Problem 87b

Extract nodes from a graph into a list in breadth-first order, without repeating any nodes.

Example

breadthFirst 'c' graph == ['c', 'b', 'f', 'g', 'k', 'h']

Unit Test

import Html
import List
import Set
import String


{-| A node has a value, degree and a color
-}
type alias Edge comparable =
    ( comparable, comparable )


type alias Graph comparable =
    ( List comparable, List (Edge comparable) )


{-| Given a graph and a starting node, return all nodes in breadth-first order.
-}
breadthFirst : comparable -> Graph comparable -> List comparable
breadthFirst start ( nodes, edges ) =
    -- your implementation goes here
    []


main : Html.Html a
main =
    Html.text
        <| case test of
            0 ->
                "Your implementation passed all tests."

            1 ->
                "Your implementation failed one test."

            x ->
                "Your implementation failed " ++ (toString x) ++ " tests."


test : Int
test =
    List.length
        <| List.filter ((==) False)
            [ String.fromList (breadthFirst 'a' graph87) == "abdc"
            , String.fromList (breadthFirst 'g' graph80) == "gh"
            , String.fromList (breadthFirst 'h' graph80) == "hg"
            , String.fromList (breadthFirst 'g' graph84) == "gbcfkh"
            , String.fromList (breadthFirst 'c' graph84) == "cbfghk"
            , String.fromList (breadthFirst 'f' graph84) == "fbcghk"
            , String.fromList (breadthFirst 'h' graph84) == "hgbcfk"
            , String.fromList (breadthFirst 'k' graph84) == "kfbcgh"
            , String.fromList (breadthFirst 'b' graph80) == "bcfk"
            , String.fromList (breadthFirst 'c' graph80) == "cbfk"
            , String.fromList (breadthFirst 'f' graph80) == "fbck"
            , String.fromList (breadthFirst 'k' graph80) == "kfbc"
            , String.fromList (breadthFirst 'k' graph80) == "kfbc"
            , String.fromList (breadthFirst 'g' graph81) == "gh"
            , String.fromList (breadthFirst 'h' graph81) == "hg"
              -- either ordering is acceptable
            , List.member (String.fromList (breadthFirst 'b' graph84)) [ "bcfgkh", "bcfghk" ]
            ]


graph80 =
    ( [ 'b', 'c', 'd', 'f', 'g', 'h', 'k' ]
    , [ ( 'b', 'c' )
      , ( 'b', 'f' )
      , ( 'c', 'f' )
      , ( 'f', 'k' )
      , ( 'g', 'h' )
      ]
    )


graph81 =
    -- has a loop and parallel edges
    ( [ 'b', 'c', 'd', 'f', 'g', 'h', 'k' ]
    , [ ( 'b', 'c' )
      , ( 'b', 'f' )
      , ( 'c', 'f' )
      , ( 'f', 'k' )
      , ( 'g', 'h' )
      , ( 'g', 'h' )
      , ( 'g', 'g' )
      ]
    )


graph84 =
    ( [ 'b', 'c', 'd', 'f', 'g', 'h', 'k' ]
    , [ ( 'b', 'c' )
      , ( 'b', 'f' )
      , ( 'b', 'g' )
      , ( 'c', 'f' )
      , ( 'f', 'g' )
      , ( 'f', 'k' )
      , ( 'g', 'h' )
      ]
    )


{-| Simple graph, in dfs - "abdc", in bfs "abcd"
-}
graph87 =
    ( [ 'a', 'b', 'c', 'd' ]
    , [ ( 'a', 'b' )
      , ( 'a', 'c' )
      , ( 'b', 'd' )
      ]
    )

Hint

1. A small modification to the depth-first graph search can resolve solve this problem.

Solution

Solution