A variable precision multigranulation rough set model and attribute reduction

Abstract

As a useful extension of rough sets, multigranulation rough sets (MGRSs) can be used to deal with a variety of complex data. Numerous significant advances have been achieved by generalizing MGRSs. However, most of the existing findings of MGRSs are sensitive to misclassification and noise in data. Furthermore, the studies of attribute reduction based on MGRSs have received little attention. To fill such gaps, this paper proposes an extended model of MGRSs named variable precision multigranulation rough sets (VPMGRSs) by introducing rough membership function and approximation parameters in variable precision rough sets (VPRSs) into the multigranulation environment. After giving some basic properties of VPMGRSs, we investigate the relationships between VPMGRSs and VPRSs, pessimistic MGRSs, and generalized MGRSs. In addition, several VPMGRSs-based attribute reductions are introduced, and it is proved that some of them are equivalent when the parameters in the model meet specific requirements. Finally, we propose a heuristic algorithm for $\alpha$-lower distribution reduct and illustrate its effectiveness and efficiency by a comparative experiment on real datasets.

Appendices

Appendix A. The proof of proposition 5

(1) For any $x\in \sim X$ and $AT_i\in \mathcal {A}$, it is obvious that $\mu ^{{AT}_i} _X(x)<1$, which leads to $\max \{\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\vert x\in \sim X\}<1$. There is $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\le \max \{\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\vert x\in \sim X\}<\alpha$. According to Eq. (8), it can be obtained $x\notin {\underline{\mathcal {A}}}_\alpha (X)$. This means $\sim X\subseteq \sim {\underline{\mathcal {A}}}_\alpha (X)$, which is equivalent to ${\underline{\mathcal {A}}}_\alpha (X)\subseteq X$.

(2) It is obvious that $\mu ^{{AT}_i} _X(x)>0$ for any $x\in X$ and $AT_i\in \mathcal {A}$, therefore, $\min \{\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\vert x\in X\}>0$. Because $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\ge \min \{\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\vert x\in X\}\ge \alpha$, there is $x\in {\underline{\mathcal {A}}}_\alpha (X)$. Thus, $X\subseteq {\underline{\mathcal {A}}}_\alpha (X)$.

(3) For any $x\in X$ and $AT_i\in \mathcal {A}$, it can be obtained that $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\ge \min \{\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\vert x\in X\}>\beta$ from the known condition; therefore, $x\in \overline{\mathcal {A}}_\beta (X)$.

(4) For any $x\in \sim X$, there is $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\le \max \{\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\vert x\in \sim X\}\le \beta$, accordingly, $x\notin \overline{\mathcal {A}}_\beta (X)$. Then, $\sim X\subseteq \sim \overline{\mathcal {A}}_\beta (X)$, which is equivalent to $\overline{\mathcal {A}}_\beta (X)\subseteq X$.

(5) From the known condition, it can be obtained that for any $x\in U$, there exists $AT_i\in \mathcal {A}$ such that $\mu ^{{AT}_i} _X(x)>0$ and $\min \{\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\vert x\in U\}>0$. There is $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i} \ge \min \{\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\vert x\in U\}\ge \alpha$, which means $x\in {\underline{\mathcal {A}}}_\alpha (X)$. Hence, $U\subseteq {\underline{\mathcal {A}}}_\alpha (X)$ and ${\underline{\mathcal {A}}}_\alpha (X)=U$ is gotten. From the property (1) in Proposition 3, $U\subseteq \overline{\mathcal {A}}_\beta (X)$, and thus, $\overline{\mathcal {A}}_\beta (X)=U$.

(6) From the known condition, for any $x\in U$, there exists $AT_i\in \mathcal {A}$ such that $\mu ^{{AT}_i} _X(x)<1$ and thus $\max \{\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\vert x\in U\}<1$. It is obvious that $\dfrac{\sum \nolimits _{i=1}^s \omega _i\mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\le \max \{\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\vert x\in U\}\le \beta$, which indicates $x\notin {\overline{\mathcal {A}}}_\alpha (X)$. From the arbitrariness of x, there is ${\overline{\mathcal {A}}}_\alpha (X)=\emptyset$. From the property (1) in Proposition 3, $\underline{\mathcal {A}}_\alpha (X)\subseteq \overline{\mathcal {A}}_\beta (X)$, and thus ${\underline{\mathcal {A}}}_\alpha (X)=\emptyset$.

Appendix B. The proof of proposition 10

(1) According to Eq. (8), if $x\in \underline{\mathcal {A}}_1(X)$, then $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}\ge 1$, which means for any $AT_i\in \mathcal {A}$, there is $\mu ^{{AT}_i} _X(x)=1$, i.e., $[x]_{{AT}_i}\subseteq X$. It can be gotten $x\in \underline{\mathcal {A}}^P(X)$ and thus $\underline{\mathcal {A}}_1(X)\subseteq \underline{\mathcal {A}}^P(X)$. And vice versa. Then $\underline{\mathcal {A}}_1(X)=\underline{\mathcal {A}}^P(X)$ is proved.

According to Eq. (9), if $x\in \overline{\mathcal {A}}_0(X)$, then $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)}{\sum \nolimits _{i=1}^s \omega _i}> 0$ and $\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _X(x)>0$. Thus, there must exist $AT_i\in \mathcal{AT}\mathcal{}$ subject to $[x]_{AT_i}\cap X\ne \emptyset$. Therefore, $x\in \overline{\mathcal {A}}^P(X)$ and it can be concluded $\overline{\mathcal {A}}_0(X)\subseteq \overline{\mathcal {A}}^P(X)$. And vice versa. ${\overline{\mathcal {A}}}_\beta (X)\subseteq {\overline{\mathcal {A}}}^P(X)$ is proved.

(2) From Proposition 10 (1), $\underline{\mathcal {A}}^P(X)=\underline{\mathcal {A}}_1(X)$ is already known. In addition, for any $\alpha \le 1$, it can be obtained ${\underline{\mathcal {A}}}_\alpha (X)\subseteq {\underline{\mathcal {A}}}_1(X)$ from Proposition 7 (1). Consequently, ${\underline{\mathcal {A}}}^P(X)\subseteq {\underline{\mathcal {A}}}_\alpha (X)$. The inclusion relationship between upper approximations can be similarly carried out.

Appendix C. The proof of proposition 13

(1) ($\Rightarrow$) Let $\mathcal {A}$ be an $\alpha$-lower distribution consistent set. For any $x\in U$, if $D_j\in \gamma _\mathcal {A}(x)$, then $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _{D_j}(x)}{\sum \nolimits _{i=1}^s \omega _i}\ge \alpha _\mathcal {A}\ge \alpha _0\ge \alpha$; therefore, $x\in \underline{\mathcal {A}}_\alpha (D_j)$. It can be obtained from $\underline{\mathcal {A}}_\alpha (D_j)=\underline{\mathcal{AT}\mathcal{}}_\alpha (D_j)$ that $x\in \underline{\mathcal{AT}\mathcal{}}_\alpha (D_j)$, and this means $\dfrac{\sum \nolimits _{i=1}^m \omega _i \mu ^{{AT}_i} _{D_j}(x)}{\sum \nolimits _{i=1}^m \omega _i}\ge \alpha >0.5$. Then, it can be obtained that, compared with other decision classes, the average membership degree of x to $D_j$ is the maximum, and there is $D_j\in \gamma _\mathcal{AT}\mathcal{}(x)$. Thus, $\gamma _\mathcal {A}(x)\subseteq \gamma _\mathcal{AT}\mathcal{}(x)$, and $\gamma _\mathcal{AT}\mathcal{}(x)\subseteq \gamma _\mathcal {A}(x)$ can proved in the same way. In conclusion, for any $x\in U$, there is $\gamma _\mathcal{AT}\mathcal{}(x)=\gamma _\mathcal {A}(x)$; namely, $\mathcal {A}$ is a maximum distribution consistent set.

($\Leftarrow$) Let $\mathcal {A}$ be a maximum distribution consistent set. For any $D_j\in U/D$, if $x\in \underline{\mathcal {A}}_\alpha (D_j)$, then there are $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _{D_j}(x)}{\sum \nolimits _{i=1}^s \omega _i}\ge \alpha >0.5$ and $D_j\in \gamma _\mathcal {A}(x)$. Because $\mathcal {A}$ is a maximum distribution consistent set, it can be obtained $\gamma _\mathcal{AT}\mathcal{}(x)=\gamma _\mathcal {A}(x)$ and $D_j\in \gamma _\mathcal{AT}\mathcal{}(x)$. Thus, $\dfrac{\sum \nolimits _{i=1}^m \omega _i \mu ^{{AT}_i} _{D_j}(x)}{\sum \nolimits _{i=1}^m \omega _i}\ge \alpha _\mathcal{AT}\mathcal{}\ge \alpha _0\ge \alpha$ and $x\in \underline{\mathcal{AT}\mathcal{}}_\alpha (D_j)$. Therefore, $\underline{\mathcal {A}}_\alpha (D_j)\subseteq \underline{\mathcal{AT}\mathcal{}}_\alpha (D_j)$. It can be proved that $\underline{\mathcal{AT}\mathcal{}}_\alpha (D_j)\subseteq \underline{\mathcal {A}}_\alpha (D_j)$ similarly. In conclusion, $\underline{\mathcal{AT}\mathcal{}}_\alpha (D_j)=\underline{\mathcal {A}}_\alpha (D_j)$ for all $D_j\in U/D$, that is $\mathcal {A}$ is an $\alpha$-lower distribution consistent set.

(2) can be easily proved by (1).

Appendix D. The proof of proposition 14

(1) ($\Rightarrow$) Let $\mathcal {A}$ be a $\beta$-upper distribution consistent set. For any $x \in U$, if $D_j\in \delta _\mathcal {A}(x)$, then $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _{D_j}(x)}{\sum \nolimits _{i=1}^s \omega _i}\ge \beta _\mathcal {A}\ge \beta _0>\beta$ and $x\in \overline{\mathcal {A}}_\alpha (D_j)$. It can be obtained that $x\in \overline{\mathcal{AT}\mathcal{}}_\beta (D_j)$ from $\overline{\mathcal{AT}\mathcal{}}_\alpha (D_j)=\overline{\mathcal {A}}_\alpha (D_j)$. Thus $\dfrac{\sum \nolimits _{i=1}^m \omega _i \mu ^{{AT}_i} _{D_j}(x)}{\sum \nolimits _{i=1}^m \omega _i}>\beta \ge 0$, and this means $D_j\in \delta _\mathcal{AT}\mathcal{}(x)$. Therefore, $\delta _\mathcal {A}(x)\subseteq \delta _\mathcal{AT}\mathcal{}(x)$. It can be similarly proved that $\delta _\mathcal{AT}\mathcal{}(x)\subseteq \delta _\mathcal {A}(x)$. In conclusion, $\delta _\mathcal {A}(x)=\delta _\mathcal{AT}\mathcal{}(x)$ for any $x\in U$, that is, $\mathcal {A}$ is a possible consistent set.

($\Leftarrow$) Let $\mathcal {A}$ be a possible consistent set. For any $D_j\in U/D$, if $x\in \overline{\mathcal {A}}_\beta (D_j)$, then $\dfrac{\sum \nolimits _{i=1}^s \omega _i \mu ^{{AT}_i} _{D_j}(x)}{\sum \nolimits _{i=1}^s \omega _i}>\beta \ge 0$ and $D_j\in \delta _\mathcal {A}(x)$. Then, it can be obtained that $D_j\in \delta _\mathcal{AT}\mathcal{}(x)$ from the known condition $\delta _\mathcal{AT}\mathcal{}(x)=\delta _\mathcal {A}(x)$, and this indicates $\dfrac{\sum \nolimits _{i=1}^m \omega _i \mu ^{{AT}_i} _{D_j}(x)}{\sum \nolimits _{i=1}^m \omega _i}>0$. Therefore, there are $\dfrac{\sum \nolimits _{i=1}^m \omega _i \mu ^{{AT}_i} _{D_j}(x)}{\sum \nolimits _{i=1}^m \omega _i}\ge \beta _\mathcal{AT}\mathcal{}\ge \beta _0>\beta$ and $x\in \overline{\mathcal{AT}\mathcal{}}_\beta (D_j)$. Thus $\overline{\mathcal {A}}_\beta (D_j)\subseteq \overline{\mathcal{AT}\mathcal{}}_\beta (D_j)$. It can be proved that $\overline{\mathcal{AT}\mathcal{}}_\beta (D_j)\subseteq \overline{\mathcal {A}}_\beta (D_j)$ in the same way. Consequently, $\overline{\mathcal{AT}\mathcal{}}_\beta (D_j)=\overline{\mathcal {A}}_\beta (D_j)$, i.e., $\mathcal {A}$ is a $\beta$-upper distribution consistent set.

(2) can be easily proved by (1).