Non loss decomposition

TerabyteUK · #1 (**permalink**) 04-27-07, 10:55

Hey,
I'm just revising for a database exam and can't remember how to work out if a decomposition is non loss:

Code:

Relation T defined over attributes A,B,C,D,E
Functional Dependancies: 
B->E
E->A
A->D
D->E

Determine if decomposition of T into AB, BCD, ADE will be non-loss.

Apparently this decomposition IS non loss.. but I can't tell how..
(What to look for.. what to match...)

According to the notes I made, the transitive dependencies, B -> A and D -> A are handy too

Thanks for any help.

iced87 · #2 (**permalink**) 05-12-08, 15:21

Hi, I am revising for the same exam as well. You are probably from Imperial College.. ;p

Anyhow,http://www.dbforums.com/db_images_v3...itor/color.gif
http://www.dbforums.com/db_images_v3...itor/color.gif

The common column of AB and BCD is B which is a determinant in AB as stated by the transitive dependency B->. So the decomposition between AB and BCD is non loss.

The same goes for BCD, ADE where D is a determinant in ADE as stated by A-> E.

Last, AD and ADE, D is a determinant in ADE since you've got A->D.

So since the common columns are determinants in one of the two decomposed parts in each case, the decomposition will be non-loss.

Hope this helps!

P.S: There is a slight chance I got all of these wrong!

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