Lec 14: How to find out Normal form of a Relation| how to identify Highest Normal Form | part1
Вставка
- Опубліковано 7 вер 2024
- In this lecture, you will learn the Simplest Way to find out Highest Normal Form of a Relation in DBMS.
Best DBMS Tutorials : • DBMS (Database Managem...
*******************************************
Connect & Contact Me:
My Second Channel Link: bit.ly/354n7C7
Facebook: / jennys-lectures-csit-n...
Quora: www.quora.com/...
Instagram: / jayantikhatrilamba
Twitter: / khatrijenny
*********************************************
More Playlists:
C Programming Tutorials: • Programming in C
C++ Programming Tutorials: • Lec 1: How to Install ...
Placement Series: • Placements Series
Data Structures and Algorithms: https: • Data Structures and Al...
Design and Analysis of Algorithms(DAA): • Design and Analysis of...
Python Full Course: • Python - Basic to Advance
Printing Pattern in C: • Printing Pattern Progr...
Dynamic Programming: • Dynamic Programming
Operating Systems: // • Operating Systems
#dbms #dbmstutorials #normalization #jennyslectures
#howtoidentifynormalformindbms
#howtofindhighestnormalformindbms
#normalizationindbms
#ugcnet
#gatecslectures
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
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.
but we know have studied if a FD is in higher normal norm it will automatically in it lower normal forms
AT 19:09 BC -> D. where B and C are prime and D is non prime ...........means. prime -> non prime , therefore it should be in 3NF
It is a proper subset of Candidate key but not the candidate key
Yes it's 3NF, but for 3NF there is a condition that it will be in 2NF, here it fails 😂 it's tricky
For 3NF, the determinant has be a super-key OR the dependant has to be a prime attribute. Here, BC as a determinant is not a super key and D is not a prime attribute too. SO it's not in 3NF.
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.
Tomorrow my DBMS exam
After visit this channel I believed that i will be clear all subjects this semester exam.....
Thanks a Lot Mam......
🙏🙏🙏
This women, Jenny, is a DBMS angel.
Thank you so much ma'am.
Thank you for being here to help us.
You cleared all my doubts
Ma'am I love that way you explain hard concept in easy way thanku so much ma'am 😘
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.
For 3nf lhs must be Super key, not prime attribute. As BC is not SK thus it is not in 3NF.
Same doubt.. Mam plz explain
@@suvankar54 for 3nf it should not follow NPA->NPA so it's 3nf
Sorry mam u npa--->npa trick fails.
I have also a same doubt.
Have u got it the answer
For this question?
Your videos really helped me a lot
Love u from jharkhand😍😍
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?
Same bro
Same bro
@@Sayan8118bro pahli condition yh hai ki esko 2NF me hona chahiye agar yh 2NF me hota hai tb hm (NPA -> NPA) check karenge uske baad decide krenge 3NF. so esse badhiya hai ki 2nd method use Karo 3nf ko check krne ke liye (super key and prime attribute wala) 🌝🌝.
Mam ,In last example
BC->D
Both b and c are prime attributes but why you are not taking it as 3NF
BC is a prime attribute and so is D
therefore, we have "prime attribute" --> "prime attribute" so it's not a 3NF
@@nasreddinemerabtene7597 D is not a prime attribute bro
There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
Forget such condition like NFA->NFA just stick on the upper condition.
@@nasreddinemerabtene7597 There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
Forget such condition like NFA->NFA just stick on the upper condition.
Answer pls
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 again..........
Same doubt... Mam plz explain
@@KhushiJain-xx2yk have you got your doubt clear??
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 the thing you are mentioning i.e NPA->NPA would certainly hold true.
World's Best Teacher..
Flawless, stunning and talkless teacher you are
💯🥰
wah, kya mast
tricky question
In case of 3NF
BC->D
Here,B and C both are prime attributes.
There is no NPA->NPA.
So no transitive dependency exist if we are taking the condition NPA - >NPA so it must be in 3NF
bt taking this condition it must not be in 3NF
NPA->NPA condition or transitivity property is only for non-prime attributes. But here BC is a prime attribute,so it is not in 3NF.( You can also check in another way that BC is not a superkey and D is not a Prime Attribute.No condition is satisfied, so not in 3NF )
There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
Forget such condition like NFA->NFA just stick on the upper condition.
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 the thing you are mentioning i.e NPA->NPA would certainly hold true.
At 19:21, this is NOT a transitive dependency. So how can this violate the 3rd normal form?
Other than just the NPA -> NPA should not be present, there is also a rule for 3NF that it should satisfy at least one of the following for the FD:
1. LHS is SK
2. RHS is a prime attribute
Since neither of these are followed, it violates 3NF (Refer the 3NF video)
@@monkeywings But when ma'am discuss about the 3NF
Bcoz of transitive dependency there updation anomaly create
So please explain the transitive dependency clearly
great for your presentation. so much thinks maam!!!!
HI mam recently i saw your video its really osm the way you are teaching it its good for understand the concept .. keep updates with new videos thanks a lot .and love you for your videos
Way of explaining is very pretty good and you are saying examples for us it's very easy to understand every thing carry on mam...............keep teaching........
9:52 here the point is clear . but i have another super shortcut, if no attribute in LHS can derive A,B,C then ABC is a candidate key.
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?
Have the same doubt bro.
📲📲📲📲📲 you are great ...
I really like your concept 😊😉
21.47 last summary is very help full for competitive exam
can u please explain its last point??
I like the way she says 'I dawn't care' and 'rawl number'
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?
yes this relation should not be in 2NF. I have same doubt too.
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.
I am also having the same problem.Jenny mam help me out
same problem with me buddy
I think use only the method that
1)left hand side super key
Or
2)right side prime variable
Then it is in 3NF
Same problem it will be in 3NF
There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
Forget such condition like NFA->NFA just stick on the upper condition.
Mam my exam is at day after tomorrow and i've watched many videos and really this type of teaching DBMS is not available any other video
Thank you
i think that BC -> D is in 3NF because
BC is prime attributes
D not prime attribute
so "prime attributes "-> "not prime attributes" is in 3NF
relation which is not in 3NF where NPA->NPA
thanks
There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
Forget such condition like NFA->NFA just stick on the upper condition.
@@bleed0P what is nfa???
@@ayushijindal3899 hey it was a little mistake over there...its NPA* instead of NFA
@@bleed0P There is NPA -> NPA but only if it is in 2NF.
For directly checking 3NF, what you said should be done.
Thanks for this lecture, Jenny Mam 🙏🙏
Starting 19:10, to check for 3NF, isnt it already in 3NF? Aren't B and C are prime attributes and condition for 3NF is that there should not be any FD between Non prime attribute --->to Non Prime attribute? Or maybe i understood something wrong in the previous video of 3NF
Ya bro,same doubt.
There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
Forget such condition like NFA->NFA just stick on the upper condition.
Very nice lecture. Keep going.
Maam your teaching is really very helpful
Why not you start teaching us java python
in the last example prime attributes are A,B,C
Non prime attribute is only D,
And last functional dependency is BC-->D
BC both are prime attribute determining non prime attribute.
confused here.
please help out.
same doubt
OK. BC are subsets of AB and AC, so BC -> D is a partial key dependence.
Thank you Jenny mam
R={A,B,C,D}
A->C
B,D->B
A->D
A->B
In the above relation, is the "B,D->B" BCNF?
NPA -> NPA fails here.
Lectures are very good thank you but it would be good if you use some good markers 😊
guys i have a doubt:
given R(ABCDE)
F:A->B
C->D
D->E
C.K={AC}
prime atr={A,C}
Non prime={B,D,E)
isnt A->B
a partial depedency?
cuz A(proper subset of CK) --> B(non prime atr)?
this mean R is not in 2Nf ryt?
but according to 21:34 1st rule , R should be in 2NF?
conflict arising
1, 4,R
1,5,B
2,4,R
2,5,B
3,4,R Mam how to find highest normal form in this problem?
Hi Jenny, I'm not able to understand why you said ABC determines EF. Which Armstrong axiom is it? At 8:25
haha it's too late but I'll still answer it, i guess it's transitive, from ABC you can determine D, thus you have A,B,C,D 4 attributes, and using all these, ABCD, you can determine E.
in last example BC was a prime attribute so why it was not in 3nf because there was no nop->nop relation. in( BC->D)
The Venn diagram is inverted.
but BCNF is a higher normal form than 2NF so i think it's all reverse
please complete this DBMS series maam..........................................................
21:50 🥳 noted
a relational schema r is in _______normal form but not ______normal form only if atleast 1 proper subset of candidate key determines proper subset of other candidate key dependencies exist. what is answer ?
in first question ABCD->EF is NPA->NPA (D,E,F are non primary attributes), which means its transitive dependency ,so how is it 3NF?
bro have you got answer of this ??
@@sauravsharma9229 have u git the answer of this ?? same doubt plz tell
In previous video you have mentioned that for a relation to be in BCNF, they should strictly satisfy two condition i.e. LHS should be a Super Key and RHS should be a Prime attribute. Here while solving the first question, you only checked the first condiion and not the second one. Kindly help me if I am getting it right and the reason for not checking the 2nd condition.
first of all, those two conditions are for 3NF
and it should satisfy ATLEAST ONE condition not both conditions.
BCNF --> for each non trivial X--> Y,
X (L.H.S) must be super key (this is what she said)
Mam aap sirf 2 subjects(DS, DBMS ) pdha rhe ho, baaki subjects bhi padhaayo, Variety bnaao apne channel pr subjects ki, Pls.
Is se aapka channel bhi kaafi jyaada subscribers se bhr jaaega.
Comment acha lga toh like kro Mam!!
Referencing to your comment, what all technologies u r looking for?
@@SonaliProgrammingHub See,for technologies there are various good channels.
But this channel should include other subjects which are related to college studies (computers).
making it more complex
At 19:20:
Tranaitivie dependency says:
If Non prime Attribute --> Non prime attribute
Then it holds transitive depency. But in this case
BC is not Non- prime, so there is no transitive dependency. This should also be in 3NF ? Not ?
yes it is in 3rd nf
Please the board is light RED & BUT WRITING IS RED PEN, SO IT IS NOT SEEN CORRECTLY TO THE STUDENTS, MAM SIR...
Plz mam algorithm and coa k lectures upload kijiye
Here in second problem B and C are prime attributes but as per rule of 3nf no non prime attribute determine non prime attribute.here the non prime attribute is D but D cannot determine non prime attribute so it is in 3nf if any body knows the answer please reply 🙏
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 the thing you are mentioning i.e NPA->NPA would certainly hold true.
❤
20:00 do we know the year it was asked?
Can somebody please explain the last point?? @22:35 ?
there is 3 options for a relation either np->np , np->p or p->np , it is a 3NF so np->np doesnt exist we have all CKs are simple so every prime attribute is a CK and so every prime attribute is a super key and so p->np in this case is in BCNF and now for the last relation np->p, we know p is prime attribute so it is a canditate key in this case and so np is determining a canditate key which means np can determin all attributes and so it is a superkey so np->p is in BCNF and with that all realtions are in BCNF so the point she lade is right.
Iam form electical but still i watch your videos
Hii ma'am, could you plzz check ven diagram of 1NF, 2NF, 3NF, BCNF As per My knowledge it is inverted so plzz check that and give me reply
prime attribute determines non prime attribute how it voilates 3NF property.
Ur sooo beautiful and Ur voice alsooo😍😍😍nice explained sis
Love u jenny
Please use & write the BLUE PEN ON THE BOARD MISS TEACHER...
madam, i am also in Haryana, lets catch up some fine day
at 12:19 is ABCD->EF not a transitive dependency?
D is NPA and E is NPA 🥲🥲
please help!
Even im confused here. If your doubt is clarified then plz help me in understanding this...
LHS is a super key
could not understand how she is cancelling not cancelling out B and D at 16:08
Gate cse 2011 rank approx 6000
Maim plz Koi professional Course Highlight kijye ga ....Ye sb Tou Hum University m b prhlyty hn. .....Taught us Some professional Languages Or subject Like Js React .or Mean stack developing. We wanna Learn these languages And Move forward In Our feild ..........
4:29 all you get to start
✌✌
20:20
7:55 anybody explain me
i didnt get it
Mam mcs22 ka video upload kro
hi
miss
In this discussion ...F.D{ABC->DE, E->GH, H->G, G->H, ABCD->EF} Here since ABC->DE in the first relation of this functional dependencies so naturally ABC->D & ABC->E by decomposition rule applying. Ok fine no issue. But since ABC->E again ABCD->EF .. So ABCD->E And ABCD->F ... So is it the fact that ABC->E also there ABCD->E So from the R.H.S ABC = ABCD????? And Since ABCD->F so it also implies ABC->F?? As a result you can discard F by existing ABC only?? Please clarify the logic ... You know discarding true is very very important for finding S.K and C.K so far... Your teaching level was not smart in this lecture... Try to video this again. Thank you....
you have mistaken, The group of attributes A,B,C determines E and also A,B,C,D determines E (A,B,C and A,B,C,D are two different groups of attributes which separately identify E). Two sets of attributes determining same attribute doesn't make them equal.
Two completely different groups of attributes or partially different groups of attributes can determine same attribute or same set of attributes.
If ABCD->F doesn't make ABC->F because we require D to be present as a member of the group to identify F. ABC->F because ABC->D and ABCD->F in place of D we can write ABC but ABC are already present in L.H.S so we can't write ABCABC->F, we can't write same attribute two times. So ABC->F.
I love you. I just keep looking you. So beautiful
mam tho gurgaon ka hai
Hello can you make some Python programming lectures.
python is so easy, you can do it on your own.
Mam vlog channel kb banare ho ??
Already bna hua h.... Channel name is Jenny Lamba
Mam i want to say that one think
If you don't angry me.,😡😡
Your face and voice is so cute.👌
I can't control say that magic word🥳
I love you mam 🥰🥰
Can you propose me mam.💞💞💞💞
I love your sister
maam app faltu ki bthye bhut kri ho video main ak bth ko hi baar baar aree itni diyyan se smj rha tha firse vahi bth bol rhi ho baar bar or fir bol rhi ho yeglt h fir se smja thi ho video short or easy explain kiya kro maam gumaaya mt kra kro
You keep on said that in the example gurgon and haryana ..do you live in that place 🙄🙄 ?is this is in your home?
She is from Haryana. 😊
@@anujmor17 how do you know ? ;)
madam, aap to niri angrezi ho
I still have doubt why i can't see u in unacadmey
Hi jeeny . I really like your all videos and also you. If i am single single then my wish to propose you. 😃
Don't mind please....
Thank you