Made Easy Test Series: Database-Normalization

Question

Consider the relation $R\left ( A,B,C,D,E \right )$ with functional dependencies

$F=${

$A\rightarrow B$

$BC\rightarrow E$

$ED\rightarrow A$

}

Number of additional relation required to convert it into lossless , dependency preserving $3NF$ decomposition is _____________

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What is meaning of additional relation (Here no table mentioned previously)??

Comments

every question of made easy is not correct.

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@aditi19

Can u try it??

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i'm trying... will update once I reach an answer

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@aditi19 did you found any answer ?

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the relation already appears to be in 3NF as candidate keys are ACD, ECD, BCD. Since all are prime attributes then the relation is already in 3NF so no additional relation required answer supposed to be ZERO.

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total number of extra table will be zero.it is already in 3NF

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agreed.

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Each and every attribute is prime attribute. so we can say directly the relation is in 3NF.

Answer 1

First, let's have a look at the candidate keys. 

In this case the candidate keys are $\rightarrow$ $CDA$, $CDB$, $CDE$.

Now, in this case, each attribute is a prime attribute. So, this realtion is already in 3NF and there is no point in decomposing this further.

Therefore, no additional relations are required.

So, the correct answer to this question will be 0.

Answer 2

4 table are required.

Comments

how this decomposition is lossless?

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@abhishekmehta4u

ans given 2 tables

Are u sure??

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how does A->B is a voilation of 3NF as B is a prime attribute ?

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Here every attribute is prime so every FD in given table is satisfying X> Y where X is a super key or  Y is prime attribute so is not in 3nf already?