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 ????
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
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.....
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 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.
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
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.
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.
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.
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❤🌸
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
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
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?
Candidate keys :- AC, AB Prime attributes :- A,B,C Non-Prime attributes :- D So there partial dependency exist So this relation is not in 2NF Am i right Ma'am ?
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 .
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 👍
In the first video of ck of playlist , she gave the same question for homework and she told answer is AB,AD,BC,CD are ck so i think therre are 4 cks not 3 ck
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?
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 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
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
