# Problem *Subgraph Isomorphism*

An instance of the subgraph isomorphism problem is defined by a pattern graph $G_p=(V_p,E_p)$ and a target graph $G_t=(V_t,E_t)$: the objective is to determine whether $G_p$ is isomorphic to some subgraph(s) in $G_t$. Finding a solution to such a problem instance means then finding a subisomorphism function, that is an injective mapping $f : V_p \rightarrow V_t$ such that all edges of $G_p$ are preserved: $\forall (v,v’) \in E_p, (f(v_p),f(v’_p)) \in E_t$. Here, we refer to the partial, and not the induced subgraph isomorphism problem.

An Instance of the Subgraph Isomorphism Problem.

To build a CSP (Constraint Satisfaction Problem) model, we need first to import the library PyCSP$^3$:

```
from pycsp3 import *
```

Then, we need some data. For example, for the two graphs shown above, we can write:

```
n = 4 # number of nodes in the pattern graph
m = 5 # number of nodes in the target graph
p_edges = [(0,1), (0,2), (0,3), (1,2), (1,3), (2,3)]
t_edges = [(0,1), (0,3), (0,4), (1,2), (1,4), (2,3), (2,4), (3,4)]
```

In the pattern graph, we have 4 nodes and 6 edges, whereas in the target graph, we have 5 nodes and 8 edges. Note that we use indexes for nodes (for example, index 0 is used for node 1 in the pattern graph and for node a in the target graph).

We start our CSP model by introducing an array $x$ of variables.

```
# x[i] is the target node to which the ith pattern node is mapped
x = VarArray(size=n, dom=range(m))
```

We can display the structure of the array, as well as the domain of the first variable (note that all variables have the same domain).

```
print("Array x: ", x)
print("Domain of any variable in x: ", x[0].dom)
```

```
Array x: [x[0], x[1], x[2], x[3]]
Domain of any variable in x: 0..4
```

Concerning the constraints, we have to post first a constraint AllDifferent.

```
satisfy(
# ensuring injectivity
AllDifferent(x)
);
```

Interestingly, by calling the function *solve()*, we can check that the problem is satisfiable (SAT). We can also display the found solution. Here, we call the function *values()* that collects the values assigned to a specified list of variables.

```
if solve() is SAT:
print(values(x))
```

```
[0, 1, 2, 3]
```

It works, but edges are not preserved. We need a table for enumerating all possible mappings of any pattern edge:

```
T = {(i, j) for (i, j) in t_edges} | {(j, i) for (i, j) in t_edges}
```

We can now post a group of constraints Extension.

```
satisfy(
# preserving edges
(x[i], x[j]) in T for (i, j) in p_edges
);
```

We can display the internal representation of the posted constraints; this way, although a little bit technical, we can see that everything we need is present.

```
print(posted())
```

```
allDifferent(list:x[])
extension(list:[x[0], x[1]], supports:(0,1)(0,3)(0,4)(1,0)(1,2)(1,4)(2,1)(2,3)(2,4)(3,0)(3,2)(3,4)(4,0)(4,1)(4,2)(4,3))
extension(list:[x[0], x[2]], supports:(0,1)(0,3)(0,4)(1,0)(1,2)(1,4)(2,1)(2,3)(2,4)(3,0)(3,2)(3,4)(4,0)(4,1)(4,2)(4,3))
extension(list:[x[0], x[3]], supports:(0,1)(0,3)(0,4)(1,0)(1,2)(1,4)(2,1)(2,3)(2,4)(3,0)(3,2)(3,4)(4,0)(4,1)(4,2)(4,3))
extension(list:[x[1], x[2]], supports:(0,1)(0,3)(0,4)(1,0)(1,2)(1,4)(2,1)(2,3)(2,4)(3,0)(3,2)(3,4)(4,0)(4,1)(4,2)(4,3))
extension(list:[x[1], x[3]], supports:(0,1)(0,3)(0,4)(1,0)(1,2)(1,4)(2,1)(2,3)(2,4)(3,0)(3,2)(3,4)(4,0)(4,1)(4,2)(4,3))
extension(list:[x[2], x[3]], supports:(0,1)(0,3)(0,4)(1,0)(1,2)(1,4)(2,1)(2,3)(2,4)(3,0)(3,2)(3,4)(4,0)(4,1)(4,2)(4,3))
```

We can run again the solver.

```
if solve() is SAT:
print(values(x))
```

Nothing is displayed. This is because the result is UNSAT, which is confirmed by executing:

```
status = solve()
print("Status: ", status)
```

```
Status: UNSAT
```

The instance is unsatisfiable because there is no 4-clique (4 nodes that are all linked each other) in the target graph.

We propose to modify the target graph by adding an edge between the nodes a and c, before updating the constraints Extension. First, we modify the table:

```
T = T | {(0,2), (2,0)}
```

We check it:

```
print(T)
```

```
{(0, 1), (2, 4), (1, 2), (0, 4), (3, 4), (4, 0), (2, 1), (4, 3), (0, 2), (1, 0), (3, 2), (4, 1), (0, 3), (2, 0), (1, 4), (4, 2), (2, 3), (3, 0)}
```

The we remove all constraints that were posted during the last call to *satsify()*. This is possible by executing:

```
unpost()
```

We control that there is only the constraint AllDifferent left.

```
print(posted())
```

```
allDifferent(list:x[])
```

Finally, we post the new table constraints:

```
satisfy(
# preserving edges
(x[i], x[j]) in T for (i, j) in p_edges
);
```

And we run the solver:

```
if solve() is SAT:
print(values(x))
```

```
[0, 1, 2, 4]
```

We can check that 48 solutions exist, as follows.

```
if solve(sols=ALL) is SAT:
print("Number of solutions: ", n_solutions())
```

```
Number of solutions: 48
```

Finally, we give below the model in one piece. Here the data is expected to be given by the user (in a command line). We also pay attention to possible self loops.

```
from pycsp3 import *
n, m, p_edges, t_edges = data
p_loops = [i for (i, j) in p_edges if i == j]
t_loops = [i for (i, j) in t_edges if i == j]
T = {(i, j) for (i, j) in t_edges} | {(j, i) for (i, j) in t_edges}
# x[i] is the target node to which the ith pattern node is mapped
x = VarArray(size=n, dom=range(m))
satisfy(
# ensuring injectivity
AllDifferent(x),
# preserving edges
[(x[i], x[j]) in T for (i, j) in p_edges],
# being careful of self-loops
[x[i] in t_loops for i in p_loops]
)
```