Solution

Data Dictionary consists of the following information:

• Name of the tables in the database
• Constraints of a table i.e. keys, relationships, etc.
• Columns of the tables that related to each other
• Owner of the table
• Last accessed information of the object
• Last updated information of the object

Solution

(0, m) means partial participation and (1, n) means total participation.

Solution

In terms of relational model, cardinality is termed as a number of tuples

Solution

Trivial − If a functional dependency X → Y holds, where Y is a subset of X, then it is called a trivial FD.

Non-trivial − If a functional dependency X → Y holds, where Y is not a subset of X, then it is called a non-trivial FD.

Semi non-trivial – Combination of both trivial and non-trivial is said to be semi non-trivial FD.

Solution

Normalization is the process of organizing a database to reduce redundancy and improve data integrity.

Solution

• (CE)+ = CEFADB ✔

CE is a Candidate key of R. Hence, BCE and CEF are Superkeys.

• (CD)+ = CDF ✖
• Hence, CD is not a key of R.

Solution

The maximum no. of super keys possible for the table with 4 attributes is

2^(4-2) = 2^2 = 4

Solution

Solution

Solution

The intervals based on the decision tree splits are:

• Interval 1: Temperature ≤ 22 → Low
• Interval 2: 22 < Temperature ≤ 32 → Medium
• Interval 3: Temperature > 32 → High

A temperature of 28 falls into Interval 2, which corresponds to Medium.

Solution

Huffman coding creates a variable-length code where more frequent characters have shorter codes. Thus, the order of encoding lengths from shortest to longest for the given frequencies would be:

• 'A' (most frequent) → shortest code
• 'B'
• 'C'
• 'D' (least frequent) → longest code

Solution

In LZW compression, we build the dictionary incrementally. For ABABABA:

1. Start with the dictionary: {A:1, B:2}.
2. Encode AB → Add AB to dictionary as 3.
3. Continue with ABAB → ABA is in the dictionary, but ABAB is not, so add ABAB as 4.
4. Continue with ABA → Encode as 5.

So, the sequence of dictionary indices is A, B, 3, 4, 5.

Solution

Solution

• For strong entities Student and Course 2 tables are required.
• The cardinality of the relationship “Enroll” is mamy-to-many. Hence, we need to create a separate table for Enroll.

Solution

All varchar, real and float are valid data types in SQL.

Solution

A View can be defined as a virtual table that contains rows and columns with fields from one or more tables.

Solution

• In SQL, a UNIQUE constraint is used to ensure that each row and column has a unique value and no value is being repeated in any other row or column.
• In SQL, CHECK constraint ensures whether the value in columns fulfils the specified condition or not.

Solution

• The basic difference in both is DELETE command is DML command and the TRUNCATE command is DDL.
• DELETE command is used to delete a specific row from the table whereas the TRUNCATE command is used to remove all rows from the table.
• We can use the DELETE command with WHERE clause but cannot use the TRUNCATE command with it.

Solution

In SQL, a table can have many foreign keys but only one primary key.

Solution

According to question:

Select Name from people where Age > 21; will return 20 rows.

Select Name from people where Height > 180; will return 17 rows.

Then Select Name from people where (Age > 21) or (Height > 180); will return 20 rows. Since there is or between the condition. That’s why it will return 20 rows.

Solution

Given X values: 3, NULL, 2, 3, NULL, 5

AVG (DISTINCT X)

= AVG (3, NULL, 2, 5)

= (3+2+5)/3 = 10/3 = 3.33

Solution

Option (b) is the correct “CREATE TABLE” statement to create the given in the question.

Solution

“SELECT class_name, AVG(age) FROM class GROUP BY class_name HAVING AVG(age)>10;”

is the correct query which returns the names and average age of all classes, the average age of students is greater than 10.

Solution

The query “SELECT Branch, SUM(Salary) FROM Employee GROUP BY Branch” returns the total salary of all employees at each branch of the bank.

Solution

The insert operator is not from a relational algebra operations set.

Solution

All are the valid join expressions.

Solution

Solution

Here every tuple in Employee can join to at most one tuple in Department since Dno is primary key in Department table. So, the maximum number of tuples is 800.

Solution

• JPEG and MPEG are lossy compression algorithms.
• MP3 is also a lossy compression algorithm.
• LZW (Lempel-Ziv-Welch) is a lossless compression algorithm used in formats like GIF and TIFF.

Solution

Run-length encoding (RLE) compresses data by representing consecutive repeated characters as a single character and its count.

• AAAA → 4A
• BBB → 3B
• CC → 2C
• D → 1D
• AA → 2A

Combining these, we get 4A3B2C1D2A.

Solution

The snowflake schema is better suited for handling updates to hierarchical data due to its normalized structure, which reduces redundancy and ensures that updates are reflected consistently across the schema.

Solution

The snowflake schema requires more complex queries due to the normalization of dimension tables, but it simplifies maintenance by reducing redundancy. The star schema is simpler to design and query but may involve more maintenance due to denormalized data.

Solution

The fact constellation schema, or galaxy schema, is most suitable for large-scale data warehouses with multiple related business processes because it allows for multiple fact tables to share dimension tables, facilitating complex queries and data integration.