I missed two weeks of database theory due to hospitalization from Covid-19, and your videos are the reason I am still able to pass this class. I wish there was more to do to thank you.
If you are wondering why BC is not a candidate key then switch to 4:49 directly. It is a Candidate Key and has been explained later to get a good understanding of 2nd N.F.
sir i think u have made a mistake,, u said all non prime attributes are fully functional dependent on ANY candidate key.... sir, i think it will be on EACH candidate key not ANY..
sir!! you said at the end to separate the functional dependency B->F into a seperate table/relation R2(B,F) and R1 will contain attributes of remaining functional dependencies then why are you putting B in R1 again.?? Please reply
sir in 2 NF the non-prime attributes must be fullyfunctional dependent on candidate key and non prime attribute can fullyfunctionaldependent on non-prime attribute
I have a doubt at 8:55 how is R1 in 2NF when D->E where both are non prime attributes and According to 2NF "All non prime attributes must be fully functionally dependent on Candidate key"? Please do correct me if i'm wrong!
The condition is that non prime attributes shud be fully functionally determined by any candidate key, it doesn't matter if it has any functional dependency on any other non prime attribute
Sir please reply.... Tables in second normal form (2NF): a) Eliminate all hidden dependencies b) Eliminate the possibility of a insertion anomalies c) Have a composite key d) Have all non key fields depend on the whole primary key
in the above example R1 is 2nf and R2 is not in 2nf right. my doubt is can i proceed the R1 individually to 3nf or the example is said to be only in 2nf
I think there was something wrong with this video, firstly candidate keys refer to the smallest subsets where all attributes can be reached from which would only be a and not bc because bc has a length of 2. if there was the case of b being b->a,c,d,e,f then there would be 2 candidate keys a and b, and either can serve as a primary key. lastly for e, e is only partially dependent on a i think(this part im not clear on) because we have the relation d->e can someone correct me if im wrong on this?
sir i have a doubt that what the main differences between the candidate, primary and the foreign key (and wanted to know whether they all have the unique nature in them or not)
i have an doubt buddy.........at the end of the lecture while we seperate the relation into r1 and r2 so there in r1 since b->f creating a problem and therfore b,f is put into r2 and in r1 you removed only f not b why..????
Sir you said that candidate key can't be derived from any of the functional dependencies but here A has been derived by BC so how can be it a candidate key, kindly help me on it.
D -> E is a FD, because A->D ( or B->D ) ( transitivity ), but for B->F , BC->F as well. to put it more simply, X->Y and Z->Y if ( Z does is not part of X , as B is of BC )
actually the second condition in 2nf is all non-key attributes are fully functional dependent on the primary key not candidate key(as I searched on google)
Am totally confused with the second example,though i had clear understanding in ur illustration of the 1NF in the previous tutorial using the students relation,..i would suggest that u better use the same example to demonstrate all the steps of normalization.....
Sir in the last minute of this video, you told that the attributes of remaining functional dependency is putting into a Relation R1 except the attributes which this (B->F) functional dependency contains.So why are you considered B also into the relation R1 as R1(A,B,C,D,E). I am getting confused bcz i think here R1(A,C,D,E) will consider. Can u please clear my doubt.
Becoz B is used as foreign key in R1 to make relation with R2. As we know, normalization is used to minimize the redundancy and in the given ques., the FD (B->F) is redundant attribute. If we remove the B from R1, we lose data of B in R1 and also lose the relation with the other attribute of R2 i.e. F. But for the given ques. all attributes must be inter-related to each other i.e. no data should be lost while normalizing or breaking the table. Hope u all understand...
it is not the matter of removing B-->F, F which is a non prime attribute is showing partial dependency(BC must only give F,but B also giving F)hence it has to be removed.Hope it helps!
as a beginner It was hard to understand how he determined his candidate keys which makes thee rest hard to follow. Is it just an example with rules set by him?
this relation will be in 1NF but if we break this relation in two diff. relations i.e. R1(DOTCRY) and R2(DN), we get R1 in 2NF form. For given ques., we get two C. Keys (DO & OCR), So, non prime attri. will be N,T,Y. In this, T & Y r fully FD on both keys but N is partial dependent on DO as D->N. So, removing this conflicting FD (D->N) from relation we get R1(DOTCRY) in 2NF form. I hope you got it...
sir give second normal form example like the first normal form. sir you give the first normal form example very clearly.if possible sir give example with tables and coloumn.
sir in this example A-> BCDEF BC->ADEF B->F D->E CK->{A,BC} NPA ->{B,D,E,F} IN THIS 'F' IS NOT FULLY DEPENDENT ON CK THEN YOUR SPLITTING IN TO TWO RELATION BUT D->E ALSO NOT FULLY DEPENDENT ON CK WHY YOUR NOT CONSIDERING
Sir ! sorry I didn't understand I think here you should Use DB table Example like as !st NF table or any other table which is use from first NF to BCNF
Aayush Joshi For a key to be partially determined it should be a subset of candidate key and since here D is not a subset of the candidate keys(i.e A,BC) it is not a violation.
+Joydip Ghosh BC is also mimimum. A is not a subset of BC. Mimimum only means that no subset of key should be a key itself. So ABC will be a superkey, A will be a candidate key, and BC will be the other candidate key. Primary key is chosen from the candidate key set, i.e. there can be more than one candidate key. You are confusing candidate key and primary key I think. Both A and BC were considered as CK's.
I missed two weeks of database theory due to hospitalization from Covid-19, and your videos are the reason I am still able to pass this class. I wish there was more to do to thank you.
If you are wondering why BC is not a candidate key then switch to 4:49 directly. It is a Candidate Key and has been explained later to get a good understanding of 2nd N.F.
Thank you! I solved my assignment after watching this. You're truly a lifesaver.
Dnt get any think. Its better if u explain its with the table example as u explain the first normal form
sir i think u have made a mistake,, u said all non prime attributes are fully functional dependent on ANY candidate key.... sir, i think it will be on EACH candidate key not ANY..
+Sajal Biswas Exactly!! Let us correct. Thanks a lot.
+Techtud Sir could you please elaborate how to find candidate key. As you said i already checked this..
why u haven't remove D->E, since it is partially dependent on candidate key A,BC and also on D ?
D is not part of the candidate key, so that dependency doesn't break the rule.
No matter how complicated the subject is there is always one Indian person explaining it in IT.
Thanks Brother. Indians are always there to save us.
sir!! you said at the end to separate the functional dependency B->F into a seperate table/relation R2(B,F) and R1 will contain attributes of remaining functional dependencies then why are you putting B in R1 again.?? Please reply
not all hero wear capes. sometimes they r holding mouse&keyboard. (y) thanks man
"ware"
wear*
Its a big deal? Hahaha lol typo btw ty
can you please explain it with an example.... much nice if we consider the 1nf table of previous video
BC is also a candidate key ,right?
sir in 2 NF the non-prime attributes must be fullyfunctional dependent on candidate key and non prime attribute can fullyfunctionaldependent on non-prime attribute
How did you find the keys?
Sir in the example, why d was not considered as a candidate key as even it determines another attribute so it must be unique.
You should continue the example of the 1st NF form so that there remains continuity.
I have a doubt at 8:55 how is R1 in 2NF when D->E where both are non prime attributes and According to 2NF "All non prime attributes must be fully functionally dependent on Candidate key"? Please do correct me if i'm wrong!
But A is present in R1 which is a candidate key.E and D both are fully determined by this key.
The condition is that non prime attributes shud be fully functionally determined by any candidate key, it doesn't matter if it has any functional dependency on any other non prime attribute
Sir please reply....
Tables in second normal form (2NF):
a) Eliminate all hidden dependencies
b) Eliminate the possibility of a insertion anomalies
c) Have a composite key
d) Have all non key fields depend on the whole primary key
does the relation R1 have dependencies of F after converting to 2nd normal ?
i.e is it A->BCDEF or A->BCDE after converting to 2nd normal form ?
in the above example R1 is 2nf and R2 is not in 2nf right. my doubt is can i proceed the R1 individually to 3nf or the example is said to be only in 2nf
I think there was something wrong with this video, firstly candidate keys refer to the smallest subsets where all attributes can be reached from which would only be a and not bc because bc has a length of 2. if there was the case of b being b->a,c,d,e,f then there would be 2 candidate keys a and b, and either can serve as a primary key. lastly for e, e is only partially dependent on a i think(this part im not clear on) because we have the relation d->e can someone correct me if im wrong on this?
siir plzz I didn't understand ... can you explain that with the table example as you explain the first normal form thaank you so muuuch ... great job
sir i have a doubt that what the main differences between the candidate, primary and the foreign key (and wanted to know whether they all have the unique nature in them or not)
i have an doubt buddy.........at the end of the lecture while we seperate the relation into r1 and r2 so there in r1 since b->f creating a problem and therfore b,f is put into r2 and in r1 you removed only f not b why..????
sir i think in the given example e is also a partial functional dependency ?
i can't even express how thankful i am
Sir you said that candidate key can't be derived from any of the functional dependencies but here A has been derived by BC so how can be it a candidate key, kindly help me on it.
well you are explaining clearly normalization topic in all videos.compare to other peoples.thanks
Good , but should be real life example with using table .
thanks
sir ,m unable to understang how
D -> E satisfing the condition of fully dependency if B -> F is not ?
D -> E is a FD, because A->D ( or B->D ) ( transitivity ), but for B->F , BC->F as well. to put it more simply, X->Y and Z->Y if ( Z does is not part of X , as B is of BC )
actually the second condition in 2nf is all non-key attributes are fully functional dependent on the primary key not candidate key(as I searched on google)
Nice. This is exactly what I was looking for...
Am totally confused with the second example,though i had clear understanding in ur illustration of the 1NF in the previous tutorial using the students relation,..i would suggest that u better use the same example to demonstrate all the steps of normalization.....
B should not be there in R1 because it is creating problem .do correct me if i am wrong
+amrit srivastav no, because then bc -->adef will have no meaning. we only remove the relation b-->f.
then what about d-->e , bc--> adef then d-->e should be seperated right??
Perfectly explained!!! Thank you so much !
your teaching class is really very helpfull.
How E is fully functional dependent since D - > E ??? like B -> F
Sir in the last minute of this video, you told that the attributes of remaining functional dependency is putting into a Relation R1 except the attributes which this (B->F) functional dependency contains.So why are you considered B also into the relation R1 as R1(A,B,C,D,E). I am getting confused bcz i think here R1(A,C,D,E) will consider. Can u please clear my doubt.
+Sonam Sharma I am confused as well. Why B is in R1?
+Ebru Kamış Same here I'm also confused
+Sami Shah the teacher is confused too.
haha!!!! ok Vikram Sharma
Becoz B is used as foreign key in R1 to make relation with R2. As we know, normalization is used to minimize the redundancy and in the given ques., the FD (B->F) is redundant attribute.
If we remove the B from R1, we lose data of B in R1 and also lose the relation with the other attribute of R2 i.e. F. But for the given ques. all attributes must be inter-related to each other i.e. no data should be lost while normalizing or breaking the table.
Hope u all understand...
thank you sir I got lot of help from tutorials.
so,the entire function to be in 2NF the relation B->F is to be removed, right ?
it is not the matter of removing B-->F, F which is a non prime attribute is showing partial dependency(BC must only give F,but B also giving F)hence it has to be removed.Hope it helps!
Thank you naga narasimha kowsik
AB can be count as superkey?
+Techtud Sir could you please elaborate how to find candidate key. As you said i already checked this..
Thank u so much... ur videos helped me a lot..
what kind of pen are you using? thanks :)
as a beginner It was hard to understand how he determined his candidate keys which makes thee rest hard to follow. Is it just an example with rules set by him?
What if more than one relation do not satisfy the fully functional dependency? @techstud
do we have any relation in 1NF and 2NF,3NF?????????????
I think you should find the ck, but also decompose R in 2nf to be clear.
it must be *EACH* key ri8..?
Q) What NF is the following R in?
DO-> NTCRY,
CR-> D,
D-> N.
plz help.
this relation will be in 1NF but if we break this relation in two diff. relations i.e. R1(DOTCRY) and R2(DN), we get R1 in 2NF form. For given ques., we get two C. Keys (DO & OCR), So, non prime attri. will be N,T,Y. In this, T & Y r fully FD on both keys but N is partial dependent on DO as D->N. So, removing this conflicting FD (D->N) from relation we get R1(DOTCRY) in 2NF form.
I hope you got it...
show in the table format so that it will be easy to understand
Awesome, helped me for the exam........
sir give second normal form example like the first normal form.
sir you give the first normal form example very clearly.if possible sir give example with tables and coloumn.
You don't state what R1 is at the start of the video, which makes all your good info a bit harder to understand.
i didn't get anything :(
Mee too dude!
and how many super key and primary key are there?
Great explanation
can you give some example for practicing?
It would have been more clear if you were have continued with the same example you created on 1NF video.
can you u make a video with table, maybe same table as first normal form, i have an exam monday.
what is the conclusion of this, we can remove B--->F from the relation reply please
,,
what will be the keys of Decomposed relations R1 and R2?
sir in this example
A-> BCDEF
BC->ADEF
B->F
D->E
CK->{A,BC}
NPA ->{B,D,E,F} IN THIS 'F' IS NOT FULLY DEPENDENT ON CK THEN YOUR SPLITTING IN TO TWO RELATION BUT D->E ALSO NOT FULLY DEPENDENT ON CK WHY YOUR NOT CONSIDERING
+harish rao
because D is not a part of the candidate key where as B is a part of the candidate key BC
E is also determined by D ....but D is not a candidate key....????but you said it satisfies 2nd NF ..how????
I didn't understand . please can you define by any example of 2NF............please sir.
Wooow, @ 7:49 so good. Thanks Techdude
why didn't u take f in r1 where as u took b in r1??
This confused me up.
Why are you finding Non-Prime Attributes of both Candidate Key A,BC.
Sir ! sorry I didn't understand I think here you should Use DB table Example like as !st NF table or any other table which is use from first NF to BCNF
But I understand 1st NF Thanks for this
Good tutorial brother. very helpful. thnx.
BC is also a candidate key.
I was wondering that why BC is not candidate key. So this is not 2NF since B-->F?
exactly! i thought i was wrong.
I have the same question.
hey you said BC determines F but B alone also determines F doesn't the same goes for E, as D alone determines E??
D is not a proper subset of any candidate key, so therefore you can't consider D determines E...
Aayush Joshi For a key to be partially determined it should be a subset of candidate key and since here D is not a subset of the candidate keys(i.e A,BC) it is not a violation.
Excellent tutorial.
bro D->E the condition is unsatied right then how u say that it is in 2NF
D is not a candidate key. 2NF means no non-prime attribute should be partially dependent on candidate key
how can we determine this in any table plz upload this
It would be better if u explain with tables..
Its ok sir but i cant understand y d and e are fully functional
Sir, bc is a candidate key break the ur rules you teach us to find the candidate key
why is only A and BC considered as candidate key, even AB is a candidate key ??
if A alone can determine all attributes then why are you combining it with another attribute
Should R1=(A,B,C,D,E,F)?
Why you again wrote B in R1 ????
B is used as a foreign key to fetch the value of F from R2.
Well Explained, thank you
Why you say candidate key???
Candidate key is the minimum super key..but here BC is not minimum,minimum is A.so why we consider A as a key?
+Joydip Ghosh BC is also mimimum. A is not a subset of BC. Mimimum only means that no subset of key should be a key itself. So ABC will be a superkey, A will be a candidate key, and BC will be the other candidate key. Primary key is chosen from the candidate key set, i.e. there can be more than one candidate key. You are confusing candidate key and primary key I think. Both A and BC were considered as CK's.
Really good job but can improve a lot in giving example to teach the topic.
I did not understand, can you explain with another example?
here d is not a canditate key and it determines e how e satisfies the condition
because form e ,there is no any edge form e.
ha s got it
thank god ,you have got it .!!!!!!!!!!!!
why did you write B in R1 again ??
What would be the final answer?
Hello beautiful friends, it is me GOFF
He should've explained it with an example of tables!
if explain with tables for 2'nd NF, we can better understand
Shouldn't there be 3 relations formed in order to be in 2NF?
R1(A,B,C,D,E)
A->BCDEF
BC->ADEF
R2(B,F)
B->F
R3(D,E)
D->E
why would AB will not be a candidate key ?
+Tarun Bhatia Because it's proper subset "A" is already a candidate key.
Please ask doubts on techtud.com
Techtud please explain me the above first normalization form example .
in second normalization
Plz explain this tutorial by using table
why F is not in R1??
try and use illustration. it will server better. good tutorial tho. thanks!
Please use the table
I cant Understand ...Please do another video completely explaining it again in an easy way
Best lecture but d holds e than is not fully fd beacuse d is not condidate key
but why B is present in R1?
B is used as a foreign key to fetch the value of F from R2