(Add an additional recursive solution to the problem)
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Latest revision as of 17:10, 4 January 2012
(**) 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]) ];