• edited by
12,210 views
14 14 votes

Consider the following table

The table is in which normal form?

  1. First Normal Form
  2. Second Normal Form
  3. Third Normal Form but not BCNF
  4. Third Normal Form and BCNF

5 Answers

14 14 votes

Answer : C

Third Normal Form but not BCNF

See the dependencies 

AB ---> CDE

C ---> B

C is not a Super key Here

9 9 votes

Answer : C


Dependency: 


AB ---> CDE

C ---> B


1NF(yes): assume all attributes are atomic

2NF(yes): no partial functional  dependency like A alone can derive C or D or E

3NF(yes): no transitive functional dependency or there is nothing like (non key)-->(non key)

BCNF(no): in non trivial FD C-->B  , C is not a SuperKey.


smileyyes

5 5 votes
AB is the key here which determines C ,D E there is no P.D  and T.D between C,D,E SO IT IS IN 3NF

Now for BCNF every determinant is a candidate key determinant here are A,B,C (as C determines B) but C is not  candidate key hence it is not in BCNF (only if C->B is removed it becomes BCNF)

AC is also key  which can determine B and so also  D , E  but B is not candidate key

hence ans is C
• edited by
2 2 votes

It is 3NF 

Bcz AB->CDE ,here we can clealy see that AB is super keys that's why it will be in 3NF and BCNF but C->B it is in 3NF not in BCNF that's why we can say that it will be in 3NF not in BCNF bcz both of them should be in BCNF not single one.

0 0 votes

Correct Option is (C)
FD =
{ AB-->C , AB-->D, AB-->E, C-->B }
CK : AB and AC

  • BCNF : Not in BCNF since C is not a SK in C-->B.
  • 3NF : It's in 3NF since there's no CK that can transitively determine any Non-Prime attribute
Answer:
Position:
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