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Question

Answer 1

Answer D)$\left ( B,C,D,E,F \right )\, \, And \, \, \left ( C,D,E,A \right )$

Given Relation $R\left ( A,B,C,D,E,F \right )$

FDs are-:

$ABC\ rightarrow DEF$

$CDE \rightarrow A$

We got candidate keys as-:

$ABC,BCDE$

Looking at FDs ,we can clearly say

• No $Non\, \, Prime\, \, Attributes \rightarrow Prime \, \, Attribute$(No Partial dependency),hence it is in $2NF$
• No $Non\, \, key\, \, Attributes \rightarrow Non \, \, Key$(No Transitive dependency),hence it is in $3NF$
• But $LHS\neq Super\, \, Key$,hence it is not in BCNF

Now Decomposing Relation $R\left ( A,B,C,D,E,F \right )$,

We have Relation $R_{1} \left ( A,C,D,E \right ) \, \, and\, \, R_{2} \left ( B,C,D,E,F \right )$

Comments

Why not C?

 

ABC is SK and CDEFA is also SK.

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@habib can u answer this....... how R(Abcdef) is decomposed ? we know that it is following 3NF but how to decompose it such that it follows BCNF. Thanks :)

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@ sourav.

R1 is in BCNF bcz of FD: CDE→A where CDE is a superkey, but how R2 is in BCNF ?