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+19 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

+35 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.

+6 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.*

–1 vote

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.

–2 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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