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Exam (elaborations)

INF2603 Exam Pack 2024

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INF2603 Exam Pack 2024 Latest exam pack questions and answers and summarized notes for exam preparation. For assistance call or W.h.a.t.s.a.p.p us on +/ 2/ 5/ 4 /7 /7 /9 /5 /4 /0 /1 /3 /2 .

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INF2603
EXAM PACK

, INF2603

October/November Exam 2024



Due Date: 31 Octoer 2024




Question 1

1. (a) True
2. (b) False
3. (a) True
4. (a) True
5. (b) False
6. (a) CHECK
7. (c) EXISTS
8. (b) HAVING
9. (d) UPDATE
10. (e) LIKE




Question 2

a. Difference between a column constraint and a table constraint:

 A column constraint applies only to a single column within a table and is used
to enforce data integrity specific to that column (e.g., NOT NULL or UNIQUE
constraints on a column).
 A table constraint applies to multiple columns within the table or even the entire
table, allowing more complex integrity checks (e.g., PRIMARY KEY, FOREIGN
KEY, or CHECK constraints that might involve multiple columns).

,b. Disjoint Subtype and Example:
A disjoint subtype is a subtype in a specialization hierarchy where an entity instance
can belong to only one subtype. If an entity is part of one subtype, it cannot be part of
another. For example, in a personnel database, an Employee could be either a
FullTimeEmployee or a PartTimeEmployee, but not both, making FullTimeEmployee
and PartTimeEmployee disjoint subtypes.

c. Three Join Types in the OUTER JOIN Classification:

1. LEFT OUTER JOIN: Returns all records from the left table and the matched
records from the right table. If no match is found, NULL values are returned for
columns from the right table.
2. RIGHT OUTER JOIN: Returns all records from the right table and the matched
records from the left table. If no match is found, NULL values are returned for
columns from the left table.
3. FULL OUTER JOIN: Returns all records when there is a match in either the left
or the right table. If no match is found, NULL values are returned for the columns
from the table without a match.

d. Strong (or Identifying) Relationship:
A strong (or identifying) relationship exists when a child entity's existence is
dependent on its parent entity, and the primary key of the parent entity is also part of the
child entity's primary key.

e. Why a table with a single-attribute primary key is automatically in 2NF when in
1NF:
When a table's primary key consists of a single attribute, it cannot have partial
dependencies (where a non-key attribute depends on part of a composite key), as
there’s only one attribute in the primary key. Thus, the table is automatically in 2NF if it
is in 1NF.

f. Metadata:
Metadata is data that describes other data. It provides information about the structure,

, organization, and characteristics of data, such as data types, data lengths, and data
relationships.

g. Physical Independence in Data Models:
Physical independence is the ability of a data model to remain unaffected by changes
in the physical storage structures or devices. This allows the database to be modified at
the storage level without impacting the logical data structure or application level.




Question 3

a. Transfer the Unnormalized Form to First Normal Form (1NF)

Unnormalized Table (Table 3.1):


StudentID StudentName CourseID CourseName Instructor EnrolmentDate

1 Thato 101 Math Dr. Moape 2024-01-15

2 Lucas 102 History Dr. Mtsweni 2024-01-16

1 Thato 103 Physics Dr. Chipa 2024-01-17

3 Sam 101 Math Dr. Moape 2024-01-18


Normalization to 1NF:
In 1NF, there should be no repeating groups, and each attribute must contain atomic
values. Since each row already contains atomic values and no repeating groups, the
table is in 1NF as shown above.




b. Transform the Relation into Second Normal Form (2NF)

Requirements for 2NF:

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