I don't want to find all the minimum spanning trees but I want to know how many of them are there, here is the method I considered:
Find one minimum spanning tree using prim's or kruskal's algorithm and then find the weights of all the spanning trees and increment the running counter when it is equal to the weight of minimum spanning tree.
I couldn't find any method to find the weights of all the spanning trees and also the number of spanning trees might be very large, so this method might not be suitable for the problem.
As the number of minimum spanning trees is exponential, counting them up wont be a good idea.
All the weights will be positive.
We may also assume that no weight will appear more than three times in the graph.
The number of vertices will be less than or equal to 40,000.
The number of edges will be less than or equal to 100,000.
There is only one minimum spanning tree in the graph where the weights of vertices are different. I think the best way of finding the number of minimum spanning tree must be something using this property.
EDIT:
I found a solution to this problem, but I am not sure, why it works. Can anyone please explain it.
Solution: The problem of finding the length of a minimal spanning tree is fairly well-known; two simplest algorithms for finding a minimum spanning tree are Prim's algorithm and Kruskal's algorithm. Of these two, Kruskal's algorithm processes edges in increasing order of their weights. There is an important key point of Kruskal's algorithm to consider, though: when considering a list of edges sorted by weight, edges can be greedily added into the spanning tree (as long as they do not connect two vertices that are already connected in some way).
Now consider a partially-formed spanning tree using Kruskal's algorithm. We have inserted some number of edges with lengths less than N, and now have to choose several edges of length N. The algorithm states that we must insert these edges, if possible, before any edges with length greater than N. However, we can insert these edges in any order that we want. Also note that, no matter which edges we insert, it does not change the connectivity of the graph at all. (Let us consider two possible graphs, one with an edge from vertex A to vertex B and one without. The second graph must have A and B as part of the same connected component; otherwise the edge from A to B would have been inserted at one point.)
These two facts together imply that our answer will be the product of the number of ways, using Kruskal's algorithm, to insert the edges of length K (for each possible value of K). Since there are at most three edges of any length, the different cases can be brute-forced, and the connected components can be determined after each step as they would be normally.
Looking at Prim's algorithm, it says to repeatedly add the edge with the lowest weight. What happens if there is more than one edge with the lowest weight that can be added? Possibly choosing one may yield a different tree than when choosing another.
If you use prim's algorithm, and run it for every edge as a starting edge, and also exercise all ties you encounter. Then you'll have a Forest containing all minimum spanning trees Prim's algorithm is able to find. I don't know if that equals the forest containing all possible minimum spanning trees.
This does still come down to finding all minimum spanning trees, but I can see no simple way to determine whether a different choice would yield the same tree or not.
MST and their count in a graph are well-studied. See for instance: http://www14.informatik.tu-muenchen.de/konferenzen/Jass08/courses/1/pieper/Pieper_Paper.pdf.
Related
Could someone explain in pretty simple words why the normalization of many network analysis indicies (graph theory) is n(n - 1), where n – the graph size? Why do we need to take into account (n - 1)?
Most network measures that focus on counting edges (e.g. clustering coefficient) are normalized by the total number of possible edges. Since every edge connects a pair of vertices, we need to know how many possible pairs of vertices we can make. There are n possible vertices we could choose as the source of our edge, and therefore there are n-1 possible vertices that could be the target of our edge (assuming no self-loops, and if undirected divide by 2 bc source and target are exchangeable). Hence, you frequently encounter $n(n-1)$ or $\binomal{n}{2}$.
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'm trying to test some models of graph partitioning (these come from the real world, where a graph slowly self-partitions). To do this, I need to be able to uniformly randomly partition this graph into contiguous components (we are given the graph is initially connected, as well). Were the contiguity criterion not required I believe this would be the problem of randomly partitioning a set, which can be combinatorially analyzed. Does anyone know of any way to randomly partition graphs into subgraphs (i.e. randomly sample one partition), or, if no such method is known, to randomly sample a set of elements? The method of randomizing the number of partitions and then randomizing membership won't work because there are different numbers of possible partitions for each partition size.
You have to differentiate edge-cut partitioning and vertex-cut partitioning, where you divide the graph along the edges or vertices. This significantly impacts your problem as the number of different vertex-cuts is much larger than the number of edge-cuts. The reason is that you exclusively assign edges to partitions in vertex-cut - as opposed to edge-cut where you assign vertices to partitions - and there are much more edges than vertices (e.g. O(n^2) edges for n vertices). Hence, the combinatorially larger vertex-cut leads to a larger number of subgraphs that have to be checked for connectivity. A naive method for randomization is to enumerate all partitionings, iteratively select one partitioning, and check connectivity of all subgraphs in the selected partitioning. Then you just take the first one. In this case, all solutions have equal probability (uniformly random).
I have come across the same problem in work I am doing. I have two solutions to randomly partition a graph into m contiguous components:
Spanning Tree Approach. Randomly choose a spanning tree of your graph (e.g. Using Wilson's algorithm which chooses uniformly amongst all spanning trees). Then randomly select m-1 edges (without replacements) and remove them from the spanning tree. This will give m components which are each connected in the original graph.
Edge contraction approach. Randomly choose an edge and contract it, renaming the (new) vertex as the union of the two previous vertices. Repeat until you have only m vertices left. Identify each vertex with the subset of (original) vertices that were contracted into it.
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.
I'm looking to build an algorithm (or reuse one) that organizes nodes and edges on a 2 dimensional canvas where edges can have corresponding weights.
Any starting material and info would be helpful.
What would the weights do to affect their placement on your canvas?
That being said, you might want to look into graphviz and, more specifically, the DOT language, which organizes nodes on a canvas.
Many graph visualization frameworks use a force-based simulation, in which all nodes exert a repulsive force against each other (with their mass being their size), and edges exert tension on the nodes they connect. This creates aesthetically-arranged graph visualizations.
Although again, I'm not sure where you want node "weights" to come into play. Do you want weighted nodes to be more in the center? To be larger? More further apart?
Many graph/network layout algorithms are implicitly capable of handling weighted networks, but you may need to do some pre-processing and tweaks to the implementation to get it to work. Usually the first step is to determine if your weights represent "similarities" (usually interpreted to mean that stronger weights should place nodes closer togeter) or "dissimilarities" (stronger weights = father apart). The most common case is the former, so you will need to translate them to dissimilarities, often done by subtracting each edge value from the maximum observed edge value in the network. The matrix of dissimilarity values for each edge can then be fed to the algorithm and interpreted as desired distances in the layout space for each edge (i.e. "spring lengths")--usually after multiplying by some constant to transform to display units (pixels).
If you tell me what language you are using, I may be able to point you to some code examples.