4 votes 4 votes If A Relation R(A,B,C,D) and CK given is (AB,C) and FD is given as A->C Then the relation is in which highest normal form? Databases candidate-key databases database-normalization + – Anirudh Pandey asked Sep 19, 2017 Anirudh Pandey 1.3k views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments Shubhanshu commented Sep 19, 2017 reply Follow Share I think the candidates key should be {A,C} because of FD A->C not {AB, C} 0 votes 0 votes srestha commented Sep 19, 2017 reply Follow Share @Anirudh U mean there must be one dependency ABC->D So, there should not be any partial dependency As both dependency from prime to prime attribute. Partial dependency valid for prime to non prime attribute So, in 3NF but not in BCNF as here D is not a prime attribute 0 votes 0 votes Rishi yadav commented Oct 21, 2017 i edited by Rishi yadav Oct 21, 2017 reply Follow Share No A and C are not Candidate Key Don't assume candidate key We have candidate key which is given But if u need to find candidate key then it should be ABD and it should be 1NF because it follows 3NF condition but not 2NF Correct me if i wrong 0 votes 0 votes Please log in or register to add a comment.
2 votes 2 votes 2NF because there is no FD for which the nonprime attribute is not fully functionally dependent on the primary key. 3NF because there is no FD for which a nonprime attribute is transitively dependent on primary key. NOT 4NF because for every non-trivial FD LHS must be a super key which is not in this case Tuhin Dutta answered Sep 19, 2017 Tuhin Dutta comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes 3 nf..... as candidate keys are ab and c so all the functional dependency would be like candidate key to non prime attributes or prime attributes to prime attributes. bandana bharti answered Sep 19, 2017 bandana bharti comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes If the candidate keys are AB, C then the dependencies would be AB->CD C->ABD And A->C (Given) Now here every non prime attribute(D) is non transitively dependent on EVERY Key attribute of R i.e. (AB , C) Therefore it is in 3NF Abbas2131 answered Sep 24, 2017 Abbas2131 comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes 1NF because no transitive partial dependency and Not 2NF Rishi yadav answered Oct 9, 2017 edited Oct 21, 2017 by Rishi yadav Rishi yadav comment Share Follow See all 5 Comments See all 5 5 Comments reply Show 2 previous comments Shubhanshu commented Oct 21, 2017 reply Follow Share CK {AB, C} FD is given as A->C since, A determine C without B, so AB should not be a candidate key instead it is super key. Consider R(A,B,C,D) with FD's = {A->C, B-> D, C->ABD} In this relation C is CK and since A->C, A is also the CK, this relation is in 2NF not in 3NF. 0 votes 0 votes Rishi yadav commented Oct 21, 2017 i edited by Rishi yadav Oct 21, 2017 reply Follow Share Candidate Key: The minimal set of attribute which can uniquely identify a tuple is known as candidate key Reference :-http://www.geeksforgeeks.org/dbms-keys-candidate-super-primary-alternate-and-foreign/ R(A,B,C,D) A->C I m assuming u r right If A is candidate key closure( A )should be ABCD but A+ = AC so A should not be key that's why candidate key is given But Suppose for ur choice if We find Candidate key Then it should be ABD not A because (ABD)+ = ABCD so According to ABD , prime attributes are A,B,D Lets Find Highest Normal Form now :- 1-Not BCNF because A is not key 2-Non prime Attribute is C and C is not dependent on Non Prime attribute so 3 NF but it is not following 2NF condition so 1NF Hence 1 NF @Shubhanshu 0 votes 0 votes Shubhanshu commented Oct 21, 2017 i edited by Shubhanshu Oct 21, 2017 reply Follow Share but A+ = AC so A should not be key that's why candidate key is given But Suppose for ur choice if We find Candidate key It is not, Closure(A) = {A,C} But it should be, Closure(A) = {A UNION Closure(C)} Beside, this question we usually use the above method to find the candidate keys of Relation. Suppose consider a relation, R(A,B,C,D,E) A->B, B->C, C->D, D->E. then closure(A) = { A UNION closure(B)} from below Closure(A) = {A, B, C, D, E} Closure(B) = {B UNION Closure(C)} from below Closure(B) = {B, C, D, E} Closure(C) = {C UNION Closure(D) from below Closure(C) = {C ,D, E} Closure(D) = {D UNION Closure(E)} from below Closure(D) = {D, E} Closure(E) = {E} this is how we calculate the closure, Coming to given question again, since, A -> C Closure(A) = {A UNION Closure(C)} Now, C is the CK of the relation so definitely its should be {A, B, C, D} and hence the Closure(A) = {A, B, C, D} That's why I am saying A is also CK of the relation and AB is not CK it is SK. Since, A and C are the only CK of the relation R, which are also simple CK { CK with one attribute}, So, R is in 2NF for sure, may or may not in 3NF. 0 votes 0 votes Please log in or register to add a comment.