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3 votes
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Say we have two relations R (a,b,c) and S (b,d,e).

Now, R has 200 tuples and S has 300 tuples. 

What will be Minimum number of tuples when we do R ⋈ S ( ⋈ = Natural Join)?

2 Answers

Best answer
3 votes
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When we take natural join, the value of common attribute (B in this case) should match.

Max= 200 when all value of b in R (a,b,c) matches with value of b in S (b,d,e).

Min= 0 when none of values matches

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7 votes

  the value of common attribute (B in this case) should match. but in relation their is no specification of key . so it may be the case that all b in R and S are different . and all values are the same and values may be indistinct bcz b is not key . so maximum will M*N   .

Max= 60000  when  all value of b in R (a,b,c) matches with all value of b in S (b,d,e). 

 Min= 0 when none of values matches

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