An organization needs to maintain database having five attributes \(A, B, C,…
2019
An organization needs to maintain database having five attributes \(A, B, C, D, E\). These attributes are functionally dependent on each other for which functionality dependency set \(𝐹\) is given as: \(F:\{A→BC,D→E,BC→D,A→D\}\). Consider a universal relation \(R(A, B, C, D, E)\) with functional dependency set \(𝐹\). Also all attributes are simple and take atomic values only.
MInimal cover \(F′\) of functional dependency set \(𝐹\) is
- A.
\(F′={A→B,A→C,BC→D,D→E} \) - B.
\(F′={A→BC,B→D,D→E} \) - C.
\(F′={A→B,A→C,A→D,D→E} \) - D.
\(F′={A→B,A→C,B→D,C→D,D→E}\)
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Correct answer: A
Solution:
Step 1 — Make the RHS atomic: decompose A→BC into A→B and A→C. Now the set is {A→B, A→C, BC→D, A→D, D→E}.
Step 2 — Remove extraneous attributes from LHS: BC→D has two attributes on the left. Test if B or C is extraneous.
Check if B is extraneous: compute C+ using the set with C→D (i.e., replacing BC→D by C→D). C+ = {C} (no D), so B is not extraneous.
Check if C is extraneous: compute B+ using the set with B→D. B+ = {B} (no D), so C is not extraneous.
Step 3 — Remove redundant dependencies: check whether any FD can be derived from the others.
A→D is redundant: from A we get B and C (A→B and A→C), and then BC→D gives D. So remove A→D.
Final minimal cover: {A→B, A→C, BC→D, D→E}.
Therefore the minimal cover is {A→B, A→C, BC→D, D→E}.
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