GA_Testseries

Question

is option B correct?
in ii) intersection of the two normalized relation is EF which is not key of either of the relations

how can option B be correct?

Comments

cn u tell how did u derive e from bc?

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first one is lossless since bc is common and bc-->e is possible

=> bc is a candidate key in bce

(

bc->bcd ( from c->df )

bc-> bcda(from bd->a)

bc ->bcdaef (from abc->def)

)

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B is the answer

as (i) is a lossless decomposition and (ii) is a lossy decomposition

Answer 1

CK ={ BC, CE}  so, PA = B,C,E

D1:  ABCDF and BCE 

ABC → DF, AC→ F, C→ DF, BD→ A goes to R(ABCDF) .CK = {BC}

Common attribute b/w them is BC and BC is a key in ABCDF. therefore, lossless.

D2: BCDEF and AEF

EF → B and C→ DF goes into R(BCDEF)  and no FD goes into AEF. 

common attribute b/w them is E and it is not a candidate key in any of the relations. therefore not lossless.

option B. only Decomposition 2 is not lossless.