Lec 14: How to find out Normal form of a Relation| how to identify Highest Normal Form | part1

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  • Опубликовано: 8 сен 2024
  • Jennys lectures DSA with Java Course Enrollment link: www.jennyslect...
    In this lecture, you will learn the Simplest Way to find out Highest Normal Form of a Relation in DBMS.
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Комментарии • 164

  • @bincymali8417
    @bincymali8417 3 года назад +27

    Maam... Your classes r awesome.. I have completed MCA in 2014 but saddest fact is dat i hav got the idea of normalization today after watching your videos.thank u so much mam.. May God bless u always

  • @randomindex9359
    @randomindex9359 Год назад +24

    For those who have a doubt that why the relation in second question was not in 3NF as it satisfies the condition of transitive dependency i.e NPA->NPA.
    For a relation to be in 3rd NF, it must be in 2nd NF and there should not be any transitive dependency.
    But here the relation is not even in 2NF. Hence, it violates the first condition of 3NF. So, there is no need to check the 2nd condition of 3NF.
    If it was in 2NF. Then this thing i.e NPA->NPA would certainly hold true.

    • @pranavmittal2976
      @pranavmittal2976 7 месяцев назад +2

      but we know have studied if a FD is in higher normal norm it will automatically in it lower normal forms

  • @qasidmubashir
    @qasidmubashir 9 месяцев назад +20

    AT 19:09 BC -> D. where B and C are prime and D is non prime ...........means. prime -> non prime , therefore it should be in 3NF

    • @sasivardhan5882
      @sasivardhan5882 3 месяца назад

      It is a proper subset of Candidate key but not the candidate key

    • @onic9623
      @onic9623 3 месяца назад +2

      Yes it's 3NF, but for 3NF there is a condition that it will be in 2NF, here it fails 😂 it's tricky

    • @nuremahmud5327
      @nuremahmud5327 2 месяца назад

      For 3NF, the determinant has be a super-key OR the dependant has to be a prime attribute. Here, BC as a determinant is not a super key and D is not a prime attribute too. SO it's not in 3NF.

  • @tambobiscuits165
    @tambobiscuits165 2 года назад +7

    This women, Jenny, is a DBMS angel.

  • @tabnashahid3784
    @tabnashahid3784 4 года назад +14

    My exams have ended but I still need a better understanding of the subject and since you're keeping your promise of completing the DBMS series, I'm really grateful to you. Thank you so much.

  • @shuklaIASaspirant
    @shuklaIASaspirant 4 года назад +9

    Tomorrow my DBMS exam
    After visit this channel I believed that i will be clear all subjects this semester exam.....
    Thanks a Lot Mam......
    🙏🙏🙏

  • @ankitchakraborty9255
    @ankitchakraborty9255 4 года назад +5

    Your videos really helped me a lot
    Love u from jharkhand😍😍

  • @rizashah2180
    @rizashah2180 4 года назад +6

    Ma'am I love that way you explain hard concept in easy way thanku so much ma'am 😘

  • @CSEDHATSHAYINIS
    @CSEDHATSHAYINIS 4 года назад +35

    Mam ,In last example
    BC->D
    Both b and c are prime attributes but why you are not taking it as 3NF

    • @nasreddinemerabtene7597
      @nasreddinemerabtene7597 4 года назад +1

      BC is a prime attribute and so is D
      therefore, we have "prime attribute" --> "prime attribute" so it's not a 3NF

    • @sumanthsaadam5138
      @sumanthsaadam5138 4 года назад +11

      @@nasreddinemerabtene7597 D is not a prime attribute bro

    • @bleed0P
      @bleed0P 4 года назад +19

      There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
      Forget such condition like NFA->NFA just stick on the upper condition.

    • @bleed0P
      @bleed0P 4 года назад +4

      @@nasreddinemerabtene7597 There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
      Forget such condition like NFA->NFA just stick on the upper condition.

    • @adeebaajnas3108
      @adeebaajnas3108 3 года назад +1

      Answer pls

  • @ArvindSingh-wj7vy
    @ArvindSingh-wj7vy 4 года назад +6

    Thank you so much ma'am.
    Thank you for being here to help us.
    You cleared all my doubts

  • @TrickstarKeshu
    @TrickstarKeshu 4 года назад +2

    World's Best Teacher..

  • @tirtharajdas2165
    @tirtharajdas2165 3 года назад +21

    I have a confusion, in the third FD (BC -> D), if i check if it is violating the 3NF or not with the rule that, if it show (NPA -> NPA) that is not in 3NF. Here BC is not NPA, so how can we say it is not in 3NF by this rule?

    • @Sayan8118
      @Sayan8118 Год назад +1

      Same bro

    • @arkayanbhowmick8463
      @arkayanbhowmick8463 Год назад +1

      Same bro

    • @AmanYadav-jf1yy
      @AmanYadav-jf1yy Год назад +7

      ​@@Sayan8118bro pahli condition yh hai ki esko 2NF me hona chahiye agar yh 2NF me hota hai tb hm (NPA -> NPA) check karenge uske baad decide krenge 3NF. so esse badhiya hai ki 2nd method use Karo 3nf ko check krne ke liye (super key and prime attribute wala) 🌝🌝.

  • @sachingoel3095
    @sachingoel3095 3 года назад +22

    The last example is both 2NF and 3NF. No subsets of CKs - AB & AC can determine D (non-prime attribute) so it's 2NF.
    Also, NPA->NPA cannot be true as there is only one NPA i.e. D, so it's in 3NF.

    • @suvankar54
      @suvankar54 2 года назад

      For 3nf lhs must be Super key, not prime attribute. As BC is not SK thus it is not in 3NF.

    • @KhushiJain-xx2yk
      @KhushiJain-xx2yk 2 года назад

      Same doubt.. Mam plz explain

    • @it026sanjanadaki2
      @it026sanjanadaki2 2 года назад +4

      @@suvankar54 for 3nf it should not follow NPA->NPA so it's 3nf

    • @khakishoiab
      @khakishoiab 2 года назад

      Sorry mam u npa--->npa trick fails.

    • @lokeshkoduri5929
      @lokeshkoduri5929 Год назад

      I have also a same doubt.
      Have u got it the answer
      For this question?

  • @dnyaneshpande23
    @dnyaneshpande23 3 года назад +3

    Flawless, stunning and talkless teacher you are
    💯🥰

  • @aplutaplut176
    @aplutaplut176 3 года назад +2

    wah, kya mast
    tricky question

  • @anshuraj6455
    @anshuraj6455 4 года назад +2

    📲📲📲📲📲 you are great ...
    I really like your concept 😊😉

  • @soubhagyalaxmisahoo6581
    @soubhagyalaxmisahoo6581 3 года назад +11

    Ma'am In R( A B C D )
    F.d are given
    You say us transitive dependency is
    NPA determine NPA
    But here in this example
    Fd BC->D
    Here B and c are prime attribute
    And D is non prime attribute
    Please explain it again..........

    • @KhushiJain-xx2yk
      @KhushiJain-xx2yk 2 года назад

      Same doubt... Mam plz explain

    • @sauravsharma9229
      @sauravsharma9229 Год назад

      @@KhushiJain-xx2yk have you got your doubt clear??

    • @randomindex9359
      @randomindex9359 Год назад

      For a relation to be in 3rd NF, it must be in 2nd NF and there should not be any transitive dependency.
      But here the relation is not even in 2NF. Hence, it violates the first condition of 3NF. So, there is no need to check the 2nd condition of 3NF.
      If it was in 2NF. Then the thing you are mentioning i.e NPA->NPA would certainly hold true.

  • @user-iu3cu5hn5z
    @user-iu3cu5hn5z Год назад +1

    great for your presentation. so much thinks maam!!!!

  • @saikiranravipati7351
    @saikiranravipati7351 4 года назад +2

    Way of explaining is very pretty good and you are saying examples for us it's very easy to understand every thing carry on mam...............keep teaching........

  • @basitaliahmed6983
    @basitaliahmed6983 4 года назад +11

    At 19:21, this is NOT a transitive dependency. So how can this violate the 3rd normal form?

    • @monkeywings
      @monkeywings 3 года назад

      Other than just the NPA -> NPA should not be present, there is also a rule for 3NF that it should satisfy at least one of the following for the FD:
      1. LHS is SK
      2. RHS is a prime attribute
      Since neither of these are followed, it violates 3NF (Refer the 3NF video)

    • @soubhagyalaxmisahoo6581
      @soubhagyalaxmisahoo6581 3 года назад +1

      @@monkeywings But when ma'am discuss about the 3NF
      Bcoz of transitive dependency there updation anomaly create
      So please explain the transitive dependency clearly

  • @surajkushwah3221
    @surajkushwah3221 4 года назад +2

    I like the way she says 'I dawn't care' and 'rawl number'

  • @RAHULKUMAR-wn8po
    @RAHULKUMAR-wn8po 3 года назад +6

    Mam I have a doubt. At 12:18 In last FD : ABCD > EF is of form Non- Prime att ---> Non-prime att. Because ABC are prime attribute but D is not. So ABCD is a non prime attribute. Hence this FD is transitive and will not satisfy 3NF. But mam you checked the tick mark there. Why?

  • @rajthacool2843
    @rajthacool2843 4 года назад +2

    HI mam recently i saw your video its really osm the way you are teaching it its good for understand the concept .. keep updates with new videos thanks a lot .and love you for your videos

  • @akshatgupta5442
    @akshatgupta5442 4 года назад +7

    In case of 3NF
    BC->D
    Here,B and C both are prime attributes.
    There is no NPA->NPA.

    • @nimisharajesh8846
      @nimisharajesh8846 4 года назад +1

      So no transitive dependency exist if we are taking the condition NPA - >NPA so it must be in 3NF
      bt taking this condition it must not be in 3NF

    • @sanyuktakhetan1977
      @sanyuktakhetan1977 4 года назад +2

      NPA->NPA condition or transitivity property is only for non-prime attributes. But here BC is a prime attribute,so it is not in 3NF.( You can also check in another way that BC is not a superkey and D is not a Prime Attribute.No condition is satisfied, so not in 3NF )

    • @bleed0P
      @bleed0P 4 года назад +3

      There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
      Forget such condition like NFA->NFA just stick on the upper condition.

    • @randomindex9359
      @randomindex9359 Год назад

      For a relation to be in 3rd NF, it must be in 2nd NF and there should not be any transitive dependency.
      But here the relation is not even in 2NF. Hence, it violates the first condition of 3NF. So, there is no need to check the 2nd condition of 3NF.
      If it was in 2NF. Then the thing you are mentioning i.e NPA->NPA would certainly hold true.

  • @meghalpatel8195
    @meghalpatel8195 4 года назад +1

    21.47 last summary is very help full for competitive exam

    • @60_v36
      @60_v36 3 года назад +1

      can u please explain its last point??

  • @renegade4413
    @renegade4413 11 месяцев назад +4

    14:31 But ABCD -> EF is a FD where the left-hand-side is proper subset of Candidate key and the right-hand-side is non prime attribute. Wouldn't that be contradict 2NF?

    • @AditiThakur-ss6zt
      @AditiThakur-ss6zt 4 месяца назад +1

      yes this relation should not be in 2NF. I have same doubt too.

  • @hardiksingh4956
    @hardiksingh4956 4 года назад +6

    9:52 here the point is clear . but i have another super shortcut, if no attribute in LHS can derive A,B,C then ABC is a candidate key.

  • @prajwalpawar6676
    @prajwalpawar6676 Год назад

    Mam my exam is at day after tomorrow and i've watched many videos and really this type of teaching DBMS is not available any other video

  • @user-sx5il1wg4g
    @user-sx5il1wg4g 7 месяцев назад +1

    Thank you

  • @subhosen4933
    @subhosen4933 4 года назад +7

    Maam in the last example of this video tutorial,The last FD BC->D, BC is Not SK and D is not Prime Attribute.So naturally it is not in 3NF.But when I tried out using the Transitive Dependency Property B,C are Prime Atrributes.So I can say there is no transitive dependency in the relation so it is in 3NF.Maam I can't understand what is wrong with my solution.So please help me out.

    • @tharmeekansivarasan5642
      @tharmeekansivarasan5642 4 года назад +2

      I am also having the same problem.Jenny mam help me out

    • @Uzair_Anwar2299
      @Uzair_Anwar2299 4 года назад +3

      same problem with me buddy

    • @tharmeekansivarasan5642
      @tharmeekansivarasan5642 4 года назад +2

      I think use only the method that
      1)left hand side super key
      Or
      2)right side prime variable
      Then it is in 3NF

    • @ayushgupta6749
      @ayushgupta6749 4 года назад

      Same problem it will be in 3NF

    • @bleed0P
      @bleed0P 4 года назад +1

      There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
      Forget such condition like NFA->NFA just stick on the upper condition.

  • @MonkeyD.3892
    @MonkeyD.3892 Год назад

    Thanks for this lecture, Jenny Mam 🙏🙏

  • @adarshchanewar4937
    @adarshchanewar4937 3 года назад +1

    Maam your teaching is really very helpful
    Why not you start teaching us java python

  • @thanikhurshid7403
    @thanikhurshid7403 3 года назад +1

    The Venn diagram is inverted.

  • @mehalapraba9959
    @mehalapraba9959 4 года назад +1

    Very nice lecture. Keep going.

  • @indervirsingh7964
    @indervirsingh7964 4 года назад +3

    Starting 19:10, to check for 3NF, isnt it already in 3NF? Aren't B and C are prime attributes and condition for 3NF is that there should not be any FD between Non prime attribute --->to Non Prime attribute? Or maybe i understood something wrong in the previous video of 3NF

    • @gurjotsingh5924
      @gurjotsingh5924 4 года назад +1

      Ya bro,same doubt.

    • @bleed0P
      @bleed0P 4 года назад +1

      There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
      Forget such condition like NFA->NFA just stick on the upper condition.

  • @abdelmalek9004
    @abdelmalek9004 4 года назад +4

    i think that BC -> D is in 3NF because
    BC is prime attributes
    D not prime attribute
    so "prime attributes "-> "not prime attributes" is in 3NF
    relation which is not in 3NF where NPA->NPA
    thanks

    • @bleed0P
      @bleed0P 4 года назад +1

      There is no such condition like NPA->NPA for 3NF either LHS must be Super key or RHS must be Prime attribute.
      Forget such condition like NFA->NFA just stick on the upper condition.

    • @ayushijindal3899
      @ayushijindal3899 3 года назад

      @@bleed0P what is nfa???

    • @bleed0P
      @bleed0P 3 года назад

      @@ayushijindal3899 hey it was a little mistake over there...its NPA* instead of NFA

    • @kumsrkhel200
      @kumsrkhel200 5 дней назад

      ​@@bleed0P There is NPA -> NPA but only if it is in 2NF.
      For directly checking 3NF, what you said should be done.

  • @dheerajswaroopsm691
    @dheerajswaroopsm691 Год назад +7

    NPA -> NPA fails here.

  • @clebsonmachado9692
    @clebsonmachado9692 6 месяцев назад

    OK. BC are subsets of AB and AC, so BC -> D is a partial key dependence.

  • @rukhu2128
    @rukhu2128 3 года назад

    Thank you Jenny mam

  • @captainamerica9576
    @captainamerica9576 Год назад

    in last example BC was a prime attribute so why it was not in 3nf because there was no nop->nop relation. in( BC->D)

  • @0xpurn
    @0xpurn 7 месяцев назад

    R={A,B,C,D}
    A->C
    B,D->B
    A->D
    A->B
    In the above relation, is the "B,D->B" BCNF?

  • @SamG-yd6ux
    @SamG-yd6ux 3 года назад +1

    Lectures are very good thank you but it would be good if you use some good markers 😊

  • @Uzair_Anwar2299
    @Uzair_Anwar2299 4 года назад +2

    in the last example prime attributes are A,B,C
    Non prime attribute is only D,
    And last functional dependency is BC-->D
    BC both are prime attribute determining non prime attribute.
    confused here.
    please help out.

  • @kruthikacg4677
    @kruthikacg4677 3 года назад +1

    1, 4,R
    1,5,B
    2,4,R
    2,5,B
    3,4,R Mam how to find highest normal form in this problem?

  • @user-sx5il1wg4g
    @user-sx5il1wg4g 9 месяцев назад

  • @myrtlefernandes716
    @myrtlefernandes716 2 года назад +2

    Hi Jenny, I'm not able to understand why you said ABC determines EF. Which Armstrong axiom is it? At 8:25

    • @yoyojuana2054
      @yoyojuana2054 2 года назад

      haha it's too late but I'll still answer it, i guess it's transitive, from ABC you can determine D, thus you have A,B,C,D 4 attributes, and using all these, ABCD, you can determine E.

  • @geekylearner3596
    @geekylearner3596 2 года назад

    please complete this DBMS series maam..........................................................

  • @isaitech4602
    @isaitech4602 3 года назад

    21:50 🥳 noted

  • @OOGWAYnewcp
    @OOGWAYnewcp 28 дней назад +1

    but BCNF is a higher normal form than 2NF so i think it's all reverse

  • @samratpatel8060
    @samratpatel8060 9 месяцев назад

    guys i have a doubt:
    given R(ABCDE)
    F:A->B
    C->D
    D->E
    C.K={AC}
    prime atr={A,C}
    Non prime={B,D,E)
    isnt A->B
    a partial depedency?
    cuz A(proper subset of CK) --> B(non prime atr)?
    this mean R is not in 2Nf ryt?
    but according to 21:34 1st rule , R should be in 2NF?
    conflict arising

  • @thanmaibhaskar7749
    @thanmaibhaskar7749 Год назад

    making it more complex

  • @1Eagler
    @1Eagler 3 года назад +1

    20:00 do we know the year it was asked?

  • @ashishjainn24
    @ashishjainn24 3 года назад

    a relational schema r is in _______normal form but not ______normal form only if atleast 1 proper subset of candidate key determines proper subset of other candidate key dependencies exist. what is answer ?

  • @LeetCodeExplained-ho7mp
    @LeetCodeExplained-ho7mp 11 месяцев назад

    In previous video you have mentioned that for a relation to be in BCNF, they should strictly satisfy two condition i.e. LHS should be a Super Key and RHS should be a Prime attribute. Here while solving the first question, you only checked the first condiion and not the second one. Kindly help me if I am getting it right and the reason for not checking the 2nd condition.

    • @nachi-03
      @nachi-03 10 месяцев назад

      first of all, those two conditions are for 3NF
      and it should satisfy ATLEAST ONE condition not both conditions.
      BCNF --> for each non trivial X--> Y,
      X (L.H.S) must be super key (this is what she said)

  • @anamshahzadi8722
    @anamshahzadi8722 Год назад

    At 19:20:
    Tranaitivie dependency says:
    If Non prime Attribute --> Non prime attribute
    Then it holds transitive depency. But in this case
    BC is not Non- prime, so there is no transitive dependency. This should also be in 3NF ? Not ?

  • @kshamaprakashkamath9677
    @kshamaprakashkamath9677 2 года назад

    in first question ABCD->EF is NPA->NPA (D,E,F are non primary attributes), which means its transitive dependency ,so how is it 3NF?

    • @sauravsharma9229
      @sauravsharma9229 Год назад +1

      bro have you got answer of this ??

    • @amlansasmal
      @amlansasmal Год назад

      @@sauravsharma9229 have u git the answer of this ?? same doubt plz tell

  • @gulamgareebnawaz7638
    @gulamgareebnawaz7638 4 года назад +1

    Plz mam algorithm and coa k lectures upload kijiye

  • @chanabasayyasindagimath9516
    @chanabasayyasindagimath9516 4 года назад +1

    Please the board is light RED & BUT WRITING IS RED PEN, SO IT IS NOT SEEN CORRECTLY TO THE STUDENTS, MAM SIR...

  • @kunal930
    @kunal930 4 года назад +1

    Mam aap sirf 2 subjects(DS, DBMS ) pdha rhe ho, baaki subjects bhi padhaayo, Variety bnaao apne channel pr subjects ki, Pls.
    Is se aapka channel bhi kaafi jyaada subscribers se bhr jaaega.
    Comment acha lga toh like kro Mam!!

    • @SonaliProgrammingHub
      @SonaliProgrammingHub 4 года назад

      Referencing to your comment, what all technologies u r looking for?

    • @kunal930
      @kunal930 4 года назад +1

      @@SonaliProgrammingHub See,for technologies there are various good channels.
      But this channel should include other subjects which are related to college studies (computers).

  • @60_v36
    @60_v36 3 года назад +1

    Can somebody please explain the last point?? @22:35 ?

    • @yassinebenjdida2394
      @yassinebenjdida2394 4 месяца назад

      there is 3 options for a relation either np->np , np->p or p->np , it is a 3NF so np->np doesnt exist we have all CKs are simple so every prime attribute is a CK and so every prime attribute is a super key and so p->np in this case is in BCNF and now for the last relation np->p, we know p is prime attribute so it is a canditate key in this case and so np is determining a canditate key which means np can determin all attributes and so it is a superkey so np->p is in BCNF and with that all realtions are in BCNF so the point she lade is right.

  • @Deepaknit_k
    @Deepaknit_k 4 года назад

    Iam form electical but still i watch your videos

  • @shivangshrivastava9939
    @shivangshrivastava9939 2 года назад

    prime attribute determines non prime attribute how it voilates 3NF property.

  • @novicecoder5753
    @novicecoder5753 3 года назад +1

    Here in second problem B and C are prime attributes but as per rule of 3nf no non prime attribute determine non prime attribute.here the non prime attribute is D but D cannot determine non prime attribute so it is in 3nf if any body knows the answer please reply 🙏

    • @randomindex9359
      @randomindex9359 Год назад

      For a relation to be in 3rd NF, it must be in 2nd NF and there should not be any transitive dependency.
      But here the relation is not even in 2NF. Hence, it violates the first condition of 3NF. So, there is no need to check the 2nd condition of 3NF.
      If it was in 2NF. Then the thing you are mentioning i.e NPA->NPA would certainly hold true.

  • @rajatbibhuty
    @rajatbibhuty 4 года назад

    Love u jenny

  • @chanabasayyasindagimath9516
    @chanabasayyasindagimath9516 4 года назад

    Please use & write the BLUE PEN ON THE BOARD MISS TEACHER...

  • @maulijadhav3393
    @maulijadhav3393 11 месяцев назад

    Hii ma'am, could you plzz check ven diagram of 1NF, 2NF, 3NF, BCNF As per My knowledge it is inverted so plzz check that and give me reply

  • @mastsher
    @mastsher 4 месяца назад

    madam, i am also in Haryana, lets catch up some fine day

  • @lavitdubey561
    @lavitdubey561 4 года назад

    ✌✌

  • @sunandabhoomolla9179
    @sunandabhoomolla9179 4 года назад

    Ur sooo beautiful and Ur voice alsooo😍😍😍nice explained sis

  • @harchitgulati3065
    @harchitgulati3065 4 месяца назад

    could not understand how she is cancelling not cancelling out B and D at 16:08

  • @bishalpoudel3586
    @bishalpoudel3586 3 года назад

    4:29 all you get to start

  • @varunkumarreddy7259
    @varunkumarreddy7259 3 года назад

    7:55 anybody explain me
    i didnt get it

  • @mohansurendar7429
    @mohansurendar7429 2 года назад

    20:20

  • @mahimamahendru477
    @mahimamahendru477 3 года назад

    at 12:19 is ABCD->EF not a transitive dependency?
    D is NPA and E is NPA 🥲🥲
    please help!

    • @Shylendra-F
      @Shylendra-F 3 года назад

      Even im confused here. If your doubt is clarified then plz help me in understanding this...

    • @mahakgawate2209
      @mahakgawate2209 3 года назад

      LHS is a super key

  • @MohitGupta-jd4bu
    @MohitGupta-jd4bu 4 года назад

    Gate cse 2011 rank approx 6000

  • @hammadali476
    @hammadali476 4 года назад +1

    Maim plz Koi professional Course Highlight kijye ga ....Ye sb Tou Hum University m b prhlyty hn. .....Taught us Some professional Languages Or subject Like Js React .or Mean stack developing. We wanna Learn these languages And Move forward In Our feild ..........

  • @pujakumari7474
    @pujakumari7474 4 года назад

    Mam mcs22 ka video upload kro

  • @janakkhanal3915
    @janakkhanal3915 4 года назад

    hi
    miss

  • @Hound_Hyena
    @Hound_Hyena Год назад

    mam tho gurgaon ka hai

  • @farahsultana1988
    @farahsultana1988 3 года назад +1

    In this discussion ...F.D{ABC->DE, E->GH, H->G, G->H, ABCD->EF} Here since ABC->DE in the first relation of this functional dependencies so naturally ABC->D & ABC->E by decomposition rule applying. Ok fine no issue. But since ABC->E again ABCD->EF .. So ABCD->E And ABCD->F ... So is it the fact that ABC->E also there ABCD->E So from the R.H.S ABC = ABCD????? And Since ABCD->F so it also implies ABC->F?? As a result you can discard F by existing ABC only?? Please clarify the logic ... You know discarding true is very very important for finding S.K and C.K so far... Your teaching level was not smart in this lecture... Try to video this again. Thank you....

    • @mohammedmahrozuddin
      @mohammedmahrozuddin Год назад

      you have mistaken, The group of attributes A,B,C determines E and also A,B,C,D determines E (A,B,C and A,B,C,D are two different groups of attributes which separately identify E). Two sets of attributes determining same attribute doesn't make them equal.
      Two completely different groups of attributes or partially different groups of attributes can determine same attribute or same set of attributes.
      If ABCD->F doesn't make ABC->F because we require D to be present as a member of the group to identify F. ABC->F because ABC->D and ABCD->F in place of D we can write ABC but ABC are already present in L.H.S so we can't write ABCABC->F, we can't write same attribute two times. So ABC->F.

  • @prabhmeetsingh6995
    @prabhmeetsingh6995 4 года назад

    Mam vlog channel kb banare ho ??

  • @Ravikant0055
    @Ravikant0055 4 года назад

    Hello can you make some Python programming lectures.

  • @sanjaymaharjann7563
    @sanjaymaharjann7563 4 года назад +3

    I love you. I just keep looking you. So beautiful

  • @shouryagupta8067
    @shouryagupta8067 Год назад +1

    maam app faltu ki bthye bhut kri ho video main ak bth ko hi baar baar aree itni diyyan se smj rha tha firse vahi bth bol rhi ho baar bar or fir bol rhi ho yeglt h fir se smja thi ho video short or easy explain kiya kro maam gumaaya mt kra kro

  • @shyamprakashm6325
    @shyamprakashm6325 4 года назад +1

    You keep on said that in the example gurgon and haryana ..do you live in that place 🙄🙄 ?is this is in your home?

    • @anujmor17
      @anujmor17 4 года назад

      She is from Haryana. 😊

    • @shashwatjha9491
      @shashwatjha9491 4 года назад

      @@anujmor17 how do you know ? ;)

  • @isaitech4602
    @isaitech4602 3 года назад

    I still have doubt why i can't see u in unacadmey

  • @mastsher
    @mastsher 4 месяца назад

    madam, aap to niri angrezi ho

  • @ahmadnawazkhan4066
    @ahmadnawazkhan4066 4 года назад

    Hi jeeny . I really like your all videos and also you. If i am single single then my wish to propose you. 😃
    Don't mind please....

  • @KaranKumar-oz9mh
    @KaranKumar-oz9mh 4 года назад

    Mam i want to say that one think
    If you don't angry me.,😡😡
    Your face and voice is so cute.👌
    I can't control say that magic word🥳
    I love you mam 🥰🥰
    Can you propose me mam.💞💞💞💞

  • @Skeybaba
    @Skeybaba Год назад

    Thank you