edited by
4,174 views
5 votes
5 votes

Consider a relation r1(ABC), r2(CDE) and r3(FG) with primary keys A, C and F respectively. Assume that r1 has 150 tupples, r2 has 100 tupples and r3 has 75 tupples. The number of resultant tuple in are ________.

edited by

1 Answer

Best answer
3 votes
3 votes
In $r_1\Join  r_2$  $ C$ is common ,in which C is foreign key ( assuming that foreign key is present) in r1 so in natural join of r1 and r2 all record of r1 will be present so $r_1\Join  r_2$ has maximum no. of tupples 150 .let say $r_1\Join  r_2$  as S ,now join of S and r3, in which no element is common( assuming all attributes have diffrent quality as in question nothing is mentioned) so $S\Join  r_3$  will become $S \textbf{*}r_3$(cross product) .now every tupple of S will relate to every tupple of $r_3$ ,so maximum no. of tupples =150*75=11250.

Related questions

1 votes
1 votes
0 answers
1
1 votes
1 votes
0 answers
2
0 votes
0 votes
0 answers
4
aditi19 asked Nov 2, 2018
297 views
in case of outer join for combining two tables, if there is no common attribute does it returns cartesian product just like natural join does?