Has anyone written up the algorithm in Sage given by Tamir Tassa in, ``Finding all maximally-matchable edges in a bipartite graph''?
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Given a undirected weighted complete graph G=(V,E) of N vertices, I want to know that if finding the smallest complete subgraph (with minimum edge weights sum) with M vertices (M <= N) is NP-hard or not.
This problem is NP-hard, because the decision form of the clique problem can be reduced to it.
The decision form of the clique problem asks, given an undirected graph and an integer k, whether there is a complete subgraph with k vertices. This problem is NP-hard. Given an instance of this problem, construct a complete weighted graph on the same number of vertices by assigning a weight of 1 to any edge present in the original graph, and a weight of 2 to any edge not present in the original graph. Then a minimum-weight complete subgraph will have every edge weight equal to 1 if and only if there is a complete subgraph of size k in the original graph.
I'm fairly new to IGraph in R.
I'm doing community detection using IGraph and have already built my communities /clusters using the walktrap technique.
Next, within each cluster, I want to count the number of vertices between each two certain vertices. The reason I want to do this is, for each vertex XX, I want to list vertices that are connected to XX via say max 3 vertices, meaning no further than 3 vertices away from XX.
Can anyone help how this can be done in R please?
making a random graph (for demonstration):
g <- erdos.renyi.game(100, 1/25)
plot(g,vertex.size=3)
get walktrap communities and save as vertex attribute:
V(g)$community<-walktrap.community(g, modularity = TRUE, membership = TRUE)$membership
V(g)$community
now make a subgraph containing only edges and vertices of one community, e.g. community 2:
sub<-induced.subgraph(g,v=V(g)$community==2)
plot(sub)
make a matrix containing all shortest paths:
shortestPs<-shortest.paths(sub)
now count the number of shortest paths smaller or equal to 3.
I also exclude shortest paths from each node to itself (shortestPaths!=0).
also divide by two because every node pair appears twice in the matrix for undirected graphs.
Number_of_shortest_paths_smaller_3 <- length(which(shortestPs<=3 & shortestPs!=0))/2
Number_of_shortest_paths_smaller_3
Hope that's close to what you need, good luck!
I have created a directed network in R. I have to find the average degree, which I think I have, the diameter and the maximum/minimum clustering. The diameter is the longest of the shortest distances between two nodes. If this makes sense to anyone, please point me in the right direction. I have what I have coded below so far.
library(igraph)
ghw <- graph.formula(1-+4:5:9:12:14, 2-+11:16:17, 3-+4:5:7,
4-+1:3:6:7:8, 5-+1:3:6:7, 6-+4:5:8,
7-+3:4:5:8:13, 8-+4:6:7, 9-+10:12:14:15,
10-+9:12:14, 11-+2:16:17, 12-+1:9:10:14,
13-+7:15:18, 14-+1:9:10:12, 15-+13:16:18,
16-+2:11:15:17:18, 17-+2:11:16:18, 18-+13:15:16:17)
plot(ghw)
get.adjacency(ghw)
Total number of directed edges
numdeg <- ecount(ghw)
Average number of edges per node
avgdeg <- numdeg / 18
How about looking at the documentation?
diameter(ghw)
I am not sure what you mean by maximum/minimum clustering, but maybe this:
range(transitivity(ghw, type="local"))
Btw. your average number of edges per node is wrong, because every edge belongs to two nodes.
Let's say I have an undirected graph with V nodes and E edges.If I represent the graph with adjacency lists,if I have a representation of an edge between x and y,I must also have a representation of the edge between y and x in the adjacency list.
I know that DFS for directed graphs has V+E complexity.For undirected graphs doesn't it have v+2*e complexity ,because you visit each edge 2 times ?Sorry if it's a noobish question..I really want to understand this think.Thank you,
The complexity is normally expressed O(|V| + |E|) and this isn't affected by the factor of 2.
But the factor of 2 is actually there. One undirected edge behaves just line 2 directed edges. E.g. the algorithm (for a connected undirected graph) is
visit(v) {
mark(v)
for each unmarked w adjacent to v, visit(w)
}
The for loop will consider each edge incident to each vertex once. Since each undirected edge is incident to 2 vertices, it will clearly be considered twice!
Note the undirected DFS doesn't have to worry about restarting from all the sources. You can pick any vertex.
I'm working on a directed graph in igraph for R. I'm trying to convert it to an undirected graph where just reciprocal edges of the former persist. Should be easy but I'm getting strange results.
first I did it like that
library(igraph)
load("dmNet.Rdata")
#http://www.unet.univie.ac.at/~a0406222/dmNet.Rdata
recNet <- as.undirected(net, mode = "mutual",edge.attr.comb="sum")
when I check E(recNet)$weight there are a lot of edges with a weight of 1, which should not be possible since the sum of two reciprocal edges has to be at least 2. Then I did it like that
recNet <- as.undirected(net, mode = "mutual",edge.attr.comb="c")
now I can see that there are actually some edges containing just one value. My new graph recNet seems to contain nonreciprocal edges of net. What am I doing wrong, or am I missunderstandig the "mutual option"?
This happens because some edges are self-loops in your graph. All the edges that end up with a weight of 1 are self-loops:
all(which(E(recNet)$weight == 1) %in% which(is.loop(recNet)))
# [1] TRUE
Apparently, a self-loop is considered as a mutual edge in a directed graph. If you want to consider self-loops as non-mutual, then you can just remove them from the graph. Be careful, though, because some vertices have multiple self-loops, and you might not want to remove these.