0 votes 0 votes R (A, B, C) Functional dependency= { A -> B, B -> C , C -> A } No of super keys are...?? dhairya asked Jun 18, 2016 dhairya 885 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 4 votes 4 votes 7 all possible combination 2^3 -1 (A,B,C,AB,AC,ABC,BC) Sanjay Sharma answered Jun 18, 2016 selected Jun 18, 2016 by rude Sanjay Sharma comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments Sanjay Sharma commented Jun 18, 2016 reply Follow Share s.key is any attribute or combination of attribute that uniquely identify a row . so each c.k is a s.k in addition any attribute added to c.k will also become s.k here a,b,c are c.k so they are also s.k now a with b can also ensure uniqueness and so is a and c and similarly a,b,c and b,c too hence ans should be 7 0 votes 0 votes dhairya commented Jun 18, 2016 reply Follow Share OK 0 votes 0 votes rameshbabu commented Jun 18, 2016 reply Follow Share see what is the definition of super key, it is any set of attributes which can uniquely identify remaining attributes.............correct.! not what is candidate key, it is minimal super key, not what do you mean my minimal, suppose ABCD is super key, but AB is sufficient to uniquely identify remaining att. so we can say AB is minimal version of ABCD. and also AB is candidate key. like AB there can be many candidate keys for relation, out of which we select one as primary key. by doing it reverse, that mean my adding any attribute to candidate key, we get super key wow.....that was too much theory.... 0 votes 0 votes Please log in or register to add a comment.
4 votes 4 votes ck for this relation are A,B,C no. of super key with A= 2^(n-1)= 2^(3-1) = 4 similarly with B again 4 super keys and with C also 4 super keys with AB it is 2^(n-2) similarly with BC and with CA it is 2^(n-2) with ABC it will be 2^(n-3) so, total no. of super keys= 2^(n-1) + 2^(n-1) + 2^(n-1) - 2^(n-2) -2^(n-2) -2^(n-2) + 2^(n-3) =7 where n=3 which is no.of attributes in relation cse23 answered Jun 18, 2016 edited Jun 18, 2016 by cse23 cse23 comment Share Follow See all 2 Comments See all 2 2 Comments reply dhairya commented Jun 18, 2016 reply Follow Share why n=4..?? n should be 3 as R (A, B, C)...?? 0 votes 0 votes cse23 commented Jun 18, 2016 reply Follow Share oops...ryt..typo mistake :) N is 3 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes there are 3 candiadate key a,b,c so all possible combination will be a,b,c,ab,ac,cb,abc rio answered Jun 18, 2016 rio comment Share Follow See all 3 Comments See all 3 3 Comments reply dhairya commented Jun 18, 2016 reply Follow Share According to formula SK=2{no of attribute - size of candidate key} here no of attribute = 3 and candidate key = closure of A={A, B , C} closure of B={ B , C, A} closure of c={ C,A,B } So candidate key=3 ryt..?? Accoring to formula, 2{3-3}= 1 so super key should be 1 ryt...? 0 votes 0 votes rio commented Jun 18, 2016 reply Follow Share yr logically soch na candidate key kya hoti h ....minimal superkey such that it contains only one candiate key...so here we have 3 candidate key ...so every combination of a,b,c will result in a superkey think ...u can distiguish a row in databse using a,b,c, ab,bc,ca,abc 0 votes 0 votes dhairya commented Jun 18, 2016 reply Follow Share ya ok...i got it now...i think so... 0 votes 0 votes Please log in or register to add a comment.