Boyce-codd normal form (bcnf), Database Management System

Assignment Help:

Boyce-Codd Normal Form (BCNF)

The relation NEWSTUDENT (Enrolmentno, Sno, Sname, Cname, Cno,) has every attributes participating in candidate keys as all the attributes are assumed to be unique. We thus had the following candidate keys.

(Enrolmentno, Cno)

(Enrolmentno, Cname)

(Sname, Cno)

(Sname, Cname)

As the relation has no non-key attributes, the relation is in 2NF and as well in 3NF. Though, the relation suffers from the anomalies (please check it yourself by making the relational instance of the NEWSTUDENT relation).

The complexity in this relation is being caused by dependence within the candidate keys.

Definition: A relation is in BCNF, if it is in 3NF and if each determinant is a candidate key.

  • A determinant is the left side of an FD
  • Most relations that are in 3NF are also in BCNF. A 3NF relation is not in

BCNF if all the following conditions apply.

(a)     The candidate keys in the relation are composite keys.

(b)     There is more than one overlapping candidate keys in the relation and a number of attributes in the keys are overlapping and some are not overlapping.

(c)      There is a FD from the non-overlapping attribute(s) of single candidate key to non-overlapping attribute(s) of other candidate key.

Let us recall the NEWSTUDENT relation:

NEWSTUDENT (Enrolmentno, Sname, Sno,  Cno, Cname) Set of FDs:

Enrolmentno     →           Sname              (1)

Sname  →                       Enrolmentno     (2)

Cno       →                      Cname              (3)

Cname  →                      Cno                   (4)

The relation even though in 3NF, but is not in BCNF and can be decomposed on any one of the FDs in (1) & (2); and any one of the FDs in (3) & (4) as:

STUD1 (Enrolmentno, Sname) COUR1 (Cno, Cname)

The third relation that will join the two relation will be:

ST_CO(Enrolmentno, Cno)

Since this is a slightly complex form, let us give one more example, for BCNF. Consider for example, the relation:

ENROL(Enrolmentno, Sname, Cno, Cname, Dateenrolled)

Let us suppose that the relation has the following candidate keys:

(Enrolmentno, Cno)

(Enrolmentno, Cname)

(Sname, Cno)

(Sname, Cname)

(We have supposed Cname and Sname are unique identifiers).

The relation has the following set of dependencies:

Enrolmentno     →    Sname

Sname             →     Enrolmentno

Cno                  →     Cname

Cname             →     Cno

Enrolmentno, Cno     →  Dateenrolled

The relation is in 3NF but not in BCNF as there are dependencies. The relation suffers from all anomalies. Please draw the relational instance and checks these troubles. The BCNF decomposition of the relation would be:

STUD1 (Enrolment no, Sname)

COU1 (Cno, Cname)

ENROL1 (Enrolmentno, Cno, Dateenrolled)

We now have a relation that only has information about students, another only about subjects and the third only about relationship enrolls.


Related Discussions:- Boyce-codd normal form (bcnf)

Higher software development cost, Drawbacks of Data Distribution: The prim...

Drawbacks of Data Distribution: The primary drawbacks of distributed database systems are the added complexity needed to ensure proper coordination between the sites. This increas

The advantages of a database management system, The advantages of a databas...

The advantages of a database management system (DBMS) include :- Data integrity and elimination of duplication.

What is derived and stored attribute, What is Derived and stored Attribute?...

What is Derived and stored Attribute? Derived and Stored Attribute - In a few cases, two or more attribute values are associated, for example, Age and BirthDate attributes of a

List some instances of collection types, List some instances of collection ...

List some instances of collection types? a) sets b) arrays c) multisets

Relational schema for person, Example: RELATIONAL SCHEMA for PERSON: ...

Example: RELATIONAL SCHEMA for PERSON: PERSON (PERSON_ID: integer, NAME: string, AGE: integer, ADDRESS: String) RELATION INSTANCE In this instance, m = 3 and n = 4

Illustrate the term- abstracting out common behaviour, Illustrate the term-...

Illustrate the term- Abstracting Out Common Behaviour Inheritance is not every time recognised during analysis phase of development, thus it is necessary to re-evaluate object

What create when a primary key combined with a foreign key, What will be cr...

What will be create when a primary key if combined with a foreign key? A primary key if combined with a foreign key it creates Parent-Child relationship between the tables that

The data of a view is not physically stored, The data of a view is not phys...

The data of a view is not physically stored, but derived from one or more tables. True the data of a view is not physically stored

Implementation of database, The ER diagram clearly showing the additional t...

The ER diagram clearly showing the additional tables you have implemented. SQL table creation scripts for the tables you have set up. SQL scripts showing the sample data you

How to create values of structured type, How to create values of structured...

How to create values of structured type? Constructor functions are used to make values of structured types. A function with the similar name as a structured type is a construct

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd