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+21 votes

Form the following instance of a relation schema $R(A,B,C)$, we can conclude that:

A |
B |
C |

$1$ | $1$ | $1$ |

$1$ | $1$ | $0$ |

$2$ | $3$ | $2$ |

$2$ | $3$ | $2$ |

- $A$ functionally determines $B$ and $B$ functionally determines $C$
- $A$ functionally determines $B$ and $B$ does not functionally determine $C$
- $B$ does not functionally determine $C$
- $A$ does not functionally determine $B$ and $B$ does not functionally determine $C$

+36 votes

Best answer

Ans. $C$

Generally Normalization is done on the schema itself.

From the relational instance given,we may strike out $FD$ s that do not hold.

e.g. $B$ does not functionally determine $C$(This is true).

But we cannot say that A functionally determines B for the entire relation itself.This is because that , $A->B$ holds for this instance,but in future there might be some tuples added to the instance that may violate $A->B$.

So overall on the relation we cannot conclude that $A->B$,from the relational instance which is just a subset of an entire relation.

Generally Normalization is done on the schema itself.

From the relational instance given,we may strike out $FD$ s that do not hold.

e.g. $B$ does not functionally determine $C$(This is true).

But we cannot say that A functionally determines B for the entire relation itself.This is because that , $A->B$ holds for this instance,but in future there might be some tuples added to the instance that may violate $A->B$.

So overall on the relation we cannot conclude that $A->B$,from the relational instance which is just a subset of an entire relation.

+7 votes

*Answer - (C)*

*If we observe carefully "Instance" of a relation Schema R(A,B,C) is given here.*

*Now as we can see A functionally determines B for the present tuples.*

*But B does not determines C. **That is clearly Visible.*

*In future there may be chances of tuples to be present where A can not determine B uniquely.*

*So,option C is most suitable.*

–2 votes

When value of A is 1,B is 1.When value of A is 2,B is 3.So A functionally determines B.

When value of B is 1,C is 1 and in another case C is 0.So B does not functionally determine C.

Hence,the answer is B.

When value of B is 1,C is 1 and in another case C is 0.So B does not functionally determine C.

Hence,the answer is B.

–3 votes

Ans will be B

Here A->B satisfies(for same value in A , B also gives unique value)

B->C not satisfies(when B is 1 , C gives two values 1,0)

Here A->B satisfies(for same value in A , B also gives unique value)

B->C not satisfies(when B is 1 , C gives two values 1,0)

For the given instance yes, but there can be another instance also for R where the FD may not hold. So, from a given instance we can only say "no FD".

** ** srestha @Arjun sir

question says **,Form the following instance of a relation** so according to that **B should ****be ****answer** .why we are considering other possibilities ?pls clear this

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