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While finding minimal cover what to do if closure of left two attributes become same as the combined attribute? I came to know that there should not be any extraneous left-hand side attribute. How to remove redundant FD if FD’s are given as:

A->B
B->A
AB->CD

I found that closure of (AB) is ABCD and similarly closure of A is also ABCD. so we can say attribute B is extraneous but the closure of B also gives ABCD. In that way A becomes extraneous. In this type of scenario can we get two sets of canonical cover? (1. when A is extraneous the Reduced FD's are- B->A and B->CD 2. When B is extraneous reduced FD's are- A->B and A->CD)

drika
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  • Right now you are effectively asking for us to rewrite a textbook with a bespoke tutorial & do your [(home)](https://meta.stackoverflow.com/q/334822/3404097)work & you show no [research effort](https://meta.stackoverflow.com/q/261592/3404097) towards answering. [ask] [Help] Show the steps of your work following a textbook/reference with justification & ask 1 specific researched non-duplicate question re the 1st place you are stuck/unsure. Quote definitions, theorems, algorithms & heuristics you rely on. All the steps are also SO faqs. Google with & without 'site:stackoverflow.com'. – philipxy Oct 11 '22 at 05:46

0 Answers0