Counting elements in a tree in Haskell - haskell

Basically I've made a polymorphic tree data type and I need a way of counting the number of elements in a given tree. Here's the declaration for my Tree data type:
data Tree a = Empty
| Leaf a
| Node (Tree a) a (Tree a)
deriving (Eq, Ord, Show)
So I can define a tree of Ints like this:
t :: Tree Int
t = Node (Leaf 5) 7 (Node (Leaf 2) 3 (Leaf 7))
However, I need a function to count the number of elements in one of these lists. I've defined this recursive function but I get the error "inferred type is not general enough":
size :: Tree a -> Int
size Empty = 0
size (Leaf n) = 1
size (Node x y z) = size x + size y + size z
Is there something here I shouldn't be doing?

I think it's just a typo when you write
size (Node x y z) = size x + size y + size z
which should just be
size (Node x y z) = size x + size z + 1
since y is no subtree but just the element stored.
Or to make it even clearer
size (Node left elem right) = size left + size right + 1
Technically, your error occurs because the term size y does only make sense if y is again a tree whose size can be computed. Therefore the type of this clause would be inferred to Tree (Tree a) -> Int, which is, compared with the actual Tree a -> Int, not general
enough.

Look at you last clause: Looking at the left hand side, at Node x y z, what is the type of y? Does size y make sense?

Related

Can't get some simple Haskell code to compile, no matter how many different ways I try it

I'm trying to write a Haskell function that takes in a tree, and replaces every node with a pair containing the height of the subtree at that node, and the original node
Depending on where I place my parenthesis in the last line of code, I get all kinds of different errors. I know my height function works because I've used it for different functions previously. I'm clearly not grouping things together correctly, because I've gotten everything from not giving enough arguments to max, to giving too many arguments to pairs. Please help! I'm really stuck here and making no progress because I'm just moving parenthesis back and forth.
data Tree a = Tip | Bin (Tree a) a (Tree a) deriving (Show, Eq)
getHeight :: Tree a -> Integer
getHeight Tip = 0
getHeight (Bin l _ r) = (max (getHeight l) (getHeight r)) +1
pairs :: Tree a -> Tree (Integer, a)
pairs Tip = Tip
pairs (Bin l x r) = (Bin (pairs l) ((max (left right)) x) (pairs r))
where left = (getHeight l)
right = (getHeight r)
The way to call a function with two arguments is my separating them with whitespace, like this:
f x y
Or in your case, this would be:
max left right
The way to construct a pair is with parens and a comma, like this:
(42, "foo")
Or in your case, this would be:
(max left right, x)
Summing all of that up, the line should be:
pairs (Bin l x r) = Bin (pairs l) (max left right, x) (pairs r)

Working with Trees in Haskell

I have this data definition for a tree:
data Tree = Leaf Int | Node Tree Int Tree
and I have to make a function, nSatisfy, to check how many items of the tree check some predicate.
Here's what I've done:
nSatisfy :: (Int->Bool) -> Tree -> Int
nSatisfy _ Leaf = 0
nSatisfy y (Node left x right)
|y x = 1 + nSatisfy y (Node left x right)
| otherwise = nSatisfy y (Node left x right)
Is this the right way to solve this problem?
In your nSatisfy function, you should add the number of nodes satisfying the condition in both subtrees with two recursive calls. The last two lines should be like this:
|x y=1+(nSatisfy y left)+(nSatisfy y right)
|otherwise=(nSatisfy y left)+(nSatisfy y right)
This way, it will call itself again on the same node but only on the subtrees.
Also, if a leaf contains an integer, as is implied in the data declaration, you should make it evaluate the condition for a leaf and return 1 if it is true, instead of always returning 0.
In addition to the main answer, I'd like to offer a slightly different way how to generalize your problem and solving it using existing libraries.
The operation you're seeking is common to many data structures - to go through all elements and perform some operation on them. Haskell defines Foldable type-class, which can be implemented by structures like yours.
First let's import some modules we'll need:
import Data.Foldable
import Data.Monoid
In order to use Foldable, we need to generalize the structure a bit, in particular parametrize its content:
data Tree a = Leaf a | Node (Tree a) a (Tree a)
In many cases this is a good idea as it separates the structure from its content and allows it to be easily reused.
Now let's define its Foldable instance. For tree-like structures it's easier to define it using foldMap, which maps each element into a monoid and then combines all values:
instance Foldable Tree where
foldMap f (Leaf x) = f x
foldMap f (Node lt x rt) = foldMap f lt <> f x <> foldMap f rt
This immediately gives us the whole library of functions in the Data.Foldable module, such as searching for an element, different kinds of folds, etc. While a function counting the number of values satisfying some predicate isn't defined there, we can easily define it for any Foldable. The idea is that we'll use the Sum:
nSatisfy :: (Foldable f) => (a -> Bool) -> f a -> Int
nSatisfy p = getSum . foldMap (\x -> Sum $ if p x then 1 else 0)
The idea behind this function is simple: Map each value to 1 if it satisfies the predicate, otherwise to 0. And then folding with the Sum monoid just adds all values up.

Assign Consecutive Numbers to Elements in a BST

So I am trying to add consecutive numbers to the elements in a BST strictly using recursion (no standard prelude functions). Here is what I have so far:
data Tree a = Empty | Node a (Tree a) (Tree a) deriving (Show)
leaf x = Node x Empty Empty
number' :: Int -> Tree a -> Tree (Int, a)
number' a Empty = Empty
number' a (Node x xl xr) = Node (a,x) (number' (a+1) xl) (number' (a+1) xr)
number :: Tree a -> Tree (Int, a)
number = number' 1
number' is an auxiliary function that carries around "a" as a counter. It should add 1 to each recursive call, so I am not sure why it is doing what it is doing.
As of now the level of the element is assigned to each element. I would like the first element to be assigned 1, the element to the left of that 2, the element to the left of that 3, etc. Each element should get a+1 assigned to it and no number should be repeated. Thanks in advance.
I want to first explain why the code in the question assigns level numbers. This will lead us directly to two different solutions, one passed on caching, one based on doing two traversals at once. Finally, I show how the second solution relates to the solutions provided by other answers.
What has to be changed in the code from the question?
The code in the question assigns the level number to each node. We can understand why the code behaves like that by looking at the recursive case of the number' function:
number' a (Node x xl xr) = Node (a,x) (number' (a+1) xl) (number' (a+1) xr)
Note that we use the same number, a + 1, for both recursive calls. So the root nodes in both subtrees will get assigned the same number. If we want each node to have a different number, we better pass different numbers to the recursive calls.
What number should we pass to the recursive call?
If we want to assign the numbers according to a left-to-right pre-order traversal, then a + 1 is correct for the recursive call on the left subtree, but not for the recursive call on the right subtree. Instead, we want to leave out enough numbers to annotate the whole left subtree, and then start annotating the right subtree with the next number.
How many numbers do we need to reserve for the left subtree? That depends on the subtree's size, as computed by this function:
size :: Tree a -> Int
size Empty = 0
size (Node _ xl xr) = 1 + size xl + size xr
Back to the recursive case of the number' function. The smallest number annotated somewhere in the left subtree is a + 1. The biggest number annotated somewhere in the left subtree is a + size xl. So the smallest number available for the right subtree is a + size xl + 1. This reasoning leads to the following implementation of the recursive case for number' that works correctly:
number' :: Int -> Tree a -> Tree (Int, a)
number' a Empty = Empty
number' a (Node x xl xr) = Node (a,x) (number' (a+1) xl) (number' (a + size xl + 1) xr)
Unfortunately, there is a problem with this solution: It is unnecessarily slow.
Why is the solution with size slow?
The function size traverses the whole tree. The function number' also traverses the whole tree, and it calls size on all left subtrees. Each of these calls will traverse the whole subtree. So overall, the function size gets executed more than once on the same node, even though it always returns the same value, of course.
How can we avoid traversing the tree when calling size?
I know two solutions: Either we avoid traversing the tree in the implementation of size by caching the sizes of all trees, or we avoid calling size in the first place by numbering the nodes and computing the size in one traversal.
How can we compute the size without traversing the tree?
We cache the size in every tree node:
data Tree a = Empty | Node Int a (Tree a) (Tree a) deriving (Show)
size :: Tree a -> Int
size Empty = 0
size (Node n _ _ _) = n
Note that in the Node case of size, we just return the cached size. So this case is not recursive, and size does not traverse the tree, and the problem with our implementation of number' above goes away.
But the information about the size has to come from somewhere! Everytime we create a Node, we have to provide the correct size to fill the cache. We can lift this task off to smart constructors:
empty :: Tree a
empty = Empty
node :: a -> Tree a -> Tree a -> Tree a
node x xl xr = Node (size xl + size xr + 1) x xl xr
leaf :: a -> Tree a
leaf x = Node 1 x Empty Empty
Only node is really necessary, but I added the other two for completeness. If we always use one of these three functions to create a tree, the cached size information will always be correct.
Here is the version of number' that works with these definitions:
number' :: Int -> Tree a -> Tree (Int, a)
number' a Empty = Empty
number' a (Node _ x xl xr) = node (a,x) (number' (a+1) xl) (number' (a + size xl + 1) xr)
We have to adjust two things: When pattern matching on Node, we ignore the size information. And when creating a Node, we use the smart constructor node.
That works fine, but it has the drawback of having to change the definition of trees. On the one hand, caching the size might be a good idea anyway, but on the other hand, it uses some memory and it forces the trees to be finite. What if we want to implement a fast number' without changing the definition of trees? This brings us to the second solution I promised.
How can we number the tree without computing the size?
We cannot. But we can number the tree and compute the size in a single traversal, avoiding the multiple size calls.
number' :: Int -> Tree a -> (Int, Tree (Int, a))
Already in the type signature, we see that this version of number' computes two pieces of information: The first component of the result tuple is the size of the tree, and the second component is the annotated tree.
number' a Empty = (0, Empty)
number' a (Node x xl xr) = (sl + sr + 1, Node (a, x) yl yr) where
(sl, yl) = number' (a + 1) xl
(sr, yr) = number' (a + sl + 1) xr
The implementation decomposes the tuples from the recursive calls and composes the components of the result. Note that sl is like size xl from the previous solution, and sr is like size xr. We also have to name the annotated subtrees: yl is the left subtree with node numbers, so it is like number' ... xl in the previous solution, and yr is the right subtree with node numbers, so it is like number' ... xr in the previous solution.
We also have to change number to only return the second component of the result of number':
number :: Tree a -> Tree (Int, a)
number = snd . number' 1
I think that in a way, this is the clearest solution.
What else could we improve?
The previous solution works by returning the size of the subtree. That information is then used to compute the next available node number. Instead, we could also return the next available node number directly.
number' a Empty = (a, Empty)
number' a (Node x xl xr) = (ar, Node (a, x) yl yr) where
(al, yl) = number' (a + 1) xl
(ar, yr) = number' al xr
Note that al is like a + sl + 1 in the previous solution, and ar is like a + sl + sr + 1. Clearly, this change avoids some additions.
This is essentially the solution from Sergey's answer, and I would expect that this is the version most Haskellers would write. You could also hide the manipulations of a, al and ar in a state monad, but I don't think that really helps for such a small example. The answer by Ankur shows how it would look like.
data Tree a = Empty | Node a (Tree a) (Tree a) deriving (Show)
number :: Tree a -> Tree (Int, a)
number = fst . number' 1
number' :: Int -> Tree a -> (Tree (Int, a), Int)
number' a Empty = (Empty, a)
number' a (Node x l r) = let (l', a') = number' (a + 1) l
(r', a'') = number' a' r
in (Node (a, x) l' r', a'')
*Tr> let t = (Node 10 (Node 20 (Node 30 Empty Empty) (Node 40 Empty Empty)) (Node 50 (Node 60 Empty Empty) Empty))
*Tr> t
Node 10 (Node 20 (Node 30 Empty Empty) (Node 40 Empty Empty)) (Node 50 (Node 60 Empty Empty) Empty)
*Tr> number t
Node (1,10) (Node (2,20) (Node (3,30) Empty Empty) (Node (4,40) Empty Empty)) (Node (5,50) (Node (6,60) Empty Empty) Empty)
As suggested by comments in your question that each call to number should return a integer also which needs to be further used for next set of nodes. This makes the signature of the function to:
Tree a -> Int -> (Tree (Int,a), Int)
Looking at the last part of it, it looks like a candidate for State monad i.e state -> (Val,state).
Below code shows how you can do this using State monad.
import Control.Monad.State
data Tree a = Empty | Node a (Tree a) (Tree a) deriving (Show)
myTree :: Tree String
myTree = Node "A" (Node "B" (Node "D" Empty Empty) (Node "E" Empty Empty)) (Node "C" (Node "F" Empty Empty) (Node "G" Empty Empty))
inc :: State Int ()
inc = do
i <- get
put $ i + 1
return ()
number :: Tree a -> State Int (Tree (Int,a))
number Empty = return Empty
number (Node x l r) = do
i <- get
inc
l' <- number l
r' <- number r
return $ Node (i,x) l' r'
main = do
putStrLn $ show (fst (runState (number myTree) 1))

Turning a tree into a heap in haskell

I need to make an implementation of a priority queue with a Heap Tree in Haskell, for example:
Given a list: [3,2,7,8,4,1,9]
3 is the main root
2 is its left leaf
7 is its right leaf
8 is the left leaf of 2
4 is the right leaf of 2
1 is the left leaf of 7
9 is the right leaf of 7
If I want to heapifiy the tree it would be like this:
7 > 3 so we exchange them
8 > 2 we exchange them
8 > 7 we exchange them
9 > 3 we exchange them
9 > 8 we exchange them
We end with a list like this: [9,7,8,2,4,1,3]
And 9 is the element with the highest number (priority) in our queue.
I will need to do this:
insert h e that inserts the element e in the heap h (in the last position)
delete h that removes the element with the highest priority (in our example 9)
heapify h that heapifies the tree.
But my problem is the heapify function, I dont even know where to start. That's why im asking for clues or advice.
module Heapify where
Let's use the tree type
data Tree a = Leaf a | Node (Tree a) a (Tree a)
deriving Show
and the example tree
ourTree = Node (Node (Leaf 8) 2 (Leaf 4)) 3 (Node (Leaf 1) 7 (Leaf 9))
And work out how to heapify it.
Your description in pictures
Top node
Left subtree
Right subtree
Heapified?
In this case, the result is indeed a heap, but this method isn't the standard way of doing heapification, and doesn't generalise (as far as I can tell) to something that does make sure you have a heap. Credit to Will Ness for pointing this out.
How should we heapify?
A tree satisfies the heap property if each parent node is no smaller than its child nodes. (It says nothing about the compararive sizes of the child nodes.)
Heapification actually works a bit like insertion sort, in that you start at the low end, and work gradually up, dragging small elements back into place as you introduce them.
Step 1: Heapify the left and right subtrees
Step 2: This node: Check if the top value should be pulled down
Step 3: If so, heapify at that side again
Steps 1,2 and 4 are just recursive calls, so let's concentrate on the top node:
Top node
We need to (a) see the value at the top of the subtrees and (b) be able to replace it.
atTop :: Tree a -> a
atTop (Leaf a) = a
atTop (Node _ a _) = a
replaceTop :: Ord a => Tree a -> a -> Tree a
replaceTop (Leaf _) a = Leaf a
replaceTop (Node l _ r) a = heapify (Node l a r)
Notice the cheeky forward reference to heapify? When we replace the top node of a tree, we need to re-heapify it to make sure it's still a tree.
Now let's see how to adjust at the left hand side if necessary.
It's necessary if the top of the left subtree, topL, is larger than the value a at the node. If it's <= we don't need to do anything, so leave the node alone.
adjustLeft :: Ord a => Tree a -> Tree a
adjustLeft (Leaf a) = Leaf a -- But we shouldn't ask to do this.
adjustLeft node#(Node l a r)
| topL <= a = node
| otherwise = Node (replaceTop l a) topL r
where topL = atTop l
and at the right:
Now let's adjust at the right hand side if necessary. This works exactly the same.
adjustRight :: Ord a => Tree a -> Tree a
adjustRight (Leaf a) = Leaf a -- But we shouldn't ask to do this.
adjustRight node#(Node l a r)
| topR <= a = node
| otherwise = Node l topR (replaceTop r a)
where topR = atTop r
Let's see some of that working:
*Heapify> ourTree
Node (Node (Leaf 8) 2 (Leaf 4)) 3 (Node (Leaf 1) 7 (Leaf 9))
*Heapify> atTop ourTree
3
Pull down to the Left or right?
If the current value belongs lower down the tree, we need to pull it down the left or the right side, by swapping it with the larger value of the two. We pick the larger value so we know it's more than the top value in the left subtree.
doTop :: Ord a => Tree a -> Tree a
doTop (Leaf a) = Leaf a
doTop node#(Node l a r)
| atTop l > atTop r = adjustLeft node
| otherwise = adjustRight node
Remember that adjustLeft and adjustRight make a recursive call to heapify.
Return of the heapification
So to heapify, we just
heapify :: Ord a => Tree a -> Tree a
heapify (Leaf a) = Leaf a
heapify (Node l a r) = doTop (Node (heapify l) a (heapify r))
OK, that was easy. Let's test it:
*Heapify> ourTree
Node (Node (Leaf 8) 2 (Leaf 4)) 3 (Node (Leaf 1) 7 (Leaf 9))
*Heapify> heapify ourTree
Node (Node (Leaf 2) 8 (Leaf 4)) 9 (Node (Leaf 1) 7 (Leaf 3))

Haskell 2-3-4 Tree

We've been asked to create a 2-3-4 tree in Haskell, as in write the data type, the insert function, and a display function.
I'm finding it very difficult to get information on this kind of tree, even in a language I'm comfortable with (Java, C++).
What I have so far -
data Tree t = Empty
| Two t (Tree t)(Tree t)
| Three t t (Tree t)(Tree t)(Tree t)
| Four t t t (Tree t)(Tree t)(Tree t)(Tree t) deriving (Eq, Ord, Show)
leaf2 a = Two a Empty Empty
leaf3 a b = Three a b Empty Empty Empty
leaf4 a b c = Four a b c Empty Empty Empty Empty
addNode::(Ord t) => t -> Tree t -> Tree t
addNode t Empty = leaf2 t
addNode x (Two t left right)
| x < t = Two t (addNode x left) right
| otherwise = Two t left (addNode x right)
This compiles but I'm not sure if it's correct, but not sure how to start writing the insert into a three node or four node.
The assignment also says that "deriving show" for the display function is not enough, that it should print out the tree in the format normally seen in diagrams. Again, unsure on the way to go with this.
Any help or direction appreciated.
I know nothing about 2-3-4 trees, but for the Three node, you would start with something like this:
addNode t (Three x y left mid right)
| cond1 = expr1
| cond2 = expr2
(etc)
What cond1, cond2, expr1, and expr2 are, exactly, is dependent on the definition of what a 2-3-4 tree is.
As for a show method, the general outline would be this:
instance (Show t) => Show (Tree t) where
show Empty = ...
show (Two x l r) = ...show x...show l...show r...
show (Three x y l m r) = ...
show (Four x y z l m n r) = ...
The implementation depends on how you want it to look, but for the non-Empty cases, you will probably invoke show on all of the components of the tree being shown. If you want to indent the nested parts of the tree, then perhaps you should create a separate method:
instance (Show t) => Show (Tree t) where
show = showTree 0
showTree :: Show t => Int -> Tree t -> String
showTree n = indent . go
where indent = (replicate n ' ' ++)
go Empty = "Empty"
go (Two x l r) = (...show x...showTree (n+1) l...showTree (n+1) r...)
(etc)
We've been asked to create a 2-3-4 tree
My condolences. I myself once had to implement one for homework. A 2-3-4 tree is a B-tree with all the disadvantages of the B-tree and none of the advantages, because writing the cases separately for each number of children as you do is as cumbersome as having a list of only 2-4 elements.
Point being: B-tree insertion algorithms should work, just fix the size. Cormen et al. have pseudocode for one in their book Introduction to algorithms (heavy imperativeness warning!).
It might still be better to have lists of data elements and children instead of the four-case algebraic data type, even if the type wouldn't enforce the size of the nodes then. At least it would make it easier to expand the node size.

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