lossless and dependency preserving decomposition #testseries

Question

The below decomposition is lossless or lossy and also dependency preserving or not?

Comments

lossy and not dependency preserving?

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yes,it will be lossy as well not f.d preserving.if R1 combined with R2 there will be redundancy .

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I understand that it is not dependency preserving.But how is it lossy? Please anyone explain in detail.

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for lossy decomposition we check whether the final combination of table contain some spurious tuple or not.if yes then the decomposition will be lossy.

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Can anybody explain this by using chase algorithm?

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@night_fury

i don't know about that algorithm but i will give how to check is it lossy or loss less?

First take Table₁ ===> R₁(A,B,C)

list out all dependencies under this table by original FD set ==> { AB -> C }

Key of this table = AB

 

Table₂ ===> R₂(B,C,D)

list out all dependencies under this table by original FD set ==> { C -> D }

Key of this Table₂ = BC

 

Table₃ ===> R₃(C,D,E)

list out all dependencies under this table by original FD set ==> { C -> D }

Key of this Table₃ = CE

 

take the intersection of attributes of Table₁ and Table₂ ==> BC, it is a key(super key or primary key) in R2, therefore you can combine Table₁ and Table₂ . let denote it as T₁₂. it's key is AB

take the intersection of attributes of T₁₂ and Table₃ ==> CD, it is neither a key in T₁₂ nor R₁₃ ==> you can combine Table₁₂ and Table₃ . ===> it is lossy.

Answer 1

Lossy Decomposition because no common attribute is a candidate key here