It is not dependency preserving as F can't determine A. Even at 1.5x speed I can understand her very easily and saves a lot of time. Thank you ma'am for your videos these are well explained.
I've watched and completed all 19 videos in "DBMS (Database Management System)" playlist. I'M SOOO READY FOR THE EXAM. BRING IT IN!!!!! BIG thanks from Russia. I've understood every single video I've watched and this happens rarely, because others don't explain some details over and over and sometimes you forget them and can't understand what's going on and what are we even doing. That's what I think you've done better than all of them. I've watched with every piece of a passion all your videos and wasn't bored once, but the main thing was - I UNDERSTOOD EVERYTHING. One suggestion if you don't mind: sometimes, when you say: "if someConditionA, someConditionB, someConditionC are true, then we can say that it's lossless decomposition" it's better to back it up with some theorem or something, because for example when I'm gonna explain it to my teacher he might ask: "and why is that so? Why if these condition are true, then it's lossless decomposition?" and I probably won't be able to explain. So, yeah. Thanks again for all your effort, appreciate it very much, one of the best teachers on RUclips!
You can derive it yourself, just use a little bit of what Madam has taught. It's there but not like Theorem. She is teaching so that we can understand a bit better, you can get it as theorems in books.
Hello Ma'am... I watched your all DBMS videos🎥. You have done a tremendous work for the respective audience. Your explanation is extremely good. Your work truly helped me to revise the concepts after 10 yrs of completing my engineering..Thanks🙏
Answer for the last question: I got { C --> AB, AB ---> C, C --> DE, D --> EC, E --> F }. Comparing with the original FDs, the one I found does not contain F --> A, either directly or indirectly.
But in second decomposition (C,D,E), the FDs are C-->DE, D-->EC and E-->CD, thus, final set should be {A->BC, B-->AC, C-->AB, C-->DE, D-->EC, E--> CD, E-->F, F-->E}. Thus, indirectly, F derives A and in this way, this is dependency preserving decomposition. As fas as I understood, please explain me if I'm wrong in any case.
thank you ma'am, it's very helpful for campus and placement purposes as I am from ECE background and I don't have any idea about DBMS, thank you as I got place in 3 companies because of you ❤️❤️❤️, love from Kolkata🤩🤩😍😍😍😍😍
I am a prefinal year student from Amity university, your videos are really helpful and easily understandable and after seeing your bloopers video I m in love with you,LOL😂
I'm new here from Pakistan, am trying to learning from your 1st videos. Hope I'll cover it soon. Your teaching way is easy to understand. Thank you. Bless you. 💐😊
bruh... F->E is also not there, but i thought if it is loseless join decomposition then it would be preserving dependency, but it is not so my answer and ur answer is same.
it is not dependency preserving because in subrelation 1= fd-null in sub relation 2=fd-null but in subrelation 3 fd-E->.F is there only so at the end we came to conculsion that it f1 U f2 U f3= f main.
Mam, thank you for such a wonderful session. I have got the ans for your last qsn as ‘ decomposition is not dependency preserving’ is it correct ?request you to please share the ans
The given example is not dependency preserving.. because F cannot determine A in R3. Hence, the union of all the decomposed relations (R1, R2, R3) won't have F->A. Hence the result is NOT DEPENDENCY PRESERVING.
Hello Mam i have been studying from you for a long time now and i want to thank you for making such helpful videos. However there is one question which is given to me as assignment in my college and i cant figure out a way to solve it. If you could help me with it.. it would be great. The question is: Assume we have one input queue, one output queue and a stack with its associated basic operations such as push and pop. No element from the output queue can be pushed into the stack. Also, once input from the input queue is pushed into stack it cannot be put back into the input queue. We can take elements from the input queue and push into the stack. At any time we can pop an element from the stack. However, if there is no element available in the input queue, then only pop operations are possible. A popped element should be always be placed into the output queue and cannot be pushed back on to the stack. The act of producing an output sequence from a given input sequence as explained is known as stack permutation. Suppose we have a sequence of k elements 1, 2, 3, ..., k available from the input queue. Generate all stack permutation of k elements. Clearly explain the nature of stack permutations and find out an expression for the number of stack permutations. i know i have commented the same comment on another post.. sorry or that ().
Your videos are great! Thank you, Small question. Can the common attribute in a two relations be a composite key? Lets say the Composite key is the candidate key in one relation. can we say that it is Lossless? Thank you
If you take first R2 and R3 then R23 is satisfy all three condition then combine with R1 and R23 all three condition is satisfy so i think this is lossless decomposition
Is the common attribute should be candidate key or it can be super key ...?cause in previous videos at some point you told it was super key and after that you are saying for candidate key all the time...
Why all are saying F->A not covered?,F->A is there right? Because in dependency EF,when finding F closure we get F->A right? Then how F->A is not there?i cant understand
What are your qualifications? Are you from any tier 1 college? How peoples are comparing you to IIT profs. ? Please tell your qualifications nation wants to know :D
Question of Dependency preserving Decomposition: R(ABCDEF) FD F(AB-C, C-D, D-EF, F-A, D-B) Decomposed(ABC, CDE, EF). This not dependency preserving Decomposition as Decomposed relation attributes can't determine F attribute. Dependency D-F of F FD is not in G FD. Kindly advise where I can see yr videos for 4NF and 5NF and decomposition methods
Umm.... NO. In G i.e. F1 U F2 U F3 = { AB->C, C->AB, C->DE, D->CE, E->F } So, clearly, by transitive property, we can conclude D->F, so, that's there in F too and not the reason for failing to preserve dependency. Rather it is F->A which is the responsible.
![Felix "Unfair" | [Stray Kids : SKZ-PLAYER]](http://i.ytimg.com/vi/Oswujxm2Ag0/mqdefault.jpg)
It is not dependency preserving as F can't determine A.
Even at 1.5x speed I can understand her very easily and saves a lot of time. Thank you ma'am for your videos these are well explained.
I watch it at 2x
@@ruqaiyyaaslammalik1792 i watch it in 3X
@@ashleylove6840 😮 flash or wot
Got that
bro i pause the video, and skip 10seconds take 2seconds to understand... that means it plays on 5x
Ma'am u literally explained everything my teacher tried to teach for 1 week in 1 video!How do you do this ma'am.Keep it Up 👏❤😊
I've watched and completed all 19 videos in "DBMS (Database Management System)" playlist. I'M SOOO READY FOR THE EXAM. BRING IT IN!!!!!
BIG thanks from Russia. I've understood every single video I've watched and this happens rarely, because others don't explain some details over and over and sometimes you forget them and can't understand what's going on and what are we even doing. That's what I think you've done better than all of them. I've watched with every piece of a passion all your videos and wasn't bored once, but the main thing was - I UNDERSTOOD EVERYTHING.
One suggestion if you don't mind: sometimes, when you say: "if someConditionA, someConditionB, someConditionC are true, then we can say that it's lossless decomposition" it's better to back it up with some theorem or something, because for example when I'm gonna explain it to my teacher he might ask: "and why is that so? Why if these condition are true, then it's lossless decomposition?" and I probably won't be able to explain.
So, yeah. Thanks again for all your effort, appreciate it very much, one of the best teachers on RUclips!
You can derive it yourself, just use a little bit of what Madam has taught. It's there but not like Theorem. She is teaching so that we can understand a bit better, you can get it as theorems in books.
Hello Ma'am... I watched your all DBMS videos🎥. You have done a tremendous work for the respective audience. Your explanation is extremely good.
Your work truly helped me to revise the concepts after 10 yrs of completing my engineering..Thanks🙏
Answer for the last question: I got { C --> AB, AB ---> C, C --> DE, D --> EC, E --> F }. Comparing with the original FDs, the one I found does not contain F --> A, either directly or indirectly.
Yes this is correct, So the R is not dependency preservation decomposition.
How did E->D come? Could you spare some time explaining
C-->A and D-->E ..but I didn't get F-->A so this is not dependency preserving ..
But in second decomposition (C,D,E), the FDs are C-->DE, D-->EC and E-->CD, thus, final set should be {A->BC, B-->AC, C-->AB, C-->DE, D-->EC, E--> CD, E-->F, F-->E}. Thus, indirectly, F derives A and in this way, this is dependency preserving decomposition. As fas as I understood, please explain me if I'm wrong in any case.
thank you ma'am, it's very helpful for campus and placement purposes as I am from ECE background and I don't have any idea about DBMS, thank you as I got place in 3 companies because of you ❤️❤️❤️, love from Kolkata🤩🤩😍😍😍😍😍
Mam, the growth of your channel itself shows how much students love your channel😊.
I am a prefinal year student from Amity university, your videos are really helpful and easily understandable and after seeing your bloopers video I m in love with you,LOL😂
Beauty + knowledge =Jenny Ma'am 🤘
I'm new here from Pakistan, am trying to learning from your 1st videos. Hope I'll cover it soon. Your teaching way is easy to understand. Thank you. Bless you. 💐😊
thank u so much mam ! tomorrow is my end sem and ur videos are really helpful
PLEASE MAM CONTINUE THIS VEDIO LECTURES ON DBMS IT IS REALLY GOOD
Mam you and your method of teaching so nice❤❤
Best teacher and beautiful also 😄
I guess this video should come before #20 (the one preceding this). Thanks for all your amazing videos!
Big fan mam from Pakistan. your lecturers are very helpful...
YOU'RE THE GOAT
You always prove to be a life saver
Your lecture are awesome ,thanks a lot mam
Best teacher ❤😊
The schema R = (A, B, C, D, E) is given. Give a lossless-join decomposition into BCNF of schema R. A → BC CD → E B → D E → A.
The answer to the question is that it is not Dependency Preserving because F->A which is not present in G.
D->B is also not covered
bruh... F->E is also not there, but i thought if it is loseless join decomposition then it would be preserving dependency, but it is not so my answer and ur answer is same.
@@asrafnizan4670 yes it is because d determines c and c determines b so by transitivity d determines b
it is not dependency preserving because in subrelation 1= fd-null in sub relation 2=fd-null but in subrelation 3 fd-E->.F is there only
so at the end we came to conculsion that it f1 U f2 U f3= f main.
not equal to
This relation is not Dependency preserving decomposition. Because F cannot derive A. IN the union of (R1 U R2 U R3).
Como te quiero Jenny!!!! Que fácil haces todo!!!
Damm short vedio and all concepts are perfectly clear... thankyou mam
Ma'am I really appreciate your efforts 💗
Mam, thank you for such a wonderful session. I have got the ans for your last qsn as ‘ decomposition is not dependency preserving’ is it correct ?request you to please share the ans
After watching this video I have decided to take computer science engineering in after finish my school
Are bhai thoda aur dekhle
Itni jldi main decision nhi lena chahiye
The given question is not dependency preserving decomposition. I am right or wrong please tell me mam... You are really good ... 👍❤️❤️
Very Excellent videos maam
Thnku so much mam for easy explanation
Thank you so much mam..this has helped me immensely
Best teacher and cute also
amazing video thank you so much
you saved me, thanks a lot
Not a dependency preserving decomposion
F->A and D->B are not preserved
Thanks didi 🙂
Tqqq mam... 🤩🤩
The given example is not dependency preserving.. because F cannot determine A in R3. Hence, the union of all the decomposed relations (R1, R2, R3) won't have F->A. Hence the result is NOT DEPENDENCY PRESERVING.
superb video mam
Hello Mam i have been studying from you for a long time now and i want to thank you for making such helpful videos. However there is one question which is given to me as assignment in my college and i cant figure out a way to solve it. If you could help me with it.. it would be great. The question is:
Assume we have one input queue, one output queue and a stack with its associated basic operations such as push and pop. No element from the output queue can be pushed into the stack. Also, once input from the input queue is pushed into stack it cannot be put back into the input queue. We can take elements from the input queue and push into the stack. At any time we can pop an element from the stack. However, if there is no element available in the input queue, then only pop operations are possible. A popped element should be always be placed into the output queue and cannot be pushed back on to the stack. The act of producing an output sequence from a given input sequence as explained is known as stack permutation. Suppose we have a sequence of k elements 1, 2, 3, ..., k available from the input queue. Generate all stack permutation of k elements. Clearly explain the nature of stack permutations and find out an expression for the number of stack permutations.
i know i have commented the same comment on another post.. sorry or that ().
Ma'am please make lectures for DCDR
Mam ,is the playlist complete? if NOT ,then please complete it as soon as possible . Placements are to become .
Your videos are great! Thank you, Small question. Can the common attribute in a two relations be a composite key? Lets say the Composite key is the candidate key in one relation. can we say that it is Lossless? Thank you
If you take first R2 and R3 then R23 is satisfy all three condition then combine with R1 and R23 all three condition is satisfy so i think this is lossless decomposition
Mam within how many days will u upload rest of the topics?
Great video
Ma'am there was a doubt why it is not a lossless decomposition ? as C is a candidate key E is not but we have to find atleast one only right?
maam plzz make a video on SHA algorithm
Thanks
It is not dependency preserving.
If anyone else has solved please let me know if I am correct or wrong.
Shi hai
Thanks madam
not necessarily candidate key.. it can be super key too
It is not dependency preserving
Thank You mam
F->A not covered...so not dependency preserving
C->D also
not covered so not dependency preserving
@@kallakurianirudh9055 C->DE here C determines D so here D->EF and F->A are not covered
Thank you man
Not dp as F->A is not present in the union of all the sub relations
Mam .. Please make videos Design and Analysis of Algorithms (DAA) after DBMS
Mam plz make more videos of DBMS
Why u stoped making videos of DBMS?
I will suffer much because of thus😥
Is the common attribute should be candidate key or it can be super key ...?cause in previous videos at some point you told it was super key and after that you are saying for candidate key all the time...
Mam please make on design and analysis algorithm
Super mam
It's Not dependency preserving
Mam when are you going to make videos on multivalued ,4 th and 5th normal form??
AB-c is also
mam if we have two common attributes in given decomposed subrelations then we should find closure for both attributes???
I'm first viewer🤗😘
MAM PLEASE UPLOAD 4TH AND 5TH NF
Mam ap relational model or ER model par BHI viedo bna do please...
Please make videos on Linux os and c++
Why all are saying F->A not covered?,F->A is there right? Because in dependency EF,when finding F closure we get F->A right? Then how F->A is not there?i cant understand
Thank you mam
Hi ,Jenny ma^m,pls make videos on System Analysis and Design.
Mam' theory of automata , kaa bhi lecture upload kar do .....plzzzz🙏🙏🙏
E is candidate key
Mam please upload all the alogo from cormen book.
Please keep data structures in order mam
Is it the end of DBMS tutorial?
No.... Will upload more soon
@@JennyslecturesCSIT K
Wow i like ur coat mam !
It's osmmm
kashmir op
Mam what happen if there is no super key?
Can someone please tell me how to get the FDs of each decomposed relations in which mam has asked us to check Dependency Preserving condition
C->D not covered so not dependency preserving
Where did f go ?
Mam I have problem in few questions...... Can you please help me with that
NET paper first ki preparation kaise ki aapne. Can u please share with us
ma'am when will you do the lecture for TOC, DAA. pls am waiting
Mam please add coa lessons
thanks mam ... nd ha kdi hss v liya kro ;)
Sweet speed is 1.5x . Thank me later.
Jenny meri jaan 🥰🥰🥰🥰🥰
maam video daliye kaha hay aap, no videos for 20 days......
I am under medical supervision and on a complete bed rest ...
@@JennyslecturesCSIT Sry maam
I didn't know about it
Take rest madam
Get well soon madam
What are your qualifications? Are you from any tier 1 college? How peoples are comparing you to IIT profs. ? Please tell your qualifications nation wants to know :D
😒😏
Question of Dependency preserving Decomposition: R(ABCDEF) FD F(AB-C, C-D, D-EF, F-A, D-B)
Decomposed(ABC, CDE, EF). This not dependency preserving Decomposition as Decomposed relation attributes can't determine F attribute. Dependency D-F of F FD is not in G FD. Kindly advise where I can see yr videos for 4NF and 5NF and decomposition methods
Umm.... NO. In G i.e. F1 U F2 U F3 = { AB->C, C->AB, C->DE, D->CE, E->F } So, clearly, by transitive property, we can conclude D->F, so, that's there in F too and not the reason for failing to preserve dependency. Rather it is F->A which is the responsible.
Not DP
This is not DP.
🥰
♥️
not dependency preserving as AB->C and F->A are not present
Ma'am PHP script Ki video banao