Hi,
No, Vertex can belong only to one partition.
Can you describe algorithm you are solving ? How many those vertexes
belonging to all partitions you have ?
Why do you need so strict partitioning ?
Regards
Lukas
On 6.4.2014 12:38, Akshay Trivedi wrote:
In order to custom partition the graph, WorkerGraphPartitioner has to
be implemented. It has a method getPartitionOwner(I vertexId) which
returns PartitionOwner of the vertex. I want that some vertices belong
to all paritions i.e all PartitionOwners. Can anyone help me with it?
Thankyou in advance
Regards
Akshay
