GATE Overflow Test Series | Databases | Test 1 | Question: 25

Question

Let $R (A, B, C)$ be a relational schema with the following functional dependencies :
$\{A \rightarrow B, B \rightarrow C, C \rightarrow A\}$

The decomposition of $R$ into $(A, B), (B, C).$

• A. Gives a lossless join, and is dependency preserving
• B. Gives a lossless join, but is not dependency preserving
• C. Does not give a lossless join, but is dependency preserving
• D. Does not give a lossless join and is not dependency preserving

Answer 1

The relation $R$ is decomposed into two relations $R_{1},R_{2}.$

$R_{1}(A,\textbf{B}):$

$A \rightarrow B,  B \to A $ are present here. $B \to A $ is implied by $B \to C, C \to A.$

$R_{2}(\textbf{B},C):$

$B \rightarrow C, C \to B$ are present here.

Whenever we decompose a relation if the common attribute is a key of one of them, then the decomposition is lossless. Here, the common attribute is B and B is the key of $R_2.$ So, the decomposition is lossless.

The FD $C \to A$ is preserved in this decomposition.

So, the correct answer is A.

Comments

How is C->A  preserved?

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$C \to B$ and $B \to A$ are preserved in the decomposition and by transitivity $C \to A$ is also preserved. We dont need to do a join to determine that.