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Functional Dependency Preserving

Concept of Functional Dependency Preserving

If F1 ∪ F2 = F then it is functional dependency preserving.

Dependency preserving and lossless or lossy decomposing are independent not related. Both are different concepts.

Another method to check dependency preserving:

So, we can use any of one but functional dependency closer is lengthy it is better to use F1 ∪ F2 = F

So, F1 ∪ F2 = F, then it is dependency preserving if it is not dependency preserving then F1 ∪ F2 ⊂ F

But F1 ∪ F2 ⊃ F is never possible because F1 and F2 are decomposed from F. So, extra FDs is never possible at F1 or F2 rather some dependency may be lost.

So, Dependency preserving happens when F1 ∪ F2 = F but how to check the equivalent of two sets

A = B (A and B is set)

1. A ⊆ B

2. B ⊆ A

If we prove 1 and 2 then we can say A = B

Now we prove

If both cases are true then we can say

But 1 F1 ∪ F2 ⊆ F is always true even it dependency preserving or not.

So, if we only prove 2 condition i.e. F ⊂ F1 ⋃ F2 then we can say it is dependency preserving or not.

For example:

Let a relation R (ABC)

FDs = {A → B, B → C, A → C}

Decomposition (D) = {AB, BC}

So, we can say

Now check F ⊂ F1 ⋃ F2, this means we check all FDs of F are present in the set of FDs (F1 ∪ F2) or not. If present then it is dependency preserving otherwise not.

F1 ∪ F2 = {A → B, B → C} and also derive A → C from the transitive property. A → B, B → C then A → C. So, all FDs of F present in F1 ∪ F2. It is dependency preserving.

Without the transitive rule, we can do another way we can calculate

Example 2:

Let a relation R (ABCD) and FDs = {A → B, B → C, A → C, D → C} and D = {AB, BCD}

From AB how many FDs we can get where AB is on left-hand side A, B, AB, ɸ.

So, here we calculate –

And take valid FDs which match with AB attribute.

We only take non-trivial FDs not consider trivial FDs like A → A, B → B, AB → A, AB → B, etc.

So, only valid FD here A → B

BCD

Here we can take

We can take BD → C but it can derive from B → C augmented both sides with D, BD → CD and by splitting properly we can get BD → C. So, only valid one B → C.