Algebraic approaches to granular computing

Abstract

We study the granular structures in granular computing from algebraic views. We model a granular structure based on an algebra that consists of a universe and a closure operator. Based on this formulation, we define the basic granules in a granular structure through the notion of subalgebras. Using formal concept analysis and rough set analysis as two examples, we demonstrate that the proposed formulation can unify a number of current studies on the related areas of granular computing. Moreover, by modelling the granular structures in rough sets with algebraic approaches, we generalize the notion of approximations in rough sets into a general granular structure. The algebraic approaches provide a mathematical formulation of granular structures, which may assist us in the construction and interpretation of granules and granular structures.

Appendices

Appendix

1. Proof of Theorem 2

Proof

we prove $X = \bigcup _{x \in X} \langle\{x\}\rangle$ by proving the following two facts:

• 1. $X \subseteq \bigcup _{x \in X} \langle\{x\}\rangle$.

  The proof is as follows:

  $$\begin{aligned} \forall \, x \in X&\Longrightarrow\{x\} \subseteq \langle\{x\}\rangle \qquad \qquad \qquad \qquad \,\,\,\, (<>\text { is a closure operator})\nonumber \\&\Longrightarrow X = \bigcup _{x \in X} \{x\} \subseteq \bigcup _{x \in X} \langle\{x\}\rangle. \end{aligned}$$

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• 2. $X \supseteq \bigcup _{x \in X} \langle\{x\}\rangle$.

  The proof is as follows:

  $$\begin{aligned} \forall \, x \in X&\Longrightarrow\{x\} \subseteq X\nonumber \\&\Longrightarrow\langle\{x\}\rangle \subseteq \langle X\rangle \qquad \qquad \qquad \qquad \,\, (<>\text { is a closure operator})\nonumber \\&\Longrightarrow\langle\{x\}\rangle \subseteq X \qquad \qquad \qquad \qquad \qquad \qquad \qquad \,\, (\langle X\rangle = X)\nonumber \\&\Longrightarrow\bigcup _{x \in X} \langle\{x\}\rangle \subseteq X. \end{aligned}$$

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2. Proof of Theorem 3

Proof

Since $<>$ is a closure operator, we have $\langle\langle\{x\}\rangle\rangle = \langle\{x\}\rangle$, that is, $\langle\{x\}\rangle$ is a granule in $G_{<>}$. Now, we prove that $\{\langle\{x\}\rangle \mid x \in U\}$ is the family of basic granules by proving the two conditions in Definition 3.

• 1. Since $\{x\}$ is a smallest nonempty subset of U, the set $\langle\{x\}\rangle$ generated from $\{x\}$ must also be a smallest set that can be generated by applying $<>$ to nonempty subsets of U. Thus, there does not exist a granule in $G_{<>}$ which is a proper subset of $\langle\{x\}\rangle$. This means that $\langle\{x\}\rangle$ satisfies the first condition of a basic granule in Definition 3.

• 2. By Definition 2, one can easily verify that the set union $\cup$ is a composition operator. By the second property in Theorem 2, any granule $X \in G_{<>}$ can be computed as the union of the set $\{\langle\{x\}\rangle \mid x \in X\} \subseteq \{\langle\{x\}\rangle \mid x \in U\}$. Thus, the family $\{\langle\{x\}\rangle \mid x \in U\}$ satisfies the second condition in Definition 3.

3. Proof of Theorem 9

Proof

Since $cl(\{x\})$, where $x \in U$ is a minimal nonempty closed set, it must also be a minimal nonempty granule in G. By Definition 3, we need to prove that any nonempty granule in G can be constructed from a subset of $\{cl(\{x\}) \mid x \in U\}$. In particular, we prove that any granule $cl(X) \in G$ can be expressed as $\bigcup _{x \in X} cl(\{x\})$. We first prove that $cl(X) = cl(\bigcup _{x \in X} cl(\{x\}))$ by the following two set-inclusion relations.

• 1. $cl(X) \subseteq cl(\bigcup _{x \in X} cl(\{x\}))$.

  By property (1) in Eq. (2), we have $\{x\} \subseteq cl(\{x\})$ for every $x \in X$. Thus, we get $\bigcup _{x \in X} \{x\} \subseteq \bigcup _{x \in X} cl(\{x\})$, that is, $X \subseteq \bigcup _{x \in X} cl(\{x\})$. By property (2) in Eq. (2), we get $cl(X) \subseteq cl(\bigcup _{x \in X} cl(\{x\}))$.

• 2. $cl(X) \supseteq cl(\bigcup _{x \in X} cl(\{x\}))$.

  By property (2) in Eq. (2), we have $cl(\{x\}) \subseteq cl(X)$ for every $x \in X$, since $\{x\} \subseteq X$. Thus, we get $\bigcup _{x \in X} cl(\{x\}) \subseteq cl(X)$. By property (2) in Eq. (2), we then have $cl(\bigcup _{x \in X} cl(\{x\})) \subseteq cl(cl(X))$. Since $cl(X) \in G$, we have $cl(cl(X)) = cl(X)$. Thus, we get $cl(\bigcup _{x \in X} cl(\{x\})) \subseteq cl(X)$.

Since cl is closed for set union, we have

$$\begin{aligned} cl(X) = cl\left( \bigcup _{x \in X} cl(\{x\})\right) = \bigcup _{x \in X} cl(cl(\{x\})) = \bigcup _{x \in X} cl(\{x\}). \end{aligned}$$

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