问题
Data.Tree includes unfoldTreeM_BF and unfoldForestM_BF functions to construct trees breadth-first using the results of monadic actions. The tree unfolder can be written easily using the forest unfolder, so I'll focus on the latter:
unfoldForestM_BF :: Monad m =>
(b -> m (a, [b])) -> [b] -> m [Tree a]
Starting with a list of seeds, it applies a function to each, generating actions that will produce the tree roots and the seeds for the next level of unfolding. The algorithm used is somewhat strict, so using unfoldForestM_BF with the Identity monad is not exactly the same as using the pure unfoldForest. I've been trying to figure out if there's a way to make it lazy without sacrificing its O(n) time bound. If (as Edward Kmett suggested to me) this is impossible, I wonder if it would be possible to do it with a more constrained type, specifically requiring MonadFix rather than Monad. The concept there would be to (somehow) set up the pointers to the results of future computations while adding those computations to the to-do list, so if they are lazy in the effects of earlier computations they will be available immediately.
回答1:
I previously claimed that the third solution presented below has the same strictness as the depth-first unfoldForest, which is not correct.
Your intuition that trees can be lazily unfolded breadth first is at least partially correct, even if we don't require a MonadFix instance. Solutions exist for the special cases when the branching factor is known to be finite and when the branching factor is known to be "large". We will start with a solution that runs in O(n) time for trees with finite branching factors including degenerate trees with only one child per node. The solution for finite branching factors will fail to terminate on trees with infinite branching factors, which we will rectify with a solution that that runs in O(n) time for trees with "large" branching factors greater than one including trees with infinite branching factor. The solution for "large" branching factors will run in O(n^2) time on degenerate trees with only one child or no children per node. When we combine the methods from both steps in an attempt to make a hybrid solution that runs in O(n) time for any branching factor we will get a solution that is lazier than the first solution for finite branching factors but cannot accommodate trees that make a rapid transition from an infinite branching factor to having no branches.
Finite Branching Factor
The general idea is that we will first build all the labels for an entire level and the seeds for the forests for the next level. We will then descend into the next level, building all of it. We will collect together the results from the deeper level to build the forests for the outer level. We will put the labels together with the forests to build the trees.
unfoldForestM_BF is fairly simple. If there are no seeds for the level it returns. After building all the labels, it takes the seeds for each forest and collects them together into one list of all the seeds to build the next level and unfolds the entire deeper level. Finally it constructs the forest for each tree from the structure of the seeds.
import Data.Tree hiding (unfoldTreeM_BF, unfoldForestM_BF)
unfoldForestM_BF :: Monad m => (b->m (a, [b])) -> [b] -> m [Tree a]
unfoldForestM_BF f [] = return []
unfoldForestM_BF f seeds = do
level <- sequence . fmap f $ seeds
let (labels, bs) = unzip level
deeper <- unfoldForestM_BF f (concat bs)
let forests = trace bs deeper
return $ zipWith Node labels forests
trace reconstructs the structure of nested lists from a flattened list.It is assumed that there is an item in [b] for each of the items anywhere in [[a]]. The use of concat ... trace to flatten all the information about ancestor levels prevents this implementation from working on trees with infinite children for a node.
trace :: [[a]] -> [b] -> [[b]]
trace [] ys = []
trace (xs:xxs) ys =
let (ys', rem) = takeRemainder xs ys
in ys':trace xxs rem
where
takeRemainder [] ys = ([], ys)
takeRemainder (x:xs) (y:ys) =
let ( ys', rem) = takeRemainder xs ys
in (y:ys', rem)
Unfolding a tree is trivial to write in terms of unfolding a forest.
unfoldTreeM_BF :: MonadFix m => (b->m (a, [b])) -> b -> m (Tree a)
unfoldTreeM_BF f = (>>= return . head) . unfoldForestMFix_BF f . (:[])
Large Branching Factor
The solution for large branching factor proceeds in much the same way as the solution for finite branching factor, except it keeps the entire structure of the tree around instead of concatenating the branches in a level to a single list and traceing that list. In addition to the imports used in the previous section, we will be using Compose to compose the functors for multiple levels of a tree together and Traversable to sequence across multi-level structures.
import Data.Tree hiding (unfoldForestM_BF, unfoldTreeM_BF)
import Data.Foldable
import Data.Traversable
import Data.Functor.Compose
import Prelude hiding (sequence, foldr)
Instead of flattening all of the ancestor structures together with concat we will wrap with Compose the ancestors and the seeds for the next level and recurse on the entire structure.
unfoldForestM_BF :: (Traversable t, Traceable t, Monad m) =>
(b->m (a, [b])) -> t b -> m (t (Tree a))
unfoldForestM_BF f seeds
| isEmpty seeds = return (fmap (const undefined) seeds)
| otherwise = do
level <- sequence . fmap f $ seeds
deeper <- unfoldForestM_BF f (Compose (fmap snd level))
return $ zipWithIrrefutable Node (fmap fst level) (getCompose deeper)
zipWithIrrefutable is a lazier version of zipWith that relies on the assumption that there is an item in the second list for each item in the first list. The Traceable structures are the Functors that can provide a zipWithIrrefutable. The laws for Traceable are for every a, xs, and ys if fmap (const a) xs == fmap (const a) ys then zipWithIrrefutable (\x _ -> x) xs ys == xs and zipWithIrrefutable (\_ y -> y) xs ys == ys. Its strictness is given for every f and xs by zipWithIrrefutable f xs ⊥ == fmap (\x -> f x ⊥) xs.
class Functor f => Traceable f where
zipWithIrrefutable :: (a -> b -> c) -> f a -> f b -> f c
We can combine two lists lazily if we already know they have the same structure.
instance Traceable [] where
zipWithIrrefutable f [] ys = []
zipWithIrrefutable f (x:xs) ~(y:ys) = f x y : zipWithIrrefutable f xs ys
We can combine the composition of two functors if we know that we can combine each functor.
instance (Traceable f, Traceable g) => Traceable (Compose f g) where
zipWithIrrefutable f (Compose xs) (Compose ys) =
Compose (zipWithIrrefutable (zipWithIrrefutable f) xs ys)
isEmpty checks for an empty structure of nodes to expand like the pattern match on [] did in the solution for finite branching factors.
isEmpty :: Foldable f => f a -> Bool
isEmpty = foldr (\_ _ -> False) True
The astute reader may notice that zipWithIrrefutable from Traceable is very similar to liftA2 which is half of the definition of Applicative.
Hybrid Solution
The hybrid solution combines the approaches of the finite solution and the "large" solution. Like the finite solution, we will compress and decompress the tree representation at each step. Like the solution for "large" branching factors we will use a data structure that allows stepping over complete branches. The finite branching factor solution used a data type that is flattened everywhere, [b]. The "large" branching factor solution used a data type that was flattened nowhere: increasingly nested lists starting with [b] then [[b]] then [[[b]]] and so on. In between these structures would be nested lists that either stop nesting and just hold a b or keep nesting and hold [b]s. That pattern of recursion is described in general by the Free monad.
data Free f a = Pure a | Free (f (Free f a))
We will be working specifically with Free [] which looks like.
data Free [] a = Pure a | Free [Free [] a]
For the hybrid solution we will repeat all of its imports and components so that the code below here should be complete working code.
import Data.Tree hiding (unfoldTreeM_BF, unfoldForestM_BF)
import Data.Traversable
import Prelude hiding (sequence, foldr)
Since we will be working with Free [], we will provide it with a zipWithIrrefutable.
class Functor f => Traceable f where
zipWithIrrefutable :: (a -> b -> c) -> f a -> f b -> f c
instance Traceable [] where
zipWithIrrefutable f [] ys = []
zipWithIrrefutable f (x:xs) ~(y:ys) = f x y : zipWithIrrefutable f xs ys
instance (Traceable f) => Traceable (Free f) where
zipWithIrrefutable f (Pure x) ~(Pure y ) = Pure (f x y)
zipWithIrrefutable f (Free xs) ~(Free ys) =
Free (zipWithIrrefutable (zipWithIrrefutable f) xs ys)
The breadth first traversal will look very similar to the original version for the finitely branching tree. We build the current labels and seeds for the current level, compress the structure of the remainder of the tree, do all the work for the remaining depths, and decompress the structure of the results to get the forests to go with the labels.
unfoldFreeM_BF :: (Monad m) => (b->m (a, [b])) -> Free [] b -> m (Free [] (Tree a))
unfoldFreeM_BF f (Free []) = return (Free [])
unfoldFreeM_BF f seeds = do
level <- sequence . fmap f $ seeds
let (compressed, decompress) = compress (fmap snd level)
deeper <- unfoldFreeM_BF f compressed
let forests = decompress deeper
return $ zipWithIrrefutable Node (fmap fst level) forests
compress takes a Free [] holding the seeds for forests [b] and flattens the [b] into the Free to get a Free [] b. It also returns a decompress function that can be used to undo the flattening to get the original structure back. We compress away branches with no remaining seeds and branches that only branch one way.
compress :: Free [] [b] -> (Free [] b, Free [] a -> Free [] [a])
compress (Pure [x]) = (Pure x, \(Pure x) -> Pure [x])
compress (Pure xs ) = (Free (map Pure xs), \(Free ps) -> Pure (map getPure ps))
compress (Free xs) = wrapList . compressList . map compress $ xs
where
compressList [] = ([], const [])
compressList ((Free [],dx):xs) = let (xs', dxs) = compressList xs
in (xs', \xs -> dx (Free []):dxs xs)
compressList ( (x,dx):xs) = let (xs', dxs) = compressList xs
in (x:xs', \(x:xs) -> dx x:dxs xs)
wrapList ([x], dxs) = (x, \x -> Free (dxs [x]))
wrapList (xs , dxs) = (Free xs, \(Free xs) -> Free (dxs xs ))
Each compression step also returns a function that will undo it when applied to a Free [] tree with the same structure. All of these functions are partially defined; what they do to Free [] trees with a different structure is undefined. For simplicity we also define partial functions for the inverses of Pure and Free.
getPure (Pure x) = x
getFree (Free xs) = xs
Both unfoldForestM_BF and unfoldTreeM_BF are defined by packaging their argument up into a Free [] b and unpackaging the results assuming they are in the same structure.
unfoldTreeM_BF :: MonadFix m => (b->m (a, [b])) -> b -> m (Tree a)
unfoldTreeM_BF f = (>>= return . getPure) . unfoldFreeM_BF f . Pure
unfoldForestM_BF :: MonadFix m => (b->m (a, [b])) -> [b] -> m [Tree a]
unfoldForestM_BF f = (>>= return . map getPure . getFree) . unfoldFreeM_BF f . Free . map Pure
A more elegant version of this algorithm can probably be made by recognizing that >>= for a Monad is grafting on trees and both Free and FreeT provide monad instances. Both compress and compressList probably have more elegant presentations.
The algorithm presented above is not lazy enough to allow querying trees that branch an infinite number of ways and then terminate. A simple counter example is the following generating function unfolded from 0.
counterExample :: Int -> (Int, [Int])
counterExample 0 = (0, [1, 2])
counterExample 1 = (1, repeat 3)
counterExample 2 = (2, [3])
counterExample 3 = (3, [])
This tree would look like
0
|
+- 1
| |
| +- 3
| |
| `- 3
| |
| ...
|
`- 2
|
+- 3
Attempting to descend the second branch (to 2) and inspect the remaining finite sub-tree will fail to terminate.
Examples
The following examples demonstrate that all implementations of unfoldForestM_BF run actions in breadth first order and that runIdentity . unfoldTreeM_BF (Identity . f) has the same strictness as unfoldTree for trees with finite branching factor. For trees with inifinite branching factor, only the solution for "large" branching factors has the same strictness as unfoldTree. To demonstrate laziness we'll define three infinite trees - a unary tree with one branch, a binary tree with two branches, and an infinitary tree with an infinite number of branches for each node.
mkUnary :: Int -> (Int, [Int])
mkUnary x = (x, [x+1])
mkBinary :: Int -> (Int, [Int])
mkBinary x = (x, [x+1,x+2])
mkInfinitary :: Int -> (Int, [Int])
mkInfinitary x = (x, [x+1..])
Together with unfoldTree, we will define unfoldTreeDF in terms of unfoldTreeM to check that unfoldTreeM really is lazy like you claimed and unfoldTreeBF in terms of unfoldTreeMFix_BF to check that the new implementation is just as lazy.
import Data.Functor.Identity
unfoldTreeDF f = runIdentity . unfoldTreeM (Identity . f)
unfoldTreeBF f = runIdentity . unfoldTreeM_BF (Identity . f)
To get finite pieces of these infinite trees, even the infinitely branching one, we'll define a way to take from a tree as long as its labels match a predicate. This could be written more succinctly in terms of the ability to apply a function to every subForest.
takeWhileTree :: (a -> Bool) -> Tree a -> Tree a
takeWhileTree p (Node label branches) = Node label (takeWhileForest p branches)
takeWhileForest :: (a -> Bool) -> [Tree a] -> [Tree a]
takeWhileForest p = map (takeWhileTree p) . takeWhile (p . rootLabel)
This lets us define nine example trees.
unary = takeWhileTree (<= 3) (unfoldTree mkUnary 0)
unaryDF = takeWhileTree (<= 3) (unfoldTreeDF mkUnary 0)
unaryBF = takeWhileTree (<= 3) (unfoldTreeBF mkUnary 0)
binary = takeWhileTree (<= 3) (unfoldTree mkBinary 0)
binaryDF = takeWhileTree (<= 3) (unfoldTreeDF mkBinary 0)
binaryBF = takeWhileTree (<= 3) (unfoldTreeBF mkBinary 0)
infinitary = takeWhileTree (<= 3) (unfoldTree mkInfinitary 0)
infinitaryDF = takeWhileTree (<= 3) (unfoldTreeDF mkInfinitary 0)
infinitaryBF = takeWhileTree (<= 3) (unfoldTreeBF mkInfinitary 0)
All five methods have the same output for the unary and binary trees. The output comes from putStrLn . drawTree . fmap show
0
|
`- 1
|
`- 2
|
`- 3
0
|
+- 1
| |
| +- 2
| | |
| | `- 3
| |
| `- 3
|
`- 2
|
`- 3
However, the breadth first traversal from the finite branching factor solution is not sufficiently lazy for a tree with an infinite branching factor. The other four methods output the entire tree
0
|
+- 1
| |
| +- 2
| | |
| | `- 3
| |
| `- 3
|
+- 2
| |
| `- 3
|
`- 3
The tree generated with unfoldTreeBF for the finite branching factor solution can never be completely drawn past its first branches.
0
|
+- 1
| |
| +- 2
| | |
| | `- 3
| |
| `- 3
The construction is definitely breadth first.
mkDepths :: Int -> IO (Int, [Int])
mkDepths d = do
print d
return (d, [d+1, d+1])
mkFiltered :: (Monad m) => (b -> Bool) -> (b -> m (a, [b])) -> (b -> m (a, [b]))
mkFiltered p f x = do
(a, bs) <- f x
return (a, filter p bs)
binaryDepths = unfoldTreeM_BF (mkFiltered (<= 2) mkDepths) 0
Running binaryDepths outputs the outer levels before the inner ones
0
1
1
2
2
2
2
From Lazy to Downright Slothful
The hybrid solution from the earlier section is not lazy enough to have the same strictness semantics as Data.Tree's unfoldTree. It is the first in a series of algorithms, each slightly lazier than their predecessor, but none lazy enough to have the same strictness semantics as unfoldTree.
The hybrid solution does not provide a guarantee that exploring a portion of a tree doesn't demand exploring other portions of the same tree. Nor will the code presented below. In one particular yet common case identified by dfeuer exploring only a log(N) sized slice of a finite tree forces the entirety of the tree. This happens when exploring the last descendant of each branch of a tree with constant depth. When compressing the tree we throw out every trivial branch with no descendants, which is necessary to avoid O(n^2) running time. We can only lazily skip over this portion of compression if we can quickly show that a branch has at least one descendant and we can therefore reject the pattern Free []. At the greatest depth of the tree with constant depth, none of the branches have any remaining descendants, so we can never skip a step of the compression. This results in exploring the entire tree to be able to visit the very last node. When the entire tree to that depth is non-finite due to infinite branching factor, exploring a portion of the tree fails to terminate when it would terminate when generated by unfoldTree.
The compression step in the hybrid solution section compresses away branches with no descendants in the first generation that they can be discovered in, which is optimal for compression but not optimal for laziness. We can make the algorithm lazier by delaying when this compression occurs. If we delay it by a single generation (or even any constant number of generations) we will maintain the O(n) upper bound on time. If we delay it by a number of generations that somehow depends on N we would necessarily sacrifice the O(N) time bound. In this section we will delay the compression by a single generation.
To control how compression happens, we will separate stuffing the innermost [] into the Free [] structure from compressing away degenerate branches with 0 or 1 descendants.
Because part of this trick doesn't work without a lot of laziness in the compression, we will adopt a paranoid level of excessively slothful laziness everywhere. If anything about a result other than the tuple constructor (,) could be determined without forcing part of its input with a pattern match we will avoid forcing it until it is necessary. For tuples, anything pattern matching on them will do so lazily. Consequently, some of the code below will look like core or worse.
bindFreeInvertible replaces Pure [b,...] with Free [Pure b,...]
bindFreeInvertible :: Free [] ([] b) -> (Free [] b, Free [] a -> Free [] ([] a))
bindFreeInvertible = wrapFree . go
where
-- wrapFree adds the {- Free -} that would have been added in both branches
wrapFree ~(xs, dxs) = (Free xs, dxs)
go (Pure xs) = ({- Free -} (map Pure xs), Pure . map getPure . getFree)
go (Free xs) = wrapList . rebuildList . map bindFreeInvertible $ xs
rebuildList = foldr k ([], const [])
k ~(x,dx) ~(xs, dxs) = (x:xs, \(~(x:xs)) -> dx x:dxs xs)
wrapList ~(xs, dxs) = ({- Free -} xs, \(~(Free xs)) -> Free (dxs xs)))
compressFreeList removes occurrences of Free [] and replaces Free [xs] with xs.
compressFreeList :: Free [] b -> (Free [] b, Free [] a -> Free [] a)
compressFreeList (Pure x) = (Pure x, id)
compressFreeList (Free xs) = wrapList . compressList . map compressFreeList $ xs
where
compressList = foldr k ([], const [])
k ~(x,dx) ~(xs', dxs) = (x', dxs')
where
x' = case x of
Free [] -> xs'
otherwise -> x:xs'
dxs' cxs = dx x'':dxs xs''
where
x'' = case x of
Free [] -> Free []
otherwise -> head cxs
xs'' = case x of
Free [] -> cxs
otherwise -> tail cxs
wrapList ~(xs, dxs) = (xs', dxs')
where
xs' = case xs of
[x] -> x
otherwise -> Free xs
dxs' cxs = Free (dxs xs'')
where
xs'' = case xs of
[x] -> [cxs]
otherwise -> getFree cxs
The overall compression will not bind the Pure []s into Frees until after the degenerate Frees have been compressed away, delaying compression of degenerate Frees introduced in one generation to the next generation's compression.
compress :: Free [] [b] -> (Free [] b, Free [] a -> Free [] [a])
compress xs = let ~(xs' , dxs' ) = compressFreeList xs
~(xs'', dxs'') = bindFreeInvertible xs'
in (xs'', dxs' . dxs'')
Out of continued paranoia, the helpers getFree and getPure are also made irrefutably lazy.
getFree ~(Free xs) = xs
getPure ~(Pure x) = x
This very quickly solves the problematic example dfeuer discovered
print . until (null . subForest) (last . subForest) $
flip unfoldTreeBF 0 (\x -> (x, if x > 5 then [] else replicate 10 (x+1)))
But since we only delayed the compression by 1 generation, we can recreate exactly the same problem if the very last node of the very last branch is 1 level deeper than all of the other branches.
print . until (null . subForest) (last . subForest) $
flip unfoldTreeBF (0,0) (\(x,y) -> ((x,y),
if x==y
then if x>5 then [] else replicate 9 (x+1, y) ++ [(x+1, y+1)]
else if x>4 then [] else replicate 10 (x+1, y)))
来源：https://stackoverflow.com/questions/27748526/is-a-lazy-breadth-first-monadic-rose-tree-unfold-possible