Minimum number of tables to represent ER-Diagram

Question

How many minimum relations required for given ER diagram ?

Comments

@Satbir
I have a few questions.
1>while converting for ER-model to relational model foreign keys can only point to candidate keys.?
2>if we have a table X(A,B) where A and B are the attributes and A and B form the compound keys,then  if we have a table Y(P,Q,R) where R is the foreign key of Y  we can have (PQ ) as the foreign key in Y which points to the compund key (AB).?

3>foreign keys cannot point to non-key attributes.right?

Please answer.

 

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foreign keys point to primary keys , not candidate keys.

A candidate key is a key which has the capability to become a primary key, but there is only 1 primary key in a table.

https://stackoverflow.com/questions/42268886/how-to-have-a-foreign-key-pointing-to-two-primary-keys

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https://csedoubts.gateoverflow.in/16478/self-doubt-dbms-er-diagrams

That mean here RELATIONAL MODEL no.1  is correct .right?

Answer 1

hope  this helps...

Comments

why we are combining $E2-R3-E3$ ?

Since E3 is not totally participating combining them will produce some redundancies

Answer 2

answer must be 5 tables

Answer 3

Ans is 5

{R1} {E1R4E2} {E5R6} {R2} {E3R3R5E2}

Answer 4

Total 8 tables are required

Let (EntityName,NumberOf Table)

(E1,1),(E2,3),(E3,1),(E4,1),(E5,2) 

E2 has 3 tables because in suppose we have 

So 3 tables required for E2 and E5 is weak entity so 2 tables required

total = 1+3+1+1+2 = 8 tables 

Reference :-http://www.edugrabs.com/entity-and-its-types/

Comments

okay . got it

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But where is it given that R4 is 1:1?

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check arrows.

if u didn't get then i ll suggest to go through basics once of ER