Consider the relation $R(P, Q, S, T, X, Y, Z, W)$ with the following functional dependencies. $$PQ \rightarrow X;\ P \rightarrow YX;\ Q \rightarrow Y;\ Y \rightarrow ZW$$ Consider the decomposition of the relation $R$ into the constituent relations according to the following two decomposition schemes. $$D_1 :\quad R = [(P, Q, S, T);\ (P, T, X);\ (Q, Y);\ (Y, Z, W)]$$ $$D_2 :\quad R = [(P, Q, S);\ (T, X);\ (Q, Y);\ (Y, Z, W)]$$ Which one of the following options is correct?

A decomposition of relation $R$ into $R_1, R_2, \dots, R_k$ is lossless if we can successively join pairs of subrelations such that the common attributes form a superkey of at least one of the subrelations.

### Analysis of Decomposition $D_1$:Constituent subrelations: $R_1(P, Q, S, T)$, $R_2(P, T, X)$, $R_3(Q, Y)$, $R_4(Y, Z, W)$.

1. Join $R_3(Q, Y)$ and $R_4(Y, Z, W)$:

   • Common attribute: $R_3 \cap R_4 = \{Y\}$.

   • Given $Y \rightarrow ZW$, $Y$ is a candidate key for $R_4$.

   • Thus, $R_3 \bowtie R_4 = R_{34}(Q, Y, Z, W)$ is a lossless join.

2. Join $R_1(P, Q, S, T)$ and $R_2(P, T, X)$:

   • Common attributes: $R_1 \cap R_2 = \{P, T\}$.

   • Since $P \rightarrow YX \implies P \rightarrow X$, it follows that $PT \rightarrow X$. Hence, $\{P, T\}$ is a superkey of $R_2$.

   • Thus, $R_1 \bowtie R_2 = R_{12}(P, Q, S, T, X)$ is a lossless join.

3. Join $R_{12}(P, Q, S, T, X)$ and $R_{34}(Q, Y, Z, W)$:

   • Common attribute: $R_{12} \cap R_{34} = \{Q\}$.

   • Attribute closure of $Q$ under the given FDs: $Q^+ = \{Q, Y, Z, W, X\}$.

   • Therefore, $Q \rightarrow \{Q, Y, Z, W\}$, meaning $Q$ is a superkey of $R_{34}$.

   • Thus, $R_{12} \bowtie R_{34} = R(P, Q, S, T, X, Y, Z, W)$ is lossless.

Hence, $D_1$ is a lossless decomposition.

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### Analysis of Decomposition $D_2$:Constituent subrelations: $R_1(P, Q, S)$, $R_2(T, X)$, $R_3(Q, Y)$, $R_4(Y, Z, W)$.

• Notice attribute $T$ appears only in $R_2(T, X)$.

• The other relations contain attributes $\{P, Q, S, Y, Z, W\}$.

• The intersection of $R_2(T, X)$ with the union of all other relations is $\{X\}$.

• For a lossless join with $R_2$, either $X \rightarrow T$ or $X \rightarrow (P, Q, S, Y, Z, W)$ must hold. However, $X^+ = \{X\}$, so $X$ cannot determine $T$ or the other attributes.

Hence, $D_2$ is a lossy decomposition.

Therefore, A is the correct option.