我想有N个顶点作为输入，然后我想返回所有路径，直到这些N个顶点中所有对之间的特定长度为止。如何在Gremlin中完成？
一些解释-具有此图（以路径表示）：
(n)-[r1]-(n1)-[r2]-(n2)-[r3]-(m)-[r5]-(n3)
(y)-[r4]-(n1)-[r2]-(n2)-[r3]-(m)-[r6]-(n4)
() node
-[]- relation
例如，这些应该是（n，m，y）中最长为3的路径
(n)-[r1]-(n1)-[r2]-(n2)-[r3]-(m)
(y)-[r4]-(n1)-[r2]-(n2)-[r3]-(m)
(n)-[r1]-(n1)-[r2]-(n1)-[r4]-(y)
这是我针对2个顶点的Gremlin示例：
g = new OrientGraph("remote:localhost/graphdb")
v = g.v('#12:110')
y = g.v('#12:109')
hops = 3
v
.as('looop')
.inE.has('label','EdgeClass')
.outV.has('@class','NodeClass')
.outE.has('label','EdgeClass')
.inV.except([v]).dedup()
.loop('looop'){it.loops<hops}{it.object.rid==y.rid}.path
谢谢
这是使用Tinkergraph玩具图的示例：
gremlin> vertices = g.v(2,3,5).toSet()
==>v[2]
==>v[3]
==>v[5]
gremlin> vertices._().as("x").bothE().bothV().simplePath().loop("x") {it.loops <= 3} {it.object in vertices}.simplePath().path() {it} {it.label}
==>[v[2], knows, v[1], created, v[3]]
==>[v[2], knows, v[1], knows, v[4], created, v[3]]
==>[v[2], knows, v[1], knows, v[4], created, v[5]]
==>[v[3], created, v[4], created, v[5]]
==>[v[3], created, v[1], knows, v[2]]
==>[v[3], created, v[4], knows, v[1], knows, v[2]]
==>[v[3], created, v[1], knows, v[4], created, v[5]]
==>[v[5], created, v[4], created, v[3]]
==>[v[5], created, v[4], knows, v[1], created, v[3]]
==>[v[5], created, v[4], knows, v[1], knows, v[2]]
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我来说两句