Why is this basic language not a regular language?Binary regular language?Is the language that accepts strings concatenated with their reverse regular?Definition of a regular languageRegular language concatenation with supersetGenerators of families of langauges?Can the definition of regular languages be simplified?Prove language is regularProve a language is regular - Regular language of 0's and 1'sDoes this proof work for infinite regular languagesProving that language is regular or not
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Why is this basic language not a regular language?
Binary regular language?Is the language that accepts strings concatenated with their reverse regular?Definition of a regular languageRegular language concatenation with supersetGenerators of families of langauges?Can the definition of regular languages be simplified?Prove language is regularProve a language is regular - Regular language of 0's and 1'sDoes this proof work for infinite regular languagesProving that language is regular or not
$begingroup$
L = x has an equal number of 0s & 1s
Based on the recursive definition of regular languages, isn't it possible to form a single regular language set over the binary alphabet 0,1 by doing the following?
- concatenating 0's and 1's to form each of the binary strings satisfying the condition, resulting in a regular language consisting of that single string
- union all the single-string regular languages together into one single regular language
regular-languages
asked 17 hours ago
ShukieShukie
61
New contributor
Shukie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
L = x has an equal number of 0s & 1s
Based on the recursive definition of regular languages, isn't it possible to form a single regular language set over the binary alphabet 0,1 by doing the following?
- concatenating 0's and 1's to form each of the binary strings satisfying the condition, resulting in a regular language consisting of that single string
- union all the single-string regular languages together into one single regular language
regular-languages
asked 17 hours ago
ShukieShukie
61
New contributor
Shukie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
$endgroup$
– David Richerby
7 hours ago
add a comment |
$begingroup$
L = x has an equal number of 0s & 1s
Based on the recursive definition of regular languages, isn't it possible to form a single regular language set over the binary alphabet 0,1 by doing the following?
- concatenating 0's and 1's to form each of the binary strings satisfying the condition, resulting in a regular language consisting of that single string
- union all the single-string regular languages together into one single regular language
regular-languages
asked 17 hours ago
ShukieShukie
61
New contributor
Shukie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
L = x has an equal number of 0s & 1s
Based on the recursive definition of regular languages, isn't it possible to form a single regular language set over the binary alphabet 0,1 by doing the following?
- concatenating 0's and 1's to form each of the binary strings satisfying the condition, resulting in a regular language consisting of that single string
- union all the single-string regular languages together into one single regular language
regular-languages
regular-languages
asked 17 hours ago
ShukieShukie
61
New contributor
Shukie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
ShukieShukie
61
New contributor
Shukie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
ShukieShukie
61
New contributor
Shukie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 17 hours ago
ShukieShukie
61
asked 17 hours ago
ShukieShukie
61
61
New contributor
Shukie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Shukie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Shukie is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
$endgroup$
– David Richerby
7 hours ago
add a comment |
$begingroup$
There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
$endgroup$
– David Richerby
7 hours ago
$begingroup$
There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
$endgroup$
– David Richerby
7 hours ago
$begingroup$
There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
$endgroup$
– David Richerby
7 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Every language is the union of regular languages:
$$
L = bigcup_x in L x .
$$
However, the set of regular languages is not closed under infinite unions. It isn't even closed under countably infinite unions, as your example demonstrates.
You can prove that your language is not regular in many ways. For example, if your language were regular, then so would its intersection with $0^*1^*$ be; yet this language is $ 0^n 1^n : n geq 0 $, the classical example of a non-regular language. You can also prove non-regularity of your language directly, using either the pumping lemma or Myhill–Nerode theory.
answered 15 hours ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
add a comment |
$begingroup$
You're on the right track! There's just one thing you're missing.
The "build all the strings and union them together" approach works great—if you have a finite number of strings. There's a theorem that says "a union of finitely many regular languages is regular", but the key is the finitely many.
In this case, there are infinitely many strings in the language, so the union-them-all trick no longer works.
If you want to prove that the language is not regular (as opposed to just failing to prove that it is regular), try a fooling set proof:
- Let $F$ be the language $1^*$. It's clearly infinite.
- Let $x$ and $y$ be two distinct strings in $F$. These can be written as $1^i$ and $1^j$ with $i neq j$.
- Now, let $z$ be $0^i$. From the definition of your language, we can see that $xz$ is in the language, but $yz$ is not.
- Therefore, all the different strings in $F$ are distinguishable as prefixes. Since the language has infinitely many distinguishable prefixes, it cannot be regular.
answered 11 hours ago
DraconisDraconis
5,762921
$endgroup$
add a comment |
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2 Answers
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oldest
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2 Answers
2
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active
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$begingroup$
Every language is the union of regular languages:
$$
L = bigcup_x in L x .
$$
However, the set of regular languages is not closed under infinite unions. It isn't even closed under countably infinite unions, as your example demonstrates.
You can prove that your language is not regular in many ways. For example, if your language were regular, then so would its intersection with $0^*1^*$ be; yet this language is $ 0^n 1^n : n geq 0 $, the classical example of a non-regular language. You can also prove non-regularity of your language directly, using either the pumping lemma or Myhill–Nerode theory.
answered 15 hours ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
add a comment |
$begingroup$
Every language is the union of regular languages:
$$
L = bigcup_x in L x .
$$
However, the set of regular languages is not closed under infinite unions. It isn't even closed under countably infinite unions, as your example demonstrates.
You can prove that your language is not regular in many ways. For example, if your language were regular, then so would its intersection with $0^*1^*$ be; yet this language is $ 0^n 1^n : n geq 0 $, the classical example of a non-regular language. You can also prove non-regularity of your language directly, using either the pumping lemma or Myhill–Nerode theory.
answered 15 hours ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
add a comment |
$begingroup$
Every language is the union of regular languages:
$$
L = bigcup_x in L x .
$$
However, the set of regular languages is not closed under infinite unions. It isn't even closed under countably infinite unions, as your example demonstrates.
You can prove that your language is not regular in many ways. For example, if your language were regular, then so would its intersection with $0^*1^*$ be; yet this language is $ 0^n 1^n : n geq 0 $, the classical example of a non-regular language. You can also prove non-regularity of your language directly, using either the pumping lemma or Myhill–Nerode theory.
answered 15 hours ago
Yuval FilmusYuval Filmus
196k15184349
$endgroup$
Every language is the union of regular languages:
$$
L = bigcup_x in L x .
$$
However, the set of regular languages is not closed under infinite unions. It isn't even closed under countably infinite unions, as your example demonstrates.
You can prove that your language is not regular in many ways. For example, if your language were regular, then so would its intersection with $0^*1^*$ be; yet this language is $ 0^n 1^n : n geq 0 $, the classical example of a non-regular language. You can also prove non-regularity of your language directly, using either the pumping lemma or Myhill–Nerode theory.
answered 15 hours ago
Yuval FilmusYuval Filmus
196k15184349
answered 15 hours ago
Yuval FilmusYuval Filmus
196k15184349
answered 15 hours ago
Yuval FilmusYuval Filmus
196k15184349
answered 15 hours ago
Yuval FilmusYuval Filmus
196k15184349
196k15184349
add a comment |
add a comment |
$begingroup$
You're on the right track! There's just one thing you're missing.
The "build all the strings and union them together" approach works great—if you have a finite number of strings. There's a theorem that says "a union of finitely many regular languages is regular", but the key is the finitely many.
In this case, there are infinitely many strings in the language, so the union-them-all trick no longer works.
If you want to prove that the language is not regular (as opposed to just failing to prove that it is regular), try a fooling set proof:
- Let $F$ be the language $1^*$. It's clearly infinite.
- Let $x$ and $y$ be two distinct strings in $F$. These can be written as $1^i$ and $1^j$ with $i neq j$.
- Now, let $z$ be $0^i$. From the definition of your language, we can see that $xz$ is in the language, but $yz$ is not.
- Therefore, all the different strings in $F$ are distinguishable as prefixes. Since the language has infinitely many distinguishable prefixes, it cannot be regular.
answered 11 hours ago
DraconisDraconis
5,762921
$endgroup$
add a comment |
$begingroup$
You're on the right track! There's just one thing you're missing.
The "build all the strings and union them together" approach works great—if you have a finite number of strings. There's a theorem that says "a union of finitely many regular languages is regular", but the key is the finitely many.
In this case, there are infinitely many strings in the language, so the union-them-all trick no longer works.
If you want to prove that the language is not regular (as opposed to just failing to prove that it is regular), try a fooling set proof:
- Let $F$ be the language $1^*$. It's clearly infinite.
- Let $x$ and $y$ be two distinct strings in $F$. These can be written as $1^i$ and $1^j$ with $i neq j$.
- Now, let $z$ be $0^i$. From the definition of your language, we can see that $xz$ is in the language, but $yz$ is not.
- Therefore, all the different strings in $F$ are distinguishable as prefixes. Since the language has infinitely many distinguishable prefixes, it cannot be regular.
answered 11 hours ago
DraconisDraconis
5,762921
$endgroup$
add a comment |
$begingroup$
You're on the right track! There's just one thing you're missing.
The "build all the strings and union them together" approach works great—if you have a finite number of strings. There's a theorem that says "a union of finitely many regular languages is regular", but the key is the finitely many.
In this case, there are infinitely many strings in the language, so the union-them-all trick no longer works.
If you want to prove that the language is not regular (as opposed to just failing to prove that it is regular), try a fooling set proof:
- Let $F$ be the language $1^*$. It's clearly infinite.
- Let $x$ and $y$ be two distinct strings in $F$. These can be written as $1^i$ and $1^j$ with $i neq j$.
- Now, let $z$ be $0^i$. From the definition of your language, we can see that $xz$ is in the language, but $yz$ is not.
- Therefore, all the different strings in $F$ are distinguishable as prefixes. Since the language has infinitely many distinguishable prefixes, it cannot be regular.
answered 11 hours ago
DraconisDraconis
5,762921
$endgroup$
You're on the right track! There's just one thing you're missing.
The "build all the strings and union them together" approach works great—if you have a finite number of strings. There's a theorem that says "a union of finitely many regular languages is regular", but the key is the finitely many.
In this case, there are infinitely many strings in the language, so the union-them-all trick no longer works.
If you want to prove that the language is not regular (as opposed to just failing to prove that it is regular), try a fooling set proof:
- Let $F$ be the language $1^*$. It's clearly infinite.
- Let $x$ and $y$ be two distinct strings in $F$. These can be written as $1^i$ and $1^j$ with $i neq j$.
- Now, let $z$ be $0^i$. From the definition of your language, we can see that $xz$ is in the language, but $yz$ is not.
- Therefore, all the different strings in $F$ are distinguishable as prefixes. Since the language has infinitely many distinguishable prefixes, it cannot be regular.
answered 11 hours ago
DraconisDraconis
5,762921
answered 11 hours ago
DraconisDraconis
5,762921
answered 11 hours ago
DraconisDraconis
5,762921
answered 11 hours ago
DraconisDraconis
5,762921
5,762921
add a comment |
add a comment |
Shukie is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
There must be something wrong with your proof attempt, since it would apply to any language and there are surely some non-regular languages. The answers explain what.
$endgroup$
– David Richerby
7 hours ago