0 votes 0 votes ben10 asked Aug 19, 2018 ben10 517 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Two minimal covers are possible : A->B, B->C, C->A AND A->C, C->B , B->A Shiv Gaur answered Aug 19, 2018 Shiv Gaur comment Share Follow See all 7 Comments See all 7 7 Comments reply ben10 commented Aug 19, 2018 reply Follow Share Can you please provide the procedure ? 0 votes 0 votes Shaik Masthan commented Aug 19, 2018 reply Follow Share @Shiv Gaur 1) A->C, B->C,C->AB 2) A->C,B->A,C->AB aren't these are minimal covers? 0 votes 0 votes Iqra Islam commented Aug 19, 2018 reply Follow Share In ypour 2nd Minimal cover , A->C,B->A,C->AB It will be A->C,B->A,C->B as from C->B if we find closure of C we will get {C,B,A}. Plus we can have more minimal covers: 1) A->BC, B->A,C->A 2) A->B,B->AC,C->B 0 votes 0 votes Shiv Gaur commented Aug 20, 2018 reply Follow Share While identifying the minimal cover three steps should be followed 1: Split the given FDs such that RHS contain only single attribute. 2: Identify the extraneous attribute in LHS and remove it. 3: Eliminate the redundant FD. 0 votes 0 votes Shaik Masthan commented Aug 20, 2018 reply Follow Share @Shiv Gaur, yes procedure is correct... but my question is A->C, B->C,C->AB A->BC, B->A,C->A ( As Iqra Islam said ) A->B,B->AC,C->B isn't these are minimal? i didn't find any redundant term or FD in that. 0 votes 0 votes Shiv Gaur commented Aug 20, 2018 reply Follow Share I don't know the exact reason but I guess , when we talk about minimal cover of a given FD F, It is a minimal set of FDs that is equivalent to the given FD F. Here {A->C, B->C,C->AB} covers the given F but it could be further split to {A->C, B->C,C->A, C->B} ie 4 FDs and we already have A->B, B->C, C->A and A->C, C->B , B->A ie 3 FDs that's why we split the RHS in first step 0 votes 0 votes Shaik Masthan commented Aug 20, 2018 reply Follow Share @Shiv Gaur, Those are also SHOULD BE minimal covers, note that minimal covers are not unique. compare this in DIGITAL LOGIC all those are irreducible, but which you mentioned are minimal expression but in DBMS we don't have two terminologies... therefore all are minimal covers only 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes Two minimal covers are possible. vikas999 answered Aug 19, 2018 vikas999 comment Share Follow See 1 comment See all 1 1 comment reply Shaik Masthan commented Aug 20, 2018 reply Follow Share @vikas999, yes procedure is correct... but my question is A->C, B->C,C->AB A->BC, B->A,C->A A->B,B->AC,C->B isn't these are minimal? i didn't find any redundant term or FD in that. 0 votes 0 votes Please log in or register to add a comment.