Imagine: everytime you produce b, you produce ( a | epsilon ) everytime you produce c, you produce ( a | epsilon ) so for any word in this language, counting occurences by stacking the yielded terminals: '( a | epsilon )' and 'c' '( a | epsilon )' and 'c' . . . '( a | epsilon )' and 'c' '( a | epsilon )' and 'b' '( a | epsilon )' and 'b' . . . '( a | epsilon )' and 'b' + ______________________ => total number of 'a'
Clear and concise explanation. You deserve more subscribers. Thank you so much!
@EasyTheory I should be really thankful to you as you helped me ace the Theory of Computer Science exam.. your lesson videos helped me a lot
Great video, really informative. Any video you might suggest about DFAs and NFAs, on drawing them with examples?
Edit: I have several examples that are really old on the channel about creating DFAs and NFAs - may want to check the "intro theory" playlist.
thank you so much. Your videos are helping me a lot
高質量的視頻內容！我還有一個問題：有人給我轉了点usdt，我有恢復短語。[pride]-[pole]-[obtain]-[together]-[second]-[when]-[future]-[mask]-[review]-[nature]-[potato]-[bulb] 我怎麼把它們變現呢？
dude my test was last friday if you had posted it earlier 😭😭😭
Should have reminded me ;)
If in the last example instead of i
no it won't , then you would have to remember the S->Sc and X->Xb productions rest all will be same
I have a question, what if the case were that n >= 1? Is it still context free?
In the first example s can also be lambda/epsilon
S -> X where X is Lambda is taking care of that my friend
Could it be that example 3 is still missing X -> X1|X0 to be able to generate all possible strings with this X rule?
thank you
Thanks !
legend
How did you know that L = {0^n1^n2^n} wasn't a context free language?
GOD
Hello, can someone let me know if my CFG for generating non palindromes is correct, thank you!
S -> 1A0 | 0A1
A -> ε | A0 | A1
In that last example the explanation could have been better. It was confusing.
Imagine:
everytime you produce b, you produce ( a | epsilon )
everytime you produce c, you produce ( a | epsilon )
so for any word in this language, counting occurences by stacking the yielded terminals:
'( a | epsilon )' and 'c'
'( a | epsilon )' and 'c'
.
.
.
'( a | epsilon )' and 'c'
'( a | epsilon )' and 'b'
'( a | epsilon )' and 'b'
.
.
.
'( a | epsilon )' and 'b'
+
______________________
=> total number of 'a'
PLease come to my college and show my professor how t ofreaking teach bruhhhh :'}
Great video