# Difference between revisions of "99 questions/Solutions/81"

From HaskellWiki

< 99 questions | Solutions

(categorize) |
(yet another solution using monadic behavior of lists) |
||

Line 52: | Line 52: | ||

path<-(paths (snd edge) sink [e|e<-edges, e/=edge]) |
path<-(paths (snd edge) sink [e|e<-edges, e/=edge]) |
||

]; |
]; |
||

+ | </haskell> |
||

+ | |||

+ | ---- |
||

+ | |||

+ | yet another solution using monadic behavior of lists |
||

+ | <haskell> |
||

+ | paths :: (Eq a) => a -> a -> [Arc a] -> [[a]] |
||

+ | paths source sink arcs |
||

+ | | source == sink = [[source]] |
||

+ | | otherwise = map (map fst) $ aux source [] |
||

+ | where |
||

+ | aux current pathSoFar = |
||

+ | let nextEdges = filter ((== current) . fst) arcs |
||

+ | notCyclic = not . (\(_,t) -> (t == source) || (elem t $ map snd pathSoFar)) |
||

+ | noCycles = filter notCyclic nextEdges |
||

+ | in noCycles >>= \(f,t) -> do |
||

+ | if (t == sink) then return $ pathSoFar ++ (f,t):[(t,t)] |
||

+ | else aux t (pathSoFar ++ [(f,t)]) |
||

</haskell> |
</haskell> |
||

## Latest revision as of 20:50, 30 July 2021

(**) Path from one node to another one

Write a function that, given two nodes a and b in a graph, returns all the acyclic paths from a to b.

```
import List (elem)
paths :: Eq a => a -> a -> [(a,a)] -> [[a]]
paths a b g = paths1 a b g []
paths1 :: Eq a => a -> a -> [(a,a)] -> [a] -> [[a]]
paths1 a b g current = paths2 a b g current [ y | (x,y) <- g, x == a ]
paths2 :: Eq a => a -> a -> [(a,a)] -> [a] -> [a] -> [[a]]
paths2 a b g current [] | a == b = [current++[b]]
| otherwise = []
paths2 a b g current (x:xs) | a == b = [current++[b]]
| elem a current = []
| otherwise = (paths1 x b g (current++[a])) ++ (paths2 a b g current xs)
```

This solution uses a representation of a (directed) graph as a list of arcs (a,b).

Here is another implementation using List's monadic behavior

```
import Data.List (partition)
pathsImpl :: Eq a => [a] -> a -> a -> [(a, a)] -> [[a]]
pathsImpl trail src dest clauses
| src == dest = [src:trail]
| otherwise = do
let (nexts, rest) = partition ((==src) . fst) clauses
next <- nexts
pathsImpl (src:trail) (snd next) dest rest
paths :: Eq a => a -> a -> [(a, a)] -> [[a]]
paths src dest clauses = map reverse (pathsImpl [] src dest clauses)
```

Here is another recursive implementation

```
paths :: Eq a =>a -> a -> [(a,a)] -> [[a]]
paths source sink edges
| source == sink = [[sink]]
| otherwise = [
source:path | edge<-edges, (fst edge) == source,
path<-(paths (snd edge) sink [e|e<-edges, e/=edge])
];
```

yet another solution using monadic behavior of lists

```
paths :: (Eq a) => a -> a -> [Arc a] -> [[a]]
paths source sink arcs
| source == sink = [[source]]
| otherwise = map (map fst) $ aux source []
where
aux current pathSoFar =
let nextEdges = filter ((== current) . fst) arcs
notCyclic = not . (\(_,t) -> (t == source) || (elem t $ map snd pathSoFar))
noCycles = filter notCyclic nextEdges
in noCycles >>= \(f,t) -> do
if (t == sink) then return $ pathSoFar ++ (f,t):[(t,t)]
else aux t (pathSoFar ++ [(f,t)])
```