Consider the states that have incoming edges but only from other unreachable states. Won't they be unreachable as well? So, to say "all states not having any incoming edge are unreachable" is correct but, "unreachable states have no incoming edge"(0:56) should be incorrect. Isn't it?
Why is g not one equivalent, we can see both E ang G have same transition states so, shouldn't we include include as one equivalence to the set A, D, E 2:59
Thank you for this course. did i miss it, or you do not touch the subject of epsilon transitions and removing epsilon transition ? if you do , can you please tell me in which lecture do you touch the subject? thanks again!
you mentioned in one of your videos that any input cannot go in 2 simultaneous states at the same time. So when we write {D, E}, {B, C} etc, are we performing a union ?
shouldn't it be {A} {D,E} {B,C} {G} in the 1- equivalence @2:58 as, state D and E goes to G by getting input 1 and 0 , where G is in another set. Thus we don't need to proceed to 3-equivalance
no .. because B C and G are all in the same set of states . We can consider two ways . Either the same states or the different states but in the same set of states. In either case, equivalence holds
Not always. Imagine a scenario when there is an entirely disjoint finite automata from the DFA that the initial state is located in. You obviously can't reduce the entire disjoint FA to nothingness.
In example 3 of minimising dfa to nfa the state A AND STATTE B ON INPUT 0. And 1 have different state are still being collected as a single state ? How it is possible
Sir in a lecture of conversion of NFA to DFA you said that in DFA we can pass only one input at a time if it isn't then it is a NFA. Same in this problem more than one input are passing through one state so why we calling it a DFA it simply maybe a NFA then . Sir please clear this confusion 😭
it seems you are little confused .... in dfa two input may lead to one state but in nfa 1 input may lead to two states that's why nfa is non deterministic.
So as far as I can think, unreachable states cannot come from the same machine, it has to be a state of a different machine which relies on the computation used by this machine.
there are many examples where an unreachable state can be reached. lets suppose you said a word "hey", an unreachable state can be when you "didn't say hi". ofcourse, this is not a real life example but a subject has to be exhaustive and consider all the possibilities so that people dont get confused when such a scenario happens. also, it never hurts to be explicit.
come in our college , we need real techers
Same. In Nepal :P
Use projectors in school.
@@adarshjaiswal7334 😂😂
Bhai tu nhi hi padhega chahe koi bhi tujhe padha le samjha
Many students are struggling to understand .....bt ur videos makes it simple .....plz make videos on computer hardwares ...
Clean nd clr explanantion👌👌 .....tqsm sir 😇😇
Consider the states that have incoming edges but only from other unreachable states. Won't they be unreachable as well?
So, to say "all states not having any incoming edge are unreachable" is correct but,
"unreachable states have no incoming edge"(0:56) should be incorrect. Isn't it?
yes, you are right.it will be sufficient condition but not necessary
i don't know why but removing F made me sad
cuz ur name sad o
The best on TH-cam❤❤❤
Why is g not one equivalent, we can see both E ang G have same transition states so, shouldn't we include include as one equivalence to the set A, D, E 2:59
You cannot combine a final and non final state
Thanks love from doon❤
Same here
State D,E and G both are going to state G for 0,1 input. Can we make it one single state like state DEG as the next state is same.
No.
{D,E} is a NON-FINAL STATE while {G} is a FINAL STATE.
Thanks for ur explanation...it is helpful...😊
Thank you for this course.
did i miss it, or you do not touch the subject of epsilon transitions and removing epsilon transition ? if you do , can you please tell me in which lecture do you touch the subject?
thanks again!
They did a playlist called theory of computation and authomata theory, you should find it there.
Thanks a lot sir ❤
So, we just remove the unreachable state and proceed as usual? Is there anything else that needs to be done?
Lol...no
Thankyou sir
you mentioned in one of your videos that any input cannot go in 2 simultaneous states at the same time. So when we write {D, E}, {B, C} etc, are we performing a union ?
watch minimising dfa first vid
Yes we are combining here with union yes
My college teacher did it without removing the unreachable state is that correct???
Thank you sir.
Please come to NIT SRINAGAR to teach us….We need You
shouldn't it be {A} {D,E} {B,C} {G} in the 1- equivalence @2:58 as, state D and E goes to G by getting input 1 and 0 , where G is in another set. Thus we don't need to proceed to 3-equivalance
no .. because B C and G are all in the same set of states . We can consider two ways . Either the same states or the different states but in the same set of states. In either case, equivalence holds
@@srayanray3815 I still dont get you...elaborate please?
@@srayanray3815 Thanks 👍
Thanks sur ❤
Thank you Sir
a 3:46,why we r finding next equvalance,,bcz in equvalance 0 and 1 ,1st set is same...why sir?
@@nesoacademy At 1 equivalence, I got {A}, {D,E}, {B,G}, {C}.
why b and c are one equivalent they reached at different states
because their transition belong to the same set in zero equivalence
+Zeeshan Bebal thnkss
@@amanpreetsehjal464 wlcm
Very....♾️ Useful......!!!
good example :)
Will we put the unreachable state once we have made the minimized DFA?
the very purpose of minimizing a DFA is to remove useless or redundant states
so placing F back destroys the aim of what we're trying to do
@@omkararora8739 if we take back the F , the minimized dfa will remain minimized
Why is {D,E,G} not possible?
All of them have the same G common in the table.
but G is a final state and according to 0 equivalence all final states are separated
can we minize a dfa with unreachable states using the table filling method?
Not always. Imagine a scenario when there is an entirely disjoint finite automata from the DFA that the initial state is located in. You obviously can't reduce the entire disjoint FA to nothingness.
In example 3 of minimising dfa to nfa the state A AND STATTE B ON INPUT 0. And 1 have different state are still being collected as a single state ?
How it is possible
DG is equivalent to each other.I have one doubt sir, canI write 2 equivalence as {A}, {B, C}, {D, G, E}.Is it correct. Else ,plz clear my doubt sir .
I think No because , both D,E are not final states from initial and not in the same set with final states.
advertisement is awesome.... so is your explanation.............
what happened to F. Should it placed back?
the very purpose of minimizing a DFA is to remove useless or redundant states
so placing F back destroys the aim of what we're trying to do
F is unreachable state from initial state. So we remove F from transition diagram.
If you're given only the transition function, how do you identify an unreachable state
it should be the one that isn't present in the rhs of the table.
In 1 -equvivalence {A D E } is equal then how in 2 -equvivalence { A D E } is not equal???
coz g got separated hence will not map
God⚡
is "G" an unreachable state?
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Sir if we not removing unreachable and we solve that normal then there is a effect in minimised states or not Please reply sir??
yes you will have 1 extra state which we dont need
Sir in a lecture of conversion of NFA to DFA you said that in DFA we can pass only one input at a time if it isn't then it is a NFA. Same in this problem more than one input are passing through one state so why we calling it a DFA it simply maybe a NFA then . Sir please clear this confusion 😭
it seems you are little confused .... in dfa two input may lead to one state but in nfa 1 input may lead to two states that's why nfa is non deterministic.
@@drkcodr thanks btw Now I had completed whole T.o.c. thinking about this dought make me laugh 😅
Where are unreachable states used in Real life?
ChildhoodDreams
A -> Starting state . eg : you
G->Final Unreachable state , eg: that girl you want to talk to in class but is way out of your state (league)
So as far as I can think, unreachable states cannot come from the same machine, it has to be a state of a different machine which relies on the computation used by this machine.
@@vimalsheoran8040 shut the *** up
@@vijoyjyotineog1896 Did I offend you brother?
isn't it a NFA as we have state E to G with 1,0?
no bro, nfa has more than 1 outputs for same input
this one has same output for 2 different inputs
My doubt is how un reacble state corresponds to final state and simply he remove that but the its not possible
i did not effect the transition so removed and proceded
Am i the only one noticing somebody scream at 00:12
what if we have a rejection state? what will be its 0th equivalence position
here it become nfa because it is going same state on two input
Don't we have to put F back
Cool story, But I wonder what scenario one would come up and draw such DFA knowing a state is unnecessary in the first place.
there are many examples where an unreachable state can be reached. lets suppose you said a word "hey", an unreachable state can be when you "didn't say hi". ofcourse, this is not a real life example but a subject has to be exhaustive and consider all the possibilities so that people dont get confused when such a scenario happens. also, it never hurts to be explicit.
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