dfa example with solution | Part-1 | TOC | Lec-10 | Bhanu Priya

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  • เผยแพร่เมื่อ 21 ก.ค. 2019
  • dfa examples : starts with 1 & ends with 0

ความคิดเห็น • 152

  • @developertiwari
    @developertiwari 4 ปีที่แล้ว +116

    Hi, Thanks for this lecture series. It's clear my concepts. There is something wrong in example 2. Please check it. The DFA doesn't fulfill conditions for 1010.

  • @siddhanttotade5702
    @siddhanttotade5702 2 ปีที่แล้ว +12

    I think the question is wrong bcoz , it says " accept the string which starts with 1 and ends with 0 ". suppose a string start with 0 or 0's , it will not accepted no matter if you put it in a dead state or the string ends with 0. There were two conditions " starts with 1 and ends with 0 " so, if the first condition is not matching or true then how will the string is accept ? if the question says " starts with 1 or ends with 0 " then there will be a dead state. (If I am wrong please correct me).

  • @waheedshahzad6288
    @waheedshahzad6288 4 ปีที่แล้ว +24

    Why you don't draw "0" input at q0 state? i think each state goes to exactly 2 states with 2 different inputs

  • @nakkasai6924
    @nakkasai6924 ปีที่แล้ว

    medam i can't understand 1010

  • @AshishSharma-jx1ki
    @AshishSharma-jx1ki 4 ปีที่แล้ว +10

    mam 2nd string you passed 1010 is incorrect

  • @pulkitsrivastava9779
    @pulkitsrivastava9779 3 ปีที่แล้ว +2

    you were explaining great but at the 1010 step you explained it wrong

  • @Mohammed.1471
    @Mohammed.1471 ปีที่แล้ว +1

    It's not complete answer ig... What if we take a qt state ( dead state ) on 0 input from q0...then self loop qt state..... to disable the strings that start with 0....

  • @syedroshan8449
    @syedroshan8449 ปีที่แล้ว +1

    This is not complete dfa because initial state doesn't have 0 input, it doesn't satisfy dfa rule

  • @shivamchhabra7135
    @shivamchhabra7135 3 ปีที่แล้ว +1

    example 2 dfa drawn is wrong, there should be a dead state going from q0 and further there should be a loop of 0,1 on that dead state, such that there is no acceptance of strings starting from 0

  • @varalaxmichalla7746
    @varalaxmichalla7746 2 ปีที่แล้ว +1

    Q0 zero input is missing give that to dead state

  • @ameyanimkar8285
    @ameyanimkar8285 4 ปีที่แล้ว +76

    Mam, Aapne q0 👉0 keliye dead state hi nahi banaya

  • @swaroopgollapati3672
    @swaroopgollapati3672 4 ปีที่แล้ว +11

    Sorry Mam

  • @aparna1487
    @aparna1487 3 ปีที่แล้ว +9

    Hi madam in dfa all input symbols transitions should be there but intial state have transition for input 1 is there don't have transition for input 0

  • @user-jq4hi3rw7m

    in the example 2 you put the string 1010 ......??? how to accepted the string???

  • @handle00725
    @handle00725 2 ปีที่แล้ว +7

    how to make the table at

  • @sharathchandra65
    @sharathchandra65 4 ปีที่แล้ว +8

    You have catched my exact problem thank you madam for the great help, it really means alot for me

  • @sachinv464
    @sachinv464 3 ปีที่แล้ว +6

    Thank u madam u have helped a lot... U explained with clarity.... Thank u so much...

  • @chulbulipriya9871
    @chulbulipriya9871 2 ชั่วโมงที่ผ่านมา

    Thanku mam , I totally understand it ,this topic has recently started in our college and i didnt understand at that time but i now understand 😊

  • @suvenduhatua_1754
    @suvenduhatua_1754 3 ปีที่แล้ว +3

    if we input 0 ,then what will happen?....so,there will be one trap state.

  • @anilkumar-alovlyfrnd7745
    @anilkumar-alovlyfrnd7745 2 ปีที่แล้ว +8

    hlo mam what about the '0'variable for q0 in example 2