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  1. #1
    Join Date
    Feb 2009
    Posts
    13

    Dependency Preserving

    Hello,

    I have a small question about dependency preserving. Example I have the following functional dependencies: A->BCD, BC->D, AD->DE. And I have the following relational scheme: R(ABCDE).

    If I decompose it in the following relation: R(ABC), R(BCD), R(ADE). Now how can I see whether this decomposition is dependency preserving?

    Can someone also give me an informal definition of what dependency preserving is?

    Thanks! =) (PS: for checks, the decomposition is dependency preserving)

  2. #2
    Join Date
    Jun 2007
    Location
    London
    Posts
    2,527
    Are you just going to post all your coursework on the forum? Wouldn't you do better trying to do it yourself. If you don't understand the work being assigned to you then you should contact your lecturer before coming here.

  3. #3
    Join Date
    Feb 2009
    Posts
    13
    No, I am not going to post all my coursework on the forum. Secondly, I tried it already and the only thing I would like to gain by posting in this forum is to get a better understanding at it. And as you may noticed, why should I post my questions as I already have the answer if I'm just "letting you guys make my coursework"?

    Back to topic, like I said I already tried. I know that dependency preserving means that you can get all the "original" functional dependencies back after the decomposition. However sometimes some functional dependencies that were "original" aren't "minimal" so for example: A->B is also A->AB. Is there a way to check whether my decomposed functional dependencies is exactly like the "given" functional dependencies?

    Thanks

  4. #4
    Join Date
    Jan 2011
    Posts
    1
    Q.Example I have the following functional dependencies: A->BCD, BC->D, AD->DE. And I have the following relational scheme: R(ABCDE).

    If I decompose it in the following relation: R(ABC), R(BCD), R(ADE). Now how can I see whether this decomposition is dependency preserving?

    yes.

    we have to check all funtional dependency holds .
    F+ =(F1+) + (F2+) ...
    if some funtional dependency not in one relation we have to check whether it is by transitivity?

    for R(ABC) A->BC hold by F+ ,but A->BCD not here ,but transitivity holds here.

    as A-> BC holds for R1 and BC->BCD holds for R(BCD)

    A->BC ---(1) from R1
    BC->BCD --(2) from R2

    so implies ,

    (1) A->BCD in that (F+) . by transitivity.

    (2) BC->D holds for R(BCD).

    (3)AD->DE holds for R(ADE).

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