Dark Mode

1,347 views

1 vote

Consider The Following Schema:-

The minimum number of tables needed to represent *E1 , **E2*and* E3 * are ______

Soln. According to me Answer should be 2 because We can merge

1. R1 and E2 in one Relation

2. E1,R2,E3 in one because its total participation on both the sides.

So total 2 Tables Should be Enough.

2 votes

NO , number of table required for above diagram will be three not two .

one table $ E_2 R_1$ [ $\underline{AD}$* $F$*]

second table $R_2 E_3$ [$\underline G H A$ ]

third table $E_1 [ \underline{A} B C]$

we cant merge $E_1$ with $R_2 E_3$ because relation $R_2$ has key G ,and A is not key of relation which will produce redudancy ,so we can't merge.total participation of ralation is given towards $E_1$ ,so participation will not arise but redudancy is still there . so we have to keep seperate $E_1$ .

0 votes

I think the answer depends on the type of relation R2 is.

If R2 is

**one to one**relation then we can merge both entities in to single entity as both the entities are**total participating**entities.(Total 2)**one to many or many to one**, then we need two tables for E1 and E3 (Total 3)**Many to many**need 3 tables for R2 E1 and E3 (Total 4)