Project ForgeCommon Data for this and the next question: Relation R has eight attributes ABCDEFGH. Fields of R contain only atomic values. F = {CH→G, A→BC, B→CFH, E→A, F→EG} is a set of functional dependencies (FDs) so that F⁺ is exactly the set of FDs that hold for R.
How many candidate keys does the relation R have?
- A3
- 4
- C5
- D6
Explanation
D never appears on the right-hand side of any FD, so no combination of the other attributes can ever derive D — D must be part of every candidate key. Computing closures of the remaining attributes individually: A⁺ = {A,B,C,E,F,G,H} (via A→BC, B→CFH, F→EG), so AD is a key. Similarly B⁺, E⁺, and F⁺ each independently expand to all of {A,B,C,E,F,G,H} too, making BD, ED, and FD each candidate keys as well. C, G, and H alone contribute nothing further (their only relevant FD, CH→G, needs both C and H together), so they can't form smaller keys. That gives exactly 4 minimal candidate keys: AD, BD, ED, FD.
Source: GATE CS 2013 (2013)
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