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+1 vote

Consider a schema R(MNPQ) and functional dependencies M→N, P→Q. Then the decomposition of R into R1(MN) and R2(PQ) 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
asked in Others by Veteran (413k points) | 1.4k views

2 Answers

+5 votes
Best answer

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

answered by Boss (10.7k points)
selected by
I think option 1 It will be not lossless join.
i answered will not be loseless
@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
+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
answered by Boss (17.9k points)
i know the method to find lossless which is so long.. by drawing tables..

is there any simpler way?
i think not..i also ude the same method..but here you can do just by visualising

For better understanding of the concept go to the article

@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.

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