Dependency Preservation
Let R is a relational schema.
Let r be the instance of R.
Let F be a set of functional dependencies on R.
Let F + is the closure of F.
Let R be decomposed into R1, R2,…Rn.
Let r be decomposed into r1, r2,…rn.
Let F1, F2,…Fn are sets of FDs applicable on R1, R2,..Rn.
Let F ‘ = F1 U F2 U ….U Fn.
Let F ‘+ is the closure of F ‘.
If F + = F ‘+, then the decomposition of R into R1, R2,…Rn is said to have the property of ‘Dependency Preservation’.
Example-1:
Let R = (A, B, C, D, E, F, G),
Let F = [ A → B, D → E, F → G]
So, F + = [ A → B, D → E, F → G]
Let R1 = (A, B, C) R2 = (C, D, E) R3 = (E, F, G)
So, F1 : A → B F2 : D → E F3 : F → G
So, F ’ = [ A → B, D → E, F → G]
So, F ‘ + = [ A → B, D → E, F → G]
It is proved that : F + = F ‘ +
Hence, this decomposition satisfies ‘Dependency Preservation’ property.
Note: In case of lossless decomposition, if the following equation is not satisfied, then it is said to be ‘lossy-decomposition’.
πR1 (r ) |X| πR2 (r) = r .
Note: If any FD which is in F, is not in F1 or F2 or… or Fn, then we can say that
F ¹ F ‘
But, in some cases, it is possible to have
F + = F ‘ +
In such a context, it can be said that the decomposition is said to follow the ‘Dependency Preservation’ property.
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