before solving read carefully

"Form the following instance of a relation schema R(A,B,C)"

Relation = one table

Instance = values at one moment ( as opposed to the schema, which is the description of all possible values)

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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 $C$

+37 votes

Best answer

Answer is $C$.

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.*

0 votes

Since for this table we can clearly see that everytime when A=1 then B=1 and everytime when A=2 then B=2

so for this particular snapshot we can say that A functionally determines B.

But observing for functional dependency of B and C,

We see that when B=1 then C has two values 1, 0 hence B fails to funtionally determine C.

Hence *option** C* can be considered to be correct.

–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)

+4

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".

0

** ** 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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