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.
Related
I try to explain my problem better.
in input I have the number of nodes N and the number of edges P.
N represent the cities while P the water pipes.
So for example in input I have 4 and 2 {0,1,2,3,4} would be my graph
and the 2 represents me the already existing water pipes
since in the following lines of the file I will have the already
existing connections so since in this case I have the 2 for example in
the file I'll have 3-1 and 2-1. so at the beginning I will have a
graph with 5 vertices and an edge that goes from node 3 to node 1 and
another arc that goes from 2 to 1.
now I have to determine how many and which connections to make to
bring water to all the cities. so for example in this case a solution
could be 0-4,0-3,0-2.
NOTE 0 represents the dam basin.
I was thinking that this problem seems to me very similar to MST but the graph I have is oriented and not weighted so I cannot apply the Prim or Kruskal algorithm.
So since it doesn't make sense to apply MST maybe I could do this via a DFS or a BFS but I don't quite understand how.
I believe that if I do a dfs from the source then from node 0 and I see all the nodes that I have not been able to reach and I connect these nodes to the source I have the solution but I have not understood how to do it.
or would it be better to try to make l mst work on this type of graph then cloning it and adding weights = 1 to the arcs?
" I know I can't apply it for the type of graph I have"
How do you know this?
Unweighted. Make your graph weighted by applying a weight of 1 to every edge.
Undirected. Make your graph directed by replacing every edge with two directed edges, one going each way.
I have a complete, weighted, undirected graph. The edge weights are the cost of a connection between two nodes, so the minimum spanning tree is the subset of the edges with the lowest total cost such that the graph remains connected.
The MST must be connected at all times, but unfortunately the connections aren't very reliable, so I would like to add redundancy to this graph/network.
Is it possible to compute a subset of edges such that the total edge cost is minimised and edge-connectivity is over a certain minimum?
I can see how it would be possible by bruteforcing, but I was looking for something more practical. I haven't been able to find much about this problem, I think mainly because I don't posses the vocabulary necessary to search.
My current idea is:
Compute the MST
While the it is still below a certain connectivity
Find a node most below that connectivity
Activate that node's edge with the lowest weight
The reason I don't find all the nodes below a certain connectivity all at once is because activating an edge may give another one enough connectivity.
I'm pretty sure this does not yield 100% provably optimal networks, because with this method, it is possible to over-connect nodes (e.g. you activate k edges for a node, then another node activates more shared edges, making some of those k redundant). I hope that makes sense.
Any tips would be much appreciated!
The Wikipedia article on edge connected graphs ends with, A related problem: finding the minimum k-edge-connected spanning subgraph of G (that is: select as few as possible edges in G that your selection is k-edge-connected) is NP-hard for k >= 2. They then cite a 1979 paper that shows it.
Therefore I'd suggest taking a greedy approach, and tip-toeing away.
Given a N*M array of 0 and 1.
A lake is a set of cells(1) which are horizontally or vertically adjacent.
We are going to connect all lakes on map by updating some cell(0) to 1.
The task is finding the way that number of updated cells is the smallest in a given time limit.
I found this similar question: What is the minimum cost to connect all the islands?
The solution on this topic get some problem:
1) It uses lib (pulp) to solve the task
2) It take time to get output
Is there an optimization solution for this problem
Thank you in advance
I think this is a tricky question but if you really draw it out and look at this matrix as a graph it makes it simpler.
Consider each cell as a node and each connection to its top/right/bottom/left to be an edge.
Start by connection the edges of the lakes to the nearby vertices. Keep doing the same for each each and only connect two vertices if it doesn't create a cycle.
At this stage carry out the same process for the immediate neighbours of the lakes. Keep doing the same and break if its creating cycles.
After this you should have a connected tree.
Once you have a connected tree you can find all the articulation point (Cut vertex) of the tree. (A vertex in an undirected connected graph is an articulation point (or cut vertex) iff removing it (and edges through it) disconnects the graph. Articulation points represent vulnerabilities in a connected network – single points whose failure would split the network into 2 or more disconnected components)
The number of cut vertex in the tree (excluding the initial lakes) would be the smallest number of cells that you need to change.
You can search there are many efficient ways to find cut vertex of a graph.
Finding articulation points takes O(V+E)
Preprocessing takes O(V+E) as it somewhat similar to a BFS.
Don't know whether you are still interested but I have an idea. What about a min cost flow algorithm.
Assume you have an m*n 2-d input array and i Islands. Create a graph where each position in the 2-d array is a node and has 4 edges to each neighbour. Each edge will be assigned a cost later on. Each edge has minimum capacity 0 and maximum capacity infinit.
Choose a random island to be the source. Create an extra node target and connect it to all other islands except the source with each new edge having maximum and minimum flow capacity 1 and cost 0.
Now assign the old edges costs, so that an edge connecting two island nodes costs nothing, but an edge between and island and a water node or an edge between two water nodes costs 1.
Calculate min cost flow over this graph. The initial graph generating can be done in nm and the min cost flow algorithm in (nm) ^3
I have a directed graph with flow volumes along edges and I would like to simplify it by removing all cyclic flows. This can be done by finding the minimum of the flow volumes along each edge in any given cycle and reducing the flows of every edge in the cycle by that minimum volume, deleting edges with zero flow. When all cyclic flows have been removed the graph will be acyclic.
For example, if I have a graph with vertices A, B and C with flows of 1 from A→B, 2 from B→C and 3 from C→A then I can rewrite this with no flow from A→B, 1 from B→C and 2 from C→A. The number of edges in the graph has reduced from 3 to 2 and the resulting graph is acyclic.
Which algorithm(s), if any, solve this problem?
You may find it useful to use the flow decomposition theorem (see §6.2 of this discussion of max-flow), which says that any flow can be broken down into a set of flow paths and flow cycles. Moreover, there can be at most m total flow paths and cycles in the graph. This means that one simple algorithm for eliminating flow cycles would be to find all of the flow paths in the graph, then remove all remaining flow in the graph since it must correspond to flow cycles.
To find a flow path or cycle, you can use a simple depth-first search from the source node. Starting at the source node, keep following edges however you'd like until either you hit the terminal node or you visit a node you've previously visited. If you hit the terminal node, then the path you've taken is a simple flow path. If you encounter some node twice, you've just found a flow cycle (formed by the loop that you just found). If you then remove the flow path/cycle from the graph and repeat, you will end up detecting all flow paths and cycles. You know that you're done when there are no flow-carrying edges leaving the source code. If each time that you find a flow path you record the total flow across all of its edges, you can eliminate cyclic flow by repeating this until no flow paths remain, clearing the flow in the network, then adding back in the flow paths.
Since each DFS takes time O(m + n) and there are at most O(m) flow paths, the total runtime of this step is O(m2 + mn), which is a polynomial in the size of the graph.
Hope this helps!
You could use topological sorting
http://en.wikipedia.org/wiki/Topological_sorting
It works great when it comes to finding a cycles in directed graphs
You can compute the value V of a flow given and then solve a min-cost flow problem for the network given and flow value V, assigning cost 1 to each edge.
Then a resulting flow should not contain any cycles, since that would be non-optimal (with respect to cost).
Have you thought about producing a minimum spanning tree? you could use Dijkstra's algorithm for that. If you want to first find out if a graph is cyclic you can determine that by using topological sorting.
I have directed graph with lot of cycles, probably strongly connected, and I need to get a minimal cycle from it. I mean I need to get cycle, which is the shortest cycle in graph, and every edge is covered at least once.
I have been searching for some algorithm or some theoretical background, but only thing I have found is Chinese postman algorithm. But this solution is not for directed graph.
Can anybody help me? Thanks
Edit>> All edges of that graph have the same cost - for instance 1
Take a look at this paper - Directed Chinese Postman Problem. That is the correct problem classification though (assuming there are no more restrictions).
If you're just reading into theory, take a good read at this page, which is from the Algorithms Design Manual.
Key quote (the second half for the directed version):
The optimal postman tour can be constructed by adding the appropriate edges to the graph G so as to make it Eulerian. Specifically, we find the shortest path between each pair of odd-degree vertices in G. Adding a path between two odd-degree vertices in G turns both of them to even-degree, thus moving us closer to an Eulerian graph. Finding the best set of shortest paths to add to G reduces to identifying a minimum-weight perfect matching in a graph on the odd-degree vertices, where the weight of edge (i,j) is the length of the shortest path from i to j. For directed graphs, this can be solved using bipartite matching, where the vertices are partitioned depending on whether they have more ingoing or outgoing edges. Once the graph is Eulerian, the actual cycle can be extracted in linear time using the procedure described above.
I doubt that it's optimal, but you could do a queue based search assuming the graph is guaranteed to have a cycle. Each queue entry would contain a list of nodes representing paths. When you take an element off the queue, add all possible next steps to the queue, ensuring you are not re-visiting nodes. If the last node is the same as the first node, you've found the minimum cycle.
what you are looking for is called "Eulerian path". You can google it to find enough info, basics are here
And about algorithm, there is an algorithm called Fleury's algorithm, google for it or take a look here
I think it might be worth while just simply writing which vertices are odd and then find which combo of them will lead to the least amount of extra time (if the weights are for times or distances) then the total length will be every edge weight plus the extra. For example, if the odd order vertices are A,B,C,D try AB&CD then AC&BD and so on. (I'm not sure if this is a specifically named method, it just worked for me).
edit: just realised this mostly only works for undirected graphs.
The special case in which the network consists entirely of directed edges can be solved in polynomial time. I think the original paper is Matching, Euler tours and the Chinese postman (1973) - a clear description of the algorithm for the directed graph problem begins on page 115 (page 28 of the pdf):
When all of the edges of a connected graph are directed and the graph
is symmetric, there is a particularly simple and attractive algorithm for
specifying an Euler tour...
The algorithm to find an Euler tour in a directed, symmetric, connected graph G is to first find a spanning arborescence of G. Then, at
any node n, except the root r of the arborescence, specify any order for
the edges directed away from n so long as the edge of the arborescence
is last in the ordering. For the root r, specify any order at all for the
edges directed away from r.
This algorithm was used by van Aardenne-Ehrenfest and de Bruin to
enumerate all Euler tours in a certain directed graph [ 1 ].