Find all paths between two graph nodes - algorithm

Question

I am working on an implementation of Dijkstra's Algorithm to retrieve the shortest path between interconnected nodes on a network of routes. I have the implementation working. It returns all the shortest paths to all the nodes when I pass the start node into the algorithm.
My question:
How does one go about retrieving all possible paths from Node A to, say, Node G or even all possible paths from Node A and back to Node A?

Answer 1

Finding all possible paths is a hard problem, since there are exponential number of simple paths. Even finding the kth shortest path [or longest path] are NP-Hard.
One possible solution to find all paths [or all paths up to a certain length] from s to t is BFS, without keeping a visited set, or for the weighted version - you might want to use uniform cost search
Note that also in every graph which has cycles [it is not a DAG] there might be infinite number of paths between s to t.

Answer 2

I've implemented a version where it basically finds all possible paths from one node to the other, but it doesn't count any possible 'cycles' (the graph I'm using is cyclical). So basically, no one node will appear twice within the same path. And if the graph were acyclical, then I suppose you could say it seems to find all the possible paths between the two nodes. It seems to be working just fine, and for my graph size of ~150, it runs almost instantly on my machine, though I'm sure the running time must be something like exponential and so it'll start to get slow quickly as the graph gets bigger.
Here is some Java code that demonstrates what I'd implemented. I'm sure there must be more efficient or elegant ways to do it as well.
Stack connectionPath = new Stack();
List<Stack> connectionPaths = new ArrayList<>();
// Push to connectionsPath the object that would be passed as the parameter 'node' into the method below
void findAllPaths(Object node, Object targetNode) {
for (Object nextNode : nextNodes(node)) {
if (nextNode.equals(targetNode)) {
Stack temp = new Stack();
for (Object node1 : connectionPath)
temp.add(node1);
connectionPaths.add(temp);
} else if (!connectionPath.contains(nextNode)) {
connectionPath.push(nextNode);
findAllPaths(nextNode, targetNode);
connectionPath.pop();
}
}
}

Answer 3

I'm gonna give you a (somewhat small) version (although comprehensible, I think) of a scientific proof that you cannot do this under a feasible amount of time.
What I'm gonna prove is that the time complexity to enumerate all simple paths between two selected and distinct nodes (say, s and t) in an arbitrary graph G is not polynomial. Notice that, as we only care about the amount of paths between these nodes, the edge costs are unimportant.
Sure that, if the graph has some well selected properties, this can be easy. I'm considering the general case though.
Suppose that we have a polynomial algorithm that lists all simple paths between s and t.
If G is connected, the list is nonempty. If G is not and s and t are in different components, it's really easy to list all paths between them, because there are none! If they are in the same component, we can pretend that the whole graph consists only of that component. So let's assume G is indeed connected.
The number of listed paths must then be polynomial, otherwise the algorithm couldn't return me them all. If it enumerates all of them, it must give me the longest one, so it is in there. Having the list of paths, a simple procedure may be applied to point me which is this longest path.
We can show (although I can't think of a cohesive way to say it) that this longest path has to traverse all vertices of G. Thus, we have just found a Hamiltonian Path with a polynomial procedure! But this is a well known NP-hard problem.
We can then conclude that this polynomial algorithm we thought we had is very unlikely to exist, unless P = NP.

Answer 4

The following functions (modified BFS with a recursive path-finding function between two nodes) will do the job for an acyclic graph:
from collections import defaultdict
# modified BFS
def find_all_parents(G, s):
Q = [s]
parents = defaultdict(set)
while len(Q) != 0:
v = Q[0]
Q.pop(0)
for w in G.get(v, []):
parents[w].add(v)
Q.append(w)
return parents
# recursive path-finding function (assumes that there exists a path in G from a to b)
def find_all_paths(parents, a, b):
return [a] if a == b else [y + b for x in list(parents[b]) for y in find_all_paths(parents, a, x)]
For example, with the following graph (DAG) G given by
G = {'A':['B','C'], 'B':['D'], 'C':['D', 'F'], 'D':['E', 'F'], 'E':['F']}
if we want to find all paths between the nodes 'A' and 'F' (using the above-defined functions as find_all_paths(find_all_parents(G, 'A'), 'A', 'F')), it will return the following paths:

Answer 5

Here is an algorithm finding and printing all paths from s to t using modification of DFS. Also dynamic programming can be used to find the count of all possible paths. The pseudo code will look like this:
AllPaths(G(V,E),s,t)
C[1...n] //array of integers for storing path count from 's' to i
TopologicallySort(G(V,E)) //here suppose 's' is at i0 and 't' is at i1 index
for i<-0 to n
if i<i0
C[i]<-0 //there is no path from vertex ordered on the left from 's' after the topological sort
if i==i0
C[i]<-1
for j<-0 to Adj(i)
C[i]<- C[i]+C[j]
return C[i1]

Answer 6

If you actually care about ordering your paths from shortest path to longest path then it would be far better to use a modified A* or Dijkstra Algorithm. With a slight modification the algorithm will return as many of the possible paths as you want in order of shortest path first. So if what you really want are all possible paths ordered from shortest to longest then this is the way to go.
If you want an A* based implementation capable of returning all paths ordered from the shortest to the longest, the following will accomplish that. It has several advantages. First off it is efficient at sorting from shortest to longest. Also it computes each additional path only when needed, so if you stop early because you dont need every single path you save some processing time. It also reuses data for subsequent paths each time it calculates the next path so it is more efficient. Finally if you find some desired path you can abort early saving some computation time. Overall this should be the most efficient algorithm if you care about sorting by path length.
import java.util.*;
public class AstarSearch {
private final Map<Integer, Set<Neighbor>> adjacency;
private final int destination;
private final NavigableSet<Step> pending = new TreeSet<>();
public AstarSearch(Map<Integer, Set<Neighbor>> adjacency, int source, int destination) {
this.adjacency = adjacency;
this.destination = destination;
this.pending.add(new Step(source, null, 0));
}
public List<Integer> nextShortestPath() {
Step current = this.pending.pollFirst();
while( current != null) {
if( current.getId() == this.destination )
return current.generatePath();
for (Neighbor neighbor : this.adjacency.get(current.id)) {
if(!current.seen(neighbor.getId())) {
final Step nextStep = new Step(neighbor.getId(), current, current.cost + neighbor.cost + predictCost(neighbor.id, this.destination));
this.pending.add(nextStep);
}
}
current = this.pending.pollFirst();
}
return null;
}
protected int predictCost(int source, int destination) {
return 0; //Behaves identical to Dijkstra's algorithm, override to make it A*
}
private static class Step implements Comparable<Step> {
final int id;
final Step parent;
final int cost;
public Step(int id, Step parent, int cost) {
this.id = id;
this.parent = parent;
this.cost = cost;
}
public int getId() {
return id;
}
public Step getParent() {
return parent;
}
public int getCost() {
return cost;
}
public boolean seen(int node) {
if(this.id == node)
return true;
else if(parent == null)
return false;
else
return this.parent.seen(node);
}
public List<Integer> generatePath() {
final List<Integer> path;
if(this.parent != null)
path = this.parent.generatePath();
else
path = new ArrayList<>();
path.add(this.id);
return path;
}
#Override
public int compareTo(Step step) {
if(step == null)
return 1;
if( this.cost != step.cost)
return Integer.compare(this.cost, step.cost);
if( this.id != step.id )
return Integer.compare(this.id, step.id);
if( this.parent != null )
this.parent.compareTo(step.parent);
if(step.parent == null)
return 0;
return -1;
}
#Override
public boolean equals(Object o) {
if (this == o) return true;
if (o == null || getClass() != o.getClass()) return false;
Step step = (Step) o;
return id == step.id &&
cost == step.cost &&
Objects.equals(parent, step.parent);
}
#Override
public int hashCode() {
return Objects.hash(id, parent, cost);
}
}
/*******************************************************
* Everything below here just sets up your adjacency *
* It will just be helpful for you to be able to test *
* It isnt part of the actual A* search algorithm *
********************************************************/
private static class Neighbor {
final int id;
final int cost;
public Neighbor(int id, int cost) {
this.id = id;
this.cost = cost;
}
public int getId() {
return id;
}
public int getCost() {
return cost;
}
}
public static void main(String[] args) {
final Map<Integer, Set<Neighbor>> adjacency = createAdjacency();
final AstarSearch search = new AstarSearch(adjacency, 1, 4);
System.out.println("printing all paths from shortest to longest...");
List<Integer> path = search.nextShortestPath();
while(path != null) {
System.out.println(path);
path = search.nextShortestPath();
}
}
private static Map<Integer, Set<Neighbor>> createAdjacency() {
final Map<Integer, Set<Neighbor>> adjacency = new HashMap<>();
//This sets up the adjacencies. In this case all adjacencies have a cost of 1, but they dont need to.
addAdjacency(adjacency, 1,2,1,5,1); //{1 | 2,5}
addAdjacency(adjacency, 2,1,1,3,1,4,1,5,1); //{2 | 1,3,4,5}
addAdjacency(adjacency, 3,2,1,5,1); //{3 | 2,5}
addAdjacency(adjacency, 4,2,1); //{4 | 2}
addAdjacency(adjacency, 5,1,1,2,1,3,1); //{5 | 1,2,3}
return Collections.unmodifiableMap(adjacency);
}
private static void addAdjacency(Map<Integer, Set<Neighbor>> adjacency, int source, Integer... dests) {
if( dests.length % 2 != 0)
throw new IllegalArgumentException("dests must have an equal number of arguments, each pair is the id and cost for that traversal");
final Set<Neighbor> destinations = new HashSet<>();
for(int i = 0; i < dests.length; i+=2)
destinations.add(new Neighbor(dests[i], dests[i+1]));
adjacency.put(source, Collections.unmodifiableSet(destinations));
}
}
The output from the above code is the following:
[1, 2, 4]
[1, 5, 2, 4]
[1, 5, 3, 2, 4]
Notice that each time you call nextShortestPath() it generates the next shortest path for you on demand. It only calculates the extra steps needed and doesnt traverse any old paths twice. Moreover if you decide you dont need all the paths and end execution early you've saved yourself considerable computation time. You only compute up to the number of paths you need and no more.
Finally it should be noted that the A* and Dijkstra algorithms do have some minor limitations, though I dont think it would effect you. Namely it will not work right on a graph that has negative weights.
Here is a link to JDoodle where you can run the code yourself in the browser and see it working. You can also change around the graph to show it works on other graphs as well: http://jdoodle.com/a/ukx

Answer 7

find_paths[s, t, d, k]
This question is now a bit old... but I'll throw my hat into the ring.
I personally find an algorithm of the form find_paths[s, t, d, k] useful, where:
s is the starting node
t is the target node
d is the maximum depth to search
k is the number of paths to find
Using your programming language's form of infinity for d and k will give you all paths§.
§ obviously if you are using a directed graph and you want all undirected paths between s and t you will have to run this both ways:
find_paths[s, t, d, k] <join> find_paths[t, s, d, k]
Helper Function
I personally like recursion, although it can difficult some times, anyway first lets define our helper function:
def find_paths_recursion(graph, current, goal, current_depth, max_depth, num_paths, current_path, paths_found)
current_path.append(current)
if current_depth > max_depth:
return
if current == goal:
if len(paths_found) <= number_of_paths_to_find:
paths_found.append(copy(current_path))
current_path.pop()
return
else:
for successor in graph[current]:
self.find_paths_recursion(graph, successor, goal, current_depth + 1, max_depth, num_paths, current_path, paths_found)
current_path.pop()
Main Function
With that out of the way, the core function is trivial:
def find_paths[s, t, d, k]:
paths_found = [] # PASSING THIS BY REFERENCE
find_paths_recursion(s, t, 0, d, k, [], paths_found)
First, lets notice a few thing:
the above pseudo-code is a mash-up of languages - but most strongly resembling python (since I was just coding in it). A strict copy-paste will not work.
[] is an uninitialized list, replace this with the equivalent for your programming language of choice
paths_found is passed by reference. It is clear that the recursion function doesn't return anything. Handle this appropriately.
here graph is assuming some form of hashed structure. There are a plethora of ways to implement a graph. Either way, graph[vertex] gets you a list of adjacent vertices in a directed graph - adjust accordingly.
this assumes you have pre-processed to remove "buckles" (self-loops), cycles and multi-edges

Answer 8

You usually don't want to, because there is an exponential number of them in nontrivial graphs; if you really want to get all (simple) paths, or all (simple) cycles, you just find one (by walking the graph), then backtrack to another.

Answer 9

I think what you want is some form of the Ford–Fulkerson algorithm which is based on BFS. Its used to calculate the max flow of a network, by finding all augmenting paths between two nodes.
http://en.wikipedia.org/wiki/Ford%E2%80%93Fulkerson_algorithm

Answer 10

There's a nice article which may answer your question /only it prints the paths instead of collecting them/.
Please note that you can experiment with the C++/Python samples in the online IDE.
http://www.geeksforgeeks.org/find-paths-given-source-destination/

Answer 11

I suppose you want to find 'simple' paths (a path is simple if no node appears in it more than once, except maybe the 1st and the last one).
Since the problem is NP-hard, you might want to do a variant of depth-first search.
Basically, generate all possible paths from A and check whether they end up in G.