WOW! perfect! the clear and consice explanation added with the amazing visuals, audio pacing being constant + composition and your way of covering a bunch of concepts within one example is perfect! loved it! Thank you!
Appreciate your clear explanation. Please continue making Computer Theory videos, such as converting regular expression to NFA and vice versa. Also, didn't mean to point this out, but because we are building DFA at the end, shouldn't our DFA have trap states? Anyway, thank you for your video, sir.
State (g,f) can be omitted, and the 0,1 transition can be from (a,s,f) to itself. I'm not sure how you got the state (g,f) in the first place, given there are no forward epsilon-transitions (indeed, no forward transitions at all) from state (g) or state (f) in the NFA. Essentially, if a newly created DFA state is a subset of any of the others, it doesn't need to be included.
RE: "not sure how you got the state (g,f) in the first place." First, make sure you understand how we got the state {a,s,f} in the first row of the table. Now, consider 0 transition out of {a, s, f} (third row of the table). Observe that we can go from a to f on a 0 transition in the NFA. We can go from s to g on a 0 transition. Hence, we obtain {g, f} from {a, s, f} on a 0 transition. Makes sense? The video has a detailed explanation, including 0 transitions. Furthermore, your comment on subset is not true in general (and needs to be further refined). In particular, there are worst case examples (see the "Dragon book" of compiler design) showing that n number of NFA states may lead to 2^n DFA states, where you MUST retain all the subsets.
This is because in the NFA, there is no transition out of g or out of f. In general, we cannot have a transition in the DFA that does not exist in the NFA.
Finally, I understand NFA to DFA before Midterms, thanks 🎉
cleanest and the clearest explanation even covers starting with an epsilon transition
WOW! perfect! the clear and consice explanation added with the amazing visuals, audio pacing being constant + composition and your way of covering a bunch of concepts within one example is perfect! loved it! Thank you!
best explanation on this topic on youtube, thank you!
Great video brother , please keep it up
this man is good af thanks life savor
Appreciate your clear explanation. Please continue making Computer Theory videos, such as converting regular expression to NFA and vice versa. Also, didn't mean to point this out, but because we are building DFA at the end, shouldn't our DFA have trap states? Anyway, thank you for your video, sir.
Amazing video, jazakAllah Khair for clarifying epsilon transition
Please more video on computation theory, and all amaizing stuff . All the best !
Excellent presentation.
You are amazing!
State (g,f) can be omitted, and the 0,1 transition can be from (a,s,f) to itself. I'm not sure how you got the state (g,f) in the first place, given there are no forward epsilon-transitions (indeed, no forward transitions at all) from state (g) or state (f) in the NFA.
Essentially, if a newly created DFA state is a subset of any of the others, it doesn't need to be included.
RE: "not sure how you got the state (g,f) in the first place." First, make sure you understand how we got the state {a,s,f} in the first row of the table. Now, consider 0 transition out of {a, s, f} (third row of the table). Observe that we can go from a to f on a 0 transition in the NFA. We can go from s to g on a 0 transition. Hence, we obtain {g, f} from {a, s, f} on a 0 transition. Makes sense? The video has a detailed explanation, including 0 transitions.
Furthermore, your comment on subset is not true in general (and needs to be further refined). In particular, there are worst case examples (see the "Dragon book" of compiler design) showing that n number of NFA states may lead to 2^n DFA states, where you MUST retain all the subsets.
shouldn't DFA cover all of the transitions(0 and 1)? g and g,f doesn't have any transitions
This is because in the NFA, there is no transition out of g or out of f. In general, we cannot have a transition in the DFA that does not exist in the NFA.
Thanks for the very helpful explanation!
should {g,f} not be {g,f,s} because u can go from a to s through epsilon.
why isnt s part of the set for {a,s,f} on 0?
best explaination
Thank you!
thank you a lot.. also i think we must create a dead state to have complete dfa.
I agree
Thank you teacher
Awesome!
please, what is the app you use?
Goodnotes
thank you so much
very nice! thx
great !
can you refer a book?
This is based on Michael Scott's Programming Languages book.
thankss
What happens when f epsilon to another state
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