Database-Management-System
August 29, 2024Database-Management-System
August 29, 2024Normalization
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Question 84
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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
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Boyce-Codd Normal form
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Third Normal form
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Second Normal form
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First Normal form
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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.
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.
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.