Normalisation

Question

R(ABCDEF)

AB->C

DC->AE

E->F

how to find the highest normalization of the above relation.......??

and also tell me the difference between prime attribute, non-prime attribute and candidate key with examples.....??

Answer 1

Candidate keys are ABD,BCD.

A candidate key is minimal super key which uniquely identifies a relation.

There can be many superkeys and out them candidate key is selected.

ABD+ = { A,B,C,D,E,F}

BCD+ = { A,B,C,D,E,F}

Now prime attributes are the one that are present in candidate key.

Non prime attributes are E,F

Here  A,B,C,D  are Prime attributes.

FD1 : AB->C ( part of key -> prime attribute) --->3NF

FD2 : DC->AE(part of key ->nonkey) ------>1NF

FD3 : E->F (nonkey->nonkey)-------->2NF.

So highest normal form of complete Relation is 1NF.

Comments

thankyou.....

also can u explain me the conditions of 2NF, 3NF and BCNF.....in short.....and how to find normalisation in fastest way possible....

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This may help you

https://www.youtube.com/playlist?list=PL6icxsSf2sRfPJIfRiwhmJxOpCYhpk0d_

https://gateoverflow.in/84635/data-base-system

Answer 2

Ck=ABD,BDC..

There exist Partial dependency as

xxAB->C

and DC->AE

so it will in 1NF