Question

Consider the relation \(R,\) which has attributes that hold schedules of courses and sections at a university; \(R=\\{\) CourseNo, SecNo, OfferingDept, Credit- Hours, CourseLevel, InstructorSSN, Semester, Year, Days_Hours, RoomNo, NoOfStudents \(\\} .\) Suppose that the following functional dependencies hold on \(R:\) \\{CourseNo \(\\} \rightarrow\\{\text { OfferingDept, CreditHours, CourseLevel }\\}\) \\{CourseNo, SecNo, Semester, Year) \(\rightarrow\) \\{Days_Hours, RoomNo, NoOfStudents, InstructorSSN \(\left\\{\mathrm{R}_{\text {oom }} \mathrm{No}, \text { Days_Hours, Semester, Year }\right\\} \rightarrow\\{\) Instructorssn, CourseNo, SecNot Try to determine which sets of attributes form keys of \(R .\) How would you normalize this relation?

Short answer

Candidate keys for the given relation R are {CourseNo,SecNo,Semester,Year,RoomNo}. After normalizing the relation to 3NF, two relations are generated: R1={CourseNo, SecNo, Semester, Year, RoomNo, Days_Hours, NoOfStudents} and R2={RoomNo, Semester, Year, InstructorSSN, CourseNo, SecNo}

Step-by-step solution

Understand the Functional Dependencies

Based on the problem, the functional dependencies are: \n\n1) CourseNo --> OfferingDept, CreditHours, CourseLevel \n2) {CourseNo, SecNo, Semester, Year} --> Days_Hours, RoomNo, NoOfStudents, InstructorSSN \n3) {RoomNo, Days_Hours, Semester, Year} --> InstructorSSN, CourseNo, SecNo

Identifying Candidate Keys

A Candidate Key is a set of attributes that uniquely identifies tuples in a relation. To find candidate keys, start with the dependencies and identify attributes that are never on the right-hand side, these are called prime attributes. Here, CourseNo,Semester, Year, SecNo, and RoomNo are identified as they're never on the right-hand side of any functional dependencies. To be a candidate key, these attributes must depend on (determine) all other attributes. Based on this assessment, the candidate keys are: {CourseNo,SecNo,Semester,Year,RoomNo}.

Normalize the Relation

Normalization is a process of organizing the data in a database to avoid redundancy and other anomalies. The goal is to decompose the Relation into two or more relations that meet the requirements of 1NF, 2NF, 3NF, and possibly BCNF. Here the relation is already in 1NF since all attributes (columns) contain atomic (indivisible) values. It's also in 2NF as there are no partial dependencies in relation R. But it's not in 3NF because there's a transitive dependency RoomNo --> InstructorSSN. So, we split it into two relations to achieve 3NF: \n\n1) R1={CourseNo, SecNo, Semester, Year, RoomNo, Days_Hours, NoOfStudents} \n2) R2={RoomNo, Semester, Year, InstructorSSN, CourseNo, SecNo} \n\nNow, the relations are in 3NF as there are no transitive dependencies.

Key concepts

Candidate Keys

Understanding candidate keys is fundamental in the design of a well-structured database. They are minimal sets of attributes that can uniquely identify each record within a table or relation. These attributes are at the core of ensuring data integrity.

To determine candidate keys, we look for attributes not dependent on any other - those not found on the right-hand side of functional dependencies. In our exercise concerning university course scheduling, attributes such as CourseNo, SecNo, Semester, Year, and RoomNo fit this criteria. However, being a candidate key requires that the combination of these attributes can determine every other attribute in the relation.

From this analysis, we deduce that the composite of {CourseNo, SecNo, Semester, Year, RoomNo} not only identifies each tuple uniquely but also satisfies all functional dependencies and thus, is a candidate key for the relation R. Recognizing these keys is a critical step to ensuring that our database design prevents duplicate records and maintains data integrity.

Relation Normalization

Normalization is like tidying a messy room; it organizes data in a way to reduce redundancy and complexity. This process structures a database according to recognized normal forms, usually starting from the first normal form (1NF) and progressing to the third (3NF) or the Boyce-Codd Normal Form (BCNF).

In our exercise, the given relation R is already in 1NF because each column holds atomic values. Advancing to 2NF, we check for and eliminate any partial dependencies - none were found in our case. However, moving to 3NF involves removing transitive dependencies, where non-prime attributes depend on other non-prime attributes.

We identify that RoomNo indirectly determines InstructorSSN, qualifying as a transitive dependency and violating the rules of 3NF. To rectify this, we decompose our relation R into two smaller relations: R1 and R2. This splitting ensures all transitive dependencies are eliminated, resulting in a streamlined, efficient database design that adheres to 3NF.

Database Design

Database design is akin to architectural planning for data structures—it requires careful thought on how tables relate and interact with one another. The goal is to produce a design that is efficient, scalable, and, above all, accessible to users as per their needs.

In the context of our example, database design begins with understanding functional dependencies which dictate how various tables in the system are linked. We apply these dependencies to structure the data effectively. As we have determined, splitting the main relation R into R1 and R2 enhances the design by adhering to the principles of normalization.

This process minimizes redundancy and potential anomalies during data operations such as insertions, deletions, and updates. When databases are normalized and dependencies are respected, the system gains flexibility and becomes less prone to errors, ultimately serving its users with more reliability and integrity. By finalizing the design with candidate keys and appropriately normalized relations, we create a database that not only holds information but also manages it proficiently.