I'm trying to solve the following problem:
I have an directed graph G = (E,V) which has a low number of edges. Now I try to find all subgraphs in it which are transitive closured and maximal which means there should be no subgraph which ist part of another subgraph.
The first idea I had was to start at every Node doing a DFS and on every step look if all edges for the closure exists but the performance is horrible. So I wonder if there is any faster algorithm
Related
My problem is a generalization of a task solved by [Blossom algorithm] by Edmonds. The original task is the following: given a complete graph with weighted undirected edges, find a set of edges such that
1) every vertex of the graph is adjacent to only one edge from this set (i.e. vertices are grouped into pairs)
2) sum over weights of edges in this set is minimal.
Now, I would like to modify the first goal into
1') vertices are grouped into sets of 3 vertices (or in general, d vertices), and leave condition 2) unchanged.
My questions:
Do you know if this 'generalised' problem has a name?
Do you know about an algorithm solving it in number of steps being polynomial of number of vertices (like Blossom algorithm for an original problem)? I don't see a straightforward generalisation of Blossom algorithm, as it is based on looking for augmenting paths on a graph compressed to a bipartite graph (and uses here Hungarian algorithm). But augmenting paths do not seem to point to groups of vertices different than pairs.
Best regards,
Paweł
I was solving this CLRS problem, which asked to find out the indegree of every vertex of a graph G(V,E). I found out the solution to be O(|E|) as we only have to scan through all the edges to find out the degrees of all vertices.
But most of the solutions, I found online, say that it is O(|V|+|E|). How come? How are the vertices accounting for the time taken?
If we suppose that the implementation of the digraph uses objects for vertices and each vertex has an associated list of successors and no additional data structures, then it will be impossible to iterate the edges directly.
If the digraph is connected, then each vertext has at least one associated edge. This means that iteration over the edges via iteration over the vertices takes O(|E|) time - the iteration over the vertices does not increase the running time, which is dominated by the number of edges.
If the digraph is not connected, then the iteration over the vertices is not necessarily dominated by the number of edges; even isolated vertices have to be processed just to find out that they have no associated outgoing arcs, which can be done in O(|V|+|E|) time.
In total, the runtime bound of O(|V|+|E|) is correct in either case; however, for a connected digraph (or an implementation which permits direct iteration over the edges regardless of the number of vertices) one can obtain a tighter runtime bound of O(|E|).
I'm doing a breadth-first search on a digraph. I'm lost at nodes c and f, and I'm not sure if and how they should be in the BF-tree or if you only go as far as reachable from the source node and don't start at another node in order to get all the vertices.
Here's what I'm getting so far. As you can see, letters mark the nodes. Distance and predecessor are marked by d and pi:
This was helpful BFS traversal of directed graph from a given node but I'm not familiar with graphs enough to understand how that applies to this situation. From what I'm getting from that question, it seems like in this case I would not include c and f at all.
In fact, it seems like I have the maximal number of nodes included already, just because I started at i. I think that the d=4 at the g node (also at k but that doesn't even connect to any other nodes), this is the greatest distance and max-depth possible in BFS in this graph.
I came across this posting from a while back:
Best algorithm to determine if an undirected graph is a tree
It says that to determine if an undirected graph is a tree, you just have to check if it has a cycle. However, don't you have to make sure the graph is connected? I was taught that a tree is connected and acyclic. How is checking only for acyclicity sufficient?
Thanks.
You're right. If the graph is acyclic, then it's a forest. In addition, if it only has one component, then it's a tree.
What the algorithm mentioned does is look for back edges. If it finds one, then the graph is not a tree. If it doesn't find one and the algorithm visited n-1 edges before running out of edges, then it IS a tree, because having visited n-1 edges means that the graph is indeed connected (a tree with n vertices has n-1 edges). If the algorithm ran out of edges but didn't reach n-1 visited edges, then it means the graph is not connected, hence it is not a tree.
I have found this paper so far. Is it outdated? Are there any faster and better implementations?
By the way, Wikipedia says that there can be n^n-2 spanning trees in a undirected graph. How many spanning trees can be in a directed graph?
If you use terms from paper you mentioned and you define spanning tree of directed graph as tree rooted in vertex r, having unique path from r to any other vertex then:
It's obvious that worst case when directed graph has the greatest number of the spanning trees is complete graph (there are a->b and b->a edges for any pair).
If we "forget" about directions we will get n^{n-2} spanning trees as in case of undirected graphs. For any of this spanning trees we have n options to choose a root, and this choice define uniquely define directions of edges we need to use. Not hard to see, that all trees we get are spanning, unique and there are no nother options. So we get n^{n-1} spanning trees. Strict proof will take time, I hope that simple explanation is enough.
So this task will take exponential time depend from vertex count in worst case. Considering the size of output (all spanning trees), I conclude that for arbitrary graph, algorithm can not be significantly faster and better. I think you need to somehow reformulate your original problem to not deal with all spanning trees, and may be search only needed by some criteria.
for undirected graph only....
n^n-2 spanning tress are possible for only complete graph....to find total number of spanning trees of any graph u can apply this method.....
find the adjacency matrix of the graph.
if column values are represented by 'i' and row entries by 'j' then...
if i=j...then the value will be the degree of vertex
suppose,there is a single edge between vertex v1 and v2 then the value of matrix entry will be -1......7 if there are two edges then it will be -2...& so on...
after constructing adjacency matrix....exclude any row and column...i.e, Nth row and Nth column....
answer will be the total number of spanning tress.