edited by
16,627 views
54 votes
54 votes

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

$$\begin{array}{|l|l|}\hline \textbf{A} & \textbf{B} & \textbf{C} \\\hline  \text{1} & \text{1} & \text{1} \\   \text{1} & \text{1} & \text{0} \\ \text{2} & \text{3} & \text{2} \\ \text{2} & \text{3} & \text{2} \\\hline \end{array}$$

  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 $C$
edited by

6 Answers

Best answer
99 votes
99 votes
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, $A\to B$ holds for this instance, but in future there might be some tuples added to the instance that may violate $A\to B$.

So, overall on the relation we cannot conclude that $A\to B$, from the relational instance which is just a subset of an entire relation.
edited by
17 votes
17 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.

6 votes
6 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.

Answer:

Related questions