Questions On Second Normal Form(2NF)

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  • čas přidán 3. 08. 2024
  • How to identify 2NF in Relational Schema

Komentáře • 10

  • @EnglsihCommByNawaz
    @EnglsihCommByNawaz Před 2 lety +1

    Thanks Sir! Learn alot from this video!

  • @b-43-pawankadwadkar96
    @b-43-pawankadwadkar96 Před 10 měsíci

    Thanks a lot ❤

  • @samandeepkaur6112
    @samandeepkaur6112 Před 10 měsíci

    Thank you

  • @EnglsihCommByNawaz
    @EnglsihCommByNawaz Před 2 lety

    Thanks sir

  • @manasvi4586
    @manasvi4586 Před 2 lety

    thankyou sir

  • @varshaeluru9816
    @varshaeluru9816 Před 9 měsíci +1

    Thank you so much sir

    • @Karma-nc3kg
      @Karma-nc3kg Před 8 měsíci

      hello bro in last sloved problem i have a doubt

  • @mnaresh3382
    @mnaresh3382 Před 3 měsíci +1

    I think the answer is not 6 but instead 5, since we don't really need a separate table for R(WY) bcz it is not violating the rule of 2NF, we can just add these attributes to the table R(PSVW XY) this is going to introduce any issue since PS is the candidate key of this relation and it can uniquely identify all other attributes of that relation and w->y is not partially depended on the candidate key PS, so the correct answer should be "NO & 5"

    • @tushaamagrawal3072
      @tushaamagrawal3072 Před 3 měsíci

      But W itself is determining Y...This is also case of partial dependency since Y is dependent on a non prime attribute which is not even a part of Candidate key. It should have been dependent on PQS. Hence we would need to make a seperate table for it too.

    • @mnaresh3382
      @mnaresh3382 Před 3 měsíci

      I agree the fact that we still have to separate the table, as you said W is not a prime attribute, but it is determining some other non prime attribute, this does not violate 2NF condition but violates 3NF condition, 2NF speaks only for partial dependency but this is a transitive dependency.