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! 
finite-automata discrete-mathematics
asked Dec 9 '18 at 18:16
bmshbmsh
92
$endgroup$
add a comment |
$begingroup$
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!
finite-automata discrete-mathematics
asked Dec 9 '18 at 18:16
bmshbmsh
92
$endgroup$
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
add a comment |
$begingroup$
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!
finite-automata discrete-mathematics
asked Dec 9 '18 at 18:16
bmshbmsh
92
$endgroup$
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!
finite-automata discrete-mathematics
finite-automata discrete-mathematics
asked Dec 9 '18 at 18:16
bmshbmsh
92
asked Dec 9 '18 at 18:16
bmshbmsh
92
edited Dec 10 '18 at 4:00
bmsh
asked Dec 9 '18 at 18:16
bmshbmsh
92
asked Dec 9 '18 at 18:16
bmshbmsh
92
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
add a comment |
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
add a comment |
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.
answered Dec 9 '18 at 19:24
shahaf findershahaf finder
312
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$begingroup$
This is much, much better than your other answer. Thanks!
$endgroup$
– David Richerby
Dec 9 '18 at 20:08
add a comment |
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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.
answered Dec 9 '18 at 19:24
shahaf findershahaf finder
312
$endgroup$
$begingroup$
This is much, much better than your other answer. Thanks!
$endgroup$
– David Richerby
Dec 9 '18 at 20:08
add a comment |
$begingroup$
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.
answered Dec 9 '18 at 19:24
shahaf findershahaf finder
312
$endgroup$
$begingroup$
This is much, much better than your other answer. Thanks!
$endgroup$
– David Richerby
Dec 9 '18 at 20:08
add a comment |
$begingroup$
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.
answered Dec 9 '18 at 19:24
shahaf findershahaf finder
312
$endgroup$
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.
answered Dec 9 '18 at 19:24
shahaf findershahaf finder
312
answered Dec 9 '18 at 19:24
shahaf findershahaf finder
312
answered Dec 9 '18 at 19:24
shahaf findershahaf finder
312
answered Dec 9 '18 at 19:24
shahaf findershahaf finder
312
312
$begingroup$
This is much, much better than your other answer. Thanks!
$endgroup$
– David Richerby
Dec 9 '18 at 20:08
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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