Minimum number of tables to represent ER-Diagram

Question

The minimum number of tables to represent ER-Diagram such that the relational database satisfies 1NF. 

Comments

The minimum number of tables are 4.

Explanation:-

Straightly we can create tables for bank,bankbranch,loan. Total tables are 3.

But in bank entity have multivalued attribute(phone),in relational model we can't store  multiple values for single attribute.

So  we need to create  one more table to store  phone numbers, with one more attribute which  reference the one of attribute uniquely indentifies the rows in bank table.

ex: phonenumbers(code,phone);

So Total minimum tables are 4. 

---

@Satbir here in top answer and in above comment by bikram sir.. most of them saying that 1 table..

i am not understanding why? Multivalued attributes should not be there for 1 nf also ..right?

so we need two tables

---

we made 2 tables one normal and one for multivalue attribute.

After doing this we can merge them into a single table . creating tables from Er daigram are by default in 1 NF

Answer 1

There is only one condition that needs to be fulfilled in 1NF that all values should be atomic no other constraint

1 table is enough.

Every 1 NF can be brought under 1 table

Comments

@Tesla, 

 There is only one condition that needs to be fulfilled in 1NF that all values should be atomic no other constraint

What will be the answer for 2 Entities related by a M:N relationship ? Can they also be brought under 1 table?

---

Yes if you want 1 NF then it doesn't matter how much complicated table is it can be brought under 1 table.

---

Redundancy or no redundancy, a relational table must be a valid relational table first, i.e. each entity in the relation must be accessible through some key. If that is not possible, then no point of 1NF, 2NF, 3NF, etc.

Answer 2

$4$ tables. 

1. {Ph_No}
2. {Branches, Bank Branch}
3. {Bank}
4. { Loan, Loans }

Comments

@reena_kandari Phone_no is multivalued attribute. If you break it without making an extra table you will have to generate 1 or more columns/attributes like Phone_no2, P_No3 etc ( based on the cardinality of multivaluedness of the column)  which are not present in the original ER diagram. This ER diagram has only 1 attribute which is Phone_no.

---

@Aghori 

https://gateoverflow.in/3717/gate2004-it-73

for 1 nf it should be 3 table right ?

---

@set2018 && @aghori for 1 NF you can merge Multivalued attribute in one single table but here you can't do it bcz if you merge it then PK of BANK will be will {Code,phone no} and in BANK_BRANCH there is FK {code} which refers to PK of BANK, but only {Code} is not the PK of this table so you have to keep a separate table for {phone_no} here..Hence we come up with 4 tables here.

Answer 3

4 tables are required

1.Bank(Code,Name,Address)

2. For multivalued attribute(Ph_No,Code)

3.Bank Branch(Addr,Branch_No,Code,Bank Name)

4.Loan(Loan_no,Amount,Type,Branch_no)

Answer 4

As phone number is multivalued attribute there should be one table for it.

Now Loan and the Loans can be combined into a single table as the relationship Loans is many-to-one relationship and participation is total in many side.There should be another one table.

Now there will be only one table needed for Bank,Branches and Bank Account.Because there is full participation between entities in the relationship.

So three tables should be there.But the question mentioned 1NF so we can combine these three tables into 1 table.So the final answer will be one I think.

Comments

See that's why I'm confused.

Everyone is saying 4 tables, I thought 3 and you've also said 3 tables. As we can combine bank branch and bank into one due to total participation and have primary key of bank branch as the primary key of this whole relation as its many.

So aside from this 1nf debacle, what would be the correct answer 3 or 4?. If we're considering separate tables for multivalued attributes.