ISI2017-PCB-CS-3-a

Question

Let $R(A,B,C)$ be a relation with primary key $(A)$ and $S(A, D, E)$ a relation with primary key $(A, D)$. Each of the relations has $n$ tuples. If the number of tuples in $R \: \text{ natural join } S$ is $m$, then determine the number of tuples in $R$ $\text{ natural left outer join } S$.

Comments

I think answer should be n to 2n-1

@Arjun Sir verify

Answer 1

Since there are $\text{m}$ rows in the natural join of R and S which contains the common elements. Left outer join of R and S will contain common elements of R, S and left over elements of R which will be $\text{(n-m)}$

Hence total no. of rows in $\text{R left outer join S = m + (n-m)} \\\ = n$

Answer 2

It's simple there will be n tuples in which m tuples will related will relation S remaining (n-m) tuples wll not be related they will contain Null in the attribute of relation S