Q7 Answer the following questions regarding schema refinement and normal forms.

Question

Q7 Answer the following questions regarding schema refinement and normal forms. Make sure that you explain your answer. No explanation means no points. (a) for R(A,B,C) with FDs A- B, B->C, can I say that A->C also holds? Explain your answer (b) for R(A,B,C) with FDs, A->B, B->C, is the decomposition of R into R1(A,B) and lossless-join (non-additive) decomposition? Explain why R2(B,C) lossless-join (non-additive) decomposition? Explain why. (c) for R(A,B,C) with FDs, A- B, B->C, is AB a key for R? Explain your answer (d) [4pts] for R(A,B,C) with FDs, A->B, B->C, sR in BCNF? Explain your answer.

Explanation / Answer

(a) Yes A-> C also holds because by A we can derive B and by B we can get C. Thus by A we can get C indirectly.

(b) 1. attribute(R1) Union attribute(R2) => attribute(R)

2.  attribute(R1) Intersection attribute(R2) != null

3. Common attribute B is key of relation R2

therefore deomposition of R into R1 and R2 is lossless-join decomposition.

(c) Yes AB is a key of R because by AB we can derive all attributes of R i.e. A,B,C

AB is here superkey not a candidate key because subset of AB i.e. A can also derive all attributes of R.

(d) Since A is the candidate key of relation R

So for R to be in BCNF A should be present at left side of each functional dependencies,but A here not present at left side of fd B->C .

Therefore R is not in BCNF.

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