1 votes 1 votes if you are given table like R(A,B,C,D,E,F)= {AB$\rightarrow$C, BC $\rightarrow$ A, AC$\rightarrow$ B , B $\rightarrow$ D , D$\rightarrow$ E } How many tables are required in each normal form A- 2NF B- 3 NF C- BCNF please illustrate 2NF only hitendra singh asked Nov 3, 2018 hitendra singh 1.5k views answer comment Share Follow See all 13 Comments See all 13 13 Comments reply Show 10 previous comments Shaik Masthan commented Nov 3, 2018 reply Follow Share @Utkarsh Joshi We can decompose R as R1(A,B,C) and R2(B,D,E) where is F ? if you add F in R1 then it will lead to PFD, isn't it? for 2NF, it requires 3 tables i.e., one of the decomposition is R1(A,B,C) , R2(B,D,E) and R3(A,C,F) for 3NF/BCNF, it requires 4 tables i.e., one of the decomposition is R1(A,B,C) , R2(B,D) R4(D,E) and R3(A,C,F) 0 votes 0 votes Utkarsh Joshi commented Nov 3, 2018 reply Follow Share Shaik Masthan the thing is i completely forgot to take F in consideration i thought we are having just ABCDE as F was not part of any FD. 0 votes 0 votes Shaik Masthan commented Nov 3, 2018 reply Follow Share i expected brother... 1 votes 1 votes Please log in or register to add a comment.
1 votes 1 votes https://gateoverflow.in/?qa=blob&qa_blobid=11466423765690194514 Maybe for bcnf. MLMukul answered Nov 3, 2018 MLMukul comment Share Follow See all 11 Comments See all 11 11 Comments reply Tejasvi96 commented Nov 3, 2018 reply Follow Share yeah exactly F should be separate for BCNF decomposition . But the decomposition may not be lossless.. As spurious tuples may be there while recombining the tables through a join 1 votes 1 votes hitendra singh commented Nov 3, 2018 reply Follow Share does seperate F make sure that loss less join occurs? i had broke the relation as ABF , ABC , BD , DE correct me if I am wrong. 2 votes 2 votes Soumya Tiwari commented Nov 3, 2018 reply Follow Share Why to create seperate table for ABF and ABC I'm unable to access that link.Someone please explain? 0 votes 0 votes hitendra singh commented Nov 3, 2018 reply Follow Share its actually a image 1 votes 1 votes MLMukul commented Nov 3, 2018 reply Follow Share Hitendra Singh. You are probably right as it makes the decompostion lossless. 1 votes 1 votes hitendra singh commented Nov 3, 2018 reply Follow Share @Soumya+Tiwari have you got ur explanation? 1 votes 1 votes Soumya Tiwari commented Nov 3, 2018 reply Follow Share Yes many thanks :) 0 votes 0 votes Gurdeep Saini commented Nov 5, 2018 reply Follow Share ABCF ,BD,DE in this decomposion why ABCF in not in bcnf help me i am not able to understand 0 votes 0 votes Tejasvi96 commented Nov 5, 2018 reply Follow Share Clearly F should be a part of the key, as F is not on the RHS of any functional dependency. So clearly either ABF, ACF, BCF should be separated which will not have any functional dependency. Thus the division as done by hitendra is perfect. 1 votes 1 votes hitendra singh commented Nov 5, 2018 reply Follow Share as given first three FD's are made of A,B,C In every AB→C, BC → A, AC→ B we will get a subset of C.K {ABF,ACF,BCF} LHS of the FD's are subset these will violate 2NF condition .so we have to split I hope you get it , revert back if u didn't. 2 votes 2 votes Gurdeep Saini commented Nov 5, 2018 reply Follow Share thanks @hitendra singh @tejasv 0 votes 0 votes Please log in or register to add a comment.