0 votes 0 votes consider R(A,B,C,D,E) with FDs AB->C,C->D,D->B,D->E if number of keys in relation R is ‘a’ and the number of relation in 3NF decomposition is ‘b’ what is the value of a-b? i am getting a=3,b=4 a-b=-1 am i correct? Gate Fever asked Jan 20, 2019 Gate Fever 266 views answer comment Share Follow See 1 comment See all 1 1 comment reply aditya333 commented Jan 20, 2019 reply Follow Share can superkeys be considered? 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Candidate keys: AB,AD,AC AB->C, (Full) C->D, (key) D->B, (key) D->E (Partial) Decompose into ABCD, DE ABCD (keys: A,B,C,D) AB->C, (Full) C->D, (key) D->B, (key) when in a relation all the attributes are prime then the minimum normal form the relation satisfies is 3NF DE D->E (Full) So the no. of Relations needed will be 2. a=3, b=2, So the answer will be 1. balchandar reddy san answered Jan 20, 2019 balchandar reddy san comment Share Follow See all 2 Comments See all 2 2 Comments reply Gate Fever commented Jan 20, 2019 reply Follow Share i also got a=3 but how u got b?? pls explain! 0 votes 0 votes balchandar reddy san commented Jan 20, 2019 reply Follow Share when in a relation all the attributes are prime then the minimum normal form the relation satisfies is 3NF. ABCD (keys: A,B,C,D) AB->C C->D D->B In this relation all the attributes are prime/key, so the Relation will be in 3-NF. 0 votes 0 votes Please log in or register to add a comment.