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Why is this basic language not a regular language?

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1

$begingroup$

L = {x in {0,1}* | 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?

1. 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


2. union all the single-string regular languages together into one single regular language

regular-languages

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asked 15 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
  6 hours ago
  
  
  

  
  
  
  

add a comment | 

1

$begingroup$

L = {x in {0,1}* | 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?

1. 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


2. union all the single-string regular languages together into one single regular language

regular-languages

share|cite|improve this question

asked 15 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
  6 hours ago
  
  
  

  
  
  
  

add a comment | 

$begingroup$

L = {x in {0,1}* | 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?

1. 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


2. union all the single-string regular languages together into one single regular language

regular-languages

share|cite|improve this question

asked 15 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.

$endgroup$

share|cite|improve this question

asked 15 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.

share|cite|improve this question

asked 15 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 15 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 15 hours ago

ShukieShukie

61

2 Answers
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3

$begingroup$

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.

share|cite|improve this answer

answered 13 hours ago

Yuval FilmusYuval Filmus

196k15184349

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 9 hours ago

DraconisDraconis

5,762921

$endgroup$

add a comment | 

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3

$begingroup$

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.

share|cite|improve this answer

answered 13 hours ago

Yuval FilmusYuval Filmus

196k15184349

$endgroup$

add a comment | 

3

$begingroup$

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.

share|cite|improve this answer

answered 13 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.

share|cite|improve this answer

answered 13 hours ago

Yuval FilmusYuval Filmus

196k15184349

$endgroup$

share|cite|improve this answer

answered 13 hours ago

Yuval FilmusYuval Filmus

196k15184349

answered 13 hours ago

Yuval FilmusYuval Filmus

196k15184349

answered 13 hours ago

Yuval FilmusYuval Filmus

196k15184349

1

$begingroup$

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.

share|cite|improve this answer

answered 9 hours ago

DraconisDraconis

5,762921

$endgroup$

add a comment | 

1

$begingroup$

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.

share|cite|improve this answer

answered 9 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.

share|cite|improve this answer

answered 9 hours ago

DraconisDraconis

5,762921

$endgroup$

share|cite|improve this answer

answered 9 hours ago

DraconisDraconis

5,762921

answered 9 hours ago

DraconisDraconis

5,762921

answered 9 hours ago

DraconisDraconis

5,762921