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I am trying to find all maximal cliques in a graph, without overlapping. the function max_cliques() returns all possible maximal cliques in the graph, but I want every vertex to be included in only one clique- in the largest clique it can be part of.

for example, if the output of the max_cliques() are the following cliques:

{A,B,C}, {A,B,D}, {A,B,J,K}, {E,F,G,H}, {E,F,G,I}

I want to remove some cliques so that all the vertexes will appear in exacly one clique, so the final set will be:

{A,B,J,K}, {E,F,G,H}

A and B are included in 3 cliques, so I want to choose the cliques so that the final set will include maximum vertexes as possible. if there are two possible cliques in the same length- take a random one. (I don't mind to not include all the vertexes)

I would really appreciate an idea to solve this problem, even without going into details of cliques- the question is basically how to remove the shortest "lists" that contain overlapping elements.

thanks in advance

Hodaya Beer
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    Given your criteria and {A,B,C}, {A,D,C}, {B,E}, one solution would be {A,B,C} and the other {A,D,C}, {B,E}. Then what would you like the output to be? Any solution? – Julius Vainora Feb 27 '18 at 18:47
  • I would like {A,D,C} and {B,E} because that way more vertexes are included in the final set. good comment – Hodaya Beer Feb 27 '18 at 20:03
  • Then the problem becomes quite a bit more difficult. And if it is {A,B,C}, {D,E,F}, then "*if there are two possible cliques in the same length- take a random one*" would suggest that {A,B,C} and {D,E,F} separately are equally good solutions even though they don't overlap. Is that the case? – Julius Vainora Feb 27 '18 at 20:31
  • if this is the case then I want both of them, because they don't overlap. – Hodaya Beer Feb 27 '18 at 20:51

1 Answers1

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This is apparently a pretty hard problem to solve as you asking about Coverage and Independent Set problems. These are NP-complete problems. What this means is that as your graph grows computational time is going to increase exponentially.

I think this what you are aiming for. My approach is as follows:

  1. Find cliques.
  2. Convert to incidence matrix (cliques by nodes).
  3. Multiply the incidence matrix by its transpose (%*%) this creates an adjacency matrix
  4. create graph of cliques from adjacency matrix (cliques are connected to other cliques if they share a node)
  5. Find all independent sets of vertices (this is the bottleneck)
  6. Retrieve original nodes for independent sets of cliques
  7. Find set with most nodes.

Code

library(igraph)  
set.seed(8675309)  
g <- graph_from_edgelist(matrix(sample(LETTERS[1:10], 50, replace=T), ncol = 2), directed = FALSE)  
plot(g, edge.arrow.size=0.5)

cliques <- max_cliques(g)

cliqueBP <- matrix(c(rep(paste0("cl", seq_along(cliques)), sapply(cliques, length)), names(unlist(cliques))), ncol=2, )
bp <- graph_from_edgelist(cliqueBP, directed = F)
V(bp)$type <- grepl("cl", V(bp)$name)
# plot(bp, layout=layout_as_bipartite)

bp.ind <- t(as_incidence_matrix(bp))
bp.adj <- bp.ind %*% t(bp.ind)

bp.adj.g <- graph_from_adjacency_matrix(bp.adj, mode = "undirected")
# plot(simplify(bp.adj.g))
bp.adj.mis <- independent.vertex.sets(bp.adj.g)

sets <- lapply(bp.adj.mis, function(x) cliqueBP[cliqueBP[,1] %in% as_ids(x), 2])
sets[which(sapply(sets, length) == max(sapply(sets, length)))]

# [[1]]
# [1] "G" "J" "E" "I" "B" "H" "F" "D"
# 
# [[2]]
# [1] "G" "J" "E" "I" "F" "C" "B" "H"
# 
# [[3]]
# [1] "G" "J" "E" "I" "F" "C" "A" "H"
# 
# [[4]]
# [1] "G" "B" "E" "I" "F" "C" "A" "H"
# 
# [[5]]
# [1] "G" "B" "E" "I" "F" "C" "H" "J"
Community
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emilliman5
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  • I tested on: m <- t(matrix(c( 0,0,0,0,0,0,0,8, 3,0,0,0,0,0,0,0, 5,0,0,5,1,0,0,0, 0,0,6,0,0,7,1,0, 0,6,2,0,0,0,0,0, 0,0,0,0,0,0,0,0, 7,4,0,0,8,0,0,3, 0,3,0,0,0,9,0,0),ncol=n)) g1 <- graph_from_adjacency_matrix(m, weighted=TRUE, mode="directed") V(g1)$name <- letters[1:n] cliques <- max_cliques(as.undirected(g1)) and result is "e" "c" "f" "d" "h" "a" "g" "b". But the expected result is: "e" "c" "d" "f" "h" "a" "g" "b". Two nodes showed be swaped f<-> d. – Nick Apr 19 '22 at 14:18