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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)?

I know, MAX no of tuples will be 200, but what will be the Minimum?
Minimum will be 0?

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

Natural join is done on common atttibute... So in this case we have b as a common attribute.

For maximum when 200 tuples have same value for b

For minimum we get zero tuples when all the 500 tuples from both the tablea have different value for b.
edited by
maximum will be 60000 cz b is not the key . so value of v can  be same for all 300 records  and all 200 records.

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

### 1 comment

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.

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

by

Yes u r right if they have not mention abut primary  key IN R :S   Then 600 tuple will be maximum

BUT if b is primary key in S then max will be 200

$S \Join_{<b=b>} R$

Then $minimum$ $number$ $of$ $tuples$ $=0$  $and$ $maximum$ $number$ $of$ $tuples$ $=300?$

please correct me if i'm wrong?
max tuple should be 300 and min 0 if all the attribute of b is same then 300 max as there is no condition mentioned
If b is fk in S referencing R then max tuples wolud be 300 in natural join as fk can contain duplicate values and min case is still same i.e 0