install number 12 answers to puzzle corner problems (installment number 7-11)

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

中正大學資管所碩一 690530023

郭溥淵

IntroductionIntroduction

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

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 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?

The DEE and DUM problem Solution:

Union

DEE DUM

DEE DEE DEE

DUM DEE DUM

OR 1 0

1 1 1

0 1 0


Intersection DEE DUM

DEE DEE DUM

DUM DUM DUM

AND 1 0

1 1 0

0 0 0


Difference DEE DUM

DEE DUM DEE

DUM DUM DUM

AND NOT

1 0

1 0 1

0 0 0


Solution of restrict and project:• 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.


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.


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 empty argument problem

Source:”Empty Bags and Identity crises” (Installment 8)

Problem statement: Give the correct “empty argument” treatment for each of the following functions. Note: In Case 1-4 the argument is intended to be a bag of number; in Case 5-6 it is a bag of tables all having the same(specified) heading.


1. Sum of the squares

Solution:

0

2.Standard deviation

Solution:

Undefined


3. Median

Solution:

Undefined

4.Geometric mean

Solution:

1


Source:”Empty Bags and Identity crises” (Installment 8)

Problem statement: Give the correct “empty argument” treatment for each of the following functions. Note: In Case 1-4 the argument is intended to be a bag of number; in Case 5-6 it is a bag of tables all having the same(specified) heading.


5. UnionSolution:

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.


SELECT CURRENT_TIME

WHERE 1=0 ;

EXTEND DUM ADD CURRENT_TIME AS X

Those two query are equivalent.

The general unification theorem

Source:”The power of keys” (Installment 9)

Problem statement: Darwen’s work on FD and key inheritance makes use of the following theorem. Let A, B, C, and D be subsets of the set of columns of relation R such that AB and CD. Then A∪(C-B)B∪D . Prove this theorem.


Solution:• Self-determination: AA

• Joint dependence: AB&AC≡AB∪C

• Transitivity: AB&BC ==> AC

• Composition: AB&CD ==>A∪CB∪D

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)


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 AB∩C。


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)


Solution:6.A∪(C-B)C (simplifying 5) why?

(B∩C)∪(C-B)就是 C 所以 A∪(C-B)C7.A∪(C-B)D (by transitivity, 6, 2) why?

因為 A∪(C-B)C ，又 CD所以 A∪(C-B)D


Solution:8. A∪(C-B)B∪D (by composition,1,7) why?

Composition: AB&CD ==>A∪CB∪D

AB ---1A∪(C-B)D ---2

所以 A∪(C-B)B∪D

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.


Solution:(a)

• ( R WHERE C1 ) WHERE C2

• R WHERE C1 AND C2

If C1 and C2 are both restriction condition for relation R, then above two expressions are obviously equivalent.


Solution:(b)

• ( R [ L1 ] ) [L2]

• R [ L2 ]

If L1 is a subset of the heading of relation R and L2 is a subset of L1,then the above two expressions are obviously equivalent.


Solution:(c)

• R [ L ] WHERE C

• ( R WHERE C ) [ L ]

If L and C are, respectively, a subset of heading of relation R and restriction condition for relation R that involves only columns mentioned in L, then the above two expressions are obviously equivalent.


SELECT E#

FROM DEPT, EMP

WHERE NOT ( DEPT.D# = EMP.D#

AND EMP.D# = ‘D1’) ; Show the answer

DEPT EMP D#

D2

E# D#

E1 --


(a) Show the correct real-world answer to this query.

DEPT EMP D#

D2

E# D#

E1 --


SELECT E#FROM DEPT, EMPWHERE NOT ( DEPT.D# = EMP.D#


DEPT EMP

The answer is E1.

D#

D2

E# D#

E1 --


(b) Show the answer delivered by the query as stated.

DEPT EMP D#

D2

E# D#

E1 --


SELECT E#FROM DEPT, EMPWHERE NOT ( DEPT.D# = EMP.D#


DEPT EMP

The answer is empty.

D#

D2

E# D#

E1 --


SELECT E#

FROM DEPT, EMP

WHERE NOT ( DEPT.D# = EMP.D#


DEPT EMP D#

D2

E# D#

E1 --


(c) Show the answer delivered if the optimizer applies “predicate transitive closure”.

DEPT EMP D#

D2

E# D#

E1 --


Optimizer will transform the expression to :SELECT E#FROM DEPT,EMPWHERE NOT ( DEPT.D# = EMP.D#

AND EMP.D# = D1AND DEPT.D# = D1 )

;

The answer is E1.

Technical correspondence

Problem:

• SUMMARIZE EMP BY ( D# )

ADD AVG ( SAL )

AS AVGSAL

Technical correspondence

Solution:• IF exp1 THEN exp2 ELSE exp3

• SUMMARIZE EMP BY ( D# )

ADD ( IF IS_EMPTY ( EMP ) THEN 0

ELSE AVG ( SAL ) )

AS AVGSAL

Conclusion

Thanks for your listening.

install number 12 answers to puzzle corner problems (installment number 7-11)

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