Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > R(A,B,C,D) is a relation. Which of the follow...
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R(A,B,C,D) is a relation. Which of the following does not have a lossless join, dependency preserving BCNF decomposition?
- a)A->B, B->CD
- b)A->B, B->C, C->D
- c)AB->C, C->AD
- d)A ->BCD
Correct answer is option 'C'. Can you explain this answer?
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Verified Answer
R(A,B,C,D) is a relation. Which of the following does not have a lossl...
Background :
- Lossless-Join Decomposition:
Decomposition of R into R1 and R2 is a lossless-join decomposition if at least one of the following functional dependencies are in F+ (Closure of functional dependencies)
R1 ∩ R2 → R1
OR
R1 ∩ R2 → R2
OR
R1 ∩ R2 → R2
- dependency preserving :
Decomposition of R into R1 and R2 is a dependency preserving decomposition if closure of functional dependencies after decomposition is same as closure of of FDs before decomposition.
A simple way is to just check whether we can derive all the original FDs from the FDs present after decomposition.
Question : We know that for lossless decomposition common attribute should be candidate key in one of the relation. A) A->B, B->CD R1(AB) and R2(BCD) B is the key of second and hence decomposition is lossless. B) A->B, B->C, C->D R1(AB) , R2(BC), R3(CD) B is the key of second and C is the key of third, hence lossless. C) AB->C, C->AD R1(ABC), R2(CD) C is key of second, but C->A violates BCNF condition in ABC as C is not a key. We cannot decompose ABC further as AB->C dependency would be lost.D) A ->BCD Already in BCNF. Therefore, Option C AB->C, C->AD is the answer.
Most Upvoted Answer
R(A,B,C,D) is a relation. Which of the following does not have a lossl...
Explanation:
Lossless join decomposition: The decomposition of a relation R is said to be lossless join if the natural join of the decomposed relations is equal to the original relation R. It is a property of a decomposition.
Dependency preserving decomposition: A decomposition is said to be dependency preserving if it satisfies all the functional dependencies present in the original relation. It is also a property of a decomposition.
BCNF decomposition: A decomposition is said to be in BCNF if all the functional dependencies in the relation are in the form of A → B, where A is a candidate key.
Now, let's analyze the given options one by one:
a) A-B, B-CD
- Functional dependencies: A → B, B → C, B → D
- Candidate keys: AB
- This decomposition is in BCNF as all the functional dependencies are in the form of A → B.
- This decomposition is also lossless join as the natural join of A-B and B-CD gives the original relation R.
b) A-B, B-C, C-D
- Functional dependencies: A → B, B → C, C → D
- Candidate keys: AB
- This decomposition is in BCNF as all the functional dependencies are in the form of A → B.
- This decomposition is also lossless join as the natural join of A-B, B-C and C-D gives the original relation R.
c) AB-C, C-AD
- Functional dependencies: A → D, C → A, AB → C
- Candidate keys: AB
- This decomposition is not in BCNF as AB → C violates the BCNF condition.
- This decomposition is also not lossless join as the natural join of AB-C and C-AD does not give the original relation R. The natural join of AB-C and C-AD gives ABCD, which is not equal to R.
d) A -BCD
- Functional dependencies: A → B, A → C, A → D
- Candidate keys: A
- This decomposition is in BCNF as all the functional dependencies are in the form of A → B, A → C and A → D.
- This decomposition is also lossless join as the natural join of A-BCD gives the original relation R.
Therefore, option 'C' does not have a lossless join, dependency preserving BCNF decomposition.
Lossless join decomposition: The decomposition of a relation R is said to be lossless join if the natural join of the decomposed relations is equal to the original relation R. It is a property of a decomposition.
Dependency preserving decomposition: A decomposition is said to be dependency preserving if it satisfies all the functional dependencies present in the original relation. It is also a property of a decomposition.
BCNF decomposition: A decomposition is said to be in BCNF if all the functional dependencies in the relation are in the form of A → B, where A is a candidate key.
Now, let's analyze the given options one by one:
a) A-B, B-CD
- Functional dependencies: A → B, B → C, B → D
- Candidate keys: AB
- This decomposition is in BCNF as all the functional dependencies are in the form of A → B.
- This decomposition is also lossless join as the natural join of A-B and B-CD gives the original relation R.
b) A-B, B-C, C-D
- Functional dependencies: A → B, B → C, C → D
- Candidate keys: AB
- This decomposition is in BCNF as all the functional dependencies are in the form of A → B.
- This decomposition is also lossless join as the natural join of A-B, B-C and C-D gives the original relation R.
c) AB-C, C-AD
- Functional dependencies: A → D, C → A, AB → C
- Candidate keys: AB
- This decomposition is not in BCNF as AB → C violates the BCNF condition.
- This decomposition is also not lossless join as the natural join of AB-C and C-AD does not give the original relation R. The natural join of AB-C and C-AD gives ABCD, which is not equal to R.
d) A -BCD
- Functional dependencies: A → B, A → C, A → D
- Candidate keys: A
- This decomposition is in BCNF as all the functional dependencies are in the form of A → B, A → C and A → D.
- This decomposition is also lossless join as the natural join of A-BCD gives the original relation R.
Therefore, option 'C' does not have a lossless join, dependency preserving BCNF decomposition.
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Community Answer
R(A,B,C,D) is a relation. Which of the following does not have a lossl...
See for lossless join you need to have some common attribute between 2 relations.
In case of option c,there is no common attribute to perform join,therefore it is lossless join.
In case of option c,there is no common attribute to perform join,therefore it is lossless join.
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R(A,B,C,D) is a relation. Which of the following does not have a lossless join, dependency preserving BCNF decomposition?a)A->B, B->CDb)A->B, B->C, C->Dc)AB->C, C->ADd)A ->BCDCorrect answer is option 'C'. Can you explain this answer?
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