Neo4j 查询性能-四种关系

我的问题是： “哪一位球员最频繁地相互较量”

我问了以下问题：

match (p:Player)-[:PLAYS_IN]->(t:Team)-[:PLAYED]->(g:Game)<-[:PLAYED]-(tt:Team)<-[:PLAYS_IN]-(pp:Player) 
USING SCAN p:Player USING SCAN pp:Player  
return p.Lastname, pp.Lastname, count(pp) order by count(pp) desc limit 100;

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match（p:Player）-[：PLAYS_IN]->（t:Team）-[：play]->（g:Game）我想你可以通过首先找到哪些球队相互踢得最厉害，然后在此基础上找到球员来简化问题。我不确定这个假设是否过分简化了问题

match (t:Team)-[:PLAYED]->(g:Game)<-[:PLAYED]-(tt:Team)
with t, tt order by count(g) desc limit 1
match  (p:Player)-[:PLAYS_IN]->(t:Team)-[:PLAYED]->(g:Game)<-[:PLAYED]-(tt:Team)<-[:PLAYS_IN]-(pp:Player) 
return p.Lastname, pp.Lastname, count(pp) order by count(pp) desc limit 100;

match（t:Team）-[：PLAYED]->（g:Game）（t:Team）-[：PLAYED]->（g:Game）感谢您的回答。但我想我可能讨论了笛卡尔积问题：neo4j sh（？）$match（g:Game）返回计数（g）；+------------+|count（g）|+-------------+|215067 |+-------------+1行450毫秒neo4j sh（？）$match（t:Team）-[：PLAYED]->（g:Game）查看google组上的答案。。。