Relation x=(d,b,c,f,g,h) {d=bc ,g=hd} is decomosed into a=(d,b,c) e=(d,f,g,h) which of the following is correct?

Question

a) lossy and dependency preserving

b)lossless and dependency preserving

c)losslsess and no dependency preserving

d)lossy and not dependency preserving

Answer 1

It is clearly Dependency preserving as the dependency D-> BC is satisfied in A and G->HD is satisfied in E.

For loseless decomposition,  intersection of (D,B,C) and (D,F,G,H) gives D which is the key in (D,B,C).

 

Hence the decomposition is both dependency preserving and loseless.

Comments

@Heinsberg .As relation e contains d,f,g,h not all are covered under fds then how it is dependency psvng

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For a decomposition to be loseless , the intersection of sets should give a set of attributes that is key for any one of the set.

When you say F is not covered, it only implies that key for e will be gf. But as D is also key of a and common in a,e . The decomposition becomes loseless.

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yaa it is lossless am not saying about lossless property ..m asking about its dependency preservation

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Yes the 2 dependencies are preserved in their respective tables. D=BC preserved in A and G=HD preserved in E. F has no role in dependency preservation.

Answer 2

Option B is correct .

Reason : It is lossless as in realtion a 'd' which is common attribute in both the tables is candidate key.

It is dependency preserving because d->bc is satisfied in relation a and g->hd is satisfied by e.

Dependency preserving states that the set of dependencies of all the decomposed tables should cover the original tables dependencies.

CORRECT ME IF I AM WRONG.