UGC NET CSE | July 2018 | Part 2 | Question: 67

Question

A many-to-one relationship exists between entity sets $r_1$ and $r_2$. How will it be represented using functional dependencies if $Pk(r)$ denotes the primary key attribute of relation $r$?

• A. $Pk(r_1) \rightarrow Pk(r_2)$
• B. $Pk(r_2) \rightarrow Pk(r_1)$
• C. $Pk(r_2) \rightarrow Pk(r_1) \text{ and } Pk(r_1) \rightarrow Pk(r_2)$
• D. $Pk(r_2) \rightarrow Pk(r_1) \text{ or } Pk(r_1) \rightarrow Pk(r_2)$

Comments

Same Question here:

https://gateoverflow.in/335229/ugc-net-cse-january-2017-part-3-question-9 

Answer 1

Answer: $A$

$Pk(r)$ denotes the primary key attribute of relation $r$.

• The functional dependency $Pk(r_1) \rightarrow  Pk(r_2)$ indicates a many-to-one relationship between $r_1,r_2,$ since any $r_1$ value which is repeated will have the same $r_2$ value, but many $r_1$ values may have the same $r_2$ value.

NOTE:

Consider the statement:

1. A many-to-one relationship set exists between entity sets student and instructor.
2. A many-to-one relationship set exists between entity sets instructor and student.

Read the above statements & think about what do they mean.

Statement 1 says that “one instructor may have many students, but one student learns from at most one instructor”.

Statement 2 says that “one student may have many instructors, but one instructor teaches at most one student”.

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Source of this UGC NET Question: Korth DBMS Book.

https://www.db-book.com/db6/practice-exer-dir/8s.pdf   (Page 10)

http://web.cs.ucla.edu/classes/fall04/cs143/solutions/ch7.pdf (Page 86)

Same Question here:

https://gateoverflow.in/335229/ugc-net-cse-january-2017-part-3-question-9 

Answer 2

Many to One relation btw entity sets r1 and r2

 

I) assume r1 is left side of relation and r2 is right side of relation.

many entities of r1 connected to one entity of r2

A  ---------->   1

B  ---------->   1

C  ---------->   1

D   ---------->   2

E  ---------->   3

by knowing the entity of r1, you can say what is entity of r2 ===> pk(r1) → pk(r2)

by knowing the entity of r2, you can not say what is entity of r1 ===> pk(r2)  → pk(r1) is not hold

∴ pk(r1) → pk(r2) ===> Answer should be A

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II) assume r2 is left side of relation and r1 is right side of relation.

many entities of r2 connected to one entity of r1

A  ---------->   1

B  ---------->   1

C  ---------->   1

D   ---------->   2

E  ---------->   3

by knowing the entity of r2, you can say what is entity of r1 ===> pk(r2) → pk(r1)

by knowing the entity of r2, you can not say what is entity of r1 ===> pk(r1)  → pk(r2) is not hold

∴ pk(r2) → pk(r1) ===> Answer should be B

Comments

ok so u r  taking relationship in either direction , then what is the final ans option C or D

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sorry sir, you are getting into wrong way....

final answer neither C nor D

it should be A ----> when r1 is left side of many-to-one relation ----> Totally different question

it should be B ----> when r2 is left side of many-to-one relation ---> Totally different question

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then  ans will be A

Answer 3

B is the right Answer

Comments

Don't give only option answers

Explanation  is mandatory

Answer 4

Option - A

Comments

Explain???