DETERMINISTIC FINITE AUTOMATA (DFA) EXAMPLE - 2 (STRINGS ENDS WITH) IN AUTOMATA THEORY || TOC

[@botzaifa]

In Question 2 (11:51) : What if we end q2 to a self loop 0. Even that gives the desired results and is a valid string. Right?

[@Juicy_18]

No,I think its because our language only accept strings that ends strictly with '10',so if u were to selfloop Q2 strings such as '100000' will be accepted of which is wrong I hope that helps

[@ezenwerekelvin2165]

@@Juicy_18 beautifully explained

[@MohammedSohail007]

man i have exam at 10am today , glad I found your channel in time , perfect explanation

[@danceflorr6758]

sir.. why can't we give 0 as the self loop to the final state "q2". It is also satisfying 10010 string....???? (check 11:50 )

[@karthikeyakallapu1274]

yes thats right

[@hashmatmohammed5807]

@@karthikeyakallapu1274 can we do that

[@mikiyasgetnet2750]

@@hashmatmohammed5807 No we cant because if u make a self loop on q2 with 0 it will be valid for 1000 which is invalid in our case

[@NGOFFICAL9]

@@mikiyasgetnet2750 👌

[@nandishnandish5467]

No you can't if input is 100100 then after 10 if the input is 0 it will stay in final state only, then it ended with 00. it is not valid language

[@sarkersaadahmed]

4:35 if you left the 1 at q1, it would still be q1 correcT?

[@naveenbanoth9642]

Sir at final state can we take loop in 1,0 when ends 10 condition

[@hashmatmohammed5807]

can we do that

[@chandugurram3444]

No you can't do that take a string 1000 if you take a self loop it ends in q2 which violates the rule given in question

[@sarkersaadahmed]

3:07 why dont you go back to q0 if the input was 0?

[@jeevannaidu1431]

for every dfa their is more than one possibility........so pls explain atleast 2 possibilities...so that we can understand easly

[@sarkersaadahmed]

i think it should like accept EVERY possible like from the set if one of them is wrong then the entire thing is wrong. Not sure really

[@MotivatedMind614]

Beautiful checkmate sir

[@PrasadPutta-n6o]

can we create q2 to dead state 0 1 we can do like that

[@vsnakshay1286]

beautifully explained sir!!

[@AadeshingaleOfficial-zl5fd]

Nice Sir 😊

[@naveen.bnaveen.b8106]

can we create a self loop 0 to the state Q2

[@mohanchandra1315]

Can we give self loop of 0,1 on initial state

[@Juicy_18]

If we were to do that,how can we transition to another state?according to what I know so far there can only be 1 transition per alphabet input when dealing with DFA's

[@adhithya2240]

Sir can we take Instead of taking transition of q2 to q0,can we place a self loop to q2 has 0

[@nandishnandish5467]

No you can't if input is 100100 then after 10 if the input is 0 it will stay in final state only, then it ended with 00. it is not valid language

[@DineshKrthakur-hf4rq]

how to decide use of self loop

[@technologyworld106]

Sir, I end (0) dead state is not use

[@kbvrama]

how q0 on 1 become q0 need clarity saradhi gaaru (10)

[@ghoshentertainment9028]

If 1001 then q2 will be the final state. Here q3 is final state , is this correct?

[@aparnaaparna9757]

No your wrong actually the question is ends with 10 your ask 1001 so first of all your question was wrong

[@chalapathi_roy6492]

1000 q1q2q0q0 ? states ends with 10 ?

[@pradyumna0073]

❤

[@zooheb9461]

You forgot to remove self transition from the state q1

[@kartikkhandelwal3518]

sir.. why can't we give 0 as the self loop to the final state "q2". It is also satisfying 10010 string....???? (check 11:50 )

[@hashmatmohammed5807]

same ...doubt .....can we do that

[@NGOFFICAL9]

We can't do self loop in final state in this string 10010