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Search paths where all nodes are in a relationship with same node


Using the below graph as an example: I am trying to determine all Persons which are_sons of Persons all born in the same country. Graph

Is there an elegant way to achieve this with Cypher?

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I'm having a hard time understanding exactly what you want from your query. Can you detail it in some sort of pseudocode instead of a sentence? (For example, do any of the nodes in this example graph match your query criteria? If ‘Ann’ is a match, is it because all of her parents which have a birth country have the same birth country? Or is it because all of her non-adoptive parents have the same birth country? Or is it that she has at least two parents with the same birth country?) r~~‭ 3 days ago

What I want is for Ann to be returned because all her Parents are from the USA. The reason why I introduced "adoptive" parents is because: 1- This is an example (I'm not actually working with a Graph with parents ;) ) 2 - I wanted to make it clear that you do not know how many parents a person has -> so has not to accept "counting 2" strategies. Maybe I should have a picked a complete abstract example. Guilherme Costa‭ 3 days ago

So should Steve be linked to USA in the graph then? r~~‭ 3 days ago

Of course... Really sorry about that. Updated the picture Guilherme Costa‭ 3 days ago

1 answer


Something like this query should work for you:

MATCH (a:Person)-[:IS_SON]->()-[:WAS_BORN]->(b:Country)
WITH a, count(DISTINCT b) AS birthplaces
WHERE birthplaces = 1

Note that this query would return people who have parents with unknown birthplaces, since those paths won't be included in the MATCH clause. If you have that case in your database and want to exclude it, then I think you lose a little elegance:

MATCH (a:Person)-[:IS_SON]->(p:Person)
OPTIONAL MATCH (p)-[:WAS_BORN]->(b:Country)
WITH a, count(p) AS parents, count(b) AS birthplaces, count(DISTINCT b) as distinct_birthplaces
WHERE parents = birthplaces AND distinct_birthplaces = 1


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