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Languages that we cannot (dis)prove to be Context-Free

Base-k representations of the co-domain of a polynomial - is it context-free?For a language to be programmable, is it mandatory that it be based on a context free grammarSufficient conditions for the regularity of a context-free languageDoes there exist an extension of regular expressions that captures the context free languages?Are deterministic context-free languages closed under outfix (or other erasing operations)Is SAT a context-free language?Is equivalence of unambiguous context-free languages decidable?Example of context-free tree language which can not be generated by monadic CFTGNumber of words of length n in a context-free languageFor which $R$ is $0^a10^b10^cmid R(a,b,c)$ context-free?Continuous mathematics and formal language theory

11

4

$begingroup$

I'm looking for languages which are "probably not Context-Free" but we are not able to (dis)prove it using known standard techniques.

Is there a recent survey on the subject or an open problem section from a recent conference ?

Probably there are not many languages which are not known to be CF, so if you know one you can also post it as an answer.

The examples I found are:

• the well known language of Primitive words $Q = > 1) $ (there's a whole nice recent book on it: Context-Free Languages and Primitive Words)


• the Base-k representations of the co-domain of a polynomial (see question "Base-k representations of the co-domain of a polynomial - is it context-free?" on cstheory)

Note: as showed by Aryeh in his answer you can build a whole class of such languages if you "link" a language to an unknown conjecture about the (non)finiteness or (non)emptiness of some sets (e.g. $L_Goldbach = 1^2n mid 2n$ cannot be expressed as a sum of two primes$$). I'm not quite interested in such examples.

reference-request big-list context-free

share|cite|improve this question

edited 11 hours ago

Marzio De Biasi

asked 14 hours ago

Marzio De BiasiMarzio De Biasi

18.5k243113

$endgroup$

add a comment | 

11

4

$begingroup$

I'm looking for languages which are "probably not Context-Free" but we are not able to (dis)prove it using known standard techniques.

Is there a recent survey on the subject or an open problem section from a recent conference ?

Probably there are not many languages which are not known to be CF, so if you know one you can also post it as an answer.

The examples I found are:

• the well known language of Primitive words $Q = > 1) $ (there's a whole nice recent book on it: Context-Free Languages and Primitive Words)


• the Base-k representations of the co-domain of a polynomial (see question "Base-k representations of the co-domain of a polynomial - is it context-free?" on cstheory)

Note: as showed by Aryeh in his answer you can build a whole class of such languages if you "link" a language to an unknown conjecture about the (non)finiteness or (non)emptiness of some sets (e.g. $L_Goldbach = 1^2n mid 2n$ cannot be expressed as a sum of two primes$$). I'm not quite interested in such examples.

reference-request big-list context-free

share|cite|improve this question

edited 11 hours ago

Marzio De Biasi

asked 14 hours ago

Marzio De BiasiMarzio De Biasi

18.5k243113

$endgroup$

add a comment | 

$begingroup$

I'm looking for languages which are "probably not Context-Free" but we are not able to (dis)prove it using known standard techniques.

Is there a recent survey on the subject or an open problem section from a recent conference ?

Probably there are not many languages which are not known to be CF, so if you know one you can also post it as an answer.

The examples I found are:

• the well known language of Primitive words $Q = > 1) $ (there's a whole nice recent book on it: Context-Free Languages and Primitive Words)


• the Base-k representations of the co-domain of a polynomial (see question "Base-k representations of the co-domain of a polynomial - is it context-free?" on cstheory)

Note: as showed by Aryeh in his answer you can build a whole class of such languages if you "link" a language to an unknown conjecture about the (non)finiteness or (non)emptiness of some sets (e.g. $L_Goldbach = 1^2n mid 2n$ cannot be expressed as a sum of two primes$$). I'm not quite interested in such examples.

reference-request big-list context-free

share|cite|improve this question

edited 11 hours ago

Marzio De Biasi

asked 14 hours ago

Marzio De BiasiMarzio De Biasi

18.5k243113

$endgroup$

share|cite|improve this question

edited 11 hours ago

Marzio De Biasi

asked 14 hours ago

Marzio De BiasiMarzio De Biasi

18.5k243113

share|cite|improve this question

edited 11 hours ago

Marzio De Biasi

asked 14 hours ago

Marzio De BiasiMarzio De Biasi

18.5k243113

asked 14 hours ago

Marzio De BiasiMarzio De Biasi

18.5k243113

asked 14 hours ago

Marzio De BiasiMarzio De Biasi

18.5k243113

2 Answers
2

active

oldest

votes

7

$begingroup$

How about the language $L_TP$ of twin primes? I.e., all pairs of natural numbers $(p,p')$ (represented, say, in unary), such that $p,p'$ are both prime and $p'=p+2$? If twin primes conjecture is true, then $L_TP$ is not context-free; otherwise, it's finite.

Edit: Let me give a quick proof sketch that the twin primes conjecture implies that $L_TP$ is not context-free. Associate to any language $L$ its length sequence $0le a_1le a_2leldots$, where the integer $ell$ appears in the sequence iff there is a word of length $ell$ in $L$. It is a consequence of the pumping lemma(s) that for $L$ that are regular or CFL, the length sequence satisfies the bounded differences property: there is an $R>0$ such that $a_n+1-a_nle R$ for all $n$. It is an easy and well-known fact in number theory that the primes do not have bounded differences. Finally, any infinite subsequence of a sequence violating the bounded differences property itself must violate it.

share|cite|improve this answer

edited 10 hours ago

answered 13 hours ago

AryehAryeh

5,84411840

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, thanks! But I'm not quite interested in languages that are linked to unknown conjectures about the (non)finiteness of some sets. BTW if those conjectures are true the resulting language is also regular :-)
  $endgroup$
  – Marzio De Biasi
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If there are infinitely many twin primes, how do you see that $L_TP$ is regular?
  $endgroup$
  – Aryeh
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean if (dis)proving a conjecture, results in a finite set. In your case - twin primes - if the conjecture is false.
  $endgroup$
  – Marzio De Biasi
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If there are infinitely many twin primes, how do you show that $L_TP$ is not context-free?
  $endgroup$
  – Emil Jeřábek
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, sorry, I didn’t notice you represent the numbers in unary. Then it is clear. (I believe that proving this for binary representation would require a considerable progress on the twin primes conjecture.)
  $endgroup$
  – Emil Jeřábek
  10 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

6

$begingroup$

Another good one is the complement of the set $S$ of contiguous subwords (aka "factors") of the Thue-Morse sequence $bf t = 0110100110010110 cdots $. To give some context, Jean Berstel proved that the complement of the set $T$ of prefixes of the Thue-Morse word is context-free (and actually something more general than that). But the corresponding result for subwords is still open.

share|cite|improve this answer

answered 8 hours ago

Jeffrey ShallitJeffrey Shallit

6,4832635

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, thanks! If you saw it stated somewhere (perhaps in one of your many papers on the Thue-Morse sequence? ;-) you can add the reference (even if stated in the iterated morphism form).
  $endgroup$
  – Marzio De Biasi
  6 hours ago
  

  
  
  
  

add a comment | 

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7

$begingroup$

How about the language $L_TP$ of twin primes? I.e., all pairs of natural numbers $(p,p')$ (represented, say, in unary), such that $p,p'$ are both prime and $p'=p+2$? If twin primes conjecture is true, then $L_TP$ is not context-free; otherwise, it's finite.

Edit: Let me give a quick proof sketch that the twin primes conjecture implies that $L_TP$ is not context-free. Associate to any language $L$ its length sequence $0le a_1le a_2leldots$, where the integer $ell$ appears in the sequence iff there is a word of length $ell$ in $L$. It is a consequence of the pumping lemma(s) that for $L$ that are regular or CFL, the length sequence satisfies the bounded differences property: there is an $R>0$ such that $a_n+1-a_nle R$ for all $n$. It is an easy and well-known fact in number theory that the primes do not have bounded differences. Finally, any infinite subsequence of a sequence violating the bounded differences property itself must violate it.

share|cite|improve this answer

edited 10 hours ago

answered 13 hours ago

AryehAryeh

5,84411840

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, thanks! But I'm not quite interested in languages that are linked to unknown conjectures about the (non)finiteness of some sets. BTW if those conjectures are true the resulting language is also regular :-)
  $endgroup$
  – Marzio De Biasi
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If there are infinitely many twin primes, how do you see that $L_TP$ is regular?
  $endgroup$
  – Aryeh
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean if (dis)proving a conjecture, results in a finite set. In your case - twin primes - if the conjecture is false.
  $endgroup$
  – Marzio De Biasi
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If there are infinitely many twin primes, how do you show that $L_TP$ is not context-free?
  $endgroup$
  – Emil Jeřábek
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, sorry, I didn’t notice you represent the numbers in unary. Then it is clear. (I believe that proving this for binary representation would require a considerable progress on the twin primes conjecture.)
  $endgroup$
  – Emil Jeřábek
  10 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

7

$begingroup$

How about the language $L_TP$ of twin primes? I.e., all pairs of natural numbers $(p,p')$ (represented, say, in unary), such that $p,p'$ are both prime and $p'=p+2$? If twin primes conjecture is true, then $L_TP$ is not context-free; otherwise, it's finite.

Edit: Let me give a quick proof sketch that the twin primes conjecture implies that $L_TP$ is not context-free. Associate to any language $L$ its length sequence $0le a_1le a_2leldots$, where the integer $ell$ appears in the sequence iff there is a word of length $ell$ in $L$. It is a consequence of the pumping lemma(s) that for $L$ that are regular or CFL, the length sequence satisfies the bounded differences property: there is an $R>0$ such that $a_n+1-a_nle R$ for all $n$. It is an easy and well-known fact in number theory that the primes do not have bounded differences. Finally, any infinite subsequence of a sequence violating the bounded differences property itself must violate it.

share|cite|improve this answer

edited 10 hours ago

answered 13 hours ago

AryehAryeh

5,84411840

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Nice, thanks! But I'm not quite interested in languages that are linked to unknown conjectures about the (non)finiteness of some sets. BTW if those conjectures are true the resulting language is also regular :-)
  $endgroup$
  – Marzio De Biasi
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If there are infinitely many twin primes, how do you see that $L_TP$ is regular?
  $endgroup$
  – Aryeh
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I mean if (dis)proving a conjecture, results in a finite set. In your case - twin primes - if the conjecture is false.
  $endgroup$
  – Marzio De Biasi
  11 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  If there are infinitely many twin primes, how do you show that $L_TP$ is not context-free?
  $endgroup$
  – Emil Jeřábek
  10 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Oh, sorry, I didn’t notice you represent the numbers in unary. Then it is clear. (I believe that proving this for binary representation would require a considerable progress on the twin primes conjecture.)
  $endgroup$
  – Emil Jeřábek
  10 hours ago
  
  

  
  
  
  

 | 
show 3 more comments

$begingroup$

How about the language $L_TP$ of twin primes? I.e., all pairs of natural numbers $(p,p')$ (represented, say, in unary), such that $p,p'$ are both prime and $p'=p+2$? If twin primes conjecture is true, then $L_TP$ is not context-free; otherwise, it's finite.

Edit: Let me give a quick proof sketch that the twin primes conjecture implies that $L_TP$ is not context-free. Associate to any language $L$ its length sequence $0le a_1le a_2leldots$, where the integer $ell$ appears in the sequence iff there is a word of length $ell$ in $L$. It is a consequence of the pumping lemma(s) that for $L$ that are regular or CFL, the length sequence satisfies the bounded differences property: there is an $R>0$ such that $a_n+1-a_nle R$ for all $n$. It is an easy and well-known fact in number theory that the primes do not have bounded differences. Finally, any infinite subsequence of a sequence violating the bounded differences property itself must violate it.

share|cite|improve this answer

edited 10 hours ago

answered 13 hours ago

AryehAryeh

5,84411840

$endgroup$

share|cite|improve this answer

edited 10 hours ago

answered 13 hours ago

AryehAryeh

5,84411840

answered 13 hours ago

AryehAryeh

5,84411840

answered 13 hours ago

AryehAryeh

5,84411840

6

$begingroup$

Another good one is the complement of the set $S$ of contiguous subwords (aka "factors") of the Thue-Morse sequence $bf t = 0110100110010110 cdots $. To give some context, Jean Berstel proved that the complement of the set $T$ of prefixes of the Thue-Morse word is context-free (and actually something more general than that). But the corresponding result for subwords is still open.

share|cite|improve this answer

answered 8 hours ago

Jeffrey ShallitJeffrey Shallit

6,4832635

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, thanks! If you saw it stated somewhere (perhaps in one of your many papers on the Thue-Morse sequence? ;-) you can add the reference (even if stated in the iterated morphism form).
  $endgroup$
  – Marzio De Biasi
  6 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

Another good one is the complement of the set $S$ of contiguous subwords (aka "factors") of the Thue-Morse sequence $bf t = 0110100110010110 cdots $. To give some context, Jean Berstel proved that the complement of the set $T$ of prefixes of the Thue-Morse word is context-free (and actually something more general than that). But the corresponding result for subwords is still open.

share|cite|improve this answer

answered 8 hours ago

Jeffrey ShallitJeffrey Shallit

6,4832635

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Great, thanks! If you saw it stated somewhere (perhaps in one of your many papers on the Thue-Morse sequence? ;-) you can add the reference (even if stated in the iterated morphism form).
  $endgroup$
  – Marzio De Biasi
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

Another good one is the complement of the set $S$ of contiguous subwords (aka "factors") of the Thue-Morse sequence $bf t = 0110100110010110 cdots $. To give some context, Jean Berstel proved that the complement of the set $T$ of prefixes of the Thue-Morse word is context-free (and actually something more general than that). But the corresponding result for subwords is still open.

share|cite|improve this answer

answered 8 hours ago

Jeffrey ShallitJeffrey Shallit

6,4832635

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Jeffrey ShallitJeffrey Shallit

6,4832635

answered 8 hours ago

Jeffrey ShallitJeffrey Shallit

6,4832635

answered 8 hours ago

Jeffrey ShallitJeffrey Shallit

6,4832635