On Attribute-Value Pairs in Information
Systems
M. Shokry
Physics and Engineering Mathematics department, Faculty of Engineering, Tanta University, Egypt,
Tanta, Zip code 3111, Seperbay, Tanta, Egypt, mohnayle@yahoo.com
Assem. Elshenawy
Physics and Engineering Mathematics department, Faculty of Engineering, Tanta University, Egypt,
Tanta, Zip code 3111, Seperbay, Tanta, Egypt, dod_assem@yahoo.com
Abstract
Complete decision tables are described by indiscernibility relations. Data sets, described by decision tables, are complete when for all cases (examples, objects) the corresponding attribute values are known, the normal technique used to get the block of attribute value pairs of any information system was the intersection between sets. We will discuss a new technique to work with decision tables using a block of an attribute-value pairs by using the union between sets. We will discuss the affect of the condition sets relation to the block of attribute value pair for the sets between each other and some new properties were discussed.
Keywords: Rough set; information system; block of attribute-value pairs.
1. Introduction
Rough set theory is approaching its maturity. Rough set theory has two facts: one logical and philosophical, and another practical. The development of the field has been related to the development of these two facets. In fact, many important developments in rough set theory are related to logical, algebraic and topological approaches. On the other side, many successful applications of rough set theory have been experimented in many fields, ranging from medicine to environmental management, from finance to marketing. However, despite the double nature of the rough set approach, we have to admit that there is a certain gap between theoretical research from one side, and the application on the other side. There is a tendency of researchers more interested in applications to consider the theory of rough sets as too esoteric. Similarly, there is also a tendency of researchers more interested in theory to consider the applications as too trivial. These tendencies deteriorate both the theoretical research and the applied research. Indeed, what is the usefulness of a theory that does not give any insight into the real life applications? And on the other hand, how one can imagine to make a good application without a well founded theory? Therefore, rough set theory needs that its two facets, the logical and the practical, start to speak to each other, permitting to the theory to give a solid background to the applications, and to the applications to avoid ad hoc solutions. In fact, the main characteristic of rough set theory is a property that is basic both in theoretical developments and in applications: simplicity. A good theory has to be elegantly simple, as well as a good application needs an approach as simple as possible The approximation space [4, 5, 7, 10] is a pair of (U, R), where U is a non-empty finite set of objects(states, patients, digits, cars, students,….,etc.) called a universe and R is an equivalence relation over U which makes a partition for U, i.e. a family C= {X1, X2, X3,…, Xn } such that Xi U, Xi , Xi Xj = for i j, i , j= 1, 2, 3,…,n and
Xi= U, the class C is called the knowledge base of (U, R). the universe U of objects with relation R play an
to the same decision value is called a concept (or a class). In most articles on rough set theory it is assumed that for all variables and all cases the corresponding values are specified. For such tables the indiscernibility relation, one of the most fundamental ideas of rough set theory, describes cases that can be distinguished from other cases [1, 2, 13]. For the set of elements U called universe and A refers to a non empty finite set of attributes,
a A, such that a: U Va . The set Va is called the domain of the attribute a. For an element x U and an
attribute a A, the pair (x, a x ) U Va indicates that x has the attribute a x . Rough set theory can be
regarded as a new mathematical tool for imperfect data analysis. The theory has found applications in many domains, such as decision support engineering, environment, banking, medicine and others. bjects characterized by the same information are indiscernible (similar) in view of the available information about them. The indiscernibility relation generated in this way is the mathematical basis of rough set theory. Any set of all indiscernible (similar) objects is called an elementary set, and forms a basic granule (atom) of knowledge about the universe. Any union of some elementary sets is referred to as a crisp (precise) set – otherwise the set is rough (imprecise, vague). Each rough set has boundary-line cases, i.e., objects which cannot be with certainty classified, by employing the available knowledge, as members of the set or its complement. Obviously rough sets, in contrast to precise sets, cannot be characterized in terms of information about their elements [11, 12, 14].
2. Preliminaries
2.1 Blocks of attribute-value pairs
The input data sets are presented in the form of a decision table, an example of a decision table is shown in Table 1. Rows of the decision table represent cases, while columns are labeled by variables. The set of all cases will be denoted by U. In Table 1, U= {1, 2, ..., 8}. Independent variables are called attributes and a dependent variable is called a decision and is denoted by d. The set of all attributes will be denoted by A. In Table 1, A= {Temperature, Headache, Nausea}. Any decision table defines a function ρ that maps the direct product of U and A into the set of all values. For example, in Table 1, ρ(1, Temperature)= high. Function ρ describing Table 1 is completely specified (total). A decision table with completely specified function ρ will be called completely specified, or, for the sake of simplicity, complete Rough set theory [8], [9] is based on the idea of an indiscernibility relation, defined for complete decision tables. Let B be a nonempty subset of the set A of all attributes. The indiscernibility relation IND (B) is a relation on U defined for , U as follows:
( , ) IND (B) if and only if ρ( , ) = ρ( , ) for all , So we have IND(A) are {1},{2},{3},{4,5} ,{6, 8} and {7}. For complete decision tables if t = ( , ) is an attribute-value pair then a block of t, denoted [t], is a set of all cases from U that for attribute a have value v then we have,
[(Temperature, high)] = {1, 3, 4, 5}, [(Temperature, very high)] = {2}, [(Temperature, normal)] = {6, 7, 8}, [(Headache, yes)] = {1, 2, 4, 5, 6, 8}, [(Headache, no)] = {3, 7}, [(Nausea, no)] = {1, 3, 6}, [(Nausea, yes)] = {2, 4, 5, 7}. The indiscernibility relation IND (B) is known when all elementary blocks of IND (B) are known. Such elementary blocks of B are intersections of the corresponding attribute-value pairs, i.e., for any case x U, [x]B
= ∩{[( , )] / , ρ( , ) = } , then we have:
[1]A = [(Temperature, high)]∩[(Headache, yes)]∩[(Nausea, no)] = {1}, [2]A = [(Temperature, very high)]
∩[(Headache, yes)]∩[(Nausea, yes)]= {2}, [3]A = [(Temperature, high)]∩[(Headache, no)]∩[(Nausea, no)]=
{3}, [4]A= [5]A= [(Temperature, high)]∩[(Headache, yes)]∩[(Nausea, yes)]= {4, 5},[6]A= [8]A= [(Temperature,
normal)]∩[(Headache, yes)]∩[(Nausea, no]= {6, 8}, [7]A = [(Temperature, normal)]∩ [(Headache,
no]∩[(Nausea, yes)]= {7}.
Table1. A complete decision table
Attributes Decision
Case Temperature Headache Nausea Flu
1 high yes no yes
2 very high yes yes yes
3 high no no no 4 high yes yes yes 5 high yes yes no 6 normal yes no no 7 normal no yes no
3. New result Definition 3.1
The [x]B can be defined as the union of blocks of attribute-value pairs ( , ) for all attributes a from B
for which ρ( , ) is specified and ρ( , ) = , i.e. [x]B = {[( , )] / , ρ( , ) = }.
Proposition 3.1
For any information system with condition attributes C, let , are sets of condition attributes, , C and , for set of objects U, x U, then [x]A [x]B ,when we consider [x] as the union between the
selected sets.
Proof:
Let = { 1, 2, ….., i}, = { 1, 2, ….., i ,…. , n}. Let [xj]B be a set induced by all unions for all attribute
and the object xj, then we have:
[xj]B = [xj] 1 [xj] 2 [xj] 3 …. [xj] i ….. [xj] n .
Since [xj]A is a part of union sets [xj]B ,
[xj]B =( ) ( ), so we have
[xi]A [xi]B ,for all xi □
Proposition 3.2
For any information system with condition attributes C, let , are sets of condition attributes, , C and , for set of objects U, x U then [x]A [x]B , when we consider [x] as the intersection between the
selected sets.
Proof: Obviously □
Example 3.1
If we have information system with five condition attributes C= { 1, 2, 3, 4, 5} and three objects
U= { 1, 2, 3}, let A= { 2, 3} and B= { 1, 2, 3, 4}, i.e. , are subsets of condition attributes C as
Table2. An example of a complete data table
Condition attributes
Objects 1 2 3 4 5
1 1 0 1 1 0
2 0 1 0 1 1
3 0 0 1 0 1
First, when we take [x] as the intersection between sets:
[x1]B = {x1} {x1, x3} {x1, x3} {x1, x2}= {x1}, [x2]B = {x2, x3} {x2} {x2} {x1, x2}= {x2} ,
[x3]B = {x2, x3} {x1, x3} {x1, x3} {x3}= {x3}, [x1]A = {x1, x3} {x1, x3}= { x1, x3},
[x2]A = {x2} {x2}= { x2} and [x3]A = {x1, x3} {x1, x3}= { x1, x3}, then we can note that for all objects, [x]A
[x]B .
Second, when we take [x] as the union between sets:
[x1]A= {x1, x3} {x1, x3}= { x1, x3}, [x2]A = {x2} {x2}= {x2} , [x3]A = {x1, x3} {x1, x3}= { x1, x3},
[x1]B = {x1} {x1, x3} {x1, x3} {x1, x2}= {x1, x2, x3}, [x2]B = {x2, x3} {x2} {x2} {x1, x2}= {x1, x2, x3} and
[x3]B = {x2, x3} {x1, x3} {x1, x3} {x3} = {x1, x2, x3}, then we can easily note that for all objects, [x]A [x]B
Proposition 3.3
For any information system having C as condition attributes and U as set of objects, for any subsets , of condition attributes, , C , x U we have the following relations when we consider [x] as the intersection between sets:
[x]A [x]B [x]A [x]B
Proof
Let = { j, j+1, j+2,…, k } then, = …
Since : = … and = … . We
have that is a part of the sets in both of and , since will decrease the sets of and , hence we have that should be a part of the intersection [x]A [x]B.
By the same idea we can proof that [x]A [x]B □
Proposition 3.4:
For any information system having C as condition attributes and U as set of objects, for any subsets , of condition attributes, , C , x U we have the following relations when we consider [x] as the intersection between sets:
Proof:
Obviously □
Proposition 3.5
For any information system having C as condition attributes and U as set of objects, for any subsets , of condition attributes, , C , x U , we have the following relations when we consider [x] as the union between sets:
[x]A [x]B [x]A [x]B [x]A , [x]B [x]A , [x]B
Proof:
Obviously □
Example 3.2:
Continued from example 3.1 if we let, = { 1, 2, 4, 5} and = { 2, 3, 5}
= { 2, 5}, = C = { 1, 2, 3, 4, 5}
First, when we take [x] as the intersection between sets:
[x1]A = {x1} {x1, x3} {x1, x2} {x1}= {x1},
[x2]A = {x2, x3} {x2} {x1, x2} {x2, x3}= {x2},
[x3]A = {x2, x3} {x1, x3} {x3} {x2, x3}= {x3},
[x1]B = {x1, x3} {x1, x3} {x1} = {x1},
[x2]B = {x2} {x2} {x2, x3} = {x2},
[x3]B = {x1, x3} {x1, x3} {x2, x3} = {x3},
= {x1, x3} {x1} = {x1},
= {x2, x3} {x2} = {x2},
= {x1, x3} {x2, x3} = {x3},
= {x1} {x1, x3} {x1, x3} {x1, x2} {x1}= {x1},
= {x2, x3} {x2} {x2} {x1, x2} {x2, x3}= {x2},
= {x2, x3} {x1, x3} {x1, x3} {x3} {x2, x3}= {x3}.
Hence it is easily can be noted the following:
2. [x]A [x]B
3. [x]A , [x]B
4. [x]A , [x]B
Second, when we take [x] as the union between sets:
[x1]A = {x1} {x1, x3} {x1, x2} {x1}= {x1, x2, x3},
[x2]A = {x2, x3} {x2} {x1, x2} {x2, x3}= {x1, x2, x3},
[x3]A = {x2, x3} {x1, x3} {x3} {x2, x3}= {x1, x2, x3},
[x1]B = {x1, x3} {x1, x3} {x1} = {x1, x3},
[x2]B = {x2} {x2} {x2, x3} = {x2, x3},
[x3]B = {x1, x3} {x1, x3} {x2, x3} = {x1, x2, x3},
1 = {x1, x3} {x1} = {x1, x3},
2 = {x2, x3} {x2} = {x2, x3},
3 = {x1, x3} {x2, x3} = {x1, x2, x3},
1 = {x1} {x1, x3} {x1, x3} {x1, x2} {x1}= {x1, x2, x3},
2 = {x2, x3} {x2} {x2} {x1, x2} {x2, x3} = {x1, x2, x3},
3 = {x2, x3} {x1, x3} {x1, x3} {x3} {x2, x3} = {x1, x2, x3}.
Also it is easily can be noted the following:
1. [x]A [x]B
2. [x]A [x]B
3. [x]A , [x]B
4. [x]A , [x]B .
Example 3.3:
Continued from example 3.1 if we let , ={ 1, 4} and ={ 1, 2, 3}
= { 1}, C = { 1, 2, 3, 4}
First, when we take [x] as the intersection between sets:
[x1]A = {x1} {x1, x2}= {x1},
[x2]A = {x2, x3} {x1, x2}= {x2},
[x3]A = {x2, x3} {x3}= {x3},
[x1]B = {x1} {x1, x3} {x1, x3}= {x1},
[x2]B = {x2, x3} {x2} {x2} = {x2},
[x3]B = {x2, x3} {x1, x3} {x1, x3} = {x3},
1 = {x1}, 2 = {x2, x3}, 3 = {x2, x3},
Easily can be noted the following:
1. [x]A [x]B
2. [x]A [x]B
3. [x]A , [x]B
4. [x]A , [x]B
Second, when we take [x] as the union between sets:
[x1]A = {x1} {x1, x2}= {x1, x2},
[x2]A = {x2, x3} {x1, x2}= {x1, x2, x3},
[x3]A = {x2, x3} {x3}= { x2, x3},
[x1]B = {x1} {x1, x3} {x1, x3}= {x1, x3},
[x2]B = {x2, x3} {x2} {x2} = {x2, x3},
[x3]B = {x2, x3} {x1, x3} {x1, x3} = {x1, x2, x3},
1 = {x1}, 2 = {x2, x3}, 3 = {x2, x3},
1 = {x1, x2, x3}, 2 ={x1, x2, x3}, 3 = {x1, x2, x3}.
Also it is easily can be noted the following:
1. [x]A [x]B
2. [x]A [x]B
3. [x]A , [x]B
4. [x]A , [x]B
Proposition 3.6
For any information system having C as condition attributes and U as set of objects, for any subsets
, of condition attributes, , C , , = C , c = is the complelement of the set , x U we have the following relation for both cases of [x] (as the intersection or the union between sets):
[x]Ac [[x]C - [x]A] but we can say [x]Ac [[x]C - [x]A].
Proof:
Obviously □
Example 3.4:
Continued from example 3.1 if we let, = { 3, 5} and = { 1, 2, 4}
= , = C = { 1, 2, 3, 4, 5}.
First, when we take [x] as the intersection between sets:
[x2]A = {x2, x3} {x2}= {x2},
[x3]A = {x1, x3} {x2, x3} = {x3},
[x1]B = {x1}, [x2]B = {x2}, [x3]B = {x3}, 1 = {x1}, 2 ={x2}, 3 = {x3}.
Note:
It is clear here that the same properties in propositions 3.3 and 3.4 in case of intersection are satisfied and also proposition 3.5 in case of the union also satisfied for , .
Since, c = [x]Ac = [x]B , = [x]C , hence we have:
For x1 [x1]C - [x1]A = {x1}- {x1}= , [x1]Ac = [x]B = {x1} {x1},
Then we have [x1]Ac [[x1]C - [x1]A] but we can say [x1]Ac [[x1]C - [x1]A] .
Same result we got for x2 and x3.
Second, when we take [x] as the union between sets:
[x1]A = {x1, x3}, [x2]A = {x2, x3}, [x3]A = {x1, x2, x3},
[x1]B = {x1, x2, x3}, [x2]B = {x1, x2, x3}, [x3]B = {x1, x2, x3},
1 = {x1, x2, x3}, 2 ={x1, x2, x3}, 3 = {x1, x2, x3}.
Since, c = [x]Ac = [x]B , = [x]C, we can find the following:
For x1 [x1]C - [x1]A = {x1, x2, x3} - {x1, x3} = {x2},
[x1]Ac = [x]B = {x1, x2, x3} {x2} {x1, x2, x3}, Then we have [x1]Ac [[x1]C - [x1]A] but we can say [x1]Ac
[[x1]C - [x1]A]. Same result we got for x2 and x3.
It is easily can be noted the following for both cases of intersection and union between sets:
[x]Ac [[x]C - [x]A] but we can say [x]Ac [[x]C - [x]A].
Example 3.5:
Continued from example 3.1 if we let, = { 1, 2} and ={ 3, 4, 5}
= , = C = { 1, 2, 3, 4, 5}.
First, when we take [x] as the intersection between sets:
[x1]A = {x1}, [x2]A = {x2}, [x3]A = {x3}, [x1]B = {x1}, [x2]B = {x2}, [x3]B = {x3},
1 = {x1}, 2 ={x2}, 3 = {x3}.
For x1 [x1]C - [x1]A = {x1}- {x1}= , [x1]Ac = [x]B = {x1} {x1},
Then we have [x1]Ac [[x1]C - [x1]A] but we can say [x1]Ac [[x1]C - [x1]A] .
Same result we got for x2 and x3.
[x1]B = {x1, x2, x3}, [x2]B = {x1, x2, x3}, [x3]B = {x1, x2, x3},
1 = {x1, x2, x3}, 2 ={x1, x2, x3}, 3 = {x1, x2, x3}.
For x1 [x1]C - [x1]A = {x1, x2, x3} - {x1, x3} = {x2},
[x1]Ac = [x]B = {x1, x2, x3} {x2} {x1, x2, x3}, Then we have [x1]Ac [[x1]C - [x1]A] but we can say [x1]Ac
[[x1]C - [x1]A]. Same result we got for x2 and x3. Then we have the same result of the previous example as the last
proposition stated.
Note:-
We need to discuss if
1. [x]Ac = U - [x]A
2. [x]Ac = U - [ U-{x}]A
First, when we take [x] as the intersection between sets:
[x1]A = {x1} U - [x1]A = {x2, x3}, since [x1]Ac = {x1} [x1]Ac U - [x1]A , and there is no relation can be
put, i.e.
[x1]Ac ( , , ) U - [x1]A. Same result we got for x2 and x3.
Second, when we take [x] as the union between sets:
[x1]A = {x1, x3} U - [x1]A = {x2}, since [x1]Ac = {x1, x2, x3} [x1]Ac U - [x1]A ,
but we can say [x1]Ac U - [x1]A , Same result we got for x2 and x3.
U- {x1} = {x2, x3}, [ U-{x1}]A = [{x2, x3}]A , now the question is how we can find [{x2, x3}]A we have two
ways to get it:
When we let [{x2, x3}]A = [x2]A [x3]A .
1- In case of union between sets we have, [x1]Ac = {x1, x2, x3}, [x2]A [x3]A ={x1, x2, x3} U - [ U-{x1}]A
= [x1]Ac U - [ U-{x1}]A , but we can say [x1]Ac U - [ U-{x1}]A. Same result we got for x2 and x3
[x]Ac U - [ U-{x}]A.
2- In case of intersection between sets we have [x1]Ac = {x1}, [x2]A [x3]A= {x2, x3}} U - [ U-{x1}]A =
{x1} [x1]Ac = U - [ U-{x1}]A , so we can say [x1]Ac = U - [ U-{x1}]A. Same result we got for x2 and x3
[x]Ac = U - [ U-{x}]A
When we let [{x2, x3}]A = [x2]A [x3]A .
1- In case of intersection between sets we have [x2]A [x3]A= U - [ U-{x1}]A = U,
since [x1]Ac = {x1} U [x1]Ac U - [ U-{x1}]A , but we can say [x1]Ac U - [ U-{x1}]A.
Same result we got for x2 and x3 [x]Ac U - [ U-{x}]A.
2- In case of union we have [{x2, x3}]A = [x2]A [x3]A= {x2, x3} U – [[x2]A [x3]A]= {x1}, since [x1]Ac =
{x1, x2, x3} U - [ U-{x1}]A [x1]Ac. Same result we got for x2 and x3 U - [ U-{x}]A [x]Ac.
4. Conclusion
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