2 votes 2 votes Databases database-normalization test-series + – ben10 asked Aug 29, 2018 • retagged Aug 29, 2018 by Sayan Bose ben10 677 views answer comment Share Follow See all 9 Comments See all 9 9 Comments reply MiNiPanda commented Aug 29, 2018 reply Follow Share I am getting B 1 votes 1 votes Shaik Masthan commented Aug 29, 2018 i edited by Shaik Masthan Aug 29, 2018 reply Follow Share 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 0 votes 0 votes MiNiPanda commented Aug 29, 2018 reply Follow Share 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) 0 votes 0 votes himgta commented Aug 29, 2018 reply Follow Share here AB,BC,AF,CF,DF,EF all are candidate keys....all the attributes are prime attributes.....how are u differentiating the options? 0 votes 0 votes MiNiPanda commented Aug 29, 2018 reply Follow Share 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.. 0 votes 0 votes ben10 commented Aug 29, 2018 reply Follow Share How you get D is correct? 0 votes 0 votes Shaik Masthan commented Aug 29, 2018 reply Follow Share @ben10 what is the problem you are facing with D? 0 votes 0 votes Shaik Masthan commented Aug 29, 2018 i edited by Shaik Masthan Aug 29, 2018 reply Follow Share 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. 0 votes 0 votes Shaik Masthan commented Aug 29, 2018 reply Follow Share @Minipanda, i edited my comments, please read it 0 votes 0 votes Please log in or register to add a comment.
–1 votes –1 votes 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 donnie answered Sep 15, 2018 donnie comment Share Follow See all 0 reply Please log in or register to add a comment.