Clear and concise explanation. You deserve more subscribers. Thank you so much!
@abhinav9936 Жыл бұрын
@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
@mrboss39522 жыл бұрын
Great video, really informative. Any video you might suggest about DFAs and NFAs, on drawing them with examples?
@EasyTheory2 жыл бұрын
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.
@dd1.d Жыл бұрын
thank you so much. Your videos are helping me a lot
@bloodthirstybutcher83652 жыл бұрын
dude my test was last friday if you had posted it earlier 😭😭😭
@EasyTheory2 жыл бұрын
Should have reminded me ;)
@sc5shout Жыл бұрын
If in the last example instead of i
@gauravghosh6562 Жыл бұрын
no it won't , then you would have to remember the S->Sc and X->Xb productions rest all will be same
@rollbacked Жыл бұрын
I have a question, what if the case were that n >= 1? Is it still context free?
@robwindey9223 Жыл бұрын
Could it be that example 3 is still missing X -> X1|X0 to be able to generate all possible strings with this X rule?
@melihathevlogger37379 ай бұрын
thank you
@pateldhyey3063 Жыл бұрын
In the first example s can also be lambda/epsilon
@saumickpradhan276711 ай бұрын
S -> X where X is Lambda is taking care of that my friend
@rocksbox156 Жыл бұрын
How did you know that L = {0^n1^n2^n} wasn't a context free language?
@vimalathithand9179 ай бұрын
Thanks !
@akshitjain29066 ай бұрын
legend
@juicewar6 ай бұрын
Hello, can someone let me know if my CFG for generating non palindromes is correct, thank you! S -> 1A0 | 0A1 A -> ε | A0 | A1
@rahuljaswal92702 ай бұрын
GOD
@CLG111 Жыл бұрын
In that last example the explanation could have been better. It was confusing.
@surferriness Жыл бұрын
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'