400M rels is not an issue, just the question what you want to do with all of 
those returned :)

Currently there might be one issue, IN is not yet supported for index lookups.

So you might have to format your query to an OR chain for the lookup, I know 
this is inconvenient. And you should probably only do it if you run into 
perf-issues.

Michael

Am 05.02.2014 um 16:36 schrieb Ben Titmarsh <[email protected]>:

> Ah.. The humble where clause! Thanks Michael this works a charm.  To answer 
> your question I will end up with about 20,000 nodes.  
> 
> Would you expect this query to be performant at that size?  Again assuming 
> each node could have a relationship to each other node, i.e. ~400,000,000 
> possible relationships.  I understand there is a cap on the number of 
> relationships though.
> 
> On Wednesday, February 5, 2014 2:19:34 PM UTC, Michael Hunger wrote:
> Depends on how many cards you have, in general you can use:
> 
>> MATCH (card1:Card)-[r:DECK_INCIDENCE]-(card2:Card)
> 
> where card1.name in {cardsA} and card2.name in {cardsB}
>>  RETURN r;
> 
> Michael
> 
> Am 05.02.2014 um 15:05 schrieb Ben Titmarsh <[email protected]>:
> 
>> Hi All,
>> 
>> I'd like your help with a bit of Cypher that I am trying to write.
>> 
>> I have two lists of Strings containing the name property of some of the 
>> nodes in my graph.  For example:
>> 
>> fromNodes: ["cardA","cardB"..."cardZ"]
>> 
>> toNodes: ["card1","card2"..."card50"]
>> 
>> I would like to run a query to find all of the relationships of type 
>> 'DECK_INCIDENCE' that exist between the nodes defined in fromNodes and 
>> toNodes, i.e. for each node in fromNodes the query should return 50 
>> relationships (assuming that there is a relationship between all nodes) * 26 
>> nodes in fromNodes = 1300 relationships.
>> 
>> I'm relatively new to Cypher, but have been using the following to retrieve 
>> one relationship between two specific nodes:
>> 
>> MATCH (card1:Card {name:"cardA"})-[r:DECK_INCIDENCE]-(card2:Card 
>> {name:"card13"}) RETURN r;
>> 
>> How can I extend this to return all of the relationships that I need without 
>> having to run 1300 queries?!
>> 
>> Thanks!
>> 
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