Question

how to find minimum no of table out of a given E-R diagram..?

Answer 1

Consider this exp 

Rules

For every entity we are going to get one table so for E1 and E2 we got 2 tables .

if the relation is either one to one OR one to many then we could represents the relation in one of the entity table .Better option is to shift the relation on many side. 

So for relation we don't need any extra table here

In case of multi-valued attributes we have to add one more table.

Here min 3 tables are required.

If you want to find out max no of tables then you can unnecessarily make a table for relation then count would be 4.

Comments

In short.. I should remember..

in many to many we will create 1 extra table ..and in rest three cardinality ratios we need not to create extra table..we can adjust them in entity set..right...??

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exactly

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thanxx..:)

Answer 2

We need just two tables for 1NF.

T1: {A11, A12, A13}

T2: {A21, A22, A23, A11}

A23 being multi-valued, {A21, A23} becomes the key for T2 as we need to repeat multiple values corresponding to the multi-valued attribute to make it 1NF. But, this causes partial FD A21 -> A22 and makes the table not in 2NF. In order to make the table in 2NF, we have to create a separate table for multi-valued attribute. Then we get

T1: {A11, A12, A13} - key is A11
T2: {A21, A22, A11} - key is A21
T3: {A21, A23} - key is {A21, A23}

Here, all determinants of all FDs are keys and hence the relation is in BCNF and so 3NF also. So, we need minimum 3 tables.