CODEWITHSUNDEEP
neo4j
Ben
Ben

Reputation: 1581

Neo4J nr of relationship hops needs to be set explicitly. Why?

I have the following NEO4J query, which doesn't return anything:

MATCH (:TimetreeRoot)-[:CHILD]->(year:Year)-[:CHILD]->(month:Month)-[:CHILD]->(day:Day)
WHERE year.value = 2016 AND month.value = 4 AND day.value = 26
MATCH (day)-[:CHILD]->(bhour:Hour)-[:CHILD]->(bminute:Minute)
WHERE bhour.value = 8 and bminute.value = 0
MATCH (day)-[:CHILD]->(ehour:Hour)-[:CHILD]->(eminute:Minute)
WHERE ehour.value = 18 AND eminute.value = 0
MATCH p=shortestPath((bminute)-[:NEXT*]->(eminute))
RETURN collect(NODES(p))

When the difference is smaller (say bhour 12 and ehour 18), i.e. less NEXT relationships to traverse, it does work.

It also works, when I set the nr of hops explicitly:

MATCH p=shortestPath((bminute)-[:NEXT*1..48]->(eminute))

So I can get it to work correctly, but my question is: why do I need to set the nr of maxHops? From the manual:

minHops and maxHops are optional and default to 1 and infinity respectively http://neo4j.com/docs/stable/query-match.html#match-variable-length-relationships

It's not necessarily a big deal, since I did figure out how to get it to work. However, I'd like to understand what's causing this problem.

Upvotes: 2

Views: 115

Answers (1)

Christophe Willemsen
Christophe Willemsen

Reputation: 20185

The shortestPath has a default max hops of 15 hops. This is why you need to specify it if you want to exceed this default.

From the documentation :

find a single shortest path between two nodes, as long as the path is max 15 relationships long

Upvotes: 3

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