Equivalence of NFAs with/without $\epsilon$-Moves

  • Context: MHB 
  • Thread starter Thread starter evinda
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around the equivalence of non-deterministic finite automata (NFAs) with epsilon-moves to those without epsilon-moves. Participants explore the implications of transitions in these automata, particularly focusing on the necessity and consequences of adding transitions based on the presence or absence of epsilon-moves.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Some participants question the need to add a transition on the symbol '1' from state $q_0$ to state $q_1$ when there is no direct transition corresponding to '1'.
  • One participant suggests that transitions can be achieved through a combination of epsilon-moves and symbol consumption, presenting examples of possible transitions.
  • Another participant proposes a correction to the transition representation, indicating that the second transition should involve a direct transition from $q_1$ to itself on '1'.
  • There is a discussion about the transformation of NFAs with epsilon-moves into those without, where participants express uncertainty about how many transitions should be retained or added.
  • Some participants express confusion about whether transitions on '0' from $q_1$ to $q_2$ should be included, leading to further exploration of the implications of such transitions.
  • Participants consider the possibility of reading multiple '0's at the beginning and the implications of removing certain transitions in the new NFA structure.
  • There is a suggestion that removing a transition could affect the ability to read specific sequences, raising questions about the overall structure of the NFAs being discussed.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and implications of specific transitions in the NFAs. There is no consensus on whether certain transitions should be added or removed, and the discussion remains unresolved regarding the best approach to transforming NFAs with epsilon-moves to those without.

Contextual Notes

Participants highlight the complexity of transitioning between NFAs with and without epsilon-moves, indicating that assumptions about the structure and behavior of these automata may vary. The discussion reflects a range of interpretations regarding the representation of transitions and the conditions under which they apply.

evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
Hello! (Wave)

I am looking at the equivalence of NFAs with $\epsilon$-moves to NFAs without $\epsilon$-moves.

View attachment 5767

View attachment 5768

Why do we have to add a transition on $1$ from $q_0$ to $q_1$ although we don't have an arrow from $q_0$ to $q_1$ that corresponds to the symbol $1$?
 

Attachments

  • ep.JPG
    ep.JPG
    43.3 KB · Views: 114
  • ep2.JPG
    ep2.JPG
    6.3 KB · Views: 118
Physics news on Phys.org
evinda said:
Hello! (Wave)

I am looking at the equivalence of NFAs with $\epsilon$-moves to NFAs without $\epsilon$-moves.Why do we have to add a transition on $1$ from $q_0$ to $q_1$ although we don't have an arrow from $q_0$ to $q_1$ that corresponds to the symbol $1$?

Hey evinda! (Smile)

Let's see how we can get from $q_0$ to $q_1$ while consuming a symbol... (Thinking)

Isn't that the following?
$$q_0 \overset 0\to q_0 \overset \varepsilon\to q_1 \\
q_0 \overset \varepsilon\to q_0 \overset 1\to q_1$$

Actually we can also pass $q_1$ by with only epsilons:
$$q_0 \overset \varepsilon\to q_1 \overset \varepsilon\to q_2$$
We'll have to consider that later to see what can happen afterwards when a symbol is consumed after all. (Sweating)
 
Last edited:
I like Serena said:
Isn't that the following?
$$q_0 \overset 0\to q_0 \overset \varepsilon\to q_1 \\
q_0 \overset \varepsilon\to q_0 \overset 1\to q_1$$

Shouldn't the second one be as follows?$$q_0 \overset \varepsilon\to q_1 \overset 1\to q_1$$

Or am I wrong?
 
evinda said:
Shouldn't the second one be as follows?$$q_0 \overset \varepsilon\to q_1 \overset 1\to q_1$$

Or am I wrong?

Yep. You're right. (Nod)
 
I like Serena said:
Yep. You're right. (Nod)

That's why I haven't understood why we have to add a transition on $1$ from $q_0$ to $q_1$... (Sweating)
 
evinda said:
That's why I haven't understood why we have to add a transition on $1$ from $q_0$ to $q_1$... (Sweating)

In the original NFA that is not possible, but we are revising it into a different NFA without $\varepsilon$-transitions.
In the new NFA we will have the same states with different transitions such that there will not be $\varepsilon$-transitions any more.
So we replace the 2 transitions that include a $\varepsilon$-transition with 1 transition. (Emo)
 
I like Serena said:
In the original NFA that is not possible, but we are revising it into a different NFA without $\varepsilon$-transitions.
In the new NFA we will have the same states with different transitions such that there will not be $\varepsilon$-transitions any more.
So we replace the 2 transitions that include a $\varepsilon$-transition with 1 transition. (Emo)

I think that I got it... But then at the following, why don't we also add a transition on $0$ from $q_1$ to $q_2$ ?

View attachment 5774
 

Attachments

  • nfa.JPG
    nfa.JPG
    19.6 KB · Views: 105
evinda said:
I think that I got it... But then at the following, why don't we also add a transition on $0$ from $q_1$ to $q_2$ ?

I see this is a slightly different NFA than in the opening post.
Once we're in $q_1$ or $q_2$ there is no transition on $0$ any more is there? (Wondering)
We do have a transition on $0$ from $q_0$ to $q_2$.
 
I like Serena said:
I see this is a slightly different NFA than in the opening post.
Once we're in $q_1$ or $q_2$ there is no transition on $0$ any more is there? (Wondering)
We do have a transition on $0$ from $q_0$ to $q_2$.

The NFA with $\epsilon$-moves that we have is the following, isn't it?

View attachment 5777

So couldn't we read 0 more than once at the beginning and then twice $\epsilon$ ?
That's why I thought that we could add a transition on 0 from $q_1$ to $q_2$.
So I am wrong, right?
 

Attachments

  • init.png
    init.png
    2.7 KB · Views: 110
  • #10
evinda said:
The NFA with $\epsilon$-moves that we have is the following, isn't it?

So couldn't we read 0 more than once at the beginning and then twice $\epsilon$ ?
That's why I thought that we could add a transition on 0 from $q_1$ to $q_2$.
So I am wrong, right?

Yes. That's the NFA with $\epsilon$-moves.

Wouldn't that then be a transition from $q_0$ to $q_2$? (Wondering)
 
  • #11
I like Serena said:
Yes. That's the NFA with $\epsilon$-moves.

Wouldn't that then be a transition from $q_0$ to $q_2$? (Wondering)

Yes, that's right... I thought so, because this case also appeared at the NFA of my first post.

So couldn't we also remove the transition on 0 from $q_1$ to $q_2$ at the corresponding new NFA of my first post? Or am I wrong? (Sweating)
 
  • #12
evinda said:
Yes, that's right... I thought so, because this case also appeared at the NFA of my first post.

So couldn't we also remove the transition on 0 from $q_1$ to $q_2$ at the corresponding new NFA of my first post? Or am I wrong? (Sweating)

But in the OP we can go $q_1 \overset\epsilon\to {q_2} \overset 0\to q_2$ can't we? (Wondering)

Suppose we remove it, can we read for instance $010$ then? (Wondering)
 
  • #13
I like Serena said:
But in the OP we can go $q_1 \overset\epsilon\to {q_2} \overset 0\to q_2$ can't we? (Wondering)

Suppose we remove it, can we read for instance $010$ then? (Wondering)

I think that I got it now... Thanks a lot! (Mmm)
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
2K
Replies
4
Views
838
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 0 ·
Replies
0
Views
2K