The inclusion degrees of skill maps and knowledge structures

1 Introduction

With the development of information technology, more and more researchers has been paid attention to the study of human cognition. At present, a complete set of theories is urgently needed for research. There is no doubt that knowledge space theory (KST) is what we need. KST provides a valuable and effective mathematical psychological framework for evaluating cognitive levels and further learning of individuals [9, 17]. In recent years, KST has been successfully applied in ALEKS (Assessment and Learning in Knowledge Space) [6, 17]. In addition, it has been applied in clinical diagnosis [3, 41], and adaptive evaluation [1, 37]. Since the proposal of KST, the question of how to accurately construct the knowledge state and knowledge structure of an individual has attracted the attention of many researchers. Some researchers have made contributions in this area, such as the expert inquiry method [15, 25], data-driven approaches, and so on [13, 41]. The introduction of basic local independent model enriches the content of KST and constructs the probabilistic knowledge structure [17, 42]. Inspired by the concepts of skill and ability in the cognitive diagnostic model [8, 18], some researchers have extended KST to competence based-knowledge space theory (Cb-KST) [2, 36]. Heller et al. investigated the association between the cognitive diagnostic model and KST [21]. Numerous studies have been done to construct knowledge structure through the relationship between items and skills [7, 22].

The relation between items and skills is expressed as a skill map. It provides the possibility of cognitive interpretation for the item-solving process. In KST, a skill map assigns a subset of skills to each item. Individual learning skills are closely related to concept-cognitive learning [39, 49]. From the perspective of cognition, skill learning is a process of concept-cognitive renewal. In concept learning, basic granules are used as the basic knowledge with which more complicated granules could be obtained by using a certain learning strategy [46]. During skill learning, we can solve more items by using acquired skills and renew the original knowledge system of individuals.

There are two common models of a skill map in KST. One is the disjunctive model, which needs at least one of several skills required for the mastery of item q. The other is the conjunctive model, which claims all the skills attached to the item q are required for mastery of q. The knowledge space and simple closure space are delineated via the disjunctive and conjunctive models by a skill map [7, 17]. Sun et al. extended the skill map to a fuzzy skill map and discussed the construction of knowledge structure by a fuzzy skill map [38]. Zhou et al. studied fuzzy knowledge structure delineated by a skill map, and discussed the skills assessment and the learning paths [50].

The above researches have greatly promoted the development of KST. However, the disjunctive and conjunctive models of a skill map fail to consider whether the number of skills attached to item q has been mastered by the individual will affect the knowledge structure. The knowledge structure of these two models may not conform to the actual knowledge structure of the individual, and thus cannot truly reflect the individual’s cognitive level. Therefore, it is necessary to study the number of skills attached to item q that have been mastered by the individual, which is the motivation for this paper.

In practical application, an item may have different solutions, and sometimes several items may be solved by the same solution. The solutions here refer to the knowledge points (i.e. skills) investigated by a certain item. Here is an example. Suppose the item set is Q = {q₁, q₂, q₃} and the skill set is S = {+ , - , × , ÷}, where

q₁: A bus with a speed of 45km/h takes 12h to travel from A to B. What is the distance from A to B?

Solution: S _AB  = 45 × 12 = 540  (km) .

q₂: Given that volleyball is $90. How much would someone have to pay if they wanted to buy three volleyballs?

Solution 1: 90 + 90 + 90 = $270 .

q₃: Please do some arithmetic with the non-repeating numbers in 2, 3, 5, 16 such that the value is 8.

Solution 1: 3 + 5 =8 .

Let the skill of student A be T _A  = {+ , ÷} and the skill of student B be T _B  = {+}. Then, the proportions of students A and B mastering the skills assigned to q₁, q₂, and q₃ are 0 , 1 2 , 2 3 and 0 , 1 2 , 1 3 , respectively. Theoretically, division is harder than addition, and student A’s knowledge level is actually higher than student B’s. Nevertheless, this conclusion can not be drawn in the disjunctive and conjunctive models of a skill map.

This problem becomes even more acute when repeated evaluations are performed on the two individuals on different occasions with the objective of monitoring knowledge levels. Thus, skill inclusion degree, as a special case of fuzzy inclusion measure [4], is the basis for the development of the theoretical work proposed in this paper. It is the proportion of skills required for an item q that is possessed by an individual with a skill set T. In fact, different individuals have different skills, so they may have different skill inclusion degrees in the same item. The main contribution of this paper is that the skill inclusion degree is proposed, and the variable precision α-model for constructing knowledge structure is given. The presented variable precision α-model can reasonably explain the phenomenon that different individuals have different knowledge structures behind solving the same items. As mentioned in the previous example, the variable precision α-model can be used in adaptive learning systems [27, 43] and become a new mechanism for evaluating individual learning.

This paper is structured as follows. In Section 2, some basic concepts of KST are reviewed. In Section 3, the concept of skill inclusion degree is given and some properties of it are discussed. In Section 4, the variable precision α-models of a skill map are specified and the relationships between the knowledge structures delineated via these models by a skill map are studied. Moreover, the algorithm for constructing knowledge structures is designed and the flaw chart for generating the knowledge structure graphs is provided. In Section 5, an experimental analysis is carried out on a dataset to verify the validity and feasibility of the designed algorithm. In Section 6, this paper is summarized.

For ease of reading and understanding, the notations of this paper are listed in Table 1.

In this section, we will review some notions from order theory and KST. For a brief introduction to order theory, we refer to [12]. For a detailed description and background of KST, we refer to [9, 17]. This paper is discussed in an ideal situation (without careless errors or lucky guesses). Q, S and 2 ^S represent the nonempty finite item set, the nonempty finite skill set and power set of S, respectively.

A quasi order (Q, ≾) is a reflexive (i.e., q ≾ q, for q ∈ Q) and transitive (i.e., if p ≾ q and q ≾ r, then p ≾ r, for p, q, r ∈ Q) relation. A quasi order satisfying antisymmetry (i.e., if p ≾ q and q ≾ p, then p = q, for p, q ∈ Q) is a partial order.

Knowledge state is the core concept of KST. In a nonempty finite item set Q, a knowledge state K ⊆ Q is represented as the item set that an individual can correctly answer in ideal conditions. The knowledge structure is an important tool to evaluate the individual cognitive level, which is denoted by a pair ( Q , K ) , where K is the family of knowledge states containing at least the empty set and Q.

When the family K of a knowledge structure ( Q , K ) is closed under finite union, i.e., for any K 1 , K 2 ∈ K , the set K 1 ∪ K 2 ∈ K , we call ( Q , K ) a knowledge space. When the family K of a knowledge structure ( Q , K ) is closed under finite intersection, that is, for any K 1 , K 2 ∈ K , the set K 1 ∩ K 2 ∈ K , we call ( Q , K ) a simple closure space. If K is a knowledge space, its dual structure K ¯ = { Q ∖ K ∣ ∀ K ∈ K } is a simple closure space. When the family K of a knowledge structure ( Q , K ) is closed under finite union and finite intersection, we call ( Q , K ) a quasi ordinal space.

A skill map is a triple (Q, S, τ), where Q is a nonempty item set, S is a nonempty skill set, and τ is a mapping from Q to 2 ^S  ∖ {∅}. For any q ∈ Q, the subset τ (q) of S will be referred to as the skill set assigned to q (by the skill map τ).

Let (Q, S, τ) be a skill map and T ⊆ S, we say that K = {q ∈ Q ∣ τ (q) ∩ T ≠ ∅} is the knowledge state delineated via the disjunctive model by T, and K = {q ∈ Q ∣ τ (q) ⊆ T} is the knowledge state delineated via the conjunctive model by T. The knowledge space and simple closure space delineated respectively via the disjunctive and conjunctive models by the same skill map are dual one to the other.

3 The skill inclusion degrees

In this section, to analyze the number of skills attached to an item that an individual has mastered, the concept of skill inclusion degree is given and some properties are discussed.

Definition 1. Let (Q, S, τ) be a skill map. For any q ∈ Q, T ⊆ S, then D ( T / τ ( q ) ) = | τ ( q ) ∩ T | | τ ( q ) |(1) is called a skill inclusion degree with respect to q and T.

For convenience, D (T/τ (q)) is denoted as τ q T .

τ q T is the proportion of skills required for an item q that is possessed by an individual with a skill set T, and this proportion is discrete. There are the following reasons:

The skill set S is a finite set, and each item has a finite number of skills.

Therefore, some finite and discrete values are used to assess the knowledge level of individuals.

Remark 1. Given a skill map (Q, S, τ), for any q ∈ Q, T ⊆ S. Then the following properties hold.

(i) If T =∅, then τ q T = 0 . By Definition 1, we have τ q ∅ = | τ ( q ) ∩ ∅ | | τ ( q ) | = 0 .

(ii) If τ (q) ⊆ T, then τ q T = 1 . For any q ∈ Q, we have τ q T = | τ ( q ) ∩ T | | τ ( q ) | = 1 .

(iii) If τ (q)∩ T ≠ ∅ and τ (q) ⊈ T, then 0 < τ q T < 1 . Since τ ( q ) ∩ T ≠ ∅ ⇔ 0 < | τ ( q ) ∩ T | < 1 , then 0 < τ q T < 1 . Thus 0 ≤ τ q T ≤ 1 .

Remark 2. Given a skill map (Q, S, τ), for any q ∈ Q, if T = {s _i , s _j } ⊆ S, then τ q T is represented as τ q s i s j ; if T = T₁ ∪ T₂ ⊆ S, then τ q T is denoted as τ q T 1 ∪ T 2 .

Example 1. Let (Q, S, τ) be a skill map, where Q = {q₁, q₂, q₃, q₄, q₅, q₆}, S = {s₁, s₂, s₃, s₄}, and τ (q₁) = {s₂, s₄}, τ (q₂) = {s₃, s₄}, τ (q₃) = {s₂, s₃}, τ (q₄) = {s₁, s₃}, τ (q₅) = {s₁, s₄}, τ (q₆) = {s₄}. Obviously, the power set of S is 2 S = { ∅ , { s 1 } , { s 2 } , { s 3 } , { s 4 } , { s 1 , s 2 } , { s 1 , s 3 } , { s 1 , s 4 } , { s 2 , s 3 } , { s 2 , s 4 } , { s 3 , s 4 } , { s 1 , s 2 , s 3 } , { s 1 , s 2 , s 4 } , { s 1 , s 3 , s 4 } , { s 2 , s 3 , s 4 } , S } . By Definition 1, τ q 1 ∅ = | τ ( q 1 ) ∩ ∅ | | τ ( q 1 ) | = 0 . Similarly, τ q 2 s 3 = | τ ( q 2 ) ∩ { s 3 } | | τ ( q 2 ) | = | { s 3 } | | { s 3 , s 4 } | = 1 2 , τ q 4 s 1 s 2 s 3 = | τ ( q 4 ) ∩ { s 1 , s 2 , s 3 } | | τ ( q 4 ) | = | { s 1 , s 3 } | | { s 1 , s 3 } | = 1 . Thus, for q ∈ Q, T ∈ 2 ^S , the skill inclusion degrees τ q T can be calculated, as shown in Table 2, where the skill subset is simply denoted as a sequence. For example, T = {s₁, s₂} is denoted as s₁s₂.

Notice that the skill subsets are represented as sequences in a skill inclusion degrees table, which we will refer to subsequently.

From Table 2, we can see τ q 4 s 1 s 2 + τ q 4 s 1 s 3 = 3 2 ≠ τ q s 1 s 2 s 3 = 1 , while τ q 4 s 1 s 2 + τ q 4 s 4 = 1 2 = τ q 4 s 1 s 2 s 4 . Thus, it is natural to get the following proposition.

Proposition 1. Let (Q, S, τ) be a skill map. For any q ∈ Q, T₁, T₂ ⊆ S, there are τ q T 1 ∪ T 2 = τ q T 1 + τ q T 2 - τ q T 1 ∩ T 2 .

Proof. According to Definition 1, for q ∈ Q, T₁, T₂ ⊆ S, we have τ q T 1 ∪ T 2 = | τ ( q ) ∩ ( T 1 ∪ T 2 ) | | τ ( q ) | = | ( τ ( q ) ∩ T 1 ) ∪ ( τ ( q ) ∩ T 2 ) | | τ ( q ) | = | τ ( q ) ∩ T 1 | + | τ ( q ) ∩ T 2 | | τ ( q ) | - | τ ( q ) ∩ ( T 1 ∩ T 2 ) | | τ ( q ) | = τ q T 1 + τ q T 2 - τ q T 1 ∩ T 2 .

In particular, if T₁∩ T₂ = ∅, then τ q T 1 ∪ T 2 = τ q T 1 + τ q T 2 . □ Given a skill map, an individual can solve an item by mastering different skills, which means that his learning is progressive. As far as the skill inclusion degree is concerned, it satisfies additivity and complementarity. Therefore, we have the following proposition.

Proposition 2. Let (Q, S, τ) be a skill map, there are the following properties:

(i) For q ∈ Q, T ⊆ S, τ q T = ∑ s ∈ T τ q s ;

(ii) For q ∈ Q, T ⊆ S, τ q T + τ q S ∖ T = 1 .

Proof. It can be proved by Proposition 1. □

For instance, in Table 2, for any q ∈ Q, we have τ q s 2 s 3 s 4 = τ q s 2 + τ q s 3 + τ q s 4 , τ q s 1 s 3 + τ q s 2 s 4 = τ q S = 1 .

Definition 2. Let (Q, S, τ) be a skill map. Then D ( τ ) = { τ q T ∣ q ∈ Q , T ⊆ S }(2) is said to be a skill inclusion degree set of τ.

As a convenience, the skill inclusion degree set of τ is denoted by D (τ) = {β₁, β₂, ⋯ , β _n } , where 0 = β₁ < β₂ < ⋯ < β _n  = 1 and β _i  + β_n-i+1 = 1.

Remark 3. Supposed that τ q T = β i , by Proposition 2, we have τ q S ∖ T = β n - i + 1 . When n is odd, then β n + 1 2 + β n + 1 2 = 1 , i.e., β n + 1 2 = 1 2 . Moreover, D ( τ q ) = { τ q T ∣ T ⊆ S } is denoted the skill inclusion degree set of q ∈ Q. Similar to D (τ), let D (τ _q ) = {γ₁, γ₂, ⋯ , γ _p } , where 0 = γ₁ < γ₂ < ⋯ < γ _p  = 1 and γ _j  + γ_p-j+1 = 1 (j = 1, 2, ⋯ , p). Obviously, D (τ) = ⋃ _q∈QD (τ _q ).

Proposition 3. Let (Q, S, τ) be a skill map, where Q = {q₁, q₂, ⋯ , q _m }. Let D (τ) = {β₁, β₂, ⋯ , β _n } be a skill inclusion degree set of τ.

(i) For q _i  ∈ Q  (i = 1, 2, ⋯ , m), |D (τ _{q i} ) | is even if and only if 1 2 ∉ D ( τ q i ) , |D (τ _{q i} ) | is odd if and only if 1 2 ∈ D ( τ q i ) ;

(ii) For any q _i  ∈ Q  (i = 1, 2, ⋯ , m), if |D (τ _{q i} ) | is even, then | D ( τ ) | = | ⋃ i = 1 m D ( τ q i ) | is even;

(iii) For some q _i  ∈ Q  (i = 1, 2, ⋯ , m), if |D (τ _{q i} ) | is odd, then | D ( τ ) | = | ⋃ i = 1 m D ( τ q i ) | is odd.

Proof. (i) For any q _i  ∈ Q, let D (τ _{q i} ) = {γ₁, γ₂, ⋯ , γ _p } .

a) If |D (τ _{q i} ) | is even, by γ _j  + γ_p-j+1 = 1  (j = 1, ⋯ , p) , when j = p 2 , we have γ p 2 + γ p 2 + 1 = 1 . Thus, we can obtain 1 2 ∉ D ( τ q i ) . Conversely, let 1 2 ∉ D ( τ q i ) . We suppose p is odd, then γ p + 1 2 + γ p + 1 2 = 1 , namely γ p + 1 2 = 1 2 . This is a contradiction. Therefore, |D (τ _{q i} ) | is even.

b) Similarly, it is easy to prove that |D (τ _{q i} ) | is odd if and only if 1 2 ∈ D ( τ q i ) .

(ii) For each q _i  ∈ Q, if |D (τ _{q i} ) | is even, by (i), then 1 2 ∉ D ( τ q i ) . Since D ( τ ) = ⋃ i = 1 m D ( τ q i ) , we then have 1 2 ∉ ⋃ i = 1 m D ( τ q i ) , i.e 1 2 ∉ D ( τ ) . Thus |D (τ) | is even.

(iii) If there exists q _i  ∈ Q such that |D (τ _{q i} ) | is odd, that means 1 2 ∈ D ( τ q i ) , then 1 2 ∈ ⋃ i = 1 m D ( τ q i ) = D ( τ ) . Thus, |D (τ) | is odd. □ Based on the above conclusions, we design an algorithm and a flowchart to calculate the skill inclusion degree set, as shown in Algorithm 1 and Figure 1.

OPEN IN VIEWER

Fig. 1

A flowchart for obtaining the skill inclusion degree set D (τ) of a skill map.

Currently, different approaches are used to study the knowledge and learning assessment of individuals. Minn [29] provided adaptive learning content to learners in an educational setting through artificial intelligent techniques. Lázaro-Cantabrana [28] et al. constructed an assessment tool to assess teacher digital competence. Pusung [34] controlled student prior knowledge to understand the impact of learning patterns and task assessment on scientific achievement. In this section, the variable precision α-models are defined based on the skill inclusion degree in order to assess the knowledge mastery of individuals more comprehensively. These models determine the individual’s knowledge state by analyzing the number of skills he has mastered assigned to each item.

Definition 3. Let (Q, S, τ) be a skill map. For q ∈ Q,  T ⊆ S,  α ∈ (0, 1], the knowledge state K T α delineated by T via the variable precision α-model is specified by K T α = { q ∈ Q ∣ τ q T ≥ α } .(3)

Remark 4. For the variable precision α-model, if there exists a skill subset T satisfying τ q T ≥ α for α ∈ (0, 1], then the item q can be solved by T. Notice that T =∅ delineates the state K ∅ α = ∅ , since τ q ∅ = | τ ( q ) ∩ ∅ | | τ ( q ) | = 0 α for any q ∈ Q, τ (q)≠ ∅. Moreover, T = S delineates the knowledge state K S α = Q , since there is τ (q) ⊆ S such that τ q S = 1 ≥ α for any q ∈ Q. Thus, a skill map via the variable precision α-model forms a knowledge structure.

Definition 4. Let (Q, S, τ) be a skill map. For α ∈ (0, 1], then the knowledge structure delineated via the variable precision α-model by (Q, S, τ) is defined as K α = { K T α ∣ T ⊆ S } .(4)

Example 2. (Continued with Example 1). The skill inclusion degree set of τ is D ( τ ) = { 0 , 1 2 , 1 } . According to Definition 3 and Table 2, the table of knowledge states K T α is shown in Table 3.

From Table 3, when α ∈ ( 0 , 1 2 ] , the knowledge structure delineated via the variable precision α-model by (Q, S, τ) is K α = { ∅ , { q 1 , q 3 } , { q 4 , q 5 } , { q 2 , q 3 , q 4 } , { q 1 , q 2 , q 3 , q 4 } , { q 1 , q 2 , q 5 , q 6 } , { q 1 , q 3 , q 4 , q 5 } , { q 2 , q 3 , q 4 , q 5 } , { q 1 , q 2 , q 3 , q 4 , q 5 } , { q 1 , q 2 , q 3 , q 5 , q 6 } , { q 1 , q 2 , q 4 , q 5 , q 6 } , Q } .

When α ∈ ( 1 2 , 1 ] , the knowledge structure delineated via the variable precision α-model by (Q, S, τ) is K α = { ∅ , { q 3 } , { q 4 } , { q 6 } , { q 1 , q 6 } , { q 2 , q 6 } , { q 3 , q 4 } , { q 5 , q 6 } , { q 1 , q 5 , q 6 } , { q 1 , q 2 , q 3 , q 6 } , { q 2 , q 4 , q 5 , q 6 } , Q } .

In this example, it can be checked that when α ∈ ( 0 , 1 2 ] , the knowledge structure delineated via the variable precision α-model by a skill map is the same as that delineated when α = 1 2 , and when α ∈ ( 1 2 , 1 ] , the knowledge structure delineated via the variable precision α-model by (Q, S, τ) is the same as that delineated when α = 1.

Theorem 1. Let (Q, S, τ) be a skill map, D (τ) = {β₁, β₂, ⋯ , β _n } be a skill inclusion degree set of τ. For any α ∈ (β _i , β_i+1] (i = 1, ⋯ , n - 1), then K α = K β i + 1 .

Proof. For α ∈ (β _i , β_i+1], let K α be a knowledge structure delineated via the variable precision α-model by τ. For any T ⊆ S, we have K T α = { q ∈ Q ∣ τ q T ≥ α } = { q ∈ Q | τ q T > β i } . Since there are no more skill inclusion degrees between β _i and β_i+1, thus K T α = { q ∈ Q ∣ τ q T ≥ β i + 1 } = K T β i + 1 . Therefore, K α = K β i + 1 . □ As mentioned above, we know that β₁ = 0,  β _n  = 1, and α ∈ (0, 1]. If β₁ is included, then α ∈ [0, 1]. According to the skill inclusion degree, we divide [0, 1] into multiple intervals, and the first one is α ∈ [0, β₂]. Then, for any T ⊆ S, we have K T α = { q ∈ Q ∣ τ q T ≥ α } = { q ∈ Q ∣ τ q T ≥ 0 } = Q . As a result, K α does not contain ∅ and is not a knowledge structure. Therefore, α ∈ (0, 1].

In D (τ) = {β₁, β₂, ⋯ , β _n }, it is easy to obtain that the knowledge structures can delineate via the variable precision α-models by a skill map except for β₁. So we have the following conclusion.

Corollary 1. Let (Q, S, τ) be a skill map, D (τ) = {β₁, β₂, ⋯ , β _n } be a skill inclusion degree set of τ. The number of knowledge structures delineated via the variable precision α-models by a skill map is n - 1.

Remark 5. In Example 2, when α ∈ ( 0 , 1 2 ] , the knowledge structure K α is closed under union. Clearly, K α is a knowledge space. When α ∈ ( 1 2 , 1 ] , the family K α is closed under intersection, then K α is a simple closure space. Therefore, it can be concluded that the variable precision α-model is more general than disjunctive and conjunctive models, and the following conclusions can be derived.

Theorem 2. Let (Q, S, τ) be a skill map, D (τ) = {β₁, β₂, ⋯ , β _n } be a skill inclusion degree set of τ.

(i) For α ∈ (β₁, β₂], the knowledge structure K α delineated via the variable precision α-model by τ is the knowledge space delineated via the disjunctive model.

(ii) For α ∈ (β_n-1, β _n ], the knowledge structure K α delineated via the variable precision α-model by τ is the simple closure space delineated via the conjunctive model.

Proof. (i) Let ( Q , K α ) be the knowledge structure delineated via the variable precision α-model by τ. For α ∈ (β₁, β₂], suppose K T 1 α and K T 2 α are any knowledge states of K α , T₁, T₂ ⊆ S. By Definition 3, we have K T 1 α = { q ∈ Q ∣ τ q T 1 ≥ α } , K T 2 α = { q ∈ Q ∣ τ q T 2 ≥ α } . Since for ∀T ⊆ S, there exists τ q T ≥ α ⇔ τ q T ≥ β 2 ⇔ τ q T > β 1 , and β₁ = 0, then K T 1 α = { q ∈ Q ∣ τ q T 1 > β 1 } = { q ∈ Q ∣ τ q T 1 > 0 } , K T 2 α = { q ∈ Q ∣ τ q T 2 > β 1 } = { q ∈ Q ∣ τ q T 2 > 0 } . According to Theorem 1, K α = K β 2 . Thus, K T 1 α = K T 1 β 2 , K T 2 α = K T 2 β 2 . First, proof K β 2 is a knowledge space. We have K T 1 β 2 ∪ K T 2 β 2 = { q ∈ Q ∣ τ q T 1 > 0 } ∪ { q ∈ Q ∣ τ q T 2 > 0 } . However, K T 1 ∪ T 2 β 2 = { q ∈ Q ∣ τ q T 1 ∪ T 2 ≥ β 2 } = { q ∈ Q ∣ τ q T 1 ∪ T 2 > 0 } = { q ∈ Q ∣ τ q T 1 + τ q T 2 - τ q T 1 ∩ T 2 > 0 } . Obviously, K T 1 ∪ T 2 β 2 ⊆ K T 1 β 2 ∪ K T 2 β 2 .

For q ∈ K T 1 β 2 ∪ K T 2 β 2 , we have q ∈ K T 1 β 2 or q ∈ K T 2 β 2 , then τ q T 1 > 0 or τ q T 2 > 0 . Whereas, τ q T 1 ∩ T 2 ≤ { τ q T 1 , τ q T 2 } , so τ q T 1 ∪ T 2 = τ q T 1 + τ q T 2 - τ q T 1 ∩ T 2 > 0 , that is q ∈ K T 1 ∪ T 2 β 2 . Therefore, K T 1 β 2 ∪ K T 2 β 2 = K T 1 ∪ T 2 β 2 ∈ K β 2 . K β 2 is a knowledge space.

Let ( Q , K ) be the knowledge space delineated via the disjunctive model by τ. Next, proof that K β 2 = K . Assume any K 1 ∈ K , K 2 ∈ K , we have K 1 ∪ K 2 = { q ∈ Q ∣ τ ( q ) ∩ T 1 ≠ ∅ } ∪ { q ∈ Q ∣ τ ( q ) ∩ T 2 ≠ ∅ } . Then K₁ ∪ K₂ = {q ∈ Q ∣ τ (q) ∩ (T₁ ∪ T₂) ≠ ∅} . Since | τ ( q ) ∩ ( T 1 ∪ T 2 ) | > 0 ⇔ τ q T 1 ∪ T 2 > 0 . Thus K β 2 = K .

(ii) In the same way, let ( Q , K β n ) be the knowledge structure delineated via the variable precision α-model by τ. First, proof K β n is a simple closure space. Suppose K T 1 β n and K T 2 β n are any knowledge states of K β n for any T₁, T₂ ⊆ S. We have K T 1 β n ∩ K T 2 β n = { q ∈ Q ∣ τ q T 1 ≥ β n } ∩ { q ∈ Q ∣ τ q T 2 ≥ β n } . Due to β _n  = 1, then K T 1 β n ∩ K T 2 β n = { q ∈ Q ∣ τ q T 1 ≥ 1 } ∩ { q ∈ Q ∣ τ q T 2 ≥ 1 } . By Remark 1, the maximum value of the skill inclusion degree is 1, so we have K T 1 β n ∩ K T 2 β n = { q ∈ Q ∣ τ q T 1 = 1 } ∩ { q ∈ Q ∣ τ q T 2 = 1 } = { q ∈ Q ∣ τ ( q ) ⊆ T 1 } ∩ { q ∈ Q ∣ τ ( q ) ⊆ T 2 } = { q ∈ Q ∣ τ ( q ) ⊆ ( T 1 ∩ T 2 ) } = { q ∈ Q ∣ τ q T 1 ∩ T 2 = 1 } = K T 1 ∩ T 2 β n . Then K T 1 β n ∩ K T 2 β n ∈ K β n . Thus K β n is a simple closure space.

Let ( Q , K ¯ ) is the simple closure space delineated via the conjunctive model by τ. Next, proof that K β n = K ¯ . Suppose L 1 ∈ K ¯ , L 2 ∈ K ¯ , we have L₁ ∩ L₂ = {q ∈ Q ∣ τ (q) ⊆ T₁} ∩ {q ∈ Q ∣ τ (q) ⊆ T₂} . Then L₁ ∩ L₂ = {q ∈ Q ∣ τ (q) ⊆ (T₁ ∩ T₂)} . Thus K β n = K ¯ . □ From Theorem 2, there must be a knowledge space and a simple closure space in the knowledge structures delineated via the variable precision α-models by a skill map. It is worth studying whether there is a quasi ordinal space in these knowledge structures.

Example 3. Let (Q, S, τ) be a skill map, where Q = {q₁, q₂, q₃} ,  S = {s₁, s₂, s₃, s₄, s₅} and τ (q₁) = {s₂, s₃, s₅} ,  τ (q₂) = {s₁, s₃, s₄} , τ (q₃) = {s₁, s₂, s₄} . From Algorithm 1, it is easy to calculate the skill inclusion degree set of τ as D ( τ ) = { 0 , 1 3 , 2 3 , 1 } . By Definition 3 and Theorem 1, for α = β_i+1 (i = 1, 2, 3), the knowledge structures delineated via the variable precision α-models by (Q, S, τ) are as follows: K 1 3 = { ∅ , { q 1 } , { q 1 , q 2 } , { q 1 , q 3 } , { q 2 , q 3 } , Q } , K 2 3 = { ∅ , { q 1 } , { q 2 } , { q 3 } , { q 1 , q 2 } , { q 1 , q 3 } , { q 2 , q 3 } , Q } , K 1 = { ∅ , { q 1 } , { q 2 } , { q 3 } , { q 2 , q 3 } , Q } .

It can be found that K 1 3 is a knowledge space, K 2 3 is a quasi order space, and K 1 is a simple closure space.

In fact, for α = β_j+1 (j = 2, ⋯ , n - 2), the knowledge structures K α delineated via the variable precision α-models by a skill map does not necessarily a knowledge space, simple closure space, or quasi ordinal space. It is not difficult to construct such an example.

Example 4. Let Q = {q₁, q₂, q₃, q₄}, S = {s₁, s₂, s₃, s₄, s₅, s₆} , and τ (q₁) = {s₁, s₂, s₃, s₄, s₆} ,  τ (q₂) = {s₂, s₃, s₅} ,  τ (q₃) = {s₁, s₅, s₆} ,  τ (q₄) = {s₄} . According to Definition 1, when T = {s _i } (i = 1, ⋯ , 6), we get a table of skill inclusion degrees τ q T as shown in Table 4.

By Table 4, we can get the skill inclusion degree set of τ as D ( τ ) = { 0 , 1 5 , 1 3 , 2 5 , 3 5 , 2 3 , 4 5 , 1 } . For α = β_i+1 (i = 1, ⋯ , 7), the knowledge structures delineated by τ are as follows:

K 1 5 = { ∅ , { q 1 , q 2 } , { q 1 , q 3 } , { q 1 , q 4 } , { q 2 , q 3 } , { q 1 , q 2 , q 3 } , { q 1 , q 2 , q 4 } , { q 1 , q 3 , q 4 } , Q } , K 1 3 = { ∅ , { q 2 } , { q 3 } , { q 4 } , { q 1 , q 2 } , { q 1 , q 3 } , { q 2 , q 3 } , { q 1 , q 2 , q 3 } , { q 1 , q 2 , q 4 } , { q 1 , q 3 , q 4 } , { q 2 , q 3 , q 4 } , Q } , K 2 5 = { ∅ , { q 1 } , { q 2 } , { q 3 } , { q 4 } , { q 1 , q 2 } , { q 1 , q 3 } , { q 1 , q 4 } , { q 1 , q 2 , q 3 } , { q 1 , q 2 , q 4 } , { q 1 , q 3 , q 4 } , Q } , K 3 5 = { ∅ , { q 2 } , { q 3 } , { q 4 } , { q 1 , q 2 } , { q 1 , q 3 } , { q 1 , q 4 } , { q 2 , q 3 } , { q 2 , q 4 } , { q 3 , q 4 } , { q 1 , q 2 , q 3 } , { q 1 , q 2 , q 4 } , { q 1 , q 3 , q 4 } , Q } , K 2 3 = { ∅ , { q 2 } , { q 3 } , { q 4 } , { q 2 , q 3 } , { q 2 , q 4 } , { q 3 , q 4 } , { q 1 , q 2 , q 3 } , { q 1 , q 2 , q 4 } , { q 1 , q 3 , q 4 } , { q 2 , q 3 , q 4 } , Q } , K 4 5 = { ∅ , { q 1 } , { q 2 } , { q 3 } , { q 4 } , { q 1 , q 4 } , { q 2 , q 4 } , { q 3 , q 4 } , { q 1 , q 2 , q 3 } , { q 1 , q 2 , q 4 } , { q 1 , q 3 , q 4 } , Q } , K 1 = { ∅ , { q 2 } , { q 3 } , { q 4 } , { q 1 , q 4 } , { q 2 , q 3 } , { q 2 , q 4 } , { q 3 , q 4 } , Q } .

Obviously, the knowledge structures K 1 3 , ⋯ , K 4 5 are not closed under union or intersection. Thus, the knowledge structures K β j + 1 (j = 2, ⋯ , n - 2) delineated via the variable precision α-models by τ are consistent with general knowledge structures (without particularity).

In KST, the knowledge structures delineated via the disjunctive and conjunctive models by the same skill map are dual one to the other [17]. In Theorem 2, we know that for a skill inclusion degree set of τ, the knowledge structure K β 2 is always a knowledge space, while the knowledge structure K β n is always a simple closure space. Therefore, the variable precision α-model has similar results to Falmagne and Doignon [17].

Corollary 2. The knowledge structures K β 2 and K β n delineated via the variable precision α-models by a skill map are dual one to the other.

More generally, the knowledge structures K β i + 1 and K β n - i + 1 (i = 1, ⋯ , n - 1) are dual. This is no coincidence because we have the following results.

Theorem 3. Let (Q, S, τ) be a skill map, D (τ) = {β₁, β₂, ⋯ , β _n } be a skill inclusion degree set of τ. The knowledge structures K β i + 1 and K β n - i + 1 ( i = 1 , ⋯ , n - 1 ) delineated via the variable precision α-models by τ are dual to each other.

Proof. For any T ⊆ S, α = β_i+1 (i = 1, ⋯ , n - 1), the elements of knowledge structure K β i + 1 is K T β i + 1 = { q ∈ Q ∣ τ q T ≥ β i + 1 } , then its dual knowledge state is Q ∖ K T β i + 1 = Q ∖ { q ∈ Q ∣ τ q T ≥ β i + 1 } = { q ∈ Q ∣ τ q T < β i + 1 } = { q ∈ Q ∣ 1 - τ q T > 1 - β i + 1 } = { q ∈ Q ∣ τ q S ∖ T > 1 - β i + 1 } . Due to there being no other skill inclusion degrees between 1 - β_i+1 and 1 - β _i , we then have Q ∖ K T β i + 1 = { q ∈ Q ∣ τ q S ∖ T ≥ 1 - β i } . Since β _i  + β_n-i+1 = 1, then Q ∖ K T β i + 1 = { q ∈ Q ∣ τ q S ∖ T ≥ β n - i + 1 } = K S ∖ T β n - i + 1 ∈ K β n - i + 1 . Thus, the knowledge structures K β i + 1 and K β n - i + 1 (i = 1, ⋯ , n - 1) are dual to each other. □

Corollary 3. Let (Q, S, τ) be a skill map, D (τ) = {β₁, β₂, ⋯ , β _n } be a skill inclusion degree set of τ. If n is odd, the knowledge structure K β i + 1 ( i = 1 , ⋯ , n - 1 ) can find a dual knowledge structure different from itself. If n is even, the knowledge structure K β n 2 + 1 is dual to itself.

We find that the knowledge structures delineated via the variable precision α-models by a skill map always have a minimal knowledge structure. Note that ‘minimal’ here refers to the minimal inclusion of knowledge states between knowledge structures. The relations between knowledge structures will be given by the following definitions.

Definition 5. Let (Q, S, τ) be a skill map, K be a collection of all knowledge structures delineated via the variable precision α-models by τ. Then the relation ≾ on K can be defined by (where β ≠ γ,  β ∈ D (τ) ,  γ ∈ D (τ) and T ⊆ S)

K γ ≾ K β ⇔ ∀ K T β ∈ K β , ∃ K T γ ∈ K γ : K T γ ⊆ K T β ( K β , K γ ∈ K ) .(5)

Definition 6. Let (Q, S, τ) be a skill map, K be a collection of all knowledge structures delineated via the variable precision α-model by τ. Then the relation ≾^∗ on K can be defined by (where β ≠ γ,  β ∈ D (τ) ,  γ ∈ D (τ), and T ⊆ S)

K γ ≾ ∗ K β ⇔ ∀ K T γ ∈ K γ , ∃ K T β ∈ K β : K T β ⊆ K T γ ( K β , K γ ∈ K ) .(6)

The relations ≾ and ≾^∗ will be referred to as the knowledge structure order on K .

Remark 6. The relation ≾ and ≾^∗ are always a quasi order, but not necessarily a partial order. Since antisymmetry is not necessarily true. For any K T β ∈ K β , there is K T γ ∈ K γ such that K T γ ⊆ K T β , but not necessarily K T β ⊆ K T γ . Likewise, for any K T γ ∈ K γ , there is K T β ∈ K β such that K T β ⊆ K T γ , but not necessarily K T γ ⊆ K T β .

Example 5. (Continued with Example 4) For the knowledge structures, we have K T 1 3 ∈ K 1 3 for any K T 1 5 ∈ K 1 5 such that K T 1 3 ⊆ K T 1 5 . For any K T 4 5 ∈ K 4 5 there exists K T 1 ∈ K 1 such that K T 1 ⊆ K T 4 5 . For instance, K s 1 1 3 = { q 3 } ⊆ { q 1 , q 3 } = K s 1 1 5 , and K s 1 s 2 s 3 s 4 1 = { q 4 } ⊆ { q 1 , q 4 } = K s 1 s 2 s 3 s 4 4 5 . Thus K 1 3 ≾ K 1 5 , K 4 5 ≾ * K 1 . Similarly, we can get K 3 5 ≾ K 2 5 ≾ K 1 3 and K 3 5 ≾ * K 2 3 ≾ * K 4 5 .

The knowledge structures delineated via the variable precision α-models by any skill map always have the order relations in Definitions 4 and 5. According to the dual property of knowledge structures in Theorem 3, we can get the following theorem.

Theorem 4. Let (Q, S, τ) be a skill map, D (τ) = {β₁, β₂, ⋯ , β _n } be a skill inclusion degree set of τ. For each α = β_i+1 (i = 1, 2, ⋯ , n - 1), the collection of all knowledge structures delineated via the variable precision α-models by τ is K . Then K β j + 1 ≾ K β j , K β n - j + 1 ≾ ∗ K β n - j + 2 (where n is even, j = 2 , ⋯ , n 2 ; n is odd, j = 2 , ⋯ , n + 1 2 - 1 ).

Proof. Take any two elements K β j , K β j + 1 of K , for any T ⊆ S, K T β j ∈ K β j , K T β j + 1 ∈ K β j + 1 , we have K T β j + 1 = { q ∈ Q ∣ τ q T ≥ β j + 1 } = { q ∈ Q ∣ τ q T > β j } , and K T β j = { q ∈ Q ∣ τ q T ≥ β j } = { q ∈ Q ∣ τ q T > β j } ∪ { q ∈ Q ∣ τ q T = β j } . Thus, for any K T β j ∈ K β j , there exists K T β j + 1 ∈ K β j + 1 such that K T β j + 1 ⊆ K T β j .