Answer: Language of Automata: DFA Q1, Q2

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Discussion Overview

The discussion revolves around determining the language accepted by a given Deterministic Finite Automaton (DFA) based on its state diagram. Participants explore the implications of the state transitions and the final states, focusing on the representation of the language in regular expression format.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • Some participants inquire about the start state of the DFA and the desired format for the language representation.
  • It is suggested that if the start state is Q1, the language could be represented as $b*a*$, where $*$ denotes zero or more occurrences.
  • Another participant proposes that the language could also be $a*b*$, questioning the initial claim.
  • Concerns are raised about the transitions leading to state Q3, which is not a final state, suggesting that once in Q3, the DFA cannot reach a final state.
  • Some participants express confusion regarding the transitions from state Q2 to both Q1 and Q3, questioning the determinism of the state machine.
  • A participant suggests that state Q3 should be removed to simplify the DFA and ensure it is deterministic.

Areas of Agreement / Disagreement

Participants express differing views on the correctness of the state transitions and the implications for the language accepted by the DFA. There is no consensus on the structure of the DFA or the language it accepts, indicating ongoing debate and uncertainty.

Contextual Notes

Participants note potential issues with the completeness of the transition table and the determinism of the state machine, but do not resolve these concerns.

mathmari
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Hi! Could you help me finding the language that is accepted by the DFA with the following state diagram?

___| a ____ b
Q1 | Q2 __ Q1
Q2 | Q2 __ Q3
Q3 | Q3 __ Q3

final states: Q1, Q2
 
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mathmari said:
Hi! Could you help me finding the language that is accepted by the DFA with the following state diagram?

___| a ____ b
Q1 | Q2 __ Q1
Q2 | Q2 __ Q3
Q3 | Q3 __ Q3

final states: Q1, Q2

Hi mathmari!

What is your start state?

If Q3 is not a final state, you can never go to Q3, since you can never go back.
Is that what you intended?

How do you want your language?
As a regular expression?
In BNF syntax? Or EBNF?

If you start in state Q1, the words you can form are $b*a*$, where $*$ denotes zero or more.
Starting from state Q2, you can only form $a*$, which is already included.
 
I like Serena said:
Hi mathmari!

What is your start state?

If Q3 is not a final state, you can never go to Q3, since you can never go back.
Is that what you intended?

How do you want your language?
As a regular expression?
In BNF syntax? Or EBNF?

If you start in state Q1, the words you can form are $b*a*$, where $*$ denotes zero or more.
Starting from state Q2, you can only form $a*$, which is already included.

The start state is Q1.
I want the language as a regular expression.
 
mathmari said:
The start state is Q1.
I want the language as a regular expression.

Then I guess I've already answered your question...
 
I like Serena said:
If you start in state Q1, the words you can form are $b*a*$, where $*$ denotes zero or more.

Starting in state Q1, couldn't the words be also $ a*b*$ ?
 
mathmari said:
Starting in state Q1, couldn't the words be also $ a*b*$ ?

After you accept the first $a$, accepting a $b$ would bring you to state Q3.
And from state Q3 you can never get to a final state anymore.

It is a bit weird though that you would have a state that doesn't go anywhere.
Are you sure that is right?
 
I like Serena said:
After you accept the first $a$, accepting a $b$ would bring you to state Q3.
And from state Q3 you can never get to a final state anymore.

It is a bit weird though that you would have a state that doesn't go anywhere.
Are you sure that is right?

Now I looked at the exercise again and realized that the state Q2 with $b$ goes to the state Q1 and to the state Q3.
 
mathmari said:
Now I looked at the exercise again and realized that the state Q2 with $b$ goes to the state Q1 and to the state Q3.

So your transition table is incomplete?
But still no other transitions starting from state Q3?
 
I like Serena said:
So your transition table is incomplete?
But still no other transitions starting from state Q3?

Starting from Q3 with $a$ or $b$ it goes to the state Q3 again...
 
  • #10
How can the state Q2 go with b to the state Q1 and also to the state Q3? :confused:
 
  • #11
mathmari said:
How can the state Q2 go with b to the state Q1 and also to the state Q3? :confused:

Now that you mention it, that's not really deterministic is it?
However, if it would go to state Q3, it can never reach a final state anymore.

In my opinion, your state machine is screwed up.
State Q3 should be deleted altogether, including all transitions toward it.
Then we're left with a nice and deterministic state machine. ;)
 

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