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Based on METIS official manual, it can partition graphs into k unequal parts with different capacities for vertices:

METIS’ graph and mesh partitioning programs and API routines are designed to partition a graph into k parts such that each part contains a pre-specified fraction of the total number of vertices/elements/nodes. In addition, in the case of multi-constraint partitioning, these pre-specified fractions are provided for each one of the vertex weights.

Now my question is how is it possible? how can we tell metis to partition graph into k unequal parts and how we can specify the capacity of these parts. I found an option -tpwgts to define Target partition weights but i don't understand how it's affecting the partitioning process and the description in manual is not very intelligible! So please can you describe how is possible to make different partitions with different sizes?

Firouziam
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  • Did you ever find a solution to your own problem? It would be useful to others if you could answer your issue! – NoseKnowsAll Apr 11 '17 at 02:52
  • Unfortunately, i haven't found a solution for this issue yet. currently i'm working on my own partitioning method. METIS is a great tool, but lack of support forced me to let it go. – Firouziam Apr 15 '17 at 11:49
  • I am facing the same kind of problem. [Here](https://stackoverflow.com/questions/45041064/interpretation-of-ufactor-on-a-toy-graph-clustering) is my question. The support of METIS is really bad, I tried to open a account to their forum. They wanted my university e-mail Id, after providing that still did not access there. This is the main reason of not becoming a popular tool although it is a great tool – sovon Jul 11 '17 at 18:21
  • yup, unfortunately there is no support for metis, so i let it go. sofware without support and manual is dead. – Firouziam Jul 12 '17 at 13:52

1 Answers1

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It is probably possible to unequally partition a graph if we take into consideration the sum of the weights of the edges contained in each partition or any metric related to that

For instance if the sum of the weights are equal for each partition that does not necessarily mean that they contain an equal number of nodes

MiniMe
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