search

RapidsAtHKUST / ContinuousSubgraphMatching

Source code and datasets of "An In-Depth Study of Continuous Subgraph Matching", accepted by VLDB'22 - By Xibo Sun, Dr. Shixuan Sun, Prof. Qiong Luo, and Prof. Bingsheng He
MIT License
40 stars 7 forks source link

Something about graphflow #1

Closed Edison0521 closed 2 years ago

Edison0521 commented 2 years ago

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?

xibosun commented 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.

Edison0521 commented 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.

Thank you for your reply , That's would be great

Edison0521 commented 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, 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?

xibosun commented 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, 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.

Edison0521 commented 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, 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?

xibosun commented 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, 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.

Edison0521 commented 2 years ago

Alright, Thank you very much!