
For example building on the previous, a tree of random height between a given upper and lower bound:
buildRandom :: (Integer, Integer) -> IO (Tree String)
buildRandom bounds = randomRIO bounds >>= return . build
This can be extended and modified to produce the behaviour you want (a fully random generated tree), but requires knowledge of monads etc., which might take some extra reading.
I want to define an infinite tree in Haskell using infinitree :: Tree, but want to set a pattern up for each node, defining what each node should be. The pattern is 1 more then then its parent. I am struggling on how to set up a tree to begin with, and how and where to define the pattern of each node?
Thank you
Infinite data structures can generally be defined by functions which call themselves but have no base case. Usually these functions don't need to pattern match on their arguments. For example, a list equal to [1..] can be written as
infiniteList :: [Int]
infiniteList = go 1 where
go n = n : go (n+1)
You can use the exact same technique for a tree:
data Tree a = Node (Tree a) a (Tree a) | Nil deriving (Show)
infiniteTree :: Tree Int
infiniteTree = go 1 where
go n = Node (go (2*n)) n (go (2*n+1))
This defines the infinite tree
1
/ \
2 3
/ \ / \
4 5 6 7
...
A type for infinite binary trees with no leaves:
data Tree a = Tree (Tree a) a (Tree a)
One general pattern for doing this sort of thing is called unfold. For this particular type:
unfold :: (a -> (a,b,a)) -> a -> Tree b
Can you see how to define this function and use it for your purpose?
I've been thinking in how to implement the equivalent of unfold for the following type:
data Tree a = Node (Tree a) (Tree a) | Leaf a | Nil
It was not immediately obvious since the standard unfold for lists returns a value and the next seed. For this datatype, it doesn't make sense, since there is no "value" until you reach a leaf node. This way, it only really makes sense to return new seeds or stop with a value. I'm using this definition:
data Drive s a = Stop | Unit a | Branch s s deriving Show
unfold :: (t -> Drive t a) -> t -> Tree a
unfold fn x = case fn x of
Branch a b -> Node (unfold fn a) (unfold fn b)
Unit a -> Leaf a
Stop -> Nil
main = print $ unfold go 5 where
go 0 = Stop
go 1 = Unit 1
go n = Branch (n - 1) (n - 2)
While this seems to work, I'm not sure this is how it is supposed to be. So, that is the question: what is the correct way to do it?
If you think of a datatype as the fixpoint of a functor then you can see that your definition is the sensible generalisation of the list case.
module Unfold where
Here we start by definition the fixpoint of a functor f: it's a layer of f followed by some more fixpoint:
newtype Fix f = InFix { outFix :: f (Fix f) }
To make things slightly clearer, here are the definitions of the functors corresponding to lists and trees. They have basically the same shape as the datatypes except that we have replace the recursive calls by an extra parameter. In other words, they describe what one layer of list / tree looks like and are generic over the possible substructures r.
data ListF a r = LNil | LCons a r
data TreeF a r = TNil | TLeaf a | TBranch r r
Lists and trees are then respectively the fixpoints of ListF and TreeF:
type List a = Fix (ListF a)
type Tree a = Fix (TreeF a)
Anyways, hopping you now have a better intuition about this fixpoint business, we can see that there is a generic way of defining an unfold function for these.
Given an original seed as well as a function taking a seed and building one layer of f where the recursive structure are new seeds, we can build a whole structure:
unfoldFix :: Functor f => (s -> f s) -> s -> Fix f
unfoldFix node = go
where go = InFix . fmap go . node
This definition specialises to the usual unfold on list or your definition for trees. In other words: your definition was indeed the right one.
Given a list of steps:
>>> let path = ["item1", "item2", "item3", "item4", "item5"]
And a labeled Tree:
>>> import Data.Tree
>>> let tree = Node "item1" [Node "itemA" [], Node "item2" [Node "item3" []]]
I'd like a function that goes through the steps in path matching the labels in tree until it can't go any further because there are no more labels matching the steps. Concretely, here it falls when stepping into "item4" (for my use case I still need to specify the last matched step):
>>> trav path tree
["item3", "item4", "item5"]
If I allow [String] -> Tree String -> [String] as the type of trav I could write a recursive function that steps in both structures at the same time until there are no labels to match the step. But I was wondering if a more general type could be used, specifically for Tree. For example: Foldable t => [String] -> t String -> [String]. If this is possible, how trav could be implemented?
I suspect there could be a way to do it using lens.
First, please let's use type Label = String. String is not exactly descriptive and might not be ideal in the end...
Now. To use Traversable, you need to pick a suitable Applicative that can contain the information you need for deciding what to do in its "structure". You only need to pass back information after a match has failed. That sounds like some Either!
A guess would thus be Either [Label] (t Label) as the pre-result. That would mean, we use the instantiation
traverse :: Traversable t
=> (Label -> Either [Label] Label) -> t Label -> Either [Label] (t Label)
So what can we pass as the argument function?
travPt0 :: [Label] -> Label -> Either [Label] Label
travPt0 ls#(l0 : _) label
| l0 /= label = Left ls
| otherwise = Right label ?
The problem is, traverse will then fail immediately and completely if any node has a non-matching label. Traversable doesn't actually have a notion of "selectively" diving down into a data structure, it just passes through everything, always. Actually, we only want to match on the topmost node at first, only that one is mandatory to match at first.
One way to circumvent immediate deep-traversal is to first split up the tree into a tree of sub-trees. Ok, so... we need to extract the topmost label. We need to split the tree in subtrees. Reminds you of anything?
trav' :: (Traversable t, Comonad t) => [Label] -> t Label -> [Label]
trav' (l0 : ls) tree
| top <- extract tree
= if top /= l0 then l0 : ls
else let subtrees = duplicate tree
in ... ?
Now amongst those subtrees, we're basically interested only in the one that matches. This can be determined from the result of trav': if the second element is passed right back again, we have a failure. Unlike normal nomenclature with Either, this means we wish to go on, but not use that branch! So we need to return Either [Label] ().
else case ls of
[] -> [l0]
l1:ls' -> let subtrees = duplicate tree
in case traverse (trav' ls >>> \case
(l1':_)
| l1'==l1 -> Right ()
ls'' -> Left ls''
) subtrees of
Left ls'' -> ls''
Right _ -> l0 : ls -- no matches further down.
I have not tested this code!
We'll take as reference the following recursive model
import Data.List (minimumBy)
import Data.Ord (comparing)
import Data.Tree
-- | Follows a path into a 'Tree' returning steps in the path which
-- are not contained in the 'Tree'
treeTail :: Eq a => [a] -> Tree a -> [a]
treeTail [] _ = []
treeTail (a:as) (Node a' trees)
| a == a' = minimumBy (comparing length)
$ (a:as) : map (treeTail as) trees
| otherwise = as
which suggests that the mechanism here is less that we're traversing through the tree accumulating (which is what a Traversable instance might do) but more that we're stepping through the tree according to some state and searching for the deepest path.
We can characterize this "step" by a Prism if we like.
import Control.Lens
step :: Eq a => a -> Prism' (Tree a) (Forest a)
step a =
prism' (Node a)
(\n -> if rootLabel n == a
then Just (subForest n)
else Nothing)
This would allow us to write the algorithm as
treeTail :: Eq a => [a] -> Tree a -> [a]
treeTail [] _ = []
treeTail pth#(a:as) t =
maybe (a:as)
(minimumBy (comparing length) . (pth:) . map (treeTail as))
(t ^? step a)
but I'm not sure that's significantly more clear.
I am impatient, looking forward to understanding catamorphism related to this SO question :)
I have only practiced the beginning of Real World Haskell tutorial. So, Maybe I'm gonna ask for way too much right now, if it was the case, just tell me the concepts I should learn.
Below, I quote the wikipedia code sample for catamorphism.
I would like to know your opinion about foldTree below, a way of traversing a Tree, compared to this other SO question and answer, also dealing with traversing a Tree n-ary tree traversal. (independantly from being binary or not, I think the catamorphism below can be written so as to manage n-ary tree)
I put in comment what I understand, and be glad if you could correct me, and clarify some things.
{-this is a binary tree definition-}
data Tree a = Leaf a
| Branch (Tree a) (Tree a)
{-I dont understand the structure between{}
however it defines two morphisms, leaf and branch
leaf take an a and returns an r, branch takes two r and returns an r-}
data TreeAlgebra a r = TreeAlgebra { leaf :: a -> r
, branch :: r -> r -> r }
{- foldTree is a morphism that takes: a TreeAlgebra for Tree a with result r, a Tree a
and returns an r -}
foldTree :: TreeAlgebra a r -> Tree a -> r
foldTree a#(TreeAlgebra {leaf = f}) (Leaf x ) = f x
foldTree a#(TreeAlgebra {branch = g}) (Branch l r) = g (foldTree a l) (foldTree a r)
at this point I am having many difficulties, I seem to guess that the morphism leaf
will be applied to any Leaf
But so as to use this code for real, foldTree needs to be fed a defined TreeAlgebra,
a TreeAlgebra that has a defined morphism leaf so as to do something ?
but in this case in the foldTree code I would expect {f = leaf} and not the contrary
Any clarification from you would be really welcome.
Not exactly sure what you're asking. But yeah, you feed a TreeAlgebra to foldTree corresponding to the computation you want to perform on the tree. For example, to sum all the elements in a tree of Ints you would use this algebra:
sumAlgebra :: TreeAlgebra Int Int
sumAlgebra = TreeAlgebra { leaf = id
, branch = (+) }
Which means, to get the sum of a leaf, apply id (do nothing) to the value in the leaf. To get the sum of a branch, add together the sums of each of the children.
The fact that we can say (+) for branch instead of, say, \x y -> sumTree x + sumTree y is the essential property of the catamorphism. It says that to compute some function f on some recursive data structure it suffices to have the values of f for its immediate children.
Haskell is a pretty unique language in that we can formalize the idea of catamorphism abstractly. Let's make a data type for a single node in your tree, parameterized over its children:
data TreeNode a child
= Leaf a
| Branch child child
See what we did there? We just replaced the recursive children with a type of our choosing. This is so that we can put the subtrees' sums there when we are folding.
Now for the really magical thing. I'm going to write this in pseudohaskell -- writing it in real Haskell is possible, but we have to add some annotations to help the typechecker which can be kind of confusing. We take the "fixed point" of a parameterized data type -- that is, constructing a data type T such that T = TreeNode a T. They call this operator Mu.
type Mu f = f (Mu f)
Look carefully here. The argument to Mu isn't a type, like Int or Foo -> Bar. It's a type constructor like Maybe or TreeNode Int -- the argument to Mu itself takes an argument. (The possibility of abstracting over type constructors is one of the things that makes Haskell's type system really stand out in its expressive power).
So the type Mu f is defined as taking f and filling in its type parameter with Mu f itself. I'm going to define a synonym to reduce some of the noise:
type IntNode = TreeNode Int
Expanding Mu IntNode, we get:
Mu IntNode = IntNode (Mu IntNode)
= Leaf Int | Branch (Mu IntNode) (Mu IntNode)
Do you see how Mu IntNode is equivalent to your Tree Int? We have just torn the recursive structure apart and then used Mu to put it back together again. This gives us the advantage that we can talk about all Mu types at once. This gives us what we need to define a catamorphism.
Let's define:
type IntTree = Mu IntNode
I said the essential property of the catamorphism is that to compute some function f, it suffices to have the values of f for its immediate children. Let's call the type of the thing we are trying to compute r, and the data structure node (IntNode would be a possible instantiation of this). So to compute r on a particular node, we need the node with its children replaced with their rs. This computation has type node r -> r. So a catamorphism says that if we have one of these computations, then we can compute r for the entire recursive structure (remember recursion is denoted explicitly here with Mu):
cata :: (node r -> r) -> Mu node -> r
Making this concrete for our example, this looks like:
cata :: (IntNode r -> r) -> IntTree -> r
Restating, if we can take a node with rs for its children and compute an r, then we can compute an r for an entire tree.
In order to actually compute this, we need node to be a Functor -- that is we need to be able to map an arbitrary function over the children of a node.
fmap :: (a -> b) -> node a -> node b
This can be done straightforwardly for IntNode.
fmap f (Leaf x) = Leaf x -- has no children, so stays the same
fmap f (Branch l r) = Branch (f l) (f r) -- apply function to each child
Now, finally, we can give a definition for cata (the Functor node constraint just says that node has a suitable fmap):
cata :: (Functor node) => (node r -> r) -> Mu node -> r
cata f t = f (fmap (cata f) t)
I used the parameter name t for the mnemonic value of "tree". This is an abstract, dense definition, but it is really very simple. It says: recursively perform cata f -- the computation we are doing over the tree -- on each of t's children (which are themselves Mu nodes) to get a node r, and then pass that result to f compute the result for t itself.
Tying this back to the beginning, the algebra you are defining is essentially a way of defining that node r -> r function. Indeed, given a TreeAlgebra, we can easily get the fold function:
foldFunction :: TreeAlgebra a r -> (TreeNode a r -> r)
foldFunction alg (Leaf a) = leaf alg a
foldFunction alg (Branch l r) = branch alg l r
Thus the tree catamorphism can be defined in terms of our generic one as follows:
type Tree a = Mu (TreeNode a)
treeCata :: TreeAlgebra a r -> (Tree a -> r)
treeCata alg = cata (foldFunction alg)
I'm out of time. I know that got really abstract really fast, but I hope it at least gave you a new viewpoint to help your learning. Good luck!
I think you were were asking a question about the {}'s. There is an earlier question with a good discussion of {}'s. Those are called Haskell's record syntax. The other question is why construct the algebra. This is a typical function paradigm where you generalize data as functions.
The most famous example is Church's construction of the Naturals, where f = + 1 and z = 0,
0 = z,
1 = f z,
2 = f (f z),
3 = f (f (f z)),
etc...
What you are seeing is essentially the same idea being applied to a tree. Work the church example and the tree will click.