2 votes 2 votes Consider the relation R(Name,courseNO,rolNO,grade) name,courseno ->grade rollno,courseno->grade name->rollno rollno->name Given answer as it is 3NF . But Im getting 1NF. is it correct ? Computer Networks database-normalization databases + – Dulqar asked Feb 1, 2017 Dulqar 940 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 6 votes 6 votes Yes it is 3NF. In this relation the candidate keys are (name,courseno) and (rollno,courseno). So the there are no partial or transitive F.Ds in the given relation. name->rollno rollno->name these two F.D s are not partial as both LHS and RHS contain only prime attributes. Kaushik.P.E answered Feb 1, 2017 selected Feb 1, 2017 by Sushant Gokhale Kaushik.P.E comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes Consider N ->Name , C-> courseNO , R-> rolNO , G->grade Candidate key : NC and RC N,C, R are prime attribute. NC -> G : BCNF RC -> G : BCNF N -> R : 3NF ( R is prime) R -> N : 3NF (N is prime) The relation is in 3NF. Arnab Bhadra answered May 24, 2017 Arnab Bhadra comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes This is definitely 3NF. But isn't it a BCNF too? rocker answered Apr 24, 2017 rocker comment Share Follow See 1 comment See all 1 1 comment reply vamp_vaibhav commented Sep 11, 2017 reply Follow Share no not in BCNF because name and roll no are not a candidate key...last two dependency will not allow BCNF 1 votes 1 votes Please log in or register to add a comment.
0 votes 0 votes So, highest normal form is 3NF srivivek95 answered Oct 23, 2017 srivivek95 comment Share Follow See all 0 reply Please log in or register to add a comment.