如何获得多个输入节点的最小关系范围？

Question

下面的语句在Neo4j中创建了一个代表性数据集。

我想返回给定输入节点集的最小关系图。

例如，节点集的最小关系图＆＃34; C＆＃34;，＆＃34; G＆＃34;，＆＃34; D＆＃34;是＆＃34; C，G，D＆＃34 ;;并且对于＆＃34; A＆＃34;，＆＃34; E＆＃34;它是＆＃34; A，G，E＆＃34;。

特定节点的数量是任意的，但要求返回的范围最小。

如何编写此查询？

样本数据

CREATE (A:Table {name:'A'})
CREATE (B:Table {name:'B'})
CREATE (C:Table {name:'C'})
CREATE (D:Table {name:'D'})
CREATE (E:Table {name:'E'})
CREATE (F:Table {name:'F'})
CREATE (G:Table {name:'G'})
CREATE
  (F)-[:Link]->(A),
  (A)-[:Link]->(G),
  (G)-[:Link]->(E),
  (G)-[:Link]->(C),
  (G)-[:Link]->(D),
  (E)-[:Link]->(B),
  (C)-[:Link]->(E)

Answer 1

这是一个简单的解决方案。生成集合中每对节点之间的最短路径的完整列表。然后将结果路径减少到不同的节点集。

// set up the input nodes as a collection of attribute values to be matched
WITH ['C','G','D'] as inputs

// build a list of pairs that are not the same 
UNWIND RANGE(0,size(inputs)-1) as i
UNWIND RANGE(0,size(inputs)-1) as j
WITH CASE 
  WHEN i > j THEN [inputs[i],inputs[j]]
  ELSE null
END as pair

// find all of the shortest paths for each pair
MATCH p=allShortestPaths((a:Table {name: pair[0]})-[:Link*]-(b:Table {name: pair[1]}))
WITH p

// recollect the distinct nodes from the shortest paths
UNWIND nodes(p) as n
RETURN COLLECT(DISTINCT n) as min_relational_scope