Multiview granular data analytics based on three-way concept analysis

Abstract

Multiview granular data analytics reflects various aspects of the knowledge embodied in data by multiple granular structures. It has been investigated in many topics closely related to granular computing, especially those researches involving formal concept analysis. Three-way concept analysis has demonstrated its usefulness for knowledge discovery in formal contexts, since it can extract positive information and negative information between objects and attributes simultaneously. Taking advantage of it, we propose a concrete model of multiview granular data analytics based on three-way concept analysis. Firstly, two hexagons of trisections in three-way concept analysis are presented. The hexagons reveal two different trisection forms in existing three-way concept lattice models. These models are then accordingly grouped into two classes, namely, orthopair-based weak tri-partition model and weak tri-covering model. Secondly, interval-set-based weak tri-partition model of three-way concept lattices is designed by reformulating the knowledge ordering of interval sets. More specifically, sufficiency-possibility three-way concept lattices and necessity-dual three-way concept lattices are defined on the basis of different combinations of modal-style operators. Finally, the transformation methods among various types of three-way concept lattices are explored by analyzing their relationships. Further interpretations of the hidden semantics in these relationships are also given in terms of trisection.

Appendix: Proofs

We just present the proofs of Proposition 4, Lemma 1, Theorems 2, 6, 8 and 10, and the other results can be proved in a similar way.

1.1 Proof of Theorem 2

Proof

We only need to prove statements (i) to (iii), since (iv) to (vi) can be proved dually. If X = Ø, then statements (i) and (ii) hold naturally based on the definition of possibility operator. Let X be a nonempty object subset, then we can know that X^◇≠Ø, $X^{\overline \diamond }\not =\O$. If there exists a x ∈ G such that X = {g ∈ G∣gI = xI}, then $X^{\diamondsuit }\cap X^{\overline \diamondsuit }=\O$ holds. Hence, we have that $X^{*}\cup X^{\overline *}$=$X^{\diamondsuit c}\cup X^{\overline \diamondsuit c}=(X^{\diamondsuit }\cap X^{\overline \diamondsuit })^{c}=M$. It follows that $X^{\diamondsuit }\cap (X^{*}\cup X^{\overline *})\not =\O$, $X^{\overline \diamondsuit }\cap (X^{*}\cup X^{\overline *})\not =\O$, $X^{\diamondsuit }\cup X^{\overline \diamondsuit }\cup (X^{*}\cup X^{\overline *})=M$. A complementary case is that there is no such element x, then we have that $X^{\diamondsuit }\cap X^{\overline \diamondsuit }\not =\O$. Thus, $X^{\diamondsuit }\cap X^{\overline \diamondsuit }\subseteq X^{\diamondsuit }$ and $X^{\diamondsuit }\cap X^{\overline \diamondsuit }\subseteq X^{\overline \diamondsuit }$ hold. It follows that $X^{\diamondsuit }\cup X^{\overline \diamondsuit }\cup (X^{*}\cup X^{\overline *})=X^{\diamondsuit }\cup X^{\overline \diamondsuit }\cup (X^{\diamondsuit }\cap X^{\overline \diamondsuit })^{c}=M$. Therefore, statements (i), (ii) and (iii) hold. □

1.2 Proof of Proposition 4

Proof

(i) For any X ∈ 2^G, $X^{\vartriangle \triangledown }=[X^{\Box }, X^{\#}]^{\triangledown }=X^{\Box \diamondsuit }\cup X^{\#\#}\subseteq X$ holds, since $X^{\Box \diamondsuit }\subseteq X$ and $X^{\#\#}\subseteq X$. We also have that $[\underline A, \overline A]^{\triangledown \vartriangle }=(\underline A^{\diamondsuit }\cup \overline A^{\#})^{\vartriangle }=[(\underline A^{\diamondsuit }\cup \overline A^{\#})^{\Box }, (\underline A^{\diamondsuit }\cup \overline A^{\#})^{\#}]$ for any interval set $[\underline A, \overline A]\in \mathbb {I}(2^{M})$. Based on Proposition 1, we can obtain that $\underline A^{\diamondsuit \Box }\cup \overline A^{\#\Box }\subseteq (\underline A^{\diamondsuit }\cup \overline A^{\#})^{\Box }$, and $(\underline A^{\diamondsuit }\cup \overline A^{\#})^{\#}\subseteq \underline A^{\diamondsuit \#}\cap \overline A^{\#\#}$. Since $\underline A\subseteq \underline A^{\diamondsuit \Box }$ and $\overline A^{\#\#}\subseteq \overline A$, then we have that $\underline A\subseteq (\underline A^{\diamondsuit }\cup \overline A^{\#})^{\Box }$ and $(\underline A^{\diamondsuit }\cup \overline A^{\#})^{\#}\subseteq \overline A$. It follows that $[\underline A, \overline A]\sqsubseteq [\underline A, \overline A]^{\triangledown \vartriangle }$.

(ii) Let X₁,X₂ be two object subsets satisfying $X_{1}\subseteq X_{2}$. Then, we have that $X_{1}^{\Box }\subseteq X_{2}^{\Box }$ and $X_{2}^{\#}\subseteq X_{1}^{\#}$. It follows that $X_{1}^{\vartriangle }\sqsubseteq X_{2}^{\vartriangle }$. For any two interval sets $[\underline {A_{1}},\overline {A_{1}}], [\underline {A_{2}},\overline {A_{2}}]$, $[\underline {A_{1}},\overline {A_{1}}]\sqsubseteq [\underline {A_{2}},\overline {A_{2}}]$ implies that $\underline {A_{1}}\subseteq \underline {A_{2}}\subseteq \overline {A_{2}}\subseteq \overline {A_{1}}$ or $[\underline {A_{2}},\overline {A_{2}}]=[\O ,\O ]$. If $[\underline {A_{2}},\overline {A_{2}}]=[\O ,\O ]$, then $[\underline {A_{1}},\overline {A_{1}}]^{\triangledown }\subseteq [\underline {A_{2}},\overline {A_{2}}]^{\triangledown }$ holds based on the definition of OE_ND-operators. If $\underline {A_{1}}\subseteq \underline {A_{2}}\subseteq \overline {A_{2}}\subseteq \overline {A_{1}}$ holds, then we have that $\underline {A_{1}}^{\diamondsuit }\subseteq \underline {A_{2}}^{\diamondsuit }$ and $\overline {A_{1}}^{\#}\subseteq \overline {A_{2}}^{\#}$. It follows that $[\underline {A_{1}},\overline {A_{1}}]^{\triangledown }=\underline {A_{1}}^{\diamondsuit }\cup \overline {A_{1}}^{\#} \subseteq \underline {A_{2}}^{\diamondsuit }\cup \overline {A_{2}}^{\#}=[\underline {A_{2}},\overline {A_{2}}]^{\triangledown }$.

(iii) For any X ∈ 2^G, to prove that $X^{\vartriangle \triangledown \vartriangle }=X^{\vartriangle }$, we need to prove $X^{\vartriangle \triangledown \vartriangle }\subseteq X^{\vartriangle }$ and $X^{\vartriangle \triangledown \vartriangle }\supseteq X^{\vartriangle }$ (that can be proved by (i) directly) hold simultaneously. From (i), we know that $X\supseteq X^{\vartriangle \triangledown }$, and we can also obtain $X^{\vartriangle }\supseteq X^{\vartriangle \triangledown \vartriangle }$ based on (ii). Thus, $X^{\vartriangle \triangledown \vartriangle }=X^{\vartriangle }$ holds. Similarly, we can prove that $[\underline A, \overline A]^{\triangledown }=[\underline {A}, \overline {A}]^{\triangledown \vartriangle \triangledown }$.

(iv) Let X₁,X₂ be two object subsets. We have that $(X_{1}\cap X_{2})^{\vartriangle }=[(X_{1}\cap X_{2})^{\Box },(X_{1}\cap X_{2})^{\#}]=[X_{1}^{\Box }\cap X_{2}^{\Box }, X_{1}^{\#}\cup X_{2}^{\#}]=[X_{1}^{\Box }, X_{1}^{\#}]\sqcap [X_{2}^{\Box }, X_{2}^{\#}]$=$X_{1}^{\vartriangle }\sqcap X_{2}^{\vartriangle }$. Supposing that $[\underline {A_{1}},\overline {A_{1}}], [\underline {A_{2}},\overline {A_{2}}]\in \mathbb {I}(2^{M})$ are two interval sets, we obatain that $([\underline {A_{1}},\overline {A_{1}}]\sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown }=[\underline {A_{1}}\cup \underline {A_{2}}, \overline {A_{1}}\cap \overline {A_{2}}]^{\triangledown }=(\underline {A_{1}}\cup \underline {A_{2}})^{\diamondsuit }\cup (\overline {A_{1}}\cap \overline {A_{2}})^{\#}=(\underline {A_{1}}^{\diamondsuit }\cup \underline {A_{2}}^{\diamondsuit })\cup (\overline {A_{1}}^{\#}\cup \overline {A_{2}}^{\#})=(\underline {A_{1}}^{\diamondsuit }\cup \overline {A_{1}}^{\#})\cup (\underline {A_{2}}^{\diamondsuit }\cup \overline {A_{2}}^{\#})=[\underline {A_{1}},\overline {A_{1}}]^{\triangledown }\cup [\underline {A_{2}},\overline {A_{2}}]^{\triangledown }$.

(v) Let X₁,X₂ be two object subsets. It follows that $(X_{1}\cup X_{2})^{\vartriangle }=[(X_{1}\cup X_{2})^{\Box },(X_{1}\cup X_{2})^{\#}]\sqsupseteq [X_{1}^{\Box }\cup X_{2}^{\Box }, X_{1}^{\#}\cap X_{2}^{\#}]=[X_{1}^{\Box }, X_{1}^{\#}]\sqcup [X_{2}^{\Box }, X_{2}^{\#}]=X_{1}^{\vartriangle }\sqcup X_{2}^{\vartriangle }$. □

1.3 Proof of Theorem 6

Proof

We only prove the equation $(X_{1}, [\underline {A_{1}},\overline {A_{1}}]) \vee (X_{2}, [\underline {A_{2}},\overline {A_{2}}])= (X_{1} \cup X_{2}, ([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle })$. The equation of infimum of any two OE_ND-concepts can be proved dually.

First, we prove that the right of the equation is an OE_ND-concept. That is, we need to prove $(X_{1} \cup X_{2})^{\vartriangle }=([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle }$ and $([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle \triangledown }=X_{1} \cup X_{2}$. We first prove that $(X_{1} \cup X_{2})^{\vartriangle }=([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle }$. Since $(X_{1}, [\underline {A_{1}},\overline {A_{1}}])$ and $(X_{2}, [\underline {A_{2}},\overline {A_{2}}])$ are two OE_ND-concepts, then $X_{1}^{\vartriangle }=[\underline {A_{1}},\overline {A_{1}}], [\underline {A_{1}},\overline {A_{1}}]^{\triangledown }= X_{1}$, and $X_{2}^{\vartriangle }=[\underline {A_{2}},\overline {A_{2}}], [\underline {A_{2}},\overline {A_{2}}]^{\triangledown }= X_{2}$ hold. Hence, $(X_{1} \cup X_{2})^{\vartriangle }=([\underline {A_{1}},\overline {A_{1}}]^{\triangledown }\cup [\underline {A_{2}},\overline {A_{2}}]^{\triangledown })^{\vartriangle }=([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle }$ holds based on (iii) in Proposition 4. Now, we prove that $([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle \triangledown }=X_{1} \cup X_{2}$. Since $X_{1} \cup X_{2}=[\underline {A_{1}}, \overline {A_{1}}]^{\triangledown } \cup [\underline {A_{2}},\overline {A_{2}}]^{\triangledown }=([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown }$, then we can know that $([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle \triangledown }=([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown }=X_{1} \cup X_{2}$ holds. These equations verify the fact that $(X_{1} \cup X_{2}, ([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle })$ is an OE_ND-concept.

Next, we prove that $(X_{1} \cup X_{2}, ([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle })$ is the least upper bound of $(X_{1}, [\underline {A_{1}},\overline {A_{1}}])$ and $(X_{2}, [\underline {A_{2}},\overline {A_{2}}])$. Evidently, we can know that $(X_{1} \cup X_{2}, ([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle })$ is an upper bound of $(X_{1}, [\underline {A_{1}},\overline {A_{1}}])$ and $(X_{2}, [\underline {A_{2}},\overline {A_{2}}])$. If there is an OE_ND-concept $(X,[\underline A, \overline A])$ being an upper bound of $(X_{1}, [\underline {A_{1}},\overline {A_{1}}])$ and $(X_{2}, [\underline {A_{2}},\overline {A_{2}}])$, then we have $X_{1}\subseteq X$ and $X_{2}\subseteq X$, then $X_{1}\cup X_{2}\subseteq X$. It indicates that $(X_{1} \cup X_{2}, ([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle })\leqslant (X,[\underline A, \overline A])$. Therefore, we can say that $(X_{1} \cup X_{2}, ([\underline {A_{1}}, \overline {A_{1}}] \sqcup [\underline {A_{2}},\overline {A_{2}}])^{\triangledown \vartriangle })$ is the least upper bound of $(X_{1}, [\underline {A_{1}},\overline {A_{1}}])$ and $(X_{2}, [\underline {A_{2}},\overline {A_{2}}])$. □

1.4 Proof of Theorem 8

Proof

We first prove the equivalence of (i) to (iv), and the equivalence of (v) to (viii) can be obtained in a similar way. The equivalence of (i) and (ii) is proved in Reference [46], then we need to prove that (i)⇔(iii) and (i)⇔(iv) hold simultaneously. Only the proof of (i)⇔(iii) is presented, since (i)⇔(iv) can be proved in a similar way.

$$ \begin{array}{@{}rcl@{}} &&(X,(A,B))\in \text{OECL}(G,M,I) \\ &\overset{\text{Definition}~4}{\Longleftrightarrow}& X^{\lessdot_{*}}=(X^{*},X^{\overline *})=(A,B), (A,B)^{\gtrdot_{*}}=A^{*}\cap B^{\overline *}=X\\ &\overset{\text{Proposition}~2}{\Longleftrightarrow}& (X^{*},X^{\diamondsuit c})=(A,B), A^{*}\cap B^{c\Box}=X \\ &\Longleftrightarrow& [X^{*},X^{\diamondsuit }]=[A,B^{c}], A^{*}\cap B^{c\Box}=X \\ &\overset{\text{Definition}~5}{\Longleftrightarrow} & X^{\triangleleft}=[A,B^{c}], [A,B^{c}]^{\triangleright}=X\\ &\overset{\text{Definition}~6}{\Longleftrightarrow} &(X,[A,B^{c}])\in \text{OEL}_{\text{SP}}(G,M,I). \end{array} $$

The proof is finished by proving (i) and (v) are equivalent.

$$ \begin{array}{@{}rcl@{}} &&(X,(A,B))\in \text{OECL}(G,M,I) \\ &\overset{\text{Definition}~4}{\Longleftrightarrow} & X^{\lessdot_{*}}=(X^{*},X^{\overline *})=(A,B), (A,B)^{\gtrdot_{*}}=A^{*}\cap B^{\overline *}=X\\ &\overset{\text{Proposition}~2}{\Longleftrightarrow} & (X^{c\overline\Box},X^{c\Box})=(A,B), A^{\overline\diamondsuit c}\cap B^{\diamondsuit c}=X \\ &\Longleftrightarrow& (X^{c\Box},X^{c\overline\Box})=(B,A), B^{\diamondsuit}\cap A^{\overline\diamondsuit}=X^{c}\\ &\overset{\text{Definition}~3}{\Longleftrightarrow} & X^{c\lessdot_{\Box}}=(B,A), (B,A)^{\gtrdot_{\diamondsuit}}=X^{c}\\ &\overset{\text{Definition}~4}{\Longleftrightarrow} &(X^{c},(B,A))\in \text{OEOL}(G,M,I). \end{array} $$

□

1.5 Proof of Lemma 1

Proof

For any X ∈ 2^G, $X^{\lessdot _{*}\gtrdot _{*}}=X^{\lessdot _{\diamondsuit }\gtrdot _{\Box }}$ has been proved in Reference [47]. Now, we prove the remaining equations. It can be known that $X^{\lessdot _{*}\gtrdot _{*}}=(X^{*},X^{\overline *})^{\gtrdot _{*}}=X^{**}\cap X^{\overline *\overline *}=X^{**}\cap X^{\diamondsuit c\overline *}=X^{**}\cap X^{\diamondsuit \Box }=X^{\triangleleft \triangleright }$ and $X^{\lessdot _{*}\gtrdot _{*}}=(X^{*},X^{\overline *})^{\gtrdot _{*}}= X^{\overline *\overline *}\cap X^{**}=X^{\overline *\overline *}\cap X^{\overline \diamondsuit c *}=X^{\overline *\overline *}\cap X^{\overline \diamondsuit \overline \Box }=X^{\overline \triangleleft \overline \triangleright }$ hold based on Proposition 2, Definitions 3 and 5. It follows that $X^{\lessdot _{*}\gtrdot _{*}}=X^{\triangleleft \triangleright }=X^{\overline \triangleleft \overline \triangleright }$. Similarly, we have that $X^{c\lessdot _\#\gtrdot _\#}=X^{c\lessdot _{\Box }\gtrdot _{\diamondsuit }}=X^{c\vartriangle \triangledown }=X^{c\overline \vartriangle \overline \triangledown }$. On the basis of Proposition 2, it can be obtained that $X^{\lessdot _{*}\gtrdot _{*}}=X^{**}\cap X^{\overline *\overline *}=X^{c\overline \Box \overline \diamondsuit c}\cap X^{c\Box \diamondsuit c}=(X^{c\overline \Box \overline \diamondsuit }\cup X^{c\Box \diamondsuit })^{c}=X^{c\lessdot _{\Box }\gtrdot _{\diamondsuit } c}$ holds. Therefore, we have that $X^{\lessdot _{*}\gtrdot _{*}}=X^{\lessdot _{\diamondsuit }\gtrdot _{\Box }}=X^{\triangleleft \triangleright }=X^{\overline \triangleleft \overline \triangleright }= X^{c\lessdot _\#\gtrdot _\# c}=X^{c\lessdot _{\Box }\gtrdot _{\diamondsuit }c}=X^{c\vartriangle \triangledown c}=X^{c\overline \vartriangle \overline \triangledown c}$. □

1.6 Proof of Theorem 10

Proof

We only prove statements (i). (ii) can be proved dually. We first prove the isomorphic relation among OECL(G,M,I), OEL_SP(G,M,I), OEL_NSP(G,M,I), and OEPL(G,M,I), the isomorphic relation among another four kinds of OE-concept lattices can obtained in a simlar way. The isomorphic relation between OECL(G,M,I) and OEPL(G,M,I) has been proved in Reference [47], then we only need to prove that OECL(G,M,I)≅OEL_SP(G,M,I) ≅OEL_NSP(G,M,I).

Let f : OECL(G,M,I)→OEL_SP(G,M,I) be a mapping defined by

$$f((X,(A,B)))=(X,[A,B^{c}]).$$

This mapping is evidently a bijection based on Theorem 8. Then, we only need to prove that f is ∨-preserving and ∧-preserving.

Assume (X, (A,B)) and (Y, (E,F)) are two OEC-concepts, then we have that

$$ \begin{array}{@{}rcl@{}} &&f((X,(A,B))\vee(Y,(E,F)))\\ &\overset{\text{Theorem}~1}{=}&f(((X\cup Y)^{\lessdot_{*}\gtrdot_{*}},(A\cap E,B\cap F))) \\ &\overset{\text{Assumption}}{=}&((X\cup Y)^{\lessdot_{*}\gtrdot_{*}},[A\cap E,(B\cap F)^{c}]) \\ &\overset{\text{Lemma}~1}{=}&((X\cup Y)^{\triangleleft\triangleright},[A\cap E, B^{c}\cup F^{c}])\\ &\overset{\text{Theorem}~4}{=}&(X,[A,B^{c}])\vee (Y,[E,F^{c}]) \\ &\overset{\text{Assumption}}{=}&f((X,(A,B)))\vee f((Y,(E,F))). \end{array} $$

It shows that f is ∨-preserving. Dually, we can prove that f((X, (A,B)) ∧ (Y, (E,F))) = f((X, (A,B))) ∧ f((Y, (E,F))). The equation states that f is also ∧-preserving. It follows that f is an isomorphic mapping between OECL(G,M,I) and OEL_SP(G,M,I). OECL(G,M,I) ≅OEL_NSP(G,M,I) can be obtained similarly.

We finish the proof by proving $\text {OECL}(G,M,I)\overset {anti}{\cong }\text {OEOL}(G,M,I)$. Let g : OECL(G,M,I)→OEOL(G,M,I) be a mapping defined by

$$g((X,(A,B)))=(X^{c},(B,A)).$$

This mapping is evidently a bijection based on Theorem 8. Then, we only need to prove that g is anti-∨-preserving and anti-∧-preserving.

Assume (X, (A,B)) and (Y, (E,F)) are two OEC-concepts, then we have that

$$ \begin{array}{@{}rcl@{}} &&g((X,(A,B))\vee(Y,(E,F)))\\ &\overset{\text{Theorem}~1}{=}&g(((X\cup Y)^{\lessdot_{*}\gtrdot_{*}},(A\cap E,B\cap F))) \\ &\overset{\text{Assumption}}{=}&((X\cup Y)^{\lessdot_{*}\gtrdot_{*}c},(B\cap F, A\cap E)) \\ &\overset{\text{Lemma}~1}{=}&((X^{c}\cap Y^{c})^{\lessdot_{\Box}\gtrdot_{\diamondsuit}},(B\cap F, A\cap E))\\ &\overset{\text{Theorem}~1}{=}&(X^{c},(B,A))\wedge (Y,(E,F)) \\ &\overset{\text{Assumption}}{=}&g((X,(A,B)))\wedge g((Y,(E,F))). \end{array} $$

It shows that g is anti-∨-preserving. Dually, we can prove that g((X, (A,B)) ∧ (Y, (E,F))) = g((X, (A,B))) ∨ g((Y, (E,F))). The equation states that g is anti-∧-preserving. It follows that g is an anti-isomorphic mapping between OECL(G,M,I) and OEOL(G,M,I). □