So, this tree is NOT a Binary Search Tree. It is in no particular order, and is just in this order for quick access to specific indices (nth element), rather than whether an element exists or not.
The form of the Tree is like so:
data Tree a = Leaf a | Node Int (Tree a) (Tree a) deriving Show
For this specific tree, the "Int" from the Node constructor is the number of elements underneath that node (or number of leaves).
Using this structure, I copied parts of the Tree functions available in a lecture I found online (that I slightly modified when trying to understand):
buildTree :: [a] -> Tree a
buildTree = growLevel . map Leaf
where
growLevel [node] = node
growLevel l = growLevel $ inner l
inner [] = []
inner (e1:e2:rest) = e1 <> e2 : inner rest
inner xs = xs
join l#(Leaf _) r#(Leaf _) = Node 2 l r
join l#(Node ct _ _) r#(Leaf _) = Node (ct+1) l r
join l#(Leaf _) r#(Node ct _ _) = Node (ct+1) l r
join l#(Node ctl _ _) r#(Node ctr _ _) = Node (ctl+ctr) l r
And I was able to create some basic functions for moving through a tree. I made one that finds the nth element and returns it. I also made a Path datatype and implemented a function to return the path (in left and rights) to a specific index, and one function that can travel through a path and return that Node/Leaf.
Now, what I would like to make is a delete function. The problem here is with the fact that the tree is "leafy", or at least that is what is causing me difficulties.
If I end up with a Leaf at the deletion path, there is no "Null" or equivalent item to replace it with. Additionally, if I try to stop at the last path (like [L]), and check if that's a Node or not, then if it's a leaf replace the whole node with the opposite side etc., I run into the problem of changing the whole tree to reflect that change, not just return the end of the deletion, and change all the numbers from the tree to reflect the change in leaves.
I would like order to be preserved when deleting an item, like if you were to use a list as a simpler example:
del 4 [1, 2, 3, 4, 5, 6, 7] = [1, 2, 3, 4, 6, 7]
If there is a simpler way to structure the Tree (that still can contain duplicate elements and preserve order) what is it?
Is there some way to delete an element using this method?
If I ... replace the whole node with the opposite side ... I run into the problem of changing the whole tree to reflect that change, not just return the end of the deletion, and change all the numbers from the tree to reflect the change in leaves.
Well, not the whole tree - just the path from the deleted node back to the root. And isn't that exactly what you want?
I guess the first step would be, define what you mean by "delete". Should the indexes of undeleted nodes remain the same after deletion, or should nodes after the deleted node have their indexes reduced by one? That is, given:
tree :: [a] -> Tree a
-- get and del both 0-indexed, as in your example
get :: Int -> Tree a -> Maybe a
del :: Int -> Tree a -> Tree a
then of course
get 5 $ tree [1..7]
should yield Just 6. But what about
get 5 . del 4 $ tree [1..7]
? If you want this to still yield Just 6 (there is a "blank" spot in your tree where 5 used to be), that is a rather tricky concept, I think. You can put Nothings in to make space, if you define Leaf (Maybe a) instead of Leaf a, but this only papers over the problem: inserts will still shift indices around.
I think it is much simpler for this to yield Just 7 instead, making del 4 $ tree [1..7] the same as tree [1,2,3,4,6,7]. If this is your goal, then you simply must renumber all the nodes on the path from the deleted node back to the root: there is no getting around the fact that they all have one fewer leaf descendant now. But the other nodes in the tree can remain untouched.
For reference, one possible implementation of del:
count :: Tree a -> Int
count (Leaf _) = 1
count (Node s _ _) = s
del :: Int -> Tree a -> Maybe (Tree a)
del n t | n < 0 || n >= size || size <= 1 = Nothing
| otherwise = go n t
where size = count t
go n (Leaf _) = Nothing
go n (Node s l r) | n < size = reparent flip l r
| otherwise = reparent id r l
where reparent k c o = pure . maybe o (k (Node (s - 1)) o) $ go n c
size = count l
If I end up with a Leaf at the deletion path, there is no "Null" or equivalent item to replace it with.
Well, make one :). This is what Maybe is for: when you delete an element from a Tree, you cannot expect to get a Tree back, because Tree is defined to be nonempty. You need to explicitly add the possibility of emptiness by wrapping in Maybe. Deletion may also fail with an out-of-bounds error, which I represent with Either Int and incorporate into the logic.
delete :: Int -> Tree a -> Either Int (Maybe (Tree a))
delete i t | i >= max = Left (i - max) where max = count t
delete _ (Leaf _) = Right Nothing
delete i (Node n l r) = case delete i l of
Left i' -> Just <$> maybe l (Node (n - 1) l) <$> delete i' r
Right l' -> Right $ Just $ maybe r (\x -> Node (n - 1) x r) l'
Where count is as I recommended in the comments:
count :: Tree a -> Int
count (Leaf _) = 1
count (Node n _ _) = n
Related
I recently started using Haskell and it will probably be for a short while. Just being asked to use it to better understand functional programming for a class I am taking at Uni.
Now I have a slight problem I am currently facing with what I am trying to do. I want to build it breadth-first but I think I got my conditions messed up or my conditions are also just wrong.
So essentially if I give it
[“A1-Gate”, “North-Region”, “South-Region”, “Convention Center”, “Rectorate”, “Academic Building1”, “Academic Building2”] and [0.0, 0.5, 0.7, 0.3, 0.6, 1.2, 1.4, 1.2], my tree should come out like
But my test run results are haha not what I expected. So an extra sharp expert in Haskell could possibly help me spot what I am doing wrong.
Output:
*Main> l1 = ["A1-Gate", "North-Region", "South-Region", "Convention Center",
"Rectorate", "Academic Building1", "Academic Building2"]
*Main> l3 = [0.0, 0.5, 0.7, 0.3, 0.6, 1.2, 1.4, 1.2]
*Main> parkingtree = createBinaryParkingTree l1 l3
*Main> parkingtree
Node "North-Region" 0.5
(Node "A1-Gate" 0.0 EmptyTree EmptyTree)
(Node "Convention Center" 0.3
(Node "South-Region" 0.7 EmptyTree EmptyTree)
(Node "Academic Building2" 1.4
(Node "Academic Building1" 1.2 EmptyTree EmptyTree)
(Node "Rectorate" 0.6 EmptyTree EmptyTree)))
A-1 Gate should be the root but it ends up being a child with no children so pretty messed up conditions.
If I could get some guidance it would help. Below is what I've written so far::
data Tree = EmptyTree | Node [Char] Float Tree Tree deriving (Show,Eq,Ord)
insertElement location cost EmptyTree =
Node location cost EmptyTree EmptyTree
insertElement newlocation newcost (Node location cost left right) =
if (left == EmptyTree && right == EmptyTree)
then Node location cost (insertElement newlocation newcost EmptyTree)
right
else if (left == EmptyTree && right /= EmptyTree)
then Node location cost (insertElement newlocation newcost EmptyTree)
right
else if (left /= EmptyTree && right == EmptyTree)
then Node location cost left
(insertElement newlocation newcost EmptyTree)
else Node newlocation newcost EmptyTree
(Node location cost left right)
buildBPT [] = EmptyTree
--buildBPT (xs:[]) = insertElement (fst xs) (snd xs) (buildBPT [])
buildBPT (x:xs) = insertElement (fst x) (snd x) (buildBPT xs)
createBinaryParkingTree a b = buildBPT (zip a b)
Thank you for any guidance that might be provided. Yes I have looked at some of the similar questions I do think my problem is different but if you think a certain post has a clear answer that will help I am willing to go and take a look at it.
Here's a corecursive solution.
{-# bft(Xs,T) :- bft( Xs, [T|Q], Q). % if you don't read Prolog, see (*)
bft( [], Nodes , []) :- maplist( =(empty), Nodes).
bft( [X|Xs], [N|Nodes], [L,R|Q]) :- N = node(X,L,R),
bft( Xs, Nodes, Q).
#-}
data Tree a = Empty | Node a (Tree a) (Tree a) deriving Show
bft :: [a] -> Tree a
bft xs = head nodes -- Breadth First Tree
where
nodes = zipWith g (map Just xs ++ repeat Nothing) -- values and
-- Empty leaves...
(pairs $ tail nodes) -- branches...
g (Just x) (lt,rt) = Node x lt rt
g Nothing _ = Empty
pairs ~(a: ~(b:c)) = (a,b) : pairs c
{-
nodes!!0 = g (Just (xs!!0)) (nodes!!1, nodes!!2) .
nodes!!1 = g (Just (xs!!1)) (nodes!!3, nodes!!4) . .
nodes!!2 = g (Just (xs!!2)) (nodes!!5, nodes!!6) . . . .
................ .................
-}
nodes is the breadth-first enumeration of all the subtrees of the result tree. The tree itself is the top subtree, i.e., the first in this list. We create Nodes from each x in the input xs, and when the input
is exhausted we create Emptys by using an indefinite number of Nothings instead (the Empty leaves' true length is length xs + 1 but we don't need to care about that).
And we didn't have to count at all.
Testing:
> bft [1..4]
Node 1 (Node 2 (Node 4 Empty Empty) Empty) (Node 3 Empty Empty)
> bft [1..10]
Node 1
(Node 2
(Node 4
(Node 8 Empty Empty)
(Node 9 Empty Empty))
(Node 5
(Node 10 Empty Empty)
Empty))
(Node 3
(Node 6 Empty Empty)
(Node 7 Empty Empty))
How does it work: the key is g's laziness, that it doesn't force lt's nor rt's value, while the tuple structure is readily served by -- very lazy in its own right -- pairs. So both are just like the not-yet-set variables in that Prolog pseudocode(*), when served as 2nd and 3rd arguments to g. But then, for the next x in xs, the node referred to by this lt becomes the next invocation of g's result.
And then it's rt's turn, etc. And when xs end, and we hit the Nothings, g stops pulling the values from pairs's output altogether. So pairs stops advancing on the nodes too, which is thus never finished though it's defined as an unending stream of Emptys past that point, just to be on the safe side.
(*) Prolog's variables are explicitly set-once: they are allowed to be in a not-yet-assigned state. Haskell's (x:xs) is Prolog's [X | Xs].
The pseudocode: maintain a queue; enqueue "unassigned pointer"; for each x in xs: { set pointer in current head of the queue to Node(x, lt, rt) where lt, rt are unassigned pointers; enqueue lt; enqueue rt; pop queue }; set all pointers remaining in queue to Empty; find resulting tree in the original head of the queue, i.e. the original first "unassigned pointer" (or "empty box" instead of "unassigned pointer" is another option).
This Prolog's "queue" is of course fully persistent: "popping" does not mutate any data structure and doesn't change any outstanding references to the queue's former head -- it just advances the current pointer into the queue. So what's left in the wake of all this queuing, is the bfs-enumeration of the built tree's nodes, with the tree itself its head element -- the tree is its top node, with the two children fully instantiated to the bottom leaves by the time the enumeration is done.
Update: #dfeuer came up with much simplified version of it which is much closer to the Prolog original (that one in the comment at the top of the post), that can be much clearer. Look for more efficient code and discussion and stuff in his post. Using the simple [] instead of dfeuer's use of the more efficient infinite stream type data IS a = a :+ IS a for the sub-trees queue, it becomes
bftree :: [a] -> Tree a
bftree xs = t
where
t : q = go xs q
go [] _ = repeat Empty
go (x:ys) ~(l : ~(r : q)) = Node x l r : go ys q
---READ-- ----READ---- ---WRITE---
{-
xs = [ x x2 x3 x4 x5 x6 x7 x8 … ]
(t:q) = [ t l r ll lr rl rr llr … Empty Empty … … ]
-}
For comparison, the opposite operation of breadth-first enumeration of a tree is
bflist :: Tree a -> [a]
bflist t = [x | Node x _ _ <- q]
where
q = t : go 1 q
go 0 _ = []
go i (Empty : q) = go (i-1) q
go i (Node _ l r : q) = l : r : go (i+1) q
-----READ------ --WRITE--
How does bftree work: t : q is the list of the tree's sub-trees in breadth-first order. A particular invocation of go (x:ys) uses l and r before they are defined by subsequent invocations of go, either with another x further down the ys, or by go [] which always returns Empty. The result t is the very first in this list, the topmost node of the tree, i.e. the tree itself.
This list of tree nodes is created by the recursive invocations of go at the same speed with which the input list of values xs is consumed, but is consumed as the input to go at twice that speed, because each node has two child nodes.
These extra nodes thus must also be defined, as Empty leaves. We don't care how many are needed and simply create an infinite list of them to fulfill any need, although the actual number of empty leaves will be one more than there were xs.
This is actually the same scheme as used in computer science for decades for array-backed trees where tree nodes are placed in breadth-first order in a linear array. Curiously, in such setting both conversions are a no-op -- only our interpretation of the same data is what's changing, our handling of it, how are we interacting with / using it.
Update: the below solution is big-O optimal and (I think) pretty easy to understand, so I'm leaving it here in case anyone's interested. However, Will Ness's solution is much more beautiful and, especially when optimized a bit, can be expected to perform better in practice. It is much more worthy of study!
I'm going to ignore the fake edge labels for now and just focus on the core of what's happening.
A common pattern in algorithm design is that it's sometimes easier to solve a more general problem. So instead of trying to build a tree, I'm going to look at how to build a forest (a list of trees) with a given number of trees. I'll make the node labels polymorphic to avoid having to think about what they look like; you can of course use the same building technique with your original tree type.
data Tree a = Empty | Node a (Tree a) (Tree a)
-- Built a tree from a breadth-first list
bft :: [a] -> Tree a
bft xs = case dff 1 xs of
[] -> Empty
[t] -> t
_ -> error "something went wrong"
-- Build a forest of nonempty trees.
-- The given number indicates the (maximum)
-- number of trees to build.
bff :: Int -> [a] -> [Tree a]
bff _ [] = []
bff n xs = case splitAt n xs of
(front, rear) -> combine front (bff (2 * n) rear)
where
combine :: [a] -> [Tree a] -> [Tree a]
-- you write this
Here's a full, industrial-strength, maximally lazy implementation. This is the most efficient version I've been able to come up with that's as lazy as possible. A slight variant is less lazy but still works for fully-defined infinite inputs; I haven't tried to test which would be faster in practice.
bft' :: [a] -> Tree a
bft' xs = case bff 1 xs of
[] -> Empty
[t] -> t
_ -> error "whoops"
bff' :: Int -> [a] -> [Tree a]
bff' !_ [] = []
bff' n xs = combine n xs (bff (2 * n) (drop n xs))
where
-- The "take" portion of the splitAt in the original
-- bff is integrated into this version of combine. That
-- lets us avoid allocating an intermediate list we don't
-- really need.
combine :: Int -> [a] -> [Tree a] -> [Tree a]
combine 0 !_ ~[] = [] -- These two lazy patterns are just documentation
combine _k [] ~[] = []
combine k (y : ys) ts = Node y l r : combine (k - 1) ys dropped
where
(l, ~(r, dropped)) = case ts of -- This lazy pattern matters.
[] -> (Empty, (Empty, []))
t1 : ts' -> (t1, case ts' of
[] -> (Empty, [])
t2 : ts'' -> (t2, ts''))
For the less-lazy variant, replace (!l, ~(!r, dropped)) with (!l, !r, dropped) and adjust the RHS accordingly.
For true industrial strength, forests should be represented using lists strict in their elements:
data SL a = Cons !a (SL a) | Nil
And the pairs in the above (l, ~(r, dropped)) should both be represented using a type like
data LSP a b = LSP !a b
This should avoid some (pretty cheap) run-time checks. More importantly, it makes it easier to see where things are and aren't getting forced.
The method that you appear to have chosen is to build the tree up backwards: from bottom-to-top, right-to-left; starting from the last element of your list. This makes your buildBPT function look nice, but requires your insertElement to be overly complex. To construct a binary tree in a breadth-first fashion this way would require some difficult pivots at every step past the first three.
Adding 8 nodes to the tree would require the following steps (see how the nodes are inserted from last to first):
. 4
6 6
8 7 8 . .
. .
3
7 4 5
8 . 6 7 8 .
6 2
7 8 3 4
5 6 7 8
5
6 7 1
8 . . . 2 3
4 5 6 7
8 . . . . . . .
If, instead, you insert the nodes left-to-right, top-to-bottom, you end up with a much simpler solution, requiring no pivoting, but instead some tree structure introspection. See the insertion order; at all times, the existing values remain where they were:
. 1
2 3
1 4 5 . .
. .
1
1 2 3
2 . 4 5 6 .
1 1
2 3 2 3
4 5 6 7
1
2 3 1
4 . . . 2 3
4 5 6 7
8 . . . . . . .
The insertion step has an asymptotic time complexity on the order of O(n^2) where n is the number of nodes to insert, as you are inserting the nodes one-by-one, and then iterating the nodes already present in the tree.
As we insert left-to-right, the trick is to check whether the left sub-tree is complete:
if it is, and the right sub-tree is not complete, then recurse to the right.
if it is, and the right sub-tree is also complete, then recurse to the left (starting a new row).
if it is not, then recurse to the left.
Here is my (more generic) solution:
data Tree a = Leaf | Node a (Tree a) (Tree a)
deriving (Eq, Show)
main = do
let l1 = ["A1-Gate", "North-Region", "South-Region", "Convention Center",
"Rectorate", "Academic Building1", "Academic Building2"]
let l2 = [0.0, 0.5, 0.7, 0.3, 0.6, 1.2, 1.4, 1.2]
print $ treeFromList $ zip l1 l2
mkNode :: a -> Tree a
mkNode x = Node x Leaf Leaf
insertValue :: Tree a -> a -> Tree a
insertValue Leaf y = mkNode y
insertValue (Node x left right) y
| isComplete left && nodeCount left /= nodeCount right = Node x left (insertValue right y)
| otherwise = Node x (insertValue left y) right
where nodeCount Leaf = 0
nodeCount (Node _ left right) = 1 + nodeCount left + nodeCount right
depth Leaf = 0
depth (Node _ left right) = 1 + max (depth left) (depth right)
isComplete n = nodeCount n == 2 ^ (depth n) - 1
treeFromList :: (Show a) => [a] -> Tree a
treeFromList = foldl insertValue Leaf
EDIT: more detailed explanation:
The idea is to remember in what order you insert nodes: left-to-right first, then top-to-bottom. I compressed the different cases in the actual function, but you can expand them into three:
Is the left side complete? If not, then insert to the left side.
Is the right side as complete as the left side, which is complete? If not, then insert to the right side.
Both sides are full, so we start a new level by inserting to the left side.
Because the function fills the nodes up from left-to-right and top-to-bottom, then we always know (it's an invariant) that the left side must fill up before the right side, and that the left side can never be more than one level deeper than the right side (nor can it be shallower than the right side).
By following the growth of the second set of example trees, you can see how the values are inserted following this invariant. This is enough to describe the process recursively, so it extrapolates to a list of any size (the recursion is the magic).
Now, how do we determine whether a tree is 'complete'? Well, it is complete if it is perfectly balanced, or if – visually – its values form a triangle. As we are working with binary trees, then the base of the triangle (when filled) must have a number of values equal to a power of two. More specifically, it must have 2^(depth-1) values. Count for yourself in the examples:
depth = 1 -> base = 1: 2^(1-1) = 1
depth = 2 -> base = 2: 2^(2-1) = 2
depth = 3 -> base = 4: 2^(3-1) = 4
depth = 4 -> base = 8: 2^(4-1) = 8
The total number of nodes above the base is one less than the width of the base: 2^(n-1) - 1. The total number of nodes in the complete tree is therefore the number of nodes above the base, plus those of the base, so:
num nodes in complete tree = 2^(depth-1) - 1 + 2^(depth-1)
= 2 × 2^(depth-1) - 1
= 2^depth - 1
So now we can say that a tree is complete if it has exactly 2^depth - 1 non-empty nodes in it.
Because we go left-to-right, top-to-bottom, when the left side is complete, we move to the right, and when the right side is just as complete as the left side (meaning that it has the same number of nodes, which is means that it is also complete because of the invariant), then we know that the whole tree is complete, and therefore a new row must be added.
I originally had three special cases in there: when both nodes are empty, when the left node is empty (and therefore so was the right) and when the right node is empty (and therefore the left could not be). These three special cases are superseded by the final case with the guards:
If both sides are empty, then countNodes left == countNodes right, so therefore we add another row (to the left).
If the left side is empty, then both sides are empty (see previous point).
If the right side is empty, then the left side must have depth 1 and node count 1, meaning that it is complete, and 1 /= 0, so we add to the right side.
I'm trying to write a bool function to return True if a binary tree is a bst using recursion, and I need a little guidance on haskell syntax.
I understand that for a binary tree to be a bst, the left subtree must always contain only nodes less than the head. and the right subtree must always contain only nodes greater than the head. I was structuring my function as such:
isBST :: Tree -> Bool --recieve Tree, return bool
isBST (Lead i) = True --return true if its only one leaf in tree
isBST (Node h l r) = if (((isBST l) < h) && ((isBST r) > h)) then True else False
--return true if left subtree < head AND right subtree > head
But this code results in the error:
Couldn't match expected type ‘Bool’ with actual type ‘Int’
Referring to the < h and > h parts specifically. Is it something wrong with my haskell formatting? Thanks in advance
Is it something wrong with my haskell formatting?
No, it is a semantical error. You write:
(isBST l) < h
So this means you ask Haskell to determine whether l is a binary search tree, which is True or False, but you can not compare True or False with h. Even if you could (some languages see True as 1 and False as 0), then it would still be incorrect, since we want to know whether all nodes in the left subtree are less than h.
So we will somehow need to define bounds. A way to do this is to pass parameters through the recursion and perform checks. A problem with this is that the root of the tree for example, has no bounds. We can fix this by using a Maybe Int is a boundary: if it is Nothing, the boundary is "inactive" so to speak, if it is Just b, then the boundary is "active" with value b.
In order to make this check more convenient, we can first write a way to check this:
checkBound :: (a -> a -> Bool) -> Maybe a -> a -> Bool
checkBound _ Nothing _ = True
checkBound f (Just b) x = f b x
So now we can make a "sandwich check" with:
sandwich :: Ord a => Maybe a -> Maybe a -> a -> Bool
sandwich low upp x = checkBound (<) low x && checkBound (>) upp x
So sandwich is given a lowerbound and an upperbound (both Maybe as), and a value, and checks the lower and upper bounds.
So we can write a function isBST' with:
isBST' :: Maybe Int -> Maybe Int -> Tree -> Bool
isBST' low upp ... = ....
There are two cases we need to take into account: the Leaf x case, in which the "sandwich constraint" should be satisfied, and the Node h l r case in which h should satisfy the "sandwich constraint" and furthermore l and r should satsify different sandwhich constraints. For the Leaf x it is thus like:
isBST' low upp (Leaf x) = sandwich low upp x
For the node case, we first check the same constraint, and then enforce a sandwich between low and h for the left part l, and a sandwich between h and upp for the right part r, so:
isBST' low upp (Node h l r) = sandwich low upp h &&
isBST' low jh l &&
isBST' jh upp r
where jh = Just h
Now the only problem we still have is to call isBST' with the root element: here we use Nothing as intial bounds, so:
isBST :: Tree -> Bool
isBST = isBST' Nothing Nothing
There are of course other ways to enforce constraints, like passing and updating functions, or by implement four variants of the isBST' function that check a subset of the constraints.
Martin, I'd recommend you to look at Willem's answer.
Another thing, you could also use your maxInt function that you asked in a previous question to define this function:
isBST (Node h l r) = ... (maxInt l) ... -- at some point we will need to use this
Taking your definition of BSTs:
I understand that for a binary tree to be a bst, the left subtree must
always contain only nodes less than the head. and the right subtree
must always contain only nodes greater than the head.
I'll add that also the subtrees of a node should be BSTs as well.
So we can define this requirement with:
isBST (Node h l r) =
((maxInt l) < h) -- the left subtree must contain nodes less than the head
&& ((minInt r) > h) -- the right must contain nodes greater than the head
&& (...) -- the left subtree should be a BST
&& (...) -- the right subtree should be a BST
Recall that you might need to define minInt :: Tree -> Int, as you probably know how to do that.
I like Willem Van Onsem's pedagogical approach in his answer.
I was going to delete my answer, but am going to post a "correction" instead, at the risk of being wrong again:
data Tree = Empty | Node Int Tree Tree deriving show
isBST :: Tree -> Bool
isBST Empty = True
isBST (Node h l r) = f (<=h) l && f (>=h) r && isBST l && isBST r
where
f _ Empty = True
f c (Node h l r) = c h && f c l && f c r
Note that I'm using Wikipedia's definition of BST, that
the key in each node must be greater than or equal to any key stored
in the left sub-tree, and less than or equal to any key stored in the
right sub-tree.
So this week we learned about union types, tail recursion and binary trees in Haskell. We defined our tree data type like so:
data BinTree a = Empty
| Node (BinTree a) a (BinTree a)
deriving (Eq, Show)
leaf :: a -> BinTree a
leaf x = Node Empty x Empty
Now we were asked to write a function to find the most left node, return it, cut it out and also return the remaining tree without the node we just cut.
We did something like this, which worked quite well:
splitleftmost :: BinTree a -> Maybe (a, BinTree a)
splitleftmost Empty = Nothing
splitleftmost (Node l a r) = case splitleftmost l of
Nothing -> Just (a, r)
Just (a',l') -> Just (a', Node l' a r)
Now I need to make this function tail recursive. I think I understood what tail recursion is about, but found it hard to apply it to this problem. I was told to write a function which calls the main function with the fitting arguments, but was still not able to solve this.
Since nodes do not have a parent link, one approach would be to maintain root-to-leaf path within a list. At the end the modified tree can be constructed using a left fold:
slm :: BinTree a -> Maybe (a, BinTree a)
slm = run []
where
run _ Empty = Nothing
run t (Node Empty x r) = Just (x, foldl go r t)
where go l (Node _ x r) = Node l x r
run t n#(Node l _ _) = run (n:t) l
As others have hinted, there is no reason, in Haskell, to make this function tail-recursive. In fact, a tail-recursive solution will almost certainly be slower than the one you have devised! The main potential inefficiencies in the code you've provided involve allocation of pair and Just constructors. I believe GHC (with optimization enabled) will be able to figure out how to avoid these. My guess is that its ultimate code will probably look something like this:
splitleftmost :: BinTree a -> Maybe (a, BinTree a)
splitleftmost Empty = Nothing
splitleftmost (Node l a r) =
case slm l a r of
(# hd, tl #) -> Just (hd, tl)
slm :: BinTree a -> a -> BinTree a
-> (# a, BinTree a #)
slm Empty a r = (# a, r #)
slm (Node ll la lr) a r =
case slm ll la lr of
(# hd, tl' #) -> (# hd, Node tl' a r #)
Those funny-looking (# ..., ... #) things are unboxed pairs, which are handled pretty much like multiple return values. In particular, no actual tuple constructor is allocated until the end. By recognizing that every invocation of splitleftmost with a non-empty tree will produce a Just result, we (and thus almost certainly GHC) can separate the empty case from the rest to avoid allocating intermediate Just constructors. So this final code only allocates stack frames to handle the recursive results. Since some representation of such a stack is inherently necessary to solve this problem, using GHC's built-in one seems pretty likely to give the best results.
Here, not to spoil anything, are some "tail recursive" definitions of functions for summing along the left and right branches, at least as I understand "tail recursion":
sumLeftBranch tree = loop 0 tree where
loop n Empty = n
loop n (Node l a r) = loop (n+a) l
sumRightBranch tree = loop 0 tree where
loop n Empty = n
loop n (Node l a r) = loop (n+a) r
You can see that all the recursive uses of loop will have the same answer as the first call loop 0 tree - the arguments just keep getting put into better and better shape, til they are in the ideal shape, loop n Empty, which is n, the desired sum.
If this is the kind of thing that is wanted, the setup for splitleftmost would be
splitLeftMost tree = loop Nothing tree
where
loop m Empty = m
loop Nothing (Node l a r) = loop ? ?
loop (Just (a',r')) (Node l a r) = loop ? ?
Here, the first use of loop is in the form of loop Nothing tree, but that's the same as loop result Empty - when we come to it, namely result. It took me a couple of tries to get the missing arguments to loop ? ? right, but, as usual, they were obvious once I got them.
This code from the answer to this question copied below quite nicely takes only O(n) space to do a depth first traversal of a tree of depth n which contains O(2^n) nodes. This is very good, the garbage collector seems to be doing a good job of cleaning up the already processed tree.
But my question I have is, how? Unlike a list, where once we process the first element we can completely forget it, we can't scrap the root node after processing the first leaf node. We have to wait until the left half the tree is processed (because eventually we'll have to traverse down the right from the root). Also, as the root node points to the nodes below it, and so on, all the way down to the leaves, which would seem to imply that we wouldn't be able to collect any of the first half of a tree until we start on the second half (as all those nodes will still have references to them starting from the still live root node). This fortunately is not the case, but could someone explain how?
import Data.List (foldl')
data Tree = Tree Int Tree Tree
tree n = Tree n (tree (2 * n)) (tree (2 * n + 1))
treeOne = tree 1
depthNTree n t = go n t [] where
go 0 (Tree x _ _) = (x:)
go n (Tree _ l r) = go (n - 1) l . go (n - 1) r
main = do
x <- getLine
print . foldl' (+) 0 . filter (\x -> x `rem` 5 == 0) $ depthNTree (read x) treeOne
Actually you don't hold on to the root while you descend the left subtree.
go n (Tree _ l r) = go (n - 1) l . go (n - 1) r
So the root is turned two thunks, composed together. One holds a reference to the left subtree, the other holds a reference to the right subtree. The root node itself is now garbage.
The left and right subtrees themselves are just thunks, because the tree is produces lazily, so they aren't consuming much space yet.
We're only evaluating go n (Tree _ l r) because we're evaluating depthNTree n t, which is go n t []. So we're immediately forcing the two composed go calls we just turned the root into:
(go (n - 1) l . go (n - 1) r) []
= (go (n - 1) l) ((go (n - 1) r) [])
And because this is lazily evaluated, we do the outermost call first, leaving ((go (n - 1) r) []) as a thunk (and so not generating any more of r).
Recursing into go will force l, so we do generate more of that. But then we do the same thing again one level down; again that tree node becomes garbage immediately, we generate two thunks holding the left and right sub sub trees, and then we force only the left one.
After n calls we'll be evaluating go 0 (Tree x _ _) = (x:). We've generated n pairs of thunks, and forced the n left ones, leaving the right ones in memory; because the right sub-trees are unevaluated thunks they're constant space each, and there are only n of them, so only O(n) space total. And all the tree nodes leading to this path are now unreferenced.
We actually have the outermost list constructor (and the first element of the list). Forcing more of the list will explore those right sub-tree thunks further down the composition chain being built up, but there will never be more than n of them.
Technically you have bound a reference to tree 1 in the globally scoped treeOne, so actually you could retain a reference to every node you ever produce, so you're relying on GHC noticing that treeOne is only ever used once and shouldn't be retained.
I wrote a little manual evaluation of a tree to depth 2. I hope it can illustrate why tree nodes can be garbage collected along the way.
Suppose we start with a tree like this:
tree =
Tree
(Tree _ -- l
(Tree a _ _) -- ll
(Tree b _ _)) -- lr
(Tree _ -- r
(Tree c _ _) -- rl
(Tree d _ _)) -- rr
Now call depthNTree 2 tree:
go 2 tree []
go 2 (Tree _ l r) []
go 1 l (go 1 r [])
go 1 (Tree _ ll lr) (go 1 r [])
go 0 ll (go 0 lr (go 1 r []))
go 0 (Tree a _ _) (go 0 lr (go 1 r []))
a : go 0 lr (go 1 r []) -- gc can collect ll
a : go 0 (Tree b _ _) (go 1 r [])
a : b : go 1 r [] -- gc can collect lr and thus l
a : b : go 1 (Tree _ rl rr) []
a : b : go 0 rl (go 0 rr [])
a : b : go 0 (Tree c _ _) (go 0 rr [])
a : b : c : go 0 rr [] -- gc can collect rl
a : b : c : go 0 (Tree d _ _) []
a : b : c : d : [] -- gc can collect rr and thus r and tree
Note that since treeOne is a static value, there has to be some extra machinery behind the scenes to allow garbage collection of it. Fortunately GHC supports GC of static values.
Let's rewrite the recursive case of go as
go n t = case t of
Tree _ l r -> go (n - 1) l . go (n - 1) r
In the right-hand side of the case alternative, the original tree t is no longer live. Only l and r are live. So, if we recurse into l first, say, there is nothing keeping the left-hand side of the tree live except l itself; r exactly keeps the right-hand side of the tree alive.
At any point in the recursion, the live nodes are exactly the roots of the subtrees cut off by the path from the original root of the tree to the node currently being inspected which have not already been processed. There are at most the length of said path of these subtrees, so the space usage is O(n).
The key is that the original tree t becomes dead before we recurse. If you write the (denotationally equivalent, but bad style for a number of reasons)
leftChild (Tree _ l r) = l
rightChild (Tree _ l r) = r
go n t = go (n - 1) (leftChild t) . go (n - 1) (rightChild t)
now when recursing into go (n - 1) (leftChild t), there is still a live reference to t in the unevaluated expression rightChild t. Hence the space usage is now exponential.
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))