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69 votes

Consider the schema $R=(S,T, U, V)$ and the dependencies $S \rightarrow T, T \rightarrow U, U \rightarrow V$ and $V \rightarrow S$. Let $R = (R1\text{ and } R2)$ be a decomposition such that $R1 \cap R2 \neq \phi$. The decomposition is

  1. not in $\textsf{2NF}$
  2. in $\textsf{2NF}$ but not $\textsf{3NF}$
  3. in $\textsf{3NF}$ but not in $\textsf{2NF}$
  4. in both $\textsf{2NF}$ and $\textsf{3NF}$
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11 Answers

Best answer
103 votes
103 votes
$R_1 \cap R_2 \neq \phi.$ This makes the decomposition lossless join, as all the attributes are keys, $R_1 \cap R_2$ will be a key of the decomposed relations (lossless condition says the common attribute must be a key in at least one of the decomposed relation).  Now, even the original relation $R$ is in $3NF$ (even $\text{BCNF}$)as all the attributes are prime attributes (in fact each attribute is a candidate key).  Hence, any decomposition will also be in $3NF$ (even $\text{BCNF}$). Option $D$.

PS: Decomposition in $3NF$ means decomposed relations are in $3NF$. But when we consider any decomposed relation, we must also include any FD which are being implied by the original relational schema. For example, in a decomposed relation $STU,$ there will be a FD $U\to S$ as well.
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25 votes
25 votes

Answer : In both 2NF and 3NF 

Dependencies are 

 S --> T, T --> U, U -->V  V --> S

S+ = STUV          U+ = UVST

T+ = UVST          V+ = VSTU

There is no Partial Dependencies here So it is in 2 NF

RHS of every Dependencies  is a Key as well as all are Prime Attributes So it is in 3 NF.

9 votes
9 votes

R1 ∩ R2 != ∅. This makes the decomposition lossless join and as all the attributes are prime, so whatever is the decomposition, it will be 2NF and 3NF. So, Ans will be option d.

6 votes
6 votes
(d) both in 2nf and 3nf

since r1 and r2 does not have any common attributes , and if we decomse it to relation having 2 attributes, then they are by default in 3nf , Suppose if we decompose r1 with 1 attribute and r2 with 3 attributes, since all of them are dependent all of them could be a prime attribute in this case also they are in 3nf, since they are in 3nf, 2nf by default satisfy
Answer:

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