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Consider the following set of functional dependency on the schema (A, B,C)
A→BC, B→C, A→B, AB→C
A→BC, B→C, A→B, AB→C
The canonical cover for this set is:
- a)A->BC and B->C
- b)A->BC and AB->C
- c)A->BC and A->B
- d)A->B and B->C
Correct answer is option 'A'. Can you explain this answer?
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Verified Answer
Consider the following set of functional dependency on the schema (A, ...
All of the FDs are implied by the FDs {A->BC, B->C}
From option b: A+=BC (hence A->BC, A->b and AB->Care implied by FDs set given in option b)
From option b: A+=BC (hence A->BC, A->b and AB->Care implied by FDs set given in option b)
B+=BC (hence B->C is implied by FDs set given in option b)
Most Upvoted Answer
Consider the following set of functional dependency on the schema (A, ...
All of the FDs are implied by the FDs {A->BC, B->C}
From option b: A+=BC (hence A->BC, A->b and AB->Care implied by FDs set given in option b)
From option b: A+=BC (hence A->BC, A->b and AB->Care implied by FDs set given in option b)
B+=BC (hence B->C is implied by FDs set given in option b)
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Community Answer
Consider the following set of functional dependency on the schema (A, ...
Canonical Cover Explanation:
Functional Dependencies:
- A→BC
- B→C
- A→B
- AB→C
Removing Redundant Dependencies:
- From A→BC and A→B, we can remove A→B as it is redundant.
- From A→BC and B→C, we can remove B→C as it is redundant.
Minimal Cover:
- A→BC
- B→C
Augmentation:
- To make A→BC and B→C unique, we need to add A→C.
Canonical Cover:
- A→BC
- A→C
Therefore, the canonical cover for the given set of functional dependencies is A→BC and A→C.
Functional Dependencies:
- A→BC
- B→C
- A→B
- AB→C
Removing Redundant Dependencies:
- From A→BC and A→B, we can remove A→B as it is redundant.
- From A→BC and B→C, we can remove B→C as it is redundant.
Minimal Cover:
- A→BC
- B→C
Augmentation:
- To make A→BC and B→C unique, we need to add A→C.
Canonical Cover:
- A→BC
- A→C
Therefore, the canonical cover for the given set of functional dependencies is A→BC and A→C.
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