Computer Science Engineering (CSE) Exam > Computer Science Engineering (CSE) Questions > Consider the schemaa r(A, B, C, D, E) with th...
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Consider the schema a r(A, B, C, D, E) with the following functional dependencies:
A → BC
CD → E
B → D
E → A
The decomposition of the schema into r1(A, B, C) and r2(A, D, E) is _________.
- a)Lossless and dependency preserving
- b)Lossy and dependency preserving
- c)Lossless and non dependency preserving
- d)Lossy and non dependency preserving
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Consider the schemaa r(A, B, C, D, E) with the following functional de...
R1 and R2 form a lossless decomposition of R if at least one of the following functional dependencies is in F:
• R1 ∩ R2 → R1
• R1 ∩ R2 → R2
In the given schema r(A, B, C, D, E), the candidate key is A because A+ = {A, B, C, D, E}
Most Upvoted Answer
Consider the schemaa r(A, B, C, D, E) with the following functional de...
Decomposition Analysis:
Lossless Decomposition:
- To check for lossless decomposition, we need to compute the intersection of the decomposed relations.
- The intersection of r1 and r2 should contain the attributes that are common to both relations.
- In this case, the intersection of r1 and r2 is on the key attribute A, which ensures lossless decomposition.
Dependency Preserving:
- To check for dependency preservation, we need to verify if the functional dependencies of the original relation can be inferred from the decomposed relations.
- Both r1 and r2 individually cannot derive all the original functional dependencies.
- However, when we join r1 and r2 on the common attribute A, we can derive all the original functional dependencies.
- Thus, the decomposition is dependency preserving.
Therefore, the decomposition of the schema into r1(A, B, C) and r2(A, D, E) is lossless and dependency preserving.
Lossless Decomposition:
- To check for lossless decomposition, we need to compute the intersection of the decomposed relations.
- The intersection of r1 and r2 should contain the attributes that are common to both relations.
- In this case, the intersection of r1 and r2 is on the key attribute A, which ensures lossless decomposition.
Dependency Preserving:
- To check for dependency preservation, we need to verify if the functional dependencies of the original relation can be inferred from the decomposed relations.
- Both r1 and r2 individually cannot derive all the original functional dependencies.
- However, when we join r1 and r2 on the common attribute A, we can derive all the original functional dependencies.
- Thus, the decomposition is dependency preserving.
Therefore, the decomposition of the schema into r1(A, B, C) and r2(A, D, E) is lossless and dependency preserving.
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Consider the schemaa r(A, B, C, D, E) with the following functional dependencies:A → BCCD → EB → DE → AThe decomposition of the schema intor1(A, B, C) and r2(A, D, E) is _________.a)Lossless and dependency preservingb)Lossy and dependency preservingc)Lossless and non dependency preservingd)Lossy and non dependency preservingCorrect answer is option 'A'. Can you explain this answer?
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