INSTALL NUMBER 12Answers to Puzzle Corner Problems(Installment Number 7-11)

0 0 0
DEE
DUM
DUM DUM
The DEE and DUM problem
Solution:
Difference DEE
DUM DEE
AND NOT
1
0
DEE
DUM
1
0
0
0
1
0
DUM
DUM
DUM
The DEE and DUM problem
Solution of restrict and project:
The general unification theorem
Solution: 1. AB (given) 2.CD (given) 3.AB∩C (by joint dependence and 1) 4.C-BC-B (self-determination) 5.A∪(C-B)(B∩C)∪(C-B) (by composition, 3, 4) 6.A∪(C-B)C (simplifying 5) 7.A∪(C-B)D (by transitivity, 6, 2) 8. A∪(C-B)B∪D (by composition, 1, 7)
完成證明
Expression transformation problem
Source:”Expression Transformation”
(Installment 10 & 11) Problem statement:Prove the following statements
• A sequence of restrictions against a given relation can be transformed into a single restriction. • A sequence of projections against a given relation can be transformed into a single projection. • A restriction of a projection can be transformed into a restriction.
The DEE and DUM problem
Solution of division:
• • • • Any table T divided by DEE yields T. Any table T divided by DUM yields an empty table with the same heading as T. DEE divided by any table T yields T. DUM divided by any tale T yields an empty table with the same heading as T. Any nonempty table divided by itself yields DEE. An empty table divided by itself yields DUM.
The empty argument problem
5. Union Solution: An empty table with the specified heading.
6.Intersection Solution: A table with the specified heading and with body equal to the Cartesian product to all underlying domains.
The general unification theorem
Solution:
1. AB (given)
2.CD (given) 3.AB∩C (by joint dependence and 1) why? 因為AB,又B∩C屬於B的一部份， 故A NhomakorabeaB∩C。
The general unification theorem
• • • Any restriction of DEE yields DEE if the restriction condition is true, DUM if it is false. Any restriction of DUM yields DUM. Projection of any_table over no columns yields DUM if the original table is empty, DEE otherwise. In particular, projection of DEE or DUM, necessarily over no columns at all, return its input.
Reading review(Installment 7-11)
Installment 7: Table with no columns
Installment 8: Empty bag and identity crises Installment 9: The power of the keys
INSTALL NUMBER 12
Answers to Puzzle Corner Problems (Installment Number 7-11)
中正大學資管所 碩一 690530023 郭 溥 淵
Introduction
Reading review(Installment 7-11) The DEE and DUM problem The empty argument problem The general unification theorem Expression transformation problem Technical correspondence Conclusion Q&A
The empty argument problem
1. Sum of the squares Solution: 0
2.Standard deviation Solution: Undefined
The empty argument problem
3. Median Solution: Undefined
The general unification theorem
Solution:
• Self-determination: AA • Joint dependence: AB&AC≡AB∪C • Transitivity: AB&BC ==> AC • Composition: AB&CD ==>A∪CB∪D
The DEE and DUM problem
Solution:
Union DEE
DUM
OR 1 0
1 1 1
0 1 0
DEE
DUM
DEE
DEE
DEE
DUM
The DEE and DUM problem
Solution:
Intersection
DEE DEE
DUM DUM
AND 1 1 0 1 0
•
The DEE and DUM problem
Solution of extend and summarize:
• Extending DEE or DUM to add a new column yields a relation of one column and the same number of rows as its input. Summarizing DEE or DUM(necessarily over no columns at all) yields a relation of one column and the same number of rows as input.
The general unification theorem
Solution: 8. A∪(C-B)B∪D (by composition,1,7) why?
Composition: AB&CD ==>A∪CB∪D
AB ---1 A∪(C-B)D ---2 所以A∪(C-B)B∪D
The general unification theorem
Solution: 6.A∪(C-B)C (simplifying 5) why? (B∩C)∪(C-B)就是C 所以A∪(C-B)C 7.A∪(C-B)D (by transitivity, 6, 2) why? 因為A∪(C-B)C ，又CD 所以A∪(C-B)D
Installment 10: Expression transformation
Installment 11: Expression transformation
The DEE and DUM problem
Source:”Tables with No Columns”
(Installment 7) Problem statement: What are the effects of DEE and DUM on the relational algebra operation union,intersection, difference, restrict, project, division, extend, and summarize?
Solution:
4.C-BC-B (self-determination) 5.A∪(C-B)(B∩C)∪(C-B) (by composition, 3, 4) why? Composition: AB&CD ==>A∪CB∪D AB∩C ---3 C-BC-B ---4 故A∪(C-B)(B∩C)∪(C-B)