Metric Logic Program Explanations for Complex Separator Functions

Abstract

There are many classifiers that treat entities to be classified as points in a high-dimensional vector space and then compute a separator S between entities in class $+1$ from those in class $-1$. However, such classifiers are usually very hard to explain in plain English to domain experts. We propose Metric Logic Programs (MLPs) which are a fragment of constraint logic programs as a new paradigm for explaining S. We present multiple measures of quality of an MLP and define the problem of finding an MLP-Explanation of S and show that it - and various related problems - are NP-hard. We present the MLP_Extract algorithm to extract MLP explanations for S. We show that while our algorithms provide more succinct, simpler, and higher fidelity explanations than association rules that are less expressive, our algorithms do require additional run-time.

Appendix: Proofs

Proof

of Theorem 1 (Sketch). Dyer [9] proved that computing the volume of the polytope $P=\{ \mathbf x \in \mathbb {R}^n \ | \ {\mathbf {a}}^T\mathbf x \le c, 0 \le \mathbf x [i] \le 1, i=1,\ldots ,n \}$, where ${\mathbf {a}}$ is a vector of integers greater than 0, is $\#P$-hard. Let us assume to have a hyper-rectangle R where $0 \le \mathbf x [i] \le 1$, $\forall i \in D$, and a hyper-plane of the form $\mathbf w ^T\phi (\mathbf x ) + b = 0$, where $\mathbf w ={\mathbf {a}}$ and $b=-c-1$. Since the volume of R is equal to 1, then, computing ${\mathbf {ov}}(R,y)=\frac{V_{\overline{y}}(R)}{V(R)}=V_{\overline{y}}(R)$ where $V_{\overline{y}}(R)$ represents the volume of ${\mathbf {ov}}(R,y)$ is the same of computing the volume of the polytope P.

Proof

of Theorem 2 (Sketch). The result is proven with a reduction from the 3-colorability problem, that is proven be NP-hard, to a decision version of the MLP-Explanation problem that is the following: verify whether there exists at most k MLRs that satisfy all the four constraints of the MLP-Explanation problem. Given a graph $G=(V,E)$, the 3-colorability problem consists into verifying whether there exists a color assignment to node s.t. two neighbors node are not colored with the same color.

The reduction is the following. We set (i) $k=3$, (ii) $n=|V|$, i.e. we have one dimension for each vertex, (iii) $T^y = \{ \mathbf u _i \in \{ 0,1\}^n \ | \ \mathbf u _i[i]=1 \wedge i\in D : \ (\forall j\in D \wedge j\ne i \Rightarrow \mathbf u _i[j]=0) \}$ to represent all vertices $v_i \in V$, (iv) $T^{\overline{y}} = \{ \mathbf u \in \{ 0,1\}^n \ | (v_i,v_j) \in E \wedge \mathbf u [i]=1 \wedge \mathbf u [j]=1 \wedge (\forall h \in D : h \ne i,j \Rightarrow \mathbf u [h]=0) \}$ to represent the set E of edges, and (v) $m=0$. Because of the first MLP-Explanation problem constraint, and $k=3$, a node must be covered by at least one rule of the three, so each rule assigns a color to each rule.

For instance, suppose we have in G only two vertices $v_1$ and $v_2$ and one edge $(v_1,v_2)$. Then, $T^y = \{[1,0],[0,1]\}$ and $T^{\overline{y}} = \{[1,1]\}$. Considering now the rule $0 \le \mathbf x [1] \le 1 \wedge 0 \le \mathbf x [2] \le 1 \rightarrow y$, we have that it covers $T^y \cup T^{\overline{y}}$.

Then, it is not possible that there exist two nodes covered by the same rule that are connected by an edge, otherwise a point in $T^{\overline{y}}$ is covered by the rule. This implies that it is not possible to use the same color to color two nodes connected by an edge.

Note that it is possible that the same node can be colored in two different ways, but to obtain a feasible coloring it is sufficient choosing one of the two color without violate the constraints. It follows that the decision version of the MLP-Explanation problem is at least hard as the 3-colorability, and this means that the MLP-Explanation problem is NP-hard.

Proof

of Theorem 3 (Sketch). We prove the theorem via a reduction from maximal independent set problem proven to be NP-hard. We recall that, given a graph $G=(V,E)$ the maximal independent set problem consists into find a set of vertices $V' \subseteq V$ s.t. $|V'| \ge k$ and there are no edges in E between any pair of vertices in $V'$. The reduction is the following. We set (i) $n=|V|$, i.e. we have one dimension for each vertex, (ii) $W = \{ \mathbf u _i \in \{ 0,1\}^n \ | \ \mathbf u _i[i]=1 \wedge i\in D : \ (\forall j\in D \wedge j\ne i \Rightarrow \mathbf u _i[j]=0) \}$ to represent all vertices $v_i \in V$, (iii) $\overline{W} = \{ \mathbf u \in \{ 0,1\}^n \ | (v_i,v_j) \in E \wedge \mathbf u [i]=1 \wedge \mathbf u [j]=1 \wedge (\forall h \in D : h \ne i,j \Rightarrow \mathbf u [h]=0) \}$ to represent the set E of edges, and (iv) $m=0$. Then, our problem is to find a rule, if there exists, that covers at least k points in W and no point in $\overline{W}$. In this reduction we do not consider the MLP-Explanation problem constraints 2 and 3. Note that if a rule covers two vertices then it automatically cover also the edge between them if it exist. For instance, suppose we have in G only two vertices $v_1$ and $v_2$ and one edge $(v_1,v_2)$. Then, $W = \{[1,0],[0,1]\}$ and $\overline{W} = \{[1,1]\}$. Considering now the rule $0 \le \mathbf x [1] \le 1 \wedge 0 \le \mathbf x [2] \le 1 \rightarrow y$, we have that it covers $W \cup \overline{W}$. It follows that finding a rule that covers at least k points in W and does not cover any point in $\overline{W}$ is at least hard as the independent set problem.