Consider the relation R={A,B, C, D, E} with the dependencies: {A->D; AB-> C; D->E}.The relation is decomposed into R1(A, B,C) and (A, D, E). The decomposition isa.Non-Lossless joinb.Non-Lossless join and Dependency Presenrvingc.Non-Lossless join and non-Dependency Presenrvingd.Lossless join
Question
Consider the relation R={A,B, C, D, E} with the dependencies: {A->D; AB-> C; D->E}.The relation is decomposed into R1(A, B,C) and (A, D, E). The decomposition isa.Non-Lossless joinb.Non-Lossless join and Dependency Presenrvingc.Non-Lossless join and non-Dependency Presenrvingd.Lossless join
Solution
The decomposition of the relation R into R1(A, B, C) and R2(A, D, E) is a Lossless join.
Here's why:
A lossless join decomposition ensures that no information is lost when the tables are joined back together. For a decomposition to be lossless, the common attribute of the decomposed relations (in this case, 'A') should be a key in at least one of the decomposed relations.
Looking at the functional dependencies, we can see that 'A' determines 'D' and 'D' determines 'E'. So, 'A' is a key for the relation R2(A, D, E).
Therefore, the decomposition is a Lossless join.
As for dependency preservation, a decomposition is dependency preserving if all the functional dependencies are preserved in the decomposed relations.
In this case, the dependencies A->D and D->E are preserved in R2, and AB->C is preserved in R1.
So, the decomposition is also Dependency Preserving.
Therefore, the correct answer is d. Lossless join.
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