8. Consider the universal relation R = {a, b, c, d, e} and the set of functional

Question

8. Consider the universal relation R = {a, b, c, d, e} and the set of functional dependencies F D = {{a, c} {b, d, e}, {b} {e}, {c} {d}}.

8a. What schema(s) would be produced by the 3NF decomposition algorithm? (Can be more than one)

A. R1={a,b,c,d}

B. R1={b,e}, R2={c,d}, R3={a,b,c}

C. R1={b,e}, R2={c,d}, R3={a,c}

D. none of the above

8b. What schema(s) would be produced by the BCNF decomposition algorithm? (Can be more than one)

A. R1={a,b,c,d}

B. R1={b,e}, R2={c,d}, R3={a,b,c}

C. R1={b,e}, R2={c,d}, R3={a,c}

D. none of the above

Explanation / Answer

BCNF because it has no functional dependency.

8(A) . A. R1={a,b,c,d}

8(B) C. R1={b,e}, R2={c,d}, R3={a,c}

A relation is in 3NF if it is in 2NF and no non-prime attribute transitively depends on the primary key. In other words, a relation R is in 3NF if for each functional dependency X A in R at least one of the following conditions are met:

3NF and BCNF is that in BCNF it is not present the second condition of the 3NF

{a,c} Super Key

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