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Let $E_1$ and $E_2$ be two entities in an $E/R$ diagram with simple-valued attributes. $R_1$ and $R_2$ are two relationships between $E_1$ and $E_2$, where $R_1$ is one-to-many and $R_2$ is many-to-many. $R_1$ and $R_2$ do not have any attributes of their own. What is the minimum number of tables required to represent this situation in the relational model?

1. $2$
2. $3$
3. $4$
4. $5$

edited | 2.2k views
+1

Notice line --> simple-valued attributes.

If it is  "simple valued attribute" then answer may change.

@Anu007 ji,

For ex. -> If $E_{1}$ has only one attribute then no need to create a separate table for it.

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explain that also how will change?
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Let E1 and E2 be two entities in an E/R diagram with simple single-valued attributes. R1 and R2 are two relationships between E1 and E2, where R1 is one-to-many and R2 is many-to-many. R1 and R2 do not have any attributes of their own. What is the minimum number of tables required to represent this situation in the relational model?

original question

thats y i am saying you are not considred multivalue attribute.

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Let E1 and E2 be two entities in an E/R diagram with simple single-valued attributes. R1 and R2 are two relationships between E1 and E2, where R1 is one-to-many and R2 is many-to-many. R1 and R2 do not have any attributes of their own. What is the minimum number of tables required to represent this situation in the relational model?

2 Tables.

+2

https://www.geeksforgeeks.org/gate-gate-cs-2005-question-75/
visit this link for better explanation

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Anu007, dont you think answer of this question depends upon the participation(either total or partial) of a entity set.

suppose $R2$ is both side with total participation and $R1$ is total participation from the $1$ side then we need two tables

$E1R2E2$ and $E2R1$

+3

Since particiation is not given , we need to assume partial only.

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@Ani_  Yeah Geeks for geeks explanation is also good

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Can someone explain why we can't combine (E1,R1) into one table ?

We need a separate table for many-to-many relation.
one-to-many relation doesn't need a separate table and can be handled using a foreign key.
So, answer is $B$ - $3$ tables.

Reference: http://web.cse.ohio-state.edu/~gurari/course/cse670/cse670Ch9.xht

by Veteran (416k points)
edited
0
Update the reference ...
0

@saurabh rai,

@Arjun sir,

Here in relation for relationship R . b3 from E2 is not related to anyone in E1 similarly a3 from E1 is not related to anyone in E3 , So they are not present in relation of R.

1) So that means when we are creating new relation for relationship , we only include entities which are participants of relationship only ?? Means there will be no NULL entries in separate relation for any relationship ??

2) Is creating New relation for Relationship like for M:N , is different from merging two relations because in merging entires for all entities will be included whether they are participant of relationship or not , Thats y they have NULL entries ??

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do we need 3 tables for even 1NF?
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@sushmita

Can you please check my above comment...is it right ?
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i think for 1 NF 2 tables would be sufficient? Isn't the question ambiguous without specifying normal forms?
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How 2 tables are sufficient ..we have to create new table for M:N ..also as no total participation on any of the side..so no chance of merging also....
+1
So that means when we are creating new relation for relationship , we only include entities which are participants of relationship only ?? Means there will be no NULL entries in separate relation for any relationship ??

yes you are right here we will only include those entries which are involved in the relationship.

If we try to establish the relationship without creating new table, we have to include many duplicates and we might encounter many null entries too.
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yes but when we are normalizing. In 1 NF we can merge everything.
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Yes but here ..which normal form is not mentioned so wht we should take...if 1NF then 1 table??
+1
as per the last years trend consider 3NF while solving questions unless specified.
+1 vote

Answer is 3 tables

by Loyal (5.4k points)
0
3 Table are

1. E1

2.E2

3..R2
+2

Let E1 has attributes (a , b)  and E2 has (c , d)

So tables will be, for relationship R1 i.e. 1 to m. T1(a , b) and T2(c , d , a).

For relationship R2 i.e. m to n.

We need one extra  table T3(a , c).

So total 3 table required.

The situation given can be expressed with following sample data.

E1
a
b
c

E2
x
y
z

R1
E1  E2
a    x
a    y
b    z

R2
E1   E2
a     x
a     y
b     y 
by (11 points)

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