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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
asked in Databases by Boss (14k points)
edited by | 85 views

4 Answers

+3 votes

 lossless and not dependency preserving.

 

 

answered by Loyal (9k points)
edited by
0
Their is no attribute F given in relation R
+1
yes i read wrong .now i am updating
0
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 

answered by Active (1.5k points)
+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.

ABCDE and ADG here common element AD is key candidate key ADG. hence It is lossless.

 

0

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

then merge  ABCDE ADG     AD is common part AD is superkey in relation ADG by using our HINT

0
Yes Your correct and its lossless @anu007 and @akshat sharma
0 votes

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.

 

answered by (235 points)
edited by
0
it is lossless
0
Yes its, lossless.
0 votes

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).

answered ago by (225 points)

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