When I first saw this questions my reaction was "this is graph matching".
Sadly our graph-matching component didn't support multiple anchor nodes. Now
it does.
So one way you would do this now is:
PatternNode p1 = new PatternNode();
p1.setAssociation( n1 );
PatternNode p2 = new PatternNode();
p2.setAssociation( n2 );
PatternNode p3 = new PatternNode();
p3.setAssociation( n3 );
PatternNode p4 = new PatternNode();
p4.setAssociation( n4 );
PatternNode p5 = new PatternNode();
p5.setAssociation( n5 );
PatternNode toFind = new PatternNode();
p1.createRelationshipTo( toFind );
p2.createRelationshipTo( toFind );
p3.createRelationshipTo( toFind );
p4.createRelationshipTo( toFind );
p5.createRelationshipTo( toFind );
Transaction tx = graphdb.beginTx();
try {
for (PatternMatch match :
PatternMatcher.getMatcher().match(p1,null,p2,p3,p4,p5)) {
Node found = match.getNodeFor( toFind );
System.out.println("found matching node: " + found);
}
tx.success();
} finally {
tx.finish();
}
You would of course need to add a dependency to the latest SNAPSHOT version
of graph-matching for this to work, with maven that would be:
<dependency>
<groupId>org.neo4j</groupId>
<artifactId>neo4j-graph-matching</artifactId>
<version>0.8-SNAPSHOT</version>
</dependency>
If you download the jar manually the latest build is here:
http://m2.neo4j.org/org/neo4j/neo4j-graph-matching/0.8-SNAPSHOT/neo4j-graph-matching-0.8-20100309.140445-6.jar
Cheers,
Tobias
On Fri, Mar 5, 2010 at 9:03 AM, Sumanth Thikka <suma...@truesparrow.com>wrote:
> Hi,
>
> I am searching for an optimal way to find a commonly connected nodes to a
> set of nodes in a graph.
>
> Consider the case where n1, n2, n3, n4 ,n5 are 5 nodes in a graph. These 5
> are connected to the other nodes.
> I need to find the nodes in the graph which are connected all above 5 nodes
> optimally.
>
> The solution which I can think of is like:
> Find all the relationships(of a type, say KNOWS) of each of the above 5
> nodes and get to know the node with minimum relationships say n3.
> For each node connected to n3, find the nodes which are connected to all
> the
> other four nodes. Which is our result set.
>
> A similar link
> http://lists.neo4j.org/pipermail/user/2010-February/002722.html provided
> earlier in the other thread.
> From what I understood from the link, it is to know the products bought by
> any of the customers who bought productX. Whereas, in my case, I need the
> products which were bought by all the customers who bought product X . I am
> not very sure how this solution can be extended to my case, if possible.
>
> Is there any other way or a graph algorithm to solve this in a more optimal
> way.
>
> Thanks,
> Sumanth
> _______________________________________________
> Neo mailing list
> User@lists.neo4j.org
> https://lists.neo4j.org/mailman/listinfo/user
>
--
Tobias Ivarsson <tobias.ivars...@neotechnology.com>
Hacker, Neo Technology
www.neotechnology.com
Cellphone: +46 706 534857
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