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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.