# Minimum Number of tables for given ER-Diagram

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4
I think Answer should be 3, but given is 4..

No of table will be 3

E1, E2,E3 we need 3 tables

but as it gives 1:n relationship , so no more tables are needed

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Please mention the primary key and foreign key(if any) of each of the tables.
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E3 will be saparate  with E is key
First E1 R will combine with key CG then E1 R P will combine with key E1 G
E2 will kept seperate with key G.
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Primary key of table E3 i.e. E works as a foreign key in the relationship set E1PR. And primary key of E1PR is CG.

Is that correct?
While converting ER Diagram to Relation Table, We create a separate table for:

1. WEAK Entity.
2. STRONG Entity.
3. M:N Relationship.
4. Composite Attribute.
5. Multivalued Attribute.

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We don't create seperate table for composite attributes.
I think the answer must be 4.

For E1,E2,E3 we will defiantly need one table each.

By putting primary key of E2 i,e G in E1's table we can specify the indentifing relationship R between them. Hence no need of table for R.

But For P we will need a different table in which we will put primary key of E3 i,e E and Primary key of E1 i,e C  as foreign keys and the primary key of the table for  P will be the combination of E and  C i,e EC

Hence we need total 4 tables( for E1, E2 ,E3 and for P)
Answere will be 3 tables as in 1:n case primary key of 1 side is taken as foreign key in N side which is used to reference 1 side table thus ER diagram is simplified to 2 tables ...same approach can be used for other side ...hence ER diagram is reduced to tables E3(E,F) , E1(C,D,E,G) , E3(E,F)

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