2 votes 2 votes Consider the relation R(ABCD) with dependencies F:{A->B, B->C , C->D ,D->A}. if decomposed into R1(ABC) and R2(BD), then the relation R1(ABC) is in ________ Normal Form and what will be the keys of R1? Nikhil Vaidya asked Jan 20, 2018 • edited Jan 23, 2018 by Nikhil Vaidya Nikhil Vaidya 490 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply hs_yadav commented Jan 20, 2018 reply Follow Share i think R(ABC) is in BCNF? 1 votes 1 votes Mk Utkarsh commented Jan 20, 2018 reply Follow Share correct me but C is transitively dependent on A? right? 0 votes 0 votes hs_yadav commented Jan 20, 2018 reply Follow Share bro according ro transitivity rule...if A->B B->c then FD A->C would be trivial... there in given set of FD B->c also a valid Fd and in R b would also be Key....therefor BCNF is the heighest NF i.... 0 votes 0 votes Mk Utkarsh commented Jan 20, 2018 reply Follow Share how A--->C is trivial? and b cannot be the key can you deduce b as key by R1 alone? 0 votes 0 votes hs_yadav commented Jan 20, 2018 reply Follow Share @Nikhil Vaidya could u please write the given answer from source? 0 votes 0 votes Hemant Parihar commented Jan 20, 2018 reply Follow Share A, B, C, D are the keys, already the given relation is BCNF. And we further decompose it. Both R1 and R2 are BCNF. Moreover the decomposition is lossless and FD's preserving. 0 votes 0 votes Nikhil Vaidya commented Jan 21, 2018 reply Follow Share The answer is BCNF but i don't know how they got it. So anyone know the perfect solution then plz post it. 0 votes 0 votes Mk Utkarsh commented Jan 21, 2018 reply Follow Share It is BCNF i do same mistake every time I will post a well formed answer after some time 0 votes 0 votes Please log in or register to add a comment.