Fuzzy entropies for class-specific and classification-based attribute reducts in three-way probabilistic rough set models

Abstract

There exist two formulations of the theory of rough sets, consisting of the conceptual formulations and the computational formulations. Class-specific and classification-based attribute reducts are two crucial notions in three-way probabilistic rough set models. In terms of conceptual formulations, the two types of attribute reducts can be defined by considering probabilistic positive or negative region preservations of a decision class and a decision classification, respectively. However, in three-way probabilistic rough set models, there are few studies on the computational formulations of the two types of attribute reducts due to the non-monotonicity of probabilistic positive and negative regions. In this paper, we examine the computational formulations of the two types of attribute reducts in three-way probabilistic rough set models based on fuzzy entropies. We construct monotonic measures based on fuzzy entropies, from which we can obtain the computational formulations of the two types of attribute reducts. On this basis, we develop algorithms for finding the two types of attribute reducts based on addition-deletion method or deletion method. Finally, the experimental results verify the monotonicity of the proposed measures with respect to the set inclusion of attributes and show that class-specific attribute reducts provide a more effective way of attribute reduction with respect to a particular decision class compared with classification-based attribute reducts.

Appendix A Proofs

Proof of Theorem 1

Since $P \supseteq Q$, $OB / P ~\preceq ~ OB / Q$. For simplicity and without any loss of generality, suppose that $OB / P = \{P_1, \ldots , P_i, \ldots , P_k, \ldots , P_t\}$ and $OB / Q = \{P_1, \ldots , P_{i-1}, P_{i+1}, \ldots , P_{k-1}, P_{k+1}, \ldots , P_t, P_i \cup P_k\}$. That is, the partition OB/Q is generated through combining equivalence classes $P_i$ and $P_k$ in the partition OB/P to $P_i \cup P_k$.

Let $Q_i = P_i \cup P_k$. Suppose that $Q_i \cap D_j \ne \emptyset$. We have

$$\begin{aligned}&\sum \limits _{x \in Q_i} \Big |\mu _{F_{\widetilde{{D_j}}}^Q}(x) - \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(x)\Big |\\&\quad = \Big |\frac{|Q_i \cap D_j|}{|Q_i|}|Q_i| - |Q_i| \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(x)\Big | \\&\quad = \Big |\frac{|Q_i \cap D_j|}{|Q_i|}|Q_i| - \sum \limits _{x \in Q_i} \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(x)\Big | \\&\quad = \Big ||P_i \cap D_j| + |P_k \cap D_j|\\&\qquad - \sum \limits _{y \in P_i} \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(y) - \sum \limits _{z \in P_k} \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(z)\Big |. \end{aligned}$$

For all $y \in P_i$ and $z \in P_k$, there are four cases.

Case (1): $\mu _{F_{\widetilde{{D_j}}}^Q}(y) = \mu _{F_{\widetilde{{D_j}}}^Q}(z) > \frac{1}{2}$, $\mu _{F_{\widetilde{{D_j}}}^P}(y) > \frac{1}{2}$ and $\mu _{F_{\widetilde{{D_j}}}^P}(z) > \frac{1}{2}$.

We have $\mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(y) = 1$, $\mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(z) = 1$, $\mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(y) = 1$ and $\mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(z) = 1$. Thus, it has $|P_i \cap D_j| - \sum \limits _{y \in P_i} \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(y) \le 0$ and $|P_k \cap D_j| - \sum \limits _{z \in P_k} \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(z) \le 0$. Hence, one has

$$\begin{aligned}&\sum \limits _{x \in Q_i} \Big |\mu _{F_{\widetilde{{D_j}}}^Q}(x) - \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(x)\Big | \\&\quad = \Big ||P_i \cap D_j| - \sum \limits _{y \in P_i} \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(y)\Big | \\&\qquad + \Big ||P_k \cap D_j| - \sum \limits _{z \in P_k} \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(z)\Big | \\&\quad = \Big |\frac{|P_i \cap D_j|}{|P_i|}|P_i| - |P_i|\Big | + \Big |\frac{|P_k \cap D_j|}{|P_k|}|P_k| - |P_k|\Big | \\&\quad = \sum \limits _{y \in P_i} \Big |\mu _{F_{\widetilde{{D_j}}}^P}(y) - \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(y)\Big | \\&\qquad + \sum \limits _{z \in P_k} \Big |\mu _{F_{\widetilde{{D_j}}}^P}(z) - \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(z)\Big |. \end{aligned}$$

As a result, $e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^Q) = e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^P)$.

Case (2): $\mu _{F_{\widetilde{{D_j}}}^Q}(y) = \mu _{F_{\widetilde{{D_j}}}^Q}(z) > \frac{1}{2}$, $\mu _{F_{\widetilde{{D_j}}}^P}(y) > \frac{1}{2}$ and $\mu _{F_{\widetilde{{D_j}}}^P}(z) \le \frac{1}{2}$ (or $\mu _{F_{\widetilde{{D_j}}}^P}(y) \le \frac{1}{2}$ and $\mu _{F_{\widetilde{{D_j}}}^P}(z) > \frac{1}{2}$).

In the following, we only prove the case $\mu _{F_{\widetilde{{D_j}}}^Q}(y) = \mu _{F_{\widetilde{{D_j}}}^Q}(z) > \frac{1}{2}$, $\mu _{F_{\widetilde{{D_j}}}^P}(y) > \frac{1}{2}$ and $\mu _{F_{\widetilde{{D_j}}}^P}(z) \le \frac{1}{2}$.

We have $\mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(y) = 1$, $\mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(z) = 1$, $\mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(y) = 1$ and $\mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(z) = 0$. Since $\mu _{F_{\widetilde{{D_j}}}^P}(z) \le \frac{1}{2}$, it has $\Big ||P_k \cap D_j| - |P_k|\Big | \ge \Big ||P_k \cap D_j|\Big |$. Hence, one has

$$\begin{aligned}&\sum \limits _{x \in Q_i} \Big |\mu _{F_{\widetilde{{D_j}}}^Q}(x) - \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(x)\Big | \\&\quad = \Big ||P_i \cap D_j| + |P_k \cap D_j| - |P_i| - |P_k|\Big | \\&\quad = \Big ||P_i \cap D_j| - |P_i|\Big | + \Big ||P_k \cap D_j| - |P_k|\Big | \\&\quad \ge \Big ||P_i \cap D_j| - |P_i|\Big | + \Big ||P_k \cap D_j| - 0\Big | \\&\quad = \sum \limits _{y \in P_i} \Big |\mu _{F_{\widetilde{{D_j}}}^P}(y) - \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(y)\Big | \\&\qquad + \sum \limits _{z \in P_k} \Big |\mu _{F_{\widetilde{{D_j}}}^P}(z) - \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(z)\Big | \end{aligned}$$

As a result, $e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^Q) \ge e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^P)$.

Case (3): $\mu _{F_{\widetilde{{D_j}}}^Q}(y) = \mu _{F_{\widetilde{{D_j}}}^Q}(z) \le \frac{1}{2}$, $\mu _{F_{\widetilde{{D_j}}}^P}(y) \le \frac{1}{2}$ and $\mu _{F_{\widetilde{{D_j}}}^P}(z) \le \frac{1}{2}$.

We have $\mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(y) = 0$, $\mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(z) = 0$, $\mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(y) = 0$ and $\mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(z) = 0$. Thus,

$$\begin{aligned}&\sum \limits _{x \in Q_i} \Big |\mu _{F_{\widetilde{{D_j}}}^Q}(x) - \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(x)\Big | \\&\quad = \Big ||P_i \cap D_j| + |P_k \cap D_j| - 0 - 0\Big |\\&\quad = \Big ||P_i \cap D_j| - 0\Big | + \Big ||P_k \cap D_j| - 0\Big | \\&\quad = \Big ||P_i \cap D_j| - \sum \limits _{y \in P_i} \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(y)\Big | \\&\qquad + \Big ||P_k \cap D_j| - \sum \limits _{z \in P_k} \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(z)\Big | \\&\quad = \sum \limits _{y \in P_i} \Big |\mu _{F_{\widetilde{{D_j}}}^P}(y) - \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(y)\Big | \\&\qquad + \sum \limits _{z \in P_k} \Big |\mu _{F_{\widetilde{{D_j}}}^P}(z) - \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(z)\Big |. \end{aligned}$$

As a result, $e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^Q) = e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^P)$.

Case (4): $\mu _{F_{\widetilde{{D_j}}}^Q}(y) = \mu _{F_{\widetilde{{D_j}}}^Q}(z) \le \frac{1}{2}$, $\mu _{F_{\widetilde{{D_j}}}^P}(y) \le \frac{1}{2}$ and $\mu _{F_{\widetilde{{D_j}}}^P}(z) > \frac{1}{2}$ (or $\mu _{F_{\widetilde{{D_j}}}^P}(y) > \frac{1}{2}$ and $\mu _{F_{\widetilde{{D_j}}}^P}(z) \le \frac{1}{2}$).

In what follows, we only prove the case $\mu _{F_{\widetilde{{D_j}}}^Q}(y) = \mu _{F_{\widetilde{{D_j}}}^Q}(z) \le \frac{1}{2}$, $\mu _{F_{\widetilde{{D_j}}}^P}(y) \le \frac{1}{2}$ and $\mu _{F_{\widetilde{{D_j}}}^P}(z) > \frac{1}{2}$.

We have $\mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(y) = 0$, $\mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(z) = 0$, $\mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(y) = 0$ and $\mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(z) = 1$. Since $\mu _{F_{\widetilde{{D_j}}}^P}(z) > \frac{1}{2}$, $\Big ||P_k \cap D_j|\Big | > \Big ||P_k \cap D_j| - |P_k|\Big |$. Thus,

$$\begin{aligned}&\sum \limits _{x \in Q_i} \Big |\mu _{F_{\widetilde{{D_j}}}^Q}(x) - \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(x)\Big | \\&\quad = \Big ||P_i \cap D_j| + |P_k \cap D_j| - 0 - 0\Big |\\&\quad = \Big ||P_i \cap D_j| - 0\Big | + \Big ||P_k \cap D_j| - 0\Big | \\&\quad > \Big ||P_i \cap D_j| - 0\Big | + \Big ||P_k \cap D_j| - |P_k|\Big | \\&\quad = \sum \limits _{y \in P_i} \Big |\mu _{F_{\widetilde{{D_j}}}^P}(y) - \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(y)\Big | \\&\qquad + \sum \limits _{z \in P_k} \Big |\mu _{F_{\widetilde{{D_j}}}^P}(z) - \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(z)\Big |. \end{aligned}$$

As a result, $e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^Q) > e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^P)$.

On the other hand, suppose that $Q_i \cap D_j = \emptyset$. We have

$$\begin{aligned}&\sum \limits _{x \in Q_i} \Big |\mu _{F_{\widetilde{{D_j}}}^Q}(x) - \mu _{{F_{\widetilde{{D_j}}}^Q}_{near}}(x)\Big | \\&\quad = \sum \limits _{y \in P_i} \Big |\mu _{F_{\widetilde{{D_j}}}^P}(y) - \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(y)\Big |\\&\qquad + \sum \limits _{z \in P_k} \Big |\mu _{F_{\widetilde{{D_j}}}^P}(z) - \mu _{{F_{\widetilde{{D_j}}}^P}_{near}}(z)\Big | = 0. \end{aligned}$$

Hence, $e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^Q) = e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^P)$.

In conclusion, the fuzzy entropy $e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^P)$ will be increased or preserved with the combining of equivalence classes. Hence, we have $P \supseteq Q \Rightarrow e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^P) \le e_{\mathrm{K}_1}(F_{\widetilde{{D_j}}}^Q)$. $\square$

Proof of Theorem 6

Since $B \subseteq C$, $\xi ({[x]_B}) = \{ {[y]_C}:{[y]_C} \subseteq {[x]_B}\}$ is a partition of $[x]_B$.

(1) “$\Rightarrow$” Since ${\mathrm{PO}}{{\mathrm{S}}_{({\alpha _j},{\beta _j})}}({D_j}{\mathrm{| }}B) = {\mathrm{PO}}{{\mathrm{S}}_{({\alpha _j},{\beta _j})}}({D_j}{\mathrm{| }}C)$, we have

$$\begin{aligned}&e_{\varDelta }\left( F_{\widetilde{{\mathrm{POS}}}_{(\alpha _j, \beta _j)}(D_j|C)}^B\right) \\&\quad = e_{\varDelta }\left( F_{\widetilde{{\mathrm{POS}}}_{(\alpha _j, \beta _j)}(D_j|B)}^B\right) = 0. \end{aligned}$$

“$\Leftarrow$” Suppose that $x \in {\mathrm{PO}}{{\mathrm{S}}_{({\alpha _j},{\beta _j})}}({D_j}{\mathrm{| }}C)$. Since $e_{\varDelta }(F_{\widetilde{{\mathrm{POS}}}_{(\alpha _j, \beta _j)}(D_j|C)}^B) = 0$, we have $\mu _{\widetilde{{\mathrm{POS}}}_{(\alpha _j, \beta _j)}(D_j|C)}^B(x) = 0$ or $\mu _{\widetilde{{\mathrm{POS}}}_{(\alpha _j, \beta _j)}(D_j|C)}^B(x) = 1$. That is, ${{ | {{[x]}_B} \cap {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C) | } \over { | {{[x]}_B} | }} = 0$ or ${{ | {{[x]}_B} \cap {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C) | } \over { | {{[x]}_B} | }} = 1$. Therefore, $[x]_B \cap {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C) = \emptyset$ or $[x]_B \subseteq {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C)$.

If $[x]_B \cap {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C) = \emptyset$, then $x \notin {\mathrm{PO}}{{\mathrm{S}}_{({\alpha _j},{\beta _j})}}({D_j}{\mathrm{| }}C)$. This statement conflicts with condition $x \in {\mathrm{PO}}{{\mathrm{S}}_{({\alpha _j},{\beta _j})}}({D_j}{\mathrm{| }}C)$. Hence, we have $[x]_B \subseteq {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C)$. Since ${[x]_B} = \cup \{ {[y]_C}:{[y]_C} \in \xi ({[x]_B})\}$, we obtain that ${[y]_C} \subseteq {[x]_B} \subseteq {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C)$ for all ${[y]_C} \in \xi ({[x]_B})$. That is, for all ${[y]_C} \in \xi ({[x]_B})$, $Pr({D_j} | {[y]_C}) \ge \alpha _j$. Thus,

$$\begin{aligned}&Pr({D_j} |{[x]_B}) \\&\quad = {{{\sum {\{ |{{[y]}_C} \cap {D_j} | :{{[y]}_C} \in \xi ({{[x]}_B})\} } }} / { |{{[x]}_B}| }} \\&\quad = \sum {\left\{ {Pr({D_j} |{{[y]}_C}) \cdot {{ |{{[y]}_C}| } \over { |{{[x]}_B}| }}:{{[y]}_C} \in \xi ({{[x]}_B})} \right\} } \\&\quad \ge \alpha _j \sum {\left\{ {{{ |{{[y]}_C}| } \over { |{{[x]}_B}| }}:{{[y]}_C} \in \xi ({{[x]}_B})} \right\} } = \alpha _j. \end{aligned}$$

As a result, $x \in {\mathrm{PO}}{{\mathrm{S}}_{({\alpha _j},{\beta _j})}}({D_j}{\mathrm{| }}B)$.

On the other hand, if $x \in {\mathrm{PO}}{{\mathrm{S}}_{({\alpha _j},{\beta _j})}}({D_j}{\mathrm{| }}B)$, then ${[x]_B} \subseteq {\mathrm{PO}}{{\mathrm{S}}_{({\alpha _j},{\beta _j})}}({D_j}{\mathrm{| }}B)$.

Since $e_{\mathrm{K}_1}(F_{\widetilde{{\mathrm{POS}}}_{(\alpha _j, \beta _j)}(D_j|C)}^B) = 0$, $[x]_B \cap {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C) = \emptyset$ or $[x]_B \in {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C)$. If $[x]_B \cap {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C) = \emptyset$, then we have ${[y]_C} \cap {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C) = \emptyset$ for all ${[y]_C} \in \xi ({[x]_B})$. That is, ${\mathrm{Pr}}({D_j} | {[y]_C}) < \alpha _j$ for all ${[y]_C} \in \xi ({[x]_B})$.

$$\begin{aligned}&Pr({D_j} |{[x]_B}) \\&\quad = {{{\sum {\{ |{{[y]}_C} \cap {D_j} | :{{[y]}_C} \in \xi ({{[x]}_B})\} } }} / { |{{[x]}_B}| }} \\&\quad = \sum {\left\{ {{\mathrm{Pr}}({D_j} |{{[y]}_C}) \cdot {{ |{{[y]}_C}| } \over { |{{[x]}_B}| }}:{{[y]}_C} \in \xi ({{[x]}_B})} \right\} } \\&\quad < \alpha _j \sum {\left\{ {{{ |{{[y]}_C}| } \over { |{{[x]}_B}| }}:{{[y]}_C} \in \xi ({{[x]}_B})} \right\} } = \alpha _j. \end{aligned}$$

As a result, $[x]_B \nsubseteq {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|B)$. It contradicts with $[x]_B \subseteq {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|B)$.

Hence, $[x]_B \subseteq {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C)$, namely, $x \in {\mathrm{POS}}_{(\alpha _j, \beta _j)}(D_j|C)$.

Therefore, we conclude that ${\mathrm{PO}}{{\mathrm{S}}_{({\alpha _j},{\beta _j})}}({D_j}{{| }}B) = {\mathrm{PO}}{{\mathrm{S}}_{({\alpha _j},{\beta _j})}}({D_j}{{| }}C)$.

The proof of (2) is similar to that of (1). $\square$