GATE IT 2008 | Question: 62

Question

Let $R (A, B, C, D, E, P, G)$ be a relational schema in which the following functional depen­dencies are known to hold: $AB \to CD, DE \to P, C \to E, P \to C$ and $B \to G.$ The relational schema $R$ is

• A. in $\text{BCNF}$
• B. in $\text{3NF}$, but not in $\text{BCNF}$
• C. in $\text{2NF}$, but not in $\text{3NF}$
• D. not in $\text{2NF}$

Comments

Detailed Video Explanation: https://www.youtube.com/watch?v=HdaHz0iY2ws&t=9293s

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DBMS Complete Summary & GATE PYQs Playlist: https://youtube.com/playlist?list=PLIPZ2_p3RNHh3otU-TnAK-GkqrvvOO33C&feature=shared 

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sol

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it is in 1NF

Answer 1

Answer: $D$

Here $AB$ is the candidate key and $B \rightarrow G$ is a partial dependency. So, $R$ is not in $\text{2NF}$.

Comments

@Anila krishna AB is the candidate key.

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Answer is right but B --> G is not partial dependency

AB --> G is partial dependency

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For a relation to be in 2NF, the non-prime attributes should be fully dependent on the super key. If it is dependent on a subset of the super key, then it violates the condition and hence is not in 2NF. (Definition of partial dependency), (Definition of 2NF)

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Here, AB is the candidate key as shown by 5a1n1amarjeet. From F.D. $B \rightarrow G$, we can see that a non-prime attribute (G) is partially dependent on a subset of the super key (B). This violates the property of 2NF, hence given relation R is not in 2NF.

Answer 2

not in 2NF because here candidate key is AB and in FD's proper subset of C.K. determine the non prime attribute i.e. B→G

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Comments

Yes, you are right. Answer is D.

Answer 3

AB is key and G is non prime attribute which is partially dependent on B.

(Ans-D)

Answer 4

(AB) forms the candidate key, so the functional dependency B→G violates 2NF.

Answer 5

AB={A,B,C,D,E,P,G}

Hence AB is candidate key

whereas, $B \rightarrow G$ is having relation of Super Key deriving Non-prime. Which means it doesn't satisfy condition of 2NF.

Comments

explanation is wrong , Here AB is candidate key , but the B -> G is not having the relation of super key deriving non-prime . instead , this a full FD , which a violation condition of 2NF where it is subset of candidate key deriving the non prime attribute .

Answer 6

BEST ANSWER

R(A, B, C, D, E, P,G)

AB------>CD

DE------>P

C------>E

P------>C

B------>G

FIND CANDIDATE KEY :- Trick --> Write all attributes which are not present on RHS of above FD.

AB not present on RHS of a relation so it must be partof CK.

Check Closre of AB---> ABCDEPG        Since all attributes comin g so it is minimal CK 

so CK=  AB

NON PRIME / NON KEY ATTRIBUTE = CDEPG

Proper subset of CK= A, B

VOILATION CONDITION OF 2NF

Proper Subset of CK----> Non Key attribute 

B------>G     (it is voilation of 2NF)

So not in 2NF 

If Relation is not in 2NF then it cannot be in 3NF and BCNF.

 

Comments

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