# Boost Users :

Subject: [Boost-users] Questions about using "mcregor_common_subgraphs" algorithm in Boost Graph Library
From: Bin Ren (rb520520_at_[hidden])
Date: 2012-04-17 23:03:35

Hi,

I am a beginner of C++ Boost Library. I am doing a project about "Maximum Common Subgraph" Algorithm, and I found there is an implementation in Boost Graph Library:

"boost/graph/mcgregor_common_subgraphs.hpp"

I use it like this: (and my question is marked as red)
----------------------------------------------------------------------------------------------------
//Compare the input graphs against the Graph Forest

for (int i = 0; i < graphInputs.size(); ++i){
graph_t * pInputGraphi = graphInputs[i];

for (int j = 0; j < graphForest.size(); ++j) {
std::cout << "Common Sub graph of input graph" << i << " and graph" << j << " :" << std::endl;
//Compare one to one among graphi and graphj
graph_t * pGraphj = graphForest[j];

//Here I am using the callback function from the example code: "mcgregor_subgraphs_example.cpp"
example_callback<graph_t> my_callback(*pInputGraphi);

property_map<graph_t, vertex_name_t>::type vname_mapi = get(vertex_name, *pInputGraphi);
property_map<graph_t, vertex_name_t>::type vname_mapj = get(vertex_name, *pGraphj);

property_map<graph_t, edge_name_t>::type ename_mapi = get(edge_name, *pInputGraphi);
property_map<graph_t, edge_name_t>::type ename_mapj = get(edge_name, *pGraphj);

//Print out Maximum connected Common Subgraphs between graphi and graphj

mcgregor_common_subgraphs_maximum_unique(*pInputGraphi, *pGraphj, true, my_callback,
//edges_equivalent(make_property_map_equivalent(ename_mapi, ename_mapj)),
//Here is my problem: if I use both edges_equivalent and vertices_equivalent, the code can not be compiled
//However, if I only use vertices_equivalent, I doubt whether the maximum common subgraph result is correct.
//If I use edges_equivalent only, the code can be compiled, however, it runs into an infinite loop.
vertices_equivalent(make_property_map_equivalent(vname_mapi, vname_mapj)));
}
}
------------------------------------------------------------------------------------------------------------------------------------------------------------------------------

Is there any way to use both the edges_equivalent and the vertices_equivalent (Or it is not necessary at all)?

BTW, it seems the performance of this algorithm is not perfect; for a small test case below, it needs around 5 seconds by vertices_equivalent, while it loops for a very long time for edges_equivalent (it might be a infinite loop...).

Thanks and have a good night,

Bin

-----------------------------------------------------------------------------------------------

Small Test Case:

For test, I only use one graph for "graphInputs", and one graph for "graphForest":

graphInputs: (v: vertex, 1st number: the vertex id, 2nd number: the vertex name; e: edge, 1st: source vertex, 2nd: target vertex, 3rd: edge name)

t # 0
v 0 0
v 1 0
v 2 0
v 3 0
v 4 0
v 5 0
v 6 1
v 7 1
v 8 1
v 9 1
v 10 2
v 11 0
v 12 0
v 13 2
v 14 0
v 15 0
v 16 0
v 17 0
v 18 1
v 19 1
v 20 1
v 21 3
v 22 3
v 23 4
v 24 5
v 25 5
e 0 1 3
e 1 2 3
e 2 3 3
e 3 4 3
e 4 5 3
e 5 0 3
e 0 6 0
e 1 7 0
e 4 8 0
e 5 9 0
e 2 10 0
e 10 11 0
e 11 12 3
e 12 13 0
e 13 3 0
e 11 14 3
e 14 15 3
e 15 16 3
e 16 17 3
e 17 12 3
e 14 18 0
e 15 19 0
e 17 20 0
e 13 21 1
e 10 22 1
e 16 23 0
e 23 24 0
e 23 25 0

graphForest:
t # 1
v 0 0
v 1 0
v 2 0
v 3 0
v 4 0
v 5 0
v 6 1
v 7 1
v 8 1
v 9 1
v 10 1
v 11 4
v 12 4
v 13 6
v 14 5
v 15 3
e 0 1 3
e 1 2 3
e 2 3 3
e 3 4 3
e 4 5 3
e 5 0 3
e 0 6 0
e 1 7 0
e 2 8 0
e 3 9 0
e 5 10 0
e 4 11 0
e 11 12 0
e 11 13 0
e 13 14 0
e 12 15 1

If we only use vertex_equivalent, we can get some result like this:

This is input Graph

0 <--> 1 5 6
1 <--> 0 2 7
2 <--> 1 3 10
3 <--> 2 4 13
4 <--> 3 5 8
5 <--> 4 0 9
6 <--> 0
7 <--> 1
8 <--> 4
9 <--> 5
10 <--> 2 11 22
11 <--> 10 12 14
12 <--> 11 13 17
13 <--> 12 3 21
14 <--> 11 15 18
15 <--> 14 16 19
16 <--> 15 17 23
17 <--> 16 12 20
18 <--> 14
19 <--> 15
20 <--> 17
21 <--> 13
22 <--> 10
23 <--> 16 24 25
24 <--> 23
25 <--> 23

This is the Graph in Forest:

0 <--> 1 5 6
1 <--> 0 2 7
2 <--> 1 3 8
3 <--> 2 4 9
4 <--> 3 5 11
5 <--> 4 0 10
6 <--> 0
7 <--> 1
8 <--> 2
9 <--> 3
10 <--> 5
11 <--> 4 12 13
12 <--> 11 15
13 <--> 11 14
14 <--> 13
15 <--> 12

Start to do Maximum Common Subgraph Search:
Found common subgraph (size 10)
0 <--> 1 5 6
1 <--> 0 2 7
2 <--> 1 3
3 <--> 2 4
4 <--> 3 5 8
5 <--> 4 0 9
6 <--> 0
7 <--> 1
8 <--> 4
9 <--> 5

....