Database-Management-System
August 29, 2024Database-Management-System
August 29, 2024Database-Management-System
| Question 818 |
Consider the schema R=(S,T,U,V) and the dependencies S→ T, T→ U, U→ V and V→ S. If R=(R1 and R2) be a decomposition such that R1 ∩R2= ∅ , then decomposition is
| not in 2 NF | |
| in 2 NF but not in 3 NF | |
| in 3 NF not in 2 NF | |
| in both 2NF and 3 NF |
Question 818 Explanation:
● R1 ∩ R2 = ∅. This makes the decomposition lossless join, as all the attributes are keys, R1 ∩ R2 will be a key of the decomposed relations (lossless condition says the common attribute must be a key in at least one of the decomposed relation).
● Now, even the original relation R is in 3NF (even BCNF) as all the attributes are prime attributes (in fact each attribute is a candidate key). Hence, any decomposition will also be in 3NF (even BCNF).
● Now, even the original relation R is in 3NF (even BCNF) as all the attributes are prime attributes (in fact each attribute is a candidate key). Hence, any decomposition will also be in 3NF (even BCNF).
Correct Answer: D
Question 818 Explanation:
● R1 ∩ R2 = ∅. This makes the decomposition lossless join, as all the attributes are keys, R1 ∩ R2 will be a key of the decomposed relations (lossless condition says the common attribute must be a key in at least one of the decomposed relation).
● Now, even the original relation R is in 3NF (even BCNF) as all the attributes are prime attributes (in fact each attribute is a candidate key). Hence, any decomposition will also be in 3NF (even BCNF).
● Now, even the original relation R is in 3NF (even BCNF) as all the attributes are prime attributes (in fact each attribute is a candidate key). Hence, any decomposition will also be in 3NF (even BCNF).