0 votes 0 votes R Decomposed into R1(BCD) and R2(CA) Sanjay Sharma asked Apr 24, 2017 • edited Apr 25, 2017 by Sanjay Sharma Sanjay Sharma 1.1k views answer comment Share Follow See all 17 Comments See all 17 17 Comments reply akash.dinkar12 commented Apr 25, 2017 reply Follow Share @sanjay sir, is there not any other information given????? 0 votes 0 votes Sanjay Sharma commented Apr 25, 2017 reply Follow Share oh yes adding 0 votes 0 votes Prashant. commented Apr 25, 2017 reply Follow Share Both dependency preseving and lossless 0 votes 0 votes akash.dinkar12 commented Apr 25, 2017 reply Follow Share how dependency preserving prashant sir??? 0 votes 0 votes akash.dinkar12 commented Apr 25, 2017 reply Follow Share AB --> CD, how will preserve this dependency??? 0 votes 0 votes Prashant. commented Apr 25, 2017 reply Follow Share See it is indirectly preserve we know A $\rightarrow$ C and C $\rightarrow$A , AB $\rightarrow$ CD can be BC$\rightarrow$ CD looks like we lost AB -> CD but since R2(AC) makes BC -> CD backs to AB -> CD by replacing A -> C 0 votes 0 votes akash.dinkar12 commented Apr 25, 2017 reply Follow Share not getting yet??? 0 votes 0 votes Prashant. commented Apr 25, 2017 reply Follow Share R2(BCD) will have BC -> CD yes or no? 0 votes 0 votes akash.dinkar12 commented Apr 25, 2017 reply Follow Share 100 %..yess...then 0 votes 0 votes Prashant. commented Apr 25, 2017 reply Follow Share Now R2(BCD) BC -> CD And R1(AC) A->C and C->A [use blue part] combination will preserve AB -> CD 0 votes 0 votes akash.dinkar12 commented Apr 25, 2017 reply Follow Share thanx sir , i got it... 1 votes 1 votes akash.dinkar12 commented Apr 25, 2017 reply Follow Share sir , for checking lossy or lossless?? have u used algorithm or any other thing??? 0 votes 0 votes Prashant. commented Apr 25, 2017 reply Follow Share dont use sir :P for lossless just check common attribute is candidate of of of the relation or not? 1 votes 1 votes akash.dinkar12 commented Apr 25, 2017 reply Follow Share ok...ok,,fine prashant 0 votes 0 votes student2018 commented May 2, 2017 reply Follow Share I dont think Dependency preserved as R2 has A and R1 has B so AB->CD is not preserved 0 votes 0 votes student2018 commented May 2, 2017 reply Follow Share i think option c is correct 0 votes 0 votes akash.dinkar12 commented May 2, 2017 reply Follow Share plz read all comments, i think u will get the answer.. 1 votes 1 votes Please log in or register to add a comment.