Question 150
Comprehension:
Answer question (146-150) based on the problem statement given below:
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as :
F : { A → BC, D → E, BC → D, A → D }.
Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.
Assume that given table R is decomposed in two tables R1(A, B, C) with functional dependency set f1 ={A → B, A → C} and R2(A, D, E) with FD set F2 ={A → D, D → E}.
Which of the following option is true w.r.t. given decomposition?
Answer : 4. R1 is in 3 NF and R2 is in 2 NF Answer question (146-150) based on the problem statement given below:
An organization needs to maintain database having five attributes A, B, C, D, E. These attributes are functionally dependent on each other for which functionally dependency set F is given as :
F : { A → BC, D → E, BC → D, A → D }.
Consider a universal relation R(A, B, C, D, E) with functional dependency set F. Also all attributes are simple and take atomic values only.
Assume that given table R is decomposed in two tables R1(A, B, C) with functional dependency set f1 ={A → B, A → C} and R2(A, D, E) with FD set F2 ={A → D, D → E}.
Which of the following option is true w.r.t. given decomposition?
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| UGC NET CS December 2019 - Question 149 |
Since, In R1 and R2 BC can’t determine BC → D of relation "R". Hence R1 and R2 are not following the Dependency preservation property.
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Candidate key of R1 is "A". And since LHS of R1 contains only "A" so R1 is in 3NF.
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Candidate key of R2 is "A" , But Since D → E neither have Super key in its LHS nor have a prime key attribute in its RHS,
Here it's a transitive dependency of A → D and D → E.
So R2 is not in 3NF but in 2NF.
So, option 4 is correct answer