Thank you so much for this video! After banging my head against countless other tutorials, yours was the one that finally made Regex conversion click for me. The clarity and detail you provided were exactly what I needed. I can't express how much I appreciate your effort-this is a game-changer for me. Keep up the awesome work! 🙌
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.
I think what Easy Theory had was correct. First, you're saying that "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" but this is equivalent to linking the last state to the beginning state because epsilon transitions don't take up any string. Second, responding to the statement "And there should be a forward path from starting state to end state with 'epsilon' read.", this is redundant because the starting state is already a final state, so we don't need the epsilon transition to the end state. Please correct me if you think I am wrong.
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?
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 ?
It doesn't. The strings accepted would be {EMPTY, c, a, aabb, aabbabb,...}. It's the union of the unitary string 'c' and the strings generated bu the regular expression a(abb)*, that is, string that start with 'a' and has any amount of 'abb's, even none.
Thank you so much for this video! After banging my head against countless other tutorials, yours was the one that finally made Regex conversion click for me. The clarity and detail you provided were exactly what I needed. I can't express how much I appreciate your effort-this is a game-changer for me. Keep up the awesome work! 🙌
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.
yes, this is very right. I was looking out for a comment that matches my thought. Thanks
I think what Easy Theory had was correct.
First, you're saying that "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" but this is equivalent to linking the last state to the beginning state because epsilon transitions don't take up any string.
Second, responding to the statement "And there should be a forward path from starting state to end state with 'epsilon' read.", this is redundant because the starting state is already a final state, so we don't need the epsilon transition to the end state.
Please correct me if you think I am wrong.
It's a standard procedure. Don't make things more complicated
If NFAs are like onions and ogres are like onions can you convert directly from NFAs to ogres?
you're asking the right questions here!
You are amazing! It`s a shame I only discovered your channel in the night before the exam :D
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?
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 ?
It doesn't. The strings accepted would be {EMPTY, c, a, aabb, aabbabb,...}. It's the union of the unitary string 'c' and the strings generated bu the regular expression a(abb)*, that is, string that start with 'a' and has any amount of 'abb's, even none.
thanks a lot thanks a lot you are my saviour i was so stressed
You are my savior.
Can we get rid of some of the epsilons?
That was easy indeed and brilliantly explained.
how would you convert this to left linear grammar? with all the Epsilons?
Thank you so much! Very good explanation and procedure!
simple and well explained
is the union and the Plus sign the same ??
yes
this made so much sense than you so much!! really helping me in my algo analysis class :))
Awesome and explained clearly, thank you!
Thank you, I'd request please more pumping lemmas problems :)
this video will be my source for compiler class lmao
Thanks I learnt a lot from your videos
Thank you! Very easy to understand ❤❤❤
ty
great explanation as always!
You're amazing, man!
Thanks, very helpful!!
Thank you!
Thank you so much ^^
Amazing thank u so so much
THANK U MAN U ROCK
Thanks! This was helpful :D
great video 😍
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Go to an indian teacher they are very good in teaching..
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