CODEWITHSUNDEEP
graphgremlintitan
Ashok Krishnamoorthy
Ashok Krishnamoorthy

Reputation: 873

Common ancestor in a titan graph..

In a titan(hbase) graph, i would like to get common ancestors for given nodes. Using gremlin is there any way we can get the common ancestors? Or any other Utils / libs available to find common ancestors in a Titan graph for given nodes?

Upvotes: 1

Views: 763

Answers (1)

stephen mallette
stephen mallette

Reputation: 46206

I'll make some assumptions on exactly what you're looking for here, but maybe this answer will help inspire you to come up with what you are looking for. I used the toy graph to demonstrate this approach and assumed the edges directionally pointed to children. Obviously this isn't a perfect tree, but there's enough data there to at least demonstrate the traversal. The following shows the "children" for whom I wanted to find ancestors:

gremlin> g = TinkerGraphFactory.createTinkerGraph()
==>tinkergraph[vertices:6 edges:6]
gremlin> children = [g.v(2),g.v(5),g.v(6)]
==>v[2]
==>v[5]
==>v[6]

I can recursively loop up through the tree through ancestors and see the paths as follows:

gremlin> children._().in.loop(1){it.loops<5}{true}.path
==>[v[2], v[1]]
==>[v[5], v[4]]
==>[v[5], v[4], v[1]]

The above line caps the loop at 5 steps away from the ancestor and the {true} closure simply means to output values at all steps of the loop. A common ancestor would be one where the each path terminates more than once:

gremlin> children._().in.loop(1){it.loops<5}{true}.groupCount.cap.next().findAll{it.value>1}
==>v[1]=2

In this sense v[1] is the only common ancestor among the three child nodes and only 2 of the 3 child nodes share that ancestor. To make it more interesting, let's connect vertex 6 to vertex 4 and re-execute the traversal:

gremlin> children._().in.loop(1){it.loops<5}{true}.groupCount.cap.next().findAll{it.value>1}
==>v[1]=3
==>v[4]=2

Now vertex 4 is included as a common ancestor. It has children of vertex 6 and 5. Given the shared nature of vertex 4, the 6 also share vertex 1 now so that ancestor is now shared by all three vertices.

UPDATE: The above answer is for TinkerPop 2.x and the following is for TinkerPop 3.x.

Given this tree:

gremlin> g.addV().property(id, 'A').as('a').
           addV().property(id, 'B').as('b').
           addV().property(id, 'C').as('c').
           addV().property(id, 'D').as('d').
           addV().property(id, 'E').as('e').
           addV().property(id, 'F').as('f').
           addV().property(id, 'G').as('g').
           addE('hasParent').from('a').to('b').
           addE('hasParent').from('b').to('c').
           addE('hasParent').from('d').to('c').
           addE('hasParent').from('c').to('e').
           addE('hasParent').from('e').to('f').
           addE('hasParent').from('g').to('f').iterate()

you can find the common ancestor with:

gremlin> g.V('A').
......1>   repeat(out('hasParent')).emit().as('x').
......2>   repeat(__.in('hasParent')).emit(hasId('D')).
......3>   select('x').limit(1)
==>v[C]

This topic is discussed more fully in the TinkerPop recipes.

Upvotes: 3

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