The Gateway to Computer Science Excellence
+1 vote

in Databases by (429 points)
retagged by | 137 views
I am getting B

F is the key of that relation ===> F is a single attribute ===> Given table itself in the 2NF

coming to the options, in Option A, by decomposing we loss AB->CDEF

remaining all options are loss-less and dependency preserving.

@MiNiPanda, i hope you should consider AB as Key

in that case option D is right, but not option B


Yes Shaik I considered AB as a CK and that is why found B->D as a PFD. Since for this FD the relation is violating 2nf i made a separate relation R(BD)

here AB,BC,AF,CF,DF,EF all are candidate keys....all the attributes are prime are u differentiating the options?

himgta F is the CK so you can't take AF,CF,DF,EF as CKs..they are superkeys..

AB,BC,F are CKs. I couldn't find more than this..

How you get D is correct?


what is the problem you are facing with D?

What is wrong with option B if we take AB as key?

if we take AB as key, Option B is in 2NF ? no it can't be in 2NF due to B->E is lead to PFD moreover it is loss-less decomposition but not dependency preserving ( D->E is not preserving )

Question is not standard... we can skip this question.... nothing is useful on discussing this question.

i edited my comments, please read it

1 Answer

–1 vote
I think option A is correct, since AB, BC, F are the candidate keys possible, B-->D and D-->E are in 1NF and 2NF respectivley. if we decompose correctly we get the relations which are in option A
by (37 points)
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
50,737 questions
57,391 answers
105,442 users