Solving Problems using Monads 1 - Reachability

— 20 October 2017 —

This is part of a series of posts on solving simple programming problems using Monads. The intention of this series is to help me (and you) learn more about Monads by solving interesting (but simple) programming problems.

Searching for a simple problem to start the series with, I stumbled upon this kata. It contains a reachability problem for a pair of nodes in a graph.

Problem Definition

Given a pair of nodes $$(s, e)$$ and a list of all edges of a unidirectional graph $$G$$, decide whether node $$e$$ is reachable from node $$s$$. Any linear graph traversal algorithm can be used to solve this problem in a simple and relatively efficient manner.

Solving the problem without the use of Monads

We will first import the Data.Set in order to keep visited nodes in an efficient data structure.

import qualified Data.Set as S


The only type definitions that we need are:

type Node = Char
type Arc  = (Node, Node)


The main function of our problem is:

solveGraph :: Node -> Node -> [Arc] -> Bool
solveGraph s e arcs = S.member e visited
where
visited = visit arcs [s] $S.singleton s  The function solveGraph visits all nodes that are reachable from $$s$$ and checks if $$e$$ is one of them. So now we have to define a function visit :: [Arc] -> [Node] -> S.Set Node -> S.Set Node that actually performs a Breadth First Search (or any other kind of traversal) on $$G$$ starting from $$s$$. visit :: [Arc] -> [Node] -> S.Set Node -> S.Set Node visit _ [] visited = visited visit arcs (n:ns) visited = visit arcs (ns ++ newOpen) newVisited where es = S.fromList$ expandOnce arcs n
newVisited = S.union visited es
newOpen = S.toList $S.difference es visited  visit performs the classic BFS algorithm. What is now left is to define expandOnce. expandOnce :: [Arc] -> Node -> [Node] expandOnce arcs s = [b | (a,b) <- arcs, a == s]  expandOnce given a node returns all the nodes that are one edge away from it. This program works as can be seen below from the interpreter output: Prelude> :l main.hs [1 of 1] Compiling Graph ( main.hs, interpreted ) Ok, modules loaded: Graph. *Graph> let arcs = [('a','b'),('b','c'),('c','a'),('c','d'),('e','a')] *Graph> solveGraph 'a' 'd' arcs True *Graph> solveGraph 'a' 'e' arcs False  Incorporating Monads in the Solution This solution is short and solves the problem, but it does not satisfy the initial requirement to use Monads. It is easy to notice that visit passes around the visited :: Set Node resembling the State monad. newtype State s a = State { runState :: s -> (a,s) }  The above newtype is provided in Control.Monad.State and we could easily think of visited as the state. So in order to not pass this around we could reimplement solveGraph and visit as follows. solveGraph :: Node -> Node -> [Arc] -> Bool solveGraph s e arcs = S.member e visited where visited = snd$ runState (visit arcs [s]) $S.singleton s visit :: [Arc] -> [Node] -> State (S.Set Node) [Node] visit _ [] = return [] visit arcs (n:ns) = do visited <- get let es = S.fromList$ expandOnce arcs n
let newVisited = S.union visited es
let newBoundary = S.toList $S.difference es visited put newVisited visit arcs$ ns ++ newBoundary


This implementation is by no means shorter than the other one, however it is now clear that the Set of visited nodes is used as state. If the program was much longer and more complicated, not having to pass the state around as an argument would significantly improve the code’s readability.

Conclusion

We tried to solve a simple programming problem using the help of Monads and we did it. However the resulting code is not more elegant than the original one and so we did not manage to showcase the greatness of the State Monad. In the next part of the series I hope that I can find a problem whose solution is greatly improved when using Monads.

Here is a link for the complete code in this post.

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