Ur lectures are best. I was always confused in normalization and functiinal dependencies. But this is the best dbms course which is free of cost and available to all.
Before watching your videos I am damn sure that I will under topics explained by you completely.......It is your power of explaining capabbility. You are Great.......
For those who are confused with C->DE.In 3NF lhs should be candidate or superkey or rhs should be prime attribute.So for C->D correct and for C->E incorrect.And we know if we get a single cross that level cancel.So he has done it in a single column. Same for 2NF As C is a subset of a candidate and E is non-prime so C->E cross.But C->D is correct. But doesn't matter as a cross already exists.
Sir ur teaching method is best to understand this normalization, due to your examples and vedio I deeply understand basic fundamental concept.thank u sir
I have a confusion, in third normal form, the only condition was that no non-prime attribute should determine another non-prime attribute, then how in this case C->D is violating this condition, as here (prime attribute)->(non-prime attribute) please explain?
@@ramanshsangal9297the rule of 3NF is , it also satisfy the creteria of 2NF ,and it voilate the condition of 2NF , while rule of 2NF is , there is no partial dependency , while here 'C' is a proper subset of candidate key and 'D' is a non prime attribute , and c->d , and it is partial dependency.
Hello i think for 3NF there is[( non prime -> non prime) AND (left side candidate key OR right side prime attribute)] these conditions must satisfy but there only one condition is satisfied. If there anything wrong please correct me
Agree with the question. If as per the video, having non-prime on RHS violates 3NF, then where will the poor Non-primes go? In any case, non-primes need to be determined by some prime even in the highest NF. I think for interpreting it correctly: 1. Decompose the C -> DE into C -> D and C -> E for clarity. 2. C -> E would pass the 3 NF check as there is Prime on both LHS and RHS. 3. Now focus on C -> D. We should take the rule as wherever there is non-Prime on RHS, the LHS shouldn't be a non-prime so that transitive dependency doesn't come in. So this C -> D would pass the test.
Im in b.tech 3rdy My sir Fully copy your video lecture wise and im one lecture ahead then Sir because of you thank you sir you can teach very simply and most important we can understand more easily ❤️❤️❤️
If any of you have a confusion in 3NF:- A functional dependency X->Y is a transitive dependency, if X is not candidate key or super key and Y is a non-key attribute(s). Hope it helps
I still don't get it, I mean 3NF has condition that a non prime attribute cannot be determined by another non prime attribute, and that is being satisfied here
exactly i have seen previous lectures and there sir told this condition(prime attribute can determine prime or non prime attribute), according to that condition it is in 3rd normal form.
see guys the problem is 'C' is a prime attribute and its a subset of candidate keys that we have and by the rule of 2NF we cannot have partial dependency so it violates 2NF therefore it cannot be 3NF. So it would be better to say prime attribute(LHS) can determine prime or non prime attribute. only when LHS is a candidate key.
Thank you so much sir Kl Mera semester exam h DBMS ka aur mne saari videos dekhli aapki DBMS ke upper aur ab mujhe pta h ki main pass ho jaonga aapki videos dekh ke Fan From Karnal, Haryana ❤️❤️
5 month college main DBMS padhe kuch samajh nahin aaya and now currently I'm in 31th lacture of this playlist aur abhi tak aisa koi topic nahin hai jo samaj na aaya ho 💙❤️💙
Step1:Find all candidate keys using closure step2:write the prime attributes and non prime attributes BCNF:Left hand side of all FD should be a candidate key or superkey 3NF:Left hand side should be a CK or right hand side should be a prime attribute 2NF:Left hand side should be a proper subset of any candidate key and right hand side is non prime attribute 1NF:We generally take its in 1NF
You actually wrote condition for partial depenedency for 2NF which when true will not make FD in 2NF. 2NF: There should be no FD such that Left hand side is a proper subset of any candidate key and right hand side is non prime attribute
@@KeshariPiyush24 Nd if LHS has been a candidate key so we will consider it as full dependency right??? Or in which case we will consider it as full dependency???
Please clear my doubt; while we check for 3rd NF, we check for the condition of Transitive Dependency right, which is in the above video, for E->F, we see if LHS is CK or RHS is PA; in this case, F is PA, and so we conclude it satisfies Transitive Depenecy (In the video, sir had ticked it). But according to 3RD NF, there should be NO Transitive Depency right, then how come we agree/tick something(E->F, F->A) that satisfies Transitive dependency and reject something (C->DE) that does not satisfy Transitive Dependency?
Normalisation is the most confusing topic. But you have explained everything in this one example. U R Great Sir G.
Correct
th-cam.com/video/awhvPD_cdRw/w-d-xo.html
Sir what if non prime attributes are null ?
good one....totally samajh aa gaya...
brilliant guruji
Ur lectures are best. I was always confused in normalization and functiinal dependencies. But this is the best dbms course which is free of cost and available to all.
This is the best clip for practice all the normal forms.
Thank u.. i have tried to explain normal form from bottom to the top.. Iske bahar kuch ni hai
@@GateSmasherswhy did you not find CK for other attributes
Before watching your videos I am damn sure that I will under topics explained by you completely.......It is your power of explaining capabbility. You are Great.......
It is amazing how I can barely understand him but somehow, it is one of the best explanations I have found in youtube,
I have always confusion with the normal form until watch ur videos. Now u have cleared my all doubts. U r the real Teacher. Thank You sir.
For those who are confused with C->DE.In 3NF lhs should be candidate or superkey or rhs should be prime attribute.So for C->D correct and for C->E incorrect.And we know if we get a single cross that level cancel.So he has done it in a single column. Same for 2NF As C is a subset of a candidate and E is non-prime so C->E cross.But C->D is correct. But doesn't matter as a cross already exists.
C->D is incorrect because D is non-prime.
Best explanation of normalisation I've ever seen so far. Seeing this at 3 am before exam
All the best
Great teaching sir.. every teacher should take a note. This is what is called teaching
thank youuuuu so much...u r a grt teacher..1st tym normalisation ka concept itne ache se smjh aaya h
Thank you sir , this question cleared all doubts of Normal Forms
Thank you sir... Your videos are helping me like god to prepare computer teacher exam.. Feeling confident for exam... Thanks sir. 🙏🙏👌
After watching ur videos normalisation is like a piece of cake now.Thank u so much sir🙏
Thank you so much Sir! You are a true lifesaver and a great teacher!
Sir ur teaching method is best to understand this normalization, due to your examples and vedio I deeply understand basic fundamental concept.thank u sir
U r my fav teacher sir ❤️thanks a lot for providing such amazing videos.
the best method for finding normal form...tq sir
Thank you sir from the bottom of my heart❤️❤️❤️❤️. God bless u sir. Aap bahut aage badho...🙏🙏🙏
Teaching at its best....Just superb explaination
Best video ever on Normalization !!!
You are great Sir.
My problem is cleared about normalisation in this video.thank you sir 😊
Thanku sir..ur way of teaching very nice....hun ni mind cho niklda sir ji...
thank you sir you are a amazing teacher i ever seen in my entier student journey love you
Great sir... Love ur lectures ♥️♥️♥️
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i have been following u for 1 year...
your videos help me a lot
This is the best way sir now i am able to solve plenty of questions thank you so much sir.
No one else explains every minute details of the topic.. I love this thing
You're great sir 👍 thank you so much for clarifying my all doubts
I have a confusion, in third normal form, the only condition was that no non-prime attribute should determine another non-prime attribute, then how in this case C->D is violating this condition, as here (prime attribute)->(non-prime attribute) please explain?
This seems to be in 3NF, i have same confusion
@@ramanshsangal9297the rule of 3NF is , it also satisfy the creteria of 2NF ,and it voilate the condition of 2NF , while rule of 2NF is , there is no partial dependency , while here 'C' is a proper subset of candidate key and 'D' is a non prime attribute , and c->d , and it is partial dependency.
Hello i think for 3NF there is[( non prime -> non prime) AND (left side candidate key OR right side prime attribute)]
these conditions must satisfy but there only one condition is satisfied. If there anything wrong please correct me
C -> E (✓) but not C -> D (×) {LHS Ck or RHS PA}
Agree with the question. If as per the video, having non-prime on RHS violates 3NF, then where will the poor Non-primes go? In any case, non-primes need to be determined by some prime even in the highest NF.
I think for interpreting it correctly:
1. Decompose the C -> DE into C -> D and C -> E for clarity.
2. C -> E would pass the 3 NF check as there is Prime on both LHS and RHS.
3. Now focus on C -> D. We should take the rule as wherever there is non-Prime on RHS, the LHS shouldn't be a non-prime so that transitive dependency doesn't come in. So this C -> D would pass the test.
ur teaching method is very gud sir.....tnku so much
Thankyou sir, this cleared all my doubts.
This type of tricks that you get in this channel are precious and very helpful no collage University tell us that
Isi ko kahte hai "Gagar Me Sagar"...ek hi example me pura normalisation cover kar diya.... thank u so much sir....🥰🥰💕💕
Ultimate, marvelous may god bless u ❤️
Im in b.tech 3rdy My sir Fully copy your video lecture wise and im one lecture ahead then Sir because of you thank you sir you can teach very simply and most important we can understand more easily ❤️❤️❤️
Your university name please?
You are amazing teacher. be happy always
Thank you so much sir.... You are the last hope of many students❤❤
Brother really good video made by you . Really helpful thanks 😊😊😊😊
If any of you have a confusion in 3NF:-
A functional dependency X->Y is a transitive dependency, if X is not candidate key or super key and Y is a non-key attribute(s).
Hope it helps
I still don't get it, I mean 3NF has condition that a non prime attribute cannot be determined by another non prime attribute, and that is being satisfied here
@@NavanshK same confusion here..😑
C is a prime attribute it does not need to be a candidate key in 3NF
Prime attribute deriving non-prime so it is in 3NF
Sir pls checkout this one
exactly i have seen previous lectures and there sir told this condition(prime attribute can determine prime or non prime attribute), according to that condition it is in 3rd normal form.
see guys the problem is 'C' is a prime attribute and its a subset of candidate keys that we have and by the rule of 2NF we cannot have partial dependency so it violates 2NF therefore it cannot be 3NF.
So it would be better to say
prime attribute(LHS) can determine prime or non prime attribute.
only when LHS is a candidate key.
Normalised...topic....I m always confuse....but you solve all dought in easy ways thanks...🙏🙏🙏🙏
Thank you sir...Because of your classes I have cleared my dbms exam....
Sir you are best 👌👌
Very thanks to u sir🙏🏻🙏🏻..my confusion has been cleared over here 👍🏻👍🏻
Amazing Explanation ☺
Thank you so much sir
Kl Mera semester exam h DBMS ka aur mne saari videos dekhli aapki DBMS ke upper aur ab mujhe pta h ki main pass ho jaonga aapki videos dekh ke
Fan From Karnal, Haryana ❤️❤️
Best explaination sir thanks alot for valuable guidance
thnkk u so much ji.....khuda aapko hr khusy de....god bless u....🙏🙏🙏
thank you sir ... explanation is superb
best video till now
Thank You Sir for such a wonderful lecture
Thank you sir ❤️❤️
thanks a lot
by far the best explaination
Great explanation bro🤓
Superb explanation... Nice video
Thankyou sir , helped me so much for last minute preparation of exam❤
Sir understood every normalization ❤
Confusing topic explained so easily 😃
one video and all my doubts are cleared
🥺🥺💥♥️♥️♥️ u made my life easy
Your teaching is really excellent
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Great sir
Thanks alot sir❤
Clear explanation
BEST BEST BEST
THANKU FOR THIS AMAZING COURSE SIR❤
thanks very much for explaining this topic sir
20 minutes of this lecture taught me more than 20 hours of my college professor's lecture 🥺 thank you 💙
Thank you! It is really helping me in my preparation.
5 month college main DBMS padhe kuch samajh nahin aaya and now currently I'm in 31th lacture of this playlist aur abhi tak aisa koi topic nahin hai jo samaj na aaya ho
💙❤️💙
Sir apke khud k concept gdbd hai
Sir, second functional dependency is in 3 NF, because C is prime attributes
3 NF says that Non prime attributes must not derive non prime attributes
U r great sir
Crystal Clear explaination
Step1:Find all candidate keys using closure
step2:write the prime attributes and non prime attributes
BCNF:Left hand side of all FD should be a candidate key or superkey
3NF:Left hand side should be a CK or right hand side should be a prime attribute
2NF:Left hand side should be a proper subset of any candidate key and right hand side is non prime attribute
1NF:We generally take its in 1NF
You actually wrote condition for partial depenedency for 2NF which when true will not make FD in 2NF.
2NF: There should be no FD such that Left hand side is a proper subset of any candidate key and right hand side is non prime attribute
@@KeshariPiyush24 Nd if LHS has been a candidate key so we will consider it as full dependency right???
Or in which case we will consider it as full dependency???
@@KeshariPiyush24 right
Thanks sir...
Best teacher ❤
14:07 but prime attribute can determine non prime attribute 🙄in 3NF
Thanks for this awesome explanation
Sir, literally you are great 😊
Thanks you sir I got good marks in DBMS in my BCA exam
Great otherwise i didnt understand hindi. I got the things. Really good
💞💞💞Awesome vedio sir
Thank u 💞💞🇮🇳🇮🇳🇮🇳🌹
You are awesome sir 🤞☺️
this one question almost coverd all the topics of database design
Thanks Sir for this lecture!
Toppaar
@@rounakchourasia2030 😝
Please clear my doubt; while we check for 3rd NF, we check for the condition of Transitive Dependency right, which is in the above video, for E->F, we see if LHS is CK or RHS is PA; in this case, F is PA, and so we conclude it satisfies Transitive Depenecy (In the video, sir had ticked it).
But according to 3RD NF, there should be NO Transitive Depency right, then how come we agree/tick something(E->F, F->A) that satisfies Transitive dependency and reject something (C->DE) that does not satisfy Transitive Dependency?
Bro transitive dependency will exist if a non prime attribute determines a non prime attribute, in your case both E and F are prime attribute
Please explain one more example. This one is really nice.
Gjb sir thanks a lot
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We don't have words to show gratitude to you Sir 🙏😊💐
you are amazing sir, you explains very well. thanks sir
I am a big fan of your way of teaching.
Awesome bro 💞💞
everything clarified
gajab sir!
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thank u sir. helpful content
Best explanation.