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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
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 lossless and not dependency preserving.

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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

{ACDE} AD->E

{ADG} AD->G

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 but not dependency preserving lossless is easy 

for dependency preserving 

B->D& E->G are not there in relation 

Answer:

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