Solution

Check dependency preservation

Given FDs:

$P \rightarrow Q$

$Q \rightarrow R$

$R \rightarrow S$

$S \rightarrow Q$

Decomposed relations:

$R_1(P,Q)$

$R_2(Q,R)$

$R_3(Q,S)$

Each dependency is preserved:

$P \rightarrow Q$ → in $R_1$

$Q \rightarrow R$ → in $R_2$

$R \rightarrow S$ → derived via $Q$ in $R_3$

$S \rightarrow Q$ → in $R_3$

Thus dependency preserving

Check lossless join:

Common attribute between relations is Q, which functionally determines other attributes through dependencies.

Hence not lossless join.