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

A B C
$1$ $1$ $1$
$1$ $1$ $0$
$2$ $3$ $2$
$2$ $3$ $2$
  1. $A$ functionally determines $B$ and $B$ functionally determines $C$
  2. $A$ functionally determines $B$ and $B$ does not functionally determine $C$
  3. $B$ does not functionally determine $C$
  4. $A$ does not functionally determine $B$ and $B$ does not functionally determine $C$
asked in Databases by Veteran (59.7k points)
edited by | 2.3k views
+15

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)
 

 
0
Why $(B)$ is not the answer$?$

6 Answers

+44 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.
answered by Active (3.6k points)
edited by
0
nice jarvis
+8 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.

answered by Active (3.7k points)
+1 vote

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.

answered by (21 points)
0 votes

ans is c

answered by (27 points)
–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.
answered by (15 points)
0
wrong answer.
0
KINDLY HIDE THIS ANSWER
–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)
answered by Veteran (103k points)
+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 

+4
Suppose a person from religion X throws a bomb. From this can we conclude that religion X is teaching terrorism?

Likewise everything depends on the definition. FD is defined on relational schema and not on any instance alone.
0
If they say $"$given relation(schema)$"$ then answer should be $(B)?$

please correct me if I'm wrong$?$
Answer:

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