She never ask for like and subscribe . She simply teaches her subject so rare these days. Lots of love from all students mam.
@yarravenkat3882
4 жыл бұрын
yes
@isaitech4602
3 жыл бұрын
🥺❤️💯
@stackritesh
2 жыл бұрын
true
@ranjiniramachandran629
2 жыл бұрын
@@stackritesh true
@shashikarkandikonda5582
2 жыл бұрын
Ok
@tirtharajdas2165
3 жыл бұрын
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.
@subhamkumarsah7885
2 жыл бұрын
To make it into 2NF we have to divide it into 2 table ABD and AC right?
@HarshYadav-qz2bm
2 жыл бұрын
@@subhamkumarsah7885 yes bro
@akashsharma2216
2 жыл бұрын
bro h b....k....l
@mohamedsahilali8809
Жыл бұрын
thank you bhai kabhi mile toh mere se 20 rupay le lena 😊😊😊😊😊😊😊😊
@ALLINONE-yu9bu
7 ай бұрын
right answer bro
@peerless3538
2 жыл бұрын
Jenny mam you are the only one who teach in a better and simple way.......still your videos worthy......thank you so much.....🥺💕
@Yashkyk
5 ай бұрын
Hey ,in 1st example we are getting. Baf,caf,daf,eaf candidate keys right
@exclusivefacts8956
3 жыл бұрын
Am lucky to found your channel,because i found all topics which i wanna learn with a fabulous explanation
@Yashkyk
5 ай бұрын
Hey ,in 1st example we are getting. Baf,caf,daf,eaf candidate keys right
@shashwatjha9491
4 жыл бұрын
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
@angelpreethi5813
4 жыл бұрын
S mam I'm also getting 4 CK's
@tejaswigopaladasu7372
4 жыл бұрын
then why in video she got 3 candidate keys
@MansiJain-qv1lz
4 жыл бұрын
Yes there will be 4 CK according to me too.
@Logan-vd1kf
4 жыл бұрын
Follow the rules.
@sayyidiskandarkhan3064
3 жыл бұрын
she missed out 1 more...
@ScottTrogam
3 жыл бұрын
This is hands down some of the best videos on databases I have come across.
@arinrahman8368
2 жыл бұрын
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
@satyamkalyane6841
2 жыл бұрын
In 2NF(partial dependency --> left hand side proper subset of candidate key not candidate key itself)
@akashsharma2216
2 жыл бұрын
I LOVE YOU
@sriramkrishnamurthy4473
2 жыл бұрын
@@akashsharma2216 Behen ke lawde 🤣🤣🤣🤣🤣🤣🤣🤣🤣🤣 comment section me bhi flirt maarne aa gaya tu bkl 😁😁🤣🤣
@suzz4668
Жыл бұрын
@@akashsharma2216 padhai pe dhyaan de
@tubakzgn1102
Жыл бұрын
Some heroes do not wear capes.
@subhashrao7996
4 жыл бұрын
Not in 2nd normal form because ck={A,C} and A->B .here is an partial dependency
@manishjoshi9737
3 жыл бұрын
Shut up
@mmrtech5495
3 жыл бұрын
correct
@aviralkhanduja5834
3 жыл бұрын
one correction AC is candidate key
@aditirai1909
3 жыл бұрын
If only one FD is partial then also it's not in 2NF?
@rishavhimmatramka7804
3 жыл бұрын
@@aditirai1909 Yes, if even one FD is partial, then its not in 2NF
@koraykara6270
4 жыл бұрын
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)
@andreiardelean5712
4 жыл бұрын
asa am calculat si eu si mi a dat bine, bafta la examen!!!
@subhamdhar8196
4 жыл бұрын
Correct
@roqayamuhammad7867
4 жыл бұрын
I'd highly appreciate it if anyone can HELP me. I have an assignment to normalize a table. The issue is I wasn't provided FDs or any keys. So how and where to start. ( I did find four FDs, not sure though if they're right or not). What to doooo ????
@mohandattabayya5584
3 жыл бұрын
yes
@heretojustvibe5760
2 жыл бұрын
Yessss
@ashankavindu2409
3 жыл бұрын
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 explained.....much love you madam.....
@Raj3486
3 жыл бұрын
CK = AC, Non Prime Attributes={B,D} A+ ={A,B,D} € Non Prime So Partial Dependency Implies R is not in 2NF
@Yashkyk
5 ай бұрын
Hey ,in 1st example we are getting. Baf,caf,daf,eaf candidate keys right
@Touay
3 жыл бұрын
Watching from Kashmir !!! Tommorow iz my Dbms viva!!! Jenny's ma'ams videos helped me alot!!!
@waseemakramkhan7093
4 жыл бұрын
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
@varunsaproo4120
3 жыл бұрын
People like you make gate preparation much easier. Thank You very much for your efforts Ma'am. Stay Safe :)
@wendixiao9275
Жыл бұрын
the way she teaches definitely way better than the prof teach in my class, love it!!!
@trinidadbosch8792
2 жыл бұрын
I LOVE YOU!! I'm 4 days away from my exam and FINALLY, I have the whole clear picture.
@tubakzgn1102
Жыл бұрын
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
@Salehalanazi-7
4 жыл бұрын
God bless you. You're an amazing teacher!
@Julia-xl7pr
Жыл бұрын
finally understood 2NF.... can't believe this can be explained in such easy to understand way. thank you!
@vaibhavsharma399
3 жыл бұрын
our country needs selfless teacher like her
@sasikalav5058
3 жыл бұрын
Your explanation is very nice madam... i am following your lectures daily... Thank you very much for providing these lectures to us...
@KayYesYouTuber
3 жыл бұрын
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.
@sivaranjanis3655
4 жыл бұрын
Your lecture is pakka...I loved it..I was searching so many video regarding normalization today I got the concept..thank you mam
@Yashkyk
5 ай бұрын
Hey ,in 1st example we are getting. Baf,caf,daf,eaf candidate keys right
@AdamHarrisongpl-projx
3 жыл бұрын
These videos are like an emerald in a coal mine.
@samsons8279
Жыл бұрын
Answer for exercise at 20:14 Partial Functional dependency exists which is A--> B Because A is a proper subset of the candidate key AC , and B is a non-prime attribute. Can't thank u enough Jenny ma'am !! 🤩🙌 The videos on Normalization are just amazing.. !! ✨🙌
@feyzanur8860
3 жыл бұрын
Thank you for your all efforts, I can easily understand all these complicate things.
@angshumanpaul999
4 жыл бұрын
The whiteboard at time brighten up way too much, I know its because of the lighting, but at times it becomes too bright that my eyes start paining. Except this, you are one of the best tutors on youtube.Thankyou for helping thousands of students in need.
@abhishekkr1133
3 жыл бұрын
2:30 start
@D1STANG3R
Жыл бұрын
As a computer engineering student, if Indian guys don't exist, probably I wouldn't learn anything about computer technologies :)
@eagle_shadow6665
2 жыл бұрын
Your explanation is great ful thank you so much mam 💫👌😊🤗
@banyaroo8691
4 жыл бұрын
The explanation is crystal clear !!!!!!!!!!! Thank You !!!!!!!!!
@rajnarayanshriwas4653
2 жыл бұрын
I think in the second example CD is also a candidate key along with AB, CB and AD.
@ITACHIitachiitachi-y7v
12 сағат бұрын
Happy teachers day mam
@keycode_302
Жыл бұрын
your videos are very helpful mam
@atj8697
4 жыл бұрын
All I can say is thank you
@mq4950
4 ай бұрын
Ans. Of hw question is Not in 2nd NF ❤❤❤ Amazing lecture
@abdullaharean257
Жыл бұрын
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
@vaibhavjaviya6100
3 жыл бұрын
Question :- R(a,b,c,d) FD={A--->B, B---->D} Answer:- not in 2nd normal phone because PA={a,c}
@khushbookumari-so8dt
4 жыл бұрын
There is only 1 Ck key i.e AC. In the question, there is a partial dependency on A->B that's why this is not in 2NF.
@mansikumari3917
5 ай бұрын
but in the second last example there is a partial dependency so it should not be in 2NF
@rizwanreshi8673
4 жыл бұрын
Ur nice mam u r so gd in teaching, lv u mam
@rizwanreshi8673
4 жыл бұрын
Sory it is u r
@meghalpatel8195
4 жыл бұрын
waooo great lecture ...i can watch on dbms last question ans : not second form
@ARK1010-cl8fd
4 ай бұрын
AC is the only candidate key. The prime attributes are A,C. It is not in 2NF because A->B also A->D(transitive) .
@EvancePatrick-y7p
Ай бұрын
I am answering the question you've asked in second normal form...my answer is, yes that relation is in second normal form
@saptarshicse0735
3 ай бұрын
Thank you so much Madam😍😍😍😍😍
@Meditation987
Жыл бұрын
Ma'am you are pretty and your teaching method is very impressive
@rahulrudra5339
4 жыл бұрын
Very nice method of teaching.....
@bhandarisoniya2580
4 жыл бұрын
There is only 1 CK i.e . AC having PA (A,C ) and it's not in 2 NF becoz of PD of (A> B ) which is PA of our AC (Candidate key ) .
@myyoutubeisthis
2 жыл бұрын
Ma'am thank you thank you thank you very very much 🙏🏻
@shivalikagupta3433
3 жыл бұрын
not in 2NF because {AC} is candidate key , hence non prime attributes are {B,D} and {A->ABD} (by transitivity), which is partial dependency . But partial dependency cant exist in 2NF.
@srujan099
4 жыл бұрын
ma'am @ 15:41 replace B with D in CB ( since FD : D --> B) we will get another C.K i.e., CD therefore total 4 C.K's... thank you ma'am .
@bense_tony
3 жыл бұрын
yup
@virendrakumarshukla8848
2 жыл бұрын
Yes
@rohitdatta4549
5 ай бұрын
Your lecture is too much good👌👌👌👌
@Yashkyk
5 ай бұрын
Hey ,in 1st example we are getting. Baf,caf,daf,eaf candidate keys right
@MonkeyD.3892
Жыл бұрын
Thanks Mam Amazing Video 🙏🙏
@FINANCIALYOGI
4 жыл бұрын
AC is only CK, PD exists as A,C are PA. B,D are NPA. A as subset of AC determines NPA B and D. Therefore PD exists and not in 2NF. Kindly advise if correct. Thank You.
@KONETI__LOKESH
6 ай бұрын
No R(A,B,C,D) Not in 2 NF Because here there is only one. Candidate key which is AC prime attributes are A,C The proper subset A determining the non prime attribute so it follows partial dependency So it is not in 2NF
@suresh.suthar.24
Жыл бұрын
best video for 2nf
@LagGamers143
2 жыл бұрын
1 million subs very soon🥳
@novaira9186
3 ай бұрын
Given a relation R( P, Q, R, S, T, U, V, W, X, Y) and Functional Dependency set FD = { PQ → R, PS → VW, QS → TU, P → X, W → Y}, determine whether the given R is in 2NF? If not convert it into 2 NF. could you solve it mam please?
@knowledgemaster5049
4 жыл бұрын
Excellent way of teaching.
@AjayThakur-zb3ee
4 жыл бұрын
Ma'am u are looking so beautiful. And thanks alot for this lacture
@babytoy2333
Жыл бұрын
Sorry ma'am! In 2nd example there are 4 cks here AB is equivalent to CD cos C=>A and D=>B that why A can be replaced with C n B with D now thus we'll get CD too
@MaheriMihirima
2 жыл бұрын
In last example R(A,B,C,D) F.D={A->B, B->D} here Ck={AC}, P.A={A,C}, non prime attribute={B,D} , partial dependency is present. so this is not 2NF. But I have a question which is partial dependency A->B or A->D ??
@starultra2863
Жыл бұрын
For 15:51 example, Won't CD be also a candidate key ??????????????????????????????????
@Mohammad_raza_01
Жыл бұрын
If there would be oscar in Teaching then mam will definetly get the award
@knowledgemaster5049
4 жыл бұрын
Not in second normal form. Because of partial dependence A->B.
@rahulvarma7257
3 жыл бұрын
Mam thank you so much the video was really helpful. And the last problem is not in second normal form,please reply for this mam it would be great help for me
@lewishoanglong1610
4 жыл бұрын
Very easy to understand, thank you Jenny
@afiraarifa1306
4 жыл бұрын
You are a wonder... Thank you ma'am
@csboi5235
2 жыл бұрын
notes : definition and example : 6:10 12:00 varaikum paaru
@SonamYadav-ux8yj
3 жыл бұрын
Best teacher ever🤩🥰
@tehlion7430
Жыл бұрын
what's the difference between the candidate key and prime attributes? (9:07)
@ApnaChanel11
3 жыл бұрын
Nice knowledge spreading spray
@virendrakumarshukla8848
2 жыл бұрын
AC is Ck but the relation is 1nf not 2nf because A->B is partially dependent and also exist non -prime attributes.
@harshverma9488
4 жыл бұрын
Ma'am in second example you said AB is a candidate key and also AB->CD, C->A, D->B then if we replace A to C and B to D then CD is also a super key? Is it a candidate key ?? As per i understand the concept it is a candidate key if i am wrong then plz clear my doubt.... also i want to tell you are an amazing teacher your lessons are helpful to me...thank you so much❤🌸
@vishaltrivedi540
3 жыл бұрын
@15:45 I think there can be four candidate keys AB, CD, CB, AD. Please correct me if I am wrong. Thanks in advance.
@HemanthKumar-if8vu
3 жыл бұрын
yeah its actually 4 CK, but she stopped because, she found out that CK={AB,CB,AD} is enough for answering it as in 2NF.. how? by that time CK={AB,CB,AD} she noticed that PrimeAttributes={A,B,C,D} - which has all the attributes in the given relation and hence no possiblity of getting non prime attributes by the subset of CK(even from following any iteration of finding CKs)
@deryasonmez2524
3 жыл бұрын
@@HemanthKumar-if8vu If CD is candidate key, it wouldn't be 2nf as it defines C-> A? I think going to this partial dependence. . Please correct me if I'm thinking wrong
@miasevda4728
3 жыл бұрын
@@deryasonmez2524 but A is a prime attribute , what you say is the rule of the 3rd forme , so the relatio is not in the 3rd forme
@vivek-rathod
3 жыл бұрын
15:54 mam we have one more CK that is CD ?????? I m confuse mam BTW thank you mam for amazing explanation I am learning so many things from your videos 👍
@fayazmd
3 жыл бұрын
Can u please share your solution? Coz I am getting only 3 ck. Either (AB,BC,CD) or (AB,BC,AD).
@neelanjanghosh1586
3 жыл бұрын
@@fayazmd bhai from B can be replaced by D as there is a FD given , so AD
@kummarguda
2 жыл бұрын
I am surprised that you did not give a practical example using a table and attributes
@shivam7164
Жыл бұрын
Ac is candidate key and not in 2nf because a->b where a is a proper subset of candidate key which implicant b ie the non prime attribute. In this case prime attributes are ac and non prime attributes b and d
@ashwinichandrachar
2 жыл бұрын
Thanks
@RitikKumar-km5io
7 ай бұрын
Thank you Mam🙏❤!
@shaunsoans6463
10 ай бұрын
God bless you mam
@selomenebit5153
7 ай бұрын
Thank you jenny 🥰
@rightwinger2709
4 жыл бұрын
Mam the question you are solving at 18:18 is not in 2nd normal form perhaps. It contain two CK {A,AB} and B is pointing to non-prime attribute....
@gyanaranjansahoo2872
2 жыл бұрын
Love you ma'am
@madhumithaa7966
4 жыл бұрын
In second example you stated as 3 Candidate keys,But we have 4 Candidate keys,AB,CB,AD,CD
@shyamprakashm6325
4 жыл бұрын
No You miss the key ..after candidate key has found you should check whether the primeattributes of the candidate key is present in the given dependencies ...in this manner , we should get .
@madhumithaa7966
4 жыл бұрын
@@shyamprakashm6325 thank you
@harman_made
2 жыл бұрын
Good explanation
@DQuranJar
3 жыл бұрын
15:50 The candidate keys will be AB , AD , BC , CD, C, D right? I mean AB -> C and AB -> D Proper subset of C is phi as well as D. Therefore, C and D are candidate keys?
@ayushigoyal6853
3 жыл бұрын
No through COMPOSITION rule,A->C and B->D
@lekesanusi751
2 жыл бұрын
The answer to the question : R(A,B,C,D) f.d> (A->B, B->D) is that it is not in second normal form. Because "A " proper subset of CK(AC) determines non prime attribute "B" i.e A->B. There exist p.d
@rajdeepdas4581
4 жыл бұрын
Mam 'A' can not be a canditate key? because closere of A is a super key and it's proper subset is an empty attribute .....?
@preetamvarun9219
11 ай бұрын
Thank you ❤
@altinarexha6134
4 жыл бұрын
In the second example you used the one from hw in Part 1 finding candidate key.. But the result is different from there, I have doubts which one is correct please
@saipavankumar4498
3 жыл бұрын
Hi I think only three candidate keys
@AkshayAnil0-1
3 жыл бұрын
AC is the CK, and A->B ; partial dependency exists: therefore,, its not 2NF.
@hudaharoon5367
2 жыл бұрын
Thanku soo much mam ❤️❤️❤️❤️
@SureshBabu-og4ys
2 жыл бұрын
In this example CD also a prime attribute is it right or wrong.
@vin5954
2 жыл бұрын
in example 3 proper subset of A is ABCD then how can it become Candidate Key??
@jay-rathod-01
4 жыл бұрын
Excuse me ma'am, how long would it take to complete the playlist of dbms as well as OS . If possible please lemme know the dates . Regards, Jay
@saikatpatra4239
4 жыл бұрын
only AC is the candidate key and there is partial dependency exists in the relation corresponds to A->B and therefore the relation is not in 2nf.
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