When a relational schema is decomposed into smaller relations, one of the central concerns in database design is whether the decomposition preserves functional dependencies. A decomposition is said to be dependency-preserving if every functional dependency (FD) in the original relation can be enforced by examining the individual decomposed relations without needing to reconstruct the original relation. If this property is not satisfied, then to verify or enforce certain dependencies, the database system must recombine the decomposed relations—typically through join operations. Consequently, joins must be executed more frequently to maintain data integrity.
Example
Consider a relation schema:
$$
R(A, B, C)
$$
with the following set of functional dependencies:
$$
\mathcal{F} = { A \rightarrow B,\ B \rightarrow C }
$$
From these, we can infer by transitivity that $ A \rightarrow C $. The attribute $ A $ is a candidate key, since $ A^+ = {A, B, C} $.
Now, decompose $ R $ into two relations:
$ R_1(A, B) $
$ R_2(B, C) $
This decomposition is lossless (because $ R_1 \cap R_2 = {B} $ and $ B \rightarrow C \in \mathcal{F} $), and each component is in Boyce-Codd Normal Form (BCNF). However, we now examine whether it is dependency-preserving.
$ A \rightarrow B $ is preserved in $ R_1 $.
$ B \rightarrow C $ is preserved in $ R_2 $.
But the derived dependency $ A \rightarrow C $ does not appear in either $ R_1 $ or $ R_2 $, and cannot be checked without combining both relations.
Thus, $ A \rightarrow C $ is not preserved, and any enforcement of this dependency requires a join.
Illustration
Assume the following valid instance of $ R $:
$$
R =
\begin{array}{|c|c|c|}
\hline
A & B & C \\
\hline
101 & \text{HR} & \text{London} \\
102 & \text{IT} & \text{Paris} \\
103 & \text{HR} & \text{London} \\
\hline
\end{array}
$$
This satisfies:
After decomposition, we store:
$$
R_1 =
\begin{array}{|c|c|}
\hline
A & B \\
\hline
101 & \text{HR} \\
102 & \text{IT} \\
103 & \text{HR} \\
\hline
\end{array}
\qquad
R_2 =
\begin{array}{|c|c|}
\hline
B & C \\
\hline
\text{HR} & \text{London} \\
\text{IT} & \text{Paris} \\
\hline
\end{array}
$$
Now, suppose a new employee is added:
Insert $ (104, \text{IT}) $ into $ R_1 $,
Insert $ (\text{IT}, \text{Berlin}) $ into $ R_2 $.
Original FD set was simply:
$$
\mathcal{F} = { A \rightarrow C }
$$
and there is no FD involving $ B $. Suppose we decompose $ R(A,B,C) $ into $ R_1(A,B) $ and $ R_2(B,C) $ for normalization (e.g., to eliminate redundancy). Now, $ A \rightarrow C $ is completely lost from both components neither relation contains both $ A $ and $ C $.
To enforce $ A \rightarrow C $ after any update, the DBMS must:
Compute the natural join:
$$
R_1 \bowtie R_2 =
\begin{array}{|c|c|c|}
\hline
A & B & C \\
\hline
101 & \text{HR} & \text{London} \\
102 & \text{IT} & \text{Paris} \\
103 & \text{HR} & \text{London} \\
104 & \text{IT} & \text{Berlin} \\
\hline
\end{array}
$$
Project onto $ (A, C) $ and check whether any $ A $ value appears with more than one $ C $ value.
If such a conflict exists (e.g., employee 102 maps to both Paris and Berlin), the dependency $ A \rightarrow C $ is violated.
Since this check cannot be performed by inspecting $ R_1 $ or $ R_2 $ alone, the join operation becomes mandatory for dependency enforcement.
Looking at the Options
Selection ($\sigma$) filters rows but cannot combine attributes from different relations.
Projection ($\pi$) reduces columns and cannot recover missing attributes.
Set union ($\cup$) combines relations with identical schemas, which is irrelevant here.
Only the join operator can reconstruct the co-occurrence of attributes that were separated during decomposition.
In a decomposition that is not dependency-preserving, certain functional dependencies span multiple relations and cannot be validated locally. To maintain data consistency, the database system must repeatedly join the decomposed relations to reconstruct the necessary attribute combinations. Therefore, the join operator is executed more frequently in such scenarios to enforce the original dependencies.
Hence, the correct answer is: $\boxed{\text{C. Join}} $.
