An Example Proof using Identities of Regular Expressions

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  • Опубликовано: 28 дек 2024

Комментарии • 95

  • @ashadas724
    @ashadas724 3 года назад +113

    My cllg teacher is also teaching me the same as this automata playlist I think my teacher is also learning from this neso academy 😂😂😂😂

  • @spoidermon4323
    @spoidermon4323 3 года назад +96

    no need to use E.R=R just see 1+00*1 take 1 out or common it becomes (E+00*)1 by {(P+Q)R=PR+QR}

  • @technicalspecial2770
    @technicalspecial2770 6 лет назад +112

    actually you are tracking 1 commen in the 5th steep,
    not the [€+00*] you are talking about.
    the process is correct but not the explanation

  • @s0ge770
    @s0ge770 6 лет назад +114

    4 hours left for exam Only hope is above

  • @sanjeevkumarsingh4198
    @sanjeevkumarsingh4198 3 года назад +21

    in 4th line if you multiply with E then it should be like this E(1+00*1)(0+10*1)* but here we only need to do is take 1 as common so the 5th step will come out.

  • @Tia-sy6zu
    @Tia-sy6zu Год назад +4

    Can anyone reconfirm my logic so in the first step we are taking (1+00*1) as common and in algebra we would have taken 1 to get the string again but here we used E or epsilon so we are basically trying to say that the common we took it has . operator of concatenation in it because e concatenated with the term would give the original term not multiplication as it seems while taking common

  • @federicobianchi5191
    @federicobianchi5191 6 лет назад +38

    how do you take epsilon and 00* out? I can't get the formal logic of it

    • @SandipHodkhasa
      @SandipHodkhasa 6 лет назад +74

      from (1+00*1) , if you take out 1 common you are left with (1)(epsilon + 00*)

    • @Harshal102
      @Harshal102 6 лет назад +4

      Sandip Hodkhasa thank you so much.

    • @javaexpertsa8947
      @javaexpertsa8947 6 лет назад +3

      Thank you so much, i was searching the whole internet, because i couldn't understand this step. That make's sense now to me.

    • @curias7
      @curias7 6 лет назад

      where is the rule about that?

    • @zyro9922
      @zyro9922 5 лет назад

      @@SandipHodkhasa Thanks.

  • @yashpandey1372
    @yashpandey1372 Год назад +1

    In my university 4th sem exam same question come. And five questions exactly same of pumping lemma, melay machine of 2'complement , nfa to dfa

  • @shaswatsoni3228
    @shaswatsoni3228 6 лет назад +6

    please explain the 5th step?

  • @chabotaluputa7665
    @chabotaluputa7665 6 месяцев назад

    Your amazing❤❤

  • @AbhishekNigam
    @AbhishekNigam 7 лет назад +1

    Thanks for the video!

  • @deepanshukumar5407
    @deepanshukumar5407 2 года назад

    Sir your teching method is very sime and intership

  • @vepakommavamsi
    @vepakommavamsi 4 года назад +2

    Can this is also called as equivalence of two regular expression?

  • @dhanushsivajaya1356
    @dhanushsivajaya1356 4 года назад +1

    Thankyou sir

  • @BetioLopezMenendez4
    @BetioLopezMenendez4 7 лет назад +4

    Hey! Hope you be ok, I have a question, so you said that E.R = R, why E.1=E?

    • @anshulgangwal4391
      @anshulgangwal4391 7 лет назад

      Think of the '.' operator as string concatenation. Joining a null string(E) to another string(R) gives you R itself.

    • @ChristianBurnsShafer
      @ChristianBurnsShafer 6 лет назад +2

      He didn't use E.1=E, rather he factored out the 1 using identity (12). You are correct in believing E.1=1.

    • @MrKyuubiJesus
      @MrKyuubiJesus 6 лет назад +1

      Well, he did factor it out, but regex is not commutative (ab != ba). To see this we can use the LHS to obtain a language that includes all of LSH.
      L = 10(0|1)* includes LSH. 01 is a string in RHS and not in L. Thus the languages are not equivalent.

  • @vepakommavamsi
    @vepakommavamsi 4 года назад +1

    Can this is also called as equivalence of two regular expression

  • @supersakib62
    @supersakib62 2 года назад +1

    Where can I get the slides? Anybody has them??

  • @keshavraghav3896
    @keshavraghav3896 3 года назад +3

    4:10 , yaha 1 ko common liya hai...

  • @kadambalapavan2280
    @kadambalapavan2280 3 года назад +4

    in step -5 which one are we taking common E+00* or 1 ?

  • @technicalspecial2770
    @technicalspecial2770 6 лет назад +1

    in 4th steep you miss to put the brackek

  • @peacecop
    @peacecop 8 лет назад +2

    How? in THE leder Side, THE furst Can behöver 1, in THE right Side 0. THE expressionism Åre not equal

  • @Explore-Fashioninlife
    @Explore-Fashioninlife 4 года назад +3

    Sir please make a another video with more examples. Please😫🙏🙏💓😫🙏🙏💓😫🙏🙏💓😫🙏🙏💓😫🙏🙏💓😫🙏🙏💓😫🙏

  • @sainikhilkura5585
    @sainikhilkura5585 2 года назад

    (a*ab+ba)*a*=(a+ab+ba)* solve ths que usng the identitties

  • @preetijha6968
    @preetijha6968 Год назад +1

    (a*ab+ba)*a* = (a+ab+ba)*.
    Prove this solution plzzz ..

  • @whateverbehappy5206
    @whateverbehappy5206 6 лет назад

    thnq sir

  • @rupakdutta8575
    @rupakdutta8575 4 месяца назад

    I was confused at E+00* part after that I realized 1+1 = 1 because R+R = R

  • @nasikchanana9658
    @nasikchanana9658 5 лет назад

    Loved u video lexture

  • @AhamedKabeer-wn1jb
    @AhamedKabeer-wn1jb 4 года назад

    Thank you..

  • @HamzaShahzadEB_
    @HamzaShahzadEB_ 2 года назад

    Where are the questions like write an re for the set of string that contains only a and b ?
    Where are these types of questions?????

  • @sameerkhatri997
    @sameerkhatri997 6 лет назад

    Where dose the one go

  • @ssm840
    @ssm840 Год назад

    is ε+r=r?? can anyone clarify?

    • @LadderVictims
      @LadderVictims 10 месяцев назад

      if r doesnt contain e then r + e is not r

  • @sajidsiddiqui4652
    @sajidsiddiqui4652 7 месяцев назад

    45 min left for the exam 😢😢

  • @karthikmandelli1612
    @karthikmandelli1612 Год назад

    6 hrs to go for a exam

  • @MyAsdfqwe
    @MyAsdfqwe 5 лет назад +8

    feels like gibberish with imaginary methods.

    • @N12458
      @N12458 4 года назад +1

      imagination is more important than knowledge-Rick Astley

  • @pashamshyamsundharreddy4438
    @pashamshyamsundharreddy4438 7 лет назад +1

    sir,€.1=1,,1+1 remains know sir taking common

  • @neiljohn2637
    @neiljohn2637 3 года назад

    49

  • @marcaironcantal515
    @marcaironcantal515 2 месяца назад

    Hello Guys

  • @omergmail9295
    @omergmail9295 6 лет назад +1

    confused

  • @sia2225
    @sia2225 Год назад

    step 4 is not logically correct

  • @vepakommavamsi
    @vepakommavamsi 4 года назад

    Plz reply

  • @shahzadabbasi5225
    @shahzadabbasi5225 3 года назад +2

    LOL. I think the logics you're following here are useless.