+1 vote
85 views

Consider the relation $R(ABCDEG)$ and $FD's$ are

$\{AB \rightarrow C,AC \rightarrow B,AD \rightarrow E,B \rightarrow D,BC \rightarrow A,E \rightarrow G\}$

then the decomposition$\{ABC,ACDE,ADG\}$

1. is lossless and not dependency preserving
2. is not lossless and dependency preserving
3. is lossless and dependency preserving
4. is not lossless and not dependency preserving
edited | 85 views

lossless and not dependency preserving.

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Their is no attribute F given in relation R
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yes i read wrong .now i am updating
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Nice explanation.
+1 vote

It is lossless but not dependency preserving lossless is easy

for dependency preserving

B->D& E->G are not there in relation

+1
can you please explain why it is lossless
+2

How you know about lossless: when Two relation R1 and R2 are made from original relation then common elements from these two relations must be a candidate key of any relation.

Now look here:

ABC and ACDE here common element AC is Candidate key of R1. Now relation will be ABCDE.

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Hint:AD->G is hidden relation bcz AD->E  E->G so we can say AD->G   merge ABC ACDE  common part AC is superkey

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Yes Your correct and its lossless @anu007 and @akshat sharma

Option (A) :It is lossless but not dependency preserving lossless.

Explaination :-

Dependency preserving: Even if there are many functional dependencies in Relation R, all of them won't be applicable, Only functional dependencies which contain only these attributes:-

{ABC} AB->C , AC->B , BC-> A

The lost functional dependencies are B->D and E->G, They cannot be derived even when the table is  joined.

lossless:-

There are two types lossy decomposition(when extra tuples are added when the tables are re-joined) and lossless decomposition(no extra tuples gets added, Original tuples are reformed)

The condition for lossless join is : If the common attribute is the key in one of the tables.

Between table {ABC and ACDE} AC is key of table {ABC} and hence between table {ACDE} and {ADC}, AD is common and key attribute of table {ADG and ACDE}

Hence, the table is lossless.

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it is lossless
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Yes its, lossless.

Ans : A.

Lossless Join decomposition Check(LJ):

Decomposition of R into R1 and R2 is lossless if: [ R1 ^ R2 ---> R1 ] or [ R1 ^ R2 ---> R2].

^ = intersection operation of set.

accordingly here:

R1(ABC) R2(ACDE) -> R1^ R2( intersection) we get, AC which is a key. hence Lossless.

R3(ACDE) R4(ADG) -> R3^R4(intersection) we get, AD which is again a key. hence, Lossless.

Now,

Dependency Preserving Check(DP):

Once we take closure say {ABC}+  dependencies like B-> D will be missing. and

DP says - if the closure of FD's after decomposition is same as the closure of FD's before decomposition. then, Dependency is preserved. Which not holding true here.

so, No Dependency Preserving.

Conclusion: Lossless and Not dependency preserving. matching Option (A).