Regex to NFA Conversion Isn't Hard! (Sipser 1.28a)

Can we get rid of some of the epsilons?

this made so much sense than you so much!! really helping me in my algo analysis class :))

Thanks! This was helpful :D

thanks a lot thanks a lot you are my saviour i was so stressed

great video 😍

If we had the regular expression a(abb)* U c and instead of b we put c in place of the corresponding b in the NFA how would the NFA accept for example the string αc ?

Amazing thank u so so much

There is a mistake in constructing the expression (abb)*. With the way you described it, you can make expressions that are of form (abb)^n where n>=1. There should be a returning loop between last state (that we get into by reading 'b' for the second time) and the state before we read 'a' with 'epsilon' read. And there should be a forward path from starting state to end state with 'epsilon' read.

Thank you so much! Very good explanation and procedure!

That was easy indeed and brilliantly explained.

is the union and the Plus sign the same ??

this video will be my source for compiler class lmao

how would you convert this to left linear grammar? with all the Epsilons?

You are amazing! It`s a shame I only discovered your channel in the night before the exam :D

You are my savior.

THANK U MAN U ROCK

Thanks I learnt a lot from your videos

Very helpful. Thank you. This converted the regex into an epsilon-NFA. Now I could convert this epsilon-NFA into an NFA but this would take even more time. Do you have any tips on how to convert this regex into an NFA without epsilon-transitions?

i'd give you my life if i could

Awesome and explained clearly, thank you!

You're amazing, man!

simple and well explained

If NFAs are like onions and ogres are like onions can you convert directly from NFAs to ogres?

Thank you, I'd request please more pumping lemmas problems :)

Thank you!