Database Storage & Design Chapter 8 Homework The Relational Algebra And Relational Calculus The Relational Algebra And

Chapter 8: The Relational Algebra and Relational Calculus

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CHAPTER 8: THE RELATIONAL ALGEBRA AND RELATIONAL CALCULUS

Answers to Selected Exercises

8.15 – Show the result of each of the sample queries in Section 6.5 as it would apply to the

database state in Figure 3.6.

Answer:

(QUERY 1) Find the name and address of all employees who work for the ‘Research’

department.

(QUERY 2) For every project located in ‘Stafford’, list the project number, the controlling

department number, and the department manager’s last name, address, and birth date.

Result:

(QUERY 3) Find the names of all employees who work on all the projects controlled by

department number 5.

Result: (empty because no tuples satisfy the result).

(QUERY 4) Make a list of project numbers for projects that involve an employee whose last

name is ‘Smith’ as a worker or as a manager of the department that controls the project.

Result: PNO

1

2

(QUERY 5) List the names of all employees with two or more dependents.

Result: LNAME FNAME

Smith John

Wong Franklin

(QUERY 6) List the names of employees who have no dependents.

Result: LNAME FNAME

Zelaya Alicia

Narayan Ramesh

English Joyce

Wallace Jennifer

Wong Franklin

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Result (empty):

LNAME FNAME

(c) WONG_SSN <— P SSN ( s FNAME=’Franklin’ AND

LNAME=’Wong’ (EMPLOYEE))

(d) PROJ_HOURS(PNO,TOT_HRS) <— PNO f SUM HOURS (WORKS_ON)

RESULT <— P PNAME,TOT_HRS ( (PROJ_HOURS) J (PNO),(PNUMBER)

(PROJECT) )

Result:

PNAME TOT_HRS

(e) PROJ_EMPS(PNO,SSN) <— P PNO,ESSN (WORKS_ON)

ALL_PROJS(PNO) <— P PNUMBER (PROJECT)

(f) ALL_EMPS <— P SSN (EMPLOYEE)

WORKING_EMPS(SSN) <— P ESSN (WORKS_ON)

NON_WORKING_EMPS <— ALL_EMPS – WORKING_EMPS (* DIFFERENCE

*)

(g) DEPT_AVG_SALS(DNUMBER,AVG_SAL) <— DNO f AVG SALARY

(EMPLOYEE)

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(h) RESULT(AVG_F_SAL) <— f AVG SALARY ( s SEX=’F’ (EMPLOYEE) )

Result:

(i) E_P_HOU(SSN) <—

P ESSN (WORKS_ON J(PNO),(PNUMBER) ( s PLOCATION=’Houston’

(PROJECT)))

D_NO_HOU <—

Result:

LNAME FNAME ADDRESS

Wallace Jennifer 291 Berry, Bellaire, TX

LNAME FNAME

Borg James

8.17 – No solution provided.

8.18 – Consider the LIBRARY relational schema shown in Figure 6.14, which is used to keep

track of books, borrowers, and book loans. Referential integrity constraints are shown as

directed arcs in Figure 6.14, as in the notation of Figure 3.7. Write down relational

expressions for the following queries on the LIBRARY database:

(a) How many copies of the book titled The Lost Tribe are owned by the library branch

whose name is “Sharpstown”?

(b) How many copies of the book titled The Lost Tribe are owned by each library branch?

(c) Retrieve the names of all borrowers who do not have any books checked out.

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(d) For each book that is loaned out from the “Sharpstown” branch and whose DueDate

is today, retrieve the book title, the borrower’s name, and the borrower’s address.

(e) For each library branch, retrieve the branch name and the total number of books

loaned out from that branch.

(f) Retrieve the names, addresses, and number of books checked out for all borrowers

who have more than five books checked out.

(g) For each book authored (or co-authored) by “Stephen King”, retrieve the title and the

number of copies owned by the library branch whose name is “Central”.

Answer:

(Note: We will use S for SELECT, P for PROJECT, * for NATURAL JOIN, – for

SET DIFFERENCE, F for AGGREGATE FUNCTION)

(a) A <— BOOKCOPIES * LIBRARY-BRANCH * BOOK

RESULT <— P No_Of_Copies ( S BranchName=‘Sharpstown’ and Title=’The Lost

(b) P BranchID,No_Of_Copies ( ( S Title=’The Lost Tribe’ (BOOK)) * BOOKCOPIES )

(c) NO_CHECKOUT_B <— P CardNo (BORROWER) – P CardNo (BOOK_LOANS)

(d) S <— P BranchId ( S BranchName=‘Sharpstown’ (LIBRARY-BRANCH) )

(e) R(BranchId,Total) <— BranchId FCOUNT(BookId,CardNo) (BOOK_LOANS)

(f) B(CardNo,TotalCheckout) <— CardNo F COUNT(BookId) (BOOK_LOANS)

(g) SK(BookId,Title) <— ( sAuthorName=’Stephen King’ ( BOOK_AUTHORS)) * BOOK

8.19 – 8.21: No solutions provided.

8.22 – Consider the two tables T1 and T2 shown in Figure 6.15. Show the results of the

following operations:

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Answers:

(a)

P Q R A B C

(b)

15 b 8 10 b 5

(c)

P Q R A B C

10 a 5 10 b 6

(d)

P Q R A B C

(e)

P Q R

10a 5

(f)

8.23 – No solution provided.

8.24 – Specify queries (a), (b), (c), (e), (f), (i), and (j) of Exercise 6.16 in both tuple and

domain relational calculus.

Answer:

(a) Retrieve the names of employees in department 5 who work more than 10 hours per

week on the ‘ProductX’ project.

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{ qs | EMPLOYEE(qrstuvwxyz) AND z=5 AND (EXISTS a) (EXISTS b) (EXISTS

(b) List the names of employees who have a dependent with the same first name as

themselves.

(c) Find the names of employees that are directly supervised by ‘Franklin Wong’.

Tuple relational Calculus:

{ e.LNAME, e.FNAME | EMPLOYEE(e) AND (EXISTS s) ( EMPLOYEE(s) AND

(e) Retrieve the names of employees who work on every project.

Tuple relational Calculus:

{ e.LNAME, e.FNAME | EMPLOYEE(e) AND (FORALL p) ( NOT(PROJECT(p))

OR

(EXISTS w) (

(f) Retrieve the names of employees who do not work on any project.

Tuple relational Calculus:

{ e.LNAME, e.FNAME | EMPLOYEE(e) AND NOT(EXISTS w) ( WORKS_ON(w)

AND

(i) Find the names and addresses of employees who work on at least one project located

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in Houston but whose department has no location in Houston.

Tuple relational Calculus:

{ e.LNAME, e.FNAME, e.ADDRESS | EMPLOYEE(e) AND (EXISTS p) (EXISTS

w) (

c=’Houston’ AND

NOT(EXISTS h) NOT(EXISTS i) ( DEPT_LOCATIONS(hi) AND z=h AND

i=’Houston’

) ) }

(j) List the last names of department managers who have no dependents.

Tuple relational Calculus:

{ e.LNAME | EMPLOYEE(e) AND (EXISTS d) ( DEPARTMENT(d) AND

e.SSN=d.MGRSSN AND

NOT(EXISTS x) (DEPENDENT(x) AND e.SSN=x.ESSN) ) }

AND NOT(EXISTS e) (DEPENDENT(efghi) AND e=t) ) }

8.25 – No solution provided.

6.26 Specify queries c, d, and e of Exercise 6.18 in both tuple and domain relational

calculus.

Answer:

(c) For each section taught by professor King, retrieve the course number, semester,

(d) Retrieve the name and transcript of each senior student (Class=5) majoring in

COSC. Transcript includes course name, course number, credit hours, semester, year, and

grade for each course completed by the student.

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s.Major=’COSC’ AND s.StudentNumber=g.StudentNumber AND

g.SectionIdentifier=t.SectionIdentifier

AND t.CourseNumber=c.CourseNumber}

Domain relational Calculus:

(e) Retrieve the names and major departments of all straight A students (students

who have a grade of A in all their courses).

Tuple relational Calculus:

{ s.Name, s.Major | STUDENT(s) AND (FORALL g) ( NOT(GRADE_REPORT(g))

OR

8.27 – In a tuple relational calculus query with n tuple variables, what would be the typical

minimum number of join conditions? Why? What is the effect of having a smaller number of

join conditions?

Answer:

8.28 – Rewrite the domain relational calculus queries that followed Q0 in Section 6.7 in the

style of the abbreviated notation of Q0A, where the objective is to minimize the number of

domain variables by writing constants in place of variables wherever possible.

Answer:

Q1A: { qsv | (EXISTS z) (EXISTS m) ( EMPLOYEE(q,r,s,t,u,v,w,x,y,z) AND

8.29 – No solution provided.

8.30 – Show how you can specify the following relational algebra operations in both tuple and

domain relational calculus.

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(a) SELECT A=c (R(A, B, C)):

(b) PROJECT <A, B> (R(A, B, C)):

(c) R(A, B, C) NATURAL JOIN S(C, D, E):

(d) R(A, B, C) UNION S(A, B, C):

(e) R(A, B, C) INTERSECT S(A, B, C):

(f) R(A, B, C) MINUS S(A, B, C):

(g) R(A, B, C) CARTESIAN PRODUCT S(D, E, F):

(h) R(A, B) DIVIDE S(A):

Answer:

For each operation, we give the tuple calculus expression followed by the domain calculus

expression.

(a) { t | R(t) AND t.A=c}, { xyz | R(xyz) AND x=c }

(b) { t.A, t.B | R(t) }, { xy | R(xyz) }

(f) { t | R(t) AND NOT(S(t)) },

{ xyz | R(xyz) AND NOT(S(xyz)) }

(g) { t.A, t.B, t.C, q.D, q.E, q.F | R(t) AND S(q) },

( xyzuvw | R(xyz) AND S(uvw) }

8.31 – No solution provided.

8.32 – A nested query is query within a query. More specifically, a nested query is a

parenthesized query that can be used as a value in a number of places, such as instead of a

relation or a selection condition. Specify the following queries on the database specified in

Figure 3.5 using the concept of nested queries and the relational operators discussed in this

chapter. Also show the result of each query as it would apply to the database state of Figure