Determine the highest possible normal form of the given relation?

Question

Given R(ABCD) and

AB → C ;

ABD → C ;

ABC → D ;

AC → D

which the highest possible normal form for the above relation ?

Answer 1

Any relation is not in 2NF iff PROPER SUBSET OF CANDIDATE KEY determines NON PRIME ATTRIBUTE..
AB is key here & there is no FD which Violate 2NF definition.
it is in 2NF but not in 3NF..

Comments

why not in 3NF?

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because AC -> D is (prime) Transitive Dependency which is not allowed in 3 NF.

using the word 'prime' as A is prime attribute, it is combined with a non prime attribute C to determine another non prime attribute D, which still makes it Transitive Dependent.

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AC is prime then.

prime -> nonprime is not a matter for 3NF

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AB is the only candidate key in this example, and thus A and B are prime attributes and C and D are non prime attributes. AC imply (prime + non-prime)

[(prime + non-prime) -> non-prime] dependency is still a Transitive Dependency just like any other [non-prime -> non-prime] dependency

This is not allowed in 3 NF

Answer 2

Above relation is in 1 NF

Here (AB)⁺ =ABCD

So, AB is the candidate key and ABD,ABC are super key

here in,AC→D , AC is nonprime attribute partially dependent on prime attribute ABC

So, it is not even 2NF, it will be 1NF

Answer 3

Answer is 2 NF

Step 1: We need to find the minimal FD set or Canonical FD set

Minimal FD set    : { AB -> C, AB -> D, AC -> D }

Canonical FD set : { AB -> CD, AC -> D }

Step 2: Finding the Candidate Keys

In this case, AB is the candidate Key

Step 3: Classify the FD to Corresponding Highest Normal Form

1. AB -> CD (Full Functional Dependency)

FD is in BC NF (Candidate Key/Super Key -> any set of attributes)

2. AC -> D (Prime Transitive Dependency)

FD is in 2 NF (Prime(key) Attributes + Non-Prime(non-key) Attributes -> Non-Prime (non-key) Attributes)

Thus, Highest Normal form for the given FD set is 2 NF

Answer 4

AC-> D is a partial functional dependency , which is not allowed in 2NF , so the relation must be in 1NF

Comments

AC -> D is (prime)Transitive Dependency and not Partial FD, so it is in 2NF

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Can you explain how can we identify partial dependencies?

If [(prime)+(non-prime)]->non-prime  transitive dependencies exists then not partial dependencies?

Pls help.

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A---->B .

A ia proper subset of key and B is non prime attribute then it is partial dependency .

example

R(ABC)

 (KEY =AB)

A---->C

HERE A IS PART OF KEY AND C IS NONPRIME ATTRIBUTE SO IT IS PARTIAL DEPENDENCY.

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Yes, you are right.But how AC->D is not partial dependencies in this question?

Since A is part of candidate key.

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No .. You take all attribute of LHS.  here AC is proper subset of key.  

Not only take A.

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Nice explanation. Doubt clear.