For complete DBMS, check out this video: kzbin.info/www/bejne/j4PRm3qbhsemfrM
@48_subhambanerjee22 Жыл бұрын
This lecture is "lossless" for sure sir. No time waste, straight to point perfect explanation!!
@KNOWLEDGEGATE_kg Жыл бұрын
Thank you so much for your valuable feedback dear @Subham ❤ Keep following & do share it with your friends👍 Stay connected for more amazing contents & upcoming videos 👍
@likithr.n96926 жыл бұрын
This probably was the only lecture that i understood about "loseless join".........
@KNOWLEDGEGATE_kg Жыл бұрын
🔥Complete DBMS by Sanchit Sir: tiny.cc/DBMS_Sanchitsir_kg 🔥🔥All Computer Science Subjects by Sanchit Sir: tiny.cc/CSbundle_dbms_kg
@yraj16522 жыл бұрын
Sir literally maza aa raha is series me. 😊 Thank you so much Sir for such content!! ❤️🙏
@KNOWLEDGEGATE_kg2 жыл бұрын
You're most welcome dear 😍Keep learning & supporting ! Do visit our website www.knowledgegate.in for more amazing courses & contents 👍😊
@Sudeshsorout6 жыл бұрын
Apne sir g meri bahut badi problem solve Kara Di ... Thank you sir g .... Amit sir toh pata nhi kya padhate h institute m .... Unhone aise padhaya iss method ko ... M toh usme ulajh Gaya really sir thank you very much
@WimpyWarlord5 жыл бұрын
I love the way u refer us as friends
@sohammaity73894 жыл бұрын
are you an idiot??
@s7s7s7s7s74 жыл бұрын
@@sohammaity7389 simp Lmao
@abirpaul90273 жыл бұрын
@@s7s7s7s7s7 best reply
@devendramahilange2 жыл бұрын
4. R1(WV) R2(WXYZ) W common hai, lekin W candidate key nhi hai. 1. R1(VWX) R2(XYZ) isme X common hai lekin X candidate key nhi hai, candidate key hone ke liye R2 me W hona chahiye na Candidate key: VW nikal rha hai
@narutoworld4k Жыл бұрын
R1 aur R2 alg tables h aur inn dono tables m se kisi m bhi common attribute candidate key bn jati h toh lossless hoga (ik my reply to your comment is useless but for future viewer it may help)
@OmitA254 жыл бұрын
We are blessed that....we got a tutor on youtube like you..... Very nice explanation....thnku....☺
@souvikchatterjee16215 жыл бұрын
sir you are really great..the way you explain all those stuffs its really beautiful and i becomes your big fan... and i wanna say something that if possible then please upload videos on 4NF and 5NF. Thank You Sir.
@AnonRavin2 жыл бұрын
Hello sir, In second example 11:10 how is X candidate key? X alone cannot find W … Please correct me if I am wrong.
@devkumar98892 жыл бұрын
Correct
@Kasperrrr2.0 Жыл бұрын
yes, also noticed the same
@toonstore1056Ай бұрын
the x should be a candidate key to either R1 or R2 since x is a candidate key for R2 we can say that the decomposition was loseless
@shadabiakram52836 жыл бұрын
I hv no word sir...u r awsommm...thank u so much sir
@terabaapfragmaster47239 ай бұрын
I have a confusion, here wv,wx,wy are the condidate keys. Then 4th option should also be lossless. Please help
@Sudeshsorout6 жыл бұрын
Wow sir g .... Dhansu trick h g ... Lossless , lossy ki
@chaitanyamogal8083 жыл бұрын
This man is god 👍
@NEHASHARMA-qz2oi6 жыл бұрын
There's a mistake in the first part - Q6, wherein we have the three decompositions R1(a,b,c),R2(b,c,d) and R3(d,e). You say that the only common attribute between R1 and R2 is b, but it's actually b and c both!
@GadgetMeta4 жыл бұрын
If B is distinct... then why we combine B with C
@PRINCE-ts3noАй бұрын
Same thought but idk why i am replying you its been almost 5 year for your watching this lecture.. 🙂😅
@jai_shree_ram2-c5bАй бұрын
To kya dikkat hai bhai b and c dono common hai check (BC) key hai ya nahi or bo hai .....
@arijitbhakta97052 жыл бұрын
Thank you Sir for clear explaination.
@ajitsafeway5 жыл бұрын
Sanchit , can we try to solve this problem , Suppose that we decompose the relation R=(A,B,C,D,E,F) into three relations- R1(C, D, E) , R2(A, B, C) and R3(A, D, F) . Find whether this decomposition is lossless, if the following set F of functional dependencies holds. F={ A->BC; C->DE; D->F}
@mgudiya4 жыл бұрын
Sir your teaching is fantabulous
@Sudeshsorout6 жыл бұрын
Guru g m toh issey bahut bade tareeke SE solve karta tha ..... Ab toh mouj aa gayi g
@preethamm.n11615 жыл бұрын
🌹💞💞Awesome teaching sir 💞💞 🌹
@OmitA256 жыл бұрын
nice explantion sir.............Sir plz start uploading videos on computer org and architecture.........
@ankitmehta41334 жыл бұрын
love your explanation
@Vedupa5 жыл бұрын
Sir really very good explanation my each nd every doubt clear regarding this topic Thankx a lot Sir!
@yash_verma2 жыл бұрын
excellent explanation
@KNOWLEDGEGATE_kg2 жыл бұрын
Thank you ❤ keep supporting & stay connected for more amazing contents👍😊 You can also check out our course for all Computer Science subjects, sharing the link below: www.knowledgegate.in/learn/Zero-to-Hero-in-Computer-Science
@pritimishra2302 Жыл бұрын
Sir ek bhi distinct hoga tab bhi 3rd property follow hogi?
@manishsinghrajput46386 жыл бұрын
sanchit sir, please upload video on dependency preserving as early as possible..
@ordinarygirlwithdreamywing45764 жыл бұрын
Nice explanation
@Gautamsingh-dy4cp6 жыл бұрын
how x is candidate key when finding x closure w is missing in last question
@ashishakash32496 жыл бұрын
proving candidate key in one relation is enough...........see previous video 3rd property!!!!!!!
@Gautamsingh-dy4cp6 жыл бұрын
i little bt confused and thanks
@Gautamsingh-dy4cp6 жыл бұрын
coommon attribute is a candidate key at least one of the table after orignal table divide itno two parts ...thanks ashish
@ashishakash32496 жыл бұрын
Gautam sonu always welcome
@parthh39634 жыл бұрын
exactly my question!!, up to the previous examples we calculated if the common attribute is the candidate key of the parent relation but in this question suddenly we are happy with the common attribute being the candidate key of either one of the child relations.
@Jyoti_Trivedi4 жыл бұрын
Thank you sir.... great 🙏
@Shrunkhla6 жыл бұрын
Hello sir, in the last option of table you calculated B as common between R1 and R2, but both of them has BC as common, still they satisfy 3rd condition?? Since C is not distinct. Please clarify.
@ujjawalmishra27025 жыл бұрын
take BC as key because R1 intersection R2=BC. He must have missed it.
@parthh39634 жыл бұрын
we are not using B and C individually instead we are taking the composite key "BC", and as we know B is distinct so it does not matter if C is distinct or not because "BC" combined will always be distinct.
@zareena604 жыл бұрын
sir in 4th ques we r getting w as candidate key in R2.i think it is lossless.Plz kindly explain whether it is right or wrong
@shriram61235 жыл бұрын
Thanks 🙏 🙏
@hsygtgh69116 жыл бұрын
mujhe achhe se samjh me aa gyi sir apka vedio
@diaworld71664 жыл бұрын
sir 6th mr R1 and R2 saare attribute cover nhi kr rhe h E nhi h dono me ...aise ke skte h kia h sir??
@Patitapaban_sahoo2 жыл бұрын
Good job
@KNOWLEDGEGATE_kg2 жыл бұрын
Thanks 😍 keep learning & supporting ! Do visit our website www.knowledgegate.in for more amazing courses & contents 👍😊
@gayatri2633 жыл бұрын
God bless you
@KNOWLEDGEGATE_kg3 жыл бұрын
Thanks a lot Gayatri..Stay blessed. Keep learning !!
@radhachranverma35053 жыл бұрын
Sir last wale quetion mai x candidate key nai hai , to ye to lossi hua na
@theIndiangirlinEngland6 жыл бұрын
Nice explanation sir.
@Ankit-we8ym6 жыл бұрын
Very very thanks sir aur iski jarurat thi .
@mbapu_art3 жыл бұрын
Love 😘you sir
@KNOWLEDGEGATE_kg3 жыл бұрын
Thanks a lot dear student.. keep learning and supporting !!
@sscknight5 жыл бұрын
Thank you so much sir... Take a bow🙌
@ankitathakur59895 жыл бұрын
Thanku sir ji
@farzanabatool27284 жыл бұрын
thank you sor🌹 🌹
@dip14106 жыл бұрын
Brilliant sir thanks
@sharvanhomi55596 жыл бұрын
Sir acha explain kya apne..... Nyc
@KNOWLEDGEGATE_kg6 жыл бұрын
Thank you my friend.. Satah dete raho padhte raho.. so that main aur video bana pau aap logo ke liye..
@Ankit-we8ym6 жыл бұрын
sanchit sir next please take on how to check fast dependency preservation .
@VikramSingh-vy3jb4 жыл бұрын
sir in the 6th question R1(ABC),R2(BCD),R3(DE). If we combine R1 and R3 we get(ABCDE) but the intersection is null since there is nothing common in between them. So it should be Lossless decomposition.
@alwayshappy85766 жыл бұрын
Great sir..
@arunavasaha70515 жыл бұрын
In the last example how you can fetch W by making X as a candidate key?
@sagnikbagchi5 жыл бұрын
X is the candidate key for the second table. Only one condition is enough for it to prove that it is distinct he said so X isn't the candidate key of the first table.
@weblocker38016 жыл бұрын
Thnku sir ji...
@mandishdahal56505 жыл бұрын
thank u
@shwetashirsat3836 жыл бұрын
Thank you
@kshitijsrivastava61486 жыл бұрын
Thank you sir!
@aradhanasaxena85616 жыл бұрын
Great
@bhupendradewangan436 жыл бұрын
Thank u so much sir...
@prof_as6 жыл бұрын
sir i have a question :- since here we can find C.K with the help of functional dependency here so 1st we find the C.K and then check that the decomposed tables have those keys in common or not
@AryanSingh-kt6ur5 жыл бұрын
No, we cant do that because if the common attribute is C.K for any of the decomposed tables then it will be lossless decomposition but it is not necessary ki wo common attribute CK prove hojaye using Functional dependencies like in last example X is not a C.K if you find it using functional dependencies but it is CK for R2 .
@narendraparmar16316 жыл бұрын
Thanks sir ji😆
@komalprajapati32436 жыл бұрын
Sir please...... upload the video of 4NF and5NF
@jagmohanagrawal76365 жыл бұрын
More questions on this topic will be helpful. Please upload soon if possible.
@Professor_fauzdar8 ай бұрын
Do them by yourself
@shrikant2820 Жыл бұрын
Sir thank you ❤
@KNOWLEDGEGATE_kg Жыл бұрын
You're welcome Shrikant! keep learning & supporting! Do visit our website www.knowledgegate.in for more amazing videos & contents👍😊
@pritimishra2302 Жыл бұрын
Sir kya agar ek bhi common attribute distinct ho to bhi vo lossless hoga?
@shankarkolkata225 жыл бұрын
Hello sir, in the last option of table of first question you calculated B as common between R1 and R2, but both of them has BC as common and C is not distinct, so it should be lossy.
@the_amankushwaha5 жыл бұрын
yes brother
@codeb3nder8624 жыл бұрын
Haat joda mat kijiye, we are blessed by you
@ashutoshraj86686 ай бұрын
But Sir candidate keys ars vw and xw so how x only be a candidate key
@sai-be2jq6 жыл бұрын
May I know why 4th option is not crct for 2nd qn...? even it contains all attributes and vw ,it is also a candidate key.
@reachrishav6 жыл бұрын
W itself is not a candidate key
@prof_as6 жыл бұрын
and one more question:- does the child tables have to have the same C.K (a table can have multiple C.K )in common to satisfy the condition of loss-less decomposition
@AryanSingh-kt6ur5 жыл бұрын
bro foreign key reference hona chaiye bas
@TheSarthak4256 жыл бұрын
How is X a candidate key? We cannot get W from X.
@spanishards6 жыл бұрын
X is candidate key in relation R2
@Gautamsingh-dy4cp6 жыл бұрын
in r1 how it is possible
@ashishakash32496 жыл бұрын
here i got the solution!!!!! common attribute X should be candidate key in either one of R1 & R2 (or) both.........refer previous video 3rd property!
@sauravjyotikalita95636 жыл бұрын
Ashish Akash I think here X is only a part of candidate key....VW and XW are two candidate key
@ashishakash32496 жыл бұрын
sauravjyoti kalita ........ you did not understand the question he raised ..........vw & xw are candidate keys for R(vwxyz) AND X is candidate key in R(xyz).....so x is candidatre key in R(xyz) and foreign key in R(vwx)!!understood?
@shanumishra24446 жыл бұрын
sir please upload videos on ER Model
@pradhyumansinghmandloi82404 жыл бұрын
common attribute can be super key?? Anyone please
@travel_with_dimpi6 жыл бұрын
Sir i want more videos on dec omposition
@ajitsafeway5 жыл бұрын
Also , just have one question , whether we need to find candidate key for whole schema R and then we need to check whether this candidate key is available in decomposed schema R1 or R2 , or Whether we verify whether common attribute find by rule 2 , is candidate key in R2 or R2 ?
@richidubey5 жыл бұрын
Just check if it is a candidate key for the particular relation R1 or not. If it is candidate key for the entire R, then it certainly would be for R1/R2.
@antarbasu53424 жыл бұрын
Last explaination was wrong.. please check sir, no lossless was found
@subhamsaha22354 жыл бұрын
maja aa gya
@KNOWLEDGEGATE_kg4 жыл бұрын
Thankyou, Visit tiny.cc/yt_kgwebsite for more such mazedar courses and content !!
@Ankit-we8ym6 жыл бұрын
sir please jaldi upload kijiega .
@ajitsafeway5 жыл бұрын
Yaar I need this reply as early as possible , can not wait for a month or many days.. plz help me
@AmitYadav-nn4lz6 жыл бұрын
opetion 4 is correct
@lalkrishnajha46816 жыл бұрын
sir i think this video u have given...i tried this on one sum bt couldn t get the answer...
@lalkrishnajha46816 жыл бұрын
if u ask me then i can send u in mail that sum if u want.....sir pls don t mind
@pratapkumar34345 жыл бұрын
Ugc net Dec 2019 m tha ye questions
@RohanSingh-ug2tv4 жыл бұрын
100th comment❤️
@souravdhar19436 жыл бұрын
"To fir chaliye suru kartehe" 🤣🤣 Inspired by Technical guruji
@aadarshmishra25046 жыл бұрын
Maybe Technical Guruji was inspired by our Sanchit Guruji.. 🤣🤣
@harshagarwal22804 жыл бұрын
Does anyone notices that he speaks very similar to technical guruji
@Sudeshsorout6 жыл бұрын
Agar aap net k saare subject padhane lage toh .... M Bhagwan ki kasam kha kar kehta hu kamjor se kamjor bachha BHI net ko aik Baar m clear kar dega ..... Sachhi sir. .. m Kisi ki khaama khaa badaayi nhi karta hu ... Ap mein mujhe kuchh lagaa h tabhi m keh Raha hu g..
@TanmayMishra-dev7 ай бұрын
06/09/2024
@aakershitsharma6 жыл бұрын
you save my ass man !!!
@Ajay51702 Жыл бұрын
bro does you test in practically! obviously no...😡
@prajwalprabhu025 Жыл бұрын
ii watch ur videos in 2x speed 😂
@anshshah49274 жыл бұрын
Who's having exams tomorrow
@farukj.s48245 жыл бұрын
SIR YOU SPEAK TOO FAST !
@elChico-TV6 жыл бұрын
Why why english tittle when you talk on another language ??? BIG DISLIKE
@krupalshah85266 жыл бұрын
Bot sale, khud GATE mein appear bhi hua hai?
@NishaSharma-nd1dk3 жыл бұрын
Thanks sir g
@sdeepa57465 жыл бұрын
Thanks😊
@askariaziz41634 жыл бұрын
thanks
@Prabhat566824 жыл бұрын
Thanks a lot sir
@harun-or-rashid87146 жыл бұрын
Thank you so much, sir.
@andistheinforitbutso75133 жыл бұрын
Thank you
@KNOWLEDGEGATE_kg3 жыл бұрын
You're most welcome dear student, keep learning & supporting !! Do visit our website www.knowledgegate.in for more courses & contents !!