0 votes 0 votes Each dealer owns one or more cars and each car is owned by at-most one dealer. If Dealer table has $20$ tuples and Cars table has $60$ tuples then owns table can have maximum _____ number of tuples Databases tbb-dbms-1 numerical-answers + – Bikram asked Nov 26, 2016 • edited Aug 21, 2019 by go_editor Bikram 766 views answer comment Share Follow See all 17 Comments See all 17 17 Comments reply ukn commented Jan 3, 2017 reply Follow Share 20 tuples –1 votes –1 votes Bikram commented Jan 4, 2017 reply Follow Share No, it is 60 tuples. see there is total participation from Dealer entity . Each car have at least and at most one dealer , there are 60 tuples for car table , so that owns table must have 60 tuples at maximum. 0 votes 0 votes Harsh181996 commented Jan 20, 2017 reply Follow Share Sir , how is the relation table constructed from the given entity table ? 0 votes 0 votes shraddha priya commented Apr 21, 2017 reply Follow Share @bikram sir, minimum and maximum both will be 60 only, right? 1 votes 1 votes Bikram commented Apr 21, 2017 reply Follow Share yes @shraddha minimum and maximum both will be 60 in owns table . 0 votes 0 votes Shubhanshu commented Jul 15, 2017 reply Follow Share @Bikram Sir, Why the minimum is 60, I think it should be 20. As each dealer will buy only one car then, suppose dealers d1 to d20 will buy car c1 to c20 and remaining cars i.e. c30 to c60 will remain as unowned cars. 0 votes 0 votes Bikram commented Jul 15, 2017 reply Follow Share Each dealer owns one or many cars and each car is owned by at-most one dealer Dealer in total participation. Car is partial participation . If we draw a table it needs 1 table only . so minimum also be 60 tuples. since each dealer can have 1 or more cars in your assumption remaining 30 cars is un-owned is completely wrong !! 0 votes 0 votes Shubhanshu commented Jul 15, 2017 reply Follow Share Sir, According to the rule of 1:M total participation at 1's side the number of tables required will be 2 then how we will merge all table into one table so that we can get 1 table only? The relational tables should be: 1. delear 2. owns_car 0 votes 0 votes Bikram commented Jul 15, 2017 i edited by Bikram Jul 16, 2017 reply Follow Share Shubhanshu According to the rule of 1:M total participation at 1's side the number of tables required will be 2 yes correct, source : https://gateoverflow.in/133641/cardinality-ratio 0 votes 0 votes Shubhanshu commented Jul 15, 2017 reply Follow Share @Bikram Sir, here is the proof please verify it. Suppose we have three relations as follow: Emp(Eid, Ename) Works(Eid, Did) Dept(Did, Dname) here is the ER Diagram:- Now consider the table of Emp:- Eid Ename E1 A E2 B Now consider the table of Dept Please remember that the participation of emp is total which means each emp must contribute in the relation works, and participation of dept is partial that's why it is not mandatory for each entry of dept to be in relation works. Did Dname D1 AA D2 BB D3 CC D4 DD Now consider the table of Works:- Eid Did E1 D1 E1 D2 E2 D3 Now for minimum relation we will join this relation works to either sides. Lets try, towards Emp side Eid Ename Did E1 A D1 E1 A D2 E2 B D3 But Eid is the candidate key of the Emp so it should uniquely identify the tuples but E1 is defining D1 and D2 both which is wrong, It means it could not merge with emp table. So lets try toward Dept table Did Dname Eid D1 AA E1 D2 BB E1 D3 CC NULL D4 DD NULL It seems perfect because here Did can uniquely identify each tuple. And Eid is the FK refering to Eid of Emp. So, from the above explanation we need two tables as follow Emp(Eid, Emp) Works_Dept(Did, Dname, Eid(FK)) Observation:- 1:M mapping total participation at either side will leads to No of Tales - 2 No of FK - 1 1 votes 1 votes Bikram commented Jul 16, 2017 reply Follow Share @ Shubhanshu yes , here in your example , it needs 2 table as minimum , it should be also mention Null is allowed. for rules check this thread : https://gateoverflow.in/133641/cardinality-ratio so how many minimum tuples it needed in this actual question 20 or 60 ? minimum also be 60 tuples. since each dealer can have 1 or more cars in your assumption remaining 30 cars is un-owned so null value will be added there right ? as it is partial dependency . 0 votes 0 votes Bikram commented Jul 16, 2017 reply Follow Share @ Shubhanshu According to the rule of 1:M total participation at 1's side the number of tables required will be 2 then how we will merge all table into one table so that we can get 1 table only? Here if Null is allowed then 60 is minimum tuple and we all merged them in single table . if not null allowed then 20 is minimum and we have 2 tables . are you agree with me ? 0 votes 0 votes Shubhanshu commented Jul 16, 2017 reply Follow Share @ Bikram Sir, Here if Null is allowed then 60 is minimum tuple and we all merged them in single table . Agree. For minimum number of tuples it is mendatory that each dealer will buy only one car not more than that, otherwise number of tuples will be more than 20 and also null is not allowed then following:- if not null allowed then 20 is minimum and we have 2 tables (If null is not allowed) AND (Mapping is 1:M) AND (total participation at 1's side) = Then (it will have 20 tuples) AND AlSO (Owns relation will merge toward dealer side) Is it correct. 1 votes 1 votes Shubhanshu commented Jul 16, 2017 reply Follow Share Moreover I conclude following things from this discussion please verify it @Bikram Sir, (Note :- Null is allowed in all tables, Considering "dealer owns car" example ) Concl 1:- a) 1:M relation b) partial participation at both side. Then number of table = 2 dealer(did, dname) owns_car(carid, did(FK), carname). Concl 2:- a) 1:M relation b) total at 1's side only Then number of table = 1 dealer_owns_car(did, dname, carid, carname) and number of tuples will be equal to number of tuples in car table. Concl 3:- a) 1:M relation b) total at 1's side only number of tables = 2 dealer(did, dname) owns_car(carid, did(FK), carname). Concl 4:- a) 1:M relation b) total at both side number of tables = 1. dealer_owns_car(did, dname, carid, carname) and number of tuples will be equal to number of tuples in car table. From all these I made one more conclusion that whenever in 1:M we have total participation at 1's side and irrespective of participation at many side, then number of tables required would be 1 only. This one is important please verify it sir.. 0 votes 0 votes Bikram commented Jul 18, 2017 reply Follow Share @ Shubhanshu Conclusion 2 1:M relation b) total at 1's side only Then number of table would be 2 and conclusion 3 you made same table, need to check here .. You can read this thread https://gateoverflow.in/123659/self-doubt 0 votes 0 votes Shubhanshu commented Jul 19, 2017 reply Follow Share @Bikram sir, it is about the self-referential relation. but I am talking about the non self-referential relation. 1 votes 1 votes Ashwani Kumar 2 commented Oct 2, 2019 reply Follow Share Dealer has $20$ tuples and cars have $60$ tuples. Since it is mentioned that dealer owns one or more cars that means we can't leave any dealer to have car. Moreover each car is owned by at most one dealer. For maximum tuples each 20 tuple of dealer associated with 20 tuple of dealer. Left with 40 more tuple of cars, we can associate each one of them by any of dealer as dealer can have one or more car. Total = 20+40 = 60 tuples 0 votes 0 votes Please log in or register to add a comment.
Best answer 1 votes 1 votes There is total participation from Dealer entity . Partial participation from Car entity . Each car have at least and at most one dealer , there are 60 tuples for Car table , so that Owns table must have 60 tuples at maximum. Bikram answered Jul 15, 2017 • selected Aug 21, 2019 by Bikram Bikram comment Share Follow See all 2 Comments See all 2 2 Comments reply hem chandra joshi commented Dec 26, 2017 reply Follow Share I think if we do full outer join than we get max(m,n) rows. 0 votes 0 votes ChandanKumar123 commented Apr 13, 2019 i edited by ChandanKumar123 Oct 29, 2019 reply Follow Share "Each dealer owns one or more cars and each car is owned by at-most one dealer." Let each dealer take 3 diff cars so owns table will have 3*20(as participation is total)=60 tuples. participation is 1's at dealers and many at cars side i.e pk of car will be pk of owns. minimum tuples will also be 60. 0 votes 0 votes Please log in or register to add a comment.