UNIT-4

SCHEMA REFINEMENT and NORMAL FORMS

Syllabus: Problems caused by redundancy, Decompositions, problem related to
decomposition, reasoning about FDS, FIRST, SECOND, THIRD Normal forms, BCNKF,
Lossless join Decomposition, Dependency preserving Decomposition, Schema refinement in
Data base Design, Multi valued Dependencies, FORTH Normal Form.

Schema Refinement: The purpose of Schema refinement is used for a refinement

approach based on decompositions. Redundant storage of information (i.e. duplication of
data) is main cause of problem. This redundancy is eliminated by decompose the relation.

Q. Problems caused by redundancy: Redundancy is a method of storing the

same information repeatedly. It means, storing the same data more than one place with in
a database is can lead several problems. Such as
1) Redundant Storage: It removes the multi-valued attribute. It means, some tuples or
information is stored repeatedly.
2) Update Anomalies: Suppose, if we update one row (or record) then DBMS will update
more than one similar row, causes update anomaly.
For example, if we update the department name those who getting the salary 40000 then
that will update more than one row of employee table is causes update anomaly.
3) Insertion Anomalies: In insertion anomaly, when allows insertion for already existed
record again, causes insertion anomaly.
4) Deletion Anomalies: In deletion anomaly, when more than one record is deleted
instead of specified or one, causes deletion anomaly.
Consider the following example,
Eno ename salary rating hourly_wages hours_worked
111 suresh 25000 8 10 40
222 eswar 30000 8 10 30
333 sankar 32000 5 7 30
444 padma 31000 5 7 32
555 aswani 35000 8 10 40

In this example,
1) Redundancy storage: The rating value 8 corresponds to the hourly_wages 10
and there is repeated three times.
2) Update anomalies: If the hourly_wages in the first tuple is updated, it does
not make changes in the corresponding second and last tuples. Because, the
key element of tuples is emp_id.
i.e. update employee set hourly_wages = 12 where emp_id = 140; But the
question is update hourly_wages from 10 to 12.
3) Insertion anomalies: If we have to inset a new tuple for an employee, we
should know both the values rating value as well as hourly_wages value.
4) Deletion anomalies: Delete all the tupes through a given rating value, causes
deletion anomaly.

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Q. Decompositions: A relation schema (table) R can be decomposition into a
collection of smaller relation schemas (tables) (i.e. R1, R2, . . . . Rm) to eliminate
anomalies caused by the redundancy in the original relation R is called
Decomposition of relation. This shown in Relational Algebra as
R1 с R for 1 ≤ i ≤ m and R1 U R2 U R3 . . . Rm = R.
(or)
A decomposition of a relation schema R consists of replacing the relation schema
by two or more relation schemas that each contains a subset of the attributes of
R and together include all attributes in R.
(or)
In simple, words, “The process of breaking larger relations into smaller relations is
known as decomposition”.
Consider the above example that can be decomposing the following hourly_emps
relation into two relations.
Hourly_emp (eno, ename, salary, rating, houly_wages, hours_worked)
This is decomposed into
Hourly_empsd( eno, ename, salary, rating, hours_worked) and
Wages(rating, houly_wages)
Eno ename salary rating hours_worked
111 suresh 25000 8 40
222 eswar 30000 8 30
333 sankar 32000 5 30
444 padma 31000 5 32
555 aswani 35000 8 40
rating hourly_wages
8 10
5 7

Q. Problems Related to Decomposition:

The use of Decomposition is to split the relation into smaller parts to
eliminate the anomalies. The question is
1. What are problems that can be caused by using decomposition?
2. When do we have to decompose a relation?
The answer for the first question is, two properties of decomposition are
considered.
1) Lossless-join Property: This property helps to identify any instance of the
original relation from the corresponding instance of the smaller relation
attained after decomposition.
2) Dependency Preservation: This property helps to enforce constraint on
smaller relation in order to enforce the constraints on the original relation.

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The answer for the second question is, number of normal forms exists. Every
relation schema is in one of these normal forms and these normal forms can help to
decide whether to decompose the relation schema further or not.
The Disadvantage of decomposition is that it enforces us to join the decomposed
relations in order to solve the queries of the original relation.

Q. What is Relation?: A relation is a named two–dimensional table of data.

Each relation consists of a set of named columns and an arbitrary number of
unnamed rows. Emp-id ename dept name salary
For example, a relation named 100 Simpson Marketing 48000
Employee contains following 140 smith Accounting 52000
110 Lucero Info-systems 43000
attributes, emp-id, ename, 190 Davis Finance 55000
dept name and salary. 150 martin marketing 42000
marketing 42000
What are the Properties of relations?:
The properties of relations are defined on two dimensional tables. They are:
 Each relation (or table) in a database has a unique name.
 An entry at the intersection of each row and column is atomic or single. These can be
no multiplied atttributes in a relation.
 Each row is unique, no two rows in a relation are identical.
 Each attribute or column within a table has a unique name.
 The sequence of columns (left to right) is insignificant the column of a relation can be
interchanged without changing the meaning use of the relation.
 The sequence of rows (top to bottom) is insignificant. As with column, the rows of
relation may be interchanged or stored in any sequence.
Q. Removing multi valued attributes from tables: the “second property of
relations” in the above is applied to this table. In this property, there is no multi valied
attributes in a relation. This rule is applied to the table or relation to eliminate the one or
more multi valued attribute. Consider the following the example, the employee table
contain 6 records. In this the course title has multi valued values/attributes. The
employee 100 taken two courses vc++ and ms-office. The record 150 did not taken any
course. So it is null.
Emp-id name dept-name salary course_title
100 Krishna cse 20000 vc++, msoffice
140 Rajasekhar cse 18000 C++, DBMS,DS

Now, this multi valued attributes are eliminated and shown in the following
employee2 table.
Emp-id name dept-name salary course_title
100 Krishna cs 20000 vc++
100 Krishna cs 20000 MSoffice
140 Rajasekhar cs 18000 C++
140 Rajasekhar cs 18000 DBMS
140 Rajasekhar cs 18000 DS

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Functional dependencies : A functional dependency is a constraint between two
attributes (or) two sets of attributes.
For example, the table EMPLOYEE has 4 columns that are Functionally dependencies on
EMP_ID.

Emp_id Ename Dept_name Salry

Partial functional dependency : It is a functional dependency in which one or more non-

key attributes are functionally dependent on part of the primary key. Consider the
following graphical representation, in that some of the attributes are partially depend on
primary key.

Emp_id Course_title Ename Dept_name Salry date completed

In this example, Ename, Dept_name, and salary are fully functionally depend on Primary
key of Emp_id. But Course_title and date_completed are partial functional dependency.
In this case, the partial functional dependency creates redundancy in that relation.
Q. What is Normal Form? What are steps in Normal Form?
NORMALIZATION: Normalization is the process of decomposing relations to produce
smaller, well-structured relation.
To produce smaller and well structured relations, the user needs to follow six
normal forms.
Steps in Normalization:
A normal form is state of relation that result from applying simple rules from
regarding functional dependencies ( relationships between attributes to that relation. The
normal form are
1. First normal form 2. Second normal form
3. Third normal form 4. Boyce/codd normal form
5. Fourth normal form 6. Fifth normal form

1) First Normal Form: Any multi-valued attributes (also called repeating groups) have
been removed,
2) Second Normal Form: Any partial functional dependencies have been removed.
3) Third Normal Form: Any transitive dependencies have been removed.
4) Boyce/Codd Normal Form: Any remaining anomalies that result from functional
dependencies have been removed.
5) Fourth Normal Form: Any multi-valued dependencies have been removed.
6) Fifth Normal Form: Any remaining anomalies have been removed.

Advantages of Normalized Relations Over the Un-normalized

Relations: The advantages of normalized relations over un-normalized relations are,

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 1) Normalized relation (table) does not contain repeating groups whereas, un-
normalized relation (table) contains one or more repeating groups.
2) Normalized relation consists of a primary key. There is no primary key presents in
un-normalized relation.
3) Normalization removes the repeating group which occurs many times in a table.
4) With the help of normalization process, we can transform un-normalized table to
First Normal Form (1NF) by removing repeating groups from un-normalized tables.
5) Normalized relations (tables) gives the more simplified result whereas un-
normalized relation gives more complicated results.
6) Normalized relations improve storage efficiency, data integrity and scalability. But
un-normalized relations cannot improvise the storage efficiency and data integrity.
7) Normalization results in database consistency, flexible data accesses.

Q. FIRST NORMAL FORM (1NF): A relation is in first normal form (1NF) contains no
multi-Valued attributes. Consider the example employee, that contain multi valued
attributes that are removing and converting into single valued attributes
Multi valued attributes in course title
Emp-id name dept-name salary course_title
100 Krishna cse 20000 vc++,
msoffice
140 Raja it 18000 C++,
DBMS,
DS
Removing the multi valued attributes and converting single valied using First NF
Emp-id name dept-name salary course_title
100 Krishna cse 20000 vc++
100 Krishna cse 20000 msoffice
140 Raja it 18000 C++
140 Raja it 18000 DBMS
140 Raja it 18000 DS

SECOND NORMAL FORM(2NF): A relation in Second Normal Form (2NF) if it is in the

1NF and if all non-key attributes are fully functionally dependent on the primary key. In
a functional dependency X  Y, the attribute on left hand side ( i.e. x) is the primary key
of the relation and right side attributes on right hand side i.e. Y is the non-key attributes.
In some situation some non key attributes are partial functional dependency on primary
key. Consider the following example for partial functional specification and also that
convert into 2 NF to decompose that into two relations.

Emp_id Course_title Ename Dept_name Salry date completed

In this example, Ename, Dept_name, and salary are fully functionally depend on Primary
key of Emp_id. But Course_title and date_completed are partial functional dependency.
In this case, the partial functional dependency creates redundancy in that relation.

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 To avoid this, convert this into Second Normal Form. The 2NF will decompose
the relation into two relations, shown in graphical representation.

EMPLOYEE Emp_id Ename Dept_name Salry

COURSE
Course_title Date_Completed Emp_id

In the above graphical representation

 the EMPLOYEE relation satisfies rule of 1 NF in Second Normal form and
 the COURSE relation satisfies rule of 2 NF by decomposing into two relation.

THIRD NORMAL FORM(3NF): A relation that is in Second Normal form and has no
transitive dependencies present.
Transitive dependency : A transitive is a functional dependency between two non key
attributes. For example, consider the relation Sales with attributes cust_id, name, sales
person and region that shown in graphical representation.

Cust_id Cname Sales_person Region

Salry

CUST_ID NAME SALESPERSON REGION

1001 Anand Smith South
1002 Sunil kiran West
1003 Govind babu rao East
1004 Manohar Smith South
1005 Madhu Somu North
In this example, to insert, delete and update any row that facing Anomaly.

a) Insertion Anomaly: A new salesperson is assigned to North Region without assign a

customer to that salesperson. This causes insertion Anomaly.
b) Deletion Anomaly: If a customer number say 1003 is deleted from the table, we lose
the information of salesperson who is assigned to that customer. This causes, Deletion
Anomaly.
c) Modification Anomaly: If salesperson Smith is reassigned to the East region, several
rows must be changed to reflect that fact. This causes, update anomaly.
To avoid this Anomaly problem, the transitive dependency can be removed by
decomposition of SALES into two relations in 3NF.

Consider the following example, that removes Anomaly by decomposing into two relations.

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CUST_ID NAME SALESPERSON SALESPERSON REGION
1001 Anand Smith Smith South
1002 Sunil kiran kiran West
1003 Govind babu rao babu rao East
1004 Manohar Smith Smith South
1005 Madhu Somu Somu North

SALES PERSON CUSTOMER

Sales_person Region Cust_id Cname Salry

Q. BOYCE/CODD NORMAL FORM(BCNF): A relation is in BCNF if it is in

3NF and every determinant is a candidate key.
+
FD in F of the form X  A where X с S and A є S, X is a super key of R.
Boyce-Codd normal form removes the remaining anomalies in 3NF that are resulting
from functional dependency, we can get the result of relation in BCNF.
For example, STUDENT-ADVIDSOR IN 3NF
STUDENT-ADVISOR

Student-id major-subject faculty-advisor

STUDENT-ADVISOR relstion with simple data.

STYDENT-ID MAJOR-SUBJECT FACULTY-ADVISOR

1 MATHS B
2 MATHS B
3 MATHS B
4 STATISTICS A
5 STATISTICS A

In the above relation the primary key in student-id and major-subject. Here the part of
the primary key major-subject is dependent upon a non key attribute faculty–advisor. So,
here the determenant the faculty-advisor. But it is not candidate key.
Here in this example there are no partial dependencies and transitive
dependencies. There is only functional dependency between part of the primary key and
non key attribute. Because of this dependency there is anomely in this relation.
Suppose that in maths subject the advisor’ B’ is replaced by X. this change must be
made in two or more rows in this relation. This is an updation anomely.

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 To convert a relation to BCNF the first step in the original relation is modified that the
determinant(non key attributes) becoms a component of the primary key of new relation.
The attribute that is dependent on determinant becomes a non key attributes .
STUDENT-ADVISOR

STUDENT-ID FACULTY-ADVIDSOR MAJOR-SUBJEST

The second step in the conversion process is decompose the relation to eleminate the
partial functional dependency. This results in two relations. these relations are in 3NF
and BCNF . since there is only one candidate key. That is determinant.
Two relations are in BCNF.
ADVISOR STUDENT

Faculty- advisor major-subject Student-id faculty-advisor

In these two relations the student relation has a composite key , which contains
attributes student-id and faculty-advisor. Here faculty–advisor a foreign key which is
referenced to the primary key of the advisor relation.
Two relations are in BCNF with simple data. Student_id Faculty_Advisor
Faculty Advisor Major_subject 1 B
2 B
B MATHS 3 A
A PHYSICS 4 A
5 A
Q.Fourth Normal Form (4 NF) : A relation is in BCNF that contain no multi-
valued dependency. In this case, 1 NF will repeated in this step. For example, R be a
relation schema, X and Y be attributes of R, and F be a set of dependencies that includes
both FDs and MVDs. (i.e. Functional Dependency and Multi-valued Dependencies). Then R is
said to be in Fourth Normal Form (4NF) if for every MVD X  Y that holds over R, one
of the following statements is true.
1) Y с X or XY = R, or 2) X is a super key.
Example: Consider a relation schema ABCD and suppose that are FD A  BCD and the
MVD B   C are given as shown in Table B C A D tuples
It shows three tuples from relation ABCD that satisfies
the given MVD B   C. From the definition of a MVD, b c1 a1 d1 - tuple t1
given tuples t1 and t2, it follows that tuples t3 must also
be included in the above relation. Now, consider b c2 a2 d2 - tuple t2
tuples t2 and t3. From the given FD A  BCD and b c1 a2 d2 - tuple t3
the fact that these tuples have the same A-value,
we can compute

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the c1 = c2. Therefore, we see that the FD B  C must hold over ABCD whenever the
FD A  BCD and the MVD B   C holds. If B  C holds, the relation is not in BCNF but
the relation is in 4 NF.
The fourth normal from is useful because it overcomes the problems of the various
approaches in which it represents the multi-valued attributes in a single relation.

Q. Fifth Normal Form (5 NF) : Any remaining anomalies from 4 NF relation

have been removed.
A relation schema R is said to be in Fifth Normal Form (5NF) if, for every join dependency
* (R1, . . . . Rn) that holds over R, one of the following statements is true.
*Ri = R for some I, or
* The JD is implied by the set of those FDs over R in which the left side is a key for R.
It deals with a property loss less joins.

Q. LOSSELESS-JOIN DECOMPOSITION:
Let R be a relation schema and let F be a set FDs (Functional Dependencies) over R. A
decomposition of R into two schemas with attribute X and Y is said to be lossless-join
decomposition with respect to F, if for every instance r of R that satisfies the
dependencies in Fr.
πx (r) πy (r) = r
In simple words, we can recover the original relation from the decomposed
relations.
In general, if we take projection of a relation and recombine them using natural join, we
obtain some additional tuples that were not in the original relation.
S P D S P P D

s1 p1 d1 s1 p1
P1 d1
s2 p2 d2 s2 p2 p2 d2

s2 p1 d3 s2 p1 p1 d3

S P D
s1 p1 d1

s1 p1 d3

s2 p2 d2

s2 p1 d1

s2 p1 d3

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The decomposition of relation schema r i.e. SPD into SP i.e. PROJECTING πsp (r
) and PD i.e., projecting πPD (r) is therefore lossless decomposition as it gains back all
original tuples of relation ‘r’ as well as with some additional tuples that were not in original
relation ‘r’.
Q. Dependency-Preserving Decomposition:
Dependency-preserving decomposition property allows us to check integrity constraints
efficiently. In simple words, a dependency-preserving decomposition allows us to enforce
all FDs by examining a single relation on each insertion or modification of a tuple.
Let R be a relation schema that is decomposed in two schemas with attributes X
and Y and let F be a set of FDs over R. The projection of F on X is the set of FD s in the
closure F+ that involves only attributes in X. We denote the projection of F on attributes
+
X as Fx. Note that a dependency U  V in F is in Fx only if all the attributes in U and Y
are in X. The decomposition of relation schema R with FDs F into schemas with attributes
X and Y is dependency-preserving if ( F x U F y)+ = F+.
That is, if we take the dependencies in Fx and Fy and compute the closure of their
union, then we get back all dependencies in the closure of F. To enforce F x. we need to
examine only relation X (that inserts on that relation) to enforce F y, we need to examine
only relation Y.
Example: Suppose that a relation R with attributes ABC is decomposed into relations with
attributes AB and BC. The set F of FDs over r includes A  B , B  C and C  A. Here, A
 B is in FAB and B  C is in FBC and both are dependency-preserving. Where as C  A is
not implied by the dependencies of FAB and FBC. Therefore C  A is not dependency-
preserving.
Consequently, FAB also contains B  A as well as A  B and FBC contains C 
B as well as B  C. Therefore FAB U FBC contain A  B , B  C, B  A and C  B.
Now, the closure of the dependencies in FAB and FBC includes C  A (because, from
C  B, B  A and transitivity rule, we compute as C  A).

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