Question: With the e-transition to the last state, doesn't it make the PDM's language = 0^n 1^m? Just as you ran 0^3 1^7 through and the machine accepted, shouldn't the machine accept any combination of 0's and 1's, where the 1's come after the 0's?
@KneckChop No I don't think it does. If you try to read more than 3 ones in the Pop state then you'll crash and go to the dead state since there are no arrows for more than 3 ones for 0^31^m. Once you hit the dead state you can't leave it.
Question: With the e-transition to the last state, doesn't it make the PDM's language = 0^n 1^m? Just as you ran 0^3 1^7 through and the machine accepted, shouldn't the machine accept any combination of 0's and 1's, where the 1's come after the 0's?
KneckChop 2 months ago
@KneckChop No I don't think it does. If you try to read more than 3 ones in the Pop state then you'll crash and go to the dead state since there are no arrows for more than 3 ones for 0^31^m. Once you hit the dead state you can't leave it.
dealerG 3 weeks ago
I've really enjoyed this series, thank you
Jaykrueger 3 months ago
thank you good sir !
FelekisGR 1 year ago
just awesome!!! thanx
ankur021190 1 year ago