3 votes 3 votes A relation $R$ with $5$ attributes $A_1,A_2,A_3,A_4,A_5$. Given the following FDs $A_1\rightarrow A_2$ $A_2A_3\rightarrow A_5$ $A_4A_5\rightarrow A_1$ the number of candidates keys that includes attribute $A_3$ is ________________. Databases databases database-normalization + – prathams asked Dec 27, 2015 edited Mar 30, 2016 by ibia prathams 947 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
3 votes 3 votes 3, A3A4A1,A3A4A2,A3A4A5 minal answered Dec 27, 2015 minal comment Share Follow See all 3 Comments See all 3 3 Comments reply prathams commented Dec 27, 2015 reply Follow Share Can you explain a bit how you derived it? 0 votes 0 votes minal commented Dec 27, 2015 reply Follow Share Given R(A1,A2,A3,A4,A5 ) Now by seeing FD , you can derive A2,A5,A1 ... so remaining attribute should also be in candidate key ... So A3A4 must be in ck .. but A3A4 + (closure of A3A4) = A3A4 , only so its not candidate key bcoz it does not derive all attribute of relation now Consider A3A4A1 +(closure of A1A3A4)= A3,A4,A1,A2,A5 SO IT derive all attribute of relation similar A3A4A2 and A3A4A5 also derive all attribute ... 5 votes 5 votes prathams commented Dec 27, 2015 reply Follow Share Thank you sonam, that helps. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes A3 and A4 is independent attribute. So A3 and A4 must be in Candidate Key. (A3A4)+ = A3A4 (A1A3A4)+ = A1A2A3A4A5 (A2A3A4)+ = A1A2A3A4A5 (A5A3A4)+ = A1A2A3A4A5 Answer is 3. Arnab Bhadra answered May 23, 2017 Arnab Bhadra comment Share Follow See all 0 reply Please log in or register to add a comment.