Try
Match (c)-[*]->(x)<-[*]-(d)
return x
And
Match p=shortestPath((a)-[*]-(b))
Return nodes(p)
If you have more rel-types, use
(c)-[:KNOWS*]->(x)
Sent from mobile device
Am 03.05.2014 um 20:54 schrieb Quazi Marufur Rahman :
> Hi,
>
> I have created a directed acyclic graph using py2neo. Pa
Hi,
I have created a directed acyclic graph using py2neo. Part of it looks like,
a->b
b->c
b->d
c->d
Here '->' represents 'knows'
I have two questions.
1. How can I find the node which knows both 'c' and 'd' and closest to
them? Something like lowest common ancestor in tree.
2. How to
