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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 __________.

- Dependency preserving but not lossless join
- Dependency preserving and lossless join
- Lossless join but not dependency preserving
- Neither dependency preserving nor lossless join.

5 votes

Best answer

@udipito lossy does not mean that some records are lost....It means that you will not be able to generate the exact relation as before ...In the above if R1 and R2 are joined to form the previous relation you will not get exact relations as many will come out of from the Cartesian product to remove unnecessary things we need a common attribute which is a key(unique) it is clearly lossy

1

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

so it is Not Loseless

and all dependency are also preserved

so 1 is correct answer here

0

For better understanding of the concept go to the article

http://www.geeksforgeeks.org/lossless-join-and-dependency-preserving-decomposition/

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@Debasmita Bhoumik

For two relations : Just see which are common attributes of the relations i.e. by looking only. Now, take closure of that common attribute if u found any one of the relations back that means it is lossless.

For more than two relations u have to follow the Chase Algorithm which is very easy to implement.

For two relations : Just see which are common attributes of the relations i.e. by looking only. Now, take closure of that common attribute if u found any one of the relations back that means it is lossless.

For more than two relations u have to follow the Chase Algorithm which is very easy to implement.

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1 vote

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.

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.