Many similar questions have been asked, but non addresses exactly my situation: Given two vertices in a simple directed unweighed graph, and an integer k, how can I find all k-tuples of edge-disjoint paths between the vertices? (In particular, I'm interested in the case that k is the outdegree of the start vertex.)
I know Suurballe's algorithm will give me k edge-disjoint paths, but it will (non-deterministically) settle on one solution, instead of giving me all of them.
It seems maximum-flow algorithms like Edmonds-Karp are related, but they don't compute paths.
Is there any algorithm already in JGraphT that does what I want?
Here's a simple recursive method:
if k is 0, output [] and return
otherwise, choose an arbitrary arc from the source vertex
for all paths p that start with that arc and end at the sink,
for all k-1 tuples T in the graph where the arcs in p are removed,
output [p] + T
This method can be improved by pruning parts of the recursion tree. The easiest idea is to remove all arcs to the source vertex, since if a path uses two arcs from the source vertex, we won't get to k before disconnecting the source and sink.
A broader version of that idea uses maximum flow to identify arcs that can be part of a solution. Compute the max flow from the source to the sink. By the flow decomposition theorem, there is at least one k-tuple of edge disjoint paths if and only if the value of the flow is at least k. If these conditions are true, then there is a solution that uses the set of arcs carrying flow, so this set needs to be retained. To test the others, compute the strongly connected components of the residual graph (arcs without flow appear forward, arcs with flow appear backward). All arcs without flow that go from one SCC to another can be safely discarded.
I have an undirected graph. Is there any efficient algorithm on how to find all independent connections between two nodes? By independent, I mean that these connections could have common nodes but cannot have common edges.
In this example, there are 2 independent connections from 0 to 8 (0-2-3-4-8 or 0-5-6-7-8). I tried using Breadth-first search algorithm continuously while "pseudo-erasing" edges which I've already seen. Problem is that I can go through graph this way: 0-5-4-8 which is not right because I can't make any other path.
Thanks for any help!
What you are interested is to solve min cut problem between a source and sink (the first of the nodes of interest for you is a source and the second is a sink).
Here you can read about the approach to this task. Basically I link to a theorem proving that the max flow between the source and the sink equals the min cut. You are interested in the min cut as this is exactly the minimum number of edges that need to be removed in order to get your source and sink disconnected.
If you run a Ford Fulkerson max flow algorithm you can reconstruct the different paths from the source to the sink considering which reverse edges have capacity after the algorithm is finished. One last note - Ford Fulkerson is classically described in directed graph. To make it work for your undirected case represent each edge as two separate directed edges facing opposite directions. All your initial capacities should be equal to 1.
Given a graph (undirected, no-weighted and all vertices are connected to eachother), I need to find the minimum number of "bad edges" I must visit to go from A to B. For example, if there's a graph with 5 vertices and the bad edges are: (0,1), (0,2), (0,3) and (0,4), to go from 0 to 4 I'll need to visit at least 1 bad edge. It could be straightfoward from 0 to 4 or from 0 to 1 and then 1 to 4. The length of the path doesn't matter at all. I'm trying a modified BFS to do the job but I'm not quite sure if this is the right way. My modification is instead of using a queue, use a list and when I find a bad edge, I put it into the back of the list, so I'll only visit this edge if really necessary, but found out that it won't minimize the number of bad edges. Any advices?
While it can indeed be solved by weighted shortest path, there is actually a more efficient (in terms of run time solutions).
First, define an auxillary graph G'=(V,E') where e is in E' iff e is "good". This step is linear in the size of the graph.
Now, you can find connected components in G' using DFS or BFS in O(|V|+|E|).
Next, all you have to do is "collapse" all nodes to a single node that represent them (this is also linear time), and add the "bad edges" (note that there is never a "good edge" that connects between two components, or they would have been in the same component).
Now, you can run BFS on the new graph, and the length of the path is the minimal number of nodes needed.
While this is significantly more complex to implement than a simple weighted shortest path, this solution offers O(|V|+|E|) (which in your graph is O(|E|)) run time, compared to O(|E|log|V|) of weighted shortest path.
I'm using Bellman-Ford to find the shortest path through a graph with some negative weights. The graph has no possibility of loops and no bi-directional connections. I'd like to find the K shortest paths through the graph, where the paths share no nodes in common. Is there an algorithm I can look up to learn how to do this? Simple implementation is more important than speed at the moment.
Added: Thanks for comments. To be clear, I'm looking for the top K ways to get from a specified start node to a specified end node, with no other nodes in common. I need a global optimum; sequentially finding the best and deleting nodes does not give a satisfactory result. This one: https://en.wikipedia.org/wiki/Yen%27s_algorithm, gives the flavor of what I'm talking about, but in this case it requires non-negative edge costs and it also allows nodes to be shared.
I think that the problem can be solved finding a Minimum Cost Flow.
Let's transform the graph in the following way:
Replace each node v other than source and sink with two nodes v1
and v2 connected by an edge of weight 0 from v1 to v2. The
incoming edges of the former v enter to v1 and the outgoing
leave from v2. With this the problem is equivalent to not using
those edges more than once.
Set capacity 1 to all the edges.
Finding a flow of value K will give you K paths that don't share a node (because of putting the capacity to 1 in those new edges). So if this flow is a minimum cost flow, you will have that those K paths also have the minimum possible sum of costs.
This is assuming that you don't have an edge connecting the source and the sink directly. Check for that corner case separately.
I have a connected, non-directed, graph with N nodes and 2N-3 edges. You can consider the graph as it is built onto an existing initial graph, which has 3 nodes and 3 edges. Every node added onto the graph and has 2 connections with the existing nodes in the graph. When all nodes are added to the graph (N-3 nodes added in total), the final graph is constructed.
Originally I'm asked, what is the maximum number of nodes in this graph that can be visited exactly once (except for the initial node), i.e., what is the maximum number of nodes contained in the largest Hamiltonian path of the given graph? (Okay, saying largest Hamiltonian path is not a valid phrase, but considering the question's nature, I need to find a max. number of nodes that are visited once and the trip ends at the initial node. I thought it can be considered as a sub-graph which is Hamiltonian, and consists max. number of nodes, thus largest possible Hamiltonian path).
Since i'm not asked to find a path, I should check if a hamiltonian path exists for given number of nodes first. I know that planar graphs and cycle graphs (Cn) are hamiltonian graphs (I also know Ore's theorem for Hamiltonian graphs, but the graph I will be working on will not be a dense graph with a great probability, thus making Ore's theorem pretty much useless in my case). Therefore I need to find an algorithm for checking if the graph is cycle graph, i.e. does there exist a cycle which contains all the nodes of the given graph.
Since DFS is used for detecting cycles, I thought some minor manipulation to the DFS can help me detect what I am looking for, as in keeping track of explored nodes, and finally checking if the last node visited has a connection to the initial node. Unfortunately
I could not succeed with that approach.
Another approach I tried was excluding a node, and then try to reach to its adjacent node starting from its other adjacent node. That algorithm may not give correct results according to the chosen adjacent nodes.
I'm pretty much stuck here. Can you help me think of another algorithm to tell me if the graph is a cycle graph?
Edit
I realized by the help of the comment (thank you for it n.m.):
A cycle graph consists of a single cycle and has N edges and N vertices. If there exist a cycle which contains all the nodes of the given graph, that's a Hamiltonian cycle. – n.m.
that I am actually searching for a Hamiltonian path, which I did not intend to do so:)
On a second thought, I think checking the Hamiltonian property of the graph while building it will be more efficient, which is I'm also looking for: time efficiency.
After some thinking, I thought whatever the number of nodes will be, the graph seems to be Hamiltonian due to node addition criteria. The problem is I can't be sure and I can't prove it. Does adding nodes in that fashion, i.e. adding new nodes with 2 edges which connect the added node to the existing nodes, alter the Hamiltonian property of the graph? If it doesn't alter the Hamiltonian property, how so? If it does alter, again, how so? Thanks.
EDIT #2
I, again, realized that building the graph the way I described might alter the Hamiltonian property. Consider an input given as follows:
1 3
2 3
1 5
1 3
these input says that 4th node is connected to node 1 and node 3, 5th to node 2 and node 3 . . .
4th and 7th node are connected to the same nodes, thus lowering the maximum number of nodes that can be visited exactly once, by 1. If i detect these collisions (NOT including an input such as 3 3, which is an example that you suggested since the problem states that the newly added edges are connected to 2 other nodes) and lower the maximum number of nodes, starting from N, I believe I can get the right result.
See, I do not choose the connections, they are given to me and I have to find the max. number of nodes.
I think counting the same connections while building the graph and subtracting the number of same connections from N will give the right result? Can you confirm this or is there a flaw with this algorithm?
What we have in this problem is a connected, non-directed graph with N nodes and 2N-3 edges. Consider the graph given below,
A
/ \
B _ C
( )
D
The Graph does not have a Hamiltonian Cycle. But the Graph is constructed conforming to your rules of adding nodes. So searching for a Hamiltonian Cycle may not give you the solution. More over even if it is possible Hamiltonian Cycle detection is an NP-Complete problem with O(2N) complexity. So the approach may not be ideal.
What I suggest is to use a modified version of Floyd's Cycle Finding algorithm (Also called the Tortoise and Hare Algorithm).
The modified algorithm is,
1. Initialize a List CYC_LIST to ∅.
2. Add the root node to the list CYC_LIST and set it as unvisited.
3. Call the function Floyd() twice with the unvisited node in the list CYC_LIST for each of the two edges. Mark the node as visited.
4. Add all the previously unvisited vertices traversed by the Tortoise pointer to the list CYC_LIST.
5. Repeat steps 3 and 4 until no more unvisited nodes remains in the list.
6. If the list CYC_LIST contains N nodes, then the Graph contains a Cycle involving all the nodes.
The algorithm calls Floyd's Cycle Finding Algorithm a maximum of 2N times. Floyd's Cycle Finding algorithm takes a linear time ( O(N) ). So the complexity of the modied algorithm is O(N2) which is much better than the exponential time taken by the Hamiltonian Cycle based approach.
One possible problem with this approach is that it will detect closed paths along with cycles unless stricter checking criteria are implemented.
Reply to Edit #2
Consider the Graph given below,
A------------\
/ \ \
B _ C \
|\ /| \
| D | F
\ / /
\ / /
E------------/
According to your algorithm this graph does not have a cycle containing all the nodes.
But there is a cycle in this graph containing all the nodes.
A-B-D-C-E-F-A
So I think there is some flaw with your approach. But suppose if your algorithm is correct, it is far better than my approach. Since mine takes O(n2) time, where as yours takes just O(n).
To add some clarification to this thread: finding a Hamiltonian Cycle is NP-complete, which implies that finding a longest cycle is also NP-complete because if we can find a longest cycle in any graph, we can find the Hamiltonian cycle of the subgraph induced by the vertices that lie on that cycle. (See also for example this paper regarding the longest cycle problem)
We can't use Dirac's criterion here: Dirac only tells us minimum degree >= n/2 -> Hamiltonian Cycle, that is the implication in the opposite direction of what we would need. The other way around is definitely wrong: take a cycle over n vertices, every vertex in it has exactly degree 2, no matter the size of the circle, but it has (is) an HC. What you can tell from Dirac is that no Hamiltonian Cycle -> minimum degree < n/2, which is of no use here since we don't know whether our graph has an HC or not, so we can't use the implication (nevertheless every graph we construct according to what OP described will have a vertex of degree 2, namely the last vertex added to the graph, so for arbitrary n, we have minimum degree 2).
The problem is that you can construct both graphs of arbitrary size that have an HC and graphs of arbitrary size that do not have an HC. For the first part: if the original triangle is A,B,C and the vertices added are numbered 1 to k, then connect the 1st added vertex to A and C and the k+1-th vertex to A and the k-th vertex for all k >= 1. The cycle is A,B,C,1,2,...,k,A. For the second part, connect both vertices 1 and 2 to A and B; that graph does not have an HC.
What is also important to note is that the property of having an HC can change from one vertex to the other during construction. You can both create and destroy the HC property when you add a vertex, so you would have to check for it every time you add a vertex. A simple example: take the graph after the 1st vertex was added, and add a second vertex along with edges to the same two vertices of the triangle that the 1st vertex was connected to. This constructs from a graph with an HC a graph without an HC. The other way around: add now a 3rd vertex and connect it to 1 and 2; this builds from a graph without an HC a graph with an HC.
Storing the last known HC during construction doesn't really help you because it may change completely. You could have an HC after the 20th vertex was added, then not have one for k in [21,2000], and have one again for the 2001st vertex added. Most likely the HC you had on 23 vertices will not help you a lot.
If you want to figure out how to solve this problem efficiently, you'll have to find criteria that work for all your graphs that can be checked for efficiently. Otherwise, your problem doesn't appear to me to be simpler than the Hamiltonian Cycle problem is in the general case, so you might be able to adjust one of the algorithms used for that problem to your variant of it.
Below I have added three extra nodes (3,4,5) in the original graph and it does seem like I can keep adding new nodes indefinitely while keeping the property of Hamiltonian cycle. For the below graph the cycle would be 0-1-3-5-4-2-0
1---3---5
/ \ / \ /
0---2---4
As there were no extra restrictions about how you can add a new node with two edges, I think by construction you can have a graph that holds the property of hamiltonian cycle.