Closed Edison0521 closed 2 years ago
Hi, the code does not explicitly store the set of all mappings from the query graph to the data graph. But one can run the code with --print-enum on
to print all the mappings to the console.
Hi, the code does not explicitly store the set of all mappings from the query graph to the data graph. But one can run the code with
--print-enum on
to print all the mappings to the console.
Thank you for your reply , That's would be great
Hi, the code does not explicitly store the set of all mappings from the query graph to the data graph. But one can run the code with
--print-enum on
to print all the mappings to the console.
Hi, could you please look after my test example ,I think here are something confused in it, My query graph is v 0 1 v 1 0 v 2 0 e 0 1 0 e 2 0 0 and data graph is v 0 0 v 1 1 v 2 1 v 3 0 v 4 2 v 5 1 v 6 0 v 7 1 v 8 3 v 9 4 v 10 1 v 11 3 e 0 1 0 e 0 4 5 e 1 9 1 e 1 2 2 e 2 8 3 e 2 9 4 e 2 3 0 e 3 4 5 e 4 11 5 e 4 5 5 e 5 10 9 e 5 11 7 e 5 6 0 e 6 7 10 e 8 10 7 e 9 10 6 e 10 11 3 And when I use the homo on and print--enmu the result is ----------- Initial Matching ---------- 1 0 0 2 3 3 5 6 6 Initial Matching: 0.431ms 3 matches. So my questing is what the 1-0-0 stand for?
Hi, the code does not explicitly store the set of all mappings from the query graph to the data graph. But one can run the code with
--print-enum on
to print all the mappings to the console.Hi, could you please look after my test example ,I think here are something confused in it, My query graph is v 0 1 v 1 0 v 2 0 e 0 1 0 e 2 0 0 and data graph is v 0 0 v 1 1 v 2 1 v 3 0 v 4 2 v 5 1 v 6 0 v 7 1 v 8 3 v 9 4 v 10 1 v 11 3 e 0 1 0 e 0 4 5 e 1 9 1 e 1 2 2 e 2 8 3 e 2 9 4 e 2 3 0 e 3 4 5 e 4 11 5 e 4 5 5 e 5 10 9 e 5 11 7 e 5 6 0 e 6 7 10 e 8 10 7 e 9 10 6 e 10 11 3 And when I use the homo on and print--enmu the result is ----------- Initial Matching ---------- 1 0 0 2 3 3 5 6 6 Initial Matching: 0.431ms 3 matches. So my questing is what the 1-0-0 stand for?
Each line is a subgraph isomorphism result. And the first line 1 0 0 represents a result, where the query vertex 0 matches the data vertex 1, the query vertex 1 matches the data vertex 0, and the query vertex 2 matches the data vertex 0.
Hi, the code does not explicitly store the set of all mappings from the query graph to the data graph. But one can run the code with
--print-enum on
to print all the mappings to the console.Hi, could you please look after my test example ,I think here are something confused in it, My query graph is v 0 1 v 1 0 v 2 0 e 0 1 0 e 2 0 0 and data graph is v 0 0 v 1 1 v 2 1 v 3 0 v 4 2 v 5 1 v 6 0 v 7 1 v 8 3 v 9 4 v 10 1 v 11 3 e 0 1 0 e 0 4 5 e 1 9 1 e 1 2 2 e 2 8 3 e 2 9 4 e 2 3 0 e 3 4 5 e 4 11 5 e 4 5 5 e 5 10 9 e 5 11 7 e 5 6 0 e 6 7 10 e 8 10 7 e 9 10 6 e 10 11 3 And when I use the homo on and print--enmu the result is ----------- Initial Matching ---------- 1 0 0 2 3 3 5 6 6 Initial Matching: 0.431ms 3 matches. So my questing is what the 1-0-0 stand for?
Each line is a subgraph isomorphism result. And the first line 1 0 0 represents a result, where the query vertex 0 matches the data vertex 1, the query vertex 1 matches the data vertex 0, and the query vertex 2 matches the data vertex 0.
Thank you for your explanation, while I suppose that each line is a subgraph homomorphism, due to the command --homo on,is that right?
Hi, the code does not explicitly store the set of all mappings from the query graph to the data graph. But one can run the code with
--print-enum on
to print all the mappings to the console.Hi, could you please look after my test example ,I think here are something confused in it, My query graph is v 0 1 v 1 0 v 2 0 e 0 1 0 e 2 0 0 and data graph is v 0 0 v 1 1 v 2 1 v 3 0 v 4 2 v 5 1 v 6 0 v 7 1 v 8 3 v 9 4 v 10 1 v 11 3 e 0 1 0 e 0 4 5 e 1 9 1 e 1 2 2 e 2 8 3 e 2 9 4 e 2 3 0 e 3 4 5 e 4 11 5 e 4 5 5 e 5 10 9 e 5 11 7 e 5 6 0 e 6 7 10 e 8 10 7 e 9 10 6 e 10 11 3 And when I use the homo on and print--enmu the result is ----------- Initial Matching ---------- 1 0 0 2 3 3 5 6 6 Initial Matching: 0.431ms 3 matches. So my questing is what the 1-0-0 stand for?
Each line is a subgraph isomorphism result. And the first line 1 0 0 represents a result, where the query vertex 0 matches the data vertex 1, the query vertex 1 matches the data vertex 0, and the query vertex 2 matches the data vertex 0.
Thank you for your explanation, while I suppose that each line is a subgraph homomorphism, due to the command --homo on,is that right?
Oh sorry. Right, it should be a homomorphism result rather than an isomorphism one.
Alright, Thank you very much!
Hello Thank for this great job, I have a problem when I implementation the code, where can I get the map between querygraph and datagraph?,for exapmle in graphflow.cpp line 245,the m[u]=v is a vertex mapping. but it seems not the mapping result form sourcedata, chould you please help me to find it?