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GATE CSE 2008 | Question: 82

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89 89 votes

Consider the following $\text{ER}$ diagram

The minimum number of tables needed to represent $M$, $N$, $P$, $R1$, $R2$ is

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

8 Answers

Best answer
147 147 votes

First strong entity types are made to tables. So, we get two tables $M$ and $P$.

I assume $R1$ is $1:1$ or $1:n$ as that would minimize the number of tables as asked in question. 

Now participation of $M$ in $R1$ is total (indicated by double arrow) meaning every entity of $M$ participate in $R1$. Since $R1$ is not having an attribute, we can simple add the primary key of $P$ to the table $M$ and add a foreign key reference to $M$. This handles $R1$ and we don't need an extra table. So, $M$ becomes ${M1, M2, M3, P1}$. 

N here is a weak entity weakly related to $P$. So, we form a new table $N$, and includes the primary key of $P (P1)$ as foreign key reference. Now $(P1, N1)$ becomes the primary key of $N$. 

Thus we get $3$ tables.

$M: {M1, M2, M3, P1}$ - $M1$ primary key, $P1$ references $P$

$P: {P1, P2} - P1$ primary key

$N: {P1, N1, N2} - (P1, N1)$ primary key, $P1$ references $P$. 

So, answers is B.

• edited by
7 7 votes

GATE | GATE CS 2008 

M, P are strong entities hence they must be represented by separate tables.

Many-to-one and one-to-many relationship sets that are total on the many-side can be represented by adding an extra attribute to the “many” side, containing the primary key of the “one” side. ( This way no extra table will be needed for Relationship sets )

M table is modified to include primary key of P side(i.e. P1). N is weak entity, and is modified to include primary key of P (i.e, P1).

Therefore there would be minimum of 3 tables with schema given below :

M ( M1, M2, M3, P1)
P ( P1, P2 )
N ( P1, N1, N2 )

Note: This modification of a table in the case of one-many or many-one to include relationship set at the many side works well, but only in the case when the relationship set doesn’t have its own attributes. If the relationship set has its own attribute then we need to make a separate table for the relationship set also.

3 3 votes

 for 82 ---> 3 table will be there ----

 for 83 ---> ans is a--> cause the R1 (M1,P1) and M(M1,M2,M3) ----both have the same primary key ,so we can merge them , so new table will be (m1,m2,m3,p1)

2 2 votes
Relationship  between M & P is N:1 so solve Accordingly .

The Side which Has "Arrow" has Cardinality = 1 always.(Remember That).
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