Lec 14: How to find out Normal form of a Relation| how to identify Highest Normal Form | part1

In this lecture, you will learn the Simplest Way to find out Highest Normal Form of a Relation in DBMS.

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[bincymali]

Maam... Your classes r awesome.. I have completed MCA in 2014 but saddest fact is dat i hav got the idea of normalization today after watching your videos.thank u so much mam.. May God bless u always

[randomindex]

For those who have a doubt that why the relation in second question was not in 3NF as it satisfies the condition of transitive dependency i.e NPA->NPA.
For a relation to be in 3rd NF, it must be in 2nd NF and there should not be any transitive dependency.
But here the relation is not even in 2NF. Hence, it violates the first condition of 3NF. So, there is no need to check the 2nd condition of 3NF.
If it was in 2NF. Then this thing i.e NPA->NPA would certainly hold true.

[qasidmubashir]

AT 19:09 BC -> D. where B and C are prime and D is non prime prime -> non prime, therefore it should be in 3NF

[tabnashahid]

My exams have ended but I still need a better understanding of the subject and since you're keeping your promise of completing the DBMS series, I'm really grateful to you. Thank you so much.

[shuklaIASaspirant]

Tomorrow my DBMS exam
After visit this channel I believed that i will be clear all subjects this semester
Thanks a Lot
🙏🙏🙏

[CSEDHATSHAYINIS]

Mam, In last example
BC->D
Both b and c are prime attributes but why you are not taking it as 3NF

[ArvindSingh-wjvy]

Thank you so much ma'am.
Thank you for being here to help us.
You cleared all my doubts

[rizashah]

Ma'am I love that way you explain hard concept in easy way thanku so much ma'am 😘

[sachingoel]

The last example is both 2NF and 3NF. No subsets of CKs - AB & AC can determine D (non-prime attribute) so it's 2NF.
Also, NPA->NPA cannot be true as there is only one NPA i.e. D, so it's in 3NF.

[tirtharajdas]

I have a confusion, in the third FD (BC -> D), if i check if it is violating the 3NF or not with the rule that, if it show (NPA -> NPA) that is not in 3NF. Here BC is not NPA, so how can we say it is not in 3NF by this rule?

[ankitchakraborty]

Your videos really helped me a lot
Love u from jharkhand😍😍

[soubhagyalaxmisahoo]

Ma'am In R( A B C D )
F.d are given
You say us transitive dependency is
NPA determine NPA
But here in this example
Fd BC->D
Here B and c are prime attribute
And D is non prime attribute
Please explain it

[RAHULKUMAR-wnpo]

Mam I have a doubt. At 12:18 In last FD : ABCD > EF is of form Non- Prime att ---> Non-prime att. Because ABC are prime attribute but D is not. So ABCD is a non prime attribute. Hence this FD is transitive and will not satisfy 3NF. But mam you checked the tick mark there. Why?

[akshatgupta]

In case of 3NF
BC->D
Here, B and C both are prime attributes.
There is no NPA->NPA.

[basitaliahmed]

At 19:21, this is NOT a transitive dependency. So how can this violate the 3rd normal form?

[dnyaneshpande]

Flawless, stunning and talkless teacher you are
💯🥰

[renegade]

14:31 But ABCD -> EF is a FD where the left-hand-side is proper subset of Candidate key and the right-hand-side is non prime attribute. Wouldn't that be contradict 2NF?

[anshuraj]

📲📲📲📲📲 you are great ...
I really like your concept 😊😉

[subhosen]

Maam in the last example of this video tutorial, The last FD BC->D, BC is Not SK and D is not Prime Attribute.So naturally it is not in 3NF.But when I tried out using the Transitive Dependency Property B, C are Prime Atrributes.So I can say there is no transitive dependency in the relation so it is in 3NF.Maam I can't understand what is wrong with my solution.So please help me out.

[saikiranravipati]

Way of explaining is very pretty good and you are saying examples for us it's very easy to understand every thing carry on