1. based on Knowledge Discovery
Chase Methods
based on Knowledge Discovery
Zbigniew W. Ras

2. Algorithm Chase GIVEN:  Incomplete Information System (IIS)
Constraints (functional dependencies,..) which IIS satisfies
[ Dept  Chair, Chair  Dept  Faculty-Name, … Dept(x1) =Dept(x2) ] X
Faculty-Name
Dept.
Chair x1
Bob x2
John
Jones x3
Mike x4 EE x5
Tom

3. Tableau System for IIS –
information system with null values replaced by variables X
Faculty Name
Department
Chair x1
Bob vd n1 x2
John
Jones x3
Mike n2 n3 x4 vE EE n4 x5
Tom n5

4. Variables in Tableaux System
distinguished variables, one for each attribute (if b is an attribute
of interest, then vb is the corresponding distinguished variable)
nondistinguished variables (there are countably many of them:
n1, n2, n3, ….)

5. Functional Dependencies:X
Faculty Name
Department
Chair x1
Bob vd n1 x2
John
Jones x3
Mike n2 n3 x4 vE EE n4 x5
Tom n5
Functional Dependencies: X
Faculty Name
Department
Chair x1
Bob vd
Jones x2
John x3
Mike n2 n3 x4
Tom EE n4 x5
[Department → Chair]
[Department *Chair
→ Faculty Name]

6. Algorithm Chase Begin S1:=S; End
Input: tableaux system S and set of functional dependencies F
Output: tableaux system CHASEF(S)
Begin
S1:=S;
while there are t1, t2  S1 and (B  b)  F
such that t1[B]= t2[B] and t1[b] < t2[b]
do change all the occurrences of the value
t2[b] in S1 to t1[b]
CHASEF(S):=S1
End

7. Algorithm Chase S1:=S; Begin End
Input: tableaux system S and set of functional dependencies F
Output: tableaux system CHASEF(S)
Begin
S1:=S;
while there are t1, t2  S1 and (B  b)  F
such that t1[B]= t2[B] and t1[b] < t2[b]
do change all the occurrences of the value
t2[b] in S1 to t1[b]
CHASEF(S):=S1
End
The algorithm always terminates if applied to a finite tableaux system. If one execution of the algorithm generates a tableaux system satisfying F, then every execution of the algorithm generates the same tableaux system.

8. Algorithm Chase 1

9. Chase supported by rules extracted
from IIS (Chase 1)
Chase 1 identifies all incomplete attributes (their values are called concepts) in IS .
2. Main Algorithm
- Extraction of rules from IS describing these concepts,
- Null values in IS are replaced by values suggested by these rules.
3. These two steps are repeated till fixpoint is reached.

10. Example (Chase1) X = {x1, x2, x3, x4, x5, x6, x7, x8, x9, x10}b c d e f g x1 b1 c1 e2 f1 x2 b2 c2 d2 e1 f2 g3 x3 d3 g1 x4 b3 c3 e3 x5 g2 x6 x7 f4 x8 x9 d1
x10
X = {x1, x2, x3, x4, x5, x6, x7, x8, x9, x10}
A = {b, c, d, e, f, g}
Attribute b
e2  b1 (support 2),
c1 f1  b1 (support 2),
g2  b2 (support 2),
c3  b3 (support 1),
c2  b2 (support 2),
g3d2  b2 (support 1),
e3d3  b3 (support 1),
f2d2  b2 (support 1).

11. Example (Chase1) Attribute b Two null values in S: b(x6), b(x8) b(x6):d e f g x1 b1 c1 e2 f1 x2 b2 c2 d2 e1 f2 g3 x3 d3 g1 x4 b3 c3 e3 x5 g2 x6 x7 f4 x8 x9 d1
x10
Attribute b
Two null values in S: b(x6), b(x8)
b(x6):
e2  b1 (support 2),
c1 f1  b1 (support 2),
g2  b2 (support 2),
c3  b3 (support 1),
c2  b2 (support 2),
g3d2  b2 (support 1),
e3d3  b3 (support 1),
f2d2  b2 (support 1).

12. Example (Chase1) Attribute b Two null values in S: b(x6), b(x8) b(x6):d e f g x1 b1 c1 e2 f1 x2 b2 c2 d2 e1 f2 g3 x3 d3 g1 x4 b3 c3 e3 x5 g2 x6 x7 f4 x8 x9 d1
x10
Attribute b
Two null values in S: b(x6), b(x8)
b(x6):
e2  b1 (support 2),
c1 f1  b1 (support 2),
g2  b2 (support 2),
c3  b3 (support 1),
c2  b2 (support 2),
g3d2  b2 (support 1),
e3d3  b3 (support 1),
f2d2  b2 (support 1).

13. Example (Chase1) Attribute b Two null values in S: b(x6), b(x8) b(x8):d e f g x1 b1 c1 e2 f1 x2 b2 c2 d2 e1 f2 g3 x3 d3 g1 x4 b3 c3 e3 x5 g2 x6 x7 f4 x8 x9 d1
x10
Attribute b
Two null values in S: b(x6), b(x8)
b(x8):
e2  b1 (support 2),
c1 f1  b1 (support 2),
g2  b2 (support 2),
c3  b3 (support 1),
c2  b2 (support 2),
g3d2  b2 (support 1),
e3d3  b3 (support 1),
f2d2  b2 (support 1).
b(x6) = b2

14. Example (Chase1) Two null values in S: c(x7), c(x8). c(x7):b c d e f g x1 b1 c1 e2 f1 x2 b2 c2 d2 e1 f2 g3 x3 d3 g1 x4 b3 c3 e3 x5 g2 x6 x7 f4 x8 x9 d1
x10
Two null values in S: c(x7), c(x8).
c(x7):
b1  c1 (support 2),
e2  c1 (support 1),
f4  c1 (support 1),
g1  c1 (support 2),
b2 d2  c2 (support 1),
b2e1  c2 (support 1),
b2f2  c2 (support 1),
b2g3  c2 (support 1),
d2e1  c2 (support 1),
d2g3  c2 (support 1).
b(x6) = b2

15. Example (Chase1) Two null values in S: c(x7), c(x8). c(x7):b c d e f g x1 b1 c1 e2 f1 x2 b2 c2 d2 e1 f2 g3 x3 d3 g1 x4 b3 c3 e3 x5 g2 x6 x7 f4 x8 x9 d1
x10
Two null values in S: c(x7), c(x8).
c(x7):
b1  c1 (support 2),
e2  c1 (support 1),
f4  c1 (support 1),
g1  c1 (support 2),
b2 d2  c2 (support 1),
b2e1  c2 (support 1),
b2f2  c2 (support 1),
b2g3  c2 (support 1),
d2e1  c2 (support 1),
d2g3  c2 (support 1).
b(x6) = b2

16. Example (Chase1) Two null values in S: c(x7), c(x8). c(x8):b c d e f g x1 b1 c1 e2 f1 x2 b2 c2 d2 e1 f2 g3 x3 d3 g1 x4 b3 c3 e3 x5 g2 x6 x7 f4 x8 x9 d1
x10
Two null values in S: c(x7), c(x8).
c(x8):
b1  c1 (support 2),
e2  c1 (support 1),
f4  c1 (support 1),
g1  c1 (support 2),
b2 d2  c2 (support 1),
b2e1  c2 (support 1),
b2f2  c2 (support 1),
b2g3  c2 (support 1),
d2e1  c2 (support 1),
d2g3  c2 (support 1).
b(x6) = b2 ,
c(x7) = c1

17. Algorithm Chase1(S, In(A), L(D))
Input: System S=(X, A, V)
Set of incomplete attributes In(A)={a1, a2, …, ak}
Set of rules L(D)
Output: System Chase1(S)
begin
j:=1; while j ≤ k do begin
Sj:=S
for all vVaj do
while
there is xX and rule (t v)L(D)
such that xNSj(t) and card(aj(x))≠1
a(x):=v;
end
j:=j+1
S:={Sj:1 ≤ j ≤ k}, Chase1 (S, In(A), L(D))

18. ZOO Database A1- “the no. of different attributes used in a query"
A2- “the percent of null values
in a queried IS"
A3- “the no. of objects returned
by QAS when IS-complete"
A4- “the no. of objects returned
by QAS when IS-incomplete"
(optimistic interpretation)
A5- “the no. of objects returned
by QAS based on rule-based
chase algorithm"
(pessimistic interpretation)
A6- “the no. of bad objects retrieved"
A7- “the no. of passes of rule-based
query A1 A2 A3 A4 A6 A7 A5 4 2 8 q1 q2 3 13 12 1 5 14 15 17 9 27 16 q3 22 25 32 30 28 23
ZOO Database

19. Rules Discovery from partially Incomplete Information Systems

20. Data (Incomplete) Information System S = ( X, A, V )
X - finite set of objects,
A - finite set of attributes,
- set of their values.
Assumption
1. For any ,
2. For any

21. Example x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X

22. Algorithm ERID for Extracting Rules
from partially Incomplete Information System
Goal:
Describe e in terms of {a,b,c,d} x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X

23. Algorithm ERID for Extracting Rules
from partially Incomplete Information System
Goal:
Describe e in terms of {a,b,c,d} x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X

24. Algorithm ERID Goal: Describe e in terms of {a,b,c,d} x8 x7 x6 x5 x4
For the values of the decision
attribute we have:

25. Algorithm ERID Goal: Describe e in terms of {a,b,c,d}. x8 x7 x6 x5 x4
Check the relationship “ ” between values of classification attributes {a,b,c,d} and values
of decision attribute e

26. Algorithm ERID Goal: Describe e in terms of {a,b,c,d} x8 x7 x6 x5 x4
Let ,
We say that:
iff
support
and confidence of the rule
are above some threshold values.

27. support and confidence
Algorithm ERID
Goal:
Describe e in terms of {a,b,c,d} x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X
Let ,
We say that:
iff
support
and confidence of the rule
are above some threshold values.
How to define
support and confidence
of a rule ?

28. Definition of Support and Confidence
(by example)
To define support and confidence of the rule a1  e3
we compute:
Support of the rule:
Support of the term a1:
Confidence of the rule:

29. Extracting Rules from partially Incomplete Information System
(Algorithm ERID(λ1, λ2)) x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X
Goal:
Describe e in terms of {a,b,c,d}
Thresholds (provided by user):
Minimal support (λ1 = 1)
Minimal confidence (λ2 = ½)
- marked negative
- marked negative
- marked positive

30. Algorithm ERID(λ1, λ2) but x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X but but

31. Algorithm ERID(λ1, λ2) but x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X but but
They all are not marked

32. Algorithm ERID(λ1, λ2) x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X
and
and

33. Algorithm ERID(λ1, λ2) x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X and and
They all are marked positive.

34. Algorithm ERID(λ1, λ2) x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X and and
They all are marked positive.
They all are marked negative.

35. Algorithm ERID(λ1, λ2) x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X
The algorithm continues for terms
of length 3, 4, … till all of them have either positive or negative marks.
Rules are automatically constructed
from relations marked positive.

36. Algorithm Chase 2
(for Partially IIS)

37. Algorithm Chase 2
partially incomplete information system of type λ, if S
is incomplete and the following three conditions hold:
is defined for any ,

38. Algorithm Chase 2
S1, S2 - partially incomplete, both of type λ and both classifying
the same sets of objects (from X) using the same sets of attributes (A)
Let
and
The pair (S1, S2)
satisfies containment relation Ψ (or Ψ(S1)= S2) if:
We also denote that fact by

39. x8 x7 x6 x5 x4 x3 x2 x1 e d c b a X
System S1
System S2 c2

40. - information system of type λ - set of all pairwise
Assumptions:
- information system of type λ
- set of all pairwise
independent rules extracted by ERID from S
NS(t) - standard interpretation of term t in S, meaning that:
, for any
where for any , we have:
In(A) = {a1, … , ak} - incomplete attributes in S

41. Algorithm Chase 2 (S, In(A), L(D))
for all do begin
if and
is a maximal subset of rules from L(D) such that
then if then
begin
end
pj:= pj +nj;
if /containment relation holds between aj(x), [bj(x)/pj]/
then

42. Incomplete Information System Sx8 x7 x6 x5 x4 x3 x2 x1 e d c b a X
ERID(λ1, λ2)
λ1=1, λ2=0.5
Incomplete Information System S
of type λ = 0.3

43. Incomplete Information System Sx8 x7 x6 x5 x4 x3 x2 x1 e d c b a X
ERID(λ1, λ2)
λ1=1, λ2=0.5
Algorithm Chase 2 will try to replace
by enew(x1) = {(e1, ), (e2, ), (e3, )}.
We will show that Ψ(e(x1)) = enew(x1)
(the value e(x1) will be changed by Chase 2).
Incomplete Information System S
of type λ = 0.3

44. Incomplete Information System Sx8 x7 x6 x5 x4 x3 x2 x1 e d c b a X
For x1:
we have:
Because the confidence assigned to e3 is below the threshold λ, then only two values remain:
(e1, 0.48), (e2, 0.97).
The value of attribute e assigned to x1 is: {(e1, 0.33), (e2, 0.67)}.
Incomplete Information System S
of type λ = 0.3

45. Distributed Chase Algorithm

46. Let:
- distributed autonomous information systems
- information system for any
(I - set of sites)
- knowledge-base at site
set of (k, i)-rules (constructed at site k and sent
to site i)

47. Strategy for constructing knowledge-base and Algorithm Chase 3
Notation:
qS=[a, b, c : d, e] - request by S for definitions of a, b, c with additional information that d, e are complete attributes in S.

48. Global Ontology S2 S1 S KB KBS g a b c b a d e a b c d rule supportqS
qS=[a, c, d : b] S a b c d a1 b2 b1 c2 a2 d2 c1
rule
support
system KB
KBS

49. Global Ontology S2 S1 S KBS g a b c b a d e a b c d rule supportqS
qS=[a, c, d : b] S a b c d a1 b2 b1 c2 a2 d2 c1
rule
support
system
b1a2 1 S
b2*d2a2
b2a2 S1
c1*b1a1 S2 r1 r2
KBS

50. Assumption:
Di - granularity level of values of attributes used in rules from Di
may differ from the granularity level of values of attribute
used in descriptions of objects in
Chase 3 algorithm to be applicable to Si has to be based on rules from
Di satisfying the following two conditions: .
attribute value used in the decision part of a rule has the granularity level
either equal to or finer than the granularity level of the corresponding
attribute in Si .
the granularity level of any attribute used in the classification part of a rule
is either equal or softer than the granularity level of the corresponding
attribute in Si

51. Example Hierarchical attributes: age, salary
Rule in Di: (age, young)  (salary, 40k)
age
young middle-aged old
salary
low medium high
18 … … … 80
10k…40k 50k 60k 70k 80k…100k

52. (Construction of new Di followed by Chase 2)
Algorithm Chase 3
(Construction of new Di followed by Chase 2)
Assumption:
tuple t in Si supports rule
Two cases:
1. An overlapping attribute between rule and the tuple is the decision attribute in
If two attributes, involved in that match, have different granularities, then the decision value d has to be replaced by a softer value which granularity will match the granularity of the corresponding attribute in Si.
2. An overlapping attribute between rule and the tuple is the
classification attribute in
If two attributes, involved in that match, have different granularities, then the value of attribute a has to be replaced by a finer value which granularity will match the granularity of a in Si.

53. (All Information Systems are equally
Chase 4
(All Information Systems are equally
involved in chase)

54. S3 S2 KB S1 qS2 qS3 qS1 qS1=[a, c, d : b] qS2=[b, a, e : d]g e d S3 S2 KB
qS2
qS3 S1
qS1
qS1=[a, c, d : b]
qS2=[b, a, e : d]
qS3=[a, b, c : g]

55. S3 S2 KB S1 qS2 qS3 qS1 qS1=[a, c, d : b] qS2=[b, a, e : d]g e d S3 S2
r1, r2
r5, r6 KB
r3, r4
qS2
qS3
r3, r4 – extracted from S3
r1, r2 – extracted from S2 S1
r5, r6 – extracted from S1
qS1
qS1=[a, c, d : b]
qS2=[b, a, e : d]
qS3=[a, b, c : g]