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 <[email protected]>: > 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. > How can I find the node which knows both 'c' and 'd' and closest to them? > Something like lowest common ancestor in tree. > How to find the shortest distance between 'a' and 'b'? > Please let me know if any tutorial is available to solve these two questions. > > Thanks > > -- > You received this message because you are subscribed to the Google Groups > "Neo4j" group. > To unsubscribe from this group and stop receiving emails from it, send an > email to [email protected]. > For more options, visit https://groups.google.com/d/optout. -- You received this message because you are subscribed to the Google Groups "Neo4j" group. To unsubscribe from this group and stop receiving emails from it, send an email to [email protected]. For more options, visit https://groups.google.com/d/optout.
