The Gateway to Computer Science Excellence
0 votes
123 views

in Databases by Active (3.2k points) | 123 views

1 Answer

+1 vote
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 (11k points)
0
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
Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true
50,833 questions
57,688 answers
199,289 comments
107,251 users