views:

47

answers:

1

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].

+1  A: 

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.

nawroth