3 3 votes Consider the following statements about relation R: S1: If a relation R is in 3NF but not in BCNF, then relation R must consist proper subset of candidate key determines proper subset of some other candidate key. S2: If a relation R is in 3NF but not BCNF, then relation R must consist atleast two over-lapped candidate keys. Which of the following statements is/are correct? (a) Both S1 and S2 (b) Only S1 © Only S2 (d) None of the above Databases + – Somoshree Datta 5 3.5k views answer comment Share Follow Print See all 19 Comments 19 19 Comments reply Magma commented Jan 20, 2019 reply Follow flag S1 is right :3 not sure about S2 --> I think S2 is not true ..let me think more .... 0 0 replyShare Somoshree Datta 5 commented Jan 20, 2019 reply Follow flag Magma Why is S1 right? 0 0 replyShare Magma commented Jan 20, 2019 reply Follow flag only S1 :3 ?? 0 0 replyShare Somoshree Datta 5 commented Jan 20, 2019 reply Follow flag if LHS is proper subset of candidate key, then isnt it violating 2NF? So it will be removed in 2 NF only right? 0 0 replyShare Somoshree Datta 5 commented Jan 20, 2019 reply Follow flag Answer is Both are true 0 0 replyShare Magma commented Jan 20, 2019 reply Follow flag R must consist proper subset of candidate key determines proper subset of some other candidate key. A ----> B A is a proper subset of candidate key B is a proper subset of candidate key Partial ----> Partial it's not BCNF but it's in 3NF ? 0 0 replyShare Magma commented Jan 20, 2019 reply Follow flag Answer is Both are true -_- 0 0 replyShare Somoshree Datta 5 commented Jan 20, 2019 reply Follow flag Partial ----> Partial ohh I overlooked this part :/ Ya so S1 is true. But S2? 0 0 replyShare Shubhanshu commented Jan 20, 2019 reply Follow flag Both are true. 0 0 replyShare Somoshree Datta 5 commented Jan 20, 2019 reply Follow flag Yup..got it :) thanks :) 0 0 replyShare himgta commented Jan 20, 2019 reply Follow flag @Somoshree Datta 5 @Shubhanshu explain the 2nd statement ..I m not getting it! 0 0 replyShare Shubhanshu commented Jan 20, 2019 i edited by Shubhanshu Jan 20, 2019 reply Follow flag @himgta Consider Relation $R(A,B,C)$ with FD's $\{ C \rightarrow A , A \rightarrow C, BC \rightarrow D \}$ have overlapped CK $AB, BC$. Now apply $S_2$ you will find it true. 1 1 replyShare garimanand commented Jan 20, 2019 reply Follow flag @Shubhanshu please give some example for s2 ? 0 0 replyShare Somoshree Datta 5 commented Jan 20, 2019 reply Follow flag https://gateoverflow.in/218551/normal-forms?show=219407 0 0 replyShare Shubhanshu commented Jan 20, 2019 reply Follow flag @garimanand check example in the comment. 0 0 replyShare garimanand commented Jan 21, 2019 reply Follow flag Thank you ,got it 0 0 replyShare ajay10 commented Jan 23, 2019 reply Follow flag this is one example, but in the S2 they have mentioned that it must have overlapped candidate keys. which i felt it need not be true because we can have so many example proving the S2 AS wrong. eg: R(ABCD) AB->CD, C->B in the given relation we don't need any overlapped keys but it is in 3NF but NOT IN BCNF. correct me , if my approach towards the question is wrong. 0 0 replyShare Shubhanshu commented Jan 23, 2019 reply Follow flag @ajay10 in your example we do have overlapped candidate keys those are $AB, AC$ where A is an overlapped prime attribute. 0 0 replyShare ajay10 commented Jan 23, 2019 reply Follow flag thank you ! i got it ,i didn't overlook it carefully 0 0 replyShare Please log in or register to add a comment.
