functional dependency

Question

1)lossless & dependency preserve

2)loss& dependency preserve

3)lossless & un preserved

4) loss & un preserved

Answer 1

R(ABCDEG)

AB->C 
AC->B
BC->A

AD->E
B->D

E->G

R1(ABC)   AB->C, AC->B, BC->A
R2(ABDE)  AD->E B->D 
R3(EG)    E->G

We got all dependencies back, hence this decomposition is Dependency Preserving.

Join R2 and R3, E is common b/w R2 and R3, and E is CK for R3(lossless condition satisfied)
=R23(ABDEG)  

now join R23 and R1, AB is common b/w both AB is CK of one relation( lossless condition satisfied)

R(ABCDEG)  

This decomposition is both lossless and D.P hence option (A) is correct.

Comments

i was just considering main primary key for all decompositions thts why I got option b tht was big mistake.

this question was from testbook series .during analysis of test i reported them for choosing b as a correct option .And the answer provided by them is wrong .Also I received mail that problem has resolve .

How can this possible ?this happens with me for 3 question , i know all 3 was  wrong from my side but after reporting all are correct

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may be the one who is correcting answers, himself doesn't know which one is correct, or there is a possibiliy that you are performing well in tests, so when you ask him, he changes the key.

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yes @manu00x they are just changing key not the reason for that answer ...well from now onwards  this will be a lesson for me ,not to blindly follow any of the reason 

Answer 2

Option 1 is correct.
Lossless and Dependency Preserving

Comments

no ,it should be option 2

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Would you explain why it is lossy decomposition.

Answer 3

lossless and dependency preserving

dependency preserving........

ABC                                        ABDE                                      EG

AB->C                                     AD->E                                      E->G

AC->B                                     B->D

BC->A

candidate  keys=AB,AC,BC                    primary key=AB                                                   primary key =E

LOSELESS..........

common attribute between ABC and ABDE= AB

AB is primary key for ABC

common attribute between ABDE and EG=E\

E is primary key for EG