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

Non prime -> Non prime also leads to transitive depedency?

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yes

$R(A,B)$

$A \rightarrow B$

$B \rightarrow C$ (Non prime $ \rightarrow $ Non prime )

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

ADB is the candidate key. Right? AB not alone can't be ck.

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(AB)+ == { A, B, G, C, E}

(ABD)+ == { A, B, D, C, E, G, P}

 

How is {AB} a Candidate Key???

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

i.e AB –> CD  : (AB)+ = {A,B,C,D}

B → G : AB+ = {A,B,C,D,G}

C → E : (AB)+ = {A,B,C,D,E,G}

DE → P: (AB)+ = {A,B,C,D,E,G,P}

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B->G is not a Partial dependency according to definition of Partial Dependency(Definition of P.D- "For any FD A->B, B is partially dependent on A if some proper subset of A itself can determine B") so here Partial dependency is in AB->G.

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.