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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3 votes

*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)?*

3 votes

Best answer

@ iarnav

Here, b is neither the prime attribute nor having the unique values.

Then max tuples will be when same value for attribute b span across both the relations i.e 300*200=60000

Min tuples is surely 0, when none of the values of b matches in the two relations.

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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

**Min= 0** when none of values matches

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