rows of $A$ in a way that contradicts the constraint $\alpha(A)=M$ in the right side of Equation (\[eq:solM\]). However, we can refine Constraint (\[eq:symbreak\], to break symmetries on the rows of $A$ only when the corresponding rows in $M$ are equal. Then $M$ can be viewed as inducing an ordered partition of $A$ and Constraint (\[eq:sbdm\]) is, in the terminology of [@DBLP:conf/ijcai/CodishMPS13], a partitioned lexicographic symmetry break. In the following, $M_i$ and $M_j$ denote the $i^{th}$ and $j^{th}$ rows of matrix $M$. $$\label{eq:sbdm}
  {\textsf{sb}}^*_\ell(A,M) = 
       \bigwedge_{i<j} \left(\begin{array}{l}
         \big(M_i=M_j\Rightarrow A_i\preceq_{\{i,j\}} A_j\big)
      \end{array}\right)$$ The following refines Equation (\[eq:solM\]) introducing the symmetry breaking predicate. $$\label{eq:scenario1}
  sol_M(\varphi(A)) = sol(\varphi(A)\wedge (\alpha(A){=}M) \wedge{\textsf{sb}}^*_\ell(A,M))$$

To justify that Equations (\[eq:solM\]) and (\[eq:scenario1\]) both compute $sol_M(\varphi(A))$, modulo weak isomorphism, we must show that if ${\textsf{sb}}^*_\ell(A,M)$ excludes a solution then there is another weakly isomorphic solution that is not excluded.

\[thm:sbl\_star\] Let $A$ be an adjacency matrix with $\alpha(A) = M$. Then, there exists $A'\approx A$ such that $\alpha(A')=M$ and ${\textsf{sb}}^{*}_\ell(A',M)$ holds.

Computing Degree Matrices for $R(3,3,3;13)$ {#sec:33313}
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This section describes how we compute a set of degree matrices that approximate those of the solutions of instance $\varphi_{(3,3,3;13)}(A)$ of Constraint (\[constraint:coloring\]). We apply a strategy mixing SAT solving with 