(c)
Either (a) or (b)
Note: There can be more
than one minimal cover for a set of functional dependencies.
Steps to find the minimal cover;
1. Ensure singleton attribute on the
right hand side of each functional dependency.
2. Remove extraneous (redundant)
attribute from the left hand side of each functional dependency.
3. Remove redundant functional
dependency if any.
In our case, the functional
dependencies given are already satisfying the first two rules.
According to rule 1, if we have any
FDs with more than one attribute on the Right Hand Side, that FD should be
decomposed using decomposition rule. We don’t have such FDs.
According to rule 2, if we have any
FDs that have more than one attribute on the Left Hand Side (determiner), that
FD must be checked for partial dependency. We don’t have such FDs.
Hence, the given set satisfies both
rules.
For the third one we need to check.
Given : F = { A → B, B → A, A → C, C →
A, B → C }
For the first FD A → B find A+
by hiding the FD A → B. A+ = AC which does not include B in it.
Hence, A → B is not redundant.
For the second FD B → A find B+
by hiding the FD B → A. B+ =
ABC which includes A in it. Hence, B → A is redundant.
We have to remove it immediately.
Now with the removal of B → A, our F
becomes,
F = { A → B, A → C, C → A, B → C }
Find A+ by hiding A → C. A+
= ABC which includes C. Hence A → C is redundant and
remove it immediately from F.
After removal of A → C, F becomes,
F = { A → B, C → A, B → C }
Find C+ by hiding C → A. C+
= C and this does not include A in the result. Hence, C → A is not redundant.
Finally find B+ by hiding B
→ C. B+ = B. Hence, B → C is mandatory.
We don’t have any more FDs. Our final
set of functional dependencies are minimal after the removal of FDs B → A and A
→ C.
Hence, the minimal cover of F is as
follows;
Fc = { A
→ B, C → A, B → C }
[Work out for the option B → C. While
applying the above said procedure, start with the FD B → C. You will get the Fc
as { A → B, B → A, A → C, C → A }]
