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3 votes
3 votes

What is minimum number of tables requires, to represent above ER diagram?

(Does total participation and partial participation will take a role in this question)?

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2 Comments

I think 4 table are enough to represent this.
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how?
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2 Answers

9 votes
9 votes
Best answer

4 tables

,

edited by

4 Comments

Lets Assume we can merge E3 AND R3 ,

as you see there is redudancy present in table of E3R3 ,thats why we cant merge,thats why in general in M-M relation always three tables are required.

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thank u sir
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 sir,

In minimized normalized form, we should have 3 criteria,

  1. PK available
  2. 1NF(not more than 1 value in a cell)
  3. number of table is minimized or not

though there is redundancy, but criteria is not about that.here we're only seeing for minimum no. of table.

So, can we make 2 table ???

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2 votes
2 votes

The Number of entity in the ER diagram are 4 E1,E2,E3,E4

According to the conversion rules of (ER to Relational )  4 tables will be there for the Entities .Also there is a Relationship R3 which is holding many-to-many relationship so 1 table will be there to maintain this relationship.

Till Now total of 5 tables we can create.

But, Optimization is possible when one to one relationship is there with atleast total participation on one entity.So Entity E1 and E2 will Merge with Each other.

So, Answer will be 4