GATE CSE 1995 | Question: 26

Question

Consider the relation scheme $R(A, B, C)$ with the following functional dependencies:

• $A, B \rightarrow C,$
• $C \rightarrow A$

• A. Show that the scheme $R$ is in $3\text{NF}$ but not in $\text{BCNF}$.
• B. Determine the minimal keys of relation $R$. 

Comments

• A, B → C, Wht does it mean ?
• is it A->C AND B->C ?@bikram sir

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@set2018

A, B → C, means AB  → C

Answer 1

BCNF; X---- >a ( X must be SK).

AB ---->C (I OK NO_PROBLEM)

C----->A (C IS NOT SK,SO BCZZ OF THIS BCNF IS FAILED)

 

in case of,

3NF X----->Y

X= MUST be SK. or Y= Must be PRIME ATTRIBUTE.

so in given question, the prime attribute is {A, B, C} Non-Prime Attribute { NULL }

so,

AB---->C (No problem )

C-----A (No problem )

so it follow 3NF but not BCNF

Answer 2

Here AB,BC both are Candidate key, So There is no non-prime attribute.

So The relation is 3NF,2NF because there is no non prime attribute

And

The relation is not BCNF because this FD’s(C-->A) LHS is not a superkey