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Mathematica Graphs and Networks Wolfram Language

How can I extract individual graph partitions as graphs?

Shawn Doyle

Posted 10 years ago

After having applied the function FindGraphPartition to an arbitary graph with weighted vertices, how can I go about making graphs of the partitions? e.g. If I cut the original graph into two parts, then how can I make two graphs (with properties preserved)?

POSTED BY: Shawn Doyle

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Bill Simpson

Posted 10 years ago

Can you adapt this? And does it preserve all your properties? In[1]:= g1 = Graph[{1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 1, 2 \[UndirectedEdge] 4, 4 \[UndirectedEdge] 5, 5 \[UndirectedEdge] 6, 6 \[UndirectedEdge] 4}]; p = FindGraphPartition[g] Out[2]= {{4, 5, 6}, {1, 2, 3}} In[3]:= g2 = Subgraph[g1, First[p]]; g3 = Subgraph[g1, Last[p]];

Can you adapt this? And does it preserve all your properties?

In[1]:= g1 = Graph[{1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 1,
   2 \[UndirectedEdge] 4, 4 \[UndirectedEdge] 5, 5 \[UndirectedEdge] 6, 6 \[UndirectedEdge] 4}];
p = FindGraphPartition[g]

Out[2]= {{4, 5, 6}, {1, 2, 3}}

In[3]:= g2 = Subgraph[g1, First[p]]; g3 = Subgraph[g1, Last[p]];

POSTED BY: Bill Simpson

Shawn Doyle

Posted 10 years ago

Thank you for the reply! The code doesn't preserve vertex weights if you add them. The following will give unweighted subgraphs: g1 = Graph[{1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 1, 2 \[UndirectedEdge] 4, 4 \[UndirectedEdge] 5, 5 \[UndirectedEdge] 6, 6 \[UndirectedEdge] 4}, VertexWeight -> {5, 7, 7, 2, 1, 5}]; p = FindGraphPartition[g1]; g2 = Subgraph[g1, First[p]]; g3 = Subgraph[g1, Last[p]]; I will continue toying with this. I thought that this link would solve the problem, but that is not the case here. Thanks again!

Thank you for the reply! The code doesn't preserve vertex weights if you add them. The following will give unweighted subgraphs:

g1 = Graph[{1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 1, 2 \[UndirectedEdge] 4,  4 \[UndirectedEdge] 5, 5 \[UndirectedEdge] 6,  6 \[UndirectedEdge] 4}, VertexWeight -> {5, 7, 7, 2, 1, 5}];
p = FindGraphPartition[g1];
g2 = Subgraph[g1, First[p]]; g3 = Subgraph[g1, Last[p]];

I will continue toying with this. I thought that this link would solve the problem, but that is not the case here. Thanks again!

POSTED BY: Shawn Doyle

Bill Simpson

Posted 10 years ago

Play the movie backwards? In[1]:= g1 = Graph[{1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 1, 2 \[UndirectedEdge] 4, 4 \[UndirectedEdge] 5, 5 \UndirectedEdge] 6, 6 \[UndirectedEdge] 4}, VertexWeight -> {5, 7, 7, 2, 1, 5}]; p = FindGraphPartition[g1]; g4 = VertexDelete[g1, First[p]]; g5 = VertexDelete[g1, Last[p]]; FullForm[g4] Out[5]//FullForm=Graph[{4 \[UndirectedEdge] 5, 5 \UndirectedEdge] 6, 6 \[UndirectedEdge] 4}, VertexWeight -> {2, 1, 5}];

Play the movie backwards?

In[1]:= g1 = Graph[{1 \[UndirectedEdge] 2, 2 \[UndirectedEdge] 3, 3 \[UndirectedEdge] 1, 2 \[UndirectedEdge] 4, 4 \[UndirectedEdge] 5, 5 \UndirectedEdge] 6, 6 \[UndirectedEdge] 4}, VertexWeight -> {5, 7, 7, 2, 1, 5}];
p = FindGraphPartition[g1];
g4 = VertexDelete[g1, First[p]];
g5 = VertexDelete[g1, Last[p]];
FullForm[g4]

Out[5]//FullForm=Graph[{4 \[UndirectedEdge] 5, 5 \UndirectedEdge] 6, 6 \[UndirectedEdge] 4}, VertexWeight -> {2, 1, 5}];

POSTED BY: Bill Simpson

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