Breadth-First Search using State monad in Haskell

Question

Recently, I've asked a question for building DFS tree from Graph in Stackoverflow and had learned that it can be simply implemented by using State Monad.

DFS in haskell

While DFS requires to track only visited nodes, so that we can use 'Set' or 'List' or some sort of linear data structure to track visited nodes, BFS requires 'visited node' and 'queue' data structure to be accomplished.

My pseudocode for BFS is

Q = empty queue
T = empty Tree
mark all nodes except u as unvisited
while Q is nonempty do
u = deq(Q)
    for each vertex v ∈ Adj(u)
        if v is not visited 
        then add edge (u,v) to T
             Mark v as visited and enq(v)

As can be inferred from pseudocode, we only have to do 3 processes per iteration.

1. dequeue point from queue
2. add all unvisited neighbors of the point to current tree's child, queue and 'visited' list
3. repeat this for next in queue

Since we are not using recursive traversal for BFS search, we need some other traversal method such as while loop. I've looked up loop-while package in hackage, but it seems somewhat deprecated.

What I assume is that I require some sort of code like this :

{-...-}
... =   evalState (bfs) ((Set.singleton start),[start])
where 
    neighbors x = Map.findWithDefault [] x adj 
    bfs =do (vis,x:queue)<-get
             map (\neighbor ->
                  if (Set.member neighbor vis)
                  then put(vis,queue) 
                  else put ((Set.insert neighbor vis), queue++[neighbor]) >> (addToTree neighbor)
                 )  neighbors x
            (vis,queue)<-get
         while (length queue > 0)

I understand that this implementation is very erroneous but this should give minimalistic view for how I think BFS should be implemented. Also, I really don't know how to circumvent using while loop for do blocks.(i.e should I use recursive algorithm to overcome it or should I think of completely different strategy)

Considering one of the answer I've found in previous question linked above, it seems like the answer should look like this :

newtype Graph a = Graph (Map.Map a [a]) deriving (Ord, Eq, Show)
data Tree a = Tree a [Tree a] deriving (Ord, Eq, Show)

bfs :: (Ord a) => Graph a -> a -> Tree a
bfs (Graph adj) start = evalState (bfs') ((Set.singleton start),[start])
    where
        bfs' = {-part where I don't know-}

Finally, if such implementation for BFS using state monad is impossible due to some reason, (which I believe not to be) please correct my false assumption.

I've seen some of the examples for BFS in Haskell without using state monad but I want to learn more about how state monad can be processed and couldn't have found any of examples of BFS implemented using state monad.

Thanks in advance.

---

EDIT: I came up with some sort of algorithm using state monad but I fall in infinite loop.

bfs :: (Ord a) => Graph a -> a -> Tree a
bfs (Graph adj) start = evalState (bfs' (Graph adj) start) (Set.singleton start)

bfs' :: (Ord a) => Graph a -> a -> State (Set.Set a) (Tree a)
bfs' (Graph adj) point= do
                        vis <- get
                        let neighbors x = Map.findWithDefault [] x adj
                        let addableNeighbors (x:xs) =   if Set.member x vis
                                                        then addableNeighbors(xs)
                                                        else x:addableNeighbors(xs)
                        let addVisited (vis) (ns) = Set.union (vis) $ Set.fromList ns
                        let newVisited = addVisited vis $ addableNeighbors $ neighbors point
                        put newVisited
                        return (Tree point $ map (flip evalState newVisited) (map (bfs' (Graph adj)) $ addableNeighbors $ neighbors point))

EDIT2: With some expense of space complexity, I've came out with a solution to get BFS graph using graph to return and queue to process. Despite it is not the optimal solution for generating BFS tree/graph, it will work.

bfs :: (Ord a) => Graph a -> a -> Graph a
bfs (Graph adj) start = evalState (bfs' (Graph adj) (Graph(Map.empty))  [start]) (Set.singleton start)


bfs':: (Ord a) => Graph a -> Graph a -> [a] -> State (Set.Set a) (Graph a)
bfs' _ (Graph ret) [] = return (Graph ret)
bfs' (Graph adj) (Graph ret) (p:points)= do
                                        vis <- get
                                        let neighbors x = Map.findWithDefault [] x adj
                                        let addableNeighbors ns
                                                | null ns = []
                                                | otherwise =   if Set.member (head ns) vis
                                                                then addableNeighbors(tail ns)
                                                                else (head ns):addableNeighbors(tail ns)
                                        let addVisited (v) (ns) = Set.union (v) $ Set.fromList ns
                                        let unVisited = addableNeighbors $ neighbors p
                                        let newVisited = addVisited vis unVisited
                                        let unionGraph (Graph g1) (Graph g2) = Graph (Map.union g1 g2)
                                        put newVisited
                                        bfs' (Graph adj) (unionGraph (Graph ret) (Graph (Map.singleton p unVisited))) (points ++ unVisited)

EDIT3: I've added convert function for graph to tree. Running function in EDIT2, and EDIT3 will yield BFS Tree. It is not the best algorithm for computation time wise, but I believe it is intuitive and easy to understand for newbies like me :)

graphToTree :: (Ord a) => Graph a -> a -> Tree a
graphToTree (Graph adj) point  = Tree point $ map (graphToTree (Graph adj)) $ neighbors point
    where neighbors x = Map.findWithDefault [] x adj

Answer 1

Converting a graph into a Tree breadth-first is a bit more difficult than simply searching the graph breadth-first. If you are searching the graph, you only ever need to return from a single branch. When converting the graph into a tree, the result needs to include results from multiple branches.

We can use a more general type than Graph a for what we can search or convert to trees. We can search or convert to trees anything with a function a -> [a]. For a Graph we'd use the function (Map.!) m, where m is the Map. Searching with a transposition table has a signature like

breadthFirstSearchUnseen:: Ord r => (a -> r) -> -- how to compare `a`s 
                           (a -> Bool) ->       -- where to stop
                           (a -> [a]) ->        -- where you can go from an `a`
                           [a] ->               -- where to start
                           Maybe [a]

Converting the function to a tree that contains each reachable node at the earliest depth has a signature like

shortestPathTree :: Ord r => (a -> r) -> -- how to compare `a`s
                    (a -> l)             -- what label to put in the tree
                    (a -> [a]) ->        -- where you can go from an `a`
                    a ->                 -- where to start
                    Tree l

We can slightly more generally start at any number of nodes and build a Forest that contains each reachable node at the earliest depth.

shortestPathTrees :: Ord r => (a -> r) -> -- how to compare `a`s
                     (a -> l)             -- what label to put in the tree
                     (a -> [a]) ->        -- where you can go from an `a`
                     [a] ->               -- where to start
                     [Tree l]

Searching

Performing the conversion to a tree doesn't really help us search, we can perform breadth first searches on the original graph.

import Data.Sequence (viewl, ViewL (..), (><))
import qualified Data.Sequence as Seq
import qualified Data.Set as Set

breadthFirstSearchUnseen:: Ord r => (a -> r) -> (a -> Bool) -> (a -> [a]) -> [a] -> Maybe [a]
breadthFirstSearchUnseen repr p expand = combine Set.empty Seq.empty []
    where
        combine seen queued ancestors unseen =
            go
                (seen  `Set.union` (Set.fromList . map repr            $ unseen))
                (queued ><         (Seq.fromList . map ((,) ancestors) $ unseen))
        go seen queue =
            case viewl queue of
                EmptyL -> Nothing
                (ancestors, a) :< queued ->
                    if p a
                    then Just . reverse $ ancestors'
                    else combine seen queued ancestors' unseen
                    where
                        ancestors' = a:ancestors
                        unseen = filter (flip Set.notMember seen . repr) . expand $ a

The state maintained in the above search algorithm is a Seq queue of what nodes to visit next and a Set of nodes that have already been seen. If we instead kept track of nodes that have already been visited, then we could visit the same node multiple times if we find multiple paths to the node at the same depth. There's a more complete explanation in the answer I wrote this breadth first search for.

We can easily write searching Graphs in terms of our general search.

import qualified Data.Map as Map

newtype Graph a = Graph (Map.Map a [a]) deriving (Ord, Eq, Show)

bfsGraph :: (Ord a) => Graph a -> (a -> Bool) -> [a] -> Maybe [a]
bfsGraph (Graph adj) test = breadthFirstSearchUnseen id test ((Map.!) adj)

We can also write how to search Trees themselves.

import Data.Tree

bfsTrees :: (Ord a) => (a -> Bool) -> [Tree a] -> Maybe [a]
bfsTrees test = fmap (map rootLabel) . breadthFirstSearchUnseen rootLabel (test . rootLabel) subForest

Building trees

Building trees breadth-first is a lot more difficult. Fortunately Data.Tree already provides ways to build Trees in breadth first order from a monadic unfold. The breadth first order will take care of the queuing, we will only need to keep track of the state for the nodes we've already seen.

unfoldTreeM_BF has the type Monad m => (b -> m (a, [b])) -> b -> m (Tree a). m is the Monad our computations will be in, b is the type of data we are going to build the tree based on, and a is the type for the labels of the tree. In order to use it to build a tree we need to make a function b -> m (a, [b]). We're going to rename a to l for label, and b to a, which is what we've been using for our nodes. We need to make an a -> m (l, [a]). For m, we'll use the State monad from transformers to keep track of some state; the state will be the Set of nodes whose representation r we've already seen; we'll be using the State (Set.Set r) monad. Overall, we need to provide a function a -> State (Set.Set r) (l, [a]).

expandUnseen :: Ord r => (a -> r) -> (a -> l) -> (a -> [a]) -> a -> State (Set.Set r) (l, [a])
expandUnseen repr label expand a = do
    seen <- get
    let unseen = filter (flip Set.notMember seen . repr) . uniqueBy repr . expand $ a
    put . Set.union seen . Set.fromList . map repr $ unseen
    return (label a, unseen)

To build the trees, we run the state computation built by unfoldForestM_BF

shortestPathTrees :: Ord r => (a -> r) -> (a -> l) -> (a -> [a]) -> [a] -> [Tree l]
shortestPathTrees repr label expand = run . unfoldForestM_BF k . uniqueBy repr
    where
        run = flip evalState Set.empty
        k = expandUnseen repr label expand

uniqueBy is a nubBy that takes advantage of an Ord instance instead of Eq.

uniqueBy :: Ord r => (a -> r) -> [a] -> [a]
uniqueBy repr = go Set.empty
    where
        go seen []     = []
        go seen (x:xs) =
            if Set.member (repr x) seen
            then go seen xs
            else x:go (Set.insert (repr x) seen) xs

We can write building shortest path trees from Graphs in terms of our general shortest path tree building

shortestPathsGraph :: Ord a => Graph a -> [a] -> [Tree a]
shortestPathsGraph (Graph adj) = shortestPathTrees id id ((Map.!) adj)

We can do the same for filtering a Forest to only the shortest paths through the Forest.

shortestPathsTree :: Ord a => [Tree a] -> [Tree a]
shortestPathsTree = shortestPathTrees rootLabel rootLabel subForest

Answer 2

My solution is based on working level-by-level (wrt. to BFS), see also this question and answer.

The general idea is: Assume we already know the sets of visited elements prior each level of our BFS as a list of sets. Then we can traverse the graph, level by level, updating our list of sets, constructing the output Tree on the way.

The trick is that after such a level-by-level traversal, we'll have the sets of visited elements after each level. And this is the same as the list before each level, just shifted by one. So by tying the knot, we can use the shifted output as the input for the procedure.

import Control.Monad.State
import qualified Data.Map as M
import Data.Maybe (fromMaybe, catMaybes)
import qualified Data.Set as S
import Data.Tree

newtype Graph a = Graph (M.Map a [a])
    deriving (Ord, Eq, Show)

tagBfs :: (Ord a) => Graph a -> a -> Maybe (Tree a)
tagBfs (Graph g) s = let (t, sets) = runState (thread s) (S.empty : sets)
                      in t
  where
    thread x = do
        sets@(s : subsets) <- get
        case M.lookup x g of
            Just vs | not (S.member x s) -> do
                -- recursively create sub-nodes and update the subsets list
                let (nodes, subsets') = runState
                                          (catMaybes `liftM` mapM thread vs) subsets
                -- put the new combined list of sets
                put (S.insert x s : subsets')
                -- .. and return the node
                return . Just $ Node x nodes
            _ -> return Nothing -- node not in the graph, or already visited

Running tagBfs example2 'b' it on the following example

example2 :: Graph Char
example2 = Graph $ M.fromList
    [ ('a', ['b', 'c', 'd'])
    , ('b', ['a'])
    , ('c', [])
    , ('d', [])
    ]

yields

Just (Node {rootLabel = 'b',
            subForest = [Node {rootLabel = 'a',
                               subForest = [Node {rootLabel = 'c',
                                                  subForest = []},
                                            Node {rootLabel = 'd',
                                                  subForest = []}
                                           ]}
                        ]}
      )