Test: Relational Algebra, Tuple Calculus & SQL- 2 - Computer Science Engineering (CSE) MCQ

The given query
SELECT DISTINCT W, X
FROM R, S
Is guaranteed to be same as R, if R has no duplicates and ‘S’ is non-empty.
Since, if R is having a duplicates, then the tuples selected by SELECT operation of the R and the given query will not be same also if ‘S’ is empty then the given query outputs null.

From the relation DEPOSIT first we have to select the those entries in which balance is over 1000. This can be obtained by  (Deposit), Now, we have to select those customers name who have balance over 1000 which can be obtained by simply projecting customer name in above subquery.

Clients of Bankers Agassi can be obtained by using the relation client_scheme while the corresponding customer_city can be obtained -using customer_scheme.
Hence cross product of client and customer must be taken. This can be achieved in two ways.
(i) First taking the Bankers Agassi clients and then checking their name in customer relation and projecting their customer city.
(ii) First taking all the customers and matching them in customer_scheme, then finding out the customers who have the Agassi as the Banker, finally projecting the customer_city of the respective customers.

We have to find all customers in the relation Borrow, this can be achieved by
 borrow (t [custome _name ])=s [customer_name]
The s here is bound variable and the condition of checking Ioan-amount more than 1200 is to be checked with each customer name condition.

All customers who have a lone amount of over 1200

The WHERE condition checks for the Grade = ‘A' and matching of course-number in the courses and Grades, where the instructor must be ‘Korth’. Overall it checks for students to whom ‘Korth’ has instructed and they have received Grade = ‘A in atleast one of the courses taught by ‘Korth’.

Student: (rollno, name, address)

• Primary Key: rollno
• Number of tuples: 120

Enroll: (rollno, courseno, coursename)

• Primary Key: (rollno, courseno)
• Number of tuples: 8

Natural Join

A natural join between the Student and Enroll tables will combine rows from both tables based on the common attribute rollno.

Minimum Number of Tuples

The minimum number of tuples in the natural join occurs when there are no matching rollno values between the two tables. However, in practice, we need at least one match for the join to yield results.

If no rollno from Enroll matches any rollno in Student, the join will produce zero tuples.

If there are matches, the minimum number of tuples is determined by the table with fewer tuples, which is Enroll in this case. Each Enroll entry needs at least one matching Student entry to form a tuple in the join.

So, the practical minimum number of tuples is 0 if no matches, otherwise at least 1, but technically, based on the fewest tuples in Enroll, the minimum could still be up to 8.

Maximum Number of Tuples

The maximum number of tuples occurs when each Enroll tuple can match multiple Student tuples, but since rollno in Student is unique, each rollno in Enroll can match only one Student tuple.

However, since we consider every tuple in Enroll can potentially join with a unique rollno in Student, and given each student (120) can enroll in multiple courses, the maximum number of tuples in the join is a product of all students and enrollments, provided the rollno in Enroll matches those in Student.

Calculations

• Minimum: 0 (if no rollno matches between Enroll and Student), but practically considering matching scenarios, up to 8.
• Maximum: If each of the 120 students is enrolled in multiple courses and each match is counted, the theoretical max is the product of the total rows in both tables (120 students * 8 enrollments each = 960), assuming all rollno in Enroll are present in Student.

Thus, the maximum number of tuples that can be present in the natural join is 960, and the practical minimum, considering enrollment existence, is up to 8.

Thus, the correct answer is: 960, 8

The given query first calculates the average salary from the relation Employee, then it selects the department name where employees who have average salary more than the average salary of the company.

Relation algebra is procedural query language while relational calculus is non-procedural query language.

The condition T. assets > S. assets and S. branch_city = “PONDICHERRY” checks for the all those branch-name who have greater assets than any branch located in Pondicherry. The keyword distinct removes the duplicacy of branch_name satisfy the given condition.

All of the given option valid:

They simply find and selects the tuples which are common to both P and Q.

The conditions c₁ and c₂ may not be same.

Find the customers who appears in r₂ with every branch name in r₁
So, it is represented by division query: 

With the help of {π, x, -} we can perform division operation.
Note: Division operator produces a relation R(X) that includes all tuples t[x] in R(z) that appear in R in combination with every tuple from R₂y) when Z = X ∪ Y.

A relational calculus expression may contain infinite tuples that are not in the relation, R₁.

Hence, the above expression is not safe.