3 3 votes Say we have two relations R (a,b,c) and S (b,d,e). Now, R has 200 tuples and S has 300 tuples. What will be Minimum number of tuples when we do R ⋈ S ( ⋈ = Natural Join)? Databases general databases relational-algebra natural-join joins + – iarnav 2.6k views answer comment Share Follow Print See all 5 Comments 5 5 Comments reply Show 2 previous comments Raushank2 commented Dec 6, 2017 reply Follow flag When we take natural join, the value of common attribute (B in this case) should match. Max= 200 when all value of b in R (a,b,c) matches with value of b in S (b,d,e). Min= 0 when none of values matches 0 0 replyShare Shivani Singhal commented Dec 15, 2017 reply Follow flag Natural join is done on common atttibute... So in this case we have b as a common attribute. For maximum when 200 tuples have same value for b For minimum we get zero tuples when all the 500 tuples from both the tablea have different value for b. 0 0 replyShare bharti commented Dec 23, 2017 i edited by bharti Dec 23, 2017 reply Follow flag maximum will be 60000 cz b is not the key . so value of v can be same for all 300 records and all 200 records. 1 1 replyShare Please log in or register to add a comment.
Best answer 4 4 votes When we take natural join, the value of common attribute (B in this case) should match. Max= 200 when all value of b in R (a,b,c) matches with value of b in S (b,d,e). Min= 0 when none of values matches Raushank2 answered Dec 6, 2017 • selected Dec 23, 2017 by iarnav Raushank2 comment Share Follow See 1 comment 1 1 comment reply srivivek95 commented Dec 23, 2017 reply Follow flag @ iarnav Here, b is neither the prime attribute nor having the unique values. Then max tuples will be when same value for attribute b span across both the relations i.e 300*200=60000 Min tuples is surely 0, when none of the values of b matches in the two relations. 5 5 replyShare Please log in or register to add a comment.
7 7 votes the value of common attribute (B in this case) should match. but in relation their is no specification of key . so it may be the case that all b in R and S are different . and all values are the same and values may be indistinct bcz b is not key . so maximum will M*N . Max= 60000 when all value of b in R (a,b,c) matches with all value of b in S (b,d,e). Min= 0 when none of values matches bharti answered Dec 23, 2017 • edited Dec 23, 2017 by bharti bharti comment Share Follow See all 4 Comments 4 4 Comments reply akshat sharma commented Dec 23, 2017 reply Follow flag Yes u r right if they have not mention abut primary key IN R :S Then 600 tuple will be maximum BUT if b is primary key in S then max will be 200 0 0 replyShare Lakshman Bhaiya commented Nov 22, 2018 reply Follow flag If they ask $S \Join_{<b=b>} R$ Then $minimum$ $number$ $of$ $tuples$ $=0$ $and$ $maximum$ $number$ $of$ $tuples$ $=300?$ please correct me if i'm wrong? 0 0 replyShare abhishek tiwary commented Apr 13, 2020 reply Follow flag max tuple should be 300 and min 0 if all the attribute of b is same then 300 max as there is no condition mentioned 0 0 replyShare Shashwat Pandey commented Oct 30, 2020 reply Follow flag If b is fk in S referencing R then max tuples wolud be 300 in natural join as fk can contain duplicate values and min case is still same i.e 0 0 0 replyShare Please log in or register to add a comment.