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If a relation with a Schema R is decomposed into two relations $R_1$ and $R_2$ such that $(R_1 \cup R_2) = R_1$ then which one of the following is to be satisfied for a lossless joint decomposition ($\rightarrow$ indicates functional independency)

- $(R_1 \cap R_2) \rightarrow R_1 \text{ or } R_1 \cap R_2 \rightarrow R_2$
- $R_1 \cap R_2 \rightarrow R_1$
- $R_1 \cap R_2 \rightarrow R_2$
- $(R_1 \cap R_2) \rightarrow R_1 \text{ and } R_1 \cap R_2 \rightarrow R_2$

0 votes

A IS THE ANS

LET R(A,B,C) AND B IS THE KEY

LET R1==(A,B) AND R2=(B) SO R1** U** R2 = R1

DEFINATION OF LOSSLESS D: SAYS THAT IF THE INTERSECTION OF TWO RELATION HAS A COMMON ATTRIBUTE AND THAT IS A KEY IN ANY OF THE RELATION THEN IT IS LOSSLESS

NOW R1 ∩ R2 =B...AND B-->R1(COZ B IS THE KEY) SO LOSSLESS

AND IT CAN ALSO BE STATED THAT

R1 ∩ R2 =B..AND B-->R2 ..SO LOSSLESS

THEREFORE IF WE HAVE ANY ONE OF THE ABOVE THEN IT IS LOSSLESS

SO A IS THE ANS

0 votes

In lossless decomposition, the ** intersection** of two decomposed relations will be the

R1 ∩ R2 --> R1 means it contains superkey of R1

R1 ∩ R2 --> R2 means it contains superkey of R2

For lossless decomposition either of these conditions is sufficent

hence Answer A

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