0 votes 0 votes Consider a relation $R$$\left ( A, B,C,D, E \right )$ with the following $FD$ set $F$: $A\rightarrow BC$ $CD\rightarrow E$ $B\rightarrow D$ $E\rightarrow A$ The canonical cover of the above $FD$ set is: $A\rightarrow BC$, $C\rightarrow E$ , $B\rightarrow D$ , $E\rightarrow A$ $A\rightarrow BC$ , $CD\rightarrow E$ , $B\rightarrow D$ , $E\rightarrow A$ $A\rightarrow B$ , $CD\rightarrow E$ , $BC\rightarrow D$ , $E\rightarrow A$ $A\rightarrow BC$ , $B\rightarrow E$ , $B\rightarrow D$ , $E\rightarrow A$ GATE tbb-mockgate-3 databases database-normalization canonical-cover + – Bikram asked Feb 9, 2017 retagged Sep 15, 2020 by ajaysoni1924 Bikram 321 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 1 votes 1 votes No extra attributes and no FD is redundant in the given FD set, hence its canonical cover is given FD set itself . Bikram answered Mar 19, 2017 selected Sep 8, 2019 by Bikram Bikram comment Share Follow See all 2 Comments See all 2 2 Comments reply bittu commented Jan 29, 2018 reply Follow Share @Bikram sir, i think (A) is also canonical cover. plz check it 1 votes 1 votes Manoja Rajalakshmi A commented Dec 12, 2018 reply Follow Share @bittu In option A, $C$ closure gives $EABC$ but in the given FD set, $C$ closure gives $C$ only. A canonical cover of a set of functional dependencies $F$ is a simplified set of functional dependencies that has the same closure as the original set $F$. 0 votes 0 votes Please log in or register to add a comment.