CIL 2017 | System | Question: 59

Question

If $\mathrm{D}=\{\mathrm{R} 1, \mathrm{R} 2\}$ is a decomposition of $\mathrm{R}$, and $\mathrm{F}$ is the set of functional dependencies on $\mathrm{R}$, then which of the following ensures that the decomposition $\mathrm{D}$ is lossless (nonadditive)?

• A. $((\mathrm{R} 1 \cap \mathrm{R} 2) \rightarrow(\mathrm{R} 1-\mathrm{R} 2)) \in \mathrm{F}^{+}$
• B. $((\mathrm{R} 1 \cap \mathrm{R} 2) \rightarrow(\mathrm{R} 1-\mathrm{R} 2)) \notin \mathrm{F}^{+}$
• C. $((\mathrm{R} 1 \cup \mathrm{R} 2) \rightarrow(\mathrm{R} 1-\mathrm{R} 2)) \in \mathrm{F}^{+}$
• D. $((\mathrm{R} 1 \cap \mathrm{R} 2) \rightarrow \mathrm{R} 1) \in \mathrm{F}^{+}$

Answer 1

The decomposition of $R=\left \{ R1,R2 \right \}$ is a lossless or non-additive if atleast one of the dependencies

• $R1\cap R2\rightarrow R1$
• $R1\cap R2\rightarrow R2$

belong to $F^{+}$ :- Closure of the set of functional dependencies .

Comments

see this question https://gateoverflow.in/61154/ugcnet-june2015-iii-11

Answer 2

The Decomposition D is lossless then it should satisfy any of the below two:

1. (R1 ∩ R2) → (R1-R2) or 
2.  R1 ∩ R2) → (R2-R1)

consider the ex: R(ABCDEF) which is decomposed into R1(ABCD) and R2(CDEF) , then this Decomposition becomes lossless when any of the below 2 satisfies:

1. CD → AB or

2. CD → EF

Therefore, I can say option A is correct.