Consider relation X(PQR) with the set of FD's [P→QR, Q→R,
Consider a relation X(PQR) with the set of FD's [P→QR, Q→R, P→Q, PQ→R]
What is number of FD present in minimal [canonical] cover?
- A.
1
- B.
2
- C.
3
- D.
4
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Correct answer: B
Final minimal cover (count = 2): {P→Q, Q→R}
Step 1: Split FDs with multiple attributes on the right-hand side. P→QR becomes two FDs: P→Q and P→R.
Step 2: Remove duplicate dependencies. If any FD appears more than once (for example P→Q), keep a single copy.
Step 3: Remove redundant dependencies by checking if a dependency can be derived from the others.
P→R is redundant because P→Q and Q→R together imply P→R (transitivity).
PQ→R is redundant because Q→R already gives R whenever Q is present, so PQ→R adds no new information.
Step 4: Keep only the essential dependencies. After removing redundancies, the minimal cover is {P→Q, Q→R}.