2 2 votes A relation $R(ABCDEFGHIJ)$ with functional dependency set $F={AB\rightarrow C , A\rightarrow DE , B\rightarrow F , F\rightarrow GH, D\rightarrow I J}$ , and the decomposition of $R$ is ${R1(ABCD) , R2(DE) ,R3(BF),R4(FGH),R5(DIJ)}.$ Which of the following is/are true? $1.$lossless $2.$lossy $3.$Not dependency preserving $4.$Dependency preserving Databases databases database-normalization dependency-preserving virtual-gate-test-series + – Turning Turing 1.4k views answer comment Share Follow Print See all 2 Comments 2 2 Comments reply himgta commented Aug 28, 2018 reply Follow flag It is not dependency preserving......A-->DE is not preserved....check again! 0 0 replyShare Ananya Jaiswal 1 commented Aug 28, 2018 reply Follow flag yeah thanks i calculated wrong closure. its not dependency preserving as A->E is not preserved. A->D is preserved in R1 0 0 replyShare Please log in or register to add a comment.
Best answer 3 3 votes A->DE FD is not satisfied in any Relation so it is not dependency preserving In R2(DE) D is not a key attribute so when combined with other relations it gives a lossy So 2.lossy 3.Not dependency preserving are true Pavan Kumar Munnam answered Sep 25, 2016 • selected Sep 25, 2016 by Turning Turing Pavan Kumar Munnam comment Share Follow See all 2 Comments 2 2 Comments reply Turning Turing commented Sep 25, 2016 reply Follow flag Yes I also thought so but the answer says lossless and dependency preserving 0 0 replyShare Prateek kumar commented Nov 2, 2016 reply Follow flag A->DE you can write it A->D which we can combine in R1(ABCD) NOW A->E doesn't combine in R2(DE) so not dependency preserving and lossy join 1 1 replyShare Please log in or register to add a comment.
0 0 votes Not dependency preserving as A->DE is not preserved lossy as in R2(DE) D is not CK 2,3 are correct Shreyshi Pandey answered Oct 31, 2019 Shreyshi Pandey comment Share Follow 0 reply Please log in or register to add a comment.