You are truly a great educator. It takes hours and hours to understand this topic but you explained in just few minutes. Your video makes me think how much garbage is thown at us in college.
Wow. So impressed. It's time to get rid of if statements. Just discovered this channel. How the explanation fits on the screen is beyond me. Thank you sir!
Thanks!! I’m studying in a graduate course for human language technology and I couldn’t quite see the picture of what was going on with this until now. Beautiful video btw
That's a great idea. I was so focused on moving toward implementation in hardware (which doesn't address epsilon transitions) that I didn't think anyone would be interested.
What if the original NDFA has multiple exit-states (let's say they are q2 and q3; q0 is our starting state and q1 is neither). Will our DFA's exit states be the ones that have BOTH/ALL of the NDFA exit states (so {q2, q3}, {q1, q2, q3}, {q0, q2, q3} and {q0, q1, q2, q3}) or if they have ANY of the NDFA's exit states (so {q1, q2}, {q2}, {q3} etc.)?
Hey i had a question. Sometimes this method doesnt work and we need to use an epsilon transition method instead. How do we know if an NFA can be solved in the way described in this video and when we need to use the epsilon method?
Well, this is out of order, but having another example NFA helped me catch a bug in my code to create a DFA from one, based on the recent Computerphile video. Of course, I might also have to re-think some things, because I don't keep around the empty-set transitions (or state e, in your final form)... instead just returning False if there's not a transition defined between two states (and, for example, I don't define a transition for starting at a and getting a 1). I'm fairly certain the effect is the same, but it looks different. Interesting.
Wouldn’t “000” be rejected even though it should be accepted"? We have a 0, then any number of 1’s (in this case 0/none), and still end in a 0. But it gets rejected as it follows states a,b,c,e (rejected)
Oh I see. Concatenation says we should start with 0, have any number of 1’s, then immediately end in 0 or 1. 000 should be rejected. As it does not immediately end in 0 but there is another 0 that follows.
This channel is gonna blow up someday. It deserves more subscribers. Too good production quality.
aaaand a legend was born...
I don't know why this channel isn't getting millions of views
You are truly a great educator. It takes hours and hours to understand this topic but you explained in just few minutes. Your video makes me think how much garbage is thown at us in college.
Wow. So impressed. It's time to get rid of if statements. Just discovered this channel. How the explanation fits on the screen is beyond me. Thank you sir!
this channel should be recognize by peoples
I generally read comments before watching a video,
so, to help others find this amazing video, here's my review - 5⭐
such a great example it cleared all doubts I had for converting from non-deterministic to dertinistic thank you !!!
This is so helpful. MILES better than my uni lecturer.
Incredibly easy thanks to you!
Such a clear and fruitful explanation! Thank you sir!
Ça ne peut être expliqué plus clairement. Bravo.
Thanks!! I’m studying in a graduate course for human language technology and I couldn’t quite see the picture of what was going on with this until now. Beautiful video btw
Beautiful explanation
Awesome. Thanks for the great explanation
Non-deterministic finite automata also has something called epsilon transitions. Can you make a video converting NFA with ε to DFA?
That's a great idea. I was so focused on moving toward implementation in hardware (which doesn't address epsilon transitions) that I didn't think anyone would be interested.
Such a clear and well presented explanation. Well done and thanks so much!
WOW, This teacher looks really like my IELTS SPEAKING TEST officer😇
BTW, Thank you for your clear explanation.
Very nice thank you so much
So so so helpful, thank you
thank you!
Thank you.
Thank you that was amazing
You're welcome! I appreciate the kind words.
Good Work thanks a lot❤
Thank you so much this helped a lot!
Nice explanation
Awesome
My only pet peeve is the squeaky sound of the marker. Other than that love the videos. Very informative!
What if the original NDFA has multiple exit-states (let's say they are q2 and q3; q0 is our starting state and q1 is neither). Will our DFA's exit states be the ones that have BOTH/ALL of the NDFA exit states (so {q2, q3}, {q1, q2, q3}, {q0, q2, q3} and {q0, q1, q2, q3}) or if they have ANY of the NDFA's exit states (so {q1, q2}, {q2}, {q3} etc.)?
Hey i had a question. Sometimes this method doesnt work and we need to use an epsilon transition method instead. How do we know if an NFA can be solved in the way described in this video and when we need to use the epsilon method?
Well, this is out of order, but having another example NFA helped me catch a bug in my code to create a DFA from one, based on the recent Computerphile video.
Of course, I might also have to re-think some things, because I don't keep around the empty-set transitions (or state e, in your final form)... instead just returning False if there's not a transition defined between two states (and, for example, I don't define a transition for starting at a and getting a 1). I'm fairly certain the effect is the same, but it looks different. Interesting.
didn't know Walter White gave CS lectures
Wouldn’t “000” be rejected even though it should be accepted"? We have a 0, then any number of 1’s (in this case 0/none), and still end in a 0. But it gets rejected as it follows states a,b,c,e (rejected)
Oh I see. Concatenation says we should start with 0, have any number of 1’s, then immediately end in 0 or 1. 000 should be rejected. As it does not immediately end in 0 but there is another 0 that follows.
Are you the god or something?
How's he writing like that 😢
Thank you soooo much!!