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Consider the following relations: $R_1(A,B,C)$ and $R_2(A,D,E)$. $R_1$ has 1000 records and $R_2$ has 2000 records. The attribute $A$ in $R_2$ is referencing attribute $A$ in $R_1$. Let $X$ be minimum number of records in $R_1$ ⨝ $R_2$ and $Y$ be the maximum number of records in $R_1$⨝$R_2$. The sum of $(X+Y)$ is _______.

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Best answer

Let **A in R2 be null** for all the tuples as foreign keys can be null , hence natural join would result in zero tuples. Hence minimum = 0 tuples

Max records will only be generated when the value in R2(A) for all the tuples is same as any value in R1(A). Here, all values of A in R1 will be unique as it must be a key due to foreign key constraint. Hence in that case , output would be 2000.

Hence sum(min+max)=2000

2 votes

Assume R1 has 3 tuples and R2 has 2 tuples

R1 R2

A B C A D E

1 2 3 1 8 9

1 4 5 1 9 10

1 6 7

R1 NATURAL JOIN R2

A B C D E

1 2 3 8 9

1 2 3 9 10

1 4 5 8 9

1 4 5 9 10

1 6 7 8 9

1 6 7 9 10

So we have 3*2 = 6 tuples.

NOTE:- Since it is not mentioned that A is key so value of A can be repeated.

So now if R1 has 1000 tuples and R2 has 2000 tuples so **R1 NATURAL JOIN R2 = 1000*2000 =2000000 tuples(max)=X**

**Minimum number of tuples= 0=Y(when all values of A in R2 is null). Value of A in R2 can be null as it is not mentioned that A in R2 is not null.**

**So X-Y= 2000000-0=2000000**

Correct me if I am wrong

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