# DBMS | How to find the highest normal form of a relation

To understand this topic, you should have basic idea about

Functional Dependency & Candidate keys

Normal forms

Steps to find the highest normal form of a relation:

1. Find all possible candidate keys of the relation.
2. Divide all attributes into two categories: prime attributes and non-prime attributes.
3. Check for 1st normal form then 2nd and so on. If it fails to satisfy nth normal form condition, highest normal form will be n-1.

Example 1. Find the highest normal form of a relation R(A,B,C,D,E) with FD set {A->D, B->A, BC->D, AC->BE}

Step 1.   As we can see, (AC)+ ={A,C,B,E,D}  but none of its subset can determine all attribute of relation, So AC will be candidate key. A can be derived from B, so we can replace A in AC by B. So BC will also be a candidate key. So there will be two candidate keys {AC, BC}.

Step 2.  Prime attribute are those attribute which are part of candidate key {A,B,C} in this example and others will be non-prime {D,E} in this example.

Step 3.  The relation R is in 1st normal form as a relational DBMS does not allow multi-valued or composite attribute.

The relation is not in 2nd Normal form because A->D is partial dependency (A which is subset of candidate key AC is determining non-prime attribute D) and 2nd normal form does not allow partial dependency.

So the highest normal form will be 1st Normal Form.

Example 2. Find the highest normal form of a relation R(A,B,C,D,E) with FD set as {BC->D, AC->BE, B->E}

Step 1.   As we can see, (AC)+ ={A,C,B,E,D}  but none of its subset can determine all attribute of relation, So AC will be candidate key. A or C can’t be derived from any other attribute of the relation, so there will be only 1 candidate key {AC}.

Step 2.  Prime attribute are those attribute which are part of candidate key {A,C} in this example and others will be non-prime {B,D,E} in this example.

Step 3.  The relation R is in 1st normal form as a relational DBMS does not allow multi-valued or composite attribute.

The relation is in 2nd normal form because BC->D is in 2nd normal form (BC is not proper subset of candidate key AC) and AC->BE is in 2nd normal form (AC is candidate key) and B->E is in 2nd normal form (B is not a proper subset of candidate key AC).

The relation is not in 3rd normal form because in BC->D (neither BC is a super key nor D is a prime attribute) and in B->E (neither B is a super key nor E is a prime attribute) but to satisfy 3rd normal for, either LHS of an FD should be super key or RHS should be prime attribute.

So the highest normal form of relation will be 2nd Normal form.

Example 3. Find the highest normal form of a relation R(A,B,C,D,E) with FD set {B->A, A->C, BC->D, AC->BE}

Step 1.   As we can see, (B)+ ={B,A,C,D,E}, so B will be candidate key. B can be derived from AC using AC->B (Decomposing AC->BE to AC->B and AC->E). So AC will be super key but (C)+ ={C} and (A)+ ={A,C,B,E,D}. So A (subset of AC) will be candidate key. So there will be two candidate keys {A,B}.

Step 2.  Prime attribute are those attribute which are part of candidate key {A,B} in this example and others will be non-prime {C,D,E} in this example.

Step 3.  The relation R is in 1st normal form as a relational DBMS does not allow multi-valued or composite attribute.

The relation is in 2nd normal form because B->A is in 2nd normal form (B is a super key) and A->C is in 2nd normal form (A is super key) and BC->D is in 2nd normal form (BC is a super key) and AC->BE is in 2nd normal form (AC is a super key).

The relation is in 3rd normal form because LHS of all FD’s are super keys. The relation is in BCNF as all LHS of all FD’s are super keys. So the highest normal form is BCNF.

Article contributed by Sonal Tuteja. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

This article is attributed to GeeksforGeeks.org

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