R is not the primary key in relation r2 therefore redundant data will result maximum selection of tuples.

The Gateway to Computer Science Excellence

First time here? Checkout the FAQ!

x

+2 votes

+1 vote

Best answer

Given R1(P, Q, R) and R2(R ,U, V) where R is primary key in R1 and U is primary key in R2.

R in R1 is primary key so it should contain unique values assume 1,2,3,......1500(1500tuples)

There is no constraint on R in R2 so it can contain duplicate values assume 1,1,1,...1(2000 tuples)

Natural join is performed on the Common attribute i.e R

1 in R( R1 table) can map with 2000 1's in R( R2 table).

Hence maximum 2000 tuples are possible in natural join

R in R1 is primary key so it should contain unique values assume 1,2,3,......1500(1500tuples)

There is no constraint on R in R2 so it can contain duplicate values assume 1,1,1,...1(2000 tuples)

Natural join is performed on the Common attribute i.e R

1 in R( R1 table) can map with 2000 1's in R( R2 table).

Hence maximum 2000 tuples are possible in natural join

0

when 1500 wud be the answer then???

plz check this too. . .https://gateoverflow.in/3553/gate2006-it-14

- All categories
- General Aptitude 1.3k
- Engineering Mathematics 5.2k
- Digital Logic 2k
- Programming & DS 3.7k
- Algorithms 3.2k
- Theory of Computation 4k
- Compiler Design 1.6k
- Databases 3k
- CO & Architecture 2.6k
- Computer Networks 3k
- Non GATE 1k
- Others 1.3k
- Admissions 488
- Exam Queries 436
- Tier 1 Placement Questions 18
- Job Queries 56
- Projects 9

36,203 questions

43,662 answers

124,116 comments

42,944 users