DFA to RE State Elimination Multiple Final and Dead State Example || Lesson 37 || Finite Automata ||

[@Touhocaloid]

I have a DFA that Has qA transfer to itself on 1 and goes to qB on 0, qB goes back to qA on 1 and goes to accepting state qC on 0, qC goes to accepting state qD on 1 and qD repeats on itself on 1 and goes back to qC on 0. I can't figure it out at all, please help!

[@programmerJourney]

professor you are professional thanks really for everything

[@LearningMonkey]

So nice of you

[@hruthikn3861]

Does the regular expression vary on different methods of solving this problem?

[@LearningMonkey]

Yes it may vary. But the language accepted by them should be same

[@thisismeagain86]

amazing work , keep it up , it really helped me in understanding this. :)

[@mohamedwelteir]

How about a regular expression for any string in 0,1 except 11 and 111

[@buzz1386]

11* is different from 1*, why did you write so?

[@LearningMonkey]

From A to B we need input 1 B is having 1* So finally 11*

[@sindhukorubilli7502]

In the end can aa* also be minimized to a*? just like how bb* was minimized?

[@LearningMonkey]

No. Epsilon or bb* can be converted to b*. aa* is not a*

[@shihabkhandoker8706]

@@LearningMonkey aa* can be written a+ right??

[@arunsuresh8547]

5:14 , why shouldn't i write epsilon+aa*b as "a*b" ?

[@LearningMonkey]

a*b means it accept b. Which is not possible using aa*b

[@tonypepper]

Sir how did epsilon+11* , become 1*

[@LearningMonkey]

Epsilon means nothing 11* means minimum single one This or that means 1* 1* means nothing or single one Two one's etc

[@tonypepper]

Thank you sir ❤. Thanks for replying

[@areejsiddiqui6706]

can the answer for 2nd question be simplified to b*aa*b* ?

[@arunsuresh8547]

Epsilon+bb* becomes b* then why aa* can't be written as a* .. ? I didn't understand this ..

[@LearningMonkey]

aa* means first one a followed by any number of a's This means atleast one a So epsilon not there

[@KamliHeer-qo9qv]

thank u

[@LearningMonkey]

Have a great learning in CSE