Database-Management-System
August 29, 2024Database-Management-System
August 29, 2024Database-Management-System
| Question 706 |
Consider two database relations R and S having 3 tuples in R and 2 tuples in S. What is the maximum number of tuples that could appear in the natural join of R and S?
| 6 | |
| 2 | |
| 3 | |
| 5 |
Question 706 Explanation:
→ If the tables R and S contains common attributes and value of that attribute in each tuple in both tables are same, then the natural join will result n*m tuples as it will return all combinations of tuples.
Consider following two tables
→ Result of natural join R * S (If domain of attribute C in the two tables are same )
→ You can see both R and S contain the attribute C whose value is 2 in each and every tuple.
Table R contains 3 tuples, Table S contains 2 tuples, where Result table contains 3*2=6 tuples.
Note: While performing a natural join, if there were no common attributes between the two relations, Natural join will behave as Cartesian Product.
Consider following two tables
→ Result of natural join R * S (If domain of attribute C in the two tables are same )
→ You can see both R and S contain the attribute C whose value is 2 in each and every tuple.
Table R contains 3 tuples, Table S contains 2 tuples, where Result table contains 3*2=6 tuples.
Note: While performing a natural join, if there were no common attributes between the two relations, Natural join will behave as Cartesian Product.
Correct Answer: A
Question 706 Explanation:
→ If the tables R and S contains common attributes and value of that attribute in each tuple in both tables are same, then the natural join will result n*m tuples as it will return all combinations of tuples.
Consider following two tables
→ Result of natural join R * S (If domain of attribute C in the two tables are same )
→ You can see both R and S contain the attribute C whose value is 2 in each and every tuple.
Table R contains 3 tuples, Table S contains 2 tuples, where Result table contains 3*2=6 tuples.
Note: While performing a natural join, if there were no common attributes between the two relations, Natural join will behave as Cartesian Product.
Consider following two tables
→ Result of natural join R * S (If domain of attribute C in the two tables are same )
→ You can see both R and S contain the attribute C whose value is 2 in each and every tuple.
Table R contains 3 tuples, Table S contains 2 tuples, where Result table contains 3*2=6 tuples.
Note: While performing a natural join, if there were no common attributes between the two relations, Natural join will behave as Cartesian Product.