Made easy workbook

Question

Consider the following ER diagram 

The minimum number of tables needed to represent E1,E2 and E3 are_______

Comments

@talha: I think 4 tables...as E1 and E2 is having many to many relationship so 3 tables and E1 and E2 is many to one, so 1 table for E2 so total 4??

What's the right answer??

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@ Abbas answer given is 3 .

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ok it's because the general assumption is we should use arrow on "one" side in many-one relationship

and weak entity relationship is one-many relationship with weak as total participation and on "many" side but there is no arrow used on "one" side i.e. E1 so I assume to be many to many.

Well test series question, so no need to worry about conventions.It will be clearly mentioned in GATE if such question is asked.

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even I got 4 tables for this;

I assumed many:many relationship b/w E1 & E2

tables are:-

E1(A,B,C)

R1(A,D)

E2(D,F)

E3(G,A,H)