Define a DFA that accepts all even length binary strings that don't contain the substring “111”?












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I think I have worked out a DFA that doesn't accept the substring "111," but I don't know how to account for accepting even length strings. Here is what I have so far. Any help would be greatly appreciated! enter image description here






finite-automata discrete-mathematics







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    I think you would have to keep track of parity. One set of states for an odd number of bits seen, another for an even number.
    $endgroup$
    – Tom Zych
    Dec 9 '18 at 19:11
















1












$begingroup$


I think I have worked out a DFA that doesn't accept the substring "111," but I don't know how to account for accepting even length strings. Here is what I have so far. Any help would be greatly appreciated! enter image description here






finite-automata discrete-mathematics







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    I think you would have to keep track of parity. One set of states for an odd number of bits seen, another for an even number.
    $endgroup$
    – Tom Zych
    Dec 9 '18 at 19:11














1












1








1





$begingroup$


I think I have worked out a DFA that doesn't accept the substring "111," but I don't know how to account for accepting even length strings. Here is what I have so far. Any help would be greatly appreciated! enter image description here






finite-automata discrete-mathematics







share|cite|improve this question











$endgroup$




I think I have worked out a DFA that doesn't accept the substring "111," but I don't know how to account for accepting even length strings. Here is what I have so far. Any help would be greatly appreciated! enter image description here






finite-automata discrete-mathematics




finite-automata discrete-mathematics






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 10 '18 at 4:00







bmsh

















asked Dec 9 '18 at 18:16









bmshbmsh

92




92








  • 1




    $begingroup$
    I think you would have to keep track of parity. One set of states for an odd number of bits seen, another for an even number.
    $endgroup$
    – Tom Zych
    Dec 9 '18 at 19:11














  • 1




    $begingroup$
    I think you would have to keep track of parity. One set of states for an odd number of bits seen, another for an even number.
    $endgroup$
    – Tom Zych
    Dec 9 '18 at 19:11








1




1




$begingroup$
I think you would have to keep track of parity. One set of states for an odd number of bits seen, another for an even number.
$endgroup$
– Tom Zych
Dec 9 '18 at 19:11




$begingroup$
I think you would have to keep track of parity. One set of states for an odd number of bits seen, another for an even number.
$endgroup$
– Tom Zych
Dec 9 '18 at 19:11










1 Answer
1






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oldest

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2












$begingroup$

Keep in mind that DFA has a "finite" memory, each state knows something about what you've read so far.
$A$ remembers that so far, you've read $w0$ for some $w in {0,1}^*$ or $epsilon$.
$B$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$ or $1$.

and so on...



Now you can duplicate the states to have the following properties:
$A_{even}$ means that so far you've read $w0$ for some $w in {0,1}^*$ or $epsilon$, and you've read even number of letters.
$A_{odd}$ means that so far you've read $w0$ for some $w in {0,1}^*$, and you've read odd number of letters ($epsilon$ has even number of letters, so it's not included here).
$B_{even}$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$, and you've read even number of letters.
$B_{odd}$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$ or $1$, and you've read odd number of letters.



And so on.

You need to re-define your transition function and and accept states to match with the definition of these new states



Edit: I misread the question as accepting '111' as a substring, so the definition of A,B that i showed are a bit off, but the answer to your question is similar in concept.






share|cite|improve this answer









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  • $begingroup$
    This is much, much better than your other answer. Thanks!
    $endgroup$
    – David Richerby
    Dec 9 '18 at 20:08











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1 Answer
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1 Answer
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$begingroup$

Keep in mind that DFA has a "finite" memory, each state knows something about what you've read so far.
$A$ remembers that so far, you've read $w0$ for some $w in {0,1}^*$ or $epsilon$.
$B$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$ or $1$.

and so on...



Now you can duplicate the states to have the following properties:
$A_{even}$ means that so far you've read $w0$ for some $w in {0,1}^*$ or $epsilon$, and you've read even number of letters.
$A_{odd}$ means that so far you've read $w0$ for some $w in {0,1}^*$, and you've read odd number of letters ($epsilon$ has even number of letters, so it's not included here).
$B_{even}$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$, and you've read even number of letters.
$B_{odd}$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$ or $1$, and you've read odd number of letters.



And so on.

You need to re-define your transition function and and accept states to match with the definition of these new states



Edit: I misread the question as accepting '111' as a substring, so the definition of A,B that i showed are a bit off, but the answer to your question is similar in concept.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is much, much better than your other answer. Thanks!
    $endgroup$
    – David Richerby
    Dec 9 '18 at 20:08
















2












$begingroup$

Keep in mind that DFA has a "finite" memory, each state knows something about what you've read so far.
$A$ remembers that so far, you've read $w0$ for some $w in {0,1}^*$ or $epsilon$.
$B$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$ or $1$.

and so on...



Now you can duplicate the states to have the following properties:
$A_{even}$ means that so far you've read $w0$ for some $w in {0,1}^*$ or $epsilon$, and you've read even number of letters.
$A_{odd}$ means that so far you've read $w0$ for some $w in {0,1}^*$, and you've read odd number of letters ($epsilon$ has even number of letters, so it's not included here).
$B_{even}$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$, and you've read even number of letters.
$B_{odd}$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$ or $1$, and you've read odd number of letters.



And so on.

You need to re-define your transition function and and accept states to match with the definition of these new states



Edit: I misread the question as accepting '111' as a substring, so the definition of A,B that i showed are a bit off, but the answer to your question is similar in concept.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    This is much, much better than your other answer. Thanks!
    $endgroup$
    – David Richerby
    Dec 9 '18 at 20:08














2












2








2





$begingroup$

Keep in mind that DFA has a "finite" memory, each state knows something about what you've read so far.
$A$ remembers that so far, you've read $w0$ for some $w in {0,1}^*$ or $epsilon$.
$B$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$ or $1$.

and so on...



Now you can duplicate the states to have the following properties:
$A_{even}$ means that so far you've read $w0$ for some $w in {0,1}^*$ or $epsilon$, and you've read even number of letters.
$A_{odd}$ means that so far you've read $w0$ for some $w in {0,1}^*$, and you've read odd number of letters ($epsilon$ has even number of letters, so it's not included here).
$B_{even}$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$, and you've read even number of letters.
$B_{odd}$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$ or $1$, and you've read odd number of letters.



And so on.

You need to re-define your transition function and and accept states to match with the definition of these new states



Edit: I misread the question as accepting '111' as a substring, so the definition of A,B that i showed are a bit off, but the answer to your question is similar in concept.






share|cite|improve this answer









$endgroup$



Keep in mind that DFA has a "finite" memory, each state knows something about what you've read so far.
$A$ remembers that so far, you've read $w0$ for some $w in {0,1}^*$ or $epsilon$.
$B$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$ or $1$.

and so on...



Now you can duplicate the states to have the following properties:
$A_{even}$ means that so far you've read $w0$ for some $w in {0,1}^*$ or $epsilon$, and you've read even number of letters.
$A_{odd}$ means that so far you've read $w0$ for some $w in {0,1}^*$, and you've read odd number of letters ($epsilon$ has even number of letters, so it's not included here).
$B_{even}$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$, and you've read even number of letters.
$B_{odd}$ remembers that so far, you've read $w01$ for some $w in {0,1}^*$ or $1$, and you've read odd number of letters.



And so on.

You need to re-define your transition function and and accept states to match with the definition of these new states



Edit: I misread the question as accepting '111' as a substring, so the definition of A,B that i showed are a bit off, but the answer to your question is similar in concept.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 19:24









shahaf findershahaf finder

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  • $begingroup$
    This is much, much better than your other answer. Thanks!
    $endgroup$
    – David Richerby
    Dec 9 '18 at 20:08


















  • $begingroup$
    This is much, much better than your other answer. Thanks!
    $endgroup$
    – David Richerby
    Dec 9 '18 at 20:08
















$begingroup$
This is much, much better than your other answer. Thanks!
$endgroup$
– David Richerby
Dec 9 '18 at 20:08




$begingroup$
This is much, much better than your other answer. Thanks!
$endgroup$
– David Richerby
Dec 9 '18 at 20:08


















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