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Question

If R1  have  120 tuples R2 have 100 tuples and R3 have 70 tuples then  R1⋈R2⋈R3 WILL BE?

Comments

ans might be 8400??

yes it can be 8400, but when?

when ⋈ means natural join, R1⋈R2⋈R3  = ( R1 ⋈ R2 ) ⋈ R3

( R1 ⋈ R2 ) = R5(A,B,C,D,E) ===> R1 and R2 have common attribute, which is C

then Minimum No.of rows from R1 ⋈ R2 = 0 ( when R1.C is NULL in every row )

Maximum No.of rows from R1 ⋈ R2 = 120 ( when R1.C is NOT NULL constraint )

 

( R1 ⋈ R2 ) ⋈ R3  ===> there is no common attribute ===> Natural join leads to Cartesian Product.

( R1 ⋈ R2 ) ⋈ R3 = R6(A,B,C,D,E,F,G) and have 120*70 = 8400 rows

 

What about keys in the resultant tables ?

it should requires, is R2 totally participated or not, R3 totally participated or not, and much more information

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I have one more query when  R1⋈R2 will be 100.    http://gateoverflow.in/2344/gate2010-43 here in this question 

R⋈S is minimum no of tuples in the relation with the logic that when all the tuples are same in both the relations then they matches and we can join them like this. but Sir please tell me how primary keys are playing roles in the question asked by me.

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By taking sample examples, you can answer your self !

let R(A,B,C) and S(C,D,E) where R have 10 tuples and S have 20 tuples... you are doing Natural join

1) if C is a key in both R and S ===> Min=0 and Max = Min(10,20) = 10

2) if C is a key in R but non-key in S ===> Min = 0 and Max = 20

3) if C is a non-key in R but key in S ===> Min = 0 and Max = 10

and many more cases, just like

4) if C is a key in R and Foreign Key in S

5) if C is a non-key in R and non-Key in S

Answer 1

Complete Question(According to  Divyanshum29):

R1 (A,B,C) R2 (C,D,E) R3 (F,G)  (bold one are keys)

If R1  have  120 tuples R2 have 100 tuples and R3 have 70 tuples then  R1⋈R2⋈R3 WILL BE?

(Note: A bit more information required but lets say asking for maximum tuples in result)

Answer:

First we will perform  R1 ⋈ R2 

R1 and R2 have common attribute, which is C and C is key of R2

So, Maximum No.of tuples in (R1 ⋈ R2) = 120. 

Now, ( R1 ⋈ R2 ) ⋈ R3  // there is no common attribute

(Note:If there are no attributes in common between two relations and you perform a natural join, it will return the Cartesian product of the two relations.)

=( R1 ⋈ R2 ) ⋈ R3

=120*70 = 8400 tuples