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Dependency preserving and lossless join decomposition

by Pritam on 8 Sep 2017

Consider the attribute set ABCDEG and the FD set
AB → C, AC → B, AD → E, B → D, BC → A, E → G
Is the following decomposition of R(ABCDEG)
(a) dependency-preserving?
(b) lossless-join?
1: AB, BC, ABDE, EG
2: ABC, ACDE, ADG
Give proper justification for your answer.

1. Decomposition into AB, BC, ABDE, EG:
(a) It's not dependency preserving. The FDs not being preserved are AB → C, AC → B, BC → A
(b) It's not lossless join. Join between AB and BC is lossy as B ↛ AB and B ↛ BC (Decomposition into R1 and R2 is lossless join, only when R1∩R2 → R1 or R1∩R2 → R2)

2. Decomposition into ABC, ACDE, ADG:
(a) It's not dependency preserving. The FDs not being preserved are B → D, E → G
(b) It's lossless join. Join between ABC and ACDE is lossless as AC → ABC and then join of the result with ADG is again lossless as AD → ADE.

1Comment
swapnilmali's picture

1. loosy and Non dependency preserving.

 

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