## 4 The Proposed Model
### Matched Vertices-Based Pruning Strategy  
The pruning strategy based on the matched vertices can effectively reduce the number of graphs or regions that execute isomorphism testing. The pruning strategy mainly contains the following two important ideas:  
1. During the isomorphism testing, some vertices in some subgraphs of data graph \(G\) are measured not to derive the subgraphs that are isomorphic to the induced subgraph. These vertices can be recorded for pruning. 2. Consider graph \(g_{1}=(V,E,l)\), query position \(q\in V\) and graph \(g_{2}=(V^{\prime},E^{\prime},l^{\prime})\). In the case of \(q^{\prime}(\in V^{\prime})\) corresponding to \(q\), if all the subgraphs containing vertex \(q^{\prime}\) of graph \(g_{2}\) are not isomorphic to \(g_{1}\), then any hypergraph of graph \(g_{1}\) and all subgraphs containing vertex \(q^{\prime}\) of graph \(g_{2}\) are not isomorphic in this case. The proposition in (2) can be proven as followings. Under the given conditions, assume that a hypergraph of graph \(g_{1}\) is isomorphic to a subgraph containing vertex \(q^{\prime}\) of graph \(g_{2}\). Since vertex \(q\) corresponds to vertex \(q^{\prime}\), graph \(g_{1}\) must be isomorphic to a subgraph containing vertex \(q^{\prime}\) of graph \(g_{2}\), which contradicts the given condition that graph \(g_{1}\) is not isomorphic to any subgraph containing vertex \(q^{\prime}\) of graph \(g_{2}\). Therefore, the assumption is not true, and thus, the proposition is true. The concept of hypergraph used here is a commonly-used concept in graph data research. The hypergraph is relative to the subgraph. If graph \(G_{1}\) is a hypergraph of graph \(G_{2}\), then graph \(G_{2}\) is a subgraph of graph \(G_{1}\). Figure 7 is an example of the above proposition. In the figure, assuming that the query position is vertex \(v_{1}\) in graph \(g_{1}\) and vertex \(v_{1}\) in graph \(g_{2}\) corresponds to vertex \(v_{1}\) in graph \(g_{1}\), it can be seen from the figure that graph \(g_{1}\) is not isomorphic to any subgraph of graph \(g_{2}\) in this case. Someone may point out that subgraph \(\{\{v_{2},v_{4}\},\{(v_{2},v_{4})\}\}\) of graph \(g_{2}\) is isomorphic to graph \(g_{1}\), but this does not satisfy the condition that vertex \(v_{1}\) in graph \(g_{2}\) corresponds to vertex \(v_{1}\) in graph \(g_{1}\). Therefore, it is not considered. The hypergraph \(g_{h}\) of graph \(g_{1}\) shown in the figure is not isomorphic to any subgraph of graph \(g_{2}\) under the condition that vertex \(v_{1}\) in graph \(g_{2}\) corresponds to vertex \(v_{1}\) in graph \(g_{1}\). However, if removing the constraint condition, the hypergraph \(g_{h}\) is isomorphic to the subgraph \(\{\{v_{1},v_{2},v_{4}\},\{(v_{1},v_{2}),(v_{2},v_{4})\}\}\) of graph \(g_{2}\). Therefore, the constraint that vertex \(v_{1}\) in the graph \(g_{2}\) corresponds to vertex \(v_{1}\) in graph \(g_{1}\) is extremely important. There is no need to enumerate all hypergraphs of graph \(g_{1}\) infinitely. This example is just for the ease of understanding the above proposition. The following shows how to use the proposition in (2) for pruning. Considering graph \(g=(V,E,I)\) and query position \(q\) and setting the query position \(q\) as the starting point of isomorphism testing (we do this in this way in our paper), the vertices corresponding to \(q\), denoted by \(V_{q}\), must be the vertices with the same label as \(q\) in the set \(V\), i.e., \(q\
otin V_{q}\) and \(V_{q}\subset V\).