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41 votes
41 votes

The following functional dependencies hold for relations $R(A, B, C)$ and $S(B, D, E).$ 

  • $ B \to A$
  • $A \to C$

The relation $R$ contains $200$ tuples and the relation $S$ contains $100$ tuples. What is the maximum number of tuples possible in the natural join  $R \bowtie S$?

  1. $100$
  2. $200$
  3. $300$
  4. $2000$
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Best answer
41 votes
41 votes

(A) $100.$

Natural join will combine tuples with the same value of the common rows(if there are two common rows then both values must be equal to get into the resultant set). So by this definition, we can get at the max only $100$ common values.

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36 votes
36 votes

From the given set of functional dependencies, it can be observed that B is a candidate key of R. So all 200 values of B must be unique in R.

There is no functional dependency given for S.

To get the maximum number of tuples in output, there can be two possibilities for S.
1) All 100 values of B in S are same and there is an entry in R that matches with this value. In this case, we get 100 tuples in output.
2) All 100 values of B in S are different and these values are present in R also. In this case also, we get 100 tuples.

11 votes
11 votes

as B is key in R and in table S, B is Foreign key that referencing to B in R

we have to find maximum number of tuples possible so there may be case that in table S every tuple of Attribute B is same

and we know natural join will combine tuples with same value

7 votes
7 votes

Every tuple in S can find atmost 1 matching tuple (with the same B value) in R...

Since asked maximum number of tuples, we can assume that every tuple in S finds 1 tuple in R. In that case natural joined table will have 100 tuples.

If we want minimum number of tuples, we can assume that every tuple in S finds 0 tuples in R. In that case natural joined table will have 0 tuples.

Answer:

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