3. Perform the test for the non-additive join property (lossless join) for the r
ID: 3603678 • Letter: 3
Question
3. Perform the test for the non-additive join property (lossless join) for the relation R(Ai, A2, A3, A4, As), and the decompositions Da, Db, De, Dd and set of functional dependencies F givern below. You can ignore attributes that are not mentioned in each particular subsection (e.g., you can ignore absence of A4 in Da, just test the join between Ri and R2): D, = { Ri(Ai, Az) , R2(A, A4, As) } D, = { Ri(Al, As) , R2(A4, As) } Da = { Ri(Ai, Az, A3) , R2(Ai, A2, As) } a. Does the decomposition D, have the non-additive join property? Explain why or why not. b. Does the decomposition Db have the non-additive join property? Explain why or why not c. Does the decomposition De have the non-additive join property? Explain why or why not. d. Does the decomposition Da have the non-additive join property? Explain why or why not.Explanation / Answer
For givem relation R(A,B,C,D,E)
F:{B ->D,CD ->E,E->A,A->BC}
Is it possible 2 decompose R into two relations R1and R2 such that
Decomposition is lossless-join ,dependency-preserving and in BCNF normal form ??
In such question we try to find candidate key first and hence prime and non prime attributes..But for knowing about BCNF , knowing candidate key is sufficient as left hand side of the FD has to be superkey for BCNF to hold.
Given FD set : F:{B ->D,
CD ->E,
E->A,
A->BC}
Hence we have candidate keys : { CD , BC , A , E }
So the FD B --> D is creating problem..So the way to remove is to separate it..We do this as :
B+ = BD
So let we have relation BD which contains the FD : B --> D .Other possible FD is D --> B..But the attribute closure of D w.r.t. original FD set gives D only, so we cannot add it in implied FD of the subrelation.
Hence let R1( B , D ) be subrelation with B as the superkey..
Now to have lossless property we place A , C , E in other subrelation as well as attribute B as for
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