Wednesday, February 18, 2015

Codd's Twelve Rules



Codd's Twelve Rules / Codd's Twelve Commandments / Relational Model Foundation Principles


Can we say “any database that supports SQL is a relational database”? No. At least, not completely. We need certain foundation principle to make this assessment. For any system to be called a relational database management system, the relational capabilities must be able to manage it completely. Edgar F. Codd, a pioneer of the relational model for databases, designed Codd’s twelve rules to define what is required from a database management system in order for it to be considered relational, i.e., a relational database management system (RDBMS).
Those rules are listed below and the links will lead you to the description pages of those rules.

Codd’s twelve rules




Monday, February 16, 2015

DBMS Basics - Quiz 5



DBMS Basics - Quiz 5

[Answers are given in Green color]
1. A relational database is said to be a group of ____.
a. common fields                                c. records
b. field values                                    d. tables

2. One of the advantages of a relational database model is ____.
a. structural dependence                     c. easier database design
b. conceptual complexity                    d. complex database design

3. Using network terminology, a relationship is called a(n) ____.
a. member                                 c. set
b. owner                                    d. table

4. The hierarchical database model is based on a ____.
a. tree structure                               c. lack of a child segment
b. lack of a parent segment                 d. Matrix

5. The hierarchical database model depicts a set of ____ relationships.
a. M:1                                               c. 1:M
b. 1:1                                                d. M:N

6. The hierarchical database is not very efficient when handling ____.
a. large amounts of data                     c. many transactions
b. few transactions                           d. 1:M relationships

7. The hierarchical database models ____.
a. have no implementation limitations          c. have very strict standards
b. promote database integrity                   d. have a simple navigational system

8. A relational database model ____.
a. does not require substantial hardware and system software overhead
b. does not promote "islands of information" problems
c. allows trained people to use a good system poorly
d. lets the user operate in a human logical environment

9. The network database models have a ____.
a. navigational system that yields simple design
b. simple system that promotes efficiency
c. relationship of owner/member that promotes database integrity
d. great deal of structural independence

10. Database models have been developed for ____.
a. modeling  real-world events or conditions     
b. depositing data within a single file
c. keeping data within multiple data repositories  
d. allowing DBMSs to maintain lose control over the database activities.

11. The RDBMS ____.
a. performs functions provided by the hierarchical and network DBMS system
b. does not manage data structures
c. allows the user/designer to ignore the logical view of the database
d. does not manage the details of physical storage

12. System’s complexity is hidden from the user and designer in RDBMS. This leads to ____.
a. it does not exhibit data independence
b. it does not exhibit structural independence
c. easier data management
d. data management is very difficult

13. Database models can be grouped into two categories:  conceptual models and ____ models.
a. implementation                             c. physical
b. logical                                   d. query

14. Every row/record in the relational table is called as an entity ____.
a. instance                                        c. attribute
b. relationship                                    d. model

15. Degree of ______ is a way to classify data models.
a. difficulty                                        c. abstraction
b. knowledge                                     d. unification


Thursday, February 12, 2015

Reduction of Weak Entity set to relation schema.

How to convert Weak entity set into relation schema?



Weak Entity set is an entity set which does not have enough attributes to form a Primary key. While converting weak entity set into relation schema, the following procedure is followed;

  • Take the Primary key of Strong entity set in which the Weak entity set depends. Then include the primary key into the relation schema of weak entity set along with other attributes of weak entity set.

Figure 1 - ER Diagram for Weak entity and relationships


Let us consider the ER diagram given below, where Apartment entity set is the Weak entity set with the Discriminator ADoor_No. Apartment stores information regarding the individual apartment details in any building. Hence, a building can have many floors with more apartments in every floor. Suppose, if we number every apartment with 1, 2, 3 and so on, then there are other buildings with same door numbers. Hence, ADoor_No cannot be the Primary key. Apartment is totally participated on Strong Entity set Building_Details through the weak relationship set BA for existence. That is, for every record in Apartment, we must have a related record in Building_details.
The relation schema Apartment will contain all the attributes of Apartment along with the Primary key of Strong entity set Building_Details. Thus, we have the following schema;
Apartment ( Building_No, ADoor_No, Floor, Rent)
The strong entity set Building_Details as follows;
Building_Details (Building_No, Building_Name, Total_Apartments, Address)

Points to note:

The relationship BA need not be converted into table. In other words, the relationship between strong and weak entity set is one-to-many. We shall see that in the post Relationship set into relation schema.


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