0 votes 0 votes The below decomposition is lossless or lossy and also dependency preserving or not? Databases dependency-preserving lossless-decomposition + – night_fury asked Sep 23, 2018 night_fury 802 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply himgta commented Sep 23, 2018 reply Follow Share lossy and not dependency preserving? 1 votes 1 votes BASANT KUMAR commented Sep 23, 2018 reply Follow Share yes,it will be lossy as well not f.d preserving.if R1 combined with R2 there will be redundancy . 0 votes 0 votes night_fury commented Sep 25, 2018 reply Follow Share I understand that it is not dependency preserving.But how is it lossy? Please anyone explain in detail. 0 votes 0 votes BASANT KUMAR commented Sep 26, 2018 reply Follow Share for lossy decomposition we check whether the final combination of table contain some spurious tuple or not.if yes then the decomposition will be lossy. 0 votes 0 votes night_fury commented Sep 26, 2018 reply Follow Share Can anybody explain this by using chase algorithm? 0 votes 0 votes Shaik Masthan commented Sep 26, 2018 reply Follow Share @night_fury i don't know about that algorithm but i will give how to check is it lossy or loss less? First take Table1 ===> R1(A,B,C) list out all dependencies under this table by original FD set ==> { AB -> C } Key of this table = AB Table2 ===> R2(B,C,D) list out all dependencies under this table by original FD set ==> { C -> D } Key of this Table2 = BC Table3 ===> R3(C,D,E) list out all dependencies under this table by original FD set ==> { C -> D } Key of this Table3 = CE take the intersection of attributes of Table1 and Table2 ==> BC, it is a key(super key or primary key) in R2, therefore you can combine Table1 and Table2 . let denote it as T12. it's key is AB take the intersection of attributes of T12 and Table3 ==> CD, it is neither a key in T12 nor R13 ==> you can combine Table12 and Table3 . ===> it is lossy. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes Lossy Decomposition because no common attribute is a candidate key here YashShah216 answered May 7, 2023 YashShah216 comment Share Follow See all 0 reply Please log in or register to add a comment.