1:1 relation with total participation at atleast one side can be merged.

The Gateway to Computer Science Excellence

+1 vote

Consider the following ER diagram:

How many number of relations are required for the above ER diagram?

- 2
- 3
- 5
- 1

Solution:

My doubt is:

Since $E_2$ isn’t involved in total participation with $E_1$ so on merging we might get 2NF violation.

Eg:

Let $E_1$ be:

<p1,q1>

<p2,q1>

<p3,q2>

$p \rightarrow q$

Let $E_2$ be:

<r1,s1>

<r2,s2>

<r3,s2>

<r4,s3>

$r \rightarrow s$

And p1 maps to r1, p2 maps to r2 and p3 maps to r3.

Merging with $E_2$ will give:

<r1,s1,p1,q1>

<r2,s2,p2,q1>

<r3,s2,p3,q2>

<r4,s3,NULL,NULL>

$pr \rightarrow sq$

$r \rightarrow s$

$p \rightarrow q$

2-NF violation.

So by default which case is to be considered when normalization form is not mentioned?? 1-NF?

+2

It is a 1:1 relationship with total participation on one side. As P is pkey of E1 and R is pkey of E2, so on merging E1, E2 and R1, the dependencies that are applicable are: P->Q, P->R, R->S. So P is the pkey of the merged relation. As it isnt a composite key, so there wont be any violation of 2NF right?

Where are you getting a violation?

- All categories
- General Aptitude 1.9k
- Engineering Mathematics 7.5k
- Digital Logic 2.9k
- Programming and DS 4.9k
- Algorithms 4.4k
- Theory of Computation 6.2k
- Compiler Design 2.1k
- Databases 4.1k
- CO and Architecture 3.4k
- Computer Networks 4.2k
- Non GATE 1.4k
- Others 1.4k
- Admissions 595
- Exam Queries 573
- Tier 1 Placement Questions 23
- Job Queries 72
- Projects 18

50,737 questions

57,331 answers

198,441 comments

105,197 users