How to implement Dijkstra Algorithm in Haskell

Question

For my studies I have to write the following function which gets the shortest route between two countries. I already have already written a function isRoute which checks if there is a connection between two countries, and a function yieldRoute which just returns a connection between two countries. Now I have to code a function which returns the shortest route between two countries.

My first approach was to get all connections between the two countries and then get the shortest one, but getting all connections is kind of annoying to programm in my opinion. Now I come up with the idea to implement a dijstra algorithm, but I actually find this kinda hard too. Can you guys give me some idea how to do this?

We have to use these types (we aren't allowed to change them but ofc we are allowed to add new types.)

type Country = String
type Countries = [Country]
type TravelTime = Integer -- Travel time in minutes
data Connection = Air Country Country TravelTime
    | Sea Country Country TravelTime
    | Rail Country Country TravelTime
    | Road Country Country TravelTime deriving (Eq,Ord,Show)
type Connections = [Connection]
data Itinerary = NoRoute | Route (Connections,TravelTime) deriving (Eq,Ord,Show)

My yield route function which is simply breadth first search: (Sry for german comments)

-- Liefert eine Route falls es eine gibt
yieldRoute :: Connections -> Country -> Country -> Connections
yieldRoute cons start goal 
            | isRoute cons start goal == False = []
            | otherwise                        = getRoute cons start [] [start] goal

getRoute :: Connections -> Country -> Connections -> Countries -> Country -> Connections
getRoute cons c gone visited target
            | (c == target) = gone 
            | otherwise  = if ( visit cons c visited ) then ( getRoute cons (deeper cons c visited) (gone ++ get_conn cons c (deeper cons c visited)) (visited ++ [(deeper cons c visited)]) target ) else ( getRoute cons (back (drop (length gone -1) gone)) (take (length gone -1) gone) visited target )

-- Geht ein Land zurück
back :: Connections -> Country
back ((Air c1 c2 _):xs) = c1
back ((Sea c1 c2 _):xs) = c1
back ((Rail c1 c2 _):xs) = c1
back ((Road c1 c2 _):xs) = c1

-- Liefert das nächste erreichbare Country
deeper :: Connections -> Country -> Countries -> Country
deeper ((Air c1 c2 _):xs) c visited
            | (c1 == c) = if ( c2 `elem` visited ) then ( deeper xs c visited ) else c2
            | (c2 == c) = if ( c1 `elem` visited ) then ( deeper xs c visited ) else c1
            | otherwise = deeper xs c visited
deeper ((Sea c1 c2 _):xs) c visited
            | (c1 == c) = if ( c2 `elem` visited ) then ( deeper xs c visited ) else c2
            | (c2 == c) = if ( c1 `elem` visited ) then ( deeper xs c visited ) else c1
            | otherwise = deeper xs c visited
deeper ((Rail c1 c2 _):xs) c visited
            | (c1 == c) = if ( c2 `elem` visited ) then ( deeper xs c visited ) else c2
            | (c2 == c) = if ( c1 `elem` visited ) then ( deeper xs c visited ) else c1
            | otherwise = deeper xs c visited
deeper ((Road c1 c2 _):xs) c visited
            | (c1 == c) = if ( c2 `elem` visited ) then ( deeper xs c visited ) else c2
            | (c2 == c) = if ( c1 `elem` visited ) then ( deeper xs c visited ) else c1
            | otherwise = deeper xs c visited

-- Liefert eine Connection zwischen zwei Countries
get_conn :: Connections -> Country -> Country -> Connections
get_conn [] _ _ = error "Something went terribly wrong"
get_conn ((Air c1 c2 t):xs) c3 c4 
            | (c1 == c3) && (c2 == c4) = [(Air c1 c2 t)]
            | (c1 == c4) && (c2 == c3) = [(Air c1 c2 t)]
            | otherwise                = get_conn xs c3 c4
get_conn ((Sea c1 c2 t):xs) c3 c4 
            | (c1 == c3) && (c2 == c4) = [(Air c1 c2 t)]
            | (c1 == c4) && (c2 == c3) = [(Air c1 c2 t)]
            | otherwise                = get_conn xs c3 c4
get_conn ((Road c1 c2 t):xs) c3 c4 
            | (c1 == c3) && (c2 == c4) = [(Air c1 c2 t)]
            | (c1 == c4) && (c2 == c3) = [(Air c1 c2 t)]
            | otherwise                = get_conn xs c3 c4
get_conn ((Rail c1 c2 t):xs) c3 c4 
            | (c1 == c3) && (c2 == c4) = [(Air c1 c2 t)]
            | (c1 == c4) && (c2 == c3) = [(Air c1 c2 t)]
            | otherwise                = get_conn xs c3 c4

-- Überprüft ob eine besuchbare Connection exestiert
visit :: Connections -> Country -> Countries -> Bool
visit [] _ _ = False
visit ((Air c1 c2 _):xs) c visited
                | (c1 == c) = if ( c2 `elem` visited) then ( visit xs c visited ) else True
                | (c2 == c) = if ( c1 `elem` visited) then ( visit xs c visited ) else True
                | otherwise = visit xs c visited
visit ((Sea c1 c2 _):xs) c visited
                | (c1 == c) = if ( c2 `elem` visited) then ( visit xs c visited ) else True
                | (c2 == c) = if ( c1 `elem` visited) then ( visit xs c visited ) else True
                | otherwise = visit xs c visited
visit ((Rail c1 c2 _):xs) c visited
                | (c1 == c) = if ( c2 `elem` visited) then ( visit xs c visited ) else True
                | (c2 == c) = if ( c1 `elem` visited) then ( visit xs c visited ) else True
                | otherwise = visit xs c visited
visit ((Road c1 c2 _):xs) c visited
                | (c1 == c) = if ( c2 `elem` visited) then ( visit xs c visited ) else True
                | (c2 == c) = if ( c1 `elem` visited) then ( visit xs c visited ) else True

This one i have to write now:

yieldFastestRoute :: Connections -> Country -> Country -> Itinerary

Dijkstra Algorithm: http://en.wikipedia.org/wiki/Dijkstra%27s_algorithm

My first approach was this: (as I said I had problems with the getallRoutes)

yieldFastestRoute :: Connections -> Country -> Country -> Itinerary
yieldFastestRoute cons start targ
            |(isRoute start targ == False) = NoRoute
            |otherwise                    = (Route (getFastest (getAllRoutes cons start targ)) (sumTT (getFastest (getAllRoutes cons start targ))))

-- Liefert alle Routen zwischen zwei Ländern
getAllRoutes :: Connections -> Country -> Country -> [Connections]

-- Liefert aus einer Reihe von Connections die schnellste zurück
getFastest :: [Connections] -> Connections
getFastest (x:xs) = if ( (sumTT x) < sumTT (getFastest xs) || null (getFastest xs) ) then x else ( getFastest xs )

sumTT :: Connections -> TravelTime
sumTT []                  = 0
sumTT ((Air _ _ t ): xs)  = t ++ sumTT xs
sumTT ((Rail _ _ t ): xs) = t ++ sumTT xs
sumTT ((Road _ _ t ): xs) = t ++ sumTT xs
sumTT ((Sea _ _ t ): xs)  = t ++ sumTT xs

I basically want to know whats the best way to implement Dijkstra in Haskell, or if there's another approach I could follow.

Answer 1

You seem to have coded great part of the Algorithm

Here is a project by Martin Erwig in Haskell that may help to give you some ideas

--  SP.hs -- Dijkstra's Shortest Path Algorithm  (c) 2000 by Martin Erwig
module SP (
   spTree,spLength,sp,      -- shortest paths
   dijkstra
) where

import qualified Heap as H
import Graph
import RootPath
expand :: Real b => b -> LPath b -> Context a b -> [H.Heap (LPath b)]
expand d p (_,_,_,s) = map (\(l,v)->H.unit ((v,l+d):p)) s
dijkstra :: Real b => H.Heap (LPath b) -> Graph a b -> LRTree b
dijkstra h g | H.isEmpty h || isEmpty g = []
dijkstra h g =
    case match v g of
         (Just c,g')  -> p:dijkstra (H.mergeAll (h':expand d p c)) g'
         (Nothing,g') -> dijkstra h' g'  
    where (p@((v,d):_),h') = H.splitMin h

spTree :: Real b => Node -> Graph a b -> LRTree b
spTree v = dijkstra (H.unit [(v,0)])
spLength :: Real b => Node -> Node -> Graph a b -> b
spLength s t = getDistance t . spTree s
sp :: Real b => Node -> Node -> Graph a b -> Path
sp s t = map fst . getLPath t . spTree s

The rest modules are here