Question

Suppose that the current instances of relations R(A, B) and S(A, C) have 100 tuples and 200 tuples, respectively.

(a) What are the minimum and maximum possible number of tuples in the resulting relation instance of R S.

(b) What is your answer to the above question if it is given that attribute A is a key for R.

(c) When attribute A is a key for R and A is a foreign key for S referencing R.

Answer 1

Solution:

(a): As no functional dependency is given for R and S in the first case, we can assume things as per our need.

Let us assume R has 100 different entries for A(1 to 100) and S has 200 values for A(1 to 200), all different values. So the maximum possible tuples, in this case, will be 100(1 to 100) after using the natural join.

For minimum let us assume all the values for A in R are 0 to 100 and in S are 101 to 300. So, in this case, none of the value of A will match and it will return zero tuples after joining using natural join.

(b): If A is given as the key attribute of R, that means R will surely have 100 distinct values of A. But S can still have any set of values for A. So for maximum again we can get 100 tuples if any of the 100 values of S.A matches with R.A.

And for the minimum again it will be zero as S.A can have all the 200 values which will be different from R.A.

(c): For this condition where S.A is a foreign key referencing R.A, the maximum will the same as 100, because all the 100 values of R.A will be there S.A.

But for minimum, it will not be zero this time. As all the values of R.A will definitely be there in S.A. So the minimum possible tuples will be 100 in this case.

** R.A means A of R

** S.A means A of S

I hope the solution helps.

Thanks!

Answer 2