Question-26
Consider a database that has the relation schema CR(StudentName, CourseName). An instance of the schema CR is as given below:
| Student Name | Course Name |
|---|---|
| SA | CA |
| SA | CB |
| SA | CC |
| SB | CB |
| SB | CC |
| SC | CA |
| SC | CB |
| SC | CC |
| SD | CA |
| SD | CB |
| SD | CC |
| SD | CD |
| SE | CD |
| SE | CA |
| SE | CB |
| SF | CA |
| SF | CB |
| SF | CC |
The following query is made on the database: \({ T1 \leftarrow \Pi_{\text{course Name}} (\sigma_{\text{Student Name} = 'SA'} (CR)) }\) \({ T2 \leftarrow CR \div T1 }\)
The number of rows in T2 is _____________.
\(4\)
The query \({ T1 \leftarrow \Pi_{\text{course Name}} (\sigma_{\text{Student Name} = 'SA'} (CR)) }\) selects all the course names where the Student Name is ‘SA’, resulting in T1 containing course names ‘CA’, ‘CB’, and ‘CC’.
The query \({ T2 \leftarrow CR \div T1 }\) performs division between CR and T1. The result T2 contains tuples from CR that have all the course names from T1.
The output steps: 1. \({ CR \div \text{CA} }\): Gives students SA, SC, SD, SE, SF. 2. \({ CR \div \text{CB} }\): Gives students SA, SC, SD, SE, SF. 3. \({ CR \div \text{CC} }\): Gives students SA, SC, SD, SF.
Taking the intersection of these outputs gives the students SA, SC, SD, SF, which results in 4 rows in T2.