In today’s Programming Praxis, our task is to implement Dijkstra’s shortest path algorithm. Let’s get started, shall we?
import Data.List import qualified Data.List.Key as K import Data.Map ((!), fromList, fromListWith, adjust, keys, Map)
In order make the rest of the algorithm simpler, we convert the list of edges to a map that lists all the neighbors of each vertex.
buildGraph :: Ord a => [(a, a, Float)] -> Map a [(a, Float)] buildGraph g = fromListWith (++) $ g >>= \(a,b,d) -> [(a,[(b,d)]), (b,[(a,d)])]
The algorithm follows the usual steps, albeit in a functional rather than the typical procedural style: start by giving all non-source vertices an infinite distance, then go through all the vertices in order of their distance from the source, relaxing all their neighbors.
dijkstra :: Ord a => a -> Map a [(a, Float)] -> Map a (Float, Maybe a) dijkstra source graph = f (fromList [(v, (if v == source then 0 else 1/0, Nothing)) | v <- keys graph]) (keys graph) where f ds  = ds f ds q = f (foldr relax ds $ graph ! m) (delete m q) where m = K.minimum (fst . (ds !)) q relax (e,d) = adjust (min (fst (ds ! m) + d, Just m)) e
Getting the shortest path is then simply a matter of tracing the path from the endpoint to the beginning.
shortestPath :: Ord a => a -> a -> Map a [(a, Float)] -> [a] shortestPath from to graph = reverse $ f to where f x = x : maybe  f (snd $ dijkstra from graph ! x)
A test to see if everything is working properly:
main :: IO () main = do let g = buildGraph [('a','c',2), ('a','d',6), ('b','a',3) ,('b','d',8), ('c','d',7), ('c','e',5) ,('d','e',10)] print $ shortestPath 'a' 'e' g == "ace"
As expected, the shortest route for the given graph is A-C-E.