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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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8 Answers

2 votes
2 votes
Give FD is lossless join as B is the key in one table and it is common attribute for the table so the maximum no. of tupples will be 100 in natural join of R & S.
2 votes
2 votes

here B is common in both relations R and S .B is key in relation R bcoz B closure determines(A,B,C) all attributes of R but non-key in relation S.

B is unique in rel R but repetetion allowed in rel S.

so maximum number of tuples possible in the natural join  R⋈S depend on non-key

so 100 is ans

0 votes
0 votes
answer (A)
 

Given the dependencies, B is the primary key in relation R. So, R has 200 unique values.

Now, S has 100 references to column B. So out of 200, there can be maximum of 100 unique values in B. So, maximum the join will return is 100(the 100 unique references of B present in S).
Answer:

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