I have a grid based graph, where nodes and edges occupy cells. Edges can cross, but cannot travel on top of each other in the same direction.
Lets say I want to optimize the graph so that the distance covered by edges is minimized.
I am currently using A* search for each connection, but the algorithm is greedy and does not plan ahead. Consider the diagram below, where the order in which connections are made is changed (note also that there can be multiple shortest paths for any given edge, see green and
purple connections).
My intuition says this is NP-Complete and that an exhaustive search is necessary, which will be extremely expensive as the size of the graph grows. However, I have no way of showing this, and it is not quite the same as other graph embedding problems which usually concern minimization of crossing.
You didn't really describe your problem and your image is gone, but your problem sounds like the minimum T-join problem.
The minimum T-join problem is defined on a graph G. You're given a set T of even size, and you're trying to find a subgraph of the graph where the vertices of T have odd degree and the other vertices have even degree. You've got weights on the edges and you're trying to minimise the sum of the weights of edges in the subgraph.
Surprisingly, the minimum T-join problem can be solved in polynomial time thanks to a very close connection with the nonbipartite matching problem. Namely, if you find all-pairs shortest paths between vertices of T, the minimum T-join is attained by the minimum-weight perfect matching of vertices in T, where there's an edge between two vertices whose length is the length of the shortest path in G.
The minimum T-join will be a collection of paths. If two distinct paths, say a->b and c->d, use the same edge uv, then they can be replaced by a->u->c and b->v->d and reduce the cost of the T-join. So it won't use the same edge twice.
Related
For an undirected graph G if possible I want to convert G to a directed graph such that in the directed graph each vertex must have an indegree of at least 2.
I worked out that for it to be possible the amount of edges will need be at least 2|V| therefore will only be possible from at least 5 vertexs.
I also worked out that each vertex will be in a cycle with at least two other vertex,
but I tried using a modification of N/F but can't seem to think of an algorithm that can actually do the conversion (if possible).
Any help / guidance will be appreciated
Given an undirected, unweighted graph with n vertices and m edges, you can decide whether an orientation with minimum indegree at least 2 exists (and find it, if so) in O(m^3) time. If there are no vertices of degree 3, you can sometimes do this fairly simply in O(m) time.
If v is a vertex of degree 1, the task is impossible. If v has degree 2, both its edges must be out-edges, so we can delete 'v' from our graph for now. We do this operation repeatedly based on the new degrees; it's possible that we end up with new degree-3 vertices, but we can now assume all vertices have degree >= 3.
If all vertices are degree 4 or greater, the orientation must exist, and you can use a standard Eulerian circuit algorithm (available in many standard graph libraries) as described in this post. You may need to modify the graph first: Eulerian circuits only exist if all vertex degrees are even. There are always an even number of odd degree vertices: match them up in pairs in any way, adding an edge for each pair. Orienting the edges on the original graph based on the circuit gives a solution for your problem.
For graphs with degree-3 vertices, you can combine a standard algorithm with a nonstandard one. The standard algorithm is maximum-cardinality matching for general graphs, usually the 'Blossom algorithm', which can be found in standard graph libraries. The non-standard (i.e., probably not found in your graph library) algorithm is found in the paper On finding orientations with the fewest number of vertices with small out-degree, which reduces the problem of 'Find an orientation with the fewest outdegree-less-than-2 vertices' to the problem of finding a maximum matching in a graph with O(m) vertices. The algorithm is quite short, and simpler since our graph has minimum degree 3. This solves your problem for fewest-indegrees-less-than-2 after you flip all edge directions.
Full question: Argue that if all edge weights of a graph are positive, then any subset of edges that connects all vertices and has minimum total weight must be a tree. Give an example to show that the same conclusion does not follow if we allow some weights to be nonpositive.
My answer: Since the edges connects all vertices, it must be a tree. In a graph, you can remove one of the edges and still connect all the vertices. Also, negative edges can be allowed in a graph (e.g. Prim and Kruskal's algorithms).
Please let me know if there's a definite answer to this and explain to me how you got the conclusion. I'm a little bit lost with this question.
First off, a tree is a type of graph. So " In a graph, you can remove one of the edges and still connect all the vertices" isn't true. A tree is a graph without cycles - i.e., with only one path between any two nodes.
Negatives weights in general can exist in either a tree or a graph.
The way to approach this problem is to show that if you have a graph that connects all components, but is NOT a tree, then it is also not of minimum weight (i.e., there is some other graph that does the same thing, with a lower total weight.) This conclusion is only true if the graph contains only positive edges, so you should also provide a counterexample - a graph which is NOT a tree, which IS of minimum weight, and which IS fully connected.
With non-negative weights, adding an edge to traverse from one node to another always results in the weight increasing, so for minimum weight you always avoid that.
If you allow negative weights, adding an edge may result in reducing the weight. If you have a cycle with negative weight overall, minimum weight demands that you stay in that cycle infinitely (leading to infinitely negative weight for the path overall).
Given a DAG (possibly not strongly connected e.i consisting of several connected components), the goal is to find the minimum number of starting vertices required to visit to fully explore the graph.
One method I thought of was to generate all permutations of the given vertices and run a topological sort in that order. The one with the minimum backtracks would be the answer.
Is there an efficient algorithm to perform the above task?
This a famous problem called minimum path cover, it's a pity that wiki says nothing about it, you can search it in google.
As methioned, the minimum path cover problem is NP-hard in normal graph. But in DAG, it can be solved with Matching.
Method:
Dividing each vertex u into two different vertex u1 and u2. For every edge (u->v) in orginal graph, adding edge (u1->v2) in new graph. Then do any matching algorithm you like. The result is n - maximum matching, n is total number of vertex in orginal graph.
I have a disconnected bipartite undirected graph. I want to completely disconnect the graph. Only operation that I can perform is to remove a node. Removing a node will automatically delete its edges. Task is to minimize the number of nodes to be removed. Each node in the graph has atmost 4 edges.
By completely disconnecting a graph, I mean that no two nodes should be connected through a link. Basically an empty edge set.
I think, you cannot prove your algorithm is optimal because, in fact, it is not optimal.
To completely disconnect your graph minimizing the number of nodes to be removed, you have to remove all the nodes belonging to the minimal vertex cover of your graph. Searching the minimal vertex cover is usually NP-complete, but for bipartite graphs there is a polynomial-time solution.
Find maximum matching in the graph (probably with Hopcroft–Karp algorithm). Then use König's theorem to get the minimal vertex cover:
Consider a bipartite graph where the vertices are partitioned into left (L) and right (R) sets. Suppose there is a maximum matching which partitions the edges into those used in the matching (E_m) and those not (E_0). Let T consist of all unmatched vertices from L, as well as all vertices reachable from those by going left-to-right along edges from E_0 and right-to-left along edges from E_m. This essentially means that for each unmatched vertex in L, we add into T all vertices that occur in a path alternating between edges from E_0 and E_m.
Then (L \ T) OR (R AND T) is a minimum vertex cover.
Here's a counter-example to your suggested algorithm.
The best solution is to remove both nodes A and B, even though they are different colors.
Since all the edges are from one set to another, find these two sets using say BFS and coloring using 2 colours. Then remove the nodes in smaller set.
Since there are no edges among themselves the rest of the nodes are disconnected as well.
[As a pre-processing step you can leave out nodes with 0 edges first.]
I have thought of an algorithm for it but am not able to prove if its optimal.
My algorithm: On each disconnected subgraph, I run a BFS and color it accordingly. Then I identify the number of nodes colored with each color and take the minimum of the two and store. I repeat the procedure for each subgraph and add up to get the required minimum. Help me prove the algorithm if it's correct.
EDIT: The above algorithm is not optimal. The accepted answer has been verified to be correct.
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 ].