Chapter 10 DISSIMILIRATY ANALYSIS

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Chapter 10

DISSIMILIRATY ANALYSIS

Presented by: Turkov. Eugene Class id : 113and Minfang Tao Class id : 112

Professor:Dr. T.Y. Lin

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Introduction In Chapter 8 and 9 we focused on decision tables in which

condition and decision attributes were distinguished:

　 CONDITIONS DECISIONS

U a b c d e

1 1 1 1 1 0

2 0 1 0 0 0

3 1 1 1 0 1

4 1 1 0 0 1

5 0 1 0 1 1

6 1 0 0 1 1

7 1 0 1 1 1

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Indroduction

In this chapter we are going to discuss Knowledge Representation Systems in which neither condition nor decision attributes are distinguished.

　 CONDITION & DECISION ATTRIBUTES

U a b c d e

1 1 1 1 1 0

2 0 1 0 0 0

3 1 1 1 0 1

4 1 1 0 0 1

5 0 1 0 1 1

6 1 0 0 1 1

7 1 0 1 1 1

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Pattern Recognition - Original Table

U a b c d e f g

1 1 1 1 1 1 1 0

2 0 1 1 0 0 0 0

3 1 1 0 1 1 0 1

4 1 1 1 1 0 0 1

5 0 1 1 0 0 1 1

6 1 0 1 1 0 1 1

7 1 0 1 1 1 1 1

8 1 1 1 0 0 0 0

9 1 1 1 1 1 1 1

10 1 1 1 1 0 1 1

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Pattern Recognition - Task

“Our task is to find a minimal description of each digit and the corresponding decision algorithms.”

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Pattern Recognition–Steps for The Task

1. Make sure that the table is consistent. If it is not, create a new consistent table from it.

2. Find the attribute(column) reduct, and create a new table that only includes the attributes which are members of the reduct.

3. Compute core and reduct values for each decision rule.

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Pattern Recognition - Consistent

Each row in the original table is unique, hence the table is consistent.

U a b c d e f g

1 1 1 1 1 1 1 0

2 0 1 1 0 0 0 0

3 1 1 0 1 1 0 1

4 1 1 1 1 0 0 1

5 0 1 1 0 0 1 1

6 1 0 1 1 0 1 1

7 1 0 1 1 1 1 1

8 1 1 1 0 0 0 0

9 1 1 1 1 1 1 1

10 1 1 1 1 0 1 1

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Pattern Recognition –Finding Core Attributes

Convert the table to seven equivalence relations:

U/a={ {1,3,4,6,7,8,9,10 } ,{2,5 } } U/b={ {1,2,3,4,5,8,9,10 } ,{6,7 } } U/c={ {1,2,4,5,6,7,8,9,10 } ,{3 } } U/d={ {1,3,4,6,7,9,10 } ,{2,5,8 } } U/e={ {1,3,7,9 } ,{2,4,5,6,8,10 } } U/f={ {1,5,6,7,9,10 } ,{2,3,4,8 } } U/g ={ {1,2,8 } ,{3,4,5,6,7,9,10 } }

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Pattern Recognition – Find Core Attributes

The process for computing the Indiscernibility for all equivalnce relations: U/IND(a,b)={ {1,3,4,8,9,10 } ,{6,7 } ,{2,5 } } U/IND(a,b,c)={ {1,4,8,9,10 } ,{3 } ,{6,7 } ,{2,5 } } U/IND(a,b,c,d)={ {1,4,9,10 } ,{8 } ,{3 } ,{6,7 } ,{2,5 } } U/IND(a,b,c,d,e)={ {1,9 } ,{4,10 } ,{8 } ,{3 } ,{7 } ,{6 } ,{2,5 } } U/IND(a,b,c,d,e,f)={ {1,9 } ,{10 } ,{4 } ,{8 } ,{3 } ,{7 } ,{6 } ,{5 } ,{2 } } U/IND(a,b,c,d,e,f,g )={ {1 } ,{9 } ,{10 } ,{4 } ,{8 } ,{3 } ,{7 } ,{6 } ,{5 } ,{2 } }

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Finally, we got:U/IND(ALL R) = U/IND(a,b,c,d,e,f,g )= { {1 } ,{9 } ,{10 } ,{4 } ,{8 } ,{3 } ,{7 } ,{6 } ,{5 } ,{2 } }

Computing U/IND(R-x) :

U/IND(R-a) = { {1 } ,{9 } ,{10 } ,{4 } ,{5 } ,{2,8 } ,{3 } ,{7 } ,{6 } } U/IND(R-a) ! = U/IND(R)

U/IND(R-b) = { {1 } ,{7,9 } ,{6,10 } ,{4 } ,{8 } ,{3 } ,{5 } ,{2 } } U/IND(R-b) ! = U/IND(R)

U/IND(R-c) = { {1 } ,{9 } ,{3 } ,{10 } ,{4 } ,{8 } ,{7 } ,{6 } ,{5 } ,{2 } } U/IND(R-c) = = U/IND(R)

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U/IND(R-d) = { {1 } ,{9 } ,{10 } ,{8 } ,{4 } ,{3 } ,{7 } ,{6 } ,{5 } ,{2 } } U/IND(R-d) = = U/IND(R)

U/IND(R-e) = { {1 } ,{9,10 } ,{4 } ,{8 } ,{3 } ,{6,7 } ,{5 } ,{2 } } U/IND(R-e) ! = U/IND(R)

U/IND(R-f) = { {1 } ,{9 } ,{4,10 } ,{8 } ,{3 } ,{7 } ,{6 } ,{2 } ,{5 } } U/IND(R-f) ! = U/IND(R)

U/IND(R-g ) = { {1,9 } ,{10 } ,{4 } ,{8 } ,{3 } ,{7 } ,{6 } ,{5 } ,{2 } } U/IND(R-g ) ! = U/IND(R)

We have determined that c and d are dispensable and our core attributes are:

{a, b, e, f, g}

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Pattern Recognition – Find Rreduct Attributes

Finding reduct attributes by using core attributes:

U/IND(a,b,e,f,g) = { {1 } ,{9 } ,{10 } ,{8 } , {4 } ,{3 } ,{7 } ,{6 } ,{5 } ,{2 } }

U/IND(a,b,e,f,g) == U/IND(a,b,c,d,e,f,g)

(a,b,e,f,g) is one and only one reduct of the original table.

(a,b,e,f,g)=>(c,d).

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U a b e f

g

1 1 1 1 1 0

2 0 1 0 0 0

3 1 1 1 0 1

4 1 1 0 0 1

5 0 1 0 1 1

6 1 0 0 1 1

7 1 0 1 1 1

8 1 1 0 0 0

9 1 1 1 1 1

10 1 1 0 1 1

Reduct Table – It is consistent

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After removing attribute {a}, the table is inconsistent

U b e f

g

1 1 1 1 0

2 1 0 0 0

3 1 1 0 1

4 1 0 0 1

5 1 0 1 1

6 0 0 1 1

7 0 1 1 1

8 1 0 0 0

9 1 1 1 1

10 1 0 1 1

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After removing attribute {b}, the table is inconsistent

U a e f

g

1 1 1 1 0

2 0 0 0 0

3 1 1 0 1

4 1 0 0 1

5 0 0 1 1

6 1 0 1 1

7 1 1 1 1

8 1 0 0 0

9 1 1 1 1

10 1 0 1 1

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After removing attribute {e}, the table is inconsistent

U a b f

g

1 1 1 1 0

2 0 1 0 0

3 1 1 0 1

4 1 1 0 1

5 0 1 1 1

6 1 0 1 1

7 1 0 1 1

8 1 1 0 0

9 1 1 1 1

10 1 1 1 1

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After removing attribute {f}, the table is inconsistent

U a b e

g

1 1 1 1 0

2 0 1 0 0

3 1 1 1 1

4 1 1 0 1

5 0 1 0 1

6 1 0 0 1

7 1 0 1 1

8 1 1 0 0

9 1 1 1 1

10 1 1 0 1

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After removing attribute {g}, the table is inconsistent U a b e f

1 1 1 1 1

2 0 1 0 0

3 1 1 1 0

4 1 1 0 0

5 0 1 0 1

6 1 0 0 1

7 1 0 1 1

8 1 1 0 0

9 1 1 1 1

10 1 1 0 1

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Pattern Recognition – Decision Rules

Conditions & DecisionsU a b e f g

1 1 1 1 1 0

2 0 1 0 0 0

3 1 1 1 0 1

4 1 1 0 0 1

5 0 1 0 1 1

6 1 0 0 1 1

7 1 0 1 1 1

8 1 1 0 0 0

9 1 1 1 1 1

10 1 1 0 1 1

Attributes {a, b, e, f, g} are not only conditions, but they are also decisions.

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For the sake of illustration, we extend this table .

Conditions Decisions

U a b e f g a’ b’ e’ f’ g’

1 1 1 1 1 0 1 1 1 1 0

2 0 1 0 0 0 0 1 0 0 0

3 1 1 1 0 1 1 1 1 0 1

4 1 1 0 0 1 1 1 0 0 1

5 0 1 0 1 1 0 1 0 1 1

6 1 0 0 1 1 1 0 0 1 1

7 1 0 1 1 1 1 0 1 1 1

8 1 1 0 0 0 1 1 0 0 0

9 1 1 1 1 1 1 1 1 1 1

10 1 1 0 1 1 1 1 0 1 1

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U/IND(a,b,e,f,g) = U/IND(a’,b’,e’,f’,g’)= {{1},{2},{3},{4}, {5},{6},{7},{8},{9}, {10}} .

To simplify the table, we use one attribute t= {1,2,3,4,5,6,7,8,9,10} to replace attribute (a’,b’,e’,f’,g’).

U/IND(t) == U/IND(a’,b’,e’,f’,g’)

{a, b, e, f, g} are conditions, the {t} is decision.

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CONDITIONS DECISIONS

U a b e f g t

1 1 1 1 1 0 1

2 0 1 0 0 0 2

3 1 1 1 0 1 3

4 1 1 0 0 1 4

5 0 1 0 1 1 5

6 1 0 0 1 1 6

7 1 0 1 1 1 7

8 1 1 0 0 0 8

9 1 1 1 1 1 9

10 1 1 0 1 1 10

Computing core and reducts values will depend on this regular decision table.

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Pattern Recognition - Decision Rules

Method 1 for finding reducts for each rule. F={{a}, {b}, {e}, {f}, {g}} All subfamilies G ⊆ to F+

G={{a}, {b}, {e}, {f}, {g}, {ab}, {ae}, {af}, .. {be}, …..{eg}, {fg}{abe}, {abf}, {abg}, {bef}, {beg}, {ebg},{abef}, {abeg}, {befg},{abefg} }.

The relationship for the elements in G is intersection :{abe}={a} {b} {e}

Using G to find reducts.

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Method 2 for finding reducts for each rule. (Pawlak’s method) Find core value for every rule Testing this core value is reduct value? If it is reduct value, we can say this rule

has only one reduct value. If it is not reduct value, we will add an

uncore value into it, then test are they reduct value?

Repeat, until find all reduct values.

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Computing core for every rule -- rule 1

In rule 1 {a={1,3,4,6,7,8,9,10}, b={1,2,3,4,5,8,9,10}, e={1,3,7,9},

f={1,5,6,7,9,10}, g={1,2,8}} and Decision for rule 1 is [1]t={1}

Removing a, Intersection (b,e,f,g) = {1} == [1]t

Removing b, Intersection (a,e,f,g) = {1} == [1]t

Removing e, Intersection (a,b,f,g) = {1} == [1]t

Removing f, Intersection (a,b,e,g) = {1} == [1]t

Removing g, Intersection (a,b,e,f) = {1,9} != [1]t

g is the core value.

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In rule 2 {a={2,5}, b={1,2,3,4,5,8,9,10}, e={2,4,5,6,8,10}, f={2,3,4,8}, g={1,2,8}} and Decision for rule 2 is [2]t ={2}

Removing a, Intersection (b,e,f,g) = {2,8} != [2]t ={2}

Removing b, Intersection (a,e,f,g) = {2} == [2]t ={2}

Removing e, Intersection (a,b,f,g) = {2} == [2]t ={2}

Removing f, Intersection (a,b,e,g) = {2} == [2]t ={2}

Removing g, Intersection (a,b,e,f) = {2} == [2]t ={2}

a is the core value.

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In rule 3 {a={1,3,4,6,7,8,9,10}, b={1,2,3,4,5,8,9,10}, e={1,3,7,9},f={2,3,4,8}, g={3,4,5,6,7,9,10}} and Decision for rule 3 is [3]t ={3}

Removing a, Intersection (b,e,f,g) = {3} == [3]t

Removing b, Intersection (a,e,f,g) = {3} == [3]t

Removing e, Intersection (a,b,f,g) = {3,4} != [3]t

Removing f, Intersection (a,b,e,g) = {3,9} != [3]t

Removing g, Intersection (a,b,e,f) = {3} == [3]t e and f are the core values.

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Computing core for every rule -- rule 4 Core values :e, f, g

Computing core for every rule -- rule 5 Core values :a

Computing core for every rule -- rule 6 Core values :b, e

Computing core for every rule -- rule 7 Core values :b, e

Computing core for every rule -- rule 8 Core values :a, g

Computing core for every rule -- rule 9 Core values :b, e, f, g

Computing core for every rule -- rule 10 Core values :a, b, e, f

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The core values table:

U a b e f g t

1 - - - - 0 1

2 0 - - - - 2

3 - - 1 0 - 3

4 - - 0 0 1 4

5 0 - - - - 5

6 - 0 0 - - 6

7 - 0 1 - - 7

8 1 - - - 0 8

9 - 1 1 1 1 9

10 1 1 0 1 - 10

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Computing reduct values by using core values -- rule 1

In rule 1 {a={1,3,4,6,7,8,9,10}, b={1,2,3,4,5,8,9,10}, e={1,3,7,9}, f={1,5,6,7,9,10}, g={1,2,8}} and Intersection(a,b,e,f,g) = [1]t ={1} and the core value is g.

g ! = [1]t , so g is not reduct value.

Intersection (a,g)={1,8}!= [1]t = {1} ;

Intersection(b,g)={1,2,8} != [1]t = {1};

Intersection(e,g)={1} == [1]t = {1} ;

Intersection(f,g)={1} == [1]t = {1} ;

Intersection (a,b,g)={1,8}!= [1]t = {1} ;

Reducts values are { {e, g}, {f, g} }

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In rule 2 {a={2,5}, b={1,2,3,4,5,8,9,10}, e={2,4,5,6,8,10}, f={2,3,4,8}, g={1,2,8}}and Intersection(a,b,e,f,g) = [2]t ={2} and the core value is a.

a ! = Intersection(a,b,e,f,g) , so a is not reduct value.

Intersection(a,b)={2,5} != [2]t = {2} ;

Intersection(a,e)={2,5} != [2]t = {2};

Intersection(a,f)={2} == [2]t = {2} ;

Intersection(a,g)={2} == [2]t = {2} ;

Intersection (a,b,e)={2, 5}!= [2]t = {2} ;

Reducts values are { {a, f}, {a, g} }

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In rule 3 {a={1,3,4,6,7,8,9,10},b={1,2,3,4,5,8,9,10},e={1,3,7,9},f={2,3,4,8}, g={3,4,5,6,7,9,10}}and Intersection(a,b,e,f,g) = [3]t ={3} and the core value is e, f.

Intersection(e,f) ={3} == [3]t ={3}

{ {e, f} } are not only core value ,but they also are reduct value and they are only on reduct value for rule 3.

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For rule 4, the reduct value is: { {e, f, g} }

For rule 5, the reduct values are: { {a, f}, {a, g} }

For rule 6, the reduct value is: { {b, e} }

For rule 7, the reduct value is: { {b, e} }

For rule 8, the reduct values are: {{a, e, g}, {a, f, g}}

For rule 9, the reduct value is: { {b, e, f, g} }

For rule 10, the reduct value is: { {a, b, e, f} }

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U a b e f g t

1(1) x x 1 x 0 1

1(2) x x x 1 0 1

2(1) 0 x x 0 x 2

2(2) 0 x x x 0 2

3(1) x x 1 0 x 3

4(1) x x 0 0 1 4

5(1) 0 x x 1 x 5

5(2) 0 x x x 1 5

6(1) x 0 0 x x 6

7(1) x 0 1 x x 7

8(1) 1 x 0 x 0 8

8(2) 1 x x 0 0 8

9(1) x 1 1 1 1 9

10(1) 1 1 0 1 x 10

For rule 8, there are two reduct values:{ {a, e, g} and {a, f, g} }

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The rule {3, 4, 6, 7, 9,10} have only one reduct value, so they are already reducted.

Because the four decision rules {1, 2, 5 ,8 } have two reduced forms, we have altogether 16 (2*2*2*2) minimal decision algorithms.

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U a b e f g t

1(1) x x 1 x 0 1

1(2) x x x 1 0 1

2(1) 0 x x 0 x 2

2(2) 0 x x x 0 2

3(1) x x 1 0 x 3

4(1) x x 0 0 1 4

5(1) 0 x x 1 x 5

5(2) 0 x x x 1 5

6(1) x 0 0 x x 6

7(1) x 0 1 x x 7

8(1) 1 x 0 x 0 8

8(2) 1 x x 0 0 8

9(1) x 1 1 1 1 9

10(1) 1 1 0 1 x 10

Getting 16 minimal decision algorithms

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U a b e f g

1(1) x x 1 x 0

2(1) 0 x x 0 x

3(1) x x 1 0 x

4(1) x x 0 0 1

5(1) 0 x x 1 x

6(1) x 0 0 x x

7(1) x 0 1 x x

8(1) 1 x 0 x 0

9(1) x 1 1 1 1

10(1) 1 1 0 1 x

{1, 2, 4 ,8 } have two reduced forms, we have altogether 16 minimal decision algorithms -- Table 1

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U a b e f g

1(2) x x x 1 0

2(1) 0 x x 0 x

3(1) x x 1 0 x

4(1) x x 0 0 1

5(1) 0 x x 1 x

6(1) x 0 0 x x

7(1) x 0 1 x x

8(1) 1 x 0 x 0

9(1) x 1 1 1 1

10(1) 1 1 0 1 x


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U a b e f g

1(2) x x x 1 0

2(2) 0 x x x 0

3(1) x x 1 0 x

4(1) x x 0 0 1

5(2) 0 x x x 1

6(1) x 0 0 x x

7(1) x 0 1 x x

8(2) 1 x x 0 0

9(1) x 1 1 1 1

10(1) 1 1 0 1 x


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U a b e f g

1(2) x x x 1 0 f1g0-->1

2(2) 0 x x x 0 a0g0-->2

3(1) x x 1 0 x e1f0-->3

4(1) x x 0 0 1 e0f0g1-->4

5(2) 0 x x x 1 a0g1-->5

6(1) x 0 0 x x b0e0-->6

7(1) x 0 1 x x b0e1-->7

8(2) 1 x x 0 0 a1f0g0-->8

9(1) x 1 1 1 1 b1e1f1g1-->9

10(1) 1 1 0 1 x a1b1e0f1-->10

Other format to represent this algorithm

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The End for Pattern Recognition

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Appendix

A: Split decision table to consistent and totally inconsisten tables

B: Using G to find reduct values

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Appendix-A

Split decision table to consistent and totally inconsisten tables

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Decision table

a b c d e

1 0 2 2 0

0 1 1 1 2

2 0 0 1 1

1 1 0 2 2

1 0 2 0 1

2 2 0 1 1

2 1 1 1 2

0 1 1 0 1

a b c are conditions , d e are decisions

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Condition Table:

a b c

1 0 2

0 1 1

2 0 0

1 1 0

1 0 2

2 2 0

2 1 1

0 1 1

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Decision Table:

d e

2 0

1 2

1 1

2 2

0 1

1 1

1 2

0 1

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Ind(Condition) and Ind(Decision)

U/IND(ALL R) =U/IND(a,b,c)= { {1,5 } ,{4 } ,{2,8 } ,{3 } ,{7 } ,{6 } }

U/IND(ALL R) =U/IND(d,e)= { {1 } ,{4 } ,{2,7 } ,{3,6 } ,{5,8 } }

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Computing POSC(D):

Computing POSc(D), Checking U/IND(C) to U/IND(D) : { {1,5 } ,{4 } ,{2,8 } ,{3 } ,{7 } ,{6 } } is belong to { {1 } ,{4 } ,{2,7 } ,{3,6 } ,{5,8 } }:

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Computing….. Check Set:{3} is belong to Set:{5,8}, it is false, {3} is throwed. Check Set:{3} is belong to Set:{3,6}, it is true.The set {3} is selected,

{3} Check Set:{3} is belong to Set:{1}, it is false, {3} is throwed. Check Set:{3} is belong to Set:{2,7}, it is false, {3} is throwed. Check Set:{3} is belong to Set:{4}, it is false, {3} is throwed. Check Set:{2,8} is belong to Set:{5,8}, it is false, {2,8} is throwed. Check Set:{2,8} is belong to Set:{3,6}, it is false, {2,8} is throwed. Check Set:{2,8} is belong to Set:{1}, it is false, {2,8} is throwed. Check Set:{2,8} is belong to Set:{2,7}, it is false, {2,8} is throwed. Check Set:{2,8} is belong to Set:{4}, it is false, {2,8} is throwed. Check Set:{4} is belong to Set:{5,8}, it is false, {4} is throwed. Check Set:{4} is belong to Set:{3,6}, it is false, {4} is throwed. Check Set:{4} is belong to Set:{1}, it is false, {4} is throwed. Check Set:{4} is belong to Set:{2,7}, it is false, {4} is throwed. Check Set:{4} is belong to Set:{4}, it is true.The set {4} is selected,

{3} U {4} Check Set:{1,5} is belong to Set:{5,8}, it is false, {1,5} is throwed.

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Computing….. Check Set:{1,5} is belong to Set:{3,6}, it is false, {1,5} is throwed. Check Set:{1,5} is belong to Set:{1}, it is false, {1,5} is throwed. Check Set:{1,5} is belong to Set:{2,7}, it is false, {1,5} is throwed. Check Set:{1,5} is belong to Set:{4}, it is false, {1,5} is throwed. Check Set:{6} is belong to Set:{5,8}, it is false, {6} is throwed. Check Set:{6} is belong to Set:{3,6}, it is true.The set {6} is selected,

{3} U {4} U {6} Check Set:{6} is belong to Set:{1}, it is false, {6} is throwed. Check Set:{6} is belong to Set:{2,7}, it is false, {6} is throwed. Check Set:{6} is belong to Set:{4}, it is false, {6} is throwed. Check Set:{7} is belong to Set:{5,8}, it is false, {7} is throwed. Check Set:{7} is belong to Set:{3,6}, it is false, {7} is throwed. Check Set:{7} is belong to Set:{1}, it is false, {7} is throwed. Check Set:{7} is belong to Set:{2,7}, it is true.The set {7} is selected,

{3} U {4} U {6} U {7} Check Set:{7} is belong to Set:{4}, it is false, {7} is throwed.

POSc(D):{3,4,6,7}

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Split Table

Using POSc(D):{3,4,6,7} to split the table.

Using 3, 4, 6,7 rows to make the consistent table.

Using 1,2,5,8 rows to make the totally inconsistent table.

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a b c d e

2 0 0 1 1

1 1 0 2 2

2 2 0 1 1

2 1 1 1 2

a b c d e

1 0 2 2 0

0 1 1 1 2

1 0 2 0 1

0 1 1 0 1

Consistent Table

Inconsistent Table:

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Appendix-B

Using G to find reduct values

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Computing reducts for every rule -- rule 1

Find reducts in {a={1,3,4,6,7,8,9,10}, b={1,2,3,4,5,8,9,10}, e={1,3,7,9}, f={1,5,6,7,9,10}, g={1,2,8}} for decision rule 1 {1}

Checking that if [1]Set(a):{1,3,4,6,7,8,9,10} is belong to decision {1}Checking that if [1]Set(b):{1,2,3,4,5,8,9,10} is belong to decision {1}Checking that if [1]Set(e):{1,3,7,9} is belong to decision {1}Checking that if [1]Set(f):{1,5,6,7,9,10} is belong to decision {1}Checking that if [1]Set(g):{1,2,8} is belong to decision {1}Checking that if [1]Intersection(a,b):{1,3,4,8,9,10} is belong to decision {1}Checking that if [1]Intersection(a,e):{1,3,7,9} is belong to decision {1}Checking that if [1]Intersection(a,f):{1,6,7,9,10} is belong to decision {1}Checking that if [1]Intersection(a,g):{1,8} is belong to decision {1}Checking that if [1]Intersection(b,e):{1,3,9} is belong to decision {1}Checking that if [1]Intersection(b,f):{1,5,9,10} is belong to decision {1}Checking that if [1]Intersection(b,g):{1,2,8} is belong to decision {1}

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Checking that if [1]Intersection(e,f):{1,7,9} is belong to decision {1}Checking that if [1]Intersection(e,g):{1} is belong to decision {1}[1]Intersection(e,g):{1} is one reductChecking that if [1]Intersection(f,g):{1} is belong to decision {1}[1]Intersection(f,g):{1} is one reductChecking that if [1]Intersection(a,b,e):{1,3,9} is belong to decision {1}Checking that if [1]Intersection(a,b,f):{1,9,10} is belong to decision {1}Checking that if [1]Intersection(a,b,g):{1,8} is belong to decision {1}Checking that if [1]Intersection(a,e,f):{1,7,9} is belong to decision {1}Checking that if [1]Intersection(a,e,g):{1} is belong to decision {1}Because Intersection(a,e,g): contains Intersection(e,g):, so skip it.Checking that if [1]Intersection(a,f,g):{1} is belong to decision {1}Because Intersection(a,f,g): contains Intersection(f,g):, so skip it.Checking that if [1]Intersection(b,e,f):{1,9} is belong to decision {1}

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Checking that if [1]Intersection(b,e,g):{1} is belong to decision {1}Because Intersection(b,e,g): contains Intersection(e,g):, so skip it.Checking that if [1]Intersection(b,f,g):{1} is belong to decision {1}Because Intersection(b,f,g): contains Intersection(f,g):, so skip it.Checking that if [1]Intersection(e,f,g):{1} is belong to decision {1}Because Intersection(e,f,g): contains Intersection(e,g):, so skip it.Checking that if [1]Intersection(a,b,e,f):{1,9} is belong to decision {1}Checking that if [1]Intersection(a,b,e,g):{1} is belong to decision {1}Because Intersection(a,b,e,g): contains Intersection(e,g):, so skip it.Checking that if [1]Intersection(a,b,f,g):{1} is belong to decision {1}Because Intersection(a,b,f,g): contains Intersection(f,g):, so skip it.Checking that if [1]Intersection(a,e,f,g):{1} is belong to decision {1}Because Intersection(a,e,f,g): contains Intersection(e,g):, so skip it.

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Checking that if [1]Intersection(b,e,f,g):{1} is belong to decision {1}Because Intersection(b,e,f,g): contains Intersection(e,g):, so skip it.Checking that if [1]Intersection(a,b,e,f,g):{1} is belong to decision {1}Because Intersection(a,b,e,f,g): contains Intersection(e,g):, so skip it.

Finally we found reducts:{ {e, g}, {f, g} }

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Find reducts in {a={2,5}, b={1,2,3,4,5,8,9,10}, e={2,4,5,6,8,10}, f={2,3,4,8}, g={1,2,8}} for decision rule 2 {2}

Checking that if [2]Set(a):{2,5} is belong to decision {2}Checking that if [2]Set(b):{1,2,3,4,5,8,9,10} is belong to decision {2}Checking that if [2]Set(e):{2,4,5,6,8,10} is belong to decision {2}Checking that if [2]Set(f):{2,3,4,8} is belong to decision {2}Checking that if [2]Set(g):{1,2,8} is belong to decision {2}Checking that if [2]Intersection(a,b):{2,5} is belong to decision {2}Checking that if [2]Intersection(a,e):{2,5} is belong to decision {2}Checking that if [2]Intersection(a,f):{2} is belong to decision {2}[2]Intersection(a,f):{2} is one reductChecking that if [2]Intersection(a,g):{2} is belong to decision {2}[2]Intersection(a,g):{2} is one reduct

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Checking that if [2]Intersection(b,e):{2,4,5,8,10} is belong to decision {2}Checking that if [2]Intersection(b,f):{2,3,4,8} is belong to decision {2}Checking that if [2]Intersection(b,g):{1,2,8} is belong to decision {2}Checking that if [2]Intersection(e,f):{2,4,8} is belong to decision {2}Checking that if [2]Intersection(e,g):{2,8} is belong to decision {2}Checking that if [2]Intersection(f,g):{2,8} is belong to decision {2}Checking that if [2]Intersection(a,b,e):{2,5} is belong to decision {2}Checking that if [2]Intersection(a,b,f):{2} is belong to decision {2}Because Intersection(a,b,f): contains Intersection(a,f):, so skip it.Checking that if [2]Intersection(a,b,g):{2} is belong to decision {2}Because Intersection(a,b,g): contains Intersection(a,g):, so skip it.Checking that if [2]Intersection(a,e,f):{2} is belong to decision {2}Because Intersection(a,e,f): contains Intersection(a,f):, so skip it.

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Checking that if [2]Intersection(a,e,g):{2} is belong to decision {2}Because Intersection(a,e,g): contains Intersection(a,g):, so skip it.Checking that if [2]Intersection(b,e,f):{2,4,8} is belong to decision {2}Checking that if [2]Intersection(b,e,g):{2,8} is belong to decision {2}Checking that if [2]Intersection(b,f,g):{2,8} is belong to decision {2}Checking that if [2]Intersection(e,f,g):{2,8} is belong to decision {2}Checking that if [2]Intersection(a,b,e,f):{2} is belong to decision {2}Because Intersection(a,b,e,f): contains Intersection(a,f):, so skip it.Checking that if [2]Intersection(a,b,e,g):{2} is belong to decision {2}Because Intersection(a,b,e,g): contains Intersection(a,g):, so skip it.Checking that if [2]Intersection(b,e,f,g):{2,8} is belong to decision {2}

Finally we found reducts:{ {a, f}, {a, g} }

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For rule 3, the reducts are: { {e, f} }

For rule 4, the reducts are: { {e, f, g} }

For rule 5, the reducts are: { {a, f}, {a, g} }

For rule 6, the reducts are: { {b, e} }

For rule 7, the reducts are: { {b, e} }

For rule 8, the reducts are: {{a, e, g}, {a, f, g}}

For rule 9, the reducts are: { {b, e, f, g} }

For rule 10, the reducts are:

{ {a, b, e, f} }

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U a b e f g w

1(1) x x 1 x 0 1

1(2) x x x 1 0 1

2(1) 0 x x 0 x 2

2(2) 0 x x x 0 2

3(1) x x 1 0 x 3

4(1) x x 0 0 1 4

5(1) 0 x x 1 x 5

5(2) 0 x x x 1 5

6(1) x 0 0 x x 6

7(1) x 0 1 x x 7

8(1) 1 x 0 x 0 8

8(2) 1 x x 0 0 8

9(1) x 1 1 1 1 9

10(1) 1 1 0 1 x 10

For rule 8, the reducts are: { {a, e, g} , {a, f, g} }

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U a b e f g w

1(1) x x 1 x 0 1

1(2) x x x 1 0 1

Core value: 1 - - - - 0 1

2(1) 0 x x 0 x 2

2(2) 0 x x x 0 2

Core value : 2 0 - - - - 2

3(1) x x 1 0 x 3

Core value : 3 - - 1 0 - 3

Intersection to get core values for all decision rules.

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The core values table:

U a b e f g w

1 - - - - 0 1

2 0 - - - - 2

3 - - 1 0 - 3

4 - - 0 0 1 4

5 0 - - - - 5

6 - 0 0 - - 6

7 - 0 1 - - 7

8 1 - - - 0 8

9 - 1 1 1 1 9

10 1 1 0 1 - 10

Chapter 10 DISSIMILIRATY ANALYSIS

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Transcript of Chapter 10 DISSIMILIRATY ANALYSIS