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Consider a schema $R(MNPQ)$ and functional dependencies $M\rightarrow N, P\rightarrow Q$. Then the decomposition of  $R$ into $R_{1} \left (MN \right )$ and  $R_{2} \left (PQ \right )$ is __________.

  1. Dependency preserving but not lossless join
  2. Dependency preserving and lossless join
  3. Lossless join but not dependency preserving
  4. Neither dependency preserving nor lossless join.
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Best answer
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5 votes

R(MNPQ) and functional dependencies M→N, P→Q. Then the decomposition of R into R1(MN) and R2(PQ) 

here clearly the dependencies are preserved but they are not lossless as both R1 and R2 cannot be formed back again as they donot have a common attribute which is a key so answer is 1

3 votes
3 votes
Here as only M->N and P->Q is there, and decomposition is also in MN and PQ

so it is Not Loseless

and all dependency are also preserved

so 1 is correct answer here
3 votes
3 votes

OPTION 1 Since there is no common attribute in the decomposed table so when we join it there will be some spurious(Extra) tuple so itwill be lossy.But the dependencies are preserved.

1 votes
1 votes
The condition for the decomposition to be lossless is that the common attribute(s) between the decomposed relation(s) should be the "key" in at least one of the decomposed relations. Since this is not true for the given decomposition, the decomposition is not lossless.

The dependencies are well preserved as  $M\rightarrow N$  is applicable for relation $R_{1}$  and  $P\rightarrow Q$  is applicable for relation  $R_{2}$ and their union gives original functional dependencies set.

$\therefore$  Option A is correct.
Answer:

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