2 2 votes Which statement correctly describes the guarantee of the standard BCNF decomposition procedure?Every relation has a decomposition that is simultaneously BCNF, lossless and dependency preserving. Every relation can be decomposed losslessly into BCNF, but some functional dependencies may no longer be preserved. BCNF decomposition always preserves dependencies, but may be lossy. If dependency preservation is impossible in BCNF, then a lossless dependency-preserving $3$NF decomposition is also impossible. Databases goclasses goclasses-da-dpp goclasses-da-dpp-day-278 goclasses-cs-dpp goclasses-cs-dpp-day-376 databases goclasses-databases-practice-questions bcnf-decomposition + – GO Classes 101 views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
1 1 vote The standard BCNF decomposition algorithm uses a BCNF-violating dependency to repeatedly split relations.Each such split is constructed to be lossless.Therefore, a lossless decomposition into BCNF relations can always be obtained.However, some original FDs may involve attributes that become separated across different relations.Those dependencies may then require a join to verify and are therefore not dependency preserved.Thus BCNF guarantees:BCNF + lossless join but not necessarily dependency preservation.So A is false and B is correct.C reverses the actual guarantee and is false.For D, the $3$NF synthesis algorithm specifically exists to guarantee a decomposition that is:in $3$NFlosslessdependency preservingfor every relation.$\therefore$ D is false. GO Classes answered Sep 19 GO Classes comment Share Follow 0 reply Please log in or register to add a comment.