FOR FREE MATERIALS
Examples on Dependency Preserving
Q.
R(ABCD)
F = {A → B, B → C, C → D}
D = {AB, BC, CD}
.png)
.png)
.png)
So, here all FDs of F present at F1 ∪ F2. So, it is dependency preserving.
Q.
R (ABCD)
F = {A → B, B → C, C → D, D → A, A → C A → D}
D = {AD, BC, CD}
.png)
.png)
.png)
Only left A → B, which can easily derive from F1 ∪ F2.
.png)
Or
.png)
.png)
So, F1 ∪ F2 ⊇ F is true and F1 ∪ F2 ⊆ F always true then F1 ∪ F2 = F (true), it is dependency preserving.
.png)
.png)
Don’t think that at F1 ∪ F2 number of FDs is larger compare to the given F, but actually, both are the same F1 ∪ F2 contain some redundant FDs,
.png)
Both are contained the same number of FDs.
Q.
R (ABCDE)
F = {AB → CD, C → D, D → E}
D = {ABC, CD, DE}
.png)
.png)
.png)
.png)
.png)
It is dependency preserving.
Q.
R (ABCDEF)
F = {AB → CD, C → D, D → E, E → F} and D = {AB, CDE, EF}
.png)
.png)
No Functional Dependencies derive or possible.
.png)
.png)
.png)
But it better to write minimize the way
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
.png)
Here AB → CD FD not present at F1 ∪ F2 so, it is not dependency preserving.
Q.
R (ABCDEG)
F = {AB → C, AC → B, AD → E, B → D, BC → A, E → G}
D = {ABC, ACDE, ADG}
.png)
.png)
.png)
Here B → D and E → G FDs are lost so, it is not dependency preserving.
Contributed by
