0 votes 0 votes Databases databases er-diagram made-easy-test-series + – dekabh asked Oct 28, 2018 edited Mar 3, 2019 by Aditi Singh dekabh 632 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Soumya Tiwari commented Oct 28, 2018 reply Follow Share Is it 2?What's the answer given?? 0 votes 0 votes Gurdeep Saini commented Oct 29, 2018 reply Follow Share here both side have total participation(given in the question) case 1 . lets suppose it is one to one or one to many (which is not given in the question ) we make a table (A,B,C,D,E,F) and ACD form a key but it have two partial dependency so to make 2NF we make table lets us try to make two table (A, B, D, E, F) Here PK is AD and (A,C). but table 1 also have partial dependency i.e. A-->B so have to make more table (A,B) (D,E,F,A) (A,C) in this no table have partial dependency and in table (D,E,F,A) A is forgien key case 2 many to many relation in this case we have to make 4 table ( A,B)(A,C)(A,D)(D,E,F) 1 votes 1 votes Please log in or register to add a comment.
1 votes 1 votes The question is given in made easy test series. They have given the correct answer as 4. I am not convinced with the explanation. As per my understanding, I got the answer as 2. Your review is needed here. dekabh answered Oct 28, 2018 dekabh comment Share Follow See all 9 Comments See all 9 9 Comments reply Show 6 previous comments Shaik Masthan commented Oct 29, 2018 reply Follow Share But by this approach we can't guarantee the loss-less property, So go with (A,B) and (D,E,F),(A,D) And (A,C) -----> minimum 4 tables required. 0 votes 0 votes Gurdeep Saini commented Oct 29, 2018 reply Follow Share @shaik masthan you said your above comment do not guarantee lossless property how? 0 votes 0 votes Shaik Masthan commented Oct 29, 2018 reply Follow Share actually in the 1 NF, itself that process is going to fail loss-less property ! from the E-R, R1(A,B,C), R2(A,D) and R3(D,E,F) but in R1 C is multi-valued For removing this, we make AC as a key in R1. given relationship in ER is M:N ===> we make AD as a key in R2. as normally D is the key in R3. the tables look like R1(A,C,B), R2(A,D) and R3(D,E,F) by observing R2 and R3, common attribute is Key ===> it can combine R23(A,D,E,F) ∴ in 1 NF ===> R1(A,C,B) and R23(A,D,E,F) ===> 2 tables required R1(A,C,B) have the FD A --> B R23(A,D,E,F) have the FD D --> E,D -->F For satisfying 2 NF, we don't have PFD R1(A,C,B) divided as R11(A,B),R12(A,C) R2(A,D,E,F) divided as R21(D,E,F),R22(A,D) ∴ in 2 NF ===> R11(A,B),R12(A,C) and R21(D,E,F),R22(A,D) ===> 4 tables required 2 votes 2 votes Please log in or register to add a comment.