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

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paths src dest clauses = map reverse (pathsImpl [] src dest clauses) |
paths src dest clauses = map reverse (pathsImpl [] src dest clauses) |
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</haskell> |
</haskell> |
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+ | |||

+ | ---- |
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+ | |||

+ | Here is another recursive implementation |
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+ | <haskell> |
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+ | paths :: Eq a =>a -> a -> [(a,a)] -> [[a]] |
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+ | paths source sink edges |
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+ | | source == sink = [[sink]] |
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+ | | otherwise = [ |
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+ | source:path | edge<-edges, (fst edge) == source, |
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+ | path<-(paths (snd edge) sink [e|e<-edges, e/=edge]) |
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+ | ]; |
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+ | </haskell> |
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+ | |||

+ | [[Category:Programming exercise spoilers]] |

## Latest revision as of 03:49, 10 January 2017

(**) 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])
];
```