Multi-granular soft rough covering sets

Abstract

This paper presents a novel model that combines several interesting features in relation with rough sets, namely multi-granularity (which extends Pawlak’s single-granular approach), soft rough sets (where the granulation structure is defined by soft sets), and coverings that induce rough sets. Optimistic and pessimistic multi-granular models are the outcome of this hybridization. Their properties, relationships, and links with existing models are thoroughly explored. Finally, an application of the model to multi-criteria group decision making is put forward. Examples and graphical discussions illustrate the performance of this criterion.

Appendices

Appendix I: Examples and graphical discussion

Example I.1 illustrates Definitions 4.1 and 4.3. It is inspired by the application in (Li et al. 2015, Section 6). Our presentation does not follow (Li et al. 2015) closely, because the goal of Li et al. is parameter reduction (and they are not concerned with decision making). In addition, the practical situation in (Li et al. 2015, Section 6) has been streamlined to emphasize the computations.

Example I.1

Let $U=\{x_1, \ldots , x_5\}$ be a set of pilots. They are evaluated by two experts who assess whether they are sufficiently well trained with respect to four attributes $A=\{a_1, a_2, a_3, a_4\}$. The respective evaluations are given in Table 1.

Table 1 Evaluation of the pilots by the first and second experts in Example I.1

The information in Table 1 naturally produces respective SCASs, namely $S_1=(U, C_{{\mathbf {G}_1}})$ and $S_2=(U, C_{{\mathbf {G}_2}})$. In formal terms, we can define ${\mathbf {G}_1}=(F_1,A)$, ${\mathbf {G}_2}=(F_2,A)$ where:

$$\begin{aligned} F_1(a_1)= & {} F_1(a_4)= \{x_1, x_2, x_4\}, \\ F_1(a_2)= & {} \{x_1, x_3\}, F_1(a_3)= \{x_5\}, \end{aligned}$$

and

$$\begin{aligned} F_2(a_1)= & {} \{x_1, x_2, x_4\}, F_2(a_2) = \{x_2, x_4\}, \\ F_2(a_3)= & {} \{x_3, x_5\}, F_2(a_4)= \{x_1, x_4, x_5\}. \end{aligned}$$

For $X=\{x_1, x_2, x_3, x_4\}$ we compute SCOMLA and SCOMUA by a routine application of Definition 4.1 as follows:

$$\begin{aligned} FS_{\sum {S_i}}^0(X) = \cup \{ F_i(a)\vert \ a\in A, F_i(a)\subseteq X \text { for some } i\} = X, \end{aligned}$$

and

$$\begin{aligned} FS_0^{\sum {S_i}}(X)= & {} \cap _{i=1,2,3,4} \left( \cup \{F_i(a)\vert \ X\cap F_i(a)\ne \varnothing , a\in A \} \right) \\= & {} (\cup _{a\in A} \{F_1(a):X\cap F_1(a)\ne \varnothing \})\\&\cap (\cup _{a\in A}\{F_2(a):X\cap F_1(a)\ne \varnothing \})=X\cap U=X \end{aligned}$$

Therefore X is not MGOSRC, because it is multi-granular optimistic SC-definable.

However, X is MGPSRC, because a routine application of Definition 4.3 shows

$$\begin{aligned} FS_{\sum {S_i}}^P(X)= & {} \cup _{a\in A} \left( \cap _{i=1,2,3,4} \{ F_i(a)\vert \, F_i(a)\subseteq X \}\right) \\= & {} (F_1(a_1)\cap F_2(a_1)) \cup (F_1(a_2)\cap F_2(a_2)) \\&\cup \varnothing \cup F_1(a_4) = \{x_1, x_2, x_4\} \\&\cup \varnothing \cup \{x_1, x_2, x_4\} \subsetneq X. \end{aligned}$$

Table 2 Computations and recommended solutions in Example I.2

Our next example illustrates the application of Algorithm 1 to the decision making problem posed by Example I.1:

Example I.2

Consider the situation of Example I.1. Let us apply the decision making mechanism explained in Sect. 5 in order to select one of the five pilots.

Some direct computations produce the choice values derived from the respective single-granulations, namely $c_1(x_1)=3$, $c_1(x_2)=2$, $c_1(x_3)=1$, $c_1(x_4)=2$, $c_1(x_5)=1$, for the first case and $c_2(x_1)=2$, $c_2(x_2)=2$, $c_2(x_3)=1$, $c_2(x_4)=3$, $c_2(x_5)=2$ for the second.

The expression of the pessimistic choice values does not depend upon the weights that we may attach to the experts. The pessimistic choice values produce the following figures by Equation (12):

$$\begin{aligned} A(x_1)=2, A(x_2)=1, A(x_3)=0, A(x_4)=2, A(x_5)=1. \end{aligned}$$

In order to compute optimistic assessments, the weights of the experts are necessary. Suppose first that the opinions of both experts are equally valuable. We represent this by fixing $\omega =(\frac{1}{2}, \frac{1}{2})$. Then, the $\omega$-optimistic choice values are computed by Equation (11) and produce the following figures:

$$\begin{aligned} B^{\omega }(x_1)= & {} B^{\omega }(x_4)= 2.5, B^{\omega }(x_2)= 2,\\ B^{\omega }(x_3)= & {} 1, \; and \; B^{\omega }(x_5)= 1.5. \end{aligned}$$

If $\omega =(\frac{3}{4}, \frac{1}{4})$ instead, so that the first opinion is 3 times as important as the second opinion, then the $\omega$-optimistic choice values are

$$\begin{aligned} B^{\omega }(x_1)= & {} 2.75, B^{\omega }(x_2)= 2, B^{\omega }(x_3)= 1,\\ B^{\omega }(x_4)= & {} 2.25, \; and \; B^{\omega }(x_5)= 1.25. \end{aligned}$$

To conclude the analysis, let us now describe the key facts when we decide to fix $\lambda =\frac{1}{4}$. This selection places 3 times more importance to the optimistic than the pessimistic evaluation.

If $\omega =(\frac{1}{2}, \frac{1}{2})$ then $Q_{\lambda }^{\omega }(x_i) = \lambda A(x_i) + (1-\lambda ) B^{\omega }(x_i)= \frac{1}{4} A(x_i) + \frac{3}{4} B^{\omega }(x_i)$ produces the following figures:

$$\begin{aligned} Q_{\lambda }^{\omega }(x_1)= & {} Q_{\lambda }^{\omega }(x_4)= 2.375, Q_{\lambda }^{\omega }(x_2)= 1.75, \\ Q_{\lambda }^{\omega }(x_3)= & {} 0.75, Q_{\lambda }^{\omega }(x_5)= 1.375. \end{aligned}$$

Henceforth, the recommended choice when $\lambda =\frac{1}{4}$ and $\omega =(\frac{1}{2}, \frac{1}{2})$ is either $x_1$ or $x_4$.

If $\omega =(\frac{3}{4}, \frac{1}{4})$ then $Q_{\lambda }^{\omega }(x_i) = \frac{1}{4} A(x_i) + \frac{3}{4} B^{\omega }(x_i)$ produces the following figures:

$$\begin{aligned} Q_{\lambda }^{\omega }(x_1)= & {} 2.5625, Q_{\lambda }^{\omega }(x_2)= 1.75, Q_{\lambda }^{\omega }(x_3)= 0.75,\\ Q_{\lambda }^{\omega }(x_4)= & {} 2.1875, Q_{\lambda }^{\omega }(x_5)= 1.1875. \end{aligned}$$

Henceforth, the recommended choice when $\lambda =\frac{1}{4}$ and $\omega =(\frac{3}{4}, \frac{1}{4})$ is $x_1$.

Table 2 summarizes the elements that produce our recommendations, as a function of the selected parameters.

Observe that we can also make a graphical exploration to know whether and to what extent the recommendations vary when we change the parameter $\lambda$. Consider the case where we have fixed $\omega =(\frac{1}{2}, \frac{1}{2})$. Then, Fig. 1 represents the values of each $Q_{\lambda }^{\omega }(x_i)$ as a function of $\lambda$, namely

$$\begin{aligned}&Q_{\lambda }^{\omega }(x_1) = Q_{\lambda }^{\omega }(x_4) = 2 \lambda + 2.5 (1-\lambda ),\\&Q_{\lambda }^{\omega }(x_2) = \lambda + 2 (1-\lambda ),\\&Q_{\lambda }^{\omega }(x_3) = 1-\lambda , \; and \\&Q_{\lambda }^{\omega }(x_5) = \lambda + 1.5 (1-\lambda ). \end{aligned}$$

As explained in Sect. 5, these values produce segments because $\lambda \in [0,1]$. We can see that except when $\lambda =1$, no matter how optimistic we are the ranking is consistently the same: pilots $x_1$ and $x_4$ are tied at the top, second is $x_2$, afterwards $x_5$ and finally $x_3$. The only difference appears when $\lambda =1$, i.e., we are totally pessimistic. Then, there is another tie at positions 3-4 (pilots $x_2$ and $x_5$).

Fig. 1

A graphical depiction of the evaluations when $\omega =(\frac{1}{2}, \frac{1}{2})$ as a function of $\lambda$. The vertical axis represents the corresponding value of $Q_{\lambda }^{\omega }(x_i)$

Figure 2 represents the values of the respective $Q_{\lambda }^{\omega }(x_i)$ as a function of $\lambda$ when $\omega =(\frac{3}{4}, \frac{1}{4})$. As in the previous case, the ranking is consistently the same when $\lambda < 1$: pilot $x_1$ is first, $x_4$ is second, third is $x_2$, afterwards $x_5$ and $x_3$ is last. But when $\lambda =1$, i.e., we assume the pessimistic evaluations and discard the optimistic figures, the ranking is the same as in the case $\omega =(\frac{1}{2}, \frac{1}{2})$: there are ties at positions 1-2 (pilots $x_1$ and $x_4$) and also at positions 3-4 (pilots $x_2$ and $x_5$).

The latter observation is consistent with the fact that when $\lambda = 1$, we only use the pessimistic evaluations which are independent of $\omega$.

Fig. 2

A graphical depiction of the evaluations when $\omega =(\frac{3}{4}, \frac{1}{4})$ as a function of $\lambda$. The vertical axis represents the corresponding value of $Q_{\lambda }^{\omega }(x_i)$

Fig. 3

A graphical depiction of the evaluations when $\lambda = \frac{1}{3}$ as a function of $\omega =(\omega _1, 1-\omega _1)$. The vertical axis represents the corresponding value of $Q_{\lambda }^{\omega }(x_i)$

Remark I.3

The examples in this “Appendix” have a feature that allows us to picture the influence of the relative weights of the granulations (for a fixed risk factor $\lambda$): the corresponding values of $Q_{\lambda }^{\omega }(x_i)$ produce segments too. The key fact is that $\omega$ can be characterized by its first component $\omega _1$ because $\omega = (\omega _1, 1-\omega _1).$ We see this in Fig. 3: when we fix $\lambda = \frac{1}{3}$, this figure represents the values of each $Q_{\lambda }^{\omega }(x_i)$ as a function of $\omega _1$, which is the weight of the first opinion.

For illustration, $Q_{\lambda }^{\omega }(x_1) = \frac{1}{3}\cdot 2 + \frac{2}{3} (3\cdot \omega _1 + 2 (1 - \omega _1))= \frac{2}{3}(3 + \omega _1)$. Similar computations produce $Q_{\lambda }^{\omega }(x_2) = \frac{5}{3}$, $Q_{\lambda }^{\omega }(x_3) = \frac{2}{3}$, $Q_{\lambda }^{\omega }(x_4) = \frac{2}{3}(4- \omega _1)$, and $Q_{\lambda }^{\omega }(x_5) = \frac{1}{3}(5 - 2 \omega _1)$.

Some conclusions may be drawn from the representation captured by Fig. 3. When $\lambda = \frac{1}{3}$, pilot $x_4$ should be selected if the first opinion weighs less than the second one. This is because $Q_{\lambda }^{\omega }(x_1)$ and $Q_{\lambda }^{\omega }(x_4)$ intersect at $\omega _1 = 0.5$. If both opinions are equally valuable (i.e., $\omega _1 = 0.5$), pilots $x_1$ and $x_4$ are deemed equally skilled, and both are strictly better than the other three pilots. Pilot $x_1$ is recommended when the first opinion weighs more than the second one.

Appendix II: Proofs

In this section, we prove the results stated in previous sections.

Proof of Theorem 4.2

Statement (1) is immediate, and it crucially depends on the fact that we are using SCASs, i.e., that the soft sets are full. It is also immediate that (2) holds true.

In order to prove (3), observe that the first inclusion is obvious. The fact that each $S_i$ is a SCAS will be helpful to prove $X \subseteq FS^{\sum {S_i}}_0(X)$. This fact implies that each $(F_i, A)$ is a full soft set. Pick any $x\in X$, then for each $i=1, \ldots , m$ there must exist $a_i\in A$ such that $x\in F_i(a_i)$. In particular, $X\cap F_i(a_i)\ne \varnothing$.

It is now clear that $x\in \cap _{i\in I} ( \cup \{F_i(a)\vert \ X\cap F_i(a)\ne \varnothing , a\in A \} )$ because for each $i=1, \ldots , m$, $x\in F_i(a_i) \subseteq \cup \{F_i(a)\vert \ X\cap F_i(a)\ne \varnothing , a\in A \}$.

Let us prove (4). In order to check that $FS_{\sum {S_i}}^0$ is monotonic, observe that when $X\subseteq Y\subseteq U$, the property $F_i(a)\subseteq X$ implies $F_i(a)\subseteq Y$. Hence,

$$\begin{aligned}&\cup \{ F_i(a)\vert \ a\in A, F_i(a)\subseteq X \text { for some } i\} \\&\quad \subseteq \cup \{ F_i(a)\vert \ a\in A, F_i(a)\subseteq Y \text { for some } i\}. \end{aligned}$$

To check that $FS^{\sum {S_i}}_0$ is monotonic, fix $X\subseteq Y\subseteq U$ and $i\in I=\{1, \ldots , m\}$. If a is such that $X\cap F_i(a)\ne \varnothing$, clearly $Y\cap F_i(a)\ne \varnothing$ holds too. Therefore for each $i\in I$,

$$\begin{aligned}&\cup \{F_i(a)\vert \ X\cap F_i(a)\ne \varnothing , a\in A \}\\&\quad \subseteq \cup \{F_i(a)\vert \ Y\cap F_i(a)\ne \varnothing , a\in A \} \end{aligned}$$

and now the conclusion is obvious by taking intersections.

In order to prove (5), observe first that $FS_{\sum {S_i}}^0(X) \subseteq X$ holds true, which in turn yields $FS_{\sum {S_i}}^0(FS_{\sum {S_i}}^0(X))\subseteq FS_{\sum {S_i}}^0(X)$ by a combination of (3) and (4). In order to prove the inclusion $FS_{\sum {S_i}}^0(X)\subseteq FS_{\sum {S_i}}^0(FS_{\sum {S_i}}^0(X))$, notice that by construction, when $x\in F_i(a)\subseteq X$ it must always be the case that $F_i(a)\subseteq FS_{\sum {S_i}}^0(X)$, for any $i\in I$ and $a\in A$. Therefore $x\in FS_{\sum {S_i}}^0(X)$ implies that there is $x\in F_i(a)\subseteq FS_{\sum {S_i}}^0(X)$, for some $i\in I$ and $a\in A$. Hence, $x\in FS_{\sum {S_i}}^0(X)$ implies $x\in FS_{\sum {S_i}}^0(FS_{\sum {S_i}}^0(X))$.

Regarding (6), observe that the fact $FS^{\sum {S_i}}_0 (X) \subseteq FS^{\sum {S_i}}_0(FS^{\sum {S_i}}_0(X))$ is again a consequence of statements (3) and (4).

Claim (7) holds by an argument in the proof of (5). When $x\in F_i(a)$ it must always be the case that $F_i(a)\subseteq FS_{\sum {S_i}}^0(F_i(a))$. And $FS_{\sum {S_i}}^0(F_i(a))\subseteq F_i(a)$ holds by (3).

In order to prove (8), first let us select $x\in \sim FS_{\sum {S_i}}^0(X)$. By the construction of $FS_{\sum {S_i}}^0$, for each $i\in I$ and $a\in A$, it must be the case that $F_i(a)\cap (\sim X) \ne \varnothing$. Therefore x verifies the condition for belonging to $FS^{\sum {S_i}}_0(\sim X)$.

Secondly, let us select $x\in \sim FS^{\sum {S_i}}_0(X)$. By the construction of $FS^{\sum {S_i}}_0$, there must be $i\in I$ such that $F_i(a)\cap X = \varnothing$ for each $a\in A$. Equivalently, there is $i\in I$ such that $F_i(a)\subseteq \sim X$ for each $a\in A$. This fact ensures $x \in FS_{\sum {S_i}}^0(\sim X)$. $\square$

Proof of Theorem 4.4

It is immediate that statements (1) and (2) hold true.

In order to prove (3), observe that the first inclusion is obvious. Proposition 4.5 and (3) in Theorem 4.2 justify the inclusion $X \subseteq FS^{\sum {S_i}}_P(X)$.

Let us prove (4). In order to check that $FS_{\sum {S_i}}^P$ is anti-monotonic, observe that when $X\subseteq Y\subseteq U$ and $x\in FS_{\sum {S_i}}^P(Y)$, there must be a with the property that $x\in F_i(a)$ whenever $F_i(a)\subseteq Y$. Hence, a verifies the property that $x\in F_i(a)$ whenever $F_i(a)\subseteq X$, which guarantees $x\in FS_{\sum {S_i}}^P(X)$.

To check that $FS^{\sum {S_i}}_P$ is monotonic, fix $X\subseteq Y\subseteq U$ and observe that $x\in FS^{\sum {S_i}}_P(X)$ entails the existence of $F_i(a)$ such that $x\in F_i(a)$ and $X\cap F_i(a)\ne \varnothing$. Clearly $Y\cap F_i(a)\ne \varnothing$ holds too, thus $x\in FS^{\sum {S_i}}_P(Y)$.

In order to justify (5), observe that it is a routine application of the first claim in (4) to the inclusions given by (3), Proposition 4.5, and (3) in Theorem 4.2.

Let us prove (6). The inclusion $FS^{\sum {S_i}}_P(FS^{\sum {S_i}}_O(X)) \subseteq FS^{\sum {S_i}}_P (X)$ is fairly immediate because $x\in FS^{\sum {S_i}}_P(FS^{\sum {S_i}}_O(X))$ means that $x\in F_i(a)$ for some $F_i(a)$ with the property $F_i(a)\cap FS^{\sum {S_i}}_O(X)\ne \varnothing$. And because $X\subseteq FS^{\sum {S_i}}_O(X)$, $F_i(a)$ verifies $F_i(a)\cap X\ne \varnothing$ thus ensuring $x\in FS^{\sum {S_i}}_P (X)$. The other inclusions in (6) are a routine application of the second claim in (4) to the inclusions given by (3), Proposition 4.5, and (3) in Theorem 4.2. $\square$