Lec 20: Problem on Lossless Join Decomposition in DBMS | Check whether Decomposition is Lossless

Best DBMS Tutorials : • DBMS (Database Managem...
In this video, I have discussed one Complex Example Practice Question of Lossless Join Decomposition with the Solution.
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#dbms #dbmstutorials #normalization #jennyslectures

[@hameed_ludba]

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.

[@ruqaiyyaaslammalik1792]

I watch it at 2x

[@ashleylove6840]

@@ruqaiyyaaslammalik1792 i watch it in 3X

[@The_attacker99]

@@ashleylove6840 😮 flash or wot

[@pratham8275]

Got that

[@lem1.23]

bro i pause the video, and skip 10seconds take 2seconds to understand... that means it plays on 5x

[@JoffreyB]

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 KZbin!

[@soumyamitra7831]

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.

[@vibhavsamaga1693]

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 👏❤😊

[@AnujKumar-tt5md]

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🙏

[@JrundBerg]

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.

[@arpitjaiswal5133]

Yes this is correct, So the R is not dependency preservation decomposition.

[@sanjaysm783]

How did E->D come? Could you spare some time explaining

[@priyankasingh6792]

C-->A and D-->E ..but I didn't get F-->A so this is not dependency preserving ..

[@Payal_Ojha]

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.

[@soumadipdas271]

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🤩🤩😍😍😍😍😍

[@sachinmishra5885]

Beauty + knowledge =Jenny Ma'am 🤘

[@shivamsahu3465]

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😂

[@anirudhdevpura2798]

Mam, the growth of your channel itself shows how much students love your channel😊.

[@rumiNITPatna]

thank u so much mam ! tomorrow is my end sem and ur videos are really helpful

[@chandankumarms4288]

PLEASE MAM CONTINUE THIS VEDIO LECTURES ON DBMS IT IS REALLY GOOD

[@zohaib166]

You always prove to be a life saver

[@ChristopherKim-vt1qg]

YOU'RE THE GOAT

[@sattiwrites3186]

Big fan mam from Pakistan. your lecturers are very helpful...

[@AhmadRazaNaveed]

Mam you and your method of teaching so nice❤❤

[@baryagameplay3757]

Best teacher and beautiful also 😄

[@castmegently55]

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. 💐😊

[@salmakd3586]

I guess this video should come before #20 (the one preceding this). Thanks for all your amazing videos!

[@eagle_shadow6665]

Your lecture are awesome ,thanks a lot mam

[@shrikant2820]

Best teacher ❤😊

[@talent_hunt2]

Damm short vedio and all concepts are perfectly clear... thankyou mam

[@pablohernanrechimon7362]

Como te quiero Jenny!!!! Que fácil haces todo!!!

[@4Explorer.]

Ma'am I really appreciate your efforts 💗

[@ramsajivankushwaha487]

Best teacher and cute also

[@_suman__]

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.

[@sunshinetheuniverse5732]

Thnku so much mam for easy explanation

[@soumadipdas6043]

After watching this video I have decided to take computer science engineering in after finish my school

[@darshu2062]

Are bhai thoda aur dekhle Itni jldi main decision nhi lena chahiye

[@abhinavpancholi4342]

superb video mam

[@153chhaviagrawal2]

The answer to the question is that it is not Dependency Preserving because F->A which is not present in G.

[@asrafnizan4670]

D->B is also not covered

[@motorway9787]

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.

[@ShahdsOneNightstand]

@@asrafnizan4670 yes it is because d determines c and c determines b so by transitivity d determines b

[@divyarana9717]

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.

[@divyarana9717]

not equal to

[@thongminh1159]

you saved me, thanks a lot

[@talentaptitude3017]

Thank you so much mam..this has helped me immensely

[@aasthajain92]

Great video

[@roniasatourian6789]

amazing video thank you so much

[@momscookbook2222]

Thanks

[@shubhamoybose2129]

Thanks didi 🙂

[@pannagabm2100]

Tqqq mam... 🤩🤩

[@A1Tech6111]

Thanks madam

[@manalirajguru438]

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

[@sumeetsingh_96]

This relation is not Dependency preserving decomposition. Because F cannot derive A. IN the union of (R1 U R2 U R3).

[@deluluatul]

Thank you man

[@DebashisSaha-xh9jy]

Thank You mam

[@imshubh17]

I'm first viewer🤗😘

[@PavanKumar-di4yy]

Super mam

[@radhikasaini9333]

The given question is not dependency preserving decomposition. I am right or wrong please tell me mam... You are really good ... 👍❤️❤️

[@vrajchokshi9731]

Ma'am please make lectures for DCDR

[@anmolsinghal484]

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 ().

[@anishjoshi1999]

Thank you mam

[@christianoronaldo1662]

Mam ,is the playlist complete? if NOT ,then please complete it as soon as possible . Placements are to become .

[@devanshsaxena8064]

maam plzz make a video on SHA algorithm

[@Anitakumari-hx4tl]

Mam within how many days will u upload rest of the topics?

[@akshaywagh9157]

AB-c is also

[@NONAME_G_R_I_D_]

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

[@sunilpoojari91]

It is not dependency preserving. If anyone else has solved please let me know if I am correct or wrong.

[@oppertunecareer5516]

Shi hai

[@uniqueversalunicorn2412]

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.

[@sanjananayak9366]

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?

[@dkdivakar699]

Not a dependency preserving decomposion F->A and D->B are not preserved

[@princepratikvines2555]

Mam please make on design and analysis algorithm

[@VamsiKrishna-ym3cr]

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...

[@debasishmishra3507]

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

[@vishalraina1898]

F->A not covered...so not dependency preserving

[@kallakurianirudh9055]

C->D also not covered so not dependency preserving

[@aravintha6166]

@@kallakurianirudh9055 C->DE here C determines D so here D->EF and F->A are not covered

[@sahithi_sribhashyam]

mam if we have two common attributes in given decomposed subrelations then we should find closure for both attributes???

[@kshitizpathak7289]

Mam .. Please make videos Design and Analysis of Algorithms (DAA) after DBMS

[@Touay]

Wow i like ur coat mam ! It's osmmm

[@259_parthpatidar9]

kashmir op

[@prernadhiman8731]

Mam when are you going to make videos on multivalued ,4 th and 5th normal form??

[@akshatchaudhary9393]

not necessarily candidate key.. it can be super key too

[@ArvindSingh-wj7vy]

Mam plz make more videos of DBMS Why u stoped making videos of DBMS? I will suffer much because of thus😥

[@anngrahdhar7400]

It is not dependency preserving

[@aviralkhanduja5834]

MAM PLEASE UPLOAD 4TH AND 5TH NF

[@ankitrastogi8719]

Hi ,Jenny ma^m,pls make videos on System Analysis and Design.

[@prathmeshsharma7628]

Please make videos on Linux os and c++

[@akshaywagh9157]

E is candidate key

[@kingmaker9082]

Sweet speed is 1.5x . Thank me later.

[@saivarun1748]

It's Not dependency preserving

[@darthdaenerys]

Not dp as F->A is not present in the union of all the sub relations

[@radhikasaini9333]

Mam ap relational model or ER model par BHI viedo bna do please...

[@lingamurthylingamurthy4083]

Please keep data structures in order mam

[@jainarayananr8679]

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

[@bongnemesis1057]

Mam what happen if there is no super key?

[@MohitRaj-do8gv]

Mam please upload all the alogo from cormen book.

[@ashubehera115]

Mam please add coa lessons

[@amasiancovers4779]

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

[@neerajprakash4271]

Jenny meri jaan 🥰🥰🥰🥰🥰

[@praveenkumar-wh1qg]

NET paper first ki preparation kaise ki aapne. Can u please share with us

[@kavitarani5891]

Mam I have problem in few questions...... Can you please help me with that

[@elaksuthra6324]

ma'am when will you do the lecture for TOC, DAA. pls am waiting

[@mahmoodtashaf]

Is it the end of DBMS tutorial?

[@JennyslecturesCSIT]

No.... Will upload more soon

[@mahmoodtashaf]

@@JennyslecturesCSIT K

[@CodeExpress99]

Mam' theory of automata , kaa bhi lecture upload kar do .....plzzzz🙏🙏🙏

[@Cookiesandnuts]

Where did f go ?

[@harshittiwari2281]

thanks mam ... nd ha kdi hss v liya kro ;)

[@Gg-pq3bq]

🥰

[@HunterAG]

♥️

[@kallakurianirudh9055]

C->D not covered so not dependency preserving

[@mytravellingtales_]

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

[@jagadeeshmandala4097]

😒😏

[@Anitakumari-hx4tl]

Not DP

[@FINANCIALYOGI]

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

[@soumyamitra7831]

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.

[@souravssg]

maam video daliye kaha hay aap, no videos for 20 days......

[@JennyslecturesCSIT]

I am under medical supervision and on a complete bed rest ...

[@souravssg]

@@JennyslecturesCSIT Sry maam I didn't know about it Take rest madam Get well soon madam

[@chetanraj7974]

Hii mam , would you like to have dinner ? Final year student at IIT Kanpur

[@KrishnaPandit1999]

This is not DP.

[@kikkarasrinivas3050]

btw who's this jenny ?

[@JennyslecturesCSIT]

I m Jenny

[@kikkarasrinivas3050]

@@JennyslecturesCSIT Aaw...good work...appreciabe🤗

[@vivekkumaryadav5977]

Ma'am PHP script Ki video banao