Minimization of DFA (Example 1)
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- Опубликовано: 15 сен 2024
- TOC: Minimization of DFA - Examples (Part 1)
This lecture shows how to minimize a DFA with an example. The steps are demonstrated using this example
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For those having confusion in 3 equivalence, both B and C belongs to different sets but with same transition(0 or 1) they belong to same set. Hence A and C are kept in one set.
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But the criteria was that the transition satate should belong to the set?
B was not the part of the set in 2 equivalence?
@@Aditya-wy4ci I think that maybe the next states of both A and C should belong to the same set.
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For those who have problem with 3 equivalence. Neso means that for the same input, the output should be stay in the same set. That does not mean both output for a state should stay in one set!!!!
You would be more clear to his logic in next video.
We start with 0-equivalence. Basically create 2 subsets, one with final states and the other with non-final states.
Now within each subset, try to find if constituent states are 1 equivalent. The way to do that is to pick two states at a time. If the two states transition to the same 0-equivalence subset on each input, then they are 1-equivalent.
If A,B and B,C are 1-equivalent, then A,C are also 1-equivalent.
If A,B are not 1-equivalent and B,C are, then A,C cannot be 1-equivalent
Repeat this process recursively till the output of n-equivalent and (n-1)-equivalent is the same
From that point, the subsets won’t change
All states within the same subset are equivalent
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My summary
We start with 0-equivalence. Basically create 2 subsets, one with final states and the other with non-final states.
Now within each subset, try to find if constituent states are 1 equivalent. The way to do that is to pick two states at a time. If the two states transition to the same 0-equivalence subset on each input, then they are 1-equivalent.
If A,B and B,C are 1-equivalent, then A,C are also 1-equivalent.
If A,B are not 1-equivalent and B,C are, then A,C cannot be 1-equivalent
Repeat this process recursively till the output of n-equivalent and (n-1)-equivalent is the same
From that point, the subsets won’t change
All states within the same subset are equivalent.
it seems like if they transition to the same exact state, then it doesn't matter if they are in the same subset or not. I feel like that wasn't explained clearly in the video, but that seems to be how he did it
two states would be equivalent if they transition to states from the same set of states from precedent equivalence or transition to the same states for each input.
in case of 1,2 equivalence the transition goes to the precedent subset
for third equivalence we have for input 0: A->B, C->B and for input 1 A->C, C->C
I think In equivalence 3- A and C doesn't go to same set as in input a B belongs to another set
I agreed 👍👍
It was to be in separate state
@Neso Academy please resolve it or admit it to all
at 10:15 A and C on getting input 0 goes to B which is not of the same set so how are the equivalent ?
Super easy explanation super easy to understand
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great explanation I didn't get the algorithm in the text book but your video clarifies everything.
So, what I get is that for n-equivalence just check, if they go to the same states with the same inputs and have the same type of being final or not, then they go together, that's it: both A and C can be joined because of the outputs, however E can't because it's a final state, if there was a F state that had the same outputs and was also final it could be joined with E, but not with A and C.
Hope i was clear.
Sir firstly thanks for your videos
here I want to tell you that you have missed concept of unreachable states as you directly applied checking state equivalence and in some questions if we dont remove unreachable states then our final state have that extra unreachable stat e and thus it will not be minimized(correct).
Although it is also advised to merge all dead states into one before applying this video's method but is optional as applying the method shown in this video will automatically do that.
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At 10:4 3rd equilence {A,C} belongs to different sets ..how can be both equilence
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how A and C are 3 equivalents?
because output of A and C have B which is in a different set in 2ND equivalence.
A and C go to same state B(taking 0) and C(taking 1)
Siddharth Y if they goes to different states then u check whether they r in same set ,if they r in same state then no worries!
got it. thanks
Zabardast hai
B is not 2 equivalent because it maps to a different set than A and C for input 1 (A and C map to {A,B,C} and B maps to {D}). A and C are three equivalent because they map to the same set regardless of the input ({B} if 0 and {A,C} if 1). Furthermore, they are n equivalent for integers >= 0 as they will always map to the same sets respectively.
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Thank you! Very clear
Why A and C are 3 equivalences ? When 0, A and C are going to B and B. But when you look at the 2 equivalences list, it is {A,C} {B} {D} {E}. B is not in the list. Are we looking for just A and C are going to the same set ?
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Good explanation sir
How, 3 equivalence of {A, C} is {A, C}. Please Explain.
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Thanks you so much for this explanation
sir i think you did a mistake in 3 equivalaence where in (A,C) when input is 0 then they are going for B and as previous equivalance they must follow the row and in the previous row their is only (A,C) which means A will be seperate and C also will be seperate. Am i wrong?, If i am wrong then correct me.
Thanks a lot!
I have a doubt that, let take if A and B are going to a same state say (non final state)on getting input 0
next, A and B are going to a same state again but this time (final state) on getting input 1 then could it be equivalent or not
please help me.. Hope it reach to NESO ACADEMY INSTRUCTOR 😇
reach to NESO ACADEMY INSTRUCTOR
If u observe the transition diagram a, c, and e also same (since e also producing output same as that of a and c)?
thanks sir appreciate the work
How come A and C are 3 equivalent to each other? On comparing A and C, A and C on getting an input 0 goes to B whereas on getting input 1 goes to C. So how come they are equivalent if they are not on the same set?
Correct
@@ujjwalbhardwaj1519no it's not correct
The output that A gets on 0 may be in a different set than the output A gets on 1, same goes with C
However, what we are checking is for 0, are the outputs of A and C in the same set, and for 1, are the outputs of A and C in the same set
On input 0, A and C go to B, which is the same set as each other
On input 1, A and C go to C, which is also the same set as each other
For 0 and 1, the set of the output state doesn't have to be the same
First condition is to check that , they goes for same state or not ,
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Nice explanation tqu sir
Since A and C have the same output (B, C) for 0 and 1, considering both are non-final states, it is guaranteed that A and C could be combined to minimize the DFA right?
In 2 equivalence b is in different state than AC and in the 3 rd equivalence a and c are going to B which is not in the same set anymore so how this became a 3 equivalence?
Simple and easy.
He says, at 1 Eq that D is separated because of the output of 1 being different between, let's say A and D, but B is the same, yet he says that B remains in the group.
At first, D is in the group, so B is accepted.
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Great explanation!
Sir .. in 3 equivalence how a and c are in 3 equivalence since a and c has comman as b so how they become same set
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Thank you so much for the help