99 questions/Solutions/81
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Trackbully (Talk  contribs) (Add an additional recursive solution to the problem) 

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(**) 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 [ee<edges, e/=edge]) ];