is it maybe faster to do it with cipher?
start n = node:idx("MyID:(1 2 3 1 1500)"),
m = node:idx("MyID:(1 2 3 1 1500)")
MATCH p=(n)-[r:RELATED_TO]-(m)
RETURN p
or 2 find the best between 2 pairs?
thank you.
On Monday, May 26, 2014 10:15:45 AM UTC+3, natali wrote:
>
> hey'
> i h
Assuming that you can identify the 100 nodes
// MATCH and/or filter the 100 nodes
MATCH (n)
WHERE
// put the nodes in a collection
WITH collect(n) AS mynodes
// get als the relations withing the context of the collection
MATCH (n)-[r]->(m)
WHERE (n IN mynodes) AND (m IN mynodes)
RETURN n,r,
