I'm trying to find the max clique in a graph (adjacency matrix), the graph can have up to 50 nodes. Currently, what I have starts taking forever when the graph sizes get to around n=40. Can anyone find any optimizations I can make?
public static void maxClique(Graph graph, List<Integer> clique) {
//Get the newest node added to the clique
int currentValue = clique.get(clique.size() - 1);
if (clique.size() > max.size()) {
max = clique;
}
// Check every node
for (int i = 0; i < graph.size; i++) {
// If the node being checked is connected to the current node, and isn't the current node
if (graph.matrix[currentValue][i] == 1 && !clique.contains(i)) {
//Check if the new clique is bigger than the current max.
//Make a new clique and add all the nodes from the current clique to it
ArrayList<Integer> newClique = new ArrayList<>();
newClique.addAll(clique);
//Add the new node
newClique.add(i);
//Repeat
if (makesNewClique(graph, clique, i)) {
maxClique(graph, newClique);
}
}
}
}
Full class contents: https://pastebin.com/fNPjvgUm Adjacency matrix: https://pastebin.com/yypN9K4L
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I think this algorithm is much slower than Bron--Kerbosch for starters (time on the order of k! versus 2^k for each k-node clique). If you just need the maximum clique, there are further optimizations.David Eisenstat– David Eisenstat2020年10月04日 03:26:01 +00:00Commented Oct 4, 2020 at 3:26
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@DavidEisenstat I implemented Bron-Kerbosh and it's way faster, thanks!Coolcodehere– Coolcodehere2020年10月04日 04:58:05 +00:00Commented Oct 4, 2020 at 4:58
1 Answer 1
David Eisenstat 's comment had the answer, here's my code for anyone that comes here later:
public void bronKerbosh(Set<Integer> p, Set<Integer> r, Set<Integer> x) {
if (x.isEmpty() && p.isEmpty() && r.size() > max.size()) {
max = new ArrayList<>();
max.addAll(r);
}
Object[] pArr = p.toArray();
for (Object vTemp : pArr) {
int v = (int)vTemp;
Set<Integer> newR = new HashSet<>();
newR.addAll(r);
newR.add(v);
bronKerbosh(intersect(p, neighbors(v)), newR, intersect(x, neighbors(v)));
p.remove(v);
x.add(v);
}
}
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