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in Databases by Active (3.1k points) | 107 views

1 Answer

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4 tables created ..

here candidate key is AC...

so A->B fails to be in bcnf ....so we divide it in table ABD....in which B->D fails so we divide in AB and BD...

C->E again fails so we divide ....CE is table...

lastly to maintain KEYS ...we have AC...

so totals tables are

(AC)(BD)(AB)(CE)
by Boss (10.9k points)
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4 tables will be created for bcnf decomposition.... AC,  AB, CE, BC..... these will be lossless as well as dependency preserving.... If it would had been asked for 2nf... Then only 3 tables will be created... AC, ABD, CE

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