2 votes 2 votes Consider relation R(A,B,C,D,E,F,G) with the following functional dependencies AB$\rightarrow$CD , D$\rightarrow$B, AF$\rightarrow$D, DE$\rightarrow$F, C$\rightarrow$G, F$\rightarrow$E, G$\rightarrow$A. What is the highest normal form. Databases database-normalization + – Rajat Agrawal007 asked Dec 20, 2018 Rajat Agrawal007 3.2k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply arvin commented Dec 20, 2018 reply Follow Share 3NF.. 0 votes 0 votes Gaurav Kadyan commented Nov 10, 2023 reply Follow Share In this how could we find all candidate keys easily 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes Here $AF,CF,GF,ADE,ABE$ are some of the candidate keys $\Rightarrow$ $A,B,C,D,E,F,G$ all are primary attributes. $D\rightarrow B$ is given in relation and $D$ is not a Super key $\Rightarrow$ The given relation is not in BCNF. In the given relation since all in the relation the R.H.S side attributes are primary attributes $\Rightarrow$ The given relation is in 3NF. Hence the relation is in 3NF. Satbir answered Dec 20, 2018 edited Nov 16, 2019 by Satbir Satbir comment Share Follow See all 6 Comments See all 6 6 Comments reply nishant_magarde commented Dec 21, 2018 reply Follow Share AF is also a primary key...... Can you please tell how to find the candidate key 0 votes 0 votes Satbir commented Dec 21, 2018 i edited by Satbir Feb 13, 2019 reply Follow Share Minimal super key is a candidate key. for eg:- AGBF is a super key coz we can derive all attributes using it i.e. $(AGBF)^{+}$ = {ABCDEFG} but it is not a candidate key because its subset ABF can also derive all attributes i.e.$ (ABF) ^{+}$ = {ABCDEFG} now ABF is also not a candidate key because its subsets AF could derive all attributes. i.e. $ (AF) ^{+}$ = {ABCDEFG} now AF is a candidate key because its subsets A or F could not derive all attributes. i.e. $ (A) ^{+}$ != {ABCDEFG} and $ (F) ^{+}$ != {ABCDEFG} ---------------------------------------------------------------------------------------------------------------------------------------------------- For deriving candidate keys u have to check for all combinations of attributes such that its closure gives the set of all attributes present in the relation and its superset is not a super key or candidate key i.e. if we know that AF is a candidate key then we do not need to check for AGF or ABFG which is its superset. 1 votes 1 votes Arjun commented Nov 16, 2019 reply Follow Share Why B is not a prime-attribute? 0 votes 0 votes Satbir commented Nov 16, 2019 reply Follow Share bcoz it is not part of any candidate key. 0 votes 0 votes Arjun commented Nov 16, 2019 reply Follow Share ABE is not a candidate key? 0 votes 0 votes Satbir commented Nov 16, 2019 reply Follow Share it is, my mistake. 0 votes 0 votes Please log in or register to add a comment.