1. COMP3030 Database Management System Final Review

2. AGENDA Ch1. Introduction Ch2. Relational Model Ch3. SQL
Ch4. Advanced SQL
Ch6. Database Design and the ER model
Ch7. Relational Database Design
Ch10. XML

3. CH2 EXERCISES
Given two relations R1 and R2, where R1 contains N1 tuples, R2 contains N2 tuples, and N2 > N1 > 0, give the min and max possible sizes for the resulting relational algebra expressions: (1) R1UR2, (2) R1∩R2, (3) R1−R2, (4) R1×R2, (5) σa=5(R1), and (6) πa(R1)
Sigma  Select rows
Pi  Projection, extract columns
UNION COMPATIBLE:
Have same number of fields AND Have corresponding fields, taken in order from left to right, have the same domains.

4. CH2 EXERCISES
Consider the Supplier-Parts-Catalog schema from the previous question. State what the following queries compute:
Find the Supplier names of the suppliers who supply a red part that costs less than 100 dollars.
This Relational Algebra statement does not return anything
Find the Supplier names of the suppliers who supply a red part that costs less than 100 dollars and a green part that costs less than 100 dollars.
**Natural joins
3… because of the sequence of projection operators. Once the sid is projected, it is the only field in the set. Therefore, projecting on sname will not return anything.

5. Ch6 EXERCISES
Consider the Supplier-Parts-Catalog schema from the previous question. State what the following queries compute:
Find the Supplier ids of the suppliers who supply a red part that costs less than 100 dollars and a green part that costs less than 100 dollars.

6. AGENDA Ch1. Introduction Ch2. Relational Model Ch3. SQL
Ch4. Advanced SQL
Ch6. Database Design and the ER model
Ch7. Relational Database Design
Ch10. XML

7. CH3 EXERCISES
What is the difference between a candidate key and the primary key for a given relation? What is a superkey?
The primary key is the key selected by the DBA from among the group of candidate keys, all of which uniquely identify a tuple. A superkey is a set of attributes that contains a key.

8. CH3 EXERCISES
Answer each of the following questions briefly. The questions are based on the following relational schema: Emp(eid: integer, ename: string, age: integer, salary: real) Works(eid: integer, did: integer, pcttime: integer) Dept(did: integer, dname: string, budget: real, managerid: integer)
1. Give an example of a foreign key constraint that involves the Dept relation. What are the options for enforcing this constraint when a user attempts to delete a Dept tuple?
Consider the following example. It is natural to require that the did field of Works should be a foreign key, and refer to Dept. CREATE TABLE Works ( eid INTEGER NOT NULL , did INTEGER NOT NULL , pcttime INTEGER, PRIMARY KEY (eid, did), FOREIGN KEY (did) REFERENCES Dept )
When a user attempts to delete a Dept tuple, There are four options:

9. CH3 EXERCISES
2. Write the SQL statements required to create the preceding relations, including appropriate versions of all primary and foreign key integrity constraints.
CREATE TABLE Emp (
eid INTEGER,
ename CHAR(10),
age INTEGER,
salary REAL,
PRIMARY KEY (eid) )
CREATE TABLE Works (
did INTEGER,
pcttime INTEGER,
PRIMARY KEY (eid, did),
FOREIGN KEY (did) REFERENCES Dept,
FOREIGN KEY (eid) REFERENCES Emp,
ON DELETE CASCADE
CREATE TABLE Dept (
budget REAL,
managerid INTEGER ,
PRIMARY KEY (did),
FOREIGN KEY (managerid) REFERENCES Emp,
ON DELETE SET NULL

10. CH3 EXERCISES
3. Define the Dept relation in SQL so that every department is guaranteed to have a manager.
CREATE TABLE Dept ( did INTEGER, budget REAL, managerid INTEGER NOT NULL , PRIMARY KEY (did), FOREIGN KEY (managerid) REFERENCES Emp )

11. CH3 EXERCISES
4. Write an SQL statement to add John Doe as an employee with eid = 101, age = 32 and salary = 15, 000.
INSERT INTO Emp (eid, ename, age, salary) VALUES (101, ’John Doe’, 32, 15000)

12. CH3 EXERCISES
5. Write an SQL statement to give every employee a 10 percent raise.
UPDATE Emp E SET E.salary = E.salary * 1.10

13. CH3 EXERCISES
6. Write an SQL statement to delete the Toy department. Given the referential integrity constraints you chose for this schema, explain what happens when this statement is executed.
DELETE FROM Dept D WHERE D.dname = ’Toy’
The did field in theWorks relation is a foreign key and references the Dept relation.
This is the referential integrity constraint chosen. By adding the action ON DELETE
CASCADE to this, when a department record is deleted, theWorks record associated
with that Dept is also deleted.
The query works as follows: The Dept relation is searched for a record with name
= ‘Toy’ and that record is deleted. The did field of that record is then used to
look in the Works relation for records with a matching did value. All such records
are then deleted from the Works relation.
sid name login age gpa
Madayan
Guldu

14. CH3 EXERCISES
Consider the following schema: Suppliers(sid: integer, sname: string, address: string) Parts(pid: integer, pname: string, color: string) Catalog(sid: integer, pid: integer, cost: real) The Catalog relation lists the prices charged for parts by Suppliers. Write the following queries in SQL:
1. Find the pnames of parts for which there is some supplier.
SELECT DISTINCT P.pname FROM Parts P, Catalog C WHERE P.pid = C.pid
5. Find the sids of suppliers who charge more for some part than the average cost of that part (averaged over all the suppliers who supply that part).
SELECT DISTINCT C.sid FROM Catalog C WHERE C.cost > ( SELECT AVG (C1.cost) FROM Catalog C1 WHERE C1.pid = C.pid )

15. Ch3 EXERCISES
8. Find the sids of Suppliers who supply a red parts and a green part
SELECT DISTINCT C.sid
FROM Catalog C, Parts P
WHERE C.pid = P.pid AND P.color = ‘Red’
INTERSECT
SELECT DISTINCT C1.sid
FROM Catalog C1, Parts P1
WHERE C1.pid = P1.pid AND P1.color = ‘Green’
9. Find the sids of Suppliers who supply a red parts or a green part
SELECT DISTINCT C.sid FROM Catalog C, Parts P
UNION
SELECT DISTINCT C1.sid FROM Catalog C1, Parts P1

16. AGENDA Ch1. Introduction Ch2. Relational Model Ch3. SQL
Ch4. Advanced SQL
Ch6. Database Design and the ER model
Ch7. Relational Database Design
Ch10. XML

17. CH6 EXERCISES
A university database contains information about professors (id. by SSN) and courses (id. by courseid). Professors teach courses; each of the following situations concerns the Teaches relationship set. For each situation, draw an ER diagram that describes it (assuming no further constraints hold).
Professors can teach the same course in several semesters, and each offering must be recorded.
WHAT ARE THE ENTITIES?
ENTITY SETS?
ATTRIBUTES?
RELATIONSHIP?

18. Professors can teach the same course in several semesters, and only the most recent such offering needs to be recorded. (Assume this condition applies in all subsequent questions.)

19. Every professor must teach some course.
Participation constraint is total for Professor in Teaches

20. Every professor teaches exactly one course (no more, no less).
Total participation
Each Professors entity appears in at most one Teaches relationship.

21. Every professor teaches exactly one course (no more, no less), and every course must be taught by some professor.

22. Certain courses can be taught by a team of professors jointly, but it is possible that no one professor in a team can teach the course. Model this situation, introducing additional entity sets and relationship sets if necessary.

23. A company database needs to store information about
2.4
A company database needs to store information about
employees (identified by ssn, with salary and phone as attributes),
departments (identified by dno, with dname and budget as attributes),
and children of employees (with name and age as attributes).
Employees work in departments; each department is managed by an
employee; a child must be identified uniquely by name when the parent
(who is an employee; assume that only one parent works for the
company) is known. We are not interested in information about a child
once the parent leaves the company.

25. Convert ER diagram to relational schema
Employee (ssn, salary, phone, mangerid)
Departments (dno, dname, budget)
Child (name, age, ssn)
Works_in (ssn, dno)

26. AGENDA Ch1. Introduction Ch2. Relational Model Ch3. SQL
Ch6. Database Design and the ER model
Ch7. Relational Database Design
Ch10. XML

27. CH7 EXERCISES A relation R is in third normal form if for
every functional dependency of the form X
 A one of the following statements is true:
A Є X that is, A is a trivial functional dependency , or (1)
X is a superkey, or (2)
A is part of some key for R (3)
A relation R is in BCNF if (1) or (2)

28. CH7 EXERCISES
7.2. Consider a relation R with five attributes ABCDE. You are given the following dependencies:
A → B, BC → E, ED → A
List all candidate keys for R.
CDE, ACD, BCD
Is R in 3NF?
R is in 3NF because B, E and A are all parts of candidate keys.
Is R in BCNF?
R is not in BCNF because none of A, BC and ED contain a key.
BCNF = Boyce-Codd Normal Form
iff X is superkey
3NF
X is superkey
A is part of some key

29. BCNF, 3NF decomposition
Canonical cover
Closure
4NF

30. AGENDA Ch1. Introduction Ch2. Relational Model Ch3. SQL
Ch6. Database Design and the ER model
Ch7. Relational Database Design
Ch10. XML

31. Give the DTD for an XML representation of the following nested-relational schema
Emp = (ename, ChildrenSet setof(Children), SkillsSet setof(Skills))
Children = (name, Birthday)
Birthday = (day, month, year)
Skills = (type, ExamsSet setof(Exams))
Exams = (year, city)

32. <!DOCTYPE db [
<!ELEMENT emp (ename, children*, skills*)>
<!ELEMENT children (name, birthday)>
<!ELEMENT birthday (day, month, year)>
<!ELEMENT skills (type, exams+)>
<!ELEMENT exams (year, city)>
<!ELEMENT ename( #PCDATA )>
<!ELEMENT name( #PCDATA )>
<!ELEMENT day( #PCDATA )>
<!ELEMENT month( #PCDATA )>
<!ELEMENT year( #PCDATA )>
<!ELEMENT type( #PCDATA )>
<!ELEMENT city( #PCDATA )>
] >

33. a. Find the names of all employees who have a child who has a birthday in March.
for $e in /db/emp,
$m in distinct($e/children/birthday/month)
where $m = ’March’
return $e/ename

34. b. Find those employees who took an examination for the skill type “typing” in the city “Dayton”.
for $e in /db/emp
$s in $e/skills[type=’typing’]
$exam in $s/exams
where $exam/city= ’Dayton’
return $e/ename

35. c. List all skill types in Emp.
for $t in /db/emp/skills/type
return $t

36. Final Exam Information
Ch1. Introduction
Ch2. Relational Model 17
Ch3. SQL 41
Ch4. Advanced SQL
Ch6. Database Design and the ER model 10
Ch7. Relational Database Design 22
Ch10. XML 10

37. GOOD LUCK!!