# 99 questions/Solutions/81

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< 99 questions | Solutions(Difference between revisions)

(paths :: Int -> Int -> [(Int , Int)] -> Int paths start end zs = let (xs,ys) = partition (\(_,z) -> z == end ) zs in map (++ [ end] ) ( concat . map (\(e, _) -> if e ==) |
(Added another solution) |
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This solution uses a representation of a (directed) graph as a list of arcs (a,b). | 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 | ||

+ | |||

+ | <haskell> | ||

+ | 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) | ||

+ | </haskell> |

## Revision as of 15:48, 28 July 2011

(**) 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)