29 votes 29 votes State True or False with reason There is always a decomposition into Boyce-Codd normal form (BCNF) that is lossless and dependency preserving. Databases gate1994 databases database-normalization easy true-false + – Kathleen asked Oct 5, 2014 recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 6.5k views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply yagneshmb commented Jan 6, 2021 reply Follow Share Can someone tell me what will be the decomposition here if we were to convert into BCNF, even if it is not dependency preserving. Can it be lossless and BCNF? 0 votes 0 votes Shiva Sagar Rao commented Jan 19, 2021 i edited by Shiva Sagar Rao Jan 19, 2021 reply Follow Share False. Let R(A,B,C,D) is a relation. F.D’s: {AB→C, C→AD} For this example, there is no lossless join, dependency preserving BCNF decomposition. For example, let’s decompose it as R1(A, B, C) and R2(C, D). This preserves all dependencies and the join is lossless too, but the relation R1 is not in BCNF. In R1 we keep ABC together otherwise preserving {AB→C} will fail, but doing so also causes {C→A} to appear in R1. {C→A} violates the condition for R1 to be in BCNF as C is not a superkey. relations formed after decomposition should be in BCNF is not satisfied here. Source: https://gateoverflow.in/741/gate2001-2-23?show=11812#a11812 0 votes 0 votes Please log in or register to add a comment.
Best answer 43 votes 43 votes False BCNF decomposition can always be lossless, but it may not be always possible to get a dependency preserving BCNF decomposition. Jarvis answered May 27, 2015 edited Apr 23, 2021 by Lakshman Bhaiya Jarvis comment Share Follow See all 0 reply Please log in or register to add a comment.
7 votes 7 votes bcnf decomposition guarantees lossless and d.p may not is correct one vijju answered Oct 6, 2016 vijju comment Share Follow See all 0 reply Please log in or register to add a comment.
6 votes 6 votes FALSE, as BCNF guarantee only lossless property. rishu_darkshadow answered Oct 6, 2017 rishu_darkshadow comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes false not every relation can decompose into BCNF with dependency preserving and lossless. you can see here: relation R is not possible to decompose into BCNF with dependency preserving and lossless. R(ABC) F={AB->C , C->A } rajoramanoj answered Aug 27, 2017 rajoramanoj comment Share Follow See all 3 Comments See all 3 3 Comments reply Puja Mishra commented Jan 18, 2018 reply Follow Share Bt example is wrong !!! 0 votes 0 votes sutanay3 commented Apr 21, 2018 reply Follow Share An example for it may be R(ABCDE) FD:(AB->CDE, C->A, D->E ).It is lossless but not fd preserving 3 votes 3 votes rohith1001 commented Oct 17, 2019 reply Follow Share The example is correct. It is in 3NF. Here the possible candidate keys are AB and BC. AB -> C (In this FD, AB is a super key. Hence in 3NF) C -> A (In this FD, A is a prime attribute. Hence in 3NF) 3NF: if X -> Y ,then X must be a super key or Y must be a prime attribute. BCNF: if X -> Y ,then X must be a super key. 2 votes 2 votes Please log in or register to add a comment.