Minimum and Maximum Number of Tuples

Consider Relations R and S where are R is having m tuples and S is having n tuples . m

• R intersection S

• R-S

• S-R

• R NATURAL JOIN S

• R Left OUTER Join S

• R/S

My Work

• R UNION S

max : n+m ( union we add all the tuples from both relations)

min: 0 ( taking m=n=null )

• R INTERSECTION S

max : m ( m

max : m ( if they are disjoint then in R-S we will get all tuples of R )

min: 0 ( if all tuples in R is also present in S)

• S-R

max : n ( as explained above )

min: 0 ( as explained above )

• R natural join S

max : n*m ( if no matching key constraints natural join will produce Cartesian product )

min: m ( m

max : m ( everything from left table will be output even if no match)

min: 0 ( when m=0 )

• R/S

max : m ( when n=0 )

min: Im Not able to make a conclusion

Yes, your answers are mostly right, except a few mistakes:

1. R UNION S

-
max : n+m ( union we add all the tuples from both relations)

Correct, when R and S have no common tuple.

-
min: 0 ( taking m=n=null )

Wrong, the minimum is n (the greatest of the two sizes, m and n). When all the tuples of R also exist in S.

And m and n cannot be null, they are the sizes of the relations, they are numbers (integers).

1. R INTERSECTION S

-
max : m ( m

-
min: 0 ( taking m=n=null if no common keys in both relations)

Correct but the reasoning is wrong. The result can be 0 because the two relations may have no common tuples).



1. R - S




-
max : m ( if they are disjoint then in R-S we will get all tuples of R )

Correct

-
min: 0 ( if all tuples in R is also present in S)

Correct



1. S - R




-
max : n ( as explained above )

Correct

-
min: 0 ( as explained above )

Wrong, the minimum is n - m.



1. R natural join S




-
max : n*m ( if no matching key constraints natural join will produce Cartesian product )

Correct

-
min: m ( m



1. R LEFT OUTER JOIN S




-
max : m ( everything from left table will be output even if no match)

Wrong, the maximum is m * n, the same as for natural join. Or just take ON TRUE.

-
min: 0 ( when m=0 )

Wrong, the minimum is m. Example can be the same as for NATURAL join above (or just take ON FALSE) but it cannot give as a result lees than the number of tuples in R (the left relation in the join).



1. R / S




-
max : m ( when n=0 )

Correct but it doesn't have to be n=0 or m=0. You can find another example.

-
min: I'm not able to make a conclusion.

Minimum is 0 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.

StackExchange Database Administrators Q#160241, answer score: 4