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Briefly explain the difference between first, second, third and fourth normal forms?

  
Total Answers and Comments: 7 Last Update: October 17, 2007     Asked by: anandm 
  
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December 15, 2005 04:30:51   #1  
Andyk Member Since: December 2005   Contribution: 2    

RE: briefly explain the difference between first ,seco...

Hi

FIrst Normal form : Attribute should be atomic.

Second Normal Form : Non-Key attribute should be fully functionally dependent on key Attribute.

Third normal Form : There is no transitivity dependency between attribute. Suppose 'y' is dependent on 'x' i.e. x->y and 'z' is dependent on 'y' i.e. y->z this is transitivity dependency So we can split table on to two tables os that result will be x->z.

Forth Normal Form : A determinant is any attribute (simple or composite) on which some other attribute is fully functionally dependent.

A relation is in BCNF is and only if every determinant is a candidate key.


 
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February 12, 2006 10:40:39   #2  
ejaita        

RE: briefly explain the difference between first ,seco...

the diffrence between the normal forms


 
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September 27, 2006 05:45:36   #3  
zukisani zaula        

RE: briefly explain the difference between first ,seco...

1)concept of funtional dependencies?

2)partial and full functional dependencies?

3)anomalies in unstructured relatioins ?

4)firt second and third normal forms ?


 
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December 08, 2006 23:33:49   #4  
sravan kumar        

RE: Briefly explain the difference between first ,seco...

1NF: A table is set to be in first NF if we identify the functional dependency.


2NF: A table is set to be in first NF if we identify and delete partial functional dependency.


3NF: A table is set to be in 3rd NF when we identify and delete transitive dependency.


Functional dependency: identify a non key attribute which is depends on key attribute.


Partial functional dependency: identify an attribute which is partially depends on key attribute.


Transitive: identify an key attribute which is independent itself.


 
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May 08, 2007 06:57:10   #5  
Nilesh Gupta        

RE: Briefly explain the difference between first, seco...

1 NF --> It has only normal form which not support functional dependency.
It has no multivalue attributes.

2 NF --> Every non key attribute should be depend on key attributes.

3 NF --> No transative dependency.

x -> y
y -> z

here no relation between x & z.


 
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July 19, 2007 08:10:10   #6  
bhawesh        

RE: Briefly explain the difference between first, seco...
First NF---Attribute should be atomic an no redundant row. (functional dependancy)

Second Normal Form : All Non-Key attribute should be fully functionally dependent on primary key.(removes partial dependancy)

Third normal Form : There is no transitivity dependency between attribute.


 
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October 17, 2007 02:15:11   #7  
anbum        

RE: Briefly explain the difference between first, seco...


Normalization is based on the functional dependency and primary key.


1NF- Attribute should be atomic value and there must be no composite attribute no multivalued attribute and no relation with in relation.
By default what ever relation we are creating in the oracle by default it will be in first normal form

2nf - There is no partial dependency. it is based on the full functional dependency. non key attribute should be fully dependent on the key attribute.

e.g
R {pno eno hours ename pname plocation}
given functional dependency
F {{pno eno}-> hours
eno->ename
pno->pname plocation}

in the above relation R ename is dependent on the eno and
pname and plocation are dependent on the pno
so the relation R can be splited in to
R1 {pno eno hours}
R2 {eno ename}

R3 {pno pname plocation}
now all the relations R1 R2 and R3 are in full functional dependency.


3NF - There is no Transitive dependency.
R {eno ename address dno dname}
functional dependency
F1 {eno->{ename address dno}
dno->dname}
if you have transitive dependency in the relation R then decompose the relation R in to R1 and R2

R1 {eno ename address dno}
R2 {dno dname}

def: x->A
x is a super key or A is a prime attribute
in the given dependency F1
dno is not a super key and dname is not a prime attribute.
so the relation R can be decompose into R1 and R2 and in relation R2 dno make it as a primary key.


 
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