Given R with n tuples S with m tuples n<m then How many minimum and maximum tuples in follwing relations . Please Justify with Reason / Examples
[Someone should verify . This is what I think]
1. R UNION S
R UNION S
max : n+m
Reason : union we add all the tuples from both relations. ie When R and S have no common tuple.
Reason : The minimum is n (the greatest of the two sizes, m and n). When all the tuples of R also exist in S.
2. R INTERSECTION S
R INTERSECTION S
max : m ( m<n )
Reason : both relation contains same tuples then we may get maximum m keys
Reason : taking m=n=null if no common tuples in both relations
3. R - S
R - S
max : m
Reason : if they are disjoint then in R-S we will get all tuples of R
Reason : if all tuples in R is also present in S
4. S - R
S - R
max : n
Reason : as explained above
Reason : m<n there will be some tuples in S after deleting the common tuples
5. R natural join S
R natural join S
max : n*m
Reason : if no matching key constraints natural join will produce Cartesian product )
Reason : Identical with case 2 (INTERSECTION).
6. R LEFT OUTER JOIN S
R LEFT OUTER JOIN S
max : m*n
Reason : if all rows in left tables matches with all rows in right table
min: m (could be 0 when m= 0)
Reason : The minimum is 1 when m=1 , minimum is 2 when m=2, minimum is 0 when m=0
7. R / S
R / S
max : m
Reason : when n=0
Reason : Consider that relational division is similar to integer division. 3 / 7 gives 0 in integer division for example. Try to convert this into relational division
3 / 7
For natural join
mn is max.
Case 1: if there is a common attribute between and , and every row of R matches with the each row of S - i.e., the join attribute has the same value in all the rows of both and ,
Case 2: If there is no common attribute between R and S .
Min is 0.
If There is a common attribute between R and S and nothing matches.