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IITM November 2016
Aboveallplayer
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Nov 25, 2016
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FD s given P1P3->P4, P1->P2, P2->P1
now
a) bcnf and p2p3->p4 holds
b)bcnf and p2p3->p4 does not hold
c)3nf not in bcnf and p2p3->p4 holds
d)3nf not in bcnf and p2p3->p4 does not hold
Aboveallplayer
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Hradesh patel
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Nov 25, 2016
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is it C???
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According to given FD's candidate keys ={P1P3, P2P3}
so p1p3->p4 is in BCNF
p1->p2 is in 3NF
p2->p1 is in 3NF
so as whole relation is in 3NF not in BCNF. and p2p3 is a candidate key so p2p3->p4 will also hold .
so option (c) is correct .
Amit Pal
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Nov 25, 2016
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