Special Edition for CSEDU
CSEDU Students

Students

TOUCH-N-PASS EXAM CRAM GUIDE SERIESDBMS – I

Prepared By

Sharafat Ibn Mollah Mosharraf
CSE, DU
12 Batch (2005-2006)
th

Includes Solutions to
DU DBMS – I Final
Exam Questions of
6 Years (2002-2007)

Table of Contents

CHAPTER 1: INTRODUCTION .................................................................................................................................................... 1
THEORIES ........................................................................................................................................................................................ 1
CHAPTER 2: ENTITY-RELATIONSHIP MODEL..................................................................................................................... 4
QUESTIONS AND ANSWERS ............................................................................................................................................................. 4
CHAPTER 3, 4: RELATIONAL MODEL & SQL ........................................................................................................................ 9
POINTS TO BE REMEMBERED........................................................................................................................................................... 9
THE TRICK OF WRITING RA EXPRESSIONS FOR COMPLEX QUERIES ................................................................................................ 9
COMPLETE CONCEPTS PROBLEM .................................................................................................................................................. 10
GENERAL STRUCTURE OF QUERY STATEMENTS ........................................................................................................................... 16
THEORIES ...................................................................................................................................................................................... 18
CHAPTER 6: INTEGRITY & SECURITY ................................................................................................................................. 29
QUESTIONS AND ANSWERS ........................................................................................................................................................... 29
CHAPTER 7: RELATIONAL DATABASE DESIGN ................................................................................................................ 33
CONCEPTS ..................................................................................................................................................................................... 33
QUESTIONS AND ANSWERS ........................................................................................................................................................... 33
CHAPTER 11: STORAGE & FILE STRUCTURE .................................................................................................................... 40
THEORIES ...................................................................................................................................................................................... 40
CHAPTER 12: INDEXING AND HASHING .............................................................................................................................. 43
CONCEPTS ..................................................................................................................................................................................... 43
QUESTIONS AND ANSWERS ........................................................................................................................................................... 45

CHAPTER 1
INTRODUCTION
Theories
1.1

What is DBMS? [In-course 2007; 2007. Marks: 1]
A database management system (DBMS) is a collection of interrelated data and a set of programsto access those data. The collection of data, usually referred to a database, contains information
relevant to an enterprise.

1.2

Mention some of the areas for database applications. [In-course 1, 2005. Marks: 2]
1. Banking
4. Credit Card Transaction
7. Sales
10. Human Resources

1.3

2. Airlines /Railways/Road Transport
5. Telecommunication
8. On-line Retailers
11. Internet

3. Universities
6. Finance
9. Manufacturing

List four significant differences between a file-processing system and a DBMS.
Some main differences between a database management system and a file-processing system are:• Both systems contain a collection of data and a set of programs which access that data. A
database management system coordinates both the physical and the logical access to the data, whereasa file-processing system coordinates only the physical access.
• A database management system reduces the amount of data duplication by ensuring that aphysical piece of data is available to all programs authorized to have access to it, whereas data written
by one program in a file-processing system may not be readable by another program.
• A database management system is designed to allow flexible access to data (i.e., queries),
whereas a file-processing system is designed to allow predetermined access to data (i.e., compiled
programs).
• A database management system is designed to coordinate multiple users accessing the same dataat the same time. A file-processing system is usually designed to allow one or more programs toaccess different data files at the same time. In a file-processing system, a file can be accessed by two
programs concurrently only if both programs have read-only access to the file.

1.4

What are the disadvantages of DBMS?
Two disadvantages associated with database systems are listed below.
a. Setup of the database system requires more knowledge, money, skills, and time.
b. The complexity of the database may result in poor performance.

1.5

Explain the difference between physical and logical data independence.
Physical data independence is the ability to modify the physical scheme without making it
necessary to rewrite application programs. Such modifications include changing from unblocked toblocked record storage, or from sequential to random access files.
Logical data independence is the ability to modify the conceptual scheme without making it
necessary to rewrite application programs. Such a modification might be adding a field to a record; anapplication program’s view hides this change from the program.

1.6

What are five main functions of a database administrator?
Five main functions of a database administrator are:
1.
2.
3.
4.

To create the scheme definition
To define the storage structure and access methods
To modify the scheme and/or physical organization when necessary
To grant authorization for data access
5. To specify integrity constraints
11.7

Classify database users. [2004. Marks: 2]
Database users are differentiated by the way they expect to interact with the system. There are fourtypes of database users:
1. Naive users – are unsophisticated users who interact with the system by invoking one of thepermanent application programs that have been written previously.
2. Application programmers – are computer professionals who write application programs.
3. Sophisticated users – interact with the system without writing programs. Instead, they form
their requests in a database query language.
4. Specialized users – are sophisticated users who write specialized database applications that do
not fit into the traditional data processing framework.

1.8

What are the jobs of a DBA? [In-course 2007; 2007; 2004. Marks: 3]
The functions of a database administrator (DBA) include:
1. Schema definition: The DBA creates the original database schema by executing a set of data
definition statements in the DDL.
2. Storage structure and access method definition: File organization (sequential, heap, hash,B+ tree), organization of records in a file (fixed length or variable length), index definition(ordered index, hash index).
3. Schema and physical-organization modification: The DBA carries out changes to theschema and physical organization to reflect the changing needs of the organization, or to alter
the physical organization to improve the performance.
4. Granting of authorization for data access: By granting different types of authorization, the
DBA can regulate which parts of the database various users can access.
5. Specifying integrity constraints: The DBA implements key declaration (primary key, foreignkey), trigger, assertion, business rules of the organization.
6. Acting as liaison with users.
7. Routine maintenance:
i. Periodically backing up the database, either onto tapes or remote servers, to prevent loss ofdata in case of disasters.
ii. Ensuring that enough disk space is available for normal operations and upgrading disk
space as required.
iii. Monitoring jobs running on the database and ensuring better performance.

1.9

What can be done using DML? What are the classes of DML? [In-course 1, 2005. Marks: 4]
DML is a language that enables users to access or manipulate data as organized by appropriate
data model. The types of accesses are:
1.
2.
3.
4.

The retrieval of information stored in the database – Query
The insertion of new information into the database – insert
The deletion of information from the database – delete
The modification of information stored in the database – update

Classes of DML:
There are basically two types:
1.

Procedural DMLs – user specifies what data are required and how to get or compute thedata. E.g. Relational Algebra.

2.

Nonprocedural / Declarative DMLs – user specifies what data are required without
specifying how to get or compute the data. E.g. SQL.
2

1.10

How do you classify query languages? Give examples of each type. [In-course 1, 2008; 2006,
Marks: 2]
Query languages can be classified into two categories:
1. Procedural: Relational Algebra
2. Non-procedural: Tuple Relational Calculus, Domain Relational Calculus

1.11

What are the differences between schema and instance? [In-course 2007, Marks: 2]
SchemaInstance

1. The overall design of a database is called the 1. The collection of information stored in a
database schema.
database at a particular moment is called an
instance of the database.
2. A relation schema is a type definition.

2. A relation is an instance of a schema.

3. Schemas are changed infrequently, if at all.

3. Instances are changed frequently.

4. This corresponds to the variable declaration 4. The values of the variable in a program at a(with type definition) of a programming
point in time correspond to an instance of thelanguage.
database schema.
1.12

What is data dictionary? [2004. Marks: 1]
A data dictionary contains metadata (data about data). The data dictionary is considered to be a
special type of table, which can only be accessed and updated by the database system itself (not a
regular user). A database system consults the data dictionary before reading or modifying actual data.

1.13

What are the components of query processor? [In-course 1, 2005. Marks: 3]
The components of query processor are:
1. DDL interpreter: Interprets DDL statements and records the definitions in the data dictionary.
2. DML compiler: Translates DML statements in a query language into an evaluation planconsisting of low-level instructions that the query evaluation engine understands.
3. Query evaluation engine: Executes low-level instructions generated by the DML compiler.

3

CHAPTER 2ENTITY-RELATIONSHIP MODEL
Questions and Answers
2.1

Why E-R model is used for data manipulation? [2002. Marks: 2]
E-R model is used for data manipulation because:
1. It can express the overall logical structure of a database graphically.
2. E-R diagrams are simple and clear.

2.2

What is the basic difference between E-R diagram and Schema diagram? [In-course 1, 2005.
Marks: 1]
The basic difference between E-R diagram and schema diagram is that E-R diagrams do not show
foreign key attributes explicitly, whereas schema diagrams show them explicitly.

2.3

Define the following:
1. Composite attribute
2. Multivalued attribute
3. Derived attribute

2.4

2.5

2.6

[2007, 2003. Marks: 1][2007, 2003. Marks: 1][2007, 2003; In-course 1, 2005. Marks: 1]

Composite attributes:
Attributes that can be divided into subparts are called composite attributes.
For example, the composite attribute address can be divided into attributes street-number, streetname and apartment-number.
Multivalued attributes:
Attributes that have multiple values for a particular entity are called multivalued attributes.
For example, an employee may have multiple telephone numbers. So, the attribute telephone-no isa multivalued attribute.
Derived attribute:
If the value of an attribute can be derived from the values of other related attributes or entities, thenthat attribute is called a derived attribute.
For example, if an entity set employee has two attributes date-of-birth and age, then the attributeage is a derived attribute as it can be derived from the attribute date-of-birth.
Explain the difference between a weak entity set and a strong entity set. [In-course 2, 2007;
2005; 2003. Marks: 2]
A strong entity set has a primary key. All tuples in the set are distinguishable by that key. A weakentity set has no primary key unless attributes of the strong entity set on which it depends are included.
Tuples in a weak entity set are partitioned according to their relationship with tuples in a strong entity
set. Tuples within each partition are distinguishable by a discriminator, which is a set of attributes.
Show with an example the association between a weak entity set and a strong entity set using
E-R diagram. [In-course 2, 2007; 2003. Marks: 1]

We can convert any weak entity set to a strong entity set by simply adding appropriate
4attributes. Why, then, do we have weak entity sets?
We have weak entities for several reasons:• We want to avoid the data duplication and consequent possible inconsistencies caused by
duplicating the key of the strong entity.
• Weak entities reflect the logical structure of an entity being dependent on another entity.
• Weak entities can be deleted automatically when their strong entity is deleted.
• Weak entities can be stored physically with their strong entities.
2.7

What is the purpose of constraints in database? [2002. Marks: 2]
The purposes of constraints in database are:
1. To implement data check.
2. To centralize and simplify the database, so to make the development of database applications
easier and more reliable.

2.8

What are the constraints used in E-R model? [In-course 2, 2007. Marks: 1]
Constraints used in E-R model:
1. Cardinality Constraints
2. Participation Constraints
3. Key Constraints

2.9

What participation constraints are used in E-R model? [2006. Marks: 1]
OR, Explain the participation constraints in E-R model. [In-course 2, 2007. Marks: 1]
OR, Explain with example the participation constraints in E-R model. [2003. Marks: 3]
The participation constraints used in E-R model are:
1. Total
2. Partial

The participation of an entity set E in a relationship set R is said to be total if every entity in E
participates in at least one relationship in R. If only some entities in E participate in relationships in R,
the participation of entity set E in relationship R is said to be partial.
For example, we expect every loan entity to be related to at least one customer through theborrower relationship. Therefore the participation of loan in the relationship set borrower is total.
In contrast, an individual can be a bank customer whether or not she has a loan with the bank.
Hence, it is possible that only some of the customer entities are related to the loan entity set throughthe borrower relationship, and the participation of customer in the borrower relationship set istherefore partial.
2.10
Explain the distinction between total and partial constraints.
In a total design constraint, each higher-level entity must belong to a lower-level entity set. The
same need not be true in a partial design constraint. For instance, some employees may belong to nowork-team.
2.11

Let R be binary relationship between A and B entity sets.
1. Show the mapping cardinalities using E-R diagrams. [In-course 1, 2005. Marks: 2]
2. How primary keys can be defined for the relationship set R for different mapping
cardinalities? [2006. Marks: 2]
3. How can you combine the tables (if possible) for different mapping cardinalities? [2004.
Marks: 3]
1. Mapping Cardinalities:
A

R

B

One-to-One
2. Primary keys for R:

A

R

B

One-to-Many

A

R

Many-to-One
5

B

A

R
Many-to-Many

B

1.
2.
3.
4.

One-to-One:
One-to-Many:
Many-to-One:
Many-to-Many:

PKA or PKB [PKA means Primary Key of the entity set A]
PKB
PKA
PKA and PKB

3. Combination of tables for different mapping cardinalities:
1.
2.
3.
4.
2.12

2.13

One-to-One:
One-to-Many:
Many-to-One:
Many-to-Many:–
B and ABA and AB–

Draw the symbols of identifying relationship set, discriminator of weak entity set, derived
attribute and multivalued attribute used in E-R model. [2004. Marks: 2]
R

A

A

A

Identifying Relationship Set

Discriminator of
Weak Entity Set

Derived Attribute

Multivalued Attribute

Draw the E-R diagram for the following relation schemas: [In-course 2, 2007. Marks: 1.5]
Worker (worker_id, worker_name, hourly_rate, skill_type, supervisor_id)
Assignment (worker_id, building_id, start_date, num_days)
Building (building_id, address, building_type)worker_id
supervisor_id

worker_nameskill_type

hourly_rate

num_days
start_date

Worker

2.14

Assignmentbuilding_id
building_type

address

Building

Give the E-R diagram for the following database: [In-course 1, 2005; 2003. Marks: 2]
person (driver-id, name, address)car (license, model, year)
accident (report-no, date, location)owns (driver-id, license)
participated (driver-id, license, report-no, damage-amount)

6

2.15

What will be the tabular representation of the following E-R diagram? [In-course 2, 2007; In-course
1, 2005. Marks: 2]

customer (customer-id, first-name, middle-initial, last-name, date-of-birth, street-number, street-name,apartment-number, city, state, zip-code)
customer-phone (phone-number, customer-id)

2.16

What will be the schema representation of the following E-R diagram? [2007. Marks: 2]

loan (loan-number, amount)loan-payment (payment-number, loan-number, payment-date, payment-amount)

2.17

We are interested to make a database for Railway Reservation System. (We will limit only for intercity train between Dhaka and Sylhet.) Generally, a passenger takes flight of inter-city train that operates
between Dhaka-Chittagong-Dhaka and Dhaka-Sylhet-Dhaka. Each train is identified by an ID and total
seating capacity. Each train is assigned a leg instance (an instance of a flight on a specific date) for which
we will keep number of compartments, number of available seats and date. Passenger reserves seat ofeach leg instance. For seat, we will keep seat ID and type. Each leg instance departs from a terminal andarrives to a terminal. We will keep departure time and arrival time; and for terminal, we will store its ID,
name and city. For each passenger, we will store name, phone and age.
i. Develop a complete E-R diagram (including cardinalities). Make reasonable assumptions during
your development phases, if needed and state them clearly.
ii. Translate the E-R diagram into relations (tables). [2005. Marks: 5 + 3]
id

phone

total_seating_capacity

id

name
Train

Passenger

no_of_available_seatsN
Leg Instance

1

N

1

Assigned

1
Reserves

no_of_compartments

date

1

age

Has

id

Seat
type

N
Arrival_time

N

Arrives_At

Departs_From1

1
Terminal

name

city

id
7

departure_time

Train (id, total_seating_capacity)
Terminal (id, name, city)
Leg_Instance (date, train_id, no_of_compartments, no_of_available_seats)
Departure (terminal_id, date, train_id, departure_time)
Arrival (terminal_id, date, train_id, arrival_time)
Reservation (seat_id, date, train_id, seat_type, passenger_id, passenger_name, age, phone)
2.18

A database is being constructed to keep track of the teams and games of a football league. A
team has a number of players. For the team, we are interested to store team id, team name,address, date established, name of manager, and name of coach. For the player, we will store
player id in team, date of birth, date joined, position etc. Each team plays games against otherteam in a round robin fashion. For each game, we will store game id, date held, score andattendance (an attribute to designate whether the participating teams have attended the game).
Games are generally taking place at various stadiums of the country. For each stadium, we willkeep its size, name and location.
i.
ii.

Develop a complete E-R diagram (including cardinalities). Make reasonable assumptions
during your development phases, if needed and state them clearly.
Translate the E-R diagram into relations (tables). [2003. Marks: 6 + 4]

player_id

position

date_joined

date_establisheddate_of_birth

team_nameteam_idmanagercoach

N

PlayerPlays_For

address

1

Team
2
Participates_In

namelocation

1
Stadium

N1

Is_Held_In

Game
score
date_heldsizeattendanceTeam (team_id, team_name, date_established, address, manager, coach)
Player (player_id, team_id, date_of_birth, date_joined, position)
Stadium (name, location, size)
Game (id, date_held, attendance, score, stadium_name)
Team_Game (team_id, game_id)

8id

CHAPTER 3, 4
RELATIONAL MODEL & SQL
Points to be Remembered
3.1

The order in which tuples or attributes appear in a relation is irrelevant, since a relation is a set oftuples – sorted or unsorted does not mater.

3.2

To represent string values, in RA, double quotes (" ") are used, whereas in SQL, single-quotes (' ')are used.
Note the difference in representation of the following operators in SQL and RA:

3.3

SQL >= <= < > and or not≠¬
RA ≥ ≤∧∨
3.4

In the projection operation, duplicate rows are eliminated in RA (as RA considers relations assets); whereas SQL retains duplicate rows by default (since duplicate elimination is time consuming).
To force the elimination of duplicate, a keyword distinct is inserted after select.

3.5

SQL does not allow the use of distinct with count(*)(however, it can be used with count for asingle attribute, e.g. count(distinct A)). distinct can be used with min and max, but result does
not change.

3.6

If a where clause and having clause appear in the same query, SQL applies the predicate in thewhere clause first. Tuples satisfying the where predicate are then placed into groups by the group byclause. SQL then applies the having clause, if it is present, to each groups; it removes the groups that
do not satisfy the having clause predicate. The select clause uses the remaining groups to generate
the tuples of the result relation.

3.7

The input to sum and avg must be a collection of numbers, but the other aggregate functions(count, min and max) can operate on collection of non-numeric data types, such as string, as well.

3.8

Aggregate functions cannot be composed in SQL. Thus, we cannot use max(avg(…)).

3.9

Every derived table must have its own alias.
Wrong: select * from (select x from y where p = q) where a = b;
Right: select * from (select x from y where p = q) as new_table where a = b;

3.10

The use of a null value in arithmetic and comparison operations causes several complications. The
result of any arithmetic expression involving null returns null. So 5 + null returns null.
Any comparison with null (other than is null and is not null) returns unknown. So, 5 < nullor null <> null or null = null returns unknown.

3.11

All aggregate functions except count(*) ignore tuples with null values on the aggregatedattributes.

3.12

If we use an arithmetic expression in the select clause, the resultant attribute does not have a name.

The Trick of Writing RA Expressions for Complex Queries
If you find that writing the RA expression of a query is getting difficult, then think of the query in termsof views. Views create multiple tables and multiple simple queries to perform a single complex query. As
you’re able to write the RA expressions for simple queries, you’ll now be able to solve the complex queries.
For applications of this trick, see the following queries:
Complete Concepts Problem – query no. 20, 21, 23.
Theory 4.10 – query no. 4
Theory 4.11 – query no. 2 and 4
9Complete Concepts Problem
Consider the database schema below:
employee (ename, street, city)
works (ename, cname, salary, jdate)company (cname, city)manages (ename, mname)Note: A manager is also an employee of a company.
Give SQL and RA expressions for the following queries:
Imp. Diff.
Level Level
0
0
1.
0
1
2.
5

2

5

2

1

1

5

5

3
0

3
0

3

5

5

5

2

2

5
5
1

4
5
5

5
2
1
5
4
5
4
4
3

5
3
1
3
2
4
4
3
5

Queries

Find the names of all employees who work for First Bank Corporation.
Find the names and cities of residence of all employees who work for First Bank
Corporation.
3. Find the names, street address, and cities of residence of all employees who work forFirst Bank Corporation and earn more than Tk. 30000.
4. Find names, street addresses and cities of residence of all employees who work undermanager Sabbir and who joined before January 01, 2009.
5. Find the names of all employees in this database who live in the same city as thecompany for which they work.
6. Find the names of all employees who live in the same city and on the same street as dotheir managers.
7. Find the names of the employees living in the same city where Rahim is residing.
8. Find the names of all employees in this database who do not work for First Bank
Corporation.
9. Find the names of all employees who earn more than every employee of Small Bank
Corporation.
10. Find the names of all employees who earn more than any employee of Small Bank
Corporation.
11. Assume the companies may be located in several cities. Find all companies located inevery city in which Small Bank Corporation is located.
12. Give all employees of First Bank Corporation a 10 percent salary raise.
13. Give all managers in the database a 10% salary raise.
14. Give all managers in this database a 10 percent salary raise, unless the salary would begreater than Tk.100,000. In such cases, give only a 3 percent raise.
15. Increase the salary of employees by 10% for the companies those are located in Bogra.
16. Modify the database so that Rahim now lives in Bhola.
17. Delete all tuples in the works relation for employees of Small Bank Corporation.
18. Delete records from works that contain employees living in Rajshahi.
19. Display the average salary of each company except Square Pharma.
20. Find the company with the most employees.
21. Find the company that has the smallest payroll.
22. Find the company with payroll less than Tk. 100000.
23. Find those companies whose employees earn a higher salary, on average, than the
average salary of Small Bank Corporation.

Note: the Imp. Level column in the above table means how much important that query is for the exam
(range: 0 – 5, where 0 means not important at all and 5 means most important); and the Diff. Level fieldmeans how difficult the problem is (range: 0-5, where 0 means very easy and 5 means very difficult).

10Sample Data
Data

Table Nameemployee

ename
BarkatJabbarJubayer
Najmun Nahar
OronnoRafique
Rahim
SabbirSalam
Sharafatstreet
x
x
u
y
z
z
w
v
y
w

works

ename
Rahim
BarkatSalam
Rafique
SharafatJabbarNajmun Nahar
OronnoJubayer
Sabbircname
First Bank Corporation
First Bank Corporation
First Bank Corporation
Small Bank Corporation
First Bank Corporation
Small Bank Corporation
Small Bank Corporation
The ONE Limited
Square Pharma
Vegabond Company

company

cname
Anonymous IT
Dream TechFirst Bank Corporation
JONS IT (Pvt.) Limited
Small Bank Corporation
Square Pharma
The ONE Limited
Unique Softs
Unknown Systems
Vegabond Company

manages

ename
Rahim
BarkatSalam
Rafique
JabbarNajmun Nahar
Jubayercity
Bogra
ComillaFaridpurSylhetDhaka
Rajshahi
Dhaka
Chittagong
ComillaDhaka

city
Chittagong
Chittagong
Dhaka
Sylhet
Dhaka
Bogra
Dhaka
Dhaka
Rajshahi
Bogra

mname
Sharafat
Sharafat
SharafatOronnoOronnoSabbir
Sabbir11

salary
50000
40000
60000
30000
80000
10000
20000
50000
15000
100000jdate
2008-01-01
2007-01-01
2009-07-01
2009-06-08
2005-06-01
2009-06-05
2009-06-30
2007-06-01
2008-01-01
2001-01-01

employee (ename, street, city)
works (ename, cname, salary, jdate)company (cname, city)manages (ename, mname)
1. Find the names of all employees who work for First Bank Corporation.
SQL: select ename from works where cname = 'First Bank Corporation';
RA: П ename (σ cname = "First Bank Corporation" (works))
2. Find the names and cities of residence of all employees who work for First Bank Corporation.
SQL: select ename, city from employee natural join works
where cname = 'First Bank Corporation';

RA: П ename, city (σ cname = "First Bank Corporation" (employee ⋈ works))
3. Find the names, street address, and cities of residence of all employees who work for FirstBank Corporation and earn more than Tk. 30000.
SQL: select ename, street, city from employee natural join works
where cname = 'First Bank Corporation' and salary > 30000;

RA: П ename, street, city (σ cname = "First Bank Corporation" ∧ salary > 30000 (employee ⋈ works))
4. Find names, street addresses and cities of residence of all employees who work under manager
Sabbir and who joined before January 01, 2009.
SQL: select ename, street, cityfrom employee natural join works natural join manages
where mname = 'Sabbir' and jdate < '01-JAN-09';

RA: П ename, street, city (σ mname = "Sabbir" ∧ jdate < "01-jan-09" (employee ⋈ works ⋈ manages))
5. Find the names of all employees in this database who live in the same city as the company forwhich they work.
SQL: select ename from employee natural join works natural join company;
RA: П ename (employee ⋈ works ⋈ company)
6. Find the names of all employees who live in the same city and on the same street as do their
managers.
SQL: select employee.ename from employee natural join manages, employee as emp
where mname = emp.ename and employee.street = emp.street and employee.city = emp.city;

RA: П employee.ename(σ mname = emp.ename ∧ employee.street = emp.street ∧ employee.city = emp.city (employee ⋈ manages × ρ emp (employee)))
7. Find the names of the employees living in the same city where Rahim is residing.
SQL: select ename from employee where city = (select city from employee where ename = 'Rahim');

RA: t ← П city (σ ename = "Rahim" (employee))П ename (employee ⋈ t)
8. Find the names of all employees in this database who do not work for First Bank Corporation.
SQL: select ename from works where cname <> 'First Bank Corporation';
RA: П ename (σ cname ≠ "First Bank Corporation" (works))
12

employee (ename, street, city)
works (ename, cname, salary, jdate)company (cname, city)manages (ename, mname)
9. Find the names of all employees who earn more than every employee of Small Bank
Corporation.
SQL: select ename from works where salary > (select max(salary) from works where cname = 'Small Bank Corporation');

RA: t ← G max(salary) as max_salary (σ cname = "Small Bank Corporation" (works))
П ename (σ salary > max_salary (works × t))
OR, t1 ← П works.salary (σ works.salary < w.salary and w.cname = "Small Bank Corporation" (works × ρ w (works)))t2 ← П salary (σ w.cname = "Small Bank Corporation" (works)) – t1
П ename (σ works.salary > t2.salary (works × t2))
10. Find the names of all employees who earn more than any employee of Small Bank
Corporation.
SQL: select ename from works where salary > (select min(salary) from works where cname = 'Small Bank Corporation');

RA: t ← G min(salary) as min_salary (σ cname = "Small Bank Corporation" (works))
П ename (σ salary > min_salary (works × t))
OR, t1 ← П works.salary (σ works.salary > w.salary and w.cname = "Small Bank Corporation" (works × ρ w (works)))t2 ← П salary (σ w.cname = "Small Bank Corporation" (works)) – t1
П ename (σ works.salary > t2.salary (works × t2))
11. Assume the companies may be located in several cities. Find all companies located in every cityin which Small Bank Corporation is located.
SQL: select cname from company where city in (select city from company where cname = 'Small Bank Corporation');

RA: city ← П city (σ cname = "Small Bank Corporation" (company))
П cname (company ⋈ city)OR, П cname (company ÷ (П city (σ cname = "Small Bank Corporation" (company))))12. Give all employees of First Bank Corporation a 10 percent salary raise.
SQL: update works set salary = salary * 1.1 where cname = 'First Bank Corporation';
RA: t ← П ename, cname, salary * 1.1, jdate (σ cname = "First Bank Corporation" (works))
works ← t ∪ (works – σ cname = "First Bank Corporation" (works))
13. Give all managers in the database a 10% salary raise.
SQL: update works set salary = salary * 1.1 where ename in (select distinct mname from manages);

RA: t1 ← П works.ename, cname, salary, jdate (σ works.ename = mname (works × manages))t2 ← П works.ename, cname, salary * 1.1, jdate (t1)works ← (works – t1) ∪ t2
13

employee (ename, street, city)
works (ename, cname, salary, jdate)company (cname, city)manages (ename, mname)
14. Give all managers in this database a 10 percent salary raise, unless the salary would be greaterthan Tk.100,000. In such cases, give only a 3 percent raise.
SQL: update works set salary = casewhen salary * 1.1 > 100000 then salary * 1.03
else salary * 1.1
end
where ename in (select distinct mname from manages);

RA: t1 ← П works.ename, cname, salary, jdate (σ works.ename = mname (works × manages))t2 ← П works.ename, cname, salary * 1.03, jdate (σ t1.salary * 1.1 > 100000 (t1))
t2 ← t2 ∪ (П works.ename, cname, salary * 1.1, jdate (σ t1.salary * 1.1 ≤ 100000 (t1)))
works ← (works – t1) ∪ t2
15. Increase the salary of employees by 10% for the companies those are located in Bogra.
SQL: update works set salary = salary * 1.1 where cname in (select cname from company where city = 'Bogra');

RA: t1 ← П ename, cname, salary, jdate (σ city = "Bogra" (works ⋈ company))t2 ← П ename, cname, salary * 1.1, jdate (t1)works ← (works – t1) ∪ t2
16. Modify the database so that Rahim now lives in Bhola.
SQL: update employee set city = 'Bhola' where ename = 'Rahim';
RA: t ← П ename, street, "Bhola" (σ ename = "Rahim" (employee))works ← (works – (σ ename = "Rahim" (employee))) ∪ t
17. Delete all tuples in the works relation for employees of Small Bank Corporation.
SQL: delete from works where cname = 'Small Bank Corporation';
RA: works ← works – (σ cname = "Small Bank Corporation" (works)))
18. Delete records from works that contain employees living in Rajshahi.
SQL: delete from works where ename in (select ename from employee where city = 'Rajshahi');

RA: t ← П ename (σ city = "Rajshahi" (employee)))works ← works – П ename, cname, salary, jdate (works ⋈ t)
19. Display the average salary of each company except Square Pharma.
SQL: select cname, avg(salary) from works where cname <> 'Square Pharma' group by cname;RA:cnameG avg(salary) (σ cname ≠ "Square Pharma" (works))

14

employee (ename, street, city)
works (ename, cname, salary, jdate)company (cname, city)manages (ename, mname)
20. Find the company with the most employees.
SQL: select cname, count(distinct ename) from works group by cname
having count(distinct ename) >= all (select count(distinct ename) from works group by cname);

RA: t1 ← cnameG count(ename) as num_employees (works)t2 ← G max(num_employees) as num_employees (t1)
П cname (t1 ⋈ t2)
21. Find the company that has the smallest payroll1. [Similar to query 20]
SQL: select cname, sum(salary) from works group by cname
having sum(salary) <= all (select sum(salary) from works group by cname);

RA: t1 ← cnameG sum(salary) as payroll (works)t2 ← G min(payroll) as payroll (t1)
П cname (t1 ⋈ t2)
22. Find the company with payroll less than Tk. 100000.
SQL: select cname, sum(salary) from works group by cname having sum(salary) < 100000;
RA: t ← cnameG sum(salary) (works)
П cname (σ payroll < 100000 (ρ c_payroll (cname, payroll) (t)))
23. Find those companies whose employees earn a higher salary, on average, than the averagesalary of Small Bank Corporation.
SQL: select cname from works group by cname
having avg(salary) > (select avg(salary)from works
where cname = 'Small Bank Corporation');

RA: t1 ← cnameG avg(salary) (works)t2 ← σ cname = "Small Bank Corporation" (t1)
П t3.cname (σ t3.avg_salary > small-bank.avg_salary (ρ t3 (cname, avg_salary) (t1) × ρ small-bank (cname, avg_salary) (t2)))

1

Payroll: The total amount of money paid by a company as salary for all the employers.
15

General Structure of Query Statements
Legend:1XYYZ[ \X]ZY^ \X]Z)optional,,,Choose either the statement above
the line or the statement below
the line at a time
You can use it or leave it
Comma-separated list

General Structure of CREATE TABLE statement:
CREATE TABLE table-name (!"#$%"& '(&0)*+,- *.//!"#$%"& '(& 789:;<=>=?<@A
B+"(#C* '(& 789:;<=>=?<@A"(B("(*1(2 D?E:@>=?<@ 789:;<=>=?<@A
1+*2,"%#*, 3456
1F(1' 7@GHI@JJK9=A

_ __);column_data_types:
1. CHAR (number_of_characters)
2. VARCHAR (maximum_number_of_characters)
3. INTEGER (number_of_digits)4.

*.$("#1
R(1#$%/
B/+%,
R+.S/( !"(1#2#+*

example: CHAR(30)example: VARCHAR(255)
example: INTEGER(10)(total_number_of_digits_including_decimals,, number_of_decimal_digits)
example: DECIMAL(5, 2) [for 999.99]

5. DATE
6. TIME
7. DATETIME
Example of CREATE TABLE statement:
create table account (account_no char(5),branch_name varchar(15),balance number(10,2)not null,
constraint a_pk primary key(account_no),constraint a_fk foreign key (branch_name) references branch(branch_name),constraint a_chk1 check (balance>=0),
constraint a_chk2 check (account_no like 'A-%'));

General Structure of DROP TABLE statement:
DROP TABLE table-name;

16

, ,,

General Structure of INSERT statement:

INSERT INTO table-name [(column-names,,,)] VALUES [(values,,,)] ;Examples of INSERT statement:
insert into account values ('a-101', 'downtown', 500);
insert into account (account_no, branch_name, balance)
values ('a-101', 'downtown', 500);

General Structure of SELECT statement:
SELECT [DISTINCTDISTINCT]DISTINCT
FROM6_4_i5 )%2 53jl6 )%2 53- ,,,D?E:@>=?<@ )%2 ?:K?J- /(B, +.,(" m+#* D?E:@>=?<@ )%2 ?:K?J+* @GHI@JJK9=
D?E:@>=?<@ )%2 ?:K?J- "#CF, +.,(" m+#* D?E:@>=?<@ )%2 ?:K?J- .2#*C 789:;<=>=?<@,,,A
D?E:@>=?<@ )%2 ?:K?J- B.// +.,(" m+#* D?E:@>=?<@ )%2 ?:K?Jno >=?<@ )%2 ?:K?J- *%,."%/ m+#* D?E:@>=?<@ )%2 ?:K?JD?E:@>=?<@ )%2 ?:K?J- *%,."%/ /(B, +.,(" m+#* D?E:@>=?<@ )%2 ?:K?JD?E:@>=?<@ )%2 ?:K?J- *%,."%/ "#CF, +.,(" m+#* D?E:@>=?<@ )%2 ?:K?J-[WHERE
WHERE expression]
GROUP BY column-name_or_alias_or_function][GROUP[HAVINGHAVING expression][ORDERORDER BY column-name_or_alias_or_function
LIMIT[LIMIT, ,,l46i64r3

%21

R(21

0]

3465s, l46i64r36i463465s];

Example of SELECT statement:
select account_no, avg(balance) as average_balancefrom account left outer join depositor using (account_no)
where balance > 1000
group by branch_namehaving count(account_no) >= 10
order by average_balance desc
limit 0, 100;

General Structure of UPDATE statement:
UPDATEl6 )%2 53- ,,,D?E:@>=?<@ )%2 ?:K?J- /(B, +.,(" m+#* D?E:@>=?<@ )%2 ?:K?J+* @GHI@JJK9=
D?E:@>=?<@ )%2 ?:K?J- "#CF, +.,(" m+#* D?E:@>=?<@ )%2 ?:K?J- .2#*C 789:;<=>=?<@,,,A
D?E:@>=?<@ )%2 ?:K?J- B.// +.,(" m+#* D?E:@>=?<@ )%2 ?:K?Jno >=?<@ )%2 ?:K?J- *%,."%/ m+#* D?E:@>=?<@ )%2 ?:K?JD?E:@>=?<@ )%2 ?:K?J- *%,."%/ /(B, +.,(" m+#* D?E:@>=?<@ )%2 ?:K?JD?E:@>=?<@ )%2 ?:K?J- *%,."%/ "#CF, +.,(" m+#* D?E:@>=?<@ )%2 ?:K?J-SET column-name = column-value ,,,[WHERE
WHERE expression] ;Example of UPDATE statement:

update account set balance = balance * 1.1 where balance >= 100000;

17

Theories
3.1

List two reasons why null values might be introduced into the database.
Nulls may be introduced into the database because the actual value is either unknown or does notexist. For example, an employee whose address has changed and whose new address is not yet known
should be retained with a null address. If employee tuples have a composite attribute dependents, anda particular employee has no dependents, then that tuple’s dependents attribute should be given a nullvalue.

3.2

List two reasons why we may choose to define a view.
1. Security conditions may require that the entire logical database be not visible to all users.
2. We may wish to create a personalized collection of relations that is better matched to a certain
user’s intuition than is the actual logical model.

3.3

List two major problems with processing update operations expressed in terms of views.
Views present significant problems if updates are expressed with them. The difficulty is that amodification to the database expressed in terms of a view must be translated to a modification to theactual relations in the logical model of the database.
1. Since the view may not have all the attributes of the underlying tables, insertion of a tuple into
the view will insert tuples into the underlying tables, with those attributes not participating in
the view getting null values. This may not be desirable, especially if the attribute in question ispart of the primary key of the table.
2. If a view is a join of several underlying tables and an insertion results in tuples with nulls in
the join columns, the desired effect of the insertion will not be achieved. In other words, an
update to a view may not be expressible at all as updates to base relations.

3.4

What are the conditions of updating a view?
1. The from clause has only one database relation.
2. The select clause contains only attribute names of the relation and does not have any
expression, aggregates or distinct specifications.
3. Any attribute not listed in the select clause can be set to null.
4. The query does not have a group by or having clause.

3.5

What is materialized view? [In-course-1, 2008; 2005. Marks: 1]
A view which makes sure that if the actual relations used in the view definition change, the view is
kept up-to-date, is called materialized view.

3.6

Define the following:
1.
2.
3.
4.

DomainAtomic Domain
Non-Atomic Domain
Tuple Variable

1. Domain:
For each attribute, there is a set of permitted values, which are called domain (D) of that
attribute. For the attribute branch-name, the domain is the set of all branch names.
2. Atomic Domain:
A domain is atomic if elements of the domain are considered to be indivisible parts. Example:
set of integers: 23, 45, 5, 78 etc.
3. Non-Atomic Domain:
If elements of a domain can be divided into several parts, the domain is called non-atomic
domain. Example: set of all sets of integers: {23, 12, 4; 5, 65, 4; 34, 23, 98}, employee-id:
HR001, IT005
18

4. Tuple Variable:
A tuple variable is a variable whose domain is the set of all tuples. For example, t[accountnumber] = “A-101”, t[branch-name] = “Mirpur”. Alternatively, t[1], t[2] denote the value oftuple t on first and second attributes and so on.
3.7

What are the fundamental operations used in Relational Algebra? What are the conditions
for set (union, set-intersect and set-difference) operations in RA? [2005. Marks: 1 + 1]
OR, What are the conditions for set operations to be valid? [In-course 1, 2008. Marks: 1]
The fundamental operations used in Relational Algebra are:
1.
2.
3.
4.
5.
6.

Select (unary)
Project (unary)Rename (unary)
Cartesian Product (binary)Union (binary)
Set-Difference (binary)

The conditions for set operations are:
1. The relations must be of the same arity. That means they must have the same number of
attributes.
2. The domains of ith attribute of the first set and the ith attribute of the second set must be thesame, for all i.
3.8What are the conditions for insertion?
The conditions for insertion are:
1. The attribute values for inserted tuples must be members of the attribute’s domain.
2. Tuples inserted must be of the same arity.

3.9

Why set-intersection operation is not included in fundamental relational algebra operations?
[In-course 2007; 2007. Marks: 1]
Because set-intersection operation can be done using fundamental operations. If r1 and r2 are twosets, then their intersection can be expressed as:
r1 ∩ r2 = r1 – (r1 – r2) = r2 – (r2 – r1)

3.10

Give an example of generalized projection. [In-course 2007, Marks: 1]
OR, Give a relational algebra expression to represent generalized projection. [In-course
2008, Marks: 1]
SQL: select student_name, marks + 5 from result;
RA: П student_name, marks + 5 (result)

3.11

Let r(R) and s(S) be two relations. Give the relational algebra expression for natural join (⋈)
and the outer joins ( ̲̅⋈, ⋈ ̲̅ , ⋈̲̅ ̲̅ ) of the said relations. [2005. Marks: 2]r ⋈ s = ∏R ∪ S (σ r.A1 = s.A1 Λ r.A2 = s.A2 ……. r.An = s.An (r × s)), where R ∩ S = {A1, A2, ……, An}r ̲̅⋈ s = (r ⋈ s) ∪ (r – ∏R (r ⋈ s)) × {(null, null, …, null)}r ⋈ ̲̅ s = (r ⋈ s) ∪ (s – ∏S (r ⋈ s)) × {(null, null, …, null)}r ̲̅⋈ ̲̅ s = (r ⋈ s) ∪ (r – ∏R (r ⋈ s)) × {(null, null, …, null)} ∪ (s – ∏S (r ⋈ s)) × {(null, null, …, null)}

3.12

The outer-join operations extend the natural-join operation so that tuples from theparticipating relations are not lost in the result of the join. Describe how the theta join operation
can be extended so that tuples from the left, right, or both relations are not lost from the resultof a theta join.
19r ̲̅⋈θ s = (r ⋈θ s) ∪ (r – ∏R (r ⋈θ s)) × {(null, null, …, null)}r ⋈ θ̲̅ s = (r ⋈θ s) ∪ (s – ∏S (r ⋈θ s)) × {(null, null, …, null)}r ̲̅⋈ θ̲̅ s = (r ⋈θ s) ∪ (r – ∏R (r ⋈θ s)) × {(null, null, …, null)} ∪ (s – ∏S (r ⋈θ s)) × {(null, null, …, null)}
3.13

With example, show the difference between Cartesian product (×) and natural join (⋈).
[2005. Marks: 2]
Let, R1 = (A, B) and R2 = (B, C) be two relation schema.
Again, let r1(R1) = {{a, 1}, {b, 2}} and r2(R2) = {{1, x}, {2, y}}Then, r1 × r2 = {{a, 1, 1, x}, {a, 1, 2, y}, {b, 2, 1, x}, {b, 2, 2, y}}
And r1 ⋈ r2 = {{a, 1, x}, {b, 2, y}}That is, the Cartesian product operation results in all the combinations of all the tuples from bothtables, whereas the natural join operation results in only the tuple combinations from both tables
where the values of the common attributes (in this example, the attribute ‘B’) are the same.

3.14

3.15

For a given relation schema, works (employee_name, company_name, salary), give a
relational algebra expression using all aggregate functions where the grouping is done oncompany name. [2007, Marks: 1]
company_nameG sum(salary), avg(salary), max(salary), min(salary), count(employee_name) (works)
Give the equivalent relational algebra expression of the following SQL form:
select A1, A2, …, An from r1, r2, …, rn where P [2005. Marks: 1]∏ A1, A2, …, An (σ P (r1 × r2 … × rn))

3.16

Write short notes on natural join, theta join and aggregate functions.
Natural Join:
The natural join is a binary operation that allows us to combine certain selection and a Cartesian
product into one operation. It is denoted by the “join” symbol ⋈.
The natural join operation forms:i) A Cartesian product of two argumentsii) Performs a selection forcing equality on those attributes that appear in both relation schemasiii) Removes duplicate attributes
Theta Join:
The theta join operation is an extension to the natural join operation that allows us to combine a
selection and a Cartesian product into a single operation. Consider relations r(R) and s(S); let θ bepredicate on attributes in the schema R ∪ S. The theta join operation is defined as follows:R ⋈θ S = σθ (r × s)

Aggregate Functions:Aggregate functions take a collection of values and return a single value as a result. It is denotedby calligraphic G, G. For a collection of values {1, 1, 3, 4, 4, 11}:
1. sum returns the sum of the values: 24
2. avg returns the average of the values: 4
3. count returns the number of the elements in the collection: 6
4. min returns the minimum value of the collection: 1
5. max returns the maximum value of the collection: 11

20

3.17

With example, explain the importance of outer joins. [In-course 2007, Marks: 2]
When joining two or more tables, if we want to keep all the records from one table and want toknow which records from the other tables don’t match with them, then outer join can be used to solve
the problem easily.
For example, if we want to know which records in two tables (e.g., x and y) do not match, then we
can write the following query using outer join:select * from x natural full outer join ywhere x.some_attribute is null or y.some_attribute is null;

3.18

Let R = (A, B, C); and let r1 and r2 both be relations on schema R. Give an expression in
SQL that is equivalent to each of the following queries. [2003. Marks: 4]
1.
2.
3.
4.
1.
2.
3.
4.

3.19

r1 ∪ r2
r1 ∩ r2
r1 – r2
ΠAB (r1) ⋈ ΠBC (r2)select * from r1 union select * from r2;
select * from r1 intersect select * from r2 ;select * from r1 minus select * from r2;
select * from (select A, B from r1) as x natural join (select B, C from r2)
as y;2Give names of the aggregate functions that ignore null values in their input collection. [2004.
Marks: 1]
sum, avg, min, max

3.20

What aggregate functions can be used for string type data? [In-course 1, 2008]
count, min, max

3.21

With examples define the terms Superkey, Candidate Key and Primary Key. [2006, Marks: 3]
Superkey:
A superkey is a set of one or more attributes that, taken collectively, allow us to identify uniquely
a tuple in the relation. For example:
Branch_schema = (branch_name, branch_city, assets)
In Branch_schema above, {branch_name}, {branch_name, branch_city}, {all attributes} are allsuperkeys.
Formal definition:3 Let R be a relation schema. If it is said that a subset K of R is a superkey of R,
it restricts consideration to relations r(R) in which no two distinct tuples have the same values on all
attributes in K. That is, if t1 and t2 are in r and t1 ≠ t2, then t1[K] ≠ t2[K].
Candidate Key:The superkey, for which no proper subset is a superkey, is a candidate key.
For example, in Branch_schema above, {branch_name} is a candidate key.
Primary Key:The primary key is a candidate key that is chosen by the database designer as the principal meansof identifying tuples within a relation.
In the Branch_schema above, {branch_name} is a primary key.

2
3

Note: Every derived table must have its own alias. So, the aliases x and y must be put to execute the query successfully.
Note: At exam, don’t write formal definitions if the marks are little, for example in this question (only 3 marks for 3 definitions).
21

3.22

Define foreign key.
A relation schema R2 may include among its attributes the primary key of another relation schema
R1. This attribute is called a foreign key from R2, referencing R1. The relation r2 is also called thereferencing relation of the foreign key dependency and r1 is called the referenced relation of theforeign key.
The attribute branch_name in Account-schema is a foreign key from Account_schema referencing
Branch_schema.
Formal definition: Let r1(R1) and r2(R2) be relations with primary keys K1 and K2, respectively. It
is said that a subset α of R2 is a foreign key referencing K1 in r1 if it is required that, for every t2 in r2,
there must be a tuple t1 in r1 such that t1[k1] = t2[α].

3.23

Identify the relations among primary key, candidate key and super key. [2003. Marks: 3]
Primary Key ⊆ Candidate Key ⊆ Super Key

3.24

Let R = (P, Q, R, S). If PQ and QS can uniquely identify a tuple in the relation r(R)separately, how many superkeys, candidate keys and primary keys are there? [In-course 1,
2008. Marks: 2]
Super Keys: 6 – {P, Q}, {P, Q, R}, {P, Q, S}, {Q, S}, {Q, R, S}, {P, Q, R, S}
Candidate Keys: 2 – {P, Q}, {Q, S}
Primary Key: 1 – either {P, Q}, or {Q, S}

3.25

Why do we need the rename operation?
1. Two relations in the from clause may have attributes with the same name, so an attribute nameis duplicated in the result.
2. If we use an arithmetic expression in the select clause, the resultant attribute does not have aname.
3. If an attribute name can be derived from the base relation, we may want to change the attributename in the result to make it more meaningful.

3.26

Give the schema diagram for the following database: [2006, Marks: 2]
book (ISBN, title, year, price)author (author-id, name, address, url)
warehouse (code, address, phone)
written-by (author-id, ISBN)stocks (code, ISBN, number)
authorauthor-id
nameaddressphone

written-by
author-id
ISBNwarehouse
code
address
phone

3.27

book
ISBN
title
year
price
stockscode
ISBNnumberDraw the schema diagram for the following part of the bank database: [In-course 1, 2008;
In-course 2, 2007. Marks: 1.5]
employee (employee-id, employee-name, street, city)
branch (branch-name, branch-city, assets)job (title, level)
works-on (employee-id, branch-name, title, salary)22employee
employee-id
employee-namestreetcity

branchbranch-namebranch-city
assetsworks-on
employee-id
branch-nametitlesalary

job
title
level

3.28

Give the schema diagram for the following part of database: [2004. Marks: 2]
person (driver-id, name, address)car (license, model, year)
accident (report-no, date, location)owns (driver-id, license)
participated (driver-id, license, report-no, damage-amount)owns
driver-id
licensepersondriver-id
nameaddress

accident
report-no
date
location

4.1

carlicensemodel
year

participateddriver-id
licensereport-no
damage-amount

What are the join types and conditions that are permitted in SQL? [2005. Marks: 2]
Join types: inner join, left outer join, right outer join, full outer join.
Join conditions: natural, on <predicate>, using (A1, A2, …, An).

4.2

Show that, in SQL, <> all is identical to not in.
Let the set S denote the result of an SQL subquery. We compare (x <> all S) with (x not in S). If aparticular value x1 satisfies (x1 <> all S) then for all elements y of S x1 ≠ y. Thus, x1 is not a member of
S and must satisfy (x1 not in S). Similarly, suppose there is a particular value x2 which satisfies (x2 notin S). It cannot be equal to any element w belonging to S, and hence (x2 <> all S) will be satisfied.
Therefore, the two expressions are equivalent.

4.3

Why duplicates are retained in SQL? [2004. Marks: 1]
Duplicates are retained in SQL because:
1. Eliminating them is costly.
2. Retaining duplicates is important in computing sum or average.

4.4

What is the difference between ‘Embedded SQL’ and ‘Dynamic SQL’? [2004. Marks: 2]
Dynamic SQL component allows programs to construct and submit SQL queries at run time. In
contrast, embedded SQL statements must be completely present at compile time; they are compiled by
the embedded SQL preprocessor.

4.5

Describe the circumstances in which you would choose to use embedded SQL rather than
SQL alone or only a general-purpose programming language.
Writing queries in SQL is typically much easier than coding the same queries in a general-purpose
programming language. However not all kinds of queries can be written in SQL. Also non-declarative
actions such as printing a report, interacting with a user, or sending the results of a query to agraphical user interface cannot be done from within SQL. Under circumstances in which we want thebest of both worlds, we can choose embedded SQL or dynamic SQL, rather than using SQL alone orusing only a general-purpose programming language.
23

Embedded SQL has the advantage of programs being less complicated since it avoids the clutter of
the ODBC or JDBC function calls, but requires a specialized preprocessor.
4.6

Consider the database schema below:
employee (ename, street, city)
emp_company (ename, cname, salary, jdate)company (cname, city)manager (ename, mname, shift)
Note: A manager is also an employee of a company.
Give SQL and RA expressions for the following queries: [In-course-1, 2007. Marks: 2.5 each.]
1. Find names, street addresses and cities of residence of all employees who work undermanager Sabbir and who joined before January 01, 2006.
2. Find the names of the employees living in the same city where Rahim is residing.
3. Display the average salary of each company except Square Pharma.
4. Increase the salary of employees by 10% for the companies those are located in Bogra.
5. Delete records from emp_company that contain employees living in Rajshahi.
1.
2.
3.
4.
5.

4.7

Similar to Complete Concepts Problem – Query No. 4
Similar to Complete Concepts Problem – Query No. 7
Similar to Complete Concepts Problem – Query No. 19
Similar to Complete Concepts Problem – Query No. 15
Similar to Complete Concepts Problem – Query No. 18

Consider the database schema below:
employee (employee-id, employee-name, street, city)
branch (branch-name, branch-city, assets)job (title, level)
works (employee-id, branch-name, title, salary)
Give SQL and RA expressions for the following queries: [2007; Marks: 2.5 each.]
1. Find names, street addresses and cities of residence and job level of all employees who
work for Dhanmondi branch and earn more than Tk. 10000.
2. Find the no. of employees and their total salaries for the branches located at Khulna city.
3. Give all Executives of Mirpur branch a 10 percent salary raise.
4. Find the branches with payroll less than Tk. 100000.
1. Similar to Complete Concepts Problem – Query No. 3
2. SQL: select count(employee-id), sum(salary) from branch natural join works
where branch-city = 'Khulna' group by branch-name;RA:branch-name G count(employee-id), sum(salary) (σ branch-city = "Khulna" (branch⋈ works))

3. Similar to Complete Concepts Problem – Query No. 12
4. Similar to Complete Concepts Problem – Query No. 22
4.8

Consider the database schema below:
employee (person-name, street, city)
works (person-name, company-name, salary)company (company-name, city)manages (person-name, manager-name)Note: A manager is also an employee of a company.
Give SQL and RA expressions for the following queries: [2006; Marks: 2.5 each.]
1. Find names, street addresses and cities of residence of all employees who work for FirstBank Corporation and earn more than Tk. 30000.
242. Find all employees in the database who earn more than any employee of Medium Bank
Corporation.
3. Give all managers of First Bank Corporation in the database a 10% salary raise.
4. Find those companies whose employees earn a higher salary, on average, than the
average salary of Small Bank Corporation.
5. Find the company that has the smallest payroll.
1. Similar to Complete Concepts Problem – Query No. 3
2. Similar to Complete Concepts Problem – Query No. 10
3. Similar to Complete Concepts Problem – Query No. 12
4. Similar to Complete Concepts Problem – Query No. 23
5. Similar to Complete Concepts Problem – Query No. 21
4.9

Consider the database schema below:
client (client-no, name, address, city)
product (product-no, description, profit-percent, qty-in-hand, reorder-level, cost-price)salesman (salesman-no, name, address, city, sale-amt)
salesorder (order-no, order-date, client-no, del-add, salesman-no, del-date, order-status)order-detail (order-no, product-no, qty-ordered, qty-delivered)
Give SQL and RA expressions for the following queries: [2005; Marks: 2.5 each.]
1. Find the list of all clients who stay in cities Dhaka or Khulna.
2. Find the products with their description whose selling price is greater than 2000 and lessthan or equal to 5000. [Selling price can be found from cost-price and profit-percent]
3. Find the total ordered and delivered quantity for each product with a product range ofP0035 to P0056.
4. Find the clients with their names and order numbers whose orders are handled by thesalesman Mr. X.
5. Find the product no and description of non-moving products, i.e., products not beingsold.
1. SQL: select name from client where city = 'Dhaka' or city = 'Khulna';
OR: select name from client where city in ('Dhaka', 'Khulna');RA:Π name (σ city = "Dhaka" ⋁ city = "Khulna" (client))

2. SQL: select product-no, description from productwhere (profit-percent / 100 * cost-price + cost-price) > 2000
and (profit-percent / 100 * cost-price + cost-price) <= 5000;

RA:Π product-no, description(σ (profit-percent / 100 * cost-price + cost-price) > 2000 ⋀ (profit-percent / 100 * cost-price + cost-price) <= 5000 (product))

3. SQL: select sum(qty-ordered), sum(qty-delivered) from order-detailwhere product-no between 'P0035' and 'P0056' group by product-no;

RA:product-noG sum(qty-ordered), sum(qty-delivered) (σ product-no >= "P0035" ⋀ product-no <= "P0056" (order-detail))

4. SQL: select client.name, order-no from client, salesman, salesorderwhere client.client-no = salesorder.client-no
and salesman.salesman-no = salesorder.salesman-noand salesman.name = 'Mr. X';RA:Π client.name, order-no(σ client.client-no = salesorder.client-no ⋀ salesman.salesman-no = salesorder.salesman-no ⋀ salesman.name = "Mr. X"(client × salesman × salesorder))5. SQL: select product-no, description from productminus
select product-no, description from product natural join order-detail;

25

RA:
OR, SQL:RA:

4.10

Π product-no, description (product) - Π product-no, description (product ⋈ order-detail)select product-no, description from product left outer join order-detailusing (product-no) where order-no = null;Π product-no, description (σ order-no = null (product ⋈̲̅ order-detail))

Consider the part of a bank database schema below:
Worker (worker-id, worker-name, hourly-rate, skill-type, supervisor-id)
Assignment (worker-id, building-id, start-date, num-days)
Building (building-id, address, building-type)
Notes: 1. skill-types are: Electric, Plumbing, Roofing, Framing etc.
2. building-types are: Office, Hospital, Residence, Warehouse etc.
3. supervisors are also workers of self-supervision.
Give SQL and RA expressions for the following queries: [In-course 1, 2008; 2004; Marks: 2.5
each.]
1.
2.
3.
4.

What are the skill types of workers assigned to building ‘B02’ (building-id)?
List the name of the workers assigned to ‘warehouse’ (building-type) buildings.
Find the no. of workers for each building where more than 5 workers are working for it.
Give 5% hourly wage increment for the workers working for ‘hospital’ buildings.

1. SQL: select skill-type from worker natural join assignment
where building-id = 'B02';

RA:Π skill-type (σ building-id = "B02" (worker ⋈ assignment))

2. SQL: select worker-namefrom worker natural join assignment natural join building
where building-type = 'warehouse';RA:Π worker-name (σ building-type = "B02" (worker ⋈ assignment ⋈ building))

3. SQL: select building-id, count(worker-id) from assignment
group by building-id having count(worker-id) > 5;RA:

t1 ← building-id G count(worker-id) as no-of-workers (assignment)
σ no-of-workers > 5 (t1)

4. SQL: update worker set hourly-rate = hourly-rate * 1.05
where worker-id in (select worker-id from assignment natural join building
where building-type = 'hospital');RA:

t1 ← П worker-id (σ building-type = "hospital" (assignment ⋈ building))
t2 ← П worker-id, worker-name, hourly-rate, skill-type, supervisor-id (worker ⋈ t1)
t3 ← П worker-id, worker-name, hourly-rate * 1.05, skill-type, supervisor-id (worker ⋈ t1)
worker ← (worker – t2) ∪ t3

4.11

Consider the part of a bank database schema below:
Branch (branch-name, branch-city, assets)
Customer (customer-name, customer-street, customer-city)
Loan (loan-no, branch-name, amount)
Borrower (customer-name, loan-no)
Account (account-no, branch-name, balance)
Depositor (customer-name, account-no)26

Give SQL and RA expressions for the following queries: [2003; Marks: 3 each.]
1. Find all customers who have either an account or a loan (but not both) at the bank.
2. Find the average account balance of those branches where the total account balance forindividual branch is greater than 160,000.
3. Find the number of depositors for each branch.
4. Find the branch that has the highest average balance.
1. SQL: select customer-namefrom depositor natural full outer join borrower
where loan-number is null or account-number is nullRA:П customer-name (σ loan-number = null ⋁ account-number = null (depositor ̲̅⋈ ̲̅ borrower))

2. SQL: select branch-name, avg(balance) from accountgroup by branch-name having sum(balance) > 16000;

OR: create view sum_bal as(select branch-name from account group by branch-namehaving sum(balance) > 16000);
create view average_bal as(select branch-name, avg(balance) as avg_bal from accountgroup by branch-name);select * from sum_bal natural join average_bal;

RA:t1 ← branch-name G sum(balance) as sum_bal (account))t2 ← σ sum_bal > 160000 (t1)t3 ← branch-name G avg(balance) as avg_bal (t2)

3. SQL: select count(distinct customer-name)from depositor natural join account group by branch-name;RA:branch-name G count(customer-name) (depositor⋈ account)

4. SQL: select branch-name from account group by branch-namehaving avg(balance) >= all(select avg(balance) from account group by branch-name);

OR: create view avg_balance as
select branch-name, avg(balance) as avg_bal
from account group by branch-name;create view max_balance as
select max(avg_bal) as max_bal from avg_bal;
select branch-name from avg_balance, max_balance where avg_bal = max_bal;

RA:

4.12

t1 ← branch-name G avg(balance) as avg_bal (account)t2 ← G max(avg_bal) as max_bal (t1)
Π branch-name (σ avg_bal = max_bal (t1 × t2))

Consider the part of a company database schema below:
employee (person-name, street, city)
works (person-name, company-name, salary)company (company-name, city)manages (person-name, manager-name)
Give SQL and RA expressions for the following queries: [In-course 1, 2005; Marks: 2.5 each.]
1. Find the names of all employees who live in the same city as the company for which theywork.
2. Find all employees in the database who earn more than any employee of Beximco Textiles
Limited.
27

3. Give all managers in the database a 10% salary raise, unless the salary would be greaterthan 100,000.
4. Find those companies whose employees earn a higher salary, on average, than the
average salary of Beximco Textiles Limited.
5. Find the company with the smallest payroll.
6. Find the names, street addresses and cities of residence of all employees who work for
Padma Textile Limited and earn more than 11,000.
1. Similar to Complete Concepts Problem – Query No. 5
2. Similar to Complete Concepts Problem – Query No. 10
3. Almost Similar to Complete Concepts Problem – Query No. 14
SQL:

update works set salary = salary * 1.1
where salary * 1.1 <= 100000 and person-name in (select distinct manager-name from manages);

RA:

t1 ← П works.person-name, company-name, salary (σ works.person-name = manager-name (works × manages))t2 ← П works.person-name, company-name, salary (σsalary * 1.1 ≤ 100000 (t1))
t3 ← П works.person-name, company-name, salary * 1.1 (t2)works ← (works – t2) ∪ t3

4. Similar to Complete Concepts Problem – Query No. 23
5. Similar to Complete Concepts Problem – Query No. 21
6. Similar to Complete Concepts Problem – Query No. 3

28

CHAPTER 6
INTEGRITY & SECURITY
Questions and Answers
6.1

What integrity constraints are used in database? [2004. Marks: 1]
The integrity constraints used in database are:
1. Domain Constraints
2. Referential Integrity Constraints

6.2

Define foreign key and dangling tuples. How foreign key defines acceptability of dangling
tuples? [2003. Marks: 4]
Consider a pair of relations r(R) and s(S), and the natural join r ⋈ s. There may be a tuple tr in rthat does not join with any tuple in s. That is, there is no ts in s such that tr[R ∩ S] = ts[R ∩ S]. Suchtuples are called dangling tuples.
Foreign key defines acceptability of dangling tuples by permitting the use of null values.
Attributes of foreign keys are allowed to be null, provided that they have not otherwise been declared
to be non-null. If all the columns of a foreign key are non-null in a given tuple, the usual definition offoreign-key constraints is used for that tuple. If any of the foreign-key columns is null, the tuple isdefined automatically to satisfy the constraint.

6.3

SQL allows a foreign-key dependency to refer to the same relation, as in the following
example:create table manager(employee-namechar(20),
manager-namechar(20),
primary key employee-name,foreign key (manager-name) references manager on delete cascade)Here, employee-name is a key to the table manager, meaning that each employee has at mostone manager. The foreign-key clause requires that every manager also be an employee. Explain
exactly what happens when a tuple in the relation manager is deleted.
The tuples of all employees of the manager, at all levels, get deleted as well.
This happens in a series of steps. The initial deletion will trigger deletion of all the tuples
corresponding to direct employees of the manager. These deletions will in turn cause deletions ofsecond level employee tuples, and so on, till all direct and indirect employee tuples are deleted.
6.4

What is a trigger and what are its parts? [2002. Marks: 3]
OR, With example define trigger. [2006, Marks: 2. 2004, Marks: 1]
A trigger is a statement that the system executes automatically as a side effect of a modification to
the database.
The parts of a trigger are:
1. An event – which causes the trigger to be checked
2. A condition – that must be satisfied for trigger execution to proceed.
3. The actions – that are to be taken when the trigger executes

29

6.5

Consider the following schemas:
Account = (Account_number, Branch name, Balance)
Depositor = (Customer_id, Account_number)
Write an SQL trigger to carry out the following action: after update on account for eachowner of the account, if the account balance is negative, delete the owner from the account anddepositor relation. [2007. Marks: 2]
create trigger check-delete-trigger after delete on Accountreferencing old row as orowfor each row
delete from depositor
where Depositor.Customer_id not in(select Customer_id from Depositor
where Account_number < > orow.Account_number)
end

6.6

Define assertion with example. [2004. Marks: 1]
An assertion is a predicate expressing a condition that we wish the database always to satisfy.
For example, the following assertion ensures that the assets value for the Perryridge branch isequal to the sum of all the amounts lent by the Perryridge branch.
create assertion perry check(not exists (select * from branch where branch-name = ′Perryridge′ andassets < > (select sum (amount) from loan where branch-name = ′Perryridge′)))6.7

Write an assertion for the bank database to ensure that the assets value for the Perryridge
branch is equal to the sum of all the amounts lent by the Perryridge branch.
See the example of assertion in Question and Answer 6.6.

6.8

What is database security? What security measures are taken to protect the database?
[2003. Marks: 4]
Database security refers to protection from malicious access.
To protect the database, the following levels of security measures are taken:
 Database system. Some database-system users may be authorized to access only a limited
portion of the database. Other users may be allowed to issue queries, but may be forbidden tomodify the data. It is the responsibility of the database system to ensure that theseauthorization restrictions are not violated.
 Operating system. No matter how secure the database system is, weakness in operating
system security may serve as a means of unauthorized access to the database.
 Network. Since almost all database systems allow remote access through terminals or
networks, software-level security within the network software is as important as physicalsecurity, both on the Internet and in private networks.
 Physical. Sites with computer systems must be physically secured against armed or
surreptitious entry by intruders.
 Human. Users must be authorized carefully to reduce the chance of any user giving access toan intruder in exchange for a bribe or other favors.

6.9

What are the different forms of authorizations used in SQL? [2005. Marks: 2]
The different forms of authorizations used in SQL are:
1. Authorization on data (instance) – read, insert, delete and update authorizations
2. Authorization on database schema – index, resource, alteration and drop authorizations
30

6.10

What is the use of role in database? [2004. Marks: 2]
Authorizations can be granted to roles, in exactly the same fashion as they are granted toindividual users. Each database user is granted a set of roles (which may be empty) that he or she isauthorized to perform.

6.11

What is authorization graph? [2006. Marks: 1]
The graph representing the passing of authorization from one user to another is called anauthorization graph.

Figure: Authorization graph.

6.12

How can you relate views with authorization? [2004. Marks: 2]

6.13

What do you understand by authentication? [2004. Marks: 1]
Authentication refers to the task of verifying the identity of a person / software connecting to a
database.

6.14

What are the different forms of authentication used in the database? [2006. Marks: 1]
The different forms of authentication used in database are:
1. Password-based authentication
2. Challenge-response system

6.15

What are the properties of a good encryption technique? [2005. Marks: 1]
The properties of a good encryption technique are:1. It is relatively simple for authorized users to encrypt and decrypt data.
2. It depends not on the secrecy of the algorithm, but rather on a parameter of the algorithm
called the encryption key.
3. Its encryption key is extremely difficult for an intruder to determine.

6.16

How public key encryption maintains security? [2005. Marks: 2]
If user U1 wants to store encrypted data, U1 encrypts them using public key E1 and decryption
requires the private key D1.
If user U1 wants to share data with U2, U1 encrypts the data using E2, the public key of U2. Since
only user U2 knows how to decrypt the data (using D2), information is transferred securely.

6.17

Describe challenge-response system for authentication. [2004. Marks: 3]
In the challenge-response system, the database system sends a challenge string to the user. The
user encrypts the challenge string using a secret password as encryption key, and then returns theresult. The database system can verify the authenticity of the user by decrypting the string with thesame secret password, and checking the result with the original challenge string. This scheme ensuresthat no passwords travel across the network.
Public-key systems can be used for encryption in challenge–response systems. The databasesystem encrypts a challenge string using the user’s public key and sends it to the user. The user
decrypts the string using her private key, and returns the result to the database system. The databasesystem then checks the response. This scheme has the added benefit of not storing the secret password
in the database, where it could potentially be seen by system administrators.
31

6.18

What are the advantages of encrypting data stored in the database?
1.

Encrypted data allows authorized users to access data without worrying about other users or
the system administrator gaining any information.

2. Encryption of data may simplify or even strengthen other authorization mechanisms. For

example, distribution of the cryptographic key amongst only trusted users is both, a simple
way to control read access, and an added layer of security above that offered by views.
6.19

Perhaps the most important data items in any database system are the passwords that
control access to the database. Suggest a scheme for the secure storage of passwords. Be surethat your scheme allows the system to test passwords supplied by users who are attempting tolog into the system.
A scheme for storing passwords would be to encrypt each password, and then use a hash index on
the user-id. The user-id can be used to easily access the encrypted password. The password being used
in a login attempt is then encrypted and compared with the stored encryption of the correct password.
An advantage of this scheme is that passwords are not stored in clear text and the code fordecryption need not even exist.

32

CHAPTER 7
RELATIONAL DATABASE DESIGN
Concepts
7.1

How to decompose a relation into BCNF with dependency preservation
Let, R be the relation and FD be the set of functional dependencies.
1. Check the first functional dependency, α → β. If α is trivial or a superkey, then R is in BCNF.
So, go for the next functional dependency.
2. If α is not a superkey, then decompose R into R1 = (α, β) and R2 = (R – β).
3. Repeat the steps 1 and 2 for each decomposed relation until it is found that each functional
dependency holds for at least one of the relations.
4. If you find one or more functional dependencies that don’t hold for any of the relations, thenstart over again by reordering the elements of FD.

Questions and Answers
7.1

Define functional dependency.
Consider a relation schema R, α and β are set of attributes of R and let α ⊆ R and β ⊆ R. The
functional dependency α → β holds on schema R if, in any legal relation r(R), for all pairs of tuples t1
and t2 in r such that t1[α] = t2[α], it is also the case that t1[β] = t2[β].

7.2Define superkey using functional dependency. [2004. Marks: 2]
A set of attributes K is a superkey of a relation schema R if K → R. That is K is a superkey if,whenever t1[K] = t2[K], it is also the case that t1[R] = t2[R] (i.e., t1 = t2).

7.3

Define trivial functional dependency. [In-course 2, 2007; In-course 2, 2008. Marks: 1]
A functional dependency of the form α → β is trivial if β ⊆ α. For example, A → A, AB → A etc.

7.4

Why certain functional dependencies are called trivial functional dependencies?
Certain functional dependencies are called trivial functional dependencies because they aresatisfied by all relations.

7.5

List all nontrivial functional dependencies (with no common attributes) satisfied by thefollowing relation: [In-course 2, 2007; In-course 2, 2008; 2005; Marks: 2]
A
a1
a1
a2
a2
A→B

7.6

C→B

B
b1
b1
b1
b1

C
c1
c2
c1
c3

AC → B

List all nontrivial functional dependencies satisfied by the following relation: [2003. Marks: 3]
A
a1
a1
a2
a2
a3
A→C
D→B

AB → CAD → BAD → CBC → A

B
b1
b2
b2
b2
b3

C
c1
c1
c2
c2
c2

CD → A
CD → B
33

D
d1
d2
d2
d3
d4
ABD → CACD → B

BCD → A

7.7

Show functional dependencies to indicate the following: [2004, 2007. Marks: 2]i)ii)

A one-to-one relationship set exists between entity sets account and customer.
A many-to-one relationship set exists between entity sets account and customer.

Where:
customer (customer-name, customer-street, customer-city)
account (account-number, balance)
Let Pk(r) denote the primary key attribute of a relation r.
1. The functional dependencies Pk(account) → Pk (customer) and Pk(customer) →Pk(account) indicate a one-to-one relationship, because any two tuples with the same value
for account must have the same value for customer, and any two tuples agreeing on customer
must have the same value for account.
2. The functional dependency Pk(account) → Pk(customer) indicates a many-to-one
relationship since any account value which is repeated will have the same customer value, butmany account values may have the same customer value.
7.8

Why Armstrong’s axioms are called ‘sound’ and ‘complete’? [2005. Marks: 1]
Armstrong’s axioms are sound, because they do not generate any incorrect functional
dependencies.
They are complete, because, for a given set F of functional dependencies, they allow us to generate
all F+.

7.9

Use the definition of functional dependency to argue that each of Armstrong’s axioms(reflexivity, augmentation, and transitivity) is sound.
The definition of functional dependency is: α → β holds on R if in any legal relation r(R), for all
pairs of tuples t1 and t2 in r such that t1[α] = t2[α], it is also the case that t1[β] = t2[β].
Reflexivity rule: If α is a set of attributes, and β ⊆ α, then α → β.
Assume ∃ t1 and t2 such that t1[α] = t2[α]z t1[β] = t2[β]zα→β[since β ⊆ α][Definition of FD]

Augmentation rule: If α → β, and γ is a set of attributes, then γα → γβ.
Assume ∃ t1,t2 such that t1[γα] = t2[γα]t1[γ] = t2[γ]t1[α] = t2[α]t1[β] = t2[β]t1[γβ] = t2[γβ]z γα → γβ[γ ⊆ γα][α ⊆ γα][Definition of α → β][γβ = γ ∪ β][Definition of FD]

Transitivity rule: If α → β and β → γ, then α → γ.
Assume ∃ t1,t2 such that t1[α]= t2[α]t1[β] = t2[β]t1[γ] = t2[γ]
zα→γ
7.10

[Definition of α → β][Definition of β → γ]
[Definition of FD]

Use Armstrong’s axioms to prove the soundness of Union, Decomposition and
Pseudotransitivity rules. [In-course 2, 2007; In-course 2, 2008; 2007; Marks: 3. 2003, Marks: 5]
Union rule: If α → β and α → γ then α → βγ.
We derive:

34

α→β
αα → αβα → αβ …(i)[Given][Augmentation rule][Union of identical sets]

Again, α → γ
αβ → γβ …(ii)z α → βγ

[Given][Augmentation rule][From (i) and (ii) using transitivity rule and set union commutativity]

Decomposition rule: If α → βγ, then α → β and α → γ.
We derive:α → βγβγ → β[Given][Reflexivity rule]zα→β[Transitivity rule]

Again, βγ → γ[Reflexive rule]zα→γ[Transitive rule]

Pseudotransitivity rule: If α → β and γβ → δ, then αγ → δ.
We derive:

7.11

α→β
αγ → γβγβ → δ[Given][Augmentation rule and set union commutativity][Given]z αγ → δ[Transitivity rule]

Consider the following proposed rule for functional dependencies: If α → β and δ → β, thenα → δ. Prove that this rule is not sound by showing a relation r that satisfies α → β and δ → β,does not satisfy α → δ. [2004. Marks: 2]
Consider the following relation r :
A
a1
a1

B
b1
b1

C
c1
c2

Let, α = A, β = B, γ = C.
From the above relation, we see that A → B and C → B (i.e., α → β and γ → β). However, it is
not the case that A → C (i.e., α → γ) since the same A (α) value is in two tuples, but the C (γ) value inthose tuples disagree.
7.12

Define closure of attribute sets, α+.
Let α be a set of attributes. We call the set of all attributes functionally determined by α under a set
F of functional dependencies the closure of α under F, and we denote it by α+.

7.13

What are uses of ‘closure of attribute sets’, α+? [In-course 2, 2007; In-course 2, 2008; 2007;
2004; Marks: 2]
1. To test if α is a superkey, we compute α+ and check if α+ contains all attributes of R. If it
contains, α is a superkey of R.
2. For a given set of F, we can check if a functional dependency α → β holds (or is in F+), bychecking if β ⊆ α+. That is, we compute α+ by using attribute closure and then check if it
contains β.
3. 3. It gives us an alternate way to compute F+: For each γ ⊆ R, we find the closure γ+, and foreach S ⊆ γ+, we output a functional dependency γ → S.

7.14

Compute the closure of the attribute/s to list the candidate key/s for relation schema R = (A, B,
C, D, E) with functional dependencies F = {A → BC, CD → E, B → D, E → A} [2005. Marks: 2]
35

7.15

Define canonical cover, FC. [2004. Marks: 2]
A canonical cover FC for F is a set of dependencies such that F logically implies all dependencies
in FC and FC logically implies all dependencies in F. Furthermore, FC must have the following
properties:
1. No functional dependency in FC contains an extraneous attribute.
2. Each left side of a functional dependency in FC is unique. That is, there are no two
dependencies α1 → β1 and α2 → β2 in FC such that α1 = α2.

7.16

What is the advantage of using canonical cover, FC? [In-course 2, 2007, Marks: 1]

The advantage of using canonical cover is that the effort spent in checking for dependency
violations can be minimized.
7.17
What are the design goals for relational database design? [In-course 2, 2007; 2005 Marks: 1]
Explain why each is desirable.
The design goals for relational database design are:
1. Lossless-join decompositions
2. Dependency preserving decompositions
3. Minimization of repetition of information
They are desirable so we can maintain an accurate database, check correctness of updates quickly,
and use the smallest amount of space possible.
7.18

Explain what is meant by repetition of information, inability to represent information and lossof information. Explain why each of these properties may indicate a bad relational database
design. [2006. Marks: 4.5]
Repetition of information is a condition in a relational database where the values of one attributeare determined by the values of another attribute in the same relation, and both values are repeatedthroughout the relation. This is a bad relational database design because it increases the storage
required for the relation and it makes updating the relation more difficult.
Inability to represent information is a condition where a relationship exists among only a proper
subset of the attributes in a relation. This is bad relational database design because all the unrelated
attributes must be filled with null values otherwise a tuple without the unrelated information cannot be
inserted into the relation.
Loss of information is a condition of a relational database which results from the decomposition of
one relation into two relations and which cannot be combined to recreate the original relation. It is abad relational database design because certain queries cannot be answered using the reconstructed
relation that could have been answered using the original relation.

7.19

Explain the condition for lossless-join decomposition. [2004. Marks: 2]
Let R be a relation schema, F be a set of functional dependencies on R; and R1 and R2 form adecomposition of R.
The decomposition is a lossless-join decomposition of R if at least one of the following functional
dependencies are in F+ :1. R1 ∩ R2 → R1
2. R1 ∩ R2 → R2
In other words, if R1 ∩ R2 (the attribute involved in the natural join) forms a superkey of either
R1 or R2, the decomposition of R is lossless-join decomposition.

7.20

Suppose that we decompose the schema R = (A, B, C, D, E) into (A, B, C) and (A, D, E).
Show that this decomposition is lossless-join decomposition if the following set F of functional
dependencies holds: [2006. Marks: 2]
A → BC
CD → E
36

B→D
E→A
A decomposition {R1, R2} is a lossless-join decomposition if
R1 ∩ R2 → R1
or R1 ∩ R2 → R2.
Let R1 = (A, B, C), R2 = (A, D, E), and R1 ∩ R2 = A.
Since A is a candidate key (because, the closure of A is R),
z R1 ∩ R2 → R1.
7.21

Suppose that we decompose the schema R = (A, B, C, D, E) into (A, B, C) and (C, D, E).
Show that this decomposition is not lossless-join decomposition if the following set F offunctional dependencies holds: [2006. Marks: 2]
A → BC
CD → E
B→D
E→A
Let, r be a relation as follows:
A
a1
a2

B
b1
b2

C
c1
c1

D
d1
d2

E
e1
e2

With R1 = (A, B, C) and R2 = (C, D, E) :
ΠR1 (r) would be:
A
B
a1
b1
a2
b2
ΠR2 (r) would be:

C
c1
c1

C
D
E
c1 d1 e1
c1 d2 e2ΠR1 (r) ⋈ ΠR2 (r) would be:
A
a1
a1
a2
a2

B
b1
b1
b2
b2

C
c1
c1
c1
c1

D
d1
d2
d1
d2

E
e1
e2
e1
e2

Clearly, ΠR1 (r) ⋈ ΠR2 (r) ≠ r.
Therefore, this is a lossy join.
7.22

Deduce the condition for dependency preservation using restrictions for decomposing agiven schema R and a set of FDs F. Decompose the schema R = (A, B, C, D, E) with functional
dependencies F = { A → B, BC → D } into BCNF with dependency preservation. [2005. Marks: 2+ 1]
Let F be a set of functional dependencies on schema R. Let R1, R2, ..., Rn be a decomposition of R.
The restriction of F to Ri is the set of all functional dependencies in F+ that include only attributesof Ri.
The set of restrictions F1, F2, ..., Fn is the set of dependencies that can be checked efficiently.
Let F' = F1 ∪ F2 ∪ ... ∪ Fn.
37

F' is a set of functional dependencies on schema R, but in general, F' ≠ F
However, it may be that F' + = F+.
If this is so, then every functional dependency in F is implied by F', and if F' is satisfied, then F
must also be satisfied.
Therefore, the condition for dependency preservation using restrictions for decomposing a given
schema R and a set of FDs F is that F' + = F+.
Decomposition of the schema R:
We change the order of the FDs in F such that F = {BC → D, A → B}.
Now, the FD BC → D holds on R, but BC is not a superkey. So, we decompose R into
R1 = (B, C, D) and R2 = (A, B, C, E)
R1 is in BCNF. However, the FD A → B holds on R2, but A is not a superkey. So, we decompose
R2 intoR3 = (A, B) and R4 = (A, C, E)
Now, R3 and R4 both are in BCNF. [R4 is in BCNF as only trivial functional dependencies exist in
R4]
So, the final decomposed relations are: R1 = (B, C, D), R3 = (A, B) and R4 = (A, C, E).
7.23

Consider a relation schema R = (A, B, C, D) and with functional dependencies F = {A → BC,
B → D, D → B}. Show the BCNF decomposition of the above schema with dependency
preservation with causes. [In-course 2, 2008; Marks: 2]
The FD A → BC holds on R, and A is a superkey (∵ A → BC and B → D, ∴ A → BCD, or, A →ABCD).
Therefore, we go for the next FD, B → D. This holds on R, but B is not a superkey. So, we
decompose R into
R1 = (B, D) and R2 = (A, B, C)
Now, both R1 and R2 are in BCNF as R1 satisfies B → D and R2 satisfies A → BC.
So, the final decomposition of R is: R1 = (B, D) and R2 = (A, B, C).

7.24

What are the differences between BCNF and 3NF? [In-course 2, 2008; 2002, 2004, 2007.
Marks: 3]
For a functional dependency α → β, 3NF allows this dependency in a relation if each attribute A inβ – α is contained in any candidate key for R. However, BCNF does not allow this condition.
It is always possible to find a dependency-preserving lossless-join decomposition that is in 3NF.
However, it is not always possible to find such decomposition that is in BCNF.
Repetition of information occurs in 3NF, whereas no repetition of information occurs in BCNF.

7.25

In designing a relational database, why might we choose a non-BCNF design?
BCNF is not always dependency preserving. Therefore, we may want to choose another normal
form (specifically, 3NF) in order to make checking dependencies easier during updates. This would
avoid joins to check dependencies and increase system performance.

7.26

What is multivalued dependency? Give an example. [2002. Marks: 4]
Let R be a relation schema and let α ⊆ R and β ⊆ R. The multivalued dependency α→→β holds
on R if in any legal relation r(R), for all pairs of tuples t1 and t2 in r such that t1[α] = t2[α], there exist
tuples t3 and t4 in r such that:t1[α] = t2[α] = t3[α] = t4[α]t3[β] = t1[β]t4[β] = t2[β]38t3[R – β] = t2[R – β]t4[R – β] = t1[R – β]Example:
Name
Tom
Tom
Tom
Tom

Address
North Road
Oak Street
North Road
Oak Street

Car
ToyotaHonda
Honda
ToyotaIn the above relation, Name →→ Address and Name →→ Car.
7.27

Give an example of a relation schema R and a set of dependencies that R is in BCNF, but not
in 4NF. [2006. Marks: 2]
R = (loan-no, amount, customer-name, customer-address)
F = {loan-no → amount}
If we assume that each customer might have more than one addresses, then the functional
dependency customer-name → customer-address cannot be enforced. Thus, R is in BCNF. However,
it is not in 4NF as it contains multivalues for customer-address and therefore there occurs repetition of
information in the loan-no and amount fields.

7.28

An employee database is to hold information about employees, the department they are inand the skills which they hold. The attributes to be stored are:(emp-id, emp-name, emp-phone, dept-name, dept-phone, dept-mgrid, skill-id, skill-name, skilldate, skill-level)
An employee may have many skills such as word-processing, typing, librarian... The date on
which the skill was last tested and the level displayed at that test are recorded for the purposesof assigning work and determining salary. An employee is attached to one department and eachdepartment has a unique manager.
i.
ii.

Derive a functional dependency set for the above database, stating clearly any
assumptions that you make.
Derive a set of BCNF relations, indicating the primary key of each relation.

[2002. Marks: 4 + 4]

39

CHAPTER 11
STORAGE & FILE STRUCTURE
Theories
11.1

Define the term: RAID [In-course 2, 2008; 2003, 2005. Marks: 1]
RAID (Redundant Array of Independent Disks) is a technology which makes use of two or more
hard drives in order to improve performance, reliability or create larger data volumes.

11.2

What is mirroring / shadowing? [2004, 2005. Marks: 1]
The simplest but most expensive approach to introduce redundancy is to duplicate every disk.
This technique is called mirroring or shadowing.

11.3

What is striping?
In context of RAID, striping means splitting and writing data across multiple drives to increase
throughput.
There are two types of striping:
1. Bit-level Striping: consists of splitting the bits of each byte across multiple disks.
2. Block-level Striping: stripes blocks across multiple disks.

11.4

How RAID improves reliability via redundancy? [2006. Marks: 2]
1. The chance that at least one disk out of a set of N disks will fail is much higher than the
chance that a specific disk will fail.
2. Redundancy stores extra information that is not needed normally, but that can be used in theevent of failure of a disk to rebuild the lost information.

11.5

How RAID improves performance via parallelism? [In-course 2, 2008; Marks: 2]
1. With disk mirroring, the rate at which read requests can be handled is doubled, since readrequests can be sent to either disk.
2. The transfer rate of each read is the same as in a single disk system, but the number of reads
per unit time is doubled.

11.6

What are the advantages of having large number of disks in a system? [2004. Marks: 2]
1. Improving the rate at which data can be read or written, if the disks are operated in parallel.
Several independent reads or writes can also be performed in parallel.
2. Improving the reliability of data storage – because redundant information can be stored on
multiple disks. Thus failure of one disk does not lead to loss of information.

11.7

What are the factors for choosing a RAID level? [2003. Marks: 2]
1.
2.
3.
4.
5.
6.

11.8

Monetary cost of extra disk storage requirements.
Performance requirements in terms of number of I/O operations.
Performance when a disk has failed.
Performance during rebuild (while the data in a failed disk is being rebuilt on a new disk).
How many disks should be in an array?
How many bits should be protected by each parity bit?

Which RAID level is used for storage of log files in a database? Justify your answer. [2007.
Marks: 2]
RAID 1 is used for storage of log files in a database, because for storing log files, fault tolerance
is needed on a limited volume of data (the limit is the capacity of 1 disk).
40

11.9

What are possible ways of organizing the records in files? What does reorganization do?
[2004. Marks: 4 + 1]
OR, Classify file organization. Why reorganization is required in sequential fileorganization? [2003. Marks: 2 + 1]
OR, What are different types of organization of records in files? What do you understandby reorganization? [2006. Marks: 3 + 1]
File organization:
1. Heap file organization: Any record can be placed anywhere in the file where there is space
for the record. There is no ordering of records. Typically, there is a single file for eachrelation.
2. Sequential file organization: Records are stored in sequential order, based on the value of a“search key” of each record.
3. Hashing file organization: A hash function is computed on some attribute of each record.
The result of the function specifies in which block of the file the record should be placed.
4. Multitable Clustering file organization: Records of several different relations can be stored
in the same file. Related records of the different relations are stored on the same block so that
one I/O operation fetches related records from all the relations.
Reorganization:
The sequential file organization will work well if relatively few records need to be stored in
overflow blocks. Eventually, however, the correspondence between search-key order and physicalorder may be totally lost. In such cases sequential processing will become mush less efficient. At thispoint, the file should be reorganized so that it is once again physically in sequential order.

11.10

Describe slotted page structure for organizing records within a single block. [In-course 2,
2008; 2004. Marks: 4]

Figure 11.10: Slotted page structure.

In slotted page structure, there is a header at the beginning of each block, containing:
1. No of record entries in the header
2. The end of free space in the block
3. An array whose entries contain the location and size of each record.
The actual records are allocated contiguously in the block, starting from the end of the block. The
free space in the block is contiguous, between the final entry in the header file and the first record.
If a record is inserted, space is allocated for it at the end of the free space and an entry containing
its size and location is added to the header.
If a record is deleted, the space it occupies is freed and its entry is set to deleted. Further, therecords in the block before the deleted record are moved, so that the free space created by deletion
gets occupied and all free space is again between the final entry in the header array and the firstrecord. The end-of-free-space pointer in the header is appropriately updated as well.
Records can be grown and shrunk by similar techniques, as long as there is space in the block. The
cost of moving the record is not so high, since the size of a block is limited: A typical value is 4KB.
41

The slotted page structure requires that there be no pointers that point directly to records. Instead,pointers must point to the entry in the header that contains the actual location of the record. Thislevel of indirection allows records to be moved to prevent fragmentation of space inside a block,while supporting indirect pointers to the record.
11.11

How variable-length records are represented by fixed-length records? [2005. Marks: 2]
Not present at sir’s lecture. Therefore, it is assumed to be not important for exam. However, theanswer is in the book – topic no. 11.6.2.2, page no. 420 (according to 4th edition).

11.12

Consider a relational database with two relations:course (course-name, room, instructor)
enrollment (course-name, student-name, grade)
Define instances of these relations for two courses, each of which enrolls three students.
Give the file structure of these relations that uses clustering. [2004. Marks: 2]
Instances of the given relations:course-name room
Java
420
OS
320

instructor
SMH
MHK

c1
c2

course relation
course-name student-name grade
Java
X
A
e1
Java
Y
B
e2
Java
Z
C
e3
OS
X
A
e4
OS
Y
B
e5
OS
Z
C
e6
enrollment relation
Clustering file structure of these relations:
Block 0 contains: c1, e1, e2 and e3
Block 1 contains: c2, e4, e5 and e6
11.13

Give one advantage and disadvantage of the following strategies for storing a relational
database: [2007. Marks: 1 + 1]
a. Store each relation in one fileb. Store multiple relations / the entire database in one filea. Advantages of storing a relation as a file include using the file system provided by the OS,thus simplifying the DBMS, but incur the disadvantage of restricting the ability of the DBMS
to increase performance by using more sophisticated storage structures.
b. By using one file for multiple relations, these complex structures can be implemented throughthe DBMS, but this increases the size and complexity of the DBMS.

11.14

Define seek time. [In-course 2, 2008. Marks: 1]
The time it takes for a disk read/write head to move to a specific data track is called seek time.

42

CHAPTER 12
INDEXING AND HASHING
Concepts
12.1
Indices

OrderedIndicesPrimary /Clustering

Secondary /Non-Clustering
Clustering

Hash
Indices

B+ Tree

B-TreeStaticDynamic[Must be dense]

Dense

12.2

SparseClosedOpen

Extendable

Linear

Search Key
An attribute or set of attributes used to look up records in a file is called a search key.
Primary / Clustering Index
If the file containing
taining the records is sequentially ordered, a primary index is an index whose search
key also defines
fines the sequential order of the file.
Secondary / Non-Clustering
Clustering Index
Indices whose search key specifies
specifies an order different from the sequential order of the file arecalled secondary indices.
Index-Sequential Files
Files that are ordered sequentially on some search key and have a primary index on thatth search
key are called index-sequential
sequential files.
files
Dense Index
Dense index is the index where an index record appears for every search-key
searchvalue in the file.
Sparse Index
Sparce index is the index where an index record appears for only some of the search-key values in
the file.
Multilevel Index
An index with two or more levels is called a multilevel index.
B+ Tree
A B+ tree is a type of index which takes the form of a balanced tree in which every path from theroot of the tree to a leaf of the tree is of the same length.
In a B+ tree, each non-leaf
leaf node
no in the tree has between n / 2 and n children, where n is fixed fora particular tree.. Each leaf has between (n – 1) / 2 and n – 1 values. The ranges of values in each leaf
do not overlap.
B-TreeA B-tree index is similar to B+ tree index except that search-key
key values in a B-treeBappear only onece.
Hashing
Hashing is a process of generating an index or address basing on some data. A hash function isused to compute the location of a record from a given search-key value.

43Hash Index
A hash index is an index which organizes the search keys, with their associated pointers, into ahash file structure.
Hash Function
A hash function is any well-defined procedure or mathematical function which converts a large,
possibly variable-sized amount of data into a small datum, usually a single integer that may serve as
an index into an array.
The values returned by a hash function are called hash values, hash codes, hash sums, or simply hashes.

12.3

Advantages of Dense Index
It is generally faster to locate a record if we have a dense index rather than a sparse index.
Advantages of Sparse Index
However, sparse indices have advantages over dense indices in that they require less space andthey impose less maintenance overhead for insertions and deletions.
Advantages of Multilevel Index
Searching for records with a multilevel index requires significantly fewer I/O operations thandoes searching for records by binary search.
Advantages of Primary Index
A sequential scan in primary index order is efficient because records in the file are stored
physically in the same order as the index order.
Advantages of Secondary Index
Secondary indices improve the performance of queries that use keys other than the search key ofthe primary index.
Disadvantages of Secondary Index
Secondary indices impose a significant overhead on modification of the database.
Disadvantages of Index-Sequential File Organization
The main disadvantage of the index-sequential file organization is that performance degrades as
the file grows, both for index lookups and for sequential scans through the data. Although thisdegradation can be remedied by reorganization of the file, frequent reorganizations are undesirable.
Disadvantages of B+ Tree
1. Imposes performance overhead on insertion and deletion.
2. Adds space overhead – Since nodes may be as much as half empty (if they have the minimum
number of children), there is some wasted space.
Advantages of B-Tree
1. May use less tree nodes than a corresponding B+ tree.
2. Sometimes it is possible to find the desired value before reaching a leaf node.
Disdvantages of B-Tree
1. Only a small fraction of desired values are found before reaching a leaf node.
2. Fewer search-keys appear in non-leaf nodes; hence, fan-out is reduced. Thus, B-trees typicallyhave greater depth than a corresponding B+ tree.
3. Insertion and deletion are more complicated than in B+ trees.
4. Implementation is harder than B+ trees, since leaf and non-leaf nodes are of different sizes.
Advantages of Hashing
1. Allows to avoid accessing an index structure.
2. Provides a way of constructing indices.
44

Questions and Answers
12.1

Define the followings:
i.
ii.
iii.
iv.
v.
vi.

Primary Index
Secondary Index
Sparse Index
Dense Index
B+ treeHashing[2006, Marks: 0.5. 2004, Marks: 1.5]
[2006. Marks: 0.5]
[2006. Marks: 0.5]
[2006. Marks: 0.5]
[2004. Marks: 1.5]
[2005, Marks: 1. 2004, Marks: 1]

See Concept 12.2.
12.2

12.3

Since indices speed query processing, why might they not be kept on several search keys?
Reasons for not keeping several search indices include:
1. Every index requires additional CPU time and disk I/O overhead during inserts and deletions.
2. Indices on non-primary keys might have to be changed on updates, although an index on the
primary key might not (this is because updates typically do not modify the primary key
attributes).
3. Each extra index requires additional storage space.
4. For queries which involve conditions on several search keys, efficiency might not be bad even
if only some of the keys have indices on them. Therefore database performance is improved
less by adding indices when many indices already exist.
What are the differences between a primary index and a secondary index? [2005, Marks: 2.
2003, Marks: 3]
Primary Index

Secondary Index

1. If the file containing the records is 1. Indices whose search key specifies an ordersequentially ordered, a primary index is andifferent from the sequential order of theindex whose search key also defines thefile are called secondary indices.
sequential order of the file.
2. There can be only one primary index for a 2. There can be many secondary indices for a
relation.
relation.
3. A sequential scan in primary index order is 3. Performance of sequential
efficient.
secondary index order is poor.
4. Can be dense or sparse.

scan

in

4. Can be only dense.

12.4

Clustering indices may allow faster access to data than a non-clustering index affords.
When must we create a non-clustering index despite the advantages of a clustering index?
Explain your answer. [2007. Marks: 2]
If we need to lookup a record using a search-key other than the search-key on which the file is stored
sequentially, then we must create a non-clustering index to improve the performance of look-up.

12.5

When is it preferable to use a dense index rather than a sparse index? Explain your answer.
[2006, Marks: 2. 2004, Marks: 3]
It is preferable to use a dense index instead of a sparse index when the file is not sorted on theindexed field (such as when the index is a secondary index) or when the index file is small comparedto the size of memory.

12.6

Why is sparse index used in database? [2002. Marks: 4]Sparse index is used in database because:
1. It requires less space.
2. It imposes less maintenance overhead for insertions and deletions.
What is the purpose of multilevel indexing? [2005. Marks: 1]
The purpose of multilevel indexing is to reduce I/O operations on indices when records are searched.

12.7

45

12.8

Is it possible in general to have two primary indices on the same relation for different
search keys? Explain your answer. [2007. Marks: 2]
In general, it is not possible to have two primary indices on the same relation for different keys
because the tuples in a relation would have to be stored in different order to have same values stored
together. We could accomplish this by storing the relation twice and duplicating all values, but for acentralized system, this is not efficient.

12.9

Consider the following dense primary index file corresponding to the sequential file Accountsorted on the attribute branch_name.
Branch_name

Pointer

AdaborC.O.
Dhanmondi
MirpurMotijheel
Index fileAccount_no

Branch_name

Balance

A-9
A-1
A-5
A-8
A-3
A-2
A-4
A-10
A-6
A-7

AdaborAdaborC. O. Bazar
Dhanmodi
Dhanmodi
MirpurMirpurMirpurMotijheel
Motijheel

300
500
560
590
420
600
520
120
600
200

Account file

Now make necessary modification to the index file after deletion of the record for theaccount no ‘A-5’ and then ‘A-2’. [2007. Marks: 1 + 1]
Index file after deletion of the record for the account no ‘A-5’:
Branch_name
AdaborDhanmondi
MirpurMotijheel

Pointer
A-9
A-8
A-2
A-6

Index file after deletion of the record for the account no ‘A-2’:
Branch_name

Pointer
A-9
A-8
A-4
A-6

AdaborDhanmondi
MirpurMotijheel

12.10

Let R = (A, B, C) is a relation schema with A as candidate key. The relation r(R) is sorted on
attribute C. Draw two secondary indices on candidate key A and non-candidate key B filling
data for different attributes. [2007. Marks: 3]
A
A-1
A-2
A-3
A-4
A-5

PointerIndex file on A
B
a
b
c
d

Pointer

A

B

C

A-4

a

100

A-1

b

200

A-5

a

300

A-2

d

400

A-3

c

500

Relation fileIndex file on B46

12.11

Consider a sequential file sorted with non-primary key. Draw a secondary index on the filefor a search key which is the primary key of the file. [In-course 2, 2005. Marks: 2]
See Question and Answer 12.10 – Index file on A.

12.12

What are the disadvantages of index-sequential file? [In-course 2, 2005. Marks: 1]
The main disadvantage of the index-sequential file organization is that performance degrades as
the file grows, both for index lookups and for sequential scans through the data. Although thisdegradation can be remedied by reorganization of the file, frequent reorganizations are undesirable.

12.13

Construct a B+ tree for the following set of key values and for the (i) four (ii) six (iii) eight
pointers that will fit in one node: [2006, Marks: 3. 2003, Marks: 2. (each)]
2, 3, 5, 7, 11, 17, 19, 23, 29, 31

(i)

(ii)

(iii)
12.14

For each B+ tree of Question and Answer 12.13, show the form of the tree after each of thefollowing series of operations:
1.
2.
3.
4.
5.

Insert 9
Insert 10
Insert 8
Delete 23
Delete 19

Structure

Operation(i)

1. Insert 9Form of Tree After Operation

2. Insert 10

3. Insert 847

4. Delete 23

5. Delete 19(ii)

1. Insert 9

2. Insert 10

3. Insert 8

4. Delete 23

5. Delete 19(iii)

1. Insert 9

2. Insert 10

3. Insert 8

4. Delete 23

5. Delete 1948

12.15Construct a B+ tree for the following set of key values:
3, 4, 6, 9, 10, 11, 12, 13, 20, 22, 23, 31, 35, 36, 38, 41
Assume that the tree is initially empty and values are added in ascending order. The
number of pointers that will fit in one node is four. [2004. Marks: 4]
12

6

3

12.16

10

4

20

10

6

35

11

12

23

13

9

38

23

20

31

22

38

35

41

36

For a B+ tree structure, search key value size = 12 bytes, pointer size = 8 bytes, block size =388 bytes and there are 1000 search key values. How many nodes are required to access for an
index lookup for the worst case? [In-course 2, 2005. Marks: 2]
Given,
Search-key value size, Sk = 12 bytes
Pointer size, Sp = 8 bytes
Block size, Sb = 388 bytes
Number of search-key values in file, K = 1000
∴ Maximum number of pointers in a node, n= Sb / (Sk + Sp)= 388 / (12 + 8)
= 19∴ Number of nodes required to access for an index lookup= log n / 2 (K)= log 19 / 2 (1000)
=3

Answer: 3.
12.17

If an index structure occupies 1000 disk blocks, how many block reads will be required in
the best case and the worst case to find a desired index entry, if no index entry is in theoverflow block? [In-course 2, 2005. Marks: 2]
In the best case, 1 block read will be required to find a desired index entry.
In the worst case, log 2 (1000) or 10 block reads will be required.

12.18

Describe the format of nodes of B+ tree. [2002. Marks: 2]
Non-leaf nodes: Each non-leaf node in the tree has between n / 2 and n pointers, where n isfixed for a particular tree.
Root node: The root node can hold fewer than n / 2 pointers, but not less than two pointers
unless the tree consists of only one node.
Leaf nodes: Each leaf has between (n – 1) / 2 and n – 1 values. The ranges of values in eachleaf do not overlap.

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Figure: B+ tree node format. [n = 3]

12.19

+

What is the fan-out of B tree indexing? [2002. Marks: 2]
OR, What is fan-out of a node? [In-course 2, 2005. Marks: 1]
The number of pointers in a node of a B+ tree is called the fan-out of the node.
A non-leaf node has a fan-out between n / 2 and n. The root node can hold fewer than n / 2pointers, but not less than two pointers unless the tree consists of only one node.

12.20

Why are the leaf nodes of a B+ tree chained together? [2007, In-course 2, 2005. Marks: 1]
OR, Why are nodes of a B+ tree at the leaf level linked? [2002. Marks: 2]
The leaf nodes of a B+ tree are chained together to allow for efficient sequential processing of the file.

12.21

What are the differences between B+ tree structure and in-memory tree structure? [2007;
In-course 2, 2005; Marks: 2]
B+ Tree Structure
In-Memory Tree Structure (Binary Tree)
1. Each node is large – typically a disk block – 1. Each node is small and has at most 2
and a node can have a large number ofpointers.
pointers – 200 to 400.
2. Fat and short.

2. Thin and tall.

3. If there are K search-key values in the file, 3. In a balanced binary tree, the path can be oflength log 2 (K).
the path for a lookup is no longer thanlog n / 2 (K).
12.22

What are the differences between B-tree and B+ tree? [2002, Marks: 3. In-course 2, 2005,
Marks: 4]
B+ Tree
1. Some search key values may appear twice.

B-Tree1. Search key values appear only once.

2. Contains redundant storage of search-key 2. Eliminates redundant storage.
values.
3. Takes more space than a corresponding B- 3. Takes less space than a corresponding B+
tree.
tree.
4. No additional pointer for search key isneeded.

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4. As search-keys in non-leaf nodes appear
nowhere else in the B-tree, an additional
pointer field for each search key in a nonleaf node must be included.

12.23What are the causes of bucket overflow in a hash file organization? [2006, Marks: 1. 2005,
2004, 2003, Marks: 2]
The causes of bucket overflow are :1. Insufficient buckets. Our estimate of the number of records that the relation will have was
too low, and hence the number of buckets allotted was not sufficient.
2. Skew in the distribution of records to buckets. This may happen either because there aremany records with the same search key value, or because the hash function chosen did nothave the desirable properties of uniformity and randomness.

12.24

What can be done to reduce the occurrence of bucket overflows? [2006. Marks: 1]
To reduce the occurrence of overflows, we can:
1. Choose the hash function more carefully, and make better estimates of the relation size.
2. If the estimated size of the relation is nr and number of records per block is fr, allocate(nr / fr) × (1 + d) buckets instead of (nr / fr) buckets. Here d is a fudge factor, typically around0.2. Some space is wasted: about 20% of the space in the buckets will be empty. But thebenefit is that some of the skew is handled and the probability of overflow is reduced.

12.25

How bucket overflows are handled? [2003. Marks: 3]
Bucket overflows can be handled using two techniques:
1. Closed Hashing. If records must be inserted into a bucket and the bucket is already full, they
are inserted into overflow buckets which are chained together in a linked list.
2. Open Hashing. In this technique, the set of buckets is fixed, and there are no overflow
chains. Instead, if a bucket is full, the system inserts records in some other bucket in the initial
set of buckets. When a new entry has to be inserted, the buckets are examined, starting withthe hashed-to slot and proceeding in some probe sequence, until an unoccupied slot is found.
The probe sequence can be any of the following:a. Liner probing. The interval between probes is fixed (usually 1).
b. Quadratic probing. The interval between probes increases by some constant (usually
1) after each probe.
c. Double hashing. The interval between probes is computed by another hash function.

12.26

What are the causes of skew? [In-course 2, 2005. Marks: 2]
Skew can occur for two reasons:
1. Multiple records may have the same search key.
2. The chosen hash function may result in non-uniform distribution of search keys.

12.27

For a customer relation, ncustomer = 40000 and fcustomer = 50. If the fudge factor is 0.2, howmany buckets will be required to reduce bucket overflow? [2007, In-course 2, 2005 (similar)
Marks: 2]
Given,
ncustomer = 40000
fcustomer = 50
Fudge factor, d = 0.2∴ Number of buckets required = (ncustomer / fcustomer) × (1 + d)
= (40000 / 50) × (1 + 0.2)
= 960
Answer: 960.

51

12.28Why is hash structure not the best choice for a search key on which range queries arelikely? [2006. Marks: 1]
A range query cannot be answered efficiently using a hash index; we will have to read all thebuckets. This is because key values in the range do not occupy consecutive locations in the buckets;
they are distributed uniformly and randomly throughout all the buckets.

12.29

Give a comparison of static hashing and dynamic hashing. [2004. Marks: 3]
In static hashing, the set of bucket addresses is fixed. As databases grow or shrink over time, useof static hashing results in degradation of performance or wastage of space.
In dynamic hashing, the hash function can be modified dynamically to accommodate the growthor shrinkage of the database.

12.30

Compare closed and open hashing. [2007, In-course 2, 2005. Marks: 2]
Closed Hashing
Open Hashing
1. On bucket overflow, records are inserted 1. Records are inserted in some other bucket ininto overflow buckets which are chained
the initial set of buckets using a probe
together in a linked list.
sequence (linear probing, quadratic probingetc.).
2. Deletion under open hashing is troublesome.
2. Deletion under closed hashing is simple.
3. Preferable in database systems as insertion- 3. Preferable in compilers and assemblers as
deletion occurs there frequently.
they perform only lookup and insertion
operations in their symbol tables.

12.31

What are the limitations of hashing? [In-course 2, 2005. Marks: 2]
The limitations of hashing are:
1. Hash function must be chosen when the system is implemented and it cannot be changed
easily thereafter if the file being indexed grows or shrinks.
2. Since the hash function maps search-key values to a fixed set of bucket addresses, space iswasted if the set of buckets is made large to handle further growth of the file.
3. If the set of buckets is too small, they will contain records of many different search-key
values and bucket overflow can occur. As the file grows, performance suffers.

12.32

Discuss the use of the hash function in identifying a bucket to search. [2002. Marks: 3]
Since it cannot be known at design time precisely which search-key values will be stored in thefile, such hash function should be chosen that assigns search-key values to buckets in such a way thatthe distribution has these qualities:
1. Uniform: The distribution is uniform. The hash function assigns each bucket the same numberof search-key values from the set of all possible search-key values.
2. Random: The distribution is random. In the average case, each bucket will have nearly thesame number of values assigned to it regardless of the actual distribution of the search-key values.
Hash functions require careful design. A bad hash function may result in lookup taking time
proportional to the number of search keys in the file. A well-designed function gives an average-caselookup time that is a small constant, independent of the number of search-keys in the file.

52