Efficient way to find node set having relationships to given nodes using neo4j

Question

Is there an efficient way with given two nodes to find a set of their common nodes (with defined relationships).

For example, having nodes A1, B1, C1-C4 connected with relationships x and y:

A1 --x--> C1
A1 --x--> C2
A1 --x--> C3
B1 --y--> C2
B1 --y--> C3
B1 --y--> C4

a common node set for A1(x) and B1(y) would be [C2, C3].

Answer 1

In Gremlin (http://gremlin.tinkerpop.com), this is expressed as such:

setA._().out('x').in('y').retain(setB).back(2)

Here is what each step does:

1. start at setA (A1, A2, A3 in your example).
2. start a Gremlin pipeline.
3. take the outgoing "x" labeled edges from those setA vertices to C1, C2, and C3.
4. take the incoming "y" labeled edges from C1, C2, and C3.
5. filter out all steps that are not in setB (thus, only C2 and C3 paths exist).
6. go back to what you saw 2 steps ago -- thus, C2 and C3.

Tada!

Good luck, Marko.

http://markorodriguez.com

Answer 2

In many cases the structure of the domain can be leveraged to improve performance. Let's say that you know that in general your A entities have less x relationships compared to the number of y relationships on the B entities. Then you could traverse two steps from the A node and see where the B node shows up, and filter out the C nodes this way. Here's some code for this approach:

Set<Node> found = new HashSet<Node>();
for ( Relationship firstRel : a1.getRelationships( Reltypes.x, Direction.OUTGOING ) )
{
    Node cNode = firstRel.getEndNode();
    for ( Relationship secondRel : cNode.getRelationships( Reltypes.y, Direction.INCOMING ) )
    {
        Node bNode = secondRel.getStartNode();
        if ( bNode.equals( b1 ) )
        {
            found.add( cNode );
            break;
        }
    }
}

Another way would be to start two threads that scan the relationships from either side.

A third approach would be to create a specialized index that would help answering this kind of queries, which would obviously hurt insert performance.