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17 votes
17 votes

Consider a relational table $R$ that is in $3NF$, but not in BCNF. Which one of the following statements is TRUE?

  1. $R$ has a nontrivial functional dependency $X \rightarrow A$, where $X$ is not a superkey and $A$ is a prime attribute.
  2. $R$ has a nontrivial functional dependency $X \rightarrow A$, where $X$ is not a superkey and $A$ is a non-prime attribute and $X$ is not a proper subset of any key.
  3. $R$ has a nontrivial functional dependency $X \rightarrow A$, where $X$ is not a superkey and $A$ is a non-prime attribute and $X$ is a proper subset of some key
  4. A cell in $R$ holds a set instead of an atomic value.
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3 Answers

Best answer
32 votes
32 votes

In $\text{3NF}$  where functional dependency is of type $X  \rightarrow Y$

$X$ can be the super key or $Y$ can be the prime attribute

Whereas in $\text{BCNF}$ where functional dependency is of type $X \rightarrow Y$

$X$ should be super key $\text{(BCNF}$ is more strict compared to $\text{3NF)}$

  • Option (C) says it has a partial dependency $($not even $\text{2NF)}$.
  • Option (D) multiple values in a cell. i.e not atomice $($not even $\text{1NF)}$.
  • Option (B)  says $X$ is not a super key and $Y$ is not a prime attribute. Therefore not $\text{3NF}$.

Ans (A): Says $X$ is not a super key but $Y$ is a prime attribute. Satisfies one of the conditions of the $\text{3NF}$ formal definition. As $X$ is not a Super Key it is not in $\text{BCNF}$.

1 votes
1 votes
OPTION A. SINCE X IS NOT A SUPER KEY BUT A  IS PRIME. ( ANY ONE CONDITION (1) X is a Super key (2) A is prime attribute ) .here option A satisfy the condition.
0 votes
0 votes

If ( X  ->  Y ) is trivial function means Y is subset of X (X ∩ Y = not-null),
example (sid+sname) -> sid,  here  X must be SuperKey.

But in question is say (X -> Y) is Non-trivial means X ∩ Y = null , means X is must be not SuperKey.

Given is in 3NF and not in BCNF.

3NF => means (L.H.S = CandidateKey or SuperKey)  or  (R.H.S = Primary Attribute)

BCNF => means  (L.H.S = CandidateKey or SuperKey)  and  (R.H.S = non Primary Attribute) .

We already know X(LHS) is not superKey , hence to fulfill it is in 3NF and not in BCNF Y(RHS) must be a Prime attribute .Since it also clear that its not in BCNF.

 

Option A is correct.

Answer:

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