# 99 questions/Solutions/81

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Revision as of 17:41, 25 April 2016 by Nadavge (talk | contribs) (Changed list concatenation to the more readable form of element appendation)

(**) 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])
];
```