On Characteristics of Information System Homomorphisms
Theory Comput Syst (2009) 44: 414–431 DOI 10.1007/s00224-007-9076-8
On Characteristics of Information System Homomorphisms Yan-Hui Zhai · Kai-She Qu
Published online: 18 October 2007 © Springer Science+Business Media, LLC 2007
Abstract The notion of information system homomorphism as a powerful tool to study the relation between two information systems was introduced by J.W. Grzymala-Busse. In this work, we will present some characteristics of information system homomorphism, which reveal the interdependence of the three mappings, namely, object mapping, attribute mapping and value domain mapping. Besides, given a partition on universe, we can derive a new information system homomorphism defining a partition on universe identical with the partition given. In the mean time, some invariant characteristics of upper approximation and lower approximation under information system homomorphism are investigated. At last, we establish a surjection between rough sets of information systems under an information system homomorphism. Keywords Information system · Rough set · Upper approximation · Lower approximation · Homomorphism 1 Introduction Rough set theory, proposed by Pawlak [1], is an extension of set theory for the study of intelligent systems characterized by inexact, uncertain or vague information. Using the concepts of lower and upper approximations in rough set theory, the knowledge hidden in the systems may be discovered. In recent years, rough algebra and rough logic have been developed [2–5] based on rough set theory and the relation between rough set theory and other mathematical theories has been studied [6–10]. Rough Y.-H. Zhai · K.-S. Qu () School of Computer and Information Technology, Shanxi University, Taiyuan 030006, China e-mail: [email protected] Y.-H. Zhai e-mail: [email protected]
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set has gained increasingly studying not only in theoretical aspects, but in applied aspects and been successfully applied to many areas including machine learning, pattern recognition, decision analysis and data mining [11–14]. In rough set theory, an information system is just a table of two dimensions. Information systems, as a very important mathematical model in artificial intelligence, were widely investigated by [15–18]. As it is well-known, the communication between two information systems is a very important topic in the field of artificial intelligence. In mathematics, it can be explained as a mapping between the given information systems, and the image system is regarded as an explanation system of the original system. The notion of information system homomorphism as a kind of tool to study the communication between two information systems was introduced by Grzymala-Busse in [16]. In [17], the authors depicted the conditions which make an information system S = (U, AT, V , f ) to be selective (i.e., for arbitrary x ∈ U, [x]AT = x) in terms of endomorphism. Afterward, the authors of [18] studied some invariant characteristics of reduct and core of information systems under some homomorphisms, and gave a sufficient condition guaranteeing the existence of an endomorphism of an information system. In this paper, our main aims are to 1. Reveal the interdependence of the three mappings, namely, object mapping, attribute mapping and value domain mapping (Theorems 2–4). Thus, under some conditions we can derive one mapping from the others. 2. Give a sufficient condition (Theorem 5) by which, we can derive a new information system homomorphism constructing a partition identical with the partition given. It can be regarded as a first step on the relation between partitions of universe and information system homomorphisms. 3. Discuss invariant characteristics of some core concepts, such as upper and lower approximation (Theorems 6–8) and rough sets (Theorems 9–11) of rough set theory under information system homomorphism. To our knowledge this paper is the first to formally investigate the invariant characteristics of upper approximation, lower approximation and rough sets under information system homomorphism. These results and theorems have not been discussed in other references and occur for the first time in this paper. The plan of the paper is as follows. In the following section, we recall some terms and definitions to be used in the paper. In Sect. 3, some properties of information system homomorphism will be discussed. Section 4 shows some invariant characteristics of upper and lower approximations under information system homomorphism, by which a surjection between rough sets of information systems is established in Sect. 5. Conclusion and discussion of further work will close the paper in Sect. 6.
2 Basic Notions Definition 1 An information system is an ordered quadruple S = (U, AT, V , f ),
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where U is a set called the universe of S—elements of U are called objects, AT is a set of attributes, V = a∈AT Va is a set of values of attributes—Va will be called the domain of a, f : U × AT → V is a description function, such that f (x, a) ∈ Va for every x ∈ U and a ∈ AT. Let S = (U, AT, V , f ) be an information system, B, C ⊆ AT and x, y ∈ U. The equivalence relation IND(B) = {(x, y)|∀a ∈ B, f (x, a) = f (y, a)} is called B-indiscernibility relation. Set [x]B = {y ∈ U |(x, y) ∈ IND(B)}. It is easy to prove that U/B = {[x]B |x ∈ U } constitutes a partition of U . For two partitions U/B and U/C, if for each element Bi ∈ U/B, there exists some Cj ∈ U/C with Bi ⊆ Cj , then we say that the partition U/C is more coarse than the partition U/B and write U/B ⊆ U/C. Definition 2 Let S = (U, AT, V , f ) be an information system, X ⊆ U and B ⊆ AT. B-lower approximation of X is defined as: B(X) = {x ∈ U |[x]B ⊆ X} and B-upper approximation of X as: B(X) = {x ∈ U |[x]B ∩ X = ∅}. Furthermore, X is a B-definable set iff B(X) = B(X), and a B-non-definable set iff B(X) = B(X). As it is well known, B(X) is a set of objects that belong to X with certainty, while B(X) is a set of objects that possibly belong to X. From the definitions of upper and lower approximations, the following theorem can be easily derived [15]. Theorem 1 Let S = (U, AT, V , f ) be an information system, X, Y ⊆ U and B ⊆ AT. Then (1): (2): (3): (4): (5):
B(X) ⊆ X ⊆ B(X); B(B(X)) = B(B(X)) = B(X); B(B(X)) = B(B(X)) = B(X); X ⊆ Y ⇒ B(X) ⊆ B(Y ); X ⊆ Y ⇒ B(X) ⊆ B(Y ); B(X ∩ Y ) = B(X) ∩ B(Y ); B(X ∪ Y ) = B(X) ∪ B(Y ).
3 Information System Homomorphism First, let us recall the definition of information system homomorphism [16]. Definition 3 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be two information systems, let hO be a mapping of U into U , let hA be a mapping of AT into AT , and
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let hD be a mapping of V into V . The triple h = (hO , hA , hD ) is called a homomorphism of S into S , iff for all x ∈ U and a ∈ AT, hD (f (x, a)) = f (hO (x), hA (a)).
(1)
Let h = (hO , hA , hD ) be a homomorphism of S into S . We denote the image set of U in U by hO (U ) ⊆ U and then for each t ∈ hO (U ), denote the primary image set of t by h−1 O (t). Then we can derive a partition of U from the mapping hO , namely, U/ hO = {h−1 O (t)|t ∈ hO (U )}. Theorem 2 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . Then, hD is uniquely determined by hO and hA . Proof Let h1 = (hO , hA , hD1 ) be an information system homomorphism of S into S . Then for every a ∈ AT and x ∈ U, by formula (1), we have hD (f (x, a)) = f (hO (x), hA (a)) and hD1 (f (x, a)) = f (hO (x), hA (a)), and therefore hD1 (f (x, a)) = hD (f (x, a)). From the arbitrariness of a and x, it follows that hD = hD1 . Hence hD is uniquely determined by hO and hA . By the above theorem, we can see that, if h = (hO , hA , hD ) is an information system homomorphism of S into S , hD depends fully on hA and hO . However, for arbitrary hO and hA , there is not always an information system homomorphism h of S into S satisfying h = (hO , hA , hD ), i.e., we can’t guarantee the existence of hD . Definition 4 Let S = (U, AT, V , f ) be an information system. If for arbitrary a1 , a2 ∈ AT and a1 = a2 , there exists x ∈ U with f (x, a1 ) = f (x, a2 ), then S is attribute-irreducible. Similarly if for arbitrary x1 , x2 ∈ U and x1 = x2 , there exists a ∈ AT with f (x1 , a) = f (x2 , a), then S is object-irreducible. Theorem 3 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . If hO is surjective and S is attribute-irreducible, then hA is uniquely determined by hO and hD . Proof Let h1 = (hO , hA1 , hD ) be an information system homomorphism of S into S . Suppose hA = hA1 . Then there exists some a ∈ AT with hA1 (a) = hA (a). Since S is attribute-irreducible, there must be x ∈ U such that f (x , hA (a)) = f (x , hA1 (a)). For hO is surjective, there is x ∈ U with hO (x) = x . Hence, we have that f (hO (x), hA (a)) = f (hO (x), hA1 (a)). To h = (hO , hA , hD ), by formula (1), hD (f (x, a)) = f (hO (x), hA (a)) . To h1 = (hO , hA1 , hD ), similarly, hD (f (x, a)) = f (hO (x), hA1 (a)). So hD (f (x, a)) = hD (f (x, a)), a contradiction with hD , a mapping of V into V . Thus hA = hA1 and hA is uniquely determined by hO and hD . Similarly, we can easily prove the following theorem.
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Theorem 4 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . If hA is surjective and S is object-irreducible, then hO is uniquely determined by hA and hD . In the above theorem, the restriction to hA is not very strict. If hA is not surjective, let
S = U , hA (AT),
V ,f ,
a
a ∈hA (AT)
where f is the restriction of f on U × hA (AT). If S is object-irreducible, the conclusion still holds between S and S . However, the condition that S is object-irreducible is necessary. See the following example: Example 1 Information systems S = (U, AT, V , f ) and S = (U , AT , V , f ) are depicted by Table 1. Obviously, information system S is not object-irreducible. The object mapping hA and the value domain mapping hD are presented in Table 2. Two object mappings hO1 and hO2 can be found in Table 3. It is easy to prove that both h1 = (hO1 , hA , hD ) and h2 = (hO2 , hA , hD ) are information system homomorphisms of S into S . Therefore, hO can not be uniquely determined by hA and hD . In fact, there are 23 = 8 mappings respecting hA and hD . Theorem 5 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . For a partition U/B generated by B ⊆ AT, if the following conditions are satisfied: Table 1 Information systems S and S
Table 2 Mappings hA and hD
Table 3 Object mappings of S into S
a
b
x1
va1
vb3
x2
va2
vb2
x3
va3
vb1
hA b
a
a
va1
x2
va1
x3
va2
va1
va2
va3
vb1
vb2
vb3
va1
va1
va1
va1
va1
va1
hO1
x1
x1
hD
a
x1
a
hO2 x2
x1
x3
x2
x1
x1
x2
x2
x3
x2
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1. U/B ⊇ U/ hO , i.e., the partition U/B is more coarse than the partition U/ hO ; 2. For each X ∈ U/B, there is some P ⊆ X, P ∈ U/ hO , satisfying ∀a ∈ AT,
{f (y, a)|y ∈ P } = {f (x, a)|x ∈ X},
then, there exists an information system homomorphism hB = (hOB , hA , hDB ) of S into S with U/ hOB = U/B, i.e., the partition generated by hOB is identical with that generated by B. Proof For each x ∈ U, since U/B is a partition of U , we have [x]B = X ∈ U/B. Next, let us construct a mapping of U into U : hO (x), h−1 O (hO (x)) ∈ U/B, hOB (x) = / U/B, hO (y), h−1 O (hO (x)) ∈ where y ∈ P and P satisfies the condition 2. If there exist more than one P satisfying the condition, then for all y ∈ [x]B = X, we appoint only one. By the definition of hOB , it is obvious that hOB is a mapping of U into U and satisfies U/ hOB = U/B. For arbitrary x ∈ U and a ∈ AT, we define: hDB (f (x, a)) = f (hOB (x), hA (a)). Since hOB is a mapping of U into U , then f (hOB , hA (a)) is determined by hOB and hA , and thus hDB is a mapping of V into V . Now, it remains to illustrate the reasonableness of the definition of hDB . We only need to prove that, for arbitrary x1 , x2 ∈ U and a ∈ AT, if f (x1 , a) = f (x2 , a), then hDB (f (x1 , a)) = hDB (f (x2 , a)). There are four cases to consider. −1 −1 Case 1. If h−1 O (hO (x1 )) ∈ U/B and hO (hO (x2 )) ∈ U/B, i.e., both hO (hO (x1 )) −1 and hO (hO (x2 )) are equivalence classes of U/B, by the definition of hOB , we have that
hDB (f (x1 , a)) = f (hOB (x1 ), hA (a)) = f (hO (x1 ), hA (a)) = hD (f (x1 , a)) and that hDB (f (x2 , a)) = f (hOB (x2 ), hA (a)) = f (hO (x2 ), hA (a)) = hD (f (x2 , a)). Since f (x1 , a) = f (x2 , a), then hD (f (x1 , a)) = hD (f (x2 , a)), and thus hDB (f (x1 , a)) = hDB (f (x2 , a)) as required. / U/B and h−1 / U/B, since h−1 Case 2. If h−1 O (hO (x1 )) ∈ O (hO (x2 )) ∈ O (hO (x1 )), −1 hO (hO (x2 )) ∈ U/ hO and U/B ⊇ U/ hO , then there must be X1 ∈ U/B with h−1 O (hO (x1 )) ⊂ X1 . By the condition 2, we have P1 ⊆ X1 and for all a ∈ AT, {f (y, a)|y ∈ P1 } = {f (x, a)|x ∈ X1 }. From h−1 O (hO (x1 )) ⊂ X1 , it follows that x1 ∈ X1 , and hence that f (x1 , a) ∈ {f (x, a)|x ∈ X1 } = {f (y, a)|y ∈ P1 }.
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So there exists y1 ∈ P1 with f (y1 , a) = f (x1 , a). By the definition of hDB and hOB , there must be y ∈ P satisfying hDB (f (x1 , a)) = f (hOB (x1 ), hA (a)) = f (hO (y), hA (a)) = f (hO (y1 ), hA (a)) = hD (f (y1 , a)). In like manner, there are X2 ∈ U/B with h−1 O (hO (x2 )) ⊂ X2 and therefore P2 ∈ U/ hO and y2 ∈ P2 respecting f (y2 , a) = f (x2 , a). Then, hDB (f (x2 , a)) = f (hOB (x2 ), hA (a)) = f (hO (y2 ), hA (a)) = hD (f (y2 , a)). Since f (x1 , a) = f (x2 , a), then f (y1 , a) = f (y2 , a). So hD (f (y2 , a)) = hD (f (y1 , a)), that is to say, hDB (f (x1 , a)) = hDB (f (x2 , a)). −1 Case 3. If h−1 / U/B and h−1 / U/B, O (hO (x1 )) ∈ O (hO (x2 )) ∈ U/B, by hO (hO (x1 )) ∈ similar to Case 2, there exists y1 ∈ P with f (y1 , a) = f (x1 , a) and hDB (f (x1 , a)) = hD (f (y1 , a)), where P satisfies the condition 2. Again, by h−1 O (hO (x2 )) ∈ U/B, similar to Case 1, hDB (f (x2 , a)) = f (hOB (x2 ), hA (a)) = f (hO (x2 ), hA (a)) = hD (f (x2 , a)). Because f (x1 , a) = f (x2 , a) and f (y1 , a) = f (x1 , a), we have that hD (f (x1 , a)) = hD (f (y1 , a)) = hD (f (x2 , a)). Thus hDB (f (x1 , a)) = hDB (f (x2 , a)). −1 Case 4. If h−1 / U/B, we can prove it similarly O (hO (x1 )) ∈ U/B and hO (hO (x2 )) ∈ to Case 3. Theorem 5 shows that, from a partition given, we can derive a new information system homomorphism constructing a partition identical with the partition given, which can be regarded as a first step on the relation between partitions of universe and information system homomorphisms.
4 Invariance of Upper and Lower Approximations In rough set theory, upper and lower approximations are two basic notions and many meaningful results are based on them. So it is very important to study the invariance of upper and lower approximations under a homomorphism between information systems. Theorem 6 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems, B ⊆ AT and h = (hO , hA , hD ) be an information system homomorphism of S into S . Then hO (B(X)) ⊆ hA (B)(hO (X)) for arbitrary X ⊆ U. If hO is a surjection and hD is one-to-one, then the inverse inclusion is available, namely, hO (B(X)) = hA (B)(hO (X)).
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Proof First of all, let us prove hO (B(X)) ⊆ hA (B)(hO (X)) for arbitrary X ⊆ U. We need to prove that if x ∈ hO (B(X)), the condition that x ∈ hA (B)(hO (X)) holds. If x ∈ hO (B(X)), since hO (B(X)) is the image set of B(X), there exists x ∈ B(X) with hO (x) = x . In this case, if x ∈ X, according to hO (x) ∈ hO (X), we have that hO (x) ∈ hA (B)(hO (X)), i.e., x ∈ hA (B)(hO (X)), and therefore the conclusion is true. If x ∈ B(X)\X, since B(X) is upper approximation of X, there must be y ∈ X with [x]B = [y]B . So for all a ∈ B, it follows that f (x, a) = f (y, a) and hence that hD (f (x, a)) = hD (f (y, a)), i.e., f (hO (x), hA (a)) = f (hO (y), hA (a)). By the arbitrariness of a, then for all a ∈ hA (B), we have that f (hO (x), a ) = f (hO (y), a ), i.e., [hO (x)]hA (B) = [hO (y)]hA (B) . So, from y ∈ X, it follows that hO (y) ∈ hO (X) and hence that hO (y) ∈ hA (B)(hO (X)), i.e., [hO (y)]hA (B) ⊆ hA (B)(hO (X)). Then [hO (x)]hA (B) ⊆ hA (B)(hO (X)) and therefore x = hO (x) ∈ hA (B)(hO (X)) as required. If hO is surjective and hD is one-to-one, let us prove that hO (B(X)) ⊇ hA (B)(hO (X)). We need to prove that if x ∈ hA (B)(hO (X)), then x ∈ hO (B(X)). If x ∈ hO (X), since X is the primary image set of hO (X), there exists x ∈ X with hO (x) = x . According to x ∈ X ⊆ B(X), we have that x = hO (x) ∈ hO (X) ⊆ hO (B(X)), and hence the conclusion is true. If x ∈ hA (B)(hO (X))\hO (X), by the definition of upper approximation, there is y ∈ hO (X) satisfying f (x , a ) = f (y , a ) for all a ∈ hA (B). Because hO (X) is the image set of X, there exists y ∈ X with hO (y) = y . Again, for hO is a surjection, there exists x ∈ U with hO (x) = x . By f (x , a ) = f (y , a ), we have f (hO (x), hA (a)) = f (hO (y), hA (a)) and by the arbitrariness of a , hD (f (x, a)) = hD (f (y, a)) for all a ∈ B. Since hD is one-toone, then for all a ∈ B, f (x, a) = f (y, a), i.e., [x]B = [y]B . So, by y ∈ X, we have [x]B = [y]B ⊆ B(X) and hence x ∈ B(X). Thus x = hO (x) ∈ hO (B(X)), which means that hO (B(X)) = hA (B)(hO (X)). Theorem 7 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems, B ⊆ AT and h = (hO , hA , hD ) be an information system homomorphism of S into S . For arbitrary X ⊆ U, if hO is surjective and hD is one-to-one, then hO (B(X)) ⊆ hA (B)(hO (X)). Proof Let x ∈ hO (B(X)), then there exists x ∈ B(X) with hO (x) = x , since B(X) is the primary image set of hO (B(X)). If x ∈ / hA (B)(hO (X)), then there must be y ∈ U \hO (X) with [x ]hA (B) = [y ]hA (B) . Now for all a ∈ hA (B), we have f (x , a ) = f (y , a ). By hO is a surjection, there is y ∈ U respecting hO (y) = y . Because of the arbitrariness of a , for all a ∈ B, then f (hO (x), hA (a)) = f (hO (y), hA (a)), i.e., hD (f (x, a)) = hD (f (y, a)). Since hD is one-to-one, for all a ∈ B, f (x, a) = f (y, a) and hence [y]B = [x]B . Thus, by x ∈ B(X), we have [y]B = [x]B ⊆ B(X) and therefore y = hO (y) ∈ hO (B(X)) ⊆ hO (X), a contradiction with y ∈ U \hO (X). So x ∈ hA (B)(hO (X)), which completes the proof.
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The above theorems express that if hO is surjective and hD is one-to-one, the invariance of upper approximation is guaranteed under a homomorphism between S and S , while that of lower approximation is not. It should be noted that, in Theorem 7 the conditions that hO is surjective and hD is one-to-one are necessary. For instance, in Example 1, if we take hA , hD and hO1 to construct an information system homomorphism and let B = {a, b}, X = {x1 , x2 }, then hA (B) = {a }, hO (B(X)) = hO ({x1 , x2 }) = {x1 }, hA (B)(hO (X)) = hA (B)({x1 }) = ∅. Thus hO (B(X)) hA (B)(hO (X)). On the other hand, in Theorem 7, the inverse inclusion cannot be concluded. The insight is suggested by Theorem 8. Example 2 We define the object mapping hO and the attribute mapping hA from S (Table 4) into S (Table 5) as Tables 6 and 7. It can be easily proved that the value domain mapping hD is one-to-one. Let B = AT, X = {x2 , x3 , x4 }, then, Table 4 Information system S
a
b
c
d
e
f
x1
1
2
2
1
1
2
x2
2
1
1
2
3
1
x3
1
3
2
1
2
2
x4
3
1
1
3
1
1
x5
1
2
2
1
1
2
x6
1
3
2
1
2
2
x7
2
2
1
2
1
1
x8
2
1
1
2
3
1
x9
1
2
2
1
1
2
x10
3
3
1
3
3
1
Table 5 Information system S x1 x2 x3 x4 x5 x6 x7 x8
a
b
c
d
e
1
2
1
1
2
2
1
2
3
1
1
3
1
2
2
3
1
3
1
1
1
3
1
2
2
2
2
2
1
1
1
2
1
1
2
3
3
3
2
1
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Table 6 The object mapping hO of S into S x1
x2
x3
x4
x5
x6
x7
x8
x9
x10
x1
x2
x3
x4
x1
x5
x6
x2
x7
x8
Table 7 The attribute mapping hA of S into S
a
b
c
d
e
f
a
b
e
c
d
e
hA (B) = hA (AT) = AT = {a , b , c , d , e }, IND(hA (B)) = {{x1 , x7 }, {x2 }, {x3 , x5 }, {x4 }, {x6 }, {x8 }}, IND(B) = {{x1 , x5 , x9 }, {x2 , x8 }, {x3 , x6 }, {x4 }, {x7 }, {x10 }}, hO (B(X)) = hO ({x4 }) = {x4 }, hA (B)(hO (X)) = hA (B)({x2 , x3 , x4 }) = {x2 , x4 }, Y = h−1 O (hO (X)) = {x2 , x3 , x4 , x8 }, hO (B(Y )) = hO ({x2 , x4 , x8 }) = {x2 , x4 } = hA (B)(hO (X)) = hA (B)(hO (Y )). Obviously, we have hO (B(X)) ⊆ hA (B)(hO (X)) and, however, hO (B(X)) = hA (B)(hO (X)). But to the primary image Y of hO (X), the invariance of lower approximation is held, i.e., hO (B(Y )) = hA (B)(hO (Y )). In general, we have the following theorem. Theorem 8 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . B ⊆ AT, hO is surjective and hD is one-to-one. For every X ⊆ U, if let Y = h−1 O (hO (X)), then the formula that hO (B(Y )) = hA (B)(hO (Y )) is held. Proof From Theorem 7, it follows that hO (B(Y )) ⊆ hA (B)(hO (Y )). So it remains to prove that hO (B(Y )) ⊇ hA (B)(hO (Y )). Let y ∈ hA (B)(hO (Y )), then y ∈ hO (Y ). Since hO (Y ) is the image set of Y , there is y ∈ Y with hO (y) = y . If y ∈ B(Y ), obviously, we have y = hO (y) ∈ hO (B(Y )) and hence the conclusion as required. If y ∈ Y \B(Y ), there is z ∈ U \Y satisfying [y]B = [z]B . In this case, for all a ∈ B, we have f (y, a) = f (z, a), and hence hD (f (y, a)) = hD (f (z, a)) i.e., f (hO (y), hA (a)) = f (hO (z), hA (a)). Since hA (B) is the image set of B, for all a ∈ hA (B), the formula that f (hO (y), a ) = f (hO (z), a ) is also held. So [hO (y)]hA (B) = [hO (z)]hA (B) . From z ∈ / Y and Y = h−1 O (hO (X)), it follows that hO (z) ∈ / hO (Y ) and therefore that [hO (y)]hA (B) = [hO (z)]hA (B) hO (Y ), i.e., [y ]hA (B) hO (Y ). Then, y ∈ / hA (B)(hO (Y )), which results in a contradiction with y ∈ hA (B)(hO (Y )). It means that the case that y ∈ Y \B(Y ) must not be held, and that the case that y ∈ B(Y ) must be held. So hO (B(Y )) ⊆ hA (B)(hO (Y )). Thus hO (B(Y )) = hA (B)(hO (Y )).
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From Theorems 7 and 8 we can see that, if hO is surjective and hD is one-to-one, then hO (B(X)) ⊆ hA (B)(hO (X)). In particular, if X = h−1 O (hO (X)), then the other inclusion can be concluded, i.e.,hO (B(X)) = hA (B)(hO (X)).
5 A Surjective Mapping between Rough Sets Definition 5 For all X, Y ⊆ U, B ⊆ AT, if B(X) = B(Y ) and B(X) = B(Y ), then we call X and Y B-rough equivalent and denote X ≈B Y or, if there is no confusion, X ≈ Y. It is obvious that ≈ is an equivalence relation on P (U ). And we call the equivalence classes of ≈ rough sets. denotes the set of rough sets, i.e., = {[X]≈ |X ⊆ U }. For the sake of convenience, X is used for denoting rough set. Then for all X, Y ∈ X , we have B(X) = B(Y )and B(X) = B(Y ). The following result shows that the intersection of images of rough equivalent sets must be a definable set. Theorem 9 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . B ⊆ AT, hO and hA are surjective and hD is one-to-one. Then, for arbitrary X ∈ , X∈X hO (X) is a definable set, i.e.,
(1): hA (B) hO (X) = hO (X), X∈X
(2):
hA (B)
X∈X
hO (X) = hO (X).
X∈X
X∈X
Proof (1): Since hA (B)( X∈X hO (X)) ⊆ X∈X hO (X) is true, we only need to prove that h A (B)( X∈X hO (X)) ⊇ X∈X hO (X). Let x ∈ X∈X hO (X). If there is y ∈ [x ]hA (B) respecting y ∈ / hO (X), X∈X / we claim that the case can not occur and let us find a contradiction. If y ∈ / hO (H ). Since hO is surX∈X hO (X), then there must be H ∈ X with y ∈ . From H ∈ X , it follows that x ∈ (y) = y jective, there is y ∈ U \H with h O X∈X hO (X) ⊆ hO (H ), so there exists x ∈ H with hO (x) = x . Because y ∈ [x ]hA (B) , then for all a ∈ B , f (x , a ) = f (y , a ). Again because hA is surjective, for all a ∈ B, f (x , hA (a)) = f (y , hA (a)), i.e., hD (f (x, a)) = hD (f (y, a)). According to the arbitrariness of a and that hD is one-to-one, we have [x]B = [y]B . Let P = (H − {x}) ∪ {y}. We claim that B(P ) = B(H ), B(P ) = B(H ).
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In fact, if z ∈ B(P ), then [z]B ⊆ P . Since x ∈ / P (obviously x = y), then x ∈ / [z]B . From [x]B = [y]B , it follows that y ∈ / [z]B and hence that {x, y} [z]B , so [z]B ⊆ P \{x, y} = H \{x, y}. Thus [z]B ⊆ H, i.e., z ∈ B(H ). If, conversely, z ∈ B(H ), then [z]B ⊆ H. It must follow that y ∈ / [z]B and x ∈ / [z]B . If not, y ∈ [z]B ⊆ H, a contradiction with y ∈ U \H , and since [x]B = [y]B , then x ∈ / [z]B . So [z]B ⊆ H \{x, y} = P \{x, y} ⊆ P and [z]B ⊆ P , i.e., z ∈ B(P ). Thus B(P ) = B(H ). Similarly, if z ∈ B(P ), then [z]B ∩ P = ∅. We can assume p ∈ [z]B and p ∈ P . If p = y, then p ∈ H and [z]B = [p]B ⊆ B(H ), namely, z ∈ B(H ). If z = y, by [z]B = [y]B = [x]B , then x ∈ [z]B and by x ∈ H then [z]B ⊆ B(H ), i.e., z ∈ B(H ). Hence B(P ) ⊆ B(H ). In like manner, B(P ) ⊇ B(H ), which means B(P ) = B(H ). Since B(P ) = B(H ) and B(P ) = B(H ), then P ∈ X . At the moment, if there still exists u ∈ P respecting hO (u) = x , due to that x ∈ X∈X hO (X) ⊆ hO (H ) and that y ∈ / hO (H ), then x = y and u = y. Similar to above, we have Q = (P \{u}) ∪ {y} = (P \{u}) ∈ X . Due to the finiteness of H , we can obtain by anal ogy S ∈ X satisfying a contradiction with the hO (s) = x for all s ∈ S. That results in ∈ [x ] condition that x ∈ h (X) ⊆ h (S). Thus for all y O hA (B) X∈X O , we can gain ] ∈ h (B)( y ∈ X∈X hO (X), i.e., [x ⊆ h (X), so x A hA (B) X∈X O X∈X hO (X)), that is to say, hA (B)( h (X)) ⊇ h (X). X∈X O X∈X O Hence, hA (B)( X∈X hO (X)) = X∈X hO (X). (2): By (1), we have
hA (B) hO (X) = hA (B) hA (B) hO (X) X∈X
X∈X
hO (X) = hO (X), = hA (B) X∈X
X∈X
so hA (B)( X∈X hO (X)) = X∈X hO (X).
By the next theorem, we can see that, the intersection of lower approximations of images of rough equivalent sets must also be a definable set. Theorem 10 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S , B ⊆ AT. If hO , hA are surjective and hD is one-to-one, then for every X ∈ , there is P ∈ X respecting hA (B)(hO (P )) = hA (B)(hO (X)). X∈X
Proof Let Y ∈ X , then we will search for P by Y. We consider the following two cases. Case 1: For arbitrary X ∈ X , if hA (B)(hO (Y )) ⊆ hA (B)(hO (X)), then hA (B)(hO (Y )) ⊆
X∈X
hA (B)(hO (X)),
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and since hA (B)(hO (Y )) ⊇
hA (B)(hO (X)),
X∈X
we have that hA (B)(hO (Y )) = X∈X hA (B)(hO (X)). Set P = Y, then the conclusion is true. / Case 2: If Case 1 does not occur, there must be y ∈ hA (B)(hO (Y )) with y ∈ hA (B)(hO (X)) for some X ∈ X . If [y ]hA (B) hO (X) = ∅, by Theorem 6, we have that y ∈ / hA (B)(hO (X)) = hO (B(X)) = hO (B(Y )) = hA (B)(hO (Y )), / hA (B)(hO (Y )), a contradiction with so y ∈ y ∈ hA (B)(hO (Y )) ⊆ hA (B)(hO (Y )). Thus [y ]hA (B) ∩ hO (X) = ∅. At the moment, since [y ]hA (B) hO (X), there must = Y \Q, where Q = be y1 , y2 ∈ [y ]hA (B) with y1 ∈ hO (X) and y2 ∈ / hO (X). Set Y {q|hO (q) = y2 , q ∈ Y }. We claim that B(Y ) = B(Y ) and B(Y ) = B(Y ). ) and B(Y ) ⊇ B(Y ). So it remains to prove ⊆ Y, we have B(Y ) ⊇ B(Y Since Y ) and B(Y ) ⊆ B(Y ). that B(Y ) ⊆ B(Y ). By B(Y ) = B(X), we have [z]B ⊆ X. If z ∈ B(Y ), we will verify that z ∈ B(Y If hO (z) = y2 , then y2 = hO (z) ∈ hO ([z]B ) ⊆ hO (X), a contradiction with y2 ∈ / hO (X). Thus hO (z) = y2 . At the moment, if there is z1 ∈ [z]B with hO (z1 ) = y2 , by [z]B ⊆ X, then z1 ∈ X and therefore y2 = hO (z1 ) ∈ hO (X), a contradiction with y2 ∈ / hO (X). So for arbitrary z1 ∈ [z]B , we have that , i.e., z ∈ B(Y ). It follows that B(Y ) ⊆ hO (z1 ) = y2 and therefore that [z]B ⊆ Y ). B(Y ). Since [z]B ∩ Y = ∅, there is z1 ∈ If z ∈ B(Y ), we will verify that z ∈ B(Y , which means that z ∈ [z]B = [z1 ]B ⊆ B(Y ). [z]B ∩ Y. If hO (z1 ) = y2 , then z1 ∈ Y If hO (z1 ) = y2 , by y ∈ hA (B)(hO (Y )), then [y ]hA (B) ⊆ hO (Y ) and we have y1 ∈ hO (Y ), since [y ]hA (B) = [y1 ]hA (B) . Then there is y1 ∈ Y with hO (y1 ) = y1 , , since y = y . By [y ]hA (B) = [y ]hA (B) , for arbitrary satisfying y1 ∈ Y \Q = Y 1 2 1 2 a ∈ B, we have that f (y1 , hA (a)) = f (y2 , hA (a)), namely, f (hO (y1 ), hA (a)) = f (hO (z1 ), hA (a)). So hD (f (y1 , a)) = hD (f (z1 , a)). , we For hD is one-to-one, thus f (y1 , a) = f (z1 , a), i.e., [y1 ]B = [z1 ]B . Since y1 ∈ Y ), from which follows that B(Y ) ⊆ B(Y ). have that z ∈ [z]B = [z1 ]B = [y1 ]B ⊆ B(Y ] ] = Y \Q, we know that y ∈ Also, by Y / h ( Y ) and by [y = [y O hA (B) 2 2 hA (B) , )). So, under Case 2, we obtain Y ∈ X , satisfying y ∈ that y ∈ / hA (B)(hO (Y )), i.e., hA (B)(hO (Y )) ⊂ hA (B)(hO (Y ). hA (B)(hO (Y )) and y ∈ / hA (B)(hO (Y )) ⊆ hA (B)(hO (X)), similar Afterward, for arbitrary X ∈ X , if hA (B)(hO (Y to Case 1, let P = Y and the conclusion is true. Otherwise, there must be y ∈
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)) with y ∈ hA (B)(hO (Y / hA (B)(hO (X)) for some X ∈ X , and similar to Case 2, we obtain Y with )) ⊂ hA (B)(hO (Y ). Y )) ⊂ hA (B)(hO (Y hA (B)(hO ( Due to the finiteness of the set hA (B)(hO (Y ), we obtain P ∈ X with hA (B)(hO (P )) = X∈X hA (B)(hO (X)). By Theorems 9 and 10, we can construct a surjection between rough sets of two information systems. The surjection shows that in some sense the two information system has almost equivalent rough sets. Theorem 11 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . Both hO and hA are surjective and hD is one-to-one. we denote rough sets of S and S by and , respectively. Then, there exists a surjective mapping of into . Proof At first, let us construct the mapping of into . Let X = {X1 , X2 , . . . , Xt } be an element of and by Theorem 10, there is P ∈ X with hA (B)(hO (P )) = hA (B)(hO (X)). X∈X
Set X = [hO (P )]≈ and we define: ϕ(X ) = X , i.e., ϕ : X → [hO (P )]≈ . To X = {X1 , X2 , . . . , Xt }, by Theorem 9, we have
hA (B)(hO (X)) = hA (B) hO (X) = hO (X). X∈X
X∈X
X∈X
So hA (B)(hO (P )) is uniquely determined by X . On the other hand, from Theorem 6, it follows that hO (B(Xi )) = hA (B)(hO (Xi )), i = 1, 2, . . . , t. Since B(Xi ) = B(Xj ), i, j = 1, 2, . . . , t, we have that hA (B)(hO (P )) = hA (B)(hO (Xi )),
i = 1, 2, . . . , t.
So hA (B)(hO (P )) is uniquely determined by X . Hence the definition of ϕ from into is reasonable. Next, let us prove that ϕ is surjective. Let X = {X1 , X2 , . . . , Xλ } ∈ and Xi = h−1 O (Xi ), i = 1, 2, . . . , λ. We claim that B(Xi ) = B(Xj ) and B(Xi ) = B(Xj ), i, j = 1, 2, . . . , λ.
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In fact, since Xi is the primary image set of Xi , by Theorem 8, we have that hO (B(Xi )) = hA (B)(hO (Xi )) = hA (B)(Xi ) and that hO (B(Xj )) = hA (B)(hO (Xj )) = hA (B)(Xj ). From Xi , Xj ∈ X , it follows that hA (B)(Xi ) = hA (B)(Xj ) and hence that hO (B(Xi )) = hO (B(Xj )). If B(Xi ) = B(Xj ), then one of the following two cases must occur: (1) There exists x ∈ U with [x]B ⊆ B(Xi ), but [x]B B(Xj ), (2) There exists x ∈ U with [x]B B(Xi ), but [x]B ⊆ B(Xj ). If case (1) occurs, from [x]B ⊆ B(Xi ), it follows that hO ([x]B ) ⊆ hO (B(Xi )) = hO (B(Xj )) = hA (B)(Xj ) ⊆ Xj , i.e., hO ([x]B ) ⊆ Xj . Since Xj is the primary image set of Xj , therefore [x]B ⊆ Xj and hence [x]B ⊆ B(Xj ), a contradiction with [x]B B(Xj ). So B(Xi ) = B(Xj ). If case (2) occurs, similarly, we have B(Xi ) = B(Xj ). It remains to prove that B(Xi ) = B(Xj ), i, j = 1, 2, . . . , λ. By Theorem 6, we have that hO (B(Xi )) = hA (B)(hO (Xi )) = hA (B)(Xi ) and that hO (B(Xj )) = hA (B)(hO (Xj )) = hA (B)(Xj ). If B(Xi ) = B(Xj ), then there exists x ∈ U with [x]B ∩ Xi = ∅ and [x]B ∩ Xj = ∅, or with [x]B ∩ Xi = ∅ and [x]B ∩ Xj = ∅. If [x]B ∩ Xi = ∅ and [x]B ∩ Xj = ∅, then [x]B ⊆ B(Xi ) and [x]B B(Xj ). So hO ([x]B ) ⊆ hO (B(Xi )) = hA (B)(hO (Xi )) = hA (B)(Xi ) = hA (B)(Xj ), which induces hO ([x]B ) ∩ Xj = ∅. Since Xj is the primary image set of Xj , we have that [x]B ∩ Xj = ∅, which contradicts with [x]B ∩ Xj = ∅. In like manner, if [x]B ∩ Xi = ∅ and [x]B ∩ Xj = ∅, then there is also a contradiction. Hence B(Xi ) = B(Xj ), i, j = 1, 2, . . . , λ. For B(Xi ) = B(Xj ), B(Xi ) = B(Xj ), i, j = 1, 2, . . . , λ, we know that Xi , i = 1, 2, . . . , λ belong to the same rough set, X , say. Let X = {X1 , X2 , . . . , Xλ , Y1 , Y2 , . . . , Yk } ∈ . By Theorem 10, there is P ∈ X respecting hA (B)(hO (P )) = hA (B)(hO (X)). X∈X
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429
Set M = {X1 , X2 , . . . , Xλ }, N = {Y1 , Y2 , . . . , Yk }, then for Xi ∈ M, i = 1, 2, . . . , λ, −1 by Xi = h−1 O (Xi ) = hO (hO (Xi )) and Theorem 8, we have that hA (B)(hO (Xi )) = hO (B(Xi )). At the mean time, for each Yj ∈ N, by B(Xi ) = B(Yj ) and Theorem 7, we have that hA (B)(hO (Xi )) = hO (B(Xi )) = hO (B(Yj )) ⊆ hA (B)(hO (Yj )). So
hA (B)(hO (Xi )) ⊆
Xi ∈M
hA (B)(hO (Yj )).
Yj ∈N
Then,
hA (B)(hO (X)) =
X∈X
hA (B)(hO (Xi )) ∩ hA (B)(hO (Yj ))
Xi ∈M
=
Yj ∈N
hA (B)(hO (Xi ))
Xi ∈M
=
hA (B)(Xi ) = hA (B)(Xi ).
Xi ∈M
Hence hA (B)(hO (P )) =
hA (B)(hO (X)) = hA (B)(Xi ).
X∈X
And by Theorem 6, hA (B)(hO (P )) = hO (B(P )) = hO (B(Xi )) = hA (B)(hO (Xi )) = hA (B)(Xi ). Thus ϕ(X ) = X , which means that ϕ is a surjection of into .
Example 3 Let us apply the above theorem to information systems S and S depicted by Tables 4 and 5: (1) Let X1 = {x2 , x3 , x4 }, B = AT, by calculating, we have: B(X1 ) = {x4 }, B(X1 ) = {x2 , x3 , x4 , x6 , x8 }, [X1 ]≈ = {{x2 , x3 , x4 }, {x2 , x4 , x6 }, {x3 , x4 , x8 }, {x4 , x6 , x8 }}. Let P1 = {x3 , x4 , x8 }, then hA (B)(hO (P1 )) = hA (B)({x2 , x3 , x4 )) =
hA (B)(hO (Z)) = {x2 , x4 },
Z∈[X1 ]≈
hA (B)(hO (P1 )) = hA (B)({x2 , x3 , x4 )) = {x2 , x3 , x4 , x5 }. According to the definition of ϕ, the image of rough set [X1 ]≈ is [hO (P1 )]≈ , that is to say, ϕ([X1 ]≈ ) = [hO (P1 )]≈ = [{x2 , x3 , x4 }]≈ .
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Similarly, Let X2 = {x2 , x3 , x4 , x8 }, in the case, then B(X2 ) = {x2 , x4 , x8 }, B(X2 ) = {x2 , x3 , x4 , x6 , x8 }, [X2 ]≈ = {{x2 , x3 , x4 , x8 }, {x2 , x4 , x6 , x8 }}. Set P2 = {x2 , x4 , x6 , x8 }, then hA (B)(hO (P2 )) = hA (B)({x2 , x4 , x5 )) =
hA (B)(hO (Z)) = {x2 , x4 },
Z∈[X2 ]≈
hA (B)(hO (P2 )) = hA (B)({x2 , x4 , x5 )) = {x2 , x3 , x4 , x5 }. We have that ϕ([X2 ]≈ ) = [hO (P2 )]≈ = [{x2 , x3 , x4 }]≈ . It is easy to see that the image of [X1 ]≈ and that of [X2 ]≈ are identical under the mapping ϕ. (2) Given a rough set of S , we will find the primary image of the rough set in S. Given X = {x1 , x2 , x3 }, B = AT, then we have: hA (B)(X ) = {x1 , x2 , x3 , x5 , x7 }, hA (B)(X ) = {x2 }, [X ]≈ = {{x1 , x2 , x3 }, {x1 , x2 , x5 }, {x2 , x3 , x7 }, {x2 , x5 , x7 }}. Noticing the proof of Theorem 11, we calculate M = {{x1 , x2 , x3 , x5 , x8 }, {x1 , x2 , x5 , x6 , x8 }, {x2 , x3 , x8 , x9 }, {x2 , x6 , x8 , x9 }}, N = {{x1 , x2 , x3 , x8 }, {x1 , x2 , x6 , x8 }, {x2 , x3 , x5 , x8 }, {x2 , x5 , x6 , x8 }, {x1 , x2 , x3 , x8 , x9 }, {x2 , x3 , x5 , x8 , x9 }, {x1 , x2 , x6 , x8 , x9 }, {x2 , x5 , x6 , x8 , x9 }}. Let X = M ∪ N and P3 = {x1 , x2 , x3 , x5 , x8 }, we compute hA (B)(hO (Z)) = {x2 }, hA (B)(hO (P3 )) = hA (B)({x1 , x2 , x3 )) = Z∈M
hA (B)(hO (P3 )) = hA (B)({x1 , x2 , x3 )) = {x1 , x2 , x3 , x5 , x7 }. So ϕ(X ) = [hO (P3 )]≈ = [X ]≈ . 6 Conclusion and Further Works We know that rough set theory deals with the approximation of an arbitrary subset of a universe by two definable or observable subsets called upper and lower approximations, which are essential notions in rough set theory. However, there has been only little work relating the invariant characteristics of upper and lower approximations under information system homomorphisms.
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In the paper, we first reveal the interdependence of the three mappings of information system homomorphism, which can be concluded as (i) under some condition, object mapping depends on attribute mapping and value domain mapping; (ii) under some condition, attribute mapping depends on object mapping and value domain mapping; and (iii) value domain mapping depends fully on object mapping and attribute mapping. Moreover, we have discussed the invariant characteristics of upper and lower approximations and these results may have potential applications in knowledge reduction, decision making and reasoning about data, especially for the case of two information systems. For example, we establish the surjection between rough sets of information systems under an information system homomorphism, which can be considered as an application of our results in Sect. 4. Two important directions for further works are to study the invariant characteristics of decision rules [15] and the interrelation between information entropies [19] of information systems under information system homomorphisms. Acknowledgements This work was supported by the national natural science foundation of China (Nos. 60275019 and 70471003) and the natural science foundation of Shanxi, China (Nos. 2007011040 and 20041040).
References 1. Pawlak, Z.: Rough sets. Int. J. Comput. Inf. Sci. 11(5), 341–356 (1982) 2. Orlowska, E.: Rough set semantics for non-classical logics. In: Ziarko, W.P. (ed.) Rough Sets, Fuzzy Sets and Knowledge Discovery, pp. 143–148. Springer, London (1994) 3. Pagliani, P.: Rough set systems and logico-algebraic structures. In: Orlowska, E. (ed.) Incomplete Information: Rough Set Analysis, pp. 109–190. Physica-Verlag, Heidelburg (1997) 4. Cattaneo, G., Ciucci, D.: Algebraic structures for rough sets. Trans. Rough Sets 208–252 (2004) 5. Yao, Y.Y., Lin, T.Y.: Generalization of rough sets using model logic. Intell. Autom. Soft Comput. 2, 103–120 (1996) 6. Dubois, D., Prade, H.: Rough fuzzy sets and fuzzy rough sets. Int. J. Gen. Syst. 17, 191–209 (1990) 7. Yao, Y.Y.: A comparative study of fuzzy sets and rough sets. Inf. Sci. 109, 227–242 (1998) 8. Cornelis, C., Cock, M.D., Kerre, E.E.: Intuitionistic fuzzy rough sets: at the crossroads of imperfect knowledge. Expert Syst. 20(5), 260–270 (2003) 9. Wu, W.-Z., Zhang, W.-X.: Connections between rough set theory and Dempster-Shafer theory of evidence. Int. J. Gen. Syst. 31(4), 405–430 (2002) 10. Yao, Y.Y.: Interpretations of belief functions in the theory of rough sets. Inf. Sci. 104, 81–106 (1998) 11. Andrzej, C.: Speaker-independent recognition of isolated words using rough sets. Inf. Sci. 104, 3–14 (1998) 12. Hamid, M., Jerzy, W.G., James, A.B.: Entropy of English text: experiments with humans and machine learning systems based on rough sets. Inf. Sci. 104, 31–47 (1998) 13. Ahn, B.S., Cho, S.S., Kim, C.Y.: The integrated methodology of rough set theory and artificial neural network for business prediction. Expert Syst. Appl. 18(2), 65–74 (2000) 14. Hong, T.P., Wang, T.T., Wang, S.L., Chien, B.C.: Learning a coverage set of maximally general fuzzy rules by rough sets. Expert Syst. Appl. 19, 97–103 (2000) 15. Pawlak, Z.: Rough Sets. Theoretical Aspects of Reasoning about Data. Kluwer Academic, Dordrecht (1991) 16. Grzymala-Busse, J.W.: Algebraic properties of knowledge representation systems. In: Proceedings of the ACM SIGART International Symposium on Methodologies for Intelligent Systems, Knoxville, pp. 432–440, 1986 17. Grzymala-Busse, J.W., Sedelow, W.A. Jr.: On rough sets, and information system homomorphisms. Bull. Pol. Acad. Sci. Tech. Sci. 36(3–4), 233–239 (1988) 18. Li, D.-Y., Ma, Y.-C.: Invariant characters of information systems under some homomorphism. Inf. Sci. Int. J. 129(1–4), 211–220 (2000) 19. Düntsch, I., Gediga, G.: Uncertainty measures of rough set prediction. Artif. Intell. 106, 109–137 (1998)
On Characteristics of Information System Homomorphisms Yan-Hui Zhai · Kai-She Qu
Published online: 18 October 2007 © Springer Science+Business Media, LLC 2007
Abstract The notion of information system homomorphism as a powerful tool to study the relation between two information systems was introduced by J.W. Grzymala-Busse. In this work, we will present some characteristics of information system homomorphism, which reveal the interdependence of the three mappings, namely, object mapping, attribute mapping and value domain mapping. Besides, given a partition on universe, we can derive a new information system homomorphism defining a partition on universe identical with the partition given. In the mean time, some invariant characteristics of upper approximation and lower approximation under information system homomorphism are investigated. At last, we establish a surjection between rough sets of information systems under an information system homomorphism. Keywords Information system · Rough set · Upper approximation · Lower approximation · Homomorphism 1 Introduction Rough set theory, proposed by Pawlak [1], is an extension of set theory for the study of intelligent systems characterized by inexact, uncertain or vague information. Using the concepts of lower and upper approximations in rough set theory, the knowledge hidden in the systems may be discovered. In recent years, rough algebra and rough logic have been developed [2–5] based on rough set theory and the relation between rough set theory and other mathematical theories has been studied [6–10]. Rough Y.-H. Zhai · K.-S. Qu () School of Computer and Information Technology, Shanxi University, Taiyuan 030006, China e-mail: [email protected] Y.-H. Zhai e-mail: [email protected]
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set has gained increasingly studying not only in theoretical aspects, but in applied aspects and been successfully applied to many areas including machine learning, pattern recognition, decision analysis and data mining [11–14]. In rough set theory, an information system is just a table of two dimensions. Information systems, as a very important mathematical model in artificial intelligence, were widely investigated by [15–18]. As it is well-known, the communication between two information systems is a very important topic in the field of artificial intelligence. In mathematics, it can be explained as a mapping between the given information systems, and the image system is regarded as an explanation system of the original system. The notion of information system homomorphism as a kind of tool to study the communication between two information systems was introduced by Grzymala-Busse in [16]. In [17], the authors depicted the conditions which make an information system S = (U, AT, V , f ) to be selective (i.e., for arbitrary x ∈ U, [x]AT = x) in terms of endomorphism. Afterward, the authors of [18] studied some invariant characteristics of reduct and core of information systems under some homomorphisms, and gave a sufficient condition guaranteeing the existence of an endomorphism of an information system. In this paper, our main aims are to 1. Reveal the interdependence of the three mappings, namely, object mapping, attribute mapping and value domain mapping (Theorems 2–4). Thus, under some conditions we can derive one mapping from the others. 2. Give a sufficient condition (Theorem 5) by which, we can derive a new information system homomorphism constructing a partition identical with the partition given. It can be regarded as a first step on the relation between partitions of universe and information system homomorphisms. 3. Discuss invariant characteristics of some core concepts, such as upper and lower approximation (Theorems 6–8) and rough sets (Theorems 9–11) of rough set theory under information system homomorphism. To our knowledge this paper is the first to formally investigate the invariant characteristics of upper approximation, lower approximation and rough sets under information system homomorphism. These results and theorems have not been discussed in other references and occur for the first time in this paper. The plan of the paper is as follows. In the following section, we recall some terms and definitions to be used in the paper. In Sect. 3, some properties of information system homomorphism will be discussed. Section 4 shows some invariant characteristics of upper and lower approximations under information system homomorphism, by which a surjection between rough sets of information systems is established in Sect. 5. Conclusion and discussion of further work will close the paper in Sect. 6.
2 Basic Notions Definition 1 An information system is an ordered quadruple S = (U, AT, V , f ),
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where U is a set called the universe of S—elements of U are called objects, AT is a set of attributes, V = a∈AT Va is a set of values of attributes—Va will be called the domain of a, f : U × AT → V is a description function, such that f (x, a) ∈ Va for every x ∈ U and a ∈ AT. Let S = (U, AT, V , f ) be an information system, B, C ⊆ AT and x, y ∈ U. The equivalence relation IND(B) = {(x, y)|∀a ∈ B, f (x, a) = f (y, a)} is called B-indiscernibility relation. Set [x]B = {y ∈ U |(x, y) ∈ IND(B)}. It is easy to prove that U/B = {[x]B |x ∈ U } constitutes a partition of U . For two partitions U/B and U/C, if for each element Bi ∈ U/B, there exists some Cj ∈ U/C with Bi ⊆ Cj , then we say that the partition U/C is more coarse than the partition U/B and write U/B ⊆ U/C. Definition 2 Let S = (U, AT, V , f ) be an information system, X ⊆ U and B ⊆ AT. B-lower approximation of X is defined as: B(X) = {x ∈ U |[x]B ⊆ X} and B-upper approximation of X as: B(X) = {x ∈ U |[x]B ∩ X = ∅}. Furthermore, X is a B-definable set iff B(X) = B(X), and a B-non-definable set iff B(X) = B(X). As it is well known, B(X) is a set of objects that belong to X with certainty, while B(X) is a set of objects that possibly belong to X. From the definitions of upper and lower approximations, the following theorem can be easily derived [15]. Theorem 1 Let S = (U, AT, V , f ) be an information system, X, Y ⊆ U and B ⊆ AT. Then (1): (2): (3): (4): (5):
B(X) ⊆ X ⊆ B(X); B(B(X)) = B(B(X)) = B(X); B(B(X)) = B(B(X)) = B(X); X ⊆ Y ⇒ B(X) ⊆ B(Y ); X ⊆ Y ⇒ B(X) ⊆ B(Y ); B(X ∩ Y ) = B(X) ∩ B(Y ); B(X ∪ Y ) = B(X) ∪ B(Y ).
3 Information System Homomorphism First, let us recall the definition of information system homomorphism [16]. Definition 3 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be two information systems, let hO be a mapping of U into U , let hA be a mapping of AT into AT , and
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let hD be a mapping of V into V . The triple h = (hO , hA , hD ) is called a homomorphism of S into S , iff for all x ∈ U and a ∈ AT, hD (f (x, a)) = f (hO (x), hA (a)).
(1)
Let h = (hO , hA , hD ) be a homomorphism of S into S . We denote the image set of U in U by hO (U ) ⊆ U and then for each t ∈ hO (U ), denote the primary image set of t by h−1 O (t). Then we can derive a partition of U from the mapping hO , namely, U/ hO = {h−1 O (t)|t ∈ hO (U )}. Theorem 2 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . Then, hD is uniquely determined by hO and hA . Proof Let h1 = (hO , hA , hD1 ) be an information system homomorphism of S into S . Then for every a ∈ AT and x ∈ U, by formula (1), we have hD (f (x, a)) = f (hO (x), hA (a)) and hD1 (f (x, a)) = f (hO (x), hA (a)), and therefore hD1 (f (x, a)) = hD (f (x, a)). From the arbitrariness of a and x, it follows that hD = hD1 . Hence hD is uniquely determined by hO and hA . By the above theorem, we can see that, if h = (hO , hA , hD ) is an information system homomorphism of S into S , hD depends fully on hA and hO . However, for arbitrary hO and hA , there is not always an information system homomorphism h of S into S satisfying h = (hO , hA , hD ), i.e., we can’t guarantee the existence of hD . Definition 4 Let S = (U, AT, V , f ) be an information system. If for arbitrary a1 , a2 ∈ AT and a1 = a2 , there exists x ∈ U with f (x, a1 ) = f (x, a2 ), then S is attribute-irreducible. Similarly if for arbitrary x1 , x2 ∈ U and x1 = x2 , there exists a ∈ AT with f (x1 , a) = f (x2 , a), then S is object-irreducible. Theorem 3 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . If hO is surjective and S is attribute-irreducible, then hA is uniquely determined by hO and hD . Proof Let h1 = (hO , hA1 , hD ) be an information system homomorphism of S into S . Suppose hA = hA1 . Then there exists some a ∈ AT with hA1 (a) = hA (a). Since S is attribute-irreducible, there must be x ∈ U such that f (x , hA (a)) = f (x , hA1 (a)). For hO is surjective, there is x ∈ U with hO (x) = x . Hence, we have that f (hO (x), hA (a)) = f (hO (x), hA1 (a)). To h = (hO , hA , hD ), by formula (1), hD (f (x, a)) = f (hO (x), hA (a)) . To h1 = (hO , hA1 , hD ), similarly, hD (f (x, a)) = f (hO (x), hA1 (a)). So hD (f (x, a)) = hD (f (x, a)), a contradiction with hD , a mapping of V into V . Thus hA = hA1 and hA is uniquely determined by hO and hD . Similarly, we can easily prove the following theorem.
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Theorem 4 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . If hA is surjective and S is object-irreducible, then hO is uniquely determined by hA and hD . In the above theorem, the restriction to hA is not very strict. If hA is not surjective, let
S = U , hA (AT),
V ,f ,
a
a ∈hA (AT)
where f is the restriction of f on U × hA (AT). If S is object-irreducible, the conclusion still holds between S and S . However, the condition that S is object-irreducible is necessary. See the following example: Example 1 Information systems S = (U, AT, V , f ) and S = (U , AT , V , f ) are depicted by Table 1. Obviously, information system S is not object-irreducible. The object mapping hA and the value domain mapping hD are presented in Table 2. Two object mappings hO1 and hO2 can be found in Table 3. It is easy to prove that both h1 = (hO1 , hA , hD ) and h2 = (hO2 , hA , hD ) are information system homomorphisms of S into S . Therefore, hO can not be uniquely determined by hA and hD . In fact, there are 23 = 8 mappings respecting hA and hD . Theorem 5 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . For a partition U/B generated by B ⊆ AT, if the following conditions are satisfied: Table 1 Information systems S and S
Table 2 Mappings hA and hD
Table 3 Object mappings of S into S
a
b
x1
va1
vb3
x2
va2
vb2
x3
va3
vb1
hA b
a
a
va1
x2
va1
x3
va2
va1
va2
va3
vb1
vb2
vb3
va1
va1
va1
va1
va1
va1
hO1
x1
x1
hD
a
x1
a
hO2 x2
x1
x3
x2
x1
x1
x2
x2
x3
x2
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1. U/B ⊇ U/ hO , i.e., the partition U/B is more coarse than the partition U/ hO ; 2. For each X ∈ U/B, there is some P ⊆ X, P ∈ U/ hO , satisfying ∀a ∈ AT,
{f (y, a)|y ∈ P } = {f (x, a)|x ∈ X},
then, there exists an information system homomorphism hB = (hOB , hA , hDB ) of S into S with U/ hOB = U/B, i.e., the partition generated by hOB is identical with that generated by B. Proof For each x ∈ U, since U/B is a partition of U , we have [x]B = X ∈ U/B. Next, let us construct a mapping of U into U : hO (x), h−1 O (hO (x)) ∈ U/B, hOB (x) = / U/B, hO (y), h−1 O (hO (x)) ∈ where y ∈ P and P satisfies the condition 2. If there exist more than one P satisfying the condition, then for all y ∈ [x]B = X, we appoint only one. By the definition of hOB , it is obvious that hOB is a mapping of U into U and satisfies U/ hOB = U/B. For arbitrary x ∈ U and a ∈ AT, we define: hDB (f (x, a)) = f (hOB (x), hA (a)). Since hOB is a mapping of U into U , then f (hOB , hA (a)) is determined by hOB and hA , and thus hDB is a mapping of V into V . Now, it remains to illustrate the reasonableness of the definition of hDB . We only need to prove that, for arbitrary x1 , x2 ∈ U and a ∈ AT, if f (x1 , a) = f (x2 , a), then hDB (f (x1 , a)) = hDB (f (x2 , a)). There are four cases to consider. −1 −1 Case 1. If h−1 O (hO (x1 )) ∈ U/B and hO (hO (x2 )) ∈ U/B, i.e., both hO (hO (x1 )) −1 and hO (hO (x2 )) are equivalence classes of U/B, by the definition of hOB , we have that
hDB (f (x1 , a)) = f (hOB (x1 ), hA (a)) = f (hO (x1 ), hA (a)) = hD (f (x1 , a)) and that hDB (f (x2 , a)) = f (hOB (x2 ), hA (a)) = f (hO (x2 ), hA (a)) = hD (f (x2 , a)). Since f (x1 , a) = f (x2 , a), then hD (f (x1 , a)) = hD (f (x2 , a)), and thus hDB (f (x1 , a)) = hDB (f (x2 , a)) as required. / U/B and h−1 / U/B, since h−1 Case 2. If h−1 O (hO (x1 )) ∈ O (hO (x2 )) ∈ O (hO (x1 )), −1 hO (hO (x2 )) ∈ U/ hO and U/B ⊇ U/ hO , then there must be X1 ∈ U/B with h−1 O (hO (x1 )) ⊂ X1 . By the condition 2, we have P1 ⊆ X1 and for all a ∈ AT, {f (y, a)|y ∈ P1 } = {f (x, a)|x ∈ X1 }. From h−1 O (hO (x1 )) ⊂ X1 , it follows that x1 ∈ X1 , and hence that f (x1 , a) ∈ {f (x, a)|x ∈ X1 } = {f (y, a)|y ∈ P1 }.
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So there exists y1 ∈ P1 with f (y1 , a) = f (x1 , a). By the definition of hDB and hOB , there must be y ∈ P satisfying hDB (f (x1 , a)) = f (hOB (x1 ), hA (a)) = f (hO (y), hA (a)) = f (hO (y1 ), hA (a)) = hD (f (y1 , a)). In like manner, there are X2 ∈ U/B with h−1 O (hO (x2 )) ⊂ X2 and therefore P2 ∈ U/ hO and y2 ∈ P2 respecting f (y2 , a) = f (x2 , a). Then, hDB (f (x2 , a)) = f (hOB (x2 ), hA (a)) = f (hO (y2 ), hA (a)) = hD (f (y2 , a)). Since f (x1 , a) = f (x2 , a), then f (y1 , a) = f (y2 , a). So hD (f (y2 , a)) = hD (f (y1 , a)), that is to say, hDB (f (x1 , a)) = hDB (f (x2 , a)). −1 Case 3. If h−1 / U/B and h−1 / U/B, O (hO (x1 )) ∈ O (hO (x2 )) ∈ U/B, by hO (hO (x1 )) ∈ similar to Case 2, there exists y1 ∈ P with f (y1 , a) = f (x1 , a) and hDB (f (x1 , a)) = hD (f (y1 , a)), where P satisfies the condition 2. Again, by h−1 O (hO (x2 )) ∈ U/B, similar to Case 1, hDB (f (x2 , a)) = f (hOB (x2 ), hA (a)) = f (hO (x2 ), hA (a)) = hD (f (x2 , a)). Because f (x1 , a) = f (x2 , a) and f (y1 , a) = f (x1 , a), we have that hD (f (x1 , a)) = hD (f (y1 , a)) = hD (f (x2 , a)). Thus hDB (f (x1 , a)) = hDB (f (x2 , a)). −1 Case 4. If h−1 / U/B, we can prove it similarly O (hO (x1 )) ∈ U/B and hO (hO (x2 )) ∈ to Case 3. Theorem 5 shows that, from a partition given, we can derive a new information system homomorphism constructing a partition identical with the partition given, which can be regarded as a first step on the relation between partitions of universe and information system homomorphisms.
4 Invariance of Upper and Lower Approximations In rough set theory, upper and lower approximations are two basic notions and many meaningful results are based on them. So it is very important to study the invariance of upper and lower approximations under a homomorphism between information systems. Theorem 6 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems, B ⊆ AT and h = (hO , hA , hD ) be an information system homomorphism of S into S . Then hO (B(X)) ⊆ hA (B)(hO (X)) for arbitrary X ⊆ U. If hO is a surjection and hD is one-to-one, then the inverse inclusion is available, namely, hO (B(X)) = hA (B)(hO (X)).
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Proof First of all, let us prove hO (B(X)) ⊆ hA (B)(hO (X)) for arbitrary X ⊆ U. We need to prove that if x ∈ hO (B(X)), the condition that x ∈ hA (B)(hO (X)) holds. If x ∈ hO (B(X)), since hO (B(X)) is the image set of B(X), there exists x ∈ B(X) with hO (x) = x . In this case, if x ∈ X, according to hO (x) ∈ hO (X), we have that hO (x) ∈ hA (B)(hO (X)), i.e., x ∈ hA (B)(hO (X)), and therefore the conclusion is true. If x ∈ B(X)\X, since B(X) is upper approximation of X, there must be y ∈ X with [x]B = [y]B . So for all a ∈ B, it follows that f (x, a) = f (y, a) and hence that hD (f (x, a)) = hD (f (y, a)), i.e., f (hO (x), hA (a)) = f (hO (y), hA (a)). By the arbitrariness of a, then for all a ∈ hA (B), we have that f (hO (x), a ) = f (hO (y), a ), i.e., [hO (x)]hA (B) = [hO (y)]hA (B) . So, from y ∈ X, it follows that hO (y) ∈ hO (X) and hence that hO (y) ∈ hA (B)(hO (X)), i.e., [hO (y)]hA (B) ⊆ hA (B)(hO (X)). Then [hO (x)]hA (B) ⊆ hA (B)(hO (X)) and therefore x = hO (x) ∈ hA (B)(hO (X)) as required. If hO is surjective and hD is one-to-one, let us prove that hO (B(X)) ⊇ hA (B)(hO (X)). We need to prove that if x ∈ hA (B)(hO (X)), then x ∈ hO (B(X)). If x ∈ hO (X), since X is the primary image set of hO (X), there exists x ∈ X with hO (x) = x . According to x ∈ X ⊆ B(X), we have that x = hO (x) ∈ hO (X) ⊆ hO (B(X)), and hence the conclusion is true. If x ∈ hA (B)(hO (X))\hO (X), by the definition of upper approximation, there is y ∈ hO (X) satisfying f (x , a ) = f (y , a ) for all a ∈ hA (B). Because hO (X) is the image set of X, there exists y ∈ X with hO (y) = y . Again, for hO is a surjection, there exists x ∈ U with hO (x) = x . By f (x , a ) = f (y , a ), we have f (hO (x), hA (a)) = f (hO (y), hA (a)) and by the arbitrariness of a , hD (f (x, a)) = hD (f (y, a)) for all a ∈ B. Since hD is one-toone, then for all a ∈ B, f (x, a) = f (y, a), i.e., [x]B = [y]B . So, by y ∈ X, we have [x]B = [y]B ⊆ B(X) and hence x ∈ B(X). Thus x = hO (x) ∈ hO (B(X)), which means that hO (B(X)) = hA (B)(hO (X)). Theorem 7 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems, B ⊆ AT and h = (hO , hA , hD ) be an information system homomorphism of S into S . For arbitrary X ⊆ U, if hO is surjective and hD is one-to-one, then hO (B(X)) ⊆ hA (B)(hO (X)). Proof Let x ∈ hO (B(X)), then there exists x ∈ B(X) with hO (x) = x , since B(X) is the primary image set of hO (B(X)). If x ∈ / hA (B)(hO (X)), then there must be y ∈ U \hO (X) with [x ]hA (B) = [y ]hA (B) . Now for all a ∈ hA (B), we have f (x , a ) = f (y , a ). By hO is a surjection, there is y ∈ U respecting hO (y) = y . Because of the arbitrariness of a , for all a ∈ B, then f (hO (x), hA (a)) = f (hO (y), hA (a)), i.e., hD (f (x, a)) = hD (f (y, a)). Since hD is one-to-one, for all a ∈ B, f (x, a) = f (y, a) and hence [y]B = [x]B . Thus, by x ∈ B(X), we have [y]B = [x]B ⊆ B(X) and therefore y = hO (y) ∈ hO (B(X)) ⊆ hO (X), a contradiction with y ∈ U \hO (X). So x ∈ hA (B)(hO (X)), which completes the proof.
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The above theorems express that if hO is surjective and hD is one-to-one, the invariance of upper approximation is guaranteed under a homomorphism between S and S , while that of lower approximation is not. It should be noted that, in Theorem 7 the conditions that hO is surjective and hD is one-to-one are necessary. For instance, in Example 1, if we take hA , hD and hO1 to construct an information system homomorphism and let B = {a, b}, X = {x1 , x2 }, then hA (B) = {a }, hO (B(X)) = hO ({x1 , x2 }) = {x1 }, hA (B)(hO (X)) = hA (B)({x1 }) = ∅. Thus hO (B(X)) hA (B)(hO (X)). On the other hand, in Theorem 7, the inverse inclusion cannot be concluded. The insight is suggested by Theorem 8. Example 2 We define the object mapping hO and the attribute mapping hA from S (Table 4) into S (Table 5) as Tables 6 and 7. It can be easily proved that the value domain mapping hD is one-to-one. Let B = AT, X = {x2 , x3 , x4 }, then, Table 4 Information system S
a
b
c
d
e
f
x1
1
2
2
1
1
2
x2
2
1
1
2
3
1
x3
1
3
2
1
2
2
x4
3
1
1
3
1
1
x5
1
2
2
1
1
2
x6
1
3
2
1
2
2
x7
2
2
1
2
1
1
x8
2
1
1
2
3
1
x9
1
2
2
1
1
2
x10
3
3
1
3
3
1
Table 5 Information system S x1 x2 x3 x4 x5 x6 x7 x8
a
b
c
d
e
1
2
1
1
2
2
1
2
3
1
1
3
1
2
2
3
1
3
1
1
1
3
1
2
2
2
2
2
1
1
1
2
1
1
2
3
3
3
2
1
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Table 6 The object mapping hO of S into S x1
x2
x3
x4
x5
x6
x7
x8
x9
x10
x1
x2
x3
x4
x1
x5
x6
x2
x7
x8
Table 7 The attribute mapping hA of S into S
a
b
c
d
e
f
a
b
e
c
d
e
hA (B) = hA (AT) = AT = {a , b , c , d , e }, IND(hA (B)) = {{x1 , x7 }, {x2 }, {x3 , x5 }, {x4 }, {x6 }, {x8 }}, IND(B) = {{x1 , x5 , x9 }, {x2 , x8 }, {x3 , x6 }, {x4 }, {x7 }, {x10 }}, hO (B(X)) = hO ({x4 }) = {x4 }, hA (B)(hO (X)) = hA (B)({x2 , x3 , x4 }) = {x2 , x4 }, Y = h−1 O (hO (X)) = {x2 , x3 , x4 , x8 }, hO (B(Y )) = hO ({x2 , x4 , x8 }) = {x2 , x4 } = hA (B)(hO (X)) = hA (B)(hO (Y )). Obviously, we have hO (B(X)) ⊆ hA (B)(hO (X)) and, however, hO (B(X)) = hA (B)(hO (X)). But to the primary image Y of hO (X), the invariance of lower approximation is held, i.e., hO (B(Y )) = hA (B)(hO (Y )). In general, we have the following theorem. Theorem 8 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . B ⊆ AT, hO is surjective and hD is one-to-one. For every X ⊆ U, if let Y = h−1 O (hO (X)), then the formula that hO (B(Y )) = hA (B)(hO (Y )) is held. Proof From Theorem 7, it follows that hO (B(Y )) ⊆ hA (B)(hO (Y )). So it remains to prove that hO (B(Y )) ⊇ hA (B)(hO (Y )). Let y ∈ hA (B)(hO (Y )), then y ∈ hO (Y ). Since hO (Y ) is the image set of Y , there is y ∈ Y with hO (y) = y . If y ∈ B(Y ), obviously, we have y = hO (y) ∈ hO (B(Y )) and hence the conclusion as required. If y ∈ Y \B(Y ), there is z ∈ U \Y satisfying [y]B = [z]B . In this case, for all a ∈ B, we have f (y, a) = f (z, a), and hence hD (f (y, a)) = hD (f (z, a)) i.e., f (hO (y), hA (a)) = f (hO (z), hA (a)). Since hA (B) is the image set of B, for all a ∈ hA (B), the formula that f (hO (y), a ) = f (hO (z), a ) is also held. So [hO (y)]hA (B) = [hO (z)]hA (B) . From z ∈ / Y and Y = h−1 O (hO (X)), it follows that hO (z) ∈ / hO (Y ) and therefore that [hO (y)]hA (B) = [hO (z)]hA (B) hO (Y ), i.e., [y ]hA (B) hO (Y ). Then, y ∈ / hA (B)(hO (Y )), which results in a contradiction with y ∈ hA (B)(hO (Y )). It means that the case that y ∈ Y \B(Y ) must not be held, and that the case that y ∈ B(Y ) must be held. So hO (B(Y )) ⊆ hA (B)(hO (Y )). Thus hO (B(Y )) = hA (B)(hO (Y )).
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From Theorems 7 and 8 we can see that, if hO is surjective and hD is one-to-one, then hO (B(X)) ⊆ hA (B)(hO (X)). In particular, if X = h−1 O (hO (X)), then the other inclusion can be concluded, i.e.,hO (B(X)) = hA (B)(hO (X)).
5 A Surjective Mapping between Rough Sets Definition 5 For all X, Y ⊆ U, B ⊆ AT, if B(X) = B(Y ) and B(X) = B(Y ), then we call X and Y B-rough equivalent and denote X ≈B Y or, if there is no confusion, X ≈ Y. It is obvious that ≈ is an equivalence relation on P (U ). And we call the equivalence classes of ≈ rough sets. denotes the set of rough sets, i.e., = {[X]≈ |X ⊆ U }. For the sake of convenience, X is used for denoting rough set. Then for all X, Y ∈ X , we have B(X) = B(Y )and B(X) = B(Y ). The following result shows that the intersection of images of rough equivalent sets must be a definable set. Theorem 9 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . B ⊆ AT, hO and hA are surjective and hD is one-to-one. Then, for arbitrary X ∈ , X∈X hO (X) is a definable set, i.e.,
(1): hA (B) hO (X) = hO (X), X∈X
(2):
hA (B)
X∈X
hO (X) = hO (X).
X∈X
X∈X
Proof (1): Since hA (B)( X∈X hO (X)) ⊆ X∈X hO (X) is true, we only need to prove that h A (B)( X∈X hO (X)) ⊇ X∈X hO (X). Let x ∈ X∈X hO (X). If there is y ∈ [x ]hA (B) respecting y ∈ / hO (X), X∈X / we claim that the case can not occur and let us find a contradiction. If y ∈ / hO (H ). Since hO is surX∈X hO (X), then there must be H ∈ X with y ∈ . From H ∈ X , it follows that x ∈ (y) = y jective, there is y ∈ U \H with h O X∈X hO (X) ⊆ hO (H ), so there exists x ∈ H with hO (x) = x . Because y ∈ [x ]hA (B) , then for all a ∈ B , f (x , a ) = f (y , a ). Again because hA is surjective, for all a ∈ B, f (x , hA (a)) = f (y , hA (a)), i.e., hD (f (x, a)) = hD (f (y, a)). According to the arbitrariness of a and that hD is one-to-one, we have [x]B = [y]B . Let P = (H − {x}) ∪ {y}. We claim that B(P ) = B(H ), B(P ) = B(H ).
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In fact, if z ∈ B(P ), then [z]B ⊆ P . Since x ∈ / P (obviously x = y), then x ∈ / [z]B . From [x]B = [y]B , it follows that y ∈ / [z]B and hence that {x, y} [z]B , so [z]B ⊆ P \{x, y} = H \{x, y}. Thus [z]B ⊆ H, i.e., z ∈ B(H ). If, conversely, z ∈ B(H ), then [z]B ⊆ H. It must follow that y ∈ / [z]B and x ∈ / [z]B . If not, y ∈ [z]B ⊆ H, a contradiction with y ∈ U \H , and since [x]B = [y]B , then x ∈ / [z]B . So [z]B ⊆ H \{x, y} = P \{x, y} ⊆ P and [z]B ⊆ P , i.e., z ∈ B(P ). Thus B(P ) = B(H ). Similarly, if z ∈ B(P ), then [z]B ∩ P = ∅. We can assume p ∈ [z]B and p ∈ P . If p = y, then p ∈ H and [z]B = [p]B ⊆ B(H ), namely, z ∈ B(H ). If z = y, by [z]B = [y]B = [x]B , then x ∈ [z]B and by x ∈ H then [z]B ⊆ B(H ), i.e., z ∈ B(H ). Hence B(P ) ⊆ B(H ). In like manner, B(P ) ⊇ B(H ), which means B(P ) = B(H ). Since B(P ) = B(H ) and B(P ) = B(H ), then P ∈ X . At the moment, if there still exists u ∈ P respecting hO (u) = x , due to that x ∈ X∈X hO (X) ⊆ hO (H ) and that y ∈ / hO (H ), then x = y and u = y. Similar to above, we have Q = (P \{u}) ∪ {y} = (P \{u}) ∈ X . Due to the finiteness of H , we can obtain by anal ogy S ∈ X satisfying a contradiction with the hO (s) = x for all s ∈ S. That results in ∈ [x ] condition that x ∈ h (X) ⊆ h (S). Thus for all y O hA (B) X∈X O , we can gain ] ∈ h (B)( y ∈ X∈X hO (X), i.e., [x ⊆ h (X), so x A hA (B) X∈X O X∈X hO (X)), that is to say, hA (B)( h (X)) ⊇ h (X). X∈X O X∈X O Hence, hA (B)( X∈X hO (X)) = X∈X hO (X). (2): By (1), we have
hA (B) hO (X) = hA (B) hA (B) hO (X) X∈X
X∈X
hO (X) = hO (X), = hA (B) X∈X
X∈X
so hA (B)( X∈X hO (X)) = X∈X hO (X).
By the next theorem, we can see that, the intersection of lower approximations of images of rough equivalent sets must also be a definable set. Theorem 10 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S , B ⊆ AT. If hO , hA are surjective and hD is one-to-one, then for every X ∈ , there is P ∈ X respecting hA (B)(hO (P )) = hA (B)(hO (X)). X∈X
Proof Let Y ∈ X , then we will search for P by Y. We consider the following two cases. Case 1: For arbitrary X ∈ X , if hA (B)(hO (Y )) ⊆ hA (B)(hO (X)), then hA (B)(hO (Y )) ⊆
X∈X
hA (B)(hO (X)),
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and since hA (B)(hO (Y )) ⊇
hA (B)(hO (X)),
X∈X
we have that hA (B)(hO (Y )) = X∈X hA (B)(hO (X)). Set P = Y, then the conclusion is true. / Case 2: If Case 1 does not occur, there must be y ∈ hA (B)(hO (Y )) with y ∈ hA (B)(hO (X)) for some X ∈ X . If [y ]hA (B) hO (X) = ∅, by Theorem 6, we have that y ∈ / hA (B)(hO (X)) = hO (B(X)) = hO (B(Y )) = hA (B)(hO (Y )), / hA (B)(hO (Y )), a contradiction with so y ∈ y ∈ hA (B)(hO (Y )) ⊆ hA (B)(hO (Y )). Thus [y ]hA (B) ∩ hO (X) = ∅. At the moment, since [y ]hA (B) hO (X), there must = Y \Q, where Q = be y1 , y2 ∈ [y ]hA (B) with y1 ∈ hO (X) and y2 ∈ / hO (X). Set Y {q|hO (q) = y2 , q ∈ Y }. We claim that B(Y ) = B(Y ) and B(Y ) = B(Y ). ) and B(Y ) ⊇ B(Y ). So it remains to prove ⊆ Y, we have B(Y ) ⊇ B(Y Since Y ) and B(Y ) ⊆ B(Y ). that B(Y ) ⊆ B(Y ). By B(Y ) = B(X), we have [z]B ⊆ X. If z ∈ B(Y ), we will verify that z ∈ B(Y If hO (z) = y2 , then y2 = hO (z) ∈ hO ([z]B ) ⊆ hO (X), a contradiction with y2 ∈ / hO (X). Thus hO (z) = y2 . At the moment, if there is z1 ∈ [z]B with hO (z1 ) = y2 , by [z]B ⊆ X, then z1 ∈ X and therefore y2 = hO (z1 ) ∈ hO (X), a contradiction with y2 ∈ / hO (X). So for arbitrary z1 ∈ [z]B , we have that , i.e., z ∈ B(Y ). It follows that B(Y ) ⊆ hO (z1 ) = y2 and therefore that [z]B ⊆ Y ). B(Y ). Since [z]B ∩ Y = ∅, there is z1 ∈ If z ∈ B(Y ), we will verify that z ∈ B(Y , which means that z ∈ [z]B = [z1 ]B ⊆ B(Y ). [z]B ∩ Y. If hO (z1 ) = y2 , then z1 ∈ Y If hO (z1 ) = y2 , by y ∈ hA (B)(hO (Y )), then [y ]hA (B) ⊆ hO (Y ) and we have y1 ∈ hO (Y ), since [y ]hA (B) = [y1 ]hA (B) . Then there is y1 ∈ Y with hO (y1 ) = y1 , , since y = y . By [y ]hA (B) = [y ]hA (B) , for arbitrary satisfying y1 ∈ Y \Q = Y 1 2 1 2 a ∈ B, we have that f (y1 , hA (a)) = f (y2 , hA (a)), namely, f (hO (y1 ), hA (a)) = f (hO (z1 ), hA (a)). So hD (f (y1 , a)) = hD (f (z1 , a)). , we For hD is one-to-one, thus f (y1 , a) = f (z1 , a), i.e., [y1 ]B = [z1 ]B . Since y1 ∈ Y ), from which follows that B(Y ) ⊆ B(Y ). have that z ∈ [z]B = [z1 ]B = [y1 ]B ⊆ B(Y ] ] = Y \Q, we know that y ∈ Also, by Y / h ( Y ) and by [y = [y O hA (B) 2 2 hA (B) , )). So, under Case 2, we obtain Y ∈ X , satisfying y ∈ that y ∈ / hA (B)(hO (Y )), i.e., hA (B)(hO (Y )) ⊂ hA (B)(hO (Y ). hA (B)(hO (Y )) and y ∈ / hA (B)(hO (Y )) ⊆ hA (B)(hO (X)), similar Afterward, for arbitrary X ∈ X , if hA (B)(hO (Y to Case 1, let P = Y and the conclusion is true. Otherwise, there must be y ∈
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)) with y ∈ hA (B)(hO (Y / hA (B)(hO (X)) for some X ∈ X , and similar to Case 2, we obtain Y with )) ⊂ hA (B)(hO (Y ). Y )) ⊂ hA (B)(hO (Y hA (B)(hO ( Due to the finiteness of the set hA (B)(hO (Y ), we obtain P ∈ X with hA (B)(hO (P )) = X∈X hA (B)(hO (X)). By Theorems 9 and 10, we can construct a surjection between rough sets of two information systems. The surjection shows that in some sense the two information system has almost equivalent rough sets. Theorem 11 Let S = (U, AT, V , f ) and S = (U , AT , V , f ) be information systems and h = (hO , hA , hD ) be an information system homomorphism of S into S . Both hO and hA are surjective and hD is one-to-one. we denote rough sets of S and S by and , respectively. Then, there exists a surjective mapping of into . Proof At first, let us construct the mapping of into . Let X = {X1 , X2 , . . . , Xt } be an element of and by Theorem 10, there is P ∈ X with hA (B)(hO (P )) = hA (B)(hO (X)). X∈X
Set X = [hO (P )]≈ and we define: ϕ(X ) = X , i.e., ϕ : X → [hO (P )]≈ . To X = {X1 , X2 , . . . , Xt }, by Theorem 9, we have
hA (B)(hO (X)) = hA (B) hO (X) = hO (X). X∈X
X∈X
X∈X
So hA (B)(hO (P )) is uniquely determined by X . On the other hand, from Theorem 6, it follows that hO (B(Xi )) = hA (B)(hO (Xi )), i = 1, 2, . . . , t. Since B(Xi ) = B(Xj ), i, j = 1, 2, . . . , t, we have that hA (B)(hO (P )) = hA (B)(hO (Xi )),
i = 1, 2, . . . , t.
So hA (B)(hO (P )) is uniquely determined by X . Hence the definition of ϕ from into is reasonable. Next, let us prove that ϕ is surjective. Let X = {X1 , X2 , . . . , Xλ } ∈ and Xi = h−1 O (Xi ), i = 1, 2, . . . , λ. We claim that B(Xi ) = B(Xj ) and B(Xi ) = B(Xj ), i, j = 1, 2, . . . , λ.
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In fact, since Xi is the primary image set of Xi , by Theorem 8, we have that hO (B(Xi )) = hA (B)(hO (Xi )) = hA (B)(Xi ) and that hO (B(Xj )) = hA (B)(hO (Xj )) = hA (B)(Xj ). From Xi , Xj ∈ X , it follows that hA (B)(Xi ) = hA (B)(Xj ) and hence that hO (B(Xi )) = hO (B(Xj )). If B(Xi ) = B(Xj ), then one of the following two cases must occur: (1) There exists x ∈ U with [x]B ⊆ B(Xi ), but [x]B B(Xj ), (2) There exists x ∈ U with [x]B B(Xi ), but [x]B ⊆ B(Xj ). If case (1) occurs, from [x]B ⊆ B(Xi ), it follows that hO ([x]B ) ⊆ hO (B(Xi )) = hO (B(Xj )) = hA (B)(Xj ) ⊆ Xj , i.e., hO ([x]B ) ⊆ Xj . Since Xj is the primary image set of Xj , therefore [x]B ⊆ Xj and hence [x]B ⊆ B(Xj ), a contradiction with [x]B B(Xj ). So B(Xi ) = B(Xj ). If case (2) occurs, similarly, we have B(Xi ) = B(Xj ). It remains to prove that B(Xi ) = B(Xj ), i, j = 1, 2, . . . , λ. By Theorem 6, we have that hO (B(Xi )) = hA (B)(hO (Xi )) = hA (B)(Xi ) and that hO (B(Xj )) = hA (B)(hO (Xj )) = hA (B)(Xj ). If B(Xi ) = B(Xj ), then there exists x ∈ U with [x]B ∩ Xi = ∅ and [x]B ∩ Xj = ∅, or with [x]B ∩ Xi = ∅ and [x]B ∩ Xj = ∅. If [x]B ∩ Xi = ∅ and [x]B ∩ Xj = ∅, then [x]B ⊆ B(Xi ) and [x]B B(Xj ). So hO ([x]B ) ⊆ hO (B(Xi )) = hA (B)(hO (Xi )) = hA (B)(Xi ) = hA (B)(Xj ), which induces hO ([x]B ) ∩ Xj = ∅. Since Xj is the primary image set of Xj , we have that [x]B ∩ Xj = ∅, which contradicts with [x]B ∩ Xj = ∅. In like manner, if [x]B ∩ Xi = ∅ and [x]B ∩ Xj = ∅, then there is also a contradiction. Hence B(Xi ) = B(Xj ), i, j = 1, 2, . . . , λ. For B(Xi ) = B(Xj ), B(Xi ) = B(Xj ), i, j = 1, 2, . . . , λ, we know that Xi , i = 1, 2, . . . , λ belong to the same rough set, X , say. Let X = {X1 , X2 , . . . , Xλ , Y1 , Y2 , . . . , Yk } ∈ . By Theorem 10, there is P ∈ X respecting hA (B)(hO (P )) = hA (B)(hO (X)). X∈X
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429
Set M = {X1 , X2 , . . . , Xλ }, N = {Y1 , Y2 , . . . , Yk }, then for Xi ∈ M, i = 1, 2, . . . , λ, −1 by Xi = h−1 O (Xi ) = hO (hO (Xi )) and Theorem 8, we have that hA (B)(hO (Xi )) = hO (B(Xi )). At the mean time, for each Yj ∈ N, by B(Xi ) = B(Yj ) and Theorem 7, we have that hA (B)(hO (Xi )) = hO (B(Xi )) = hO (B(Yj )) ⊆ hA (B)(hO (Yj )). So
hA (B)(hO (Xi )) ⊆
Xi ∈M
hA (B)(hO (Yj )).
Yj ∈N
Then,
hA (B)(hO (X)) =
X∈X
hA (B)(hO (Xi )) ∩ hA (B)(hO (Yj ))
Xi ∈M
=
Yj ∈N
hA (B)(hO (Xi ))
Xi ∈M
=
hA (B)(Xi ) = hA (B)(Xi ).
Xi ∈M
Hence hA (B)(hO (P )) =
hA (B)(hO (X)) = hA (B)(Xi ).
X∈X
And by Theorem 6, hA (B)(hO (P )) = hO (B(P )) = hO (B(Xi )) = hA (B)(hO (Xi )) = hA (B)(Xi ). Thus ϕ(X ) = X , which means that ϕ is a surjection of into .
Example 3 Let us apply the above theorem to information systems S and S depicted by Tables 4 and 5: (1) Let X1 = {x2 , x3 , x4 }, B = AT, by calculating, we have: B(X1 ) = {x4 }, B(X1 ) = {x2 , x3 , x4 , x6 , x8 }, [X1 ]≈ = {{x2 , x3 , x4 }, {x2 , x4 , x6 }, {x3 , x4 , x8 }, {x4 , x6 , x8 }}. Let P1 = {x3 , x4 , x8 }, then hA (B)(hO (P1 )) = hA (B)({x2 , x3 , x4 )) =
hA (B)(hO (Z)) = {x2 , x4 },
Z∈[X1 ]≈
hA (B)(hO (P1 )) = hA (B)({x2 , x3 , x4 )) = {x2 , x3 , x4 , x5 }. According to the definition of ϕ, the image of rough set [X1 ]≈ is [hO (P1 )]≈ , that is to say, ϕ([X1 ]≈ ) = [hO (P1 )]≈ = [{x2 , x3 , x4 }]≈ .
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Similarly, Let X2 = {x2 , x3 , x4 , x8 }, in the case, then B(X2 ) = {x2 , x4 , x8 }, B(X2 ) = {x2 , x3 , x4 , x6 , x8 }, [X2 ]≈ = {{x2 , x3 , x4 , x8 }, {x2 , x4 , x6 , x8 }}. Set P2 = {x2 , x4 , x6 , x8 }, then hA (B)(hO (P2 )) = hA (B)({x2 , x4 , x5 )) =
hA (B)(hO (Z)) = {x2 , x4 },
Z∈[X2 ]≈
hA (B)(hO (P2 )) = hA (B)({x2 , x4 , x5 )) = {x2 , x3 , x4 , x5 }. We have that ϕ([X2 ]≈ ) = [hO (P2 )]≈ = [{x2 , x3 , x4 }]≈ . It is easy to see that the image of [X1 ]≈ and that of [X2 ]≈ are identical under the mapping ϕ. (2) Given a rough set of S , we will find the primary image of the rough set in S. Given X = {x1 , x2 , x3 }, B = AT, then we have: hA (B)(X ) = {x1 , x2 , x3 , x5 , x7 }, hA (B)(X ) = {x2 }, [X ]≈ = {{x1 , x2 , x3 }, {x1 , x2 , x5 }, {x2 , x3 , x7 }, {x2 , x5 , x7 }}. Noticing the proof of Theorem 11, we calculate M = {{x1 , x2 , x3 , x5 , x8 }, {x1 , x2 , x5 , x6 , x8 }, {x2 , x3 , x8 , x9 }, {x2 , x6 , x8 , x9 }}, N = {{x1 , x2 , x3 , x8 }, {x1 , x2 , x6 , x8 }, {x2 , x3 , x5 , x8 }, {x2 , x5 , x6 , x8 }, {x1 , x2 , x3 , x8 , x9 }, {x2 , x3 , x5 , x8 , x9 }, {x1 , x2 , x6 , x8 , x9 }, {x2 , x5 , x6 , x8 , x9 }}. Let X = M ∪ N and P3 = {x1 , x2 , x3 , x5 , x8 }, we compute hA (B)(hO (Z)) = {x2 }, hA (B)(hO (P3 )) = hA (B)({x1 , x2 , x3 )) = Z∈M
hA (B)(hO (P3 )) = hA (B)({x1 , x2 , x3 )) = {x1 , x2 , x3 , x5 , x7 }. So ϕ(X ) = [hO (P3 )]≈ = [X ]≈ . 6 Conclusion and Further Works We know that rough set theory deals with the approximation of an arbitrary subset of a universe by two definable or observable subsets called upper and lower approximations, which are essential notions in rough set theory. However, there has been only little work relating the invariant characteristics of upper and lower approximations under information system homomorphisms.
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In the paper, we first reveal the interdependence of the three mappings of information system homomorphism, which can be concluded as (i) under some condition, object mapping depends on attribute mapping and value domain mapping; (ii) under some condition, attribute mapping depends on object mapping and value domain mapping; and (iii) value domain mapping depends fully on object mapping and attribute mapping. Moreover, we have discussed the invariant characteristics of upper and lower approximations and these results may have potential applications in knowledge reduction, decision making and reasoning about data, especially for the case of two information systems. For example, we establish the surjection between rough sets of information systems under an information system homomorphism, which can be considered as an application of our results in Sect. 4. Two important directions for further works are to study the invariant characteristics of decision rules [15] and the interrelation between information entropies [19] of information systems under information system homomorphisms. Acknowledgements This work was supported by the national natural science foundation of China (Nos. 60275019 and 70471003) and the natural science foundation of Shanxi, China (Nos. 2007011040 and 20041040).
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