The Candidate keys are AE, then BE, then CE, then DE. So, all the attributes in the relation are prime attributes. Hence, it can be said that the relation is in 3NF. We have to check only for BCNF. None of the Functional Dependencies have super keys at its Left Hand Side, so, clearly the relation is not in BCNF. Hence, the highest Normal Form of this relation R(A,B,C,D,E) is 3NF.
I really appreciate your hardworking towards us , you explained it in very well and convenient manner even I never found such a great video as well as teacher on TH-cam and even in my college.
For the Relation R(A, B, C, D, E) and Functional Dependencies are FD = {A->B, B->C, C->D, D->A} the highest normal form in 3NF , and the prime attributes are {A, E, D, C, B}
Candidate key are AE, DE, CE, BE and prime attribute are A, B, C, D, E. If all attribute of a relation are in prime attribute then it would be in 3NF. so the highest normal form is 3NF.
How did she wrote/derived the F.D from the table? Can anyone plz tell me in which video she had explained about this because i'm continuous at watching her video from the very first video of this topic. I'm talking about 2:51
think of it as a function, look at every input/combination of inputs. If there are two of the same inputs say C=1, look at the output, for example A; we can see the same input of 1 for C outputs 2 different values for A, 1 and 3, therefore it can't be a FD. However, for A, every value is different therefore the output can be anything even the same for every input, so A -> B and A-> C are both FD's. This is a rough explination, watch the video of lecture 7 from her playlist to better understand.
Highest normal form is 3NF, because NPA is null , and condition for transitive dependency is NPA->NPA . For 3NF it should not be in transitive dependency.
Great lectures mam! I read in a text that a decomposition is dependency preserving if we can compute all FDs without joining R1 and R2, but here it is not possible, we cannot compute D --> A without joining R1 and R2.
Your vedios are excellent with full concept but it is a humble request to make short length vedios as it is quite difficult to watch whole content in exam days
Hello ma'am I have one doubt..Tomorrow have my exam, please solve my doubt. Ma'am for R1 why we are not take (AC) clouser..!?? But at R2 we take (CD) (DE) (CE) clouser... But R1 we only take (AB) clouser and (BC) clouser... If I take (AC) clouser then any problem..!??
R(ABCDE). FD(A-B,B-C,C-D,D-A). CK(as E not determined by any of attributes, AE,BE,CE,DE are CK. All attribute are Prime, therefore in 2NF. As all attributes are Primary, the RHS of SK-PA is PA for deciding 3NF, so R is in 3NF. BCNF requires SK-PA or NPA but none FD HV SK as determinants, not in BCNF. HNF is 3NF. Jesus bless U.
P.A(A,E,C,D,B)... no N.P.A..and the C.K (AE,DE,CE and BE) this relation in a 3rd NF bcoz, if no any N.P.A then definitely that relation it's in a 3rd NF and that left side of FD's not contained any S.K so BCNF is not possible.
Candidate keys are AE,DE,CE and BE. Prime attributes are A,B,C,D,E So, the relation is 3 NF but not BCNF.
Same answer brother
All keys are prime attributes, so we can blindly say this relation is in 3rd nf
@@ujithadharshana7397 but we must check for BCNF , if All keys are prime attributes
Yeah it should be 3NF.
Same answer
The Candidate keys are AE, then BE, then CE, then DE. So, all the attributes in the relation are prime attributes. Hence, it can be said that the relation is in 3NF. We have to check only for BCNF. None of the Functional Dependencies have super keys at its Left Hand Side, so, clearly the relation is not in BCNF. Hence, the highest Normal Form of this relation R(A,B,C,D,E) is 3NF.
very detailed and well explained, thats what i call a high quality stuff
I really appreciate your hardworking towards us , you explained it in very well and convenient manner even I never found such a great video as well as teacher on TH-cam and even in my college.
Very good way of teaching. Keep posting your knowledgable videos as they are helping us a lot. Thanks
Love your lectures, like lots of your fans, you are my Secret potion of power,
Really makes the lectures more easy❤️Thank you sooo much
simp
Thank you Soo much for ur lecture's mam ...... 😊Really you are genius mam 👍 .... Aap jaisi lecturer mere clg hona chahiye tha mam
Thank you Madam...Because of your classes I have cleared my dbms exam
Ma'am you teach so naturally that I feels like I am sitting in front of you to learn DBMS.
For the Relation R(A, B, C, D, E) and Functional Dependencies are FD = {A->B, B->C, C->D, D->A} the highest normal form in 3NF , and the prime attributes are {A, E, D, C, B}
Same answer
yes
Candidate key are AE, DE, CE, BE and prime attribute are A, B, C, D, E. If all attribute of a relation are in prime attribute then it would be in 3NF. so the highest normal form is 3NF.
Your so smart madame I like your lectures and thanks a lot because your videos are very helpful to me👋
Thank u so much❤️...Really catching the clear ideas behind every topic
Detailed explanation of the topic🔥🔥😀
Mam You teach very well👌👏😍
This time u taught better than Gate Smasher! , he did not cover decomposition neatly .
Thank you mam for making this video,you are the best 👌😊
mam your teaching skills are very good.
Thanks Mam for explaining such a Tough concept in a well Simplified & Clear manner.
Mam you teach very well. So, please apply for data structures in nit delhi.
candidate keys are AE,DE,CE,BE and prime attributes are A,E,D,C,B . and highest normal forms is 3NF.
Thanks mam😌, your the best
Ck--》 AE,CE,BE,DE AND PRIME ATTRIBUTES ARE A,B,C,D,E
AND THE HIGHEST NORMAL FORM IS 3NF.✅
Thank you for carrying us the 4CHIF!
ois oke heisem?
@@peppi44 Fühl mich gecarried
Excellent teaching mam...!!
How did she wrote/derived the F.D from the table? Can anyone plz tell me in which video she had explained about this because i'm continuous at watching her video from the very first video of this topic. I'm talking about 2:51
think of it as a function, look at every input/combination of inputs. If there are two of the same inputs say C=1, look at the output, for example A; we can see the same input of 1 for C outputs 2 different values for A, 1 and 3, therefore it can't be a FD. However, for A, every value is different therefore the output can be anything even the same for every input, so A -> B and A-> C are both FD's. This is a rough explination, watch the video of lecture 7 from her playlist to better understand.
Really very very helpful lectures
i dont have an words to appreciate you🥰🥰🥰🥰
Awesome explanation mam!!
Great explanation
Thank you Ma'am 😊
The Relation is in 1NF.
Thanku mam . Mam answer is 3NF
Highest normal form is 3NF, because NPA is null , and condition for transitive dependency is NPA->NPA . For 3NF it should not be in transitive dependency.
Nice now I can cross check my ans
Ck-AE,BE,CE,DE
It is in 3NF
You are the best...
Great lectures mam! I read in a text that a decomposition is dependency preserving if we can compute all FDs without joining R1 and R2, but here it is not possible, we cannot compute D --> A without joining R1 and R2.
the second one is also not Dependency Preserving. Good notice
Very good video for cheching preserv fd thanku mam
The highest NF is 3NF in the relation .
Thank u ma'am
Thanks from kerala🥰
Your vedios are excellent with full concept but it is a humble request to make short length vedios as it is quite difficult to watch whole content in exam days
Thank you mam
i'from ETHIOPIA...i'm here not to attend this lecture,but because of missing u my love jenny.......lov u forever
Thank you.
Respect from pakistan
ma'am please also teach ER diagram with example that topic is very important for interview.
should we need to understand all deeper concepts in DBMS??
Mam how would u divide this realtion
Have you ever thought about home tutions??
The highest normal form of R is 3NF.
Please make videos on Linux os
at 4:52 why didn't you take BC->A and AC->B in functional dependency
Mam what is the answer of this question:
R(ABCDEG) F:{AB->C , AC->B ,AD->E,B->D,BC->A,E->G} R1(ABC) R2(ABDE) R3(EG) Dependency is preserved or not?
yes preserved bro
Yes 👍
Thank you !
ck DE,CE,BE,AE .All attribute is prime attribute then direcktly we say this is in 3NF no need to check
mam plz make more video on DBMS
Given example answer is
Candidate key : AE BE CE DE
ITS NOT IN BCNF
ITS IN 3NF , 2NF , 1NF
HIGEST FORM IS 3NF
THanks a lot for the lecture. Suppose I have a dataset that has more than 20 columns. How can I decompose the dataset?
pretty and teach so well
Mam please make videos on 4nf 5nf
luv u vadhinA
It is in 3NF
What is that previous video that mam told in this video to derive dependency from given relation.I didn't find that video
Did you? Bcoz i'm still at search. Plz share it.
Just checkout the ibutton
thanks mam
First 😌🤭
Ma'am how did you write F.D.(BC->A ) ....i couldn't understand , Can you plz tell me in which video you explained about this.
this lady is CRAZY
Mam why you can't find the AC+ closure
Smile❤️
dude ur outfits are fire
Hello ma'am I have one doubt..Tomorrow have my exam, please solve my doubt.
Ma'am for R1 why we are not take (AC) clouser..!??
But at R2 we take (CD) (DE) (CE) clouser... But R1 we only take (AB) clouser and (BC) clouser... If I take (AC) clouser then any problem..!??
3rd normal form
highets normal form is 3NF.
3nf is the answer.
3NF
Watch lecture 4 is anyone is confused in deriving functional dep from tables
What are those numbers?
Tabah deya g
AE, DE, CE, BE ARE CK SO HIGHEST NORMAL FORM IS 3NF
R(ABCDE). FD(A-B,B-C,C-D,D-A). CK(as E not determined by any of attributes, AE,BE,CE,DE are CK. All attribute are Prime, therefore in 2NF. As all attributes are Primary, the RHS of SK-PA is PA for deciding 3NF, so R is in 3NF. BCNF requires SK-PA or NPA but none FD HV SK as determinants, not in BCNF. HNF is 3NF. Jesus bless U.
💛💛💛💛💛
3rd nf
SQL database video
Khatri + lamba ? Omg you are married 😣😣😣😣
Mam cutie 💓
😘
H.N.F.=>3NF
It is 3nf
SPECIAL PLACE IN HELL RESERVED FOR THOSE WHO HAVE DISLIKED THIS VIDEO😡🤬
Amazing video but you looks like Jerry from Tom and Jerry
3nf
I love you
Sedkay
mam lgta hai aaj ap nahaaye nhi... ya fir winters me glow kam ho gaya?
nice to listen the class mam
mam are u got married
😍😍😍😍😍😘😘😘😘😘😘😘😘😘
P.A(A,E,C,D,B)... no N.P.A..and the C.K (AE,DE,CE and BE) this relation in a 3rd NF bcoz, if no any N.P.A then definitely that relation it's in a 3rd NF and that left side of FD's not contained any S.K so BCNF is not possible.
Don't promote white hat jr
KUCH SAMAJH NAHI AAYA MUJHE, VIDEO DUBARA DEKHNI PADEGI🤦♂️,AAP ITNI SUNDAR KYON HO?????