在Erlang中查找有向循环图中一个顶点的所有可能路径

Question

I would like to implement a function which finds all possible paths to all possible vertices from a source vertex V in a directed cyclic graph G.

The performance doesn't matter now, I just would like to understand the algorithm. I have read the definition of the Depth-first search algorithm, but I don't have full comprehension of what to do.

I don't have any completed piece of code to provide here, because I am not sure how to:

• store the results (along with A->B->C-> we should also store A->B and A->B->C);
• represent the graph (digraph? list of tuples?);
• how many recursions to use (work with each adjacent vertex?).

How can I find all possible paths form one given source vertex in a directed cyclic graph in Erlang?

UPD: Based on the answers so far I have to redefine the graph definition: it is a non-acyclic graph. I know that if my recursive function hits a cycle it is an indefinite loop. To avoid that, I can just check if a current vertex is in the list of the resulting path - if yes, I stop traversing and return the path.

---

UPD2: Thanks for thought provoking comments! Yes, I need to find all simple paths that do not have loops from one source vertex to all the others.

In a graph like this:

with the source vertex A the algorithm should find the following paths:

• A,B
• A,B,C
• A,B,C,D
• A,D
• A,D,C
• A,D,C,B

The following code does the job, but it is unusable with graphs that have more that 20 vertices (I guess it is something wrong with recursion - takes too much memory, never ends):

dfs(Graph,Source) ->
    ?DBG("Started to traverse graph~n", []),
            Neighbours = digraph:out_neighbours(Graph,Source),
    ?DBG("Entering recursion for source vertex ~w~n", [Source]),
            dfs(Neighbours,[Source],[],Graph,Source),
ok.


dfs([],Paths,Result,_Graph,Source) ->
    ?DBG("There are no more neighbours left for vertex ~w~n", [Source]),
    Result;

dfs([Neighbour|Other_neighbours],Paths,Result,Graph,Source) ->
    ?DBG("///The neighbour to check is ~w, other neighbours are: ~w~n",[Neighbour,Other_neighbours]),
    ?DBG("***Current result: ~w~n",[Result]),
    New_result = relax_neighbours(Neighbour,Paths,Result,Graph,Source),

        dfs(Other_neighbours,Paths,New_result,Graph,Source).


relax_neighbours(Neighbour,Paths,Result,Graph,Source) ->
     case lists:member(Neighbour,Paths) of 
        false ->
            ?DBG("Found an unvisited neighbour ~w, path is: ~w~n",[Neighbour,Paths]),
            Neighbours = digraph:out_neighbours(Graph,Neighbour),
            ?DBG("The neighbours of the unvisited vertex ~w are ~w, path is:
                ~w~n",[Neighbour,Neighbours,[Neighbour|Paths]]),
                dfs(Neighbours,[Neighbour|Paths],Result,Graph,Source);
            true ->
                [Paths|Result]

        end.

---

UPD3:

The problem is that the regular depth-first search algorithm will go one of the to paths first: (A,B,C,D) or (A,D,C,B) and will never go the second path.

In either case it will be the only path - for example, when the regular DFS backtracks from (A,B,C,D) it goes back up to A and checks if D (the second neighbour of A) is visited. And since the regular DFS maintains a global state for each vertex, D would have 'visited' state.

So, we have to introduce a recursion-dependent state - if we backtrack from (A,B,C,D) up to A, we should have (A,B,C,D) in the list of the results and we should have D marked as unvisited as at the very beginning of the algorithm.

I have tried to optimize the solution to tail-recursive one, but still the running time of the algorithm is unfeasible - it takes about 4 seconds to traverse a tiny graph of 16 vertices with 3 edges per vertex:

dfs(Graph,Source) ->
    ?DBG("Started to traverse graph~n", []),
            Neighbours = digraph:out_neighbours(Graph,Source),
    ?DBG("Entering recursion for source vertex ~w~n", [Source]),
    Result = ets:new(resulting_paths, [bag]),
Root = Source,
            dfs(Neighbours,[Source],Result,Graph,Source,[],Root).


dfs([],Paths,Result,_Graph,Source,_,_) ->
    ?DBG("There are no more neighbours left for vertex ~w, paths are ~w, result is ~w~n", [Source,Paths,Result]),
    Result;

dfs([Neighbour|Other_neighbours],Paths,Result,Graph,Source,Recursion_list,Root) ->
    ?DBG("~w *Current source is ~w~n",[Recursion_list,Source]),
    ?DBG("~w Checking neighbour _~w_ of _~w_, other neighbours are: ~w~n",[Recursion_list,Neighbour,Source,Other_neighbours]),



?    DBG("~w Ready to check for visited: ~w~n",[Recursion_list,Neighbour]),

 case lists:member(Neighbour,Paths) of 
        false ->
            ?DBG("~w Found an unvisited neighbour ~w, path is: ~w~n",[Recursion_list,Neighbour,Paths]),
New_paths = [Neighbour|Paths],
?DBG("~w Added neighbour to paths: ~w~n",[Recursion_list,New_paths]),
ets:insert(Result,{Root,Paths}),

            Neighbours = digraph:out_neighbours(Graph,Neighbour),
            ?DBG("~w The neighbours of the unvisited vertex ~w are ~w, path is: ~w, recursion:~n",[Recursion_list,Neighbour,Neighbours,[Neighbour|Paths]]),
                dfs(Neighbours,New_paths,Result,Graph,Neighbour,[[[]]|Recursion_list],Root);
            true -> 
            ?DBG("~w The neighbour ~w is: already visited, paths: ~w, backtracking to other neighbours:~n",[Recursion_list,Neighbour,Paths]),
ets:insert(Result,{Root,Paths})

end,

        dfs(Other_neighbours,Paths,Result,Graph,Source,Recursion_list,Root).

Any ideas to run this in acceptable time?

Answer 1

I don't understand question. If I have graph G = (V, E) = ({A,B}, {(A,B),(B,A)}), there is infinite paths from A to B {[A,B], [A,B,A,B], [A,B,A,B,A,B], ...}. How I can find all possible paths to any vertex in cyclic graph?

Edit:

Did you even tried compute or guess growing of possible paths for some graphs? If you have fully connected graph you will get

• 2 - 1
• 3 - 4
• 4 - 15
• 5 - 64
• 6 - 325
• 7 - 1956
• 8 - 13699
• 9 - 109600
• 10 - 986409
• 11 - 9864100
• 12 - 108505111
• 13 - 1302061344
• 14 - 16926797485
• 15 - 236975164804
• 16 - 3554627472075
• 17 - 56874039553216
• 18 - 966858672404689
• 19 - 17403456103284420
• 20 - 330665665962403999

Are you sure you would like find all paths for all nodes? It means if you compute one milion paths in one second it would take 10750 years to compute all paths to all nodes in fully connected graph with 20 nodes. It is upper bound for your task so I think you don't would like do it. I think you want something else.

Answer 2

Edit: Okay I understand now, you want to find all simple paths from a vertex in a directed graph. So a depth-first search with backtracking would be suitable, as you have realised. The general idea is to go to a neighbour, then go to another one (not one which you've visited), and keep going until you hit a dead end. Then backtrack to the last vertex you were at and pick a different neighbour, etc. You need to get the fiddly bits right, but it shouldn't be too hard. Eg at every step you need to label the vertices 'explored' or 'unexplored' depending on whether you've already visited them before. The performance shouldn't be an issue, a properly implemented algorithm should take maybe O(n^2) time. So I don't know what you are doing wrong, perhaps you are visiting too many neighbours? Eg maybe you are revisiting neighbours that you've already visited, and going round in loops or something.

I haven't really read your program, but the Wiki page on Depth-first Search has a short, simple pseudocode program which you can try to copy in your language. Store the graphs as Adjacency Lists to make it easier.

Edit: Yes, sorry, you are right, the standard DFS search won't work as it stands, you need to adjust it slightly so that does revisit vertices it has visited before. So you are allowed to visit any vertices except the ones you have already stored in your current path. This of course means my running time was completely wrong, the complexity of your algorithm will be through the roof. If the average complexity of your graph is d+1, then there will be approximately d*d*d*...*d = d^n possible paths. So even if every vertex has only 3 neighbours, there's still quite a few paths when you get above 20 vertices. There's no way around that really, because if you want your program to output all possible paths then indeed you will have to output all d^n of them.

I'm interested to know whether you need this for a specific task, or are just trying to program this out of interest. If the latter, you will just have to be happy with small, sparsely connected graphs.

Answer 3

Not an improved algorithmic solution by any means, but you can often improve performance by spawning multiple worker threads, potentially here one for each first level node and then aggregating the results. This can often improve naive brute force algorithms relatively easily.

You can see an example here: Some Erlang Matrix Functions , in the maximise_assignment function (comments starting on line 191 as of today). Again, the underlying algorithm there is fairly naive and brute force, but the parallelisation speeds it up quite well for many forms of matrices.

I have used a similar approach in the past to find the number of Hamiltonian Paths in a graph.