查找通过 2 个关系连接到至少 2 个相同节点但中间节点不同的节点

Question

I would like to find out all the couples of Persons and Accidents in the following case:

• Persons who are involved into at least 2 same Accidents, when they were in 2 different Cars.

I always have the following relationships: Person --> Car -- > Accident

I made a simplified and reproducible example:

CREATE
  (p1:Person {name: 'Paul'})-[:DRIVES]->(c1:Car {name: 'Car A'}),
  (p2:Person {name: 'John'})-[:DRIVES]->(c2:Car {name: 'Car B'}),
  (p3:Person {name: 'Mike'})-[:DRIVES]->(c3:Car {name: 'Car C'}),
  (p4:Person {name: 'Joe'})-[:OCCUPANT]->(c1),
  (p4)-[:OCCUPANT]->(c3),
  (c1)-[:IS_DOER]->(a1: Accident {name: 'Crash 1'}),
  (c2)-[:IS_VICTIM]->(a1),
  (c2)-[:IS_DOER]->(a2: Accident {name: 'Crash 2'}),
  (c3)-[:IS_VICTIM]->(a2) ;

And here is what the graph looks like:

In this example, I would like to return the couples ("Joe", "John") and ("Crash 1", "Crash 2"), because Joe and John are both involved into at least 2 same accidents while being in different cars.

Thank you for your help.

Answer 1

Actually I think I found the answer:

MATCH
  (p1:Person)-[r1]->(c1:Car)-[r2]->(a:Accident)<-[r3]-(c2:Car)<-[r4]-(p2:Person)
WHERE c1<>c2
WITH p1, p2, COUNT(DISTINCT a) AS count_accident
WHERE count_accident>= 2
RETURN p1, p2