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GO Classes DPP | GATE CS, DA | DBMS | Decomposition

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1 1 vote

Consider $R(A,B,C,D,E)$ with $F=\{A\to E,\ EC\to BD,\ D\to C\}$. The candidate keys are $AC$ and $AD$. Which decomposition correctly removes the $\text{2NF}$ violation?

  1. $R_1(A,E)$ and $R_2(A,B,C,D)$
     
  2. $R_1(A,B,E)$ and $R_2(C,D)$
     
  3. $R_1(A,C,E)$ and $R_2(A,B,D)$
     
  4. No decomposition is required

1 Answer

1 1 vote

The candidate keys are $AC$ and $AD$.

Therefore, $A$ is a proper subset of both candidate keys.

Now consider $A\to E$.

$E$ does not belong to either candidate key, so $E$ is non-prime.

Thus, $A\to E$ is a partial dependency of a non-prime attribute on a proper subset of candidate keys.

This violates $\text{2NF}$.

Separate this dependency into $R_1(A,E)$.

Keep $A$ in the remaining relation so that the decomposition is $R_2(A,B,C,D)$.

Hence, $R_1(A,E)$ and $R_2(A,B,C,D)$ correctly remove the 2NF violation.

Therefore, the correct answer is A.

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