# 99 questions/Solutions/81

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< 99 questions | Solutions(Difference between revisions)

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| otherwise = (paths1 x b g (current++[a])) ++ (paths2 a b g current xs) | | otherwise = (paths1 x b g (current++[a])) ++ (paths2 a b g current xs) | ||

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

− | + | 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 == start then [[start]] else paths start e ys) $ xs ) | ||

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). |

## Revision as of 05:41, 18 April 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)

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 == start then start else paths start e ys) $ xs )

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