ISRO CSE 2014 | Question: 22

Question

Consider the schema $R (A, B, C, D)$ and the functional dependencies $A\to B$ and $C\to D$.  If the decomposition is made as $R_1(A, B)$ and $R_2 (C, D)$, then which of the following is TRUE?

• A. Preserves dependency but cannot perform lossless join
• B. Preserves dependency and performs lossless join
• C. Does not perform dependency and cannot perform lossless join
• D. Does not preserve dependency but perform lossless join

Comments

Option A ??

Answer 1

Decomposition is dependency preserving.

It is not lossless as there is no common attribute on which two sub relations can be  joined to get original one

So ans is a

Comments

Thank you so much Pooja

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Yes you are correct answer is ' A '

While decomposing a relational table we must verify the following properties:

i) Dependency Preserving Property: A decomposition is said to be  dependency preserving if F+=(F1 ∪ F2 ∪ .. Fn)+, Where F+=total functional dependencies(FDs) on universal relation R, F1 = set of FDs of R1, and F2 = set of FDs of R2.
For the above question R1 preserves A->B and R2 preserves C->D. Since the FDs of universal relation R is preserved by R1 and R2, the decomposition is dependency preserving.

ii) Lossless-Join Property:
The decomposition is a lossless-join decomposition of R if at least one of the following functional dependencies are in F+:-
a) R1 ∩ R2 -> R1
b) R1 ∩ R2 -> R2
It ensures that the attributes involved in the natural join (  ) are a candidate key for at least one of the two relations.In the above question schema R is decomposed into R1 (A, B) and R2(C, D), and R1 ∩ R2 is empty. So, the decomposition is not lossless.

 

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awesome explanation

Answer 2

for lossless, after decomposition of relation into 2, if there is any common attribute in the decomposed relations and that attribute is a key in any of the relation then it is lossless - (any decomposition should have a common attribute(s) which is a key in one of the relation)

here no attributes are common in R1 and R2

so this one is not lossless.

for dependency preserving= if we are able to derive the original  dependency from the dependencies of the splited schemas then it is dependency preserving.

here from R1===A->B and from R2====c->D can be derived..

so it is DP

ANS=OPTION A

Comments

nice explanation...

Answer 3

option a because before decomposition fd's are equal to after decomposition fd's .but it is not lossless because r1 insection r2=phi

Answer 4

• R(A, B, C, D)

• Functional dependencies: A → B, C → D

• Decomposed into:

  • R₁(A, B)       R₂(C, D)

By Using Lossless Join using Chase Test

• R₁ and R₂ have no common attribute

• So, using chase test, we cannot merge any data back

• So the decomposition is NOT lossles

Dependency Preservation

• R₁ = (A, B) ⇒ preserves A → B 

• R₂ = (C, D) ⇒ preserves C → D 

• Together, the decomposition preserves all FDs given in the question

Not Lossless and  Dependency Preserving