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GATE | GATE CS 2013 | Question 55

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  • Difficulty Level : Easy
  • Last Updated : 28 Jun, 2021
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For the relation R(ABCDEFGH) with FD’s= {CH->G, A->BC, B->CHF, E->A, F->EG such that F+ is exactly the set of FDs that hold for R.} Consider the FDs given in above question. The relation R is


in 1NF, but not in 2NF.


in 2NF, but not in 3NF.


in 3NF, but not in BCNF.



Answer: (A)


The table is not in 2nd Normal Form as the non-prime attributes are dependent on subsets of candidate keys. The candidate keys are AD, BD, ED and FD. In all of the following FDs, the non-prime attributes are dependent on a partial candidate key. A -> BC B -> CFH F -> EG

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