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.
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
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 ????
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.....
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
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
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.. !! ✨🙌
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.
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
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.
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.
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.
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
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?
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 ??
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
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❤🌸
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)
@@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
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
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
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?
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
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
In first example..AF is candidate key and A is it's proper subset and A is determining non-prime attribute then their will be partial dependency exist, but you have said that "only the part of c.k is determing non-prime attribute "..? I don't understand
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 👍
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 .
Mam, in second question, inorder to find ck, from AB is possible to derive B directly through the transitive rule AB_c AB-d &d-bto get ab-b,?is IT possible mam
Can someone please guide how to "Convert" it into 2NF?? I got the Answer of the homework ; AC is a C.key but relation is not in 2NF.. How to convert it??
She never ask for like and subscribe .
She simply teaches her subject so rare these days.
Lots of love from all students mam.
yes
🥺❤️💯
true
@@stackritesh true
Ok
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.
To make it into 2NF we have to divide it into 2 table
ABD and AC right?
@@subhamkumarsah7885 yes bro
bro h b....k....l
thank you bhai kabhi mile toh mere se 20 rupay le lena 😊😊😊😊😊😊😊😊
right answer bro
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
S mam I'm also getting 4 CK's
then why in video she got 3 candidate keys
Yes there will be 4 CK according to me too.
Follow the rules.
she missed out 1 more...
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
In 2NF(partial dependency --> left hand side proper subset of candidate key not candidate key itself)
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@@akashsharma2216 padhai pe dhyaan de
Some heroes do not wear capes.
Jenny mam you are the only one who teach in a better and simple way.......still your videos worthy......thank you so much.....🥺💕
Hey ,in 1st example we are getting. Baf,caf,daf,eaf candidate keys right
Am lucky to found your channel,because i found all topics which i wanna learn with a fabulous explanation
Hey ,in 1st example we are getting. Baf,caf,daf,eaf candidate keys right
This is hands down some of the best videos on databases I have come across.
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)
asa am calculat si eu si mi a dat bine, bafta la examen!!!
Correct
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 ????
yes
Yessss
Not in 2nd normal form because ck={A,C} and A->B .here is an partial dependency
Shut up
correct
one correction AC is candidate key
If only one FD is partial then also it's not in 2NF?
@@aditirai1909 Yes, if even one FD is partial, then its not in 2NF
CK = AC,
Non Prime Attributes={B,D}
A+ ={A,B,D} € Non Prime
So Partial Dependency
Implies R is not in 2NF
Hey ,in 1st example we are getting. Baf,caf,daf,eaf candidate keys right
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.....
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
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Tommorow iz my Dbms viva!!!
Jenny's ma'ams videos helped me alot!!!
I LOVE YOU!! I'm 4 days away from my exam and FINALLY, I have the whole clear picture.
People like you make gate preparation much easier. Thank You very much for your efforts Ma'am. Stay Safe :)
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
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.. !! ✨🙌
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.
the way she teaches definitely way better than the prof teach in my class, love it!!!
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
our country needs selfless teacher like her
God bless you. You're an amazing teacher!
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.
finally understood 2NF.... can't believe this can be explained in such easy to understand way. thank you!
These videos are like an emerald in a coal mine.
Your explanation is very nice madam... i am following your lectures daily... Thank you very much for providing these lectures to us...
Question :- R(a,b,c,d)
FD={A--->B, B---->D}
Answer:- not in 2nd normal phone because PA={a,c}
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.
Your lecture is pakka...I loved it..I was searching so many video regarding normalization today I got the concept..thank you mam
Hey ,in 1st example we are getting. Baf,caf,daf,eaf candidate keys right
As a computer engineering student, if Indian guys don't exist, probably I wouldn't learn anything about computer technologies :)
2:30 start
I think in the second example CD is also a candidate key along with AB, CB and AD.
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.
The explanation is crystal clear !!!!!!!!!!! Thank You !!!!!!!!!
Thank you for your all efforts, I can easily understand all these complicate things.
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.
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
All I can say is thank you
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 ) .
AC is the only candidate key. The prime attributes are A,C. It is not in 2NF because A->B also A->D(transitive) .
Your explanation is great ful thank you so much mam 💫👌😊🤗
Ans. Of hw question is
Not in 2nd NF
❤❤❤ Amazing lecture
but in the second last example there is a partial dependency so it should not be in 2NF
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?
your videos are very helpful mam
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 ??
I am answering the question you've asked in second normal form...my answer is, yes that relation is in second normal form
Not in second normal form. Because of partial dependence A->B.
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
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 .
yup
Yes
Happy teachers day mam
waooo great lecture ...i can watch on dbms last question ans : not second form
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❤🌸
If there would be oscar in Teaching then mam will definetly get the award
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....
Thank you so much Madam😍😍😍😍😍
Ma'am you are pretty and your teaching method is very impressive
Ur nice mam u r so gd in teaching, lv u mam
Sory it is u r
@15:45 I think there can be four candidate keys AB, CD, CB, AD.
Please correct me if I am wrong. Thanks in advance.
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)
@@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
@@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
For 15:51 example, Won't CD be also a candidate key ??????????????????????????????????
Very nice method of teaching.....
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
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
Best teacher ever🤩🥰
AC is the CK, and A->B ; partial dependency exists:
therefore,, its not 2NF.
notes : definition and example : 6:10
12:00 varaikum paaru
what's the difference between the candidate key and prime attributes? (9:07)
AC is Ck but the relation is 1nf not 2nf because A->B is partially dependent and also exist non -prime attributes.
Ma'am u are looking so beautiful. And thanks alot for this lacture
best video for 2nf
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?
No through COMPOSITION rule,A->C and B->D
Excellent way of teaching.
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
Hi I think only three candidate keys
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Your lecture is too much good👌👌👌👌
Hey ,in 1st example we are getting. Baf,caf,daf,eaf candidate keys right
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
in example 3 proper subset of A is ABCD then how can it become Candidate Key??
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
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 .....?
Very easy to understand, thank you Jenny
Ma'am thank you thank you thank you very very much 🙏🏻
In first example..AF is candidate key and A is it's proper subset and A is determining non-prime attribute then their will be partial dependency exist, but you have said that "only the part of c.k is determing non-prime attribute "..? I don't understand
can anyone explain what is the point of creating the normal forms ? Like what is the significance of this is real scenario?
Thanks Mam
Amazing Video
🙏🙏
You are a wonder... Thank you ma'am
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 👍
Can u please share your solution? Coz I am getting only 3 ck. Either (AB,BC,CD) or (AB,BC,AD).
@@fayazmd bhai from B can be replaced by D as there is a FD given , so AD
I am surprised that you did not give a practical example using a table and attributes
In second example you stated as 3 Candidate keys,But we have 4 Candidate keys,AB,CB,AD,CD
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 .
@@shyamprakashm6325 thank you
Love you ma'am
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.
In this example CD also a prime attribute is it right or wrong.
Mam, in second question, inorder to find ck, from AB is possible to derive B directly through the transitive rule AB_c AB-d &d-bto get ab-b,?is IT possible mam
what if there exists no proper subset of ck in FD? Can we consider it as in 2NF?
Thanks
Can someone please guide how to "Convert" it into 2NF?? I got the Answer of the homework ; AC is a C.key but relation is not in 2NF.. How to convert it??
Mam, what if we get these two situations 1. Non prime-> Non prime 2. Non prime-> Prime
Thank you jenny 🥰
Why not CBB and AAB while replacing the prime attributes??
Nice knowledge spreading spray
God bless you mam