Language involving irrational number is not a CFL












8












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I am working through a hard exercise in a textbook, and I just can't figure out how to proceed. Here is the problem. Suppose we have the language $L = {a^ib^j: i leq j gamma, igeq 0, jgeq 1}$ where $gamma$ is some irrational number. How would I prove that $L$ is not a context-free language?



In the case when $gamma$ is rational, it's pretty easy to construct a grammar that accepts the language. But because $gamma$ is irrational, I don't really know what to do. It doesn't look like any of the pumping lemmas would work here. Maybe Parikh's theorem would work here, since it would intuitively seem like this language doesn't have an accompanying semilinear Parikh image.



This exercise is from "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit, Exercise 25 of Chapter 4.



I would really appreciate any help, or nudges in the right direction. Thank you!










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  • $begingroup$
    Have you tried applying Parikh’s theorem?
    $endgroup$
    – Yuval Filmus
    9 hours ago












  • $begingroup$
    Why not show that it’s not semilinear directly? Use the definition.
    $endgroup$
    – Yuval Filmus
    9 hours ago






  • 3




    $begingroup$
    Just in time for my homework! Thanks. CS 462/662 Formal Languages and Parsing Winter 2019, Problem Set 9, exercise 3. Due Friday, March 22 2019.
    $endgroup$
    – Hendrik Jan
    6 hours ago












  • $begingroup$
    @ChenyiShiwen, so the pumping lemma does work.
    $endgroup$
    – Apass.Jack
    5 hours ago










  • $begingroup$
    @HendrikJan I'm selfstudying from the textbook "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit. It is Exercise 25 of Chapter 4 fyi. Would it be possible to hide this post until the assignment is due?
    $endgroup$
    – ChenyiShiwen
    2 hours ago
















8












$begingroup$


I am working through a hard exercise in a textbook, and I just can't figure out how to proceed. Here is the problem. Suppose we have the language $L = {a^ib^j: i leq j gamma, igeq 0, jgeq 1}$ where $gamma$ is some irrational number. How would I prove that $L$ is not a context-free language?



In the case when $gamma$ is rational, it's pretty easy to construct a grammar that accepts the language. But because $gamma$ is irrational, I don't really know what to do. It doesn't look like any of the pumping lemmas would work here. Maybe Parikh's theorem would work here, since it would intuitively seem like this language doesn't have an accompanying semilinear Parikh image.



This exercise is from "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit, Exercise 25 of Chapter 4.



I would really appreciate any help, or nudges in the right direction. Thank you!










share|cite|improve this question









New contributor




ChenyiShiwen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Have you tried applying Parikh’s theorem?
    $endgroup$
    – Yuval Filmus
    9 hours ago












  • $begingroup$
    Why not show that it’s not semilinear directly? Use the definition.
    $endgroup$
    – Yuval Filmus
    9 hours ago






  • 3




    $begingroup$
    Just in time for my homework! Thanks. CS 462/662 Formal Languages and Parsing Winter 2019, Problem Set 9, exercise 3. Due Friday, March 22 2019.
    $endgroup$
    – Hendrik Jan
    6 hours ago












  • $begingroup$
    @ChenyiShiwen, so the pumping lemma does work.
    $endgroup$
    – Apass.Jack
    5 hours ago










  • $begingroup$
    @HendrikJan I'm selfstudying from the textbook "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit. It is Exercise 25 of Chapter 4 fyi. Would it be possible to hide this post until the assignment is due?
    $endgroup$
    – ChenyiShiwen
    2 hours ago














8












8








8





$begingroup$


I am working through a hard exercise in a textbook, and I just can't figure out how to proceed. Here is the problem. Suppose we have the language $L = {a^ib^j: i leq j gamma, igeq 0, jgeq 1}$ where $gamma$ is some irrational number. How would I prove that $L$ is not a context-free language?



In the case when $gamma$ is rational, it's pretty easy to construct a grammar that accepts the language. But because $gamma$ is irrational, I don't really know what to do. It doesn't look like any of the pumping lemmas would work here. Maybe Parikh's theorem would work here, since it would intuitively seem like this language doesn't have an accompanying semilinear Parikh image.



This exercise is from "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit, Exercise 25 of Chapter 4.



I would really appreciate any help, or nudges in the right direction. Thank you!










share|cite|improve this question









New contributor




ChenyiShiwen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




I am working through a hard exercise in a textbook, and I just can't figure out how to proceed. Here is the problem. Suppose we have the language $L = {a^ib^j: i leq j gamma, igeq 0, jgeq 1}$ where $gamma$ is some irrational number. How would I prove that $L$ is not a context-free language?



In the case when $gamma$ is rational, it's pretty easy to construct a grammar that accepts the language. But because $gamma$ is irrational, I don't really know what to do. It doesn't look like any of the pumping lemmas would work here. Maybe Parikh's theorem would work here, since it would intuitively seem like this language doesn't have an accompanying semilinear Parikh image.



This exercise is from "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit, Exercise 25 of Chapter 4.



I would really appreciate any help, or nudges in the right direction. Thank you!







formal-languages automata context-free






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edited 2 hours ago









D.W.

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asked 10 hours ago









ChenyiShiwenChenyiShiwen

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New contributor





ChenyiShiwen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






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Check out our Code of Conduct.












  • $begingroup$
    Have you tried applying Parikh’s theorem?
    $endgroup$
    – Yuval Filmus
    9 hours ago












  • $begingroup$
    Why not show that it’s not semilinear directly? Use the definition.
    $endgroup$
    – Yuval Filmus
    9 hours ago






  • 3




    $begingroup$
    Just in time for my homework! Thanks. CS 462/662 Formal Languages and Parsing Winter 2019, Problem Set 9, exercise 3. Due Friday, March 22 2019.
    $endgroup$
    – Hendrik Jan
    6 hours ago












  • $begingroup$
    @ChenyiShiwen, so the pumping lemma does work.
    $endgroup$
    – Apass.Jack
    5 hours ago










  • $begingroup$
    @HendrikJan I'm selfstudying from the textbook "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit. It is Exercise 25 of Chapter 4 fyi. Would it be possible to hide this post until the assignment is due?
    $endgroup$
    – ChenyiShiwen
    2 hours ago


















  • $begingroup$
    Have you tried applying Parikh’s theorem?
    $endgroup$
    – Yuval Filmus
    9 hours ago












  • $begingroup$
    Why not show that it’s not semilinear directly? Use the definition.
    $endgroup$
    – Yuval Filmus
    9 hours ago






  • 3




    $begingroup$
    Just in time for my homework! Thanks. CS 462/662 Formal Languages and Parsing Winter 2019, Problem Set 9, exercise 3. Due Friday, March 22 2019.
    $endgroup$
    – Hendrik Jan
    6 hours ago












  • $begingroup$
    @ChenyiShiwen, so the pumping lemma does work.
    $endgroup$
    – Apass.Jack
    5 hours ago










  • $begingroup$
    @HendrikJan I'm selfstudying from the textbook "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit. It is Exercise 25 of Chapter 4 fyi. Would it be possible to hide this post until the assignment is due?
    $endgroup$
    – ChenyiShiwen
    2 hours ago
















$begingroup$
Have you tried applying Parikh’s theorem?
$endgroup$
– Yuval Filmus
9 hours ago






$begingroup$
Have you tried applying Parikh’s theorem?
$endgroup$
– Yuval Filmus
9 hours ago














$begingroup$
Why not show that it’s not semilinear directly? Use the definition.
$endgroup$
– Yuval Filmus
9 hours ago




$begingroup$
Why not show that it’s not semilinear directly? Use the definition.
$endgroup$
– Yuval Filmus
9 hours ago




3




3




$begingroup$
Just in time for my homework! Thanks. CS 462/662 Formal Languages and Parsing Winter 2019, Problem Set 9, exercise 3. Due Friday, March 22 2019.
$endgroup$
– Hendrik Jan
6 hours ago






$begingroup$
Just in time for my homework! Thanks. CS 462/662 Formal Languages and Parsing Winter 2019, Problem Set 9, exercise 3. Due Friday, March 22 2019.
$endgroup$
– Hendrik Jan
6 hours ago














$begingroup$
@ChenyiShiwen, so the pumping lemma does work.
$endgroup$
– Apass.Jack
5 hours ago




$begingroup$
@ChenyiShiwen, so the pumping lemma does work.
$endgroup$
– Apass.Jack
5 hours ago












$begingroup$
@HendrikJan I'm selfstudying from the textbook "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit. It is Exercise 25 of Chapter 4 fyi. Would it be possible to hide this post until the assignment is due?
$endgroup$
– ChenyiShiwen
2 hours ago




$begingroup$
@HendrikJan I'm selfstudying from the textbook "A Second Course in Formal Languages and Automata Theory" by Jeffrey Shallit. It is Exercise 25 of Chapter 4 fyi. Would it be possible to hide this post until the assignment is due?
$endgroup$
– ChenyiShiwen
2 hours ago










2 Answers
2






active

oldest

votes


















6












$begingroup$

According to Parikh's theorem, if $L$ were context-free then the set $M = {(a,b) : a leq gamma b }$ would be semilinear, that is, it would be the union of finitely many sets of the form $S = u_0 + mathbb{N} u_1 + cdots + mathbb{N} u_ell$, for some $u_i = (a_i,b_i)$.



Obviously $u_0 in M$, and moreover $u_i in M$ for each $i > 0$, since otherwise $u_0 + N u_i notin M$ for large enough $N$. Therefore $g(S) := max(a_0/b_0,ldots,a_ell/b_ell) < gamma$ (since $g(S)$ is rational). This means that every $(a,b) in S$ satisfies $a/b leq g(S)$.



Now suppose that $M$ is the union of $S^{(1)},ldots,S^{(m)}$, and define $g = max(g(S^{(1)}),ldots,g(S^{(m)})) < gamma$. The foregoing shows that every $(a,b)$ in the union satisfies $a/b leq g < gamma$, and we obtain a contradiction, since $sup { a/b : (a,b) in M } = gamma$.





When $gamma$ is rational, the proof fails, and indeed $M$ is semilinear:
$$
{ (a,b) : a leq tfrac{s}{t} b } = bigcup_{a=0}^{s-1} (a,lceil tfrac{t}{s} a rceil) + mathbb{N} (s,t) + mathbb{N} (0,1).
$$

Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfrac{s}{t} b$ (since $s = tfrac{s}{t} t$). Conversely, suppose that $a leq frac{s}{t} b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq frac{s}{t}b < s$). Since $a leq frac{s}{t} b$, necessarily $b geq lceil tfrac{t}{s} a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfrac{t}{s} a rceil)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice answer. Just a clarification, the logic behind "every $(a,b) in S$ satisfies $a/b leq g(S)$" is that otherwise if there was an $(a,b)> g(S)$, then we could build an $(x,y)in S$ such that $x/y$ is as large as wanted and therefore larger than $gamma$?
    $endgroup$
    – ChenyiShiwen
    6 hours ago












  • $begingroup$
    No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
    $endgroup$
    – Yuval Filmus
    6 hours ago



















4












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Every variable except $gamma$ in this answer stands for a positive integer. It is well-known that given an irrational $gamma>0$, there is a sequence of rational numbers $dfrac{a_1}{b_1}ltdfrac{a_2}{b_2}ltdfrac{a_3}{b_3}ltcdots ltgamma$ such that $dfrac{a_i}{b_i}$ is nearer to $gamma$ than any other rational number smaller than $gamma$ whose denominator is less than $b_i$.





It turns out that the pumping lemma does work!



For the sake of contradiction, let $p$ be the pumping length of $L$ as a context-free language. Let $s=a^{a_p}b^{b_p}$, a word that is $L$ but "barely". Note that $|s|>b_pge p$. Consider
$s=uvwxy$, where $|vx|> 1$ and $s_n=uv^nwx^nyin L$ for all $nge0$.



Let $t_a$ and $t_b$ be the number of $a$s and $b$s in $vx$ respectively.




  • If $t_b=0$ or $dfrac{t_a}{t_b}gtgamma$, for $n$ large enough, the ratio of the number of $a$s to that of $b$s in $s_n$ will be larger than $gamma$, i.e., $s_nnotin L$.

  • Otherwise, $dfrac{t_a}{t_b}ltgamma$. Since $t_b<b_p$, $dfrac{t_a}{t_b}lt dfrac{a_p}{b_p}$. Hence,
    $dfrac{a_p-t_a}{b_p-t_b}>dfrac{a_p}{b_p}$
    Since $b_p-t_b<b_p$, $dfrac{a_p-t_a}{b_p-t_b}>gamma,$
    which says that $s_0notin L$.


The above contradiction shows that $L$ cannot be context-free.





Here are two related easier exercises.



Exercise 1. Show that $L_gamma={a^{lfloor i gammarfloor}: iinBbb N}$ is not context-free where $gamma$ is an irrational number.



Exercise 2. Show that $L_gamma={a^ib^j: i leq j gamma, i ge0, jge 0}$ is context-free where $gamma$ is a rational number.






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$endgroup$













  • $begingroup$
    The property in the answer can be proved simply by selecting all rational numbers that is nearer to $gamma$ than all previous numbers in the list of all rational numbers that are smaller than $gamma$ in the order of increasing denominators and, for the same denominators, in increasing order.
    $endgroup$
    – Apass.Jack
    5 hours ago












  • $begingroup$
    The usual construction is to take convergents of the continued fraction.
    $endgroup$
    – Yuval Filmus
    5 hours ago










  • $begingroup$
    @YuvalFilmus Yes, I agree. On the other hand, that almost-one-line proof is much simpler and accessible. (the "increasing order" in my last message should be "decreasing order".)
    $endgroup$
    – Apass.Jack
    5 hours ago













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2 Answers
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2 Answers
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active

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active

oldest

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6












$begingroup$

According to Parikh's theorem, if $L$ were context-free then the set $M = {(a,b) : a leq gamma b }$ would be semilinear, that is, it would be the union of finitely many sets of the form $S = u_0 + mathbb{N} u_1 + cdots + mathbb{N} u_ell$, for some $u_i = (a_i,b_i)$.



Obviously $u_0 in M$, and moreover $u_i in M$ for each $i > 0$, since otherwise $u_0 + N u_i notin M$ for large enough $N$. Therefore $g(S) := max(a_0/b_0,ldots,a_ell/b_ell) < gamma$ (since $g(S)$ is rational). This means that every $(a,b) in S$ satisfies $a/b leq g(S)$.



Now suppose that $M$ is the union of $S^{(1)},ldots,S^{(m)}$, and define $g = max(g(S^{(1)}),ldots,g(S^{(m)})) < gamma$. The foregoing shows that every $(a,b)$ in the union satisfies $a/b leq g < gamma$, and we obtain a contradiction, since $sup { a/b : (a,b) in M } = gamma$.





When $gamma$ is rational, the proof fails, and indeed $M$ is semilinear:
$$
{ (a,b) : a leq tfrac{s}{t} b } = bigcup_{a=0}^{s-1} (a,lceil tfrac{t}{s} a rceil) + mathbb{N} (s,t) + mathbb{N} (0,1).
$$

Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfrac{s}{t} b$ (since $s = tfrac{s}{t} t$). Conversely, suppose that $a leq frac{s}{t} b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq frac{s}{t}b < s$). Since $a leq frac{s}{t} b$, necessarily $b geq lceil tfrac{t}{s} a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfrac{t}{s} a rceil)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice answer. Just a clarification, the logic behind "every $(a,b) in S$ satisfies $a/b leq g(S)$" is that otherwise if there was an $(a,b)> g(S)$, then we could build an $(x,y)in S$ such that $x/y$ is as large as wanted and therefore larger than $gamma$?
    $endgroup$
    – ChenyiShiwen
    6 hours ago












  • $begingroup$
    No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
    $endgroup$
    – Yuval Filmus
    6 hours ago
















6












$begingroup$

According to Parikh's theorem, if $L$ were context-free then the set $M = {(a,b) : a leq gamma b }$ would be semilinear, that is, it would be the union of finitely many sets of the form $S = u_0 + mathbb{N} u_1 + cdots + mathbb{N} u_ell$, for some $u_i = (a_i,b_i)$.



Obviously $u_0 in M$, and moreover $u_i in M$ for each $i > 0$, since otherwise $u_0 + N u_i notin M$ for large enough $N$. Therefore $g(S) := max(a_0/b_0,ldots,a_ell/b_ell) < gamma$ (since $g(S)$ is rational). This means that every $(a,b) in S$ satisfies $a/b leq g(S)$.



Now suppose that $M$ is the union of $S^{(1)},ldots,S^{(m)}$, and define $g = max(g(S^{(1)}),ldots,g(S^{(m)})) < gamma$. The foregoing shows that every $(a,b)$ in the union satisfies $a/b leq g < gamma$, and we obtain a contradiction, since $sup { a/b : (a,b) in M } = gamma$.





When $gamma$ is rational, the proof fails, and indeed $M$ is semilinear:
$$
{ (a,b) : a leq tfrac{s}{t} b } = bigcup_{a=0}^{s-1} (a,lceil tfrac{t}{s} a rceil) + mathbb{N} (s,t) + mathbb{N} (0,1).
$$

Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfrac{s}{t} b$ (since $s = tfrac{s}{t} t$). Conversely, suppose that $a leq frac{s}{t} b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq frac{s}{t}b < s$). Since $a leq frac{s}{t} b$, necessarily $b geq lceil tfrac{t}{s} a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfrac{t}{s} a rceil)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice answer. Just a clarification, the logic behind "every $(a,b) in S$ satisfies $a/b leq g(S)$" is that otherwise if there was an $(a,b)> g(S)$, then we could build an $(x,y)in S$ such that $x/y$ is as large as wanted and therefore larger than $gamma$?
    $endgroup$
    – ChenyiShiwen
    6 hours ago












  • $begingroup$
    No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
    $endgroup$
    – Yuval Filmus
    6 hours ago














6












6








6





$begingroup$

According to Parikh's theorem, if $L$ were context-free then the set $M = {(a,b) : a leq gamma b }$ would be semilinear, that is, it would be the union of finitely many sets of the form $S = u_0 + mathbb{N} u_1 + cdots + mathbb{N} u_ell$, for some $u_i = (a_i,b_i)$.



Obviously $u_0 in M$, and moreover $u_i in M$ for each $i > 0$, since otherwise $u_0 + N u_i notin M$ for large enough $N$. Therefore $g(S) := max(a_0/b_0,ldots,a_ell/b_ell) < gamma$ (since $g(S)$ is rational). This means that every $(a,b) in S$ satisfies $a/b leq g(S)$.



Now suppose that $M$ is the union of $S^{(1)},ldots,S^{(m)}$, and define $g = max(g(S^{(1)}),ldots,g(S^{(m)})) < gamma$. The foregoing shows that every $(a,b)$ in the union satisfies $a/b leq g < gamma$, and we obtain a contradiction, since $sup { a/b : (a,b) in M } = gamma$.





When $gamma$ is rational, the proof fails, and indeed $M$ is semilinear:
$$
{ (a,b) : a leq tfrac{s}{t} b } = bigcup_{a=0}^{s-1} (a,lceil tfrac{t}{s} a rceil) + mathbb{N} (s,t) + mathbb{N} (0,1).
$$

Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfrac{s}{t} b$ (since $s = tfrac{s}{t} t$). Conversely, suppose that $a leq frac{s}{t} b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq frac{s}{t}b < s$). Since $a leq frac{s}{t} b$, necessarily $b geq lceil tfrac{t}{s} a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfrac{t}{s} a rceil)$.






share|cite|improve this answer









$endgroup$



According to Parikh's theorem, if $L$ were context-free then the set $M = {(a,b) : a leq gamma b }$ would be semilinear, that is, it would be the union of finitely many sets of the form $S = u_0 + mathbb{N} u_1 + cdots + mathbb{N} u_ell$, for some $u_i = (a_i,b_i)$.



Obviously $u_0 in M$, and moreover $u_i in M$ for each $i > 0$, since otherwise $u_0 + N u_i notin M$ for large enough $N$. Therefore $g(S) := max(a_0/b_0,ldots,a_ell/b_ell) < gamma$ (since $g(S)$ is rational). This means that every $(a,b) in S$ satisfies $a/b leq g(S)$.



Now suppose that $M$ is the union of $S^{(1)},ldots,S^{(m)}$, and define $g = max(g(S^{(1)}),ldots,g(S^{(m)})) < gamma$. The foregoing shows that every $(a,b)$ in the union satisfies $a/b leq g < gamma$, and we obtain a contradiction, since $sup { a/b : (a,b) in M } = gamma$.





When $gamma$ is rational, the proof fails, and indeed $M$ is semilinear:
$$
{ (a,b) : a leq tfrac{s}{t} b } = bigcup_{a=0}^{s-1} (a,lceil tfrac{t}{s} a rceil) + mathbb{N} (s,t) + mathbb{N} (0,1).
$$

Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfrac{s}{t} b$ (since $s = tfrac{s}{t} t$). Conversely, suppose that $a leq frac{s}{t} b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq frac{s}{t}b < s$). Since $a leq frac{s}{t} b$, necessarily $b geq lceil tfrac{t}{s} a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfrac{t}{s} a rceil)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 7 hours ago









Yuval FilmusYuval Filmus

195k14183347




195k14183347












  • $begingroup$
    Nice answer. Just a clarification, the logic behind "every $(a,b) in S$ satisfies $a/b leq g(S)$" is that otherwise if there was an $(a,b)> g(S)$, then we could build an $(x,y)in S$ such that $x/y$ is as large as wanted and therefore larger than $gamma$?
    $endgroup$
    – ChenyiShiwen
    6 hours ago












  • $begingroup$
    No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
    $endgroup$
    – Yuval Filmus
    6 hours ago


















  • $begingroup$
    Nice answer. Just a clarification, the logic behind "every $(a,b) in S$ satisfies $a/b leq g(S)$" is that otherwise if there was an $(a,b)> g(S)$, then we could build an $(x,y)in S$ such that $x/y$ is as large as wanted and therefore larger than $gamma$?
    $endgroup$
    – ChenyiShiwen
    6 hours ago












  • $begingroup$
    No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
    $endgroup$
    – Yuval Filmus
    6 hours ago
















$begingroup$
Nice answer. Just a clarification, the logic behind "every $(a,b) in S$ satisfies $a/b leq g(S)$" is that otherwise if there was an $(a,b)> g(S)$, then we could build an $(x,y)in S$ such that $x/y$ is as large as wanted and therefore larger than $gamma$?
$endgroup$
– ChenyiShiwen
6 hours ago






$begingroup$
Nice answer. Just a clarification, the logic behind "every $(a,b) in S$ satisfies $a/b leq g(S)$" is that otherwise if there was an $(a,b)> g(S)$, then we could build an $(x,y)in S$ such that $x/y$ is as large as wanted and therefore larger than $gamma$?
$endgroup$
– ChenyiShiwen
6 hours ago














$begingroup$
No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
$endgroup$
– Yuval Filmus
6 hours ago




$begingroup$
No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
$endgroup$
– Yuval Filmus
6 hours ago











4












$begingroup$

Every variable except $gamma$ in this answer stands for a positive integer. It is well-known that given an irrational $gamma>0$, there is a sequence of rational numbers $dfrac{a_1}{b_1}ltdfrac{a_2}{b_2}ltdfrac{a_3}{b_3}ltcdots ltgamma$ such that $dfrac{a_i}{b_i}$ is nearer to $gamma$ than any other rational number smaller than $gamma$ whose denominator is less than $b_i$.





It turns out that the pumping lemma does work!



For the sake of contradiction, let $p$ be the pumping length of $L$ as a context-free language. Let $s=a^{a_p}b^{b_p}$, a word that is $L$ but "barely". Note that $|s|>b_pge p$. Consider
$s=uvwxy$, where $|vx|> 1$ and $s_n=uv^nwx^nyin L$ for all $nge0$.



Let $t_a$ and $t_b$ be the number of $a$s and $b$s in $vx$ respectively.




  • If $t_b=0$ or $dfrac{t_a}{t_b}gtgamma$, for $n$ large enough, the ratio of the number of $a$s to that of $b$s in $s_n$ will be larger than $gamma$, i.e., $s_nnotin L$.

  • Otherwise, $dfrac{t_a}{t_b}ltgamma$. Since $t_b<b_p$, $dfrac{t_a}{t_b}lt dfrac{a_p}{b_p}$. Hence,
    $dfrac{a_p-t_a}{b_p-t_b}>dfrac{a_p}{b_p}$
    Since $b_p-t_b<b_p$, $dfrac{a_p-t_a}{b_p-t_b}>gamma,$
    which says that $s_0notin L$.


The above contradiction shows that $L$ cannot be context-free.





Here are two related easier exercises.



Exercise 1. Show that $L_gamma={a^{lfloor i gammarfloor}: iinBbb N}$ is not context-free where $gamma$ is an irrational number.



Exercise 2. Show that $L_gamma={a^ib^j: i leq j gamma, i ge0, jge 0}$ is context-free where $gamma$ is a rational number.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The property in the answer can be proved simply by selecting all rational numbers that is nearer to $gamma$ than all previous numbers in the list of all rational numbers that are smaller than $gamma$ in the order of increasing denominators and, for the same denominators, in increasing order.
    $endgroup$
    – Apass.Jack
    5 hours ago












  • $begingroup$
    The usual construction is to take convergents of the continued fraction.
    $endgroup$
    – Yuval Filmus
    5 hours ago










  • $begingroup$
    @YuvalFilmus Yes, I agree. On the other hand, that almost-one-line proof is much simpler and accessible. (the "increasing order" in my last message should be "decreasing order".)
    $endgroup$
    – Apass.Jack
    5 hours ago


















4












$begingroup$

Every variable except $gamma$ in this answer stands for a positive integer. It is well-known that given an irrational $gamma>0$, there is a sequence of rational numbers $dfrac{a_1}{b_1}ltdfrac{a_2}{b_2}ltdfrac{a_3}{b_3}ltcdots ltgamma$ such that $dfrac{a_i}{b_i}$ is nearer to $gamma$ than any other rational number smaller than $gamma$ whose denominator is less than $b_i$.





It turns out that the pumping lemma does work!



For the sake of contradiction, let $p$ be the pumping length of $L$ as a context-free language. Let $s=a^{a_p}b^{b_p}$, a word that is $L$ but "barely". Note that $|s|>b_pge p$. Consider
$s=uvwxy$, where $|vx|> 1$ and $s_n=uv^nwx^nyin L$ for all $nge0$.



Let $t_a$ and $t_b$ be the number of $a$s and $b$s in $vx$ respectively.




  • If $t_b=0$ or $dfrac{t_a}{t_b}gtgamma$, for $n$ large enough, the ratio of the number of $a$s to that of $b$s in $s_n$ will be larger than $gamma$, i.e., $s_nnotin L$.

  • Otherwise, $dfrac{t_a}{t_b}ltgamma$. Since $t_b<b_p$, $dfrac{t_a}{t_b}lt dfrac{a_p}{b_p}$. Hence,
    $dfrac{a_p-t_a}{b_p-t_b}>dfrac{a_p}{b_p}$
    Since $b_p-t_b<b_p$, $dfrac{a_p-t_a}{b_p-t_b}>gamma,$
    which says that $s_0notin L$.


The above contradiction shows that $L$ cannot be context-free.





Here are two related easier exercises.



Exercise 1. Show that $L_gamma={a^{lfloor i gammarfloor}: iinBbb N}$ is not context-free where $gamma$ is an irrational number.



Exercise 2. Show that $L_gamma={a^ib^j: i leq j gamma, i ge0, jge 0}$ is context-free where $gamma$ is a rational number.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The property in the answer can be proved simply by selecting all rational numbers that is nearer to $gamma$ than all previous numbers in the list of all rational numbers that are smaller than $gamma$ in the order of increasing denominators and, for the same denominators, in increasing order.
    $endgroup$
    – Apass.Jack
    5 hours ago












  • $begingroup$
    The usual construction is to take convergents of the continued fraction.
    $endgroup$
    – Yuval Filmus
    5 hours ago










  • $begingroup$
    @YuvalFilmus Yes, I agree. On the other hand, that almost-one-line proof is much simpler and accessible. (the "increasing order" in my last message should be "decreasing order".)
    $endgroup$
    – Apass.Jack
    5 hours ago
















4












4








4





$begingroup$

Every variable except $gamma$ in this answer stands for a positive integer. It is well-known that given an irrational $gamma>0$, there is a sequence of rational numbers $dfrac{a_1}{b_1}ltdfrac{a_2}{b_2}ltdfrac{a_3}{b_3}ltcdots ltgamma$ such that $dfrac{a_i}{b_i}$ is nearer to $gamma$ than any other rational number smaller than $gamma$ whose denominator is less than $b_i$.





It turns out that the pumping lemma does work!



For the sake of contradiction, let $p$ be the pumping length of $L$ as a context-free language. Let $s=a^{a_p}b^{b_p}$, a word that is $L$ but "barely". Note that $|s|>b_pge p$. Consider
$s=uvwxy$, where $|vx|> 1$ and $s_n=uv^nwx^nyin L$ for all $nge0$.



Let $t_a$ and $t_b$ be the number of $a$s and $b$s in $vx$ respectively.




  • If $t_b=0$ or $dfrac{t_a}{t_b}gtgamma$, for $n$ large enough, the ratio of the number of $a$s to that of $b$s in $s_n$ will be larger than $gamma$, i.e., $s_nnotin L$.

  • Otherwise, $dfrac{t_a}{t_b}ltgamma$. Since $t_b<b_p$, $dfrac{t_a}{t_b}lt dfrac{a_p}{b_p}$. Hence,
    $dfrac{a_p-t_a}{b_p-t_b}>dfrac{a_p}{b_p}$
    Since $b_p-t_b<b_p$, $dfrac{a_p-t_a}{b_p-t_b}>gamma,$
    which says that $s_0notin L$.


The above contradiction shows that $L$ cannot be context-free.





Here are two related easier exercises.



Exercise 1. Show that $L_gamma={a^{lfloor i gammarfloor}: iinBbb N}$ is not context-free where $gamma$ is an irrational number.



Exercise 2. Show that $L_gamma={a^ib^j: i leq j gamma, i ge0, jge 0}$ is context-free where $gamma$ is a rational number.






share|cite|improve this answer











$endgroup$



Every variable except $gamma$ in this answer stands for a positive integer. It is well-known that given an irrational $gamma>0$, there is a sequence of rational numbers $dfrac{a_1}{b_1}ltdfrac{a_2}{b_2}ltdfrac{a_3}{b_3}ltcdots ltgamma$ such that $dfrac{a_i}{b_i}$ is nearer to $gamma$ than any other rational number smaller than $gamma$ whose denominator is less than $b_i$.





It turns out that the pumping lemma does work!



For the sake of contradiction, let $p$ be the pumping length of $L$ as a context-free language. Let $s=a^{a_p}b^{b_p}$, a word that is $L$ but "barely". Note that $|s|>b_pge p$. Consider
$s=uvwxy$, where $|vx|> 1$ and $s_n=uv^nwx^nyin L$ for all $nge0$.



Let $t_a$ and $t_b$ be the number of $a$s and $b$s in $vx$ respectively.




  • If $t_b=0$ or $dfrac{t_a}{t_b}gtgamma$, for $n$ large enough, the ratio of the number of $a$s to that of $b$s in $s_n$ will be larger than $gamma$, i.e., $s_nnotin L$.

  • Otherwise, $dfrac{t_a}{t_b}ltgamma$. Since $t_b<b_p$, $dfrac{t_a}{t_b}lt dfrac{a_p}{b_p}$. Hence,
    $dfrac{a_p-t_a}{b_p-t_b}>dfrac{a_p}{b_p}$
    Since $b_p-t_b<b_p$, $dfrac{a_p-t_a}{b_p-t_b}>gamma,$
    which says that $s_0notin L$.


The above contradiction shows that $L$ cannot be context-free.





Here are two related easier exercises.



Exercise 1. Show that $L_gamma={a^{lfloor i gammarfloor}: iinBbb N}$ is not context-free where $gamma$ is an irrational number.



Exercise 2. Show that $L_gamma={a^ib^j: i leq j gamma, i ge0, jge 0}$ is context-free where $gamma$ is a rational number.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 5 hours ago

























answered 6 hours ago









Apass.JackApass.Jack

13.1k1939




13.1k1939












  • $begingroup$
    The property in the answer can be proved simply by selecting all rational numbers that is nearer to $gamma$ than all previous numbers in the list of all rational numbers that are smaller than $gamma$ in the order of increasing denominators and, for the same denominators, in increasing order.
    $endgroup$
    – Apass.Jack
    5 hours ago












  • $begingroup$
    The usual construction is to take convergents of the continued fraction.
    $endgroup$
    – Yuval Filmus
    5 hours ago










  • $begingroup$
    @YuvalFilmus Yes, I agree. On the other hand, that almost-one-line proof is much simpler and accessible. (the "increasing order" in my last message should be "decreasing order".)
    $endgroup$
    – Apass.Jack
    5 hours ago




















  • $begingroup$
    The property in the answer can be proved simply by selecting all rational numbers that is nearer to $gamma$ than all previous numbers in the list of all rational numbers that are smaller than $gamma$ in the order of increasing denominators and, for the same denominators, in increasing order.
    $endgroup$
    – Apass.Jack
    5 hours ago












  • $begingroup$
    The usual construction is to take convergents of the continued fraction.
    $endgroup$
    – Yuval Filmus
    5 hours ago










  • $begingroup$
    @YuvalFilmus Yes, I agree. On the other hand, that almost-one-line proof is much simpler and accessible. (the "increasing order" in my last message should be "decreasing order".)
    $endgroup$
    – Apass.Jack
    5 hours ago


















$begingroup$
The property in the answer can be proved simply by selecting all rational numbers that is nearer to $gamma$ than all previous numbers in the list of all rational numbers that are smaller than $gamma$ in the order of increasing denominators and, for the same denominators, in increasing order.
$endgroup$
– Apass.Jack
5 hours ago






$begingroup$
The property in the answer can be proved simply by selecting all rational numbers that is nearer to $gamma$ than all previous numbers in the list of all rational numbers that are smaller than $gamma$ in the order of increasing denominators and, for the same denominators, in increasing order.
$endgroup$
– Apass.Jack
5 hours ago














$begingroup$
The usual construction is to take convergents of the continued fraction.
$endgroup$
– Yuval Filmus
5 hours ago




$begingroup$
The usual construction is to take convergents of the continued fraction.
$endgroup$
– Yuval Filmus
5 hours ago












$begingroup$
@YuvalFilmus Yes, I agree. On the other hand, that almost-one-line proof is much simpler and accessible. (the "increasing order" in my last message should be "decreasing order".)
$endgroup$
– Apass.Jack
5 hours ago






$begingroup$
@YuvalFilmus Yes, I agree. On the other hand, that almost-one-line proof is much simpler and accessible. (the "increasing order" in my last message should be "decreasing order".)
$endgroup$
– Apass.Jack
5 hours ago












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