Lec 11: Second Normal Form in DBMS | 2NF in DBMS | Normalization in DBMS

In this lecture, We will learn:
What is Second Normal Form in DBMS.
What is Partial Dependency
What are the rules and requirements for a relation to be in Second Normal Form

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[rakeshpatel]

She never ask for like and subscribe .
She simply teaches her subject so rare these days.
Lots of love from all students mam.

[tirtharajdas]

the answer for the homework-->
AC is the candidate key but relation is not in 2NF, because A is proper subset of AC(candidate key)and A->B, where B is non prime attribute.

[shashwatjha]

The question with R(A, B, C, D) FD:{AB->CD, C->A, D->B } There will be 4 candidate keys AB, AD, BC, CD. Timestamp : 15:50

[exclusivefacts]

Am lucky to found your channel, because i found all topics which i wanna learn with a fabulous explanation

[peerless]

Jenny mam you are the only one who teach in a better and simple your videos you so

[ScottTrogam]

This is hands down some of the best videos on databases I have come across.

[arinrahman]

1NF:
Each attribute should contain atomic values
A column should contai value from the same domain
Each column should have unique name
No ordering to rows and columns.
No duplicate rows.

2NF:
It must be 1NF
No Patial dependency in the relation (Partial dependency occurs when the left hand side of a candidate key points non-prime attributes)

3NF:
It is in 2NF
No transitive dependency for non-prime attributes
(To be non transitive and 3NF atleast one of these must be true: Either the left handside of funtional dependency is superkey or the right handside points to a prime attribute)

BCNF:
A relation is BCNF if it is 3NF
For each functional dependency there must be a super key

[koraykara]

Only AC is a candidate key, prime attributes are {A, C} and the relation is in 1NF but not in 2NF because of the partial dependency (A -> B)

[subhashrao]

Not in 2nd normal form because ck={A, C} and A->B .here is an partial dependency

[waseemakramkhan]

Good Evening Ma'am, ur explaination is very helpful and understandable, Thanks alot.
Plz make complete videos on SQL's numericals, Transaction management & Concurrency and File Structure


Last Question which is given by you in 2nd Normal Form Lecture ....The answer would be the given relation, R(A, B, C, D) & F.D.={A->B, B->D} is not in 2nd Normal Form

[Raj]

CK = AC,
Non Prime Attributes={B, D}
A+ ={A, B, D} € Non Prime
So Partial Dependency
Implies R is not in 2NF

[Touay]

Watching from Kashmir !!!
Tommorow iz my Dbms viva!!!
Jenny's ma'ams videos helped me alot!!!

[ashankavindu]

madam obviously I don't know how I thankful to you....I'm speechless about this service....God bless you madam...great English pronunciation and well love you

[varunsaproo]

People like you make gate preparation much easier. Thank You very much for your efforts Ma'am. Stay Safe :)

[KayYesYouTuber]

Hi Jenny, you take so much effort to explain things. I like your videos very much. But for this particular video, you can take a practical example like orders database to explain partial dependence. That would have made things easier to understand.

[Salehalanazi-]

God bless you. You're an amazing teacher!

[tubakzgn]

Answer of last example question: We obtain AC closure as candidate key which shows us the FD is not in the 2nd normal form. Perfect explanation, thank you very much

[trinidadbosch]

I LOVE YOU!! I'm 4 days away from my exam and FINALLY, I have the whole clear picture.

[abdullaharean]

The question with R(A, B, C, D) FD:{AB->CD, C->A, D->B } There will be 4 candidate keys AB, AD, BC, CD. Timestamp : 15:50
Details Explaination:
Classification of the attributes:

| I(isolated) | L(left) | B(both) | R(right) |

| - | - | A, B, C, D | - |


Union of I and L:
Computation of the closure of the attributes from Step 4
Attributes on the both side of FD: A, B, C, D
Compute the closure of the combination of the power set of B(both) and .
A⁺ = A ⊂ R
B⁺ = B ⊂ R
C⁺ = AC ⊂ R
D⁺ = BD ⊂ R
We have not found any candidate keys by adding one-element sets to
AB⁺ = ABCD = R (candidate key)
AC⁺= AC ⊂ R
AD⁺ = ABCD = R (candidate key)
BC⁺ = ABCD = R (candidate key)
BD⁺= BD ⊂ R
CD⁺ = ABCD = R (candidate key)
Adding any other attribute leads to a superkey.
Hence, ['AB', 'AD', 'BC', 'CD'] are the (only) candidate keys.
AB, AD, BC, CD

[sasikalav]

Your explanation is very nice madam... i am following your lectures daily... Thank you very much for providing these lectures to us...