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Database-Management-System
August 29, 2024
Database-Management-System
August 29, 2024
Database-Management-System
August 29, 2024
Database-Management-System
August 29, 2024

Normalization

Question 84
The best normal form of relation scheme R(A, B, C, D) along with the set of functional dependencies F = {AB → C, AB → D, C → A, D → B} is
A
Boyce-Codd Normal form
B
Third Normal form
C
Second Normal form
D
First Normal form
Question 84 Explanation: 
AB → C
AB → D
C → A
D → B
A​ +​ ={A}
B​ +​ ={B}
C​ +​ ={C,A}
D​ +​ ={D,B}
Since single attributes can’t determine all the attributes of relation. Find closure of combination of two attributes to check whey could be the key for the relation or not.
AB​ +​ ={A,B,C,D}
AC​ +​ ={A,C}
AD​ +​ ={A,D,B,C}
BC​ +​ ={B,C,A,D}
AB​ +​ ={A,B,C,D}
AC​ +​ ={A,C}
AD​ +​ ={A,D,B,C}
BC​ +​ ={B,C,A,D}
BD​ +​ ={B,D}
CD​ +​ ={C,D,B,A}
So, AB,AD,BC,CD are the candidate keys for the given relation. All attributes are key attributes and since last two functional dependencies are violating BCNF property (that LHS if a functional
dependency should be a super key) so the given relation is in 3NF only.
Correct Answer: B
Question 84 Explanation: 
AB → C
AB → D
C → A
D → B
A​ +​ ={A}
B​ +​ ={B}
C​ +​ ={C,A}
D​ +​ ={D,B}
Since single attributes can’t determine all the attributes of relation. Find closure of combination of two attributes to check whey could be the key for the relation or not.
AB​ +​ ={A,B,C,D}
AC​ +​ ={A,C}
AD​ +​ ={A,D,B,C}
BC​ +​ ={B,C,A,D}
AB​ +​ ={A,B,C,D}
AC​ +​ ={A,C}
AD​ +​ ={A,D,B,C}
BC​ +​ ={B,C,A,D}
BD​ +​ ={B,D}
CD​ +​ ={C,D,B,A}
So, AB,AD,BC,CD are the candidate keys for the given relation. All attributes are key attributes and since last two functional dependencies are violating BCNF property (that LHS if a functional
dependency should be a super key) so the given relation is in 3NF only.
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