Give a regular expression
Let Σ be {0, 1} Give a regular expression generating words over Σ containing an even number of 1’s or with a length which is multiple of 3.
I came up with this solution:
ε ( ((0*(10*10*)) + ((0+1) (0+1) (0+1)) )
and that's from this automata I drew up 
but I am unsure if it's correct...
automata
edited Nov 30 at 13:52
Glorfindel
3,41981830
asked Apr 29 '16 at 14:49
Paul
1616
add a comment |
Let Σ be {0, 1} Give a regular expression generating words over Σ containing an even number of 1’s or with a length which is multiple of 3.
I came up with this solution:
ε ( ((0*(10*10*)) + ((0+1) (0+1) (0+1)) )
and that's from this automata I drew up
but I am unsure if it's correct...
automata
edited Nov 30 at 13:52
Glorfindel
3,41981830
asked Apr 29 '16 at 14:49
Paul
1616
You would be better served posting your question on cs.stackexchange.com . (But - in short - the NFA you constructed is not correct, as it accepts the string "1" which it shouldn't.)
– Aky
Apr 29 '16 at 15:00
@Aky This question would be immediately closed on http://cs.stackexchange.com.
– J.-E. Pin
Apr 29 '16 at 16:57
add a comment |
Let Σ be {0, 1} Give a regular expression generating words over Σ containing an even number of 1’s or with a length which is multiple of 3.
I came up with this solution:
ε ( ((0*(10*10*)) + ((0+1) (0+1) (0+1)) )
and that's from this automata I drew up
but I am unsure if it's correct...
automata
edited Nov 30 at 13:52
Glorfindel
3,41981830
asked Apr 29 '16 at 14:49
Paul
1616
Let Σ be {0, 1} Give a regular expression generating words over Σ containing an even number of 1’s or with a length which is multiple of 3.
I came up with this solution:
ε ( ((0*(10*10*)) + ((0+1) (0+1) (0+1)) )
and that's from this automata I drew up
but I am unsure if it's correct...
automata
automata
edited Nov 30 at 13:52
Glorfindel
3,41981830
asked Apr 29 '16 at 14:49
Paul
1616
edited Nov 30 at 13:52
Glorfindel
3,41981830
asked Apr 29 '16 at 14:49
Paul
1616
edited Nov 30 at 13:52
Glorfindel
3,41981830
edited Nov 30 at 13:52
Glorfindel
3,41981830
edited Nov 30 at 13:52
Glorfindel
3,41981830
3,41981830
asked Apr 29 '16 at 14:49
Paul
1616
asked Apr 29 '16 at 14:49
Paul
1616
asked Apr 29 '16 at 14:49
Paul
1616
1616
You would be better served posting your question on cs.stackexchange.com . (But - in short - the NFA you constructed is not correct, as it accepts the string "1" which it shouldn't.)
– Aky
Apr 29 '16 at 15:00
@Aky This question would be immediately closed on http://cs.stackexchange.com.
– J.-E. Pin
Apr 29 '16 at 16:57
add a comment |
You would be better served posting your question on cs.stackexchange.com . (But - in short - the NFA you constructed is not correct, as it accepts the string "1" which it shouldn't.)
– Aky
Apr 29 '16 at 15:00
@Aky This question would be immediately closed on http://cs.stackexchange.com.
– J.-E. Pin
Apr 29 '16 at 16:57
You would be better served posting your question on cs.stackexchange.com . (But - in short - the NFA you constructed is not correct, as it accepts the string "1" which it shouldn't.)
– Aky
Apr 29 '16 at 15:00
You would be better served posting your question on cs.stackexchange.com . (But - in short - the NFA you constructed is not correct, as it accepts the string "1" which it shouldn't.)
– Aky
Apr 29 '16 at 15:00
@Aky This question would be immediately closed on http://cs.stackexchange.com.
– J.-E. Pin
Apr 29 '16 at 16:57
@Aky This question would be immediately closed on http://cs.stackexchange.com.
– J.-E. Pin
Apr 29 '16 at 16:57
add a comment |
2 Answers
2
active
oldest
votes
I think in the even number of $1$'s category, your NFA will miss strings of the form $110$ since you do not have an ending $0*$ at the end of your regex.
The other part of your regex only detects strings of length $3$, not multiples of 3. The correct thing would be to allow the second part of your regex to be repeated as many times as you want.
So, doing this step-by-step:
even number of ones: $ (0^*(10^*)(10^*)0^*)^*$ The inner part of the regex allows for any number of zeroes before, after, and in between two $1$s. The outer $*$ allows for the two 1's matching regex to repeat as many times as desired, thereby allowing for an even number of 1's
strings of lengths of multiples of 3: $((0 + 1)(0 + 1)(0 + 1))^*$ By allowing the entire chunk to repeat, we allow strings of length $ 0, 3, ldots, 3n, n in mathbb{N}$.
Full regex: $$R = ((0^*(10^*)(10^*)0^*)^*) + ((0 + 1)(0 + 1)(0 + 1))^*$$
answered Apr 29 '16 at 14:56
Siddharth Bhat
2,8361918
edited the regex to work for even number of zeroes
– Siddharth Bhat
Apr 29 '16 at 15:03
Thank you so much! , but do the regex also need to work for even number of zeros for the question? i mean is it right for this question also?
– Paul
Apr 29 '16 at 15:05
I'm not entirely sure what you mean, but the regex is a solution to your question if that's what you're asking.
– Siddharth Bhat
Apr 29 '16 at 15:09
@Paul - yes, it is definitely "inefficient" in the sense that there are some states that could be pruned. But, well, I think this is the most intuitive way to go about it. Efficiency is an exercise for the reader :)
– Siddharth Bhat
Apr 29 '16 at 15:09
add a comment |
Hint. Your language is the union of two (regular) languages: the langage
$L_1$ of all words containing an even number of $1$'s and the language $L_2$
of the words of length multiple of 3. Now just find a regular expression for $L_1$ and a regular expression for $L_2$ (this is easy in both cases) and just take the union of your two regular expressions.
answered Apr 29 '16 at 17:07
J.-E. Pin
18.3k21754
add a comment |
Your Answer
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
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oldest
votes
I think in the even number of $1$'s category, your NFA will miss strings of the form $110$ since you do not have an ending $0*$ at the end of your regex.
The other part of your regex only detects strings of length $3$, not multiples of 3. The correct thing would be to allow the second part of your regex to be repeated as many times as you want.
So, doing this step-by-step:
even number of ones: $ (0^*(10^*)(10^*)0^*)^*$ The inner part of the regex allows for any number of zeroes before, after, and in between two $1$s. The outer $*$ allows for the two 1's matching regex to repeat as many times as desired, thereby allowing for an even number of 1's
strings of lengths of multiples of 3: $((0 + 1)(0 + 1)(0 + 1))^*$ By allowing the entire chunk to repeat, we allow strings of length $ 0, 3, ldots, 3n, n in mathbb{N}$.
Full regex: $$R = ((0^*(10^*)(10^*)0^*)^*) + ((0 + 1)(0 + 1)(0 + 1))^*$$
answered Apr 29 '16 at 14:56
Siddharth Bhat
2,8361918
edited the regex to work for even number of zeroes
– Siddharth Bhat
Apr 29 '16 at 15:03
Thank you so much! , but do the regex also need to work for even number of zeros for the question? i mean is it right for this question also?
– Paul
Apr 29 '16 at 15:05
I'm not entirely sure what you mean, but the regex is a solution to your question if that's what you're asking.
– Siddharth Bhat
Apr 29 '16 at 15:09
@Paul - yes, it is definitely "inefficient" in the sense that there are some states that could be pruned. But, well, I think this is the most intuitive way to go about it. Efficiency is an exercise for the reader :)
– Siddharth Bhat
Apr 29 '16 at 15:09
add a comment |
I think in the even number of $1$'s category, your NFA will miss strings of the form $110$ since you do not have an ending $0*$ at the end of your regex.
The other part of your regex only detects strings of length $3$, not multiples of 3. The correct thing would be to allow the second part of your regex to be repeated as many times as you want.
So, doing this step-by-step:
even number of ones: $ (0^*(10^*)(10^*)0^*)^*$ The inner part of the regex allows for any number of zeroes before, after, and in between two $1$s. The outer $*$ allows for the two 1's matching regex to repeat as many times as desired, thereby allowing for an even number of 1's
strings of lengths of multiples of 3: $((0 + 1)(0 + 1)(0 + 1))^*$ By allowing the entire chunk to repeat, we allow strings of length $ 0, 3, ldots, 3n, n in mathbb{N}$.
Full regex: $$R = ((0^*(10^*)(10^*)0^*)^*) + ((0 + 1)(0 + 1)(0 + 1))^*$$
answered Apr 29 '16 at 14:56
Siddharth Bhat
2,8361918
edited the regex to work for even number of zeroes
– Siddharth Bhat
Apr 29 '16 at 15:03
Thank you so much! , but do the regex also need to work for even number of zeros for the question? i mean is it right for this question also?
– Paul
Apr 29 '16 at 15:05
I'm not entirely sure what you mean, but the regex is a solution to your question if that's what you're asking.
– Siddharth Bhat
Apr 29 '16 at 15:09
@Paul - yes, it is definitely "inefficient" in the sense that there are some states that could be pruned. But, well, I think this is the most intuitive way to go about it. Efficiency is an exercise for the reader :)
– Siddharth Bhat
Apr 29 '16 at 15:09
add a comment |
I think in the even number of $1$'s category, your NFA will miss strings of the form $110$ since you do not have an ending $0*$ at the end of your regex.
The other part of your regex only detects strings of length $3$, not multiples of 3. The correct thing would be to allow the second part of your regex to be repeated as many times as you want.
So, doing this step-by-step:
even number of ones: $ (0^*(10^*)(10^*)0^*)^*$ The inner part of the regex allows for any number of zeroes before, after, and in between two $1$s. The outer $*$ allows for the two 1's matching regex to repeat as many times as desired, thereby allowing for an even number of 1's
strings of lengths of multiples of 3: $((0 + 1)(0 + 1)(0 + 1))^*$ By allowing the entire chunk to repeat, we allow strings of length $ 0, 3, ldots, 3n, n in mathbb{N}$.
Full regex: $$R = ((0^*(10^*)(10^*)0^*)^*) + ((0 + 1)(0 + 1)(0 + 1))^*$$
answered Apr 29 '16 at 14:56
Siddharth Bhat
2,8361918
I think in the even number of $1$'s category, your NFA will miss strings of the form $110$ since you do not have an ending $0*$ at the end of your regex.
The other part of your regex only detects strings of length $3$, not multiples of 3. The correct thing would be to allow the second part of your regex to be repeated as many times as you want.
So, doing this step-by-step:
even number of ones: $ (0^*(10^*)(10^*)0^*)^*$ The inner part of the regex allows for any number of zeroes before, after, and in between two $1$s. The outer $*$ allows for the two 1's matching regex to repeat as many times as desired, thereby allowing for an even number of 1's
strings of lengths of multiples of 3: $((0 + 1)(0 + 1)(0 + 1))^*$ By allowing the entire chunk to repeat, we allow strings of length $ 0, 3, ldots, 3n, n in mathbb{N}$.
Full regex: $$R = ((0^*(10^*)(10^*)0^*)^*) + ((0 + 1)(0 + 1)(0 + 1))^*$$
answered Apr 29 '16 at 14:56
Siddharth Bhat
2,8361918
edited Apr 29 '16 at 15:03
answered Apr 29 '16 at 14:56
Siddharth Bhat
2,8361918
answered Apr 29 '16 at 14:56
Siddharth Bhat
2,8361918
answered Apr 29 '16 at 14:56
Siddharth Bhat
2,8361918
2,8361918
edited the regex to work for even number of zeroes
– Siddharth Bhat
Apr 29 '16 at 15:03
Thank you so much! , but do the regex also need to work for even number of zeros for the question? i mean is it right for this question also?
– Paul
Apr 29 '16 at 15:05
I'm not entirely sure what you mean, but the regex is a solution to your question if that's what you're asking.
– Siddharth Bhat
Apr 29 '16 at 15:09
@Paul - yes, it is definitely "inefficient" in the sense that there are some states that could be pruned. But, well, I think this is the most intuitive way to go about it. Efficiency is an exercise for the reader :)
– Siddharth Bhat
Apr 29 '16 at 15:09
add a comment |
edited the regex to work for even number of zeroes
– Siddharth Bhat
Apr 29 '16 at 15:03
Thank you so much! , but do the regex also need to work for even number of zeros for the question? i mean is it right for this question also?
– Paul
Apr 29 '16 at 15:05
I'm not entirely sure what you mean, but the regex is a solution to your question if that's what you're asking.
– Siddharth Bhat
Apr 29 '16 at 15:09
@Paul - yes, it is definitely "inefficient" in the sense that there are some states that could be pruned. But, well, I think this is the most intuitive way to go about it. Efficiency is an exercise for the reader :)
– Siddharth Bhat
Apr 29 '16 at 15:09
edited the regex to work for even number of zeroes
– Siddharth Bhat
Apr 29 '16 at 15:03
edited the regex to work for even number of zeroes
– Siddharth Bhat
Apr 29 '16 at 15:03
Thank you so much! , but do the regex also need to work for even number of zeros for the question? i mean is it right for this question also?
– Paul
Apr 29 '16 at 15:05
Thank you so much! , but do the regex also need to work for even number of zeros for the question? i mean is it right for this question also?
– Paul
Apr 29 '16 at 15:05
I'm not entirely sure what you mean, but the regex is a solution to your question if that's what you're asking.
– Siddharth Bhat
Apr 29 '16 at 15:09
I'm not entirely sure what you mean, but the regex is a solution to your question if that's what you're asking.
– Siddharth Bhat
Apr 29 '16 at 15:09
@Paul - yes, it is definitely "inefficient" in the sense that there are some states that could be pruned. But, well, I think this is the most intuitive way to go about it. Efficiency is an exercise for the reader :)
– Siddharth Bhat
Apr 29 '16 at 15:09
@Paul - yes, it is definitely "inefficient" in the sense that there are some states that could be pruned. But, well, I think this is the most intuitive way to go about it. Efficiency is an exercise for the reader :)
– Siddharth Bhat
Apr 29 '16 at 15:09
add a comment |
Hint. Your language is the union of two (regular) languages: the langage
$L_1$ of all words containing an even number of $1$'s and the language $L_2$
of the words of length multiple of 3. Now just find a regular expression for $L_1$ and a regular expression for $L_2$ (this is easy in both cases) and just take the union of your two regular expressions.
answered Apr 29 '16 at 17:07
J.-E. Pin
18.3k21754
add a comment |
Hint. Your language is the union of two (regular) languages: the langage
$L_1$ of all words containing an even number of $1$'s and the language $L_2$
of the words of length multiple of 3. Now just find a regular expression for $L_1$ and a regular expression for $L_2$ (this is easy in both cases) and just take the union of your two regular expressions.
answered Apr 29 '16 at 17:07
J.-E. Pin
18.3k21754
add a comment |
Hint. Your language is the union of two (regular) languages: the langage
$L_1$ of all words containing an even number of $1$'s and the language $L_2$
of the words of length multiple of 3. Now just find a regular expression for $L_1$ and a regular expression for $L_2$ (this is easy in both cases) and just take the union of your two regular expressions.
answered Apr 29 '16 at 17:07
J.-E. Pin
18.3k21754
Hint. Your language is the union of two (regular) languages: the langage
$L_1$ of all words containing an even number of $1$'s and the language $L_2$
of the words of length multiple of 3. Now just find a regular expression for $L_1$ and a regular expression for $L_2$ (this is easy in both cases) and just take the union of your two regular expressions.
answered Apr 29 '16 at 17:07
J.-E. Pin
18.3k21754
answered Apr 29 '16 at 17:07
J.-E. Pin
18.3k21754
answered Apr 29 '16 at 17:07
J.-E. Pin
18.3k21754
answered Apr 29 '16 at 17:07
J.-E. Pin
18.3k21754
18.3k21754
add a comment |
add a comment |
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You would be better served posting your question on cs.stackexchange.com . (But - in short - the NFA you constructed is not correct, as it accepts the string "1" which it shouldn't.)
– Aky
Apr 29 '16 at 15:00
@Aky This question would be immediately closed on http://cs.stackexchange.com.
– J.-E. Pin
Apr 29 '16 at 16:57