Question 43

Given schema $R(P, Q, S, T, X, Y, Z, W)$ with FDs:

For decomposition $D_1 = \{R_1(P, Q, S, T), R_2(P, T, X), R_3(Q, Y), R_4(Y, Z, W)\}$:

1. Consider $R_3(Q, Y)$ and $R_4(Y, Z, W)$: $R_3 \cap R_4 = Y$. The FD $Y \rightarrow ZW$ means $Y$ is a key of $R_4$. Merging them gives $R_{34}(Q, Y, Z, W)$ losslessly.
2. Consider $R_1(P, Q, S, T)$ and $R_2(P, T, X)$: $R_1 \cap R_2 = PT$. Since $P \rightarrow YX$, we have $P \rightarrow X$, so $PT \rightarrow X$. Hence $PT$ is a key of $R_2$. Merging them gives $R_{12}(P, Q, S, T, X)$ losslessly.
3. Consider $R_{12}$ and $R_{34}$: $R_{12} \cap R_{34} = Q$. Since $Q \rightarrow Y$ and $Y \rightarrow ZW$, $Q^+ = QYZW$. Thus $Q$ is a key of $R_{34}$. Merging them produces $R$ losslessly.
   Therefore, $D_1$ is a lossless decomposition.

For decomposition $D_2 = \{R_1(P, Q, S), R_2(T, X), R_3(Q, Y), R_4(Y, Z, W)\}$:
In $D_2$, attribute $T$ appears only in $R_2(T, X)$. None of the given FDs have $T$ on the right side, so $T$ cannot be derived from any attributes outside $R_2$. Furthermore, the intersection of $R_2$ with $(R_1 \cup R_3 \cup R_4)$ contains only $X$, but $X$ does not functionally determine $T$ or any other attribute. Hence $R_2$ cannot be losslessly rejoined. Thus $D_2$ is lossy.

Therefore, $D_1$ is lossless, but $D_2$ is lossy (option A).