A Review on Rough Set-Based Interrelationship Mining

Abstract

Interrelationship mining, proposed by the authors, aims at extracting characteristics of objects based on interrelationships between attributes. Interrelationship mining is an extension of rough set-based data mining, which enables us to extract characteristics based on comparison of values of two different attributes such that “the value of attribute a is higher than the value of attribute b.” In this paper, we mainly review theoretical aspects of rough set-based interrelationship mining.

Proofs of Theoretical Properties

Proposition 1

Let S be a decision table, $\mathsf{a},\mathsf{b}\in C$ be condition attributes in S, $R\subseteq V_{\mathsf{a}}\times V_{\mathsf{b}}$ be a binary relation, and $S_{int}$ be an information table $S_{int}$ for interrelationship mining with respect to S such that $R\in \{R_{\mathsf{a}\times \mathsf{b}}\}$. The following equality holds:

$$ IND_{S}(\mathsf{a}R \mathsf{b})=IND_{S_{int}}(\{\mathsf{a}\mathsf{R}\mathsf{b}\}), $$

where $IND_{S}(\mathsf{a}R \mathsf{b})$ is the indiscernibility relation for S defined by (12), and $IND_{S_{int}}(\{\mathsf{a}\mathsf{R}\mathsf{b}\})$ is the indiscernibility relation for $S_{int}$ by a singleton $\{\mathsf{a}\mathsf{R}\mathsf{b}\}$ of an interrelated condition attribute defined by (2).

Proof

Suppose $(x,y)\in IND_{S}(\mathsf{a}R \mathsf{b})$ holds. By the definition of $IND_{S}(\mathsf{a}R \mathsf{b})$ by (12), $x\in R(\mathsf{a},\mathsf{b})$ holds if and only if $y\in R(\mathsf{a},\mathsf{b})$ holds. Therefore, for the interrelated condition attribute $\mathsf{a}\mathsf{R}\mathsf{b}$, it implies that either $\rho (x,\mathsf{a}\mathsf{R}\mathsf{b})=\rho (y,\mathsf{a}\mathsf{R}\mathsf{b})=1$ or $\rho (x,\mathsf{a}\mathsf{R}\mathsf{b})=\rho (y,\mathsf{a}\mathsf{R}\mathsf{b})=0$ holds, which concludes $(x,y)\in IND_{S_{int}}(\{\mathsf{a}\mathsf{R}\mathsf{b}\})$. The converse is also proved similarly.

Proposition 2

Let S be a decision table and $S_{int}$ be an information table for interrelationship mining with respect to S. The following equality holds:

$$ IND_{S_{int}}(AT)=IND_{S_{int}}(AT_{int}), $$

where $IND_{S_{int}}(AT)$ is the indiscernibility relation for $S_{int}$ by using all attributes in AT of the original decision table S, and $IND_{S_{int}}(AT_{int})$ is the indiscernibility relation for $S_{int}$ by using all attributes of $S_{int}$ including interrelated condition attributes.

Proof

Because $AT\subseteq AT_{int}$ by the definition of the information table $S_{int}$ for interrelationship mining, $IND_{S_{int}}(AT)\supseteq IND_{S_{int}}(AT_{int})$ holds trivially. We show the converse set inclusion. Suppose $(x,y)\in IND_{S_{int}}(AT)$ holds. For any two attributes $\mathsf{a},\mathsf{b}\in AT$ and any binary relation $R\in \{R_{\mathsf{a}\times \mathsf{b}}\}$ such that $R(\mathsf{a},\mathsf{b})\ne \emptyset$, the assumption $(x,y)\in IND_{S_{int}}(AT)$ implies that $\rho (x,\mathsf{a})=\rho (y,\mathsf{a})$ and $\rho (x,\mathsf{b})=\rho (y,\mathsf{b})$ hold. This implies that $(\rho (x,\mathsf{a}),\rho (x,\mathsf{b}))\in R$ holds if and only if $(\rho (y,\mathsf{a}),\rho (y,\mathsf{b}))\in R$ holds, i.e., $x\in R(\mathsf{a},\mathsf{b})$ holds if and only if $y\in R(\mathsf{a},\mathsf{b})$ holds. By the definition of attribute value assignment for interrelated attributes by (16), it concludes that $\rho (x,\mathsf{a}\mathsf{R}\mathsf{b})=\rho (y,\mathsf{a}\mathsf{R}\mathsf{b})$ holds. Therefore, x is indiscernible from y by any interrelated attribute $\mathsf{a}\mathsf{R}\mathsf{b}$ and $(x,y)\in IND_{S_{int}}(AT_{int})$ holds. This concludes the equality $IND_{S_{int}}(AT)=IND_{S_{int}}(AT_{int}).$

Corollary 1

Let S be a decision table and $S_{int}$ be an information table for interrelationship mining with respect to S. For every decision class $D\in \mathcal{D}$ of $S_{int}$, The following equality holds:

$$ \underline{C}(D)=\underline{C_{int}}(D), $$

where C is the set of all condition attributes of S and $C_{int}=AT_{int}\setminus \{\mathsf{d}\}$ is the set of all condition attributes of $S_{int}$ including interrelated condition attributes.

Proof

From the definitions of the decision table S, $C=AT\setminus \{\mathsf{d}\}$ holds. By Proposition 2, it is easily confirmed that $IND_{S_{int}}(C)=IND_{S_{int}}(C_{int})$ holds, which concludes $\underline{C}(D)=\underline{C_{int}}(D)$ for every decision class $D\in \mathcal{D}$ of $S_{int}$.

Proposition 3

([7]) Let $S_{int}$ be a decision table for interrelationship mining. If there exists a relative reduct $A\subseteq AT_{int}$ of $S_{int}$ that contains an interrelated attribute $\mathsf{aRb}$, then there exists at least one relative reduct $B\subseteq AT_{int}$ such that either $\mathsf{a}\in B$ or $\mathsf{b}\in B$ holds.

Proof

([7]) Suppose that the interrelated attribute $\mathsf{aRb}$ by a binary relation R on $V_{\mathsf{a}}\times V_{\mathsf{b}}$ appears in a relative reduct A. From the definition of relative reducts, there exist two elements $x,y\in U$ such that x and y are discernible each other by A, however, x and y are not discernible by $A'\mathop {=}\limits ^\mathrm{def}A\setminus \{\mathsf{aRb}\}$. This means that the values of x and y at $\mathsf{aRb}$ are different each other, and without losing generality, we assume that $\rho _{int}(x,\mathsf{aRb})=1$ and $\rho _{int}(y,\mathsf{aRb})=0$. This assumption means $(\rho _{int}(x,\mathsf{a}), \rho _{int}(x,\mathsf{b}))\in R$ and $(\rho _{int}(y,\mathsf{a}), \rho _{int}(y,\mathsf{b}))\not \in R$ hold, respectively, which implies that either the values of x and y at $\mathsf{a}$ are different or the values of x and y at $\mathsf{b}$ are different. Again, without loosing generality, we assume that the values of x and y at $\mathsf{a}$ are different. By this assumption, the subset $A'\cup \{\mathsf{a}\}$ satisfies the condition 1) of relative reducts; the attribute $\mathsf{a}$ can discern x from y and the subset $A'$ can discern other elements that belong to different decision class each other. Moreover, the set $B\subseteq A'\cup \{\mathsf{a}\}$ with no redundant attributes is a relative reduct and it is easily confirmed that $\mathsf{a}\in B$. It concludes the proof.

Corollary 2

([7]) Let S be a decision table, and $\mathsf{a}$ and $\mathsf{b}$ be two condition attributes of S that do not appear in any relative reduct of S. Then, for any binary relation R on $V_{\mathsf{a}}\times V_{\mathsf{b}}$ and the interrelated attribute $\mathsf{a}\mathsf{R}\mathsf{b}$ by the binary relation R, the interrelated attribute $\mathsf{a}\mathsf{R}\mathsf{b}$ does not appear in any relative reduct of the decision table $S_{int}$ that are induced from the table S.

Proof

It is obvious from Proposition 3.

Proposition 4

([9]) Let $B=\{\mathsf{a},\mathsf{b}\}$ be any set of two condition attributes in C, $\mathsf{a}\mathsf{R}\mathsf{b}\in AT_{int}$ be an interrelated attribute based on the attributes $\mathsf{a}, \mathsf{b}\in C$ and a binary relation $R\subseteq V_{\mathsf{a}}\times V_{\mathsf{b}}$, and $B'\subseteq AT_{int}$ be a set of attributes generated by replacing either the attribute $\mathsf{a}$ or the attribute $\mathsf{b}$ in B by the interrelated attribute $\mathsf{a}\mathsf{R}\mathsf{b}$. The following equation holds:

$$\begin{aligned}{}[x]_B\subseteq [x]_{B'},\;\forall x\in U. \end{aligned}$$

(22)

Proof

Suppose that $y\in [x]_B$ holds and let $\rho (x,\mathsf{a})=v$ and $\rho (x,\mathsf{b})=w$ be the values of x at the attributes $\mathsf{a}$ and $\mathsf{b}$, respectively. Because $y\in [x]_B$, $\rho (y,\mathsf{a})=v$ and $\rho (y,\mathsf{b})=w$ also hold. In the binary relation $R\subseteq V_{\mathsf{a}}\times V_{\mathsf{b}}$ that is used for constructing the interrelated attribute $\mathsf{a}\mathsf{R}\mathsf{b}\in AT_{int}$, if $(v,w)\in R$ holds, it implies $\rho (x,\mathsf{a}\mathsf{R}\mathsf{b}) = \rho (y,\mathsf{a}\mathsf{R}\mathsf{b}) =1$; otherwise, if $(v,w)\not \in R$ holds, it implies $\rho (x,\mathsf{a}\mathsf{R}\mathsf{b}) = \rho (y,\mathsf{a}\mathsf{R}\mathsf{b}) =0$. Therefore, the object y is still indiscernible from x by replacing either $\mathsf{a}$ or $\mathsf{b}$ by $\mathsf{a}\mathsf{R}\mathsf{b}$, which concludes $y\in [x]_{B'}$.

Proposition 5

([9]) Suppose that $B\subseteq AT_{int}$ is a subset of condition attributes in the information table $S_{int}$ for interrelationship mining. For any two condition attributes $\mathsf{a},\mathsf{b}\in B$, let a set $B'$ be either $B'=(B\setminus \{\mathsf{a}\})\cup \{\mathsf{a}\mathsf{R}\mathsf{b}\}$ or $B'=(B\setminus \{\mathsf{b}\})\cup \{\mathsf{a}\mathsf{R}\mathsf{b}\}$. then the following inequality holds:

$$ ACov(B)\le ACov(B'). $$

Proof

Let $N_B$ and $N_{B'}$ be the denominators of $ACov(B)$ and $ACov(B')$, respectively. By the definition of $ACov(\cdot )$ by (9), it is sufficient to show $N_{B'}\le N_B$. Let $U/IND(B)$ and $U/IND(B')$ be the sets of equivalence classes by the indiscernibility relations $IND(B)$ and $IND(B')$, respectively, and suppose that there are m equivalence classes in $U/IND(B')$, i.e, there are m objects $x_1,\ldots ,x_m\in U$ and $U/IND(B')=\{[x_1]_{B'},\ldots ,[x_m]_{B'}\}$. By Proposition 4, for each equivalence class $[x_i]_{B'}\in U/IND(B')$, there exist $k_i(\ge 1)$ equivalence classes $[y_{i_1}]_B,\ldots ,[y_{i_{k_i}}]_B\in U/IND(B)$ such that $[x_i]_{B'}=\bigcup _{j=i_1}^{i_{k_i}}[y_j]_B$ holds. Therefore, $U/IND(B)=\{[y_{1_1}]_{B},$ $\ldots ,[y_{1_{k_1}}]_{B},$ $\ldots ,[y_{m_1}]_{B},$ $\ldots ,[y_{m_{k_m}}]_{B}\}$. For each decision class $D\in \mathcal{D}$, it is clear that $D\cap [x_i]_{B'}\ne \emptyset$ holds if and only if there is at least one equivalence class $[y_{i_l}]_B$ such that $D\cap [y_{i_l}]_B\ne \emptyset$ holds. This fact implies that $|\{D\in \mathcal{D}\;|\;D\cap [x_i]_{B'}\ne \emptyset \}| \le \sum _{j=i_1}^{i_{k_i}}|\{D\in \mathcal{D}\;|\;D\cap [y_{j}]_B\ne \emptyset \}|$ holds, which concludes $N_{B'}\le N_B$.

Corollary 3

([9]) Let B and $B'$ in Proposition 5 be both relative reducts in the information table $S_{int}$ for interrelationship mining such that $ACov(B)\le ACov(B')$. The number of decision rules generated from $B'$ is smaller than (or at least equal to) the number of decision rules generated from B.

Proof

It is obvious from Proposition 5.