## 2 Background and Related Work
### Number of Interrelated Attributes  
In this subsection, we discuss the number of interrelated attributes constructed in the proposed approach. The proposed approach has an assumption: the interrelated attributes constructed from \(AT\) depend on the order of attributes. For any two attributes \(a_{i},a_{j}\in AT\) and any binary relation \(R\in\big{\{}<_{le},=_{eq},>_{mo}\big{\}}\), the index \(i\) of \(a_{i}\) must be smaller than the index \(j\) of \(a_{j}\) in order to be able to construct the interrelated attributes \(a_{i}Ra_{j}\). Hence, if we exchange the order of attributes \(a_{i}\) and \(a_{j}\), interrelated attributes \(a_{i}Ra_{i+1}\),..., \(a_{i}Ra_{j}\) and \(a_{i+1}Ra_{j}\),..., \(a_{j-1}Ra_{j}\) are lost and \(a_{j}Ra_{i+1}\),..., \(a_{j}Ra_{j-1}\) and \(a_{j}Ra_{i}\) and \(a_{i+1}Ra_{i}\),..., \(a_{j-1}Ra_{i}\) are newly obtained. Then, even though the order of the attributes \(a_{i}\) and \(a_{j}\) is exchanged, the number of interrelated attributes that have the value \(1\) does not change. It is based on the following property. **Lemma 1**. : _Let \(a_{i},a_{j}\in AT\) be any two attributes in which \(i<j\) holds, and \(i_{x}\in I\) be an arbitrary item. If the order of the attributes \(a_{i}\) and \(a_{j}\) is changed, there is no change in the number of interrelated attributes that the item \(i_{x}\) owns._  
Proof. : Let \(AT=\{a_{1},\dots,a_{l}\}\) be a set of attributes that comprises \(l\) attribute. Suppose the order of the attributes \(a_{i}\) and \(a_{j}\) is changed, i.e., the order \(a_{1}\),..., \(a_{i-1}\), \(a_{i}\), \(a_{i+1}\),..., \(a_{j-1}\), \(a_{j}\), \(a_{j+1}\),..., \(a_{l}\) is changed to \(a_{1}\),..., \(a_{i-1}\), \(a_{j}\), \(a_{i+1}\),..., \(a_{j-1}\), \(a_{i}\), \(a_{j+1}\),..., \(a_{l}\). It is obvious that there is no influence of change between \(a_{i}\) and \(a_{j}\) in the case of \(\rho(x,a_{i})=\rho\Big{(}x,a_{j}\Big{)}=1\) or \(\rho(x,a_{i})=\rho\Big{(}x,a_{j}\Big{)}=0\). Further, suppose \(\rho(x,a_{i})=1\)and \(\rho\Big{(}x,a_{i}\Big{)}=0\) holds. Then, before changing the order of \(a_{i}\)_and_\(a_{j}\), it implies that for any attribute \(a_{k}(i<k<j)\), the following three statements can easily be confirmed:  
\[ \begin{array}{lcl}\rho_{int}\Big{(}x,a_{i}=_{eq}a_{k}\Big{)}=1&iff&\rho(x,a_{k })=1,\\ \rho_{int}(x,a_{i}>_{mo}a_{k})=1&iff&\rho(x,a_{k})=0,\\ \rho_{int}\Big{(}x,a_{k}>_{mo}a_{j}\Big{)}=1&iff&\rho(x,a_{k})=1.\end{array} \]  
Hence, if \(\rho(x,a_{k})=1\) holds, item \(i_{x}\) owns two interrelated attributes \(a_{i}=_{eq}a_{k}\)_and_\(a_{k}>_{mo}a_{j}\). Otherwise, item \(i_{x}\) owns one interrelated attributes \(a_{i}>_{mo}a_{k}\). After changing the order of \(a_{i}\) and \(a_{j}\), the following three expressions are obtained:  
\[ \begin{array}{lcl}\rho_{int}\Big{(}x,a_{k}=_{eq}a_{i}\Big{)}=1&iff&\rho(x,a_ {k})=1,\\ \rho_{int}\big{(}x,a_{k}<_{le}a_{i}\big{)}=1&iff&\rho(x,a_{k})=0,\\ \rho_{int}\Big{(}x,a_{j}<_{le}a_{k}\big{)}=1&iff&\rho(x,a_{k})=1.\end{array} \]  
Therefore, by changing the order of \(a_{i}\) and \(a_{j}\), if \(\rho(x,a_{k})=1\) holds, the two interrelated attributes \(a_{i}=_{eq}a_{k}\) and \(a_{k}>_{mo}a_{j}\) are lost, however, item \(i_{x}\) owns two the newly obtained interrelated attributes \(a_{k}=_{eq}a_{i}\) and \(a_{j}<_{le}a_{k}\). Otherwise, if \(\rho(x,a_{k})=0\) holds, the interrelated attribute \(a_{i}>_{mo}a_{k}\) is lost, and item \(i_{x}\) owns a new interrelated attribute \(a_{k}<_{le}a_{i}\). This implies that changing the order of \(a_{i}\) and \(a_{j}\) does not change the number of interrelated attributes that have a value of \(1\) in the case of \(\rho(x,a_{i})=1\) and \(\rho\Big{(}x,a_{j}\Big{)}=0\). Similarly, another case of \(\rho(x,a_{i})=0\) and \(\rho\Big{(}x,a_{j}\Big{)}=1\) can be proven. Based on this lemma, the order of attributes can be freely changed without changing the number of interrelated attributes from the original situation. So, without loss of generality, the following situation can be assumed; let \(i_{x}\in I\) be an item.