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Canonical Cover

Posted on 2012-09-12
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Last Modified: 2016-09-22
I need help finding canonical cover for the following dependencies:

AB->CD
BCD->EF
BC->D
EB->A
D->B

This is for a practice assignment, however, the only thing we've learned so far are the armstrong axioms, and it doesn't seem like I can apply those here. The book talks about eliminating extraneous attributes, however the algorithms they use in the book is complicated and I don' understand it. Hoping someone here is able to provide a solution, and an english version interpretation of how to solve future canonical covers.

Thank you.
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Question by:pzozulka
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DcpKing earned 2000 total points
ID: 38394493
What are you studying? How to write relational database management systems? Most people I know write database systems that run on RDBMS!
This might help

http://www.cs.sfu.ca/CourseCentral/354/zaiane/material/notes/Chapter6/node14.html

and, of course, there's always the minimally explained

http://www.koffeinhaltig.com/fds/ueberdeckung.php
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by:pzozulka
ID: 38395580
I'm studying relational database design. I've already researched both links.

The first link you sent is exactly the part of the textbook that I needed help interpreting into regular english. It just doesn't make any sense to me. I just don't seem to understand how to check for extraneous attributes. For example in the example I posted above, BCD->EF, how do you check if B is extraneous in the left hand side. And how do you check if D is extraneous on the left hand side?

The second link is what I've used over and over to double check my work, and so far I guess I'm not doing it right.
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Expert Comment

by:DcpKing
ID: 38397201
Well, I suppose I should understand it all, as I've been using relational databases since about 1987, but they weren't on the curriculum when I took Comp Sci :( In fact, they didn't exist! I'll be honest - I happened to know those refs, but that's all, unless I take the course too!

Good luck

Mike
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Expert Comment

by:Rohan Menon
ID: 41811335
The minimal cover for the problem is

AB->C
BCD->E
BCD->F
BC->D
EB->A
D->B
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