Language involving irrational number is not a CFLHow to prove that a language is not regular?Why is $L= n geq 1 $ not regular language?Prove that the language of unary not-prime numbers satisfies the Pumping LemmaAlgorithm to test whether a language is context-freeUsing the Pumping Lemma to show that the language $a^n b a^n$ is not regularA non-regular language satisfying the pumping lemmaProving non-regularity of $u u^R v$?Prove using pumping free lemma for context-free languagesProve that $L_1 = , 0^m 1^k 2^n ,vert, lvert m - n rvert = k ,$ is not regular using Pumping lemmaPumping lemma for context-free languages - Am I doing it right?
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Language involving irrational number is not a CFL
How to prove that a language is not regular?Why is $L= 0^n 1^n $ not regular language?Prove that the language of unary not-prime numbers satisfies the Pumping LemmaAlgorithm to test whether a language is context-freeUsing the Pumping Lemma to show that the language $a^n b a^n$ is not regularA non-regular language satisfying the pumping lemmaProving non-regularity of $u u^R v$?Prove using pumping free lemma for context-free languagesProve that $L_1 = , 0^m 1^k 2^n ,vert, lvert m - n rvert = k ,$ is not regular using Pumping lemmaPumping lemma for context-free languages - Am I doing it right?
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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!
formal-languages automata context-free
New contributor
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show 1 more comment
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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!
formal-languages automata context-free
New contributor
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Have you tried applying Parikh’s theorem?
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– Yuval Filmus
8 hours ago
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Why not show that it’s not semilinear directly? Use the definition.
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– Yuval Filmus
7 hours ago
3
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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.
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– Hendrik Jan
4 hours ago
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@ChenyiShiwen, so the pumping lemma does work.
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– Apass.Jack
4 hours ago
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@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
1 hour ago
|
show 1 more comment
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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!
formal-languages automata context-free
New contributor
$endgroup$
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
formal-languages automata context-free
New contributor
New contributor
edited 39 mins ago
D.W.♦
102k12127291
102k12127291
New contributor
asked 8 hours ago
ChenyiShiwenChenyiShiwen
434
434
New contributor
New contributor
$begingroup$
Have you tried applying Parikh’s theorem?
$endgroup$
– Yuval Filmus
8 hours ago
$begingroup$
Why not show that it’s not semilinear directly? Use the definition.
$endgroup$
– Yuval Filmus
7 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
4 hours ago
$begingroup$
@ChenyiShiwen, so the pumping lemma does work.
$endgroup$
– Apass.Jack
4 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
1 hour ago
|
show 1 more comment
$begingroup$
Have you tried applying Parikh’s theorem?
$endgroup$
– Yuval Filmus
8 hours ago
$begingroup$
Why not show that it’s not semilinear directly? Use the definition.
$endgroup$
– Yuval Filmus
7 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
4 hours ago
$begingroup$
@ChenyiShiwen, so the pumping lemma does work.
$endgroup$
– Apass.Jack
4 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
1 hour ago
$begingroup$
Have you tried applying Parikh’s theorem?
$endgroup$
– Yuval Filmus
8 hours ago
$begingroup$
Have you tried applying Parikh’s theorem?
$endgroup$
– Yuval Filmus
8 hours ago
$begingroup$
Why not show that it’s not semilinear directly? Use the definition.
$endgroup$
– Yuval Filmus
7 hours ago
$begingroup$
Why not show that it’s not semilinear directly? Use the definition.
$endgroup$
– Yuval Filmus
7 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
4 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
4 hours ago
$begingroup$
@ChenyiShiwen, so the pumping lemma does work.
$endgroup$
– Apass.Jack
4 hours ago
$begingroup$
@ChenyiShiwen, so the pumping lemma does work.
$endgroup$
– Apass.Jack
4 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
1 hour 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
1 hour ago
|
show 1 more comment
2 Answers
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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 $S = u_0 + mathbbN u_1 + cdots + mathbbN u_ell$, 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:
$$
(a,b) : a leq tfracst b = bigcup_a=0^s-1 (a,lceil tfracts a rceil) + mathbbN (s,t) + mathbbN (0,1).
$$
Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfracst b$ (since $s = tfracst t$). Conversely, suppose that $a leq fracst b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq fracstb < s$). Since $a leq fracst b$, necessarily $b geq lceil tfracts a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfracts a rceil)$.
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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 $gamma$?
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– ChenyiShiwen
4 hours ago
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No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
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– Yuval Filmus
4 hours ago
add a comment |
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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 $dfraca_1b_1ltdfraca_2b_2ltdfraca_3b_3ltcdots ltgamma$ such that $dfraca_ib_i$ 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 $s=a^a_pb^b_p$, 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 $dfract_at_bgtgamma$, 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, $dfract_at_bltgamma$. Since $t_b<b_p$, $dfract_at_blt dfraca_pb_p$. Hence,
$dfraca_p-t_ab_p-t_b>dfraca_pb_p$
Since $b_p-t_b<b_p$, $dfraca_p-t_ab_p-t_b>gamma,$
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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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.
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– Apass.Jack
4 hours ago
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The usual construction is to take convergents of the continued fraction.
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– Yuval Filmus
3 hours ago
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@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".)
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– Apass.Jack
3 hours ago
add a comment |
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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 $S = u_0 + mathbbN u_1 + cdots + mathbbN u_ell$, 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:
$$
(a,b) : a leq tfracst b = bigcup_a=0^s-1 (a,lceil tfracts a rceil) + mathbbN (s,t) + mathbbN (0,1).
$$
Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfracst b$ (since $s = tfracst t$). Conversely, suppose that $a leq fracst b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq fracstb < s$). Since $a leq fracst b$, necessarily $b geq lceil tfracts a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfracts a rceil)$.
$endgroup$
$begingroup$
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
4 hours ago
$begingroup$
No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
$endgroup$
– Yuval Filmus
4 hours ago
add a comment |
$begingroup$
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 $S = u_0 + mathbbN u_1 + cdots + mathbbN u_ell$, 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:
$$
(a,b) : a leq tfracst b = bigcup_a=0^s-1 (a,lceil tfracts a rceil) + mathbbN (s,t) + mathbbN (0,1).
$$
Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfracst b$ (since $s = tfracst t$). Conversely, suppose that $a leq fracst b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq fracstb < s$). Since $a leq fracst b$, necessarily $b geq lceil tfracts a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfracts a rceil)$.
$endgroup$
$begingroup$
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
4 hours ago
$begingroup$
No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
$endgroup$
– Yuval Filmus
4 hours ago
add a comment |
$begingroup$
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 $S = u_0 + mathbbN u_1 + cdots + mathbbN u_ell$, 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:
$$
(a,b) : a leq tfracst b = bigcup_a=0^s-1 (a,lceil tfracts a rceil) + mathbbN (s,t) + mathbbN (0,1).
$$
Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfracst b$ (since $s = tfracst t$). Conversely, suppose that $a leq fracst b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq fracstb < s$). Since $a leq fracst b$, necessarily $b geq lceil tfracts a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfracts a rceil)$.
$endgroup$
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 $S = u_0 + mathbbN u_1 + cdots + mathbbN u_ell$, 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:
$$
(a,b) : a leq tfracst b = bigcup_a=0^s-1 (a,lceil tfracts a rceil) + mathbbN (s,t) + mathbbN (0,1).
$$
Indeed, by construction, any pair $(a,b)$ in the right-hand side satisfies $a leq tfracst b$ (since $s = tfracst t$). Conversely, suppose that $a leq fracst b$. While $a geq s$ and $b geq t$, subtract $(s,t)$ from $(a,b)$. Eventually $a < s$ (since $b < t$ implies $a leq fracstb < s$). Since $a leq fracst b$, necessarily $b geq lceil tfracts a rceil$. Hence we can subtract $(0,1)$ from $(a,b)$ until we reach $(a,lceil tfracts a rceil)$.
answered 5 hours ago
Yuval FilmusYuval Filmus
195k14183347
195k14183347
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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 $gamma$?
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– ChenyiShiwen
4 hours ago
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No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
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– Yuval Filmus
4 hours ago
add a comment |
$begingroup$
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
4 hours ago
$begingroup$
No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
$endgroup$
– Yuval Filmus
4 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
4 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
4 hours ago
$begingroup$
No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
$endgroup$
– Yuval Filmus
4 hours ago
$begingroup$
No, this follows directly from the definition of $g(S)$. Your argument explains why $g(S) < gamma$.
$endgroup$
– Yuval Filmus
4 hours ago
add a comment |
$begingroup$
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 $dfraca_1b_1ltdfraca_2b_2ltdfraca_3b_3ltcdots ltgamma$ such that $dfraca_ib_i$ 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 $s=a^a_pb^b_p$, 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 $dfract_at_bgtgamma$, 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, $dfract_at_bltgamma$. Since $t_b<b_p$, $dfract_at_blt dfraca_pb_p$. Hence,
$dfraca_p-t_ab_p-t_b>dfraca_pb_p$
Since $b_p-t_b<b_p$, $dfraca_p-t_ab_p-t_b>gamma,$
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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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.
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– Apass.Jack
4 hours ago
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The usual construction is to take convergents of the continued fraction.
$endgroup$
– Yuval Filmus
3 hours ago
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@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
3 hours ago
add a comment |
$begingroup$
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 $dfraca_1b_1ltdfraca_2b_2ltdfraca_3b_3ltcdots ltgamma$ such that $dfraca_ib_i$ 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 $s=a^a_pb^b_p$, 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 $dfract_at_bgtgamma$, 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, $dfract_at_bltgamma$. Since $t_b<b_p$, $dfract_at_blt dfraca_pb_p$. Hence,
$dfraca_p-t_ab_p-t_b>dfraca_pb_p$
Since $b_p-t_b<b_p$, $dfraca_p-t_ab_p-t_b>gamma,$
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.
$endgroup$
$begingroup$
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
4 hours ago
$begingroup$
The usual construction is to take convergents of the continued fraction.
$endgroup$
– Yuval Filmus
3 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
3 hours ago
add a comment |
$begingroup$
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 $dfraca_1b_1ltdfraca_2b_2ltdfraca_3b_3ltcdots ltgamma$ such that $dfraca_ib_i$ 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 $s=a^a_pb^b_p$, 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 $dfract_at_bgtgamma$, 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, $dfract_at_bltgamma$. Since $t_b<b_p$, $dfract_at_blt dfraca_pb_p$. Hence,
$dfraca_p-t_ab_p-t_b>dfraca_pb_p$
Since $b_p-t_b<b_p$, $dfraca_p-t_ab_p-t_b>gamma,$
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.
$endgroup$
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 $dfraca_1b_1ltdfraca_2b_2ltdfraca_3b_3ltcdots ltgamma$ such that $dfraca_ib_i$ 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 $s=a^a_pb^b_p$, 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 $dfract_at_bgtgamma$, 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, $dfract_at_bltgamma$. Since $t_b<b_p$, $dfract_at_blt dfraca_pb_p$. Hence,
$dfraca_p-t_ab_p-t_b>dfraca_pb_p$
Since $b_p-t_b<b_p$, $dfraca_p-t_ab_p-t_b>gamma,$
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.
edited 3 hours ago
answered 4 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
4 hours ago
$begingroup$
The usual construction is to take convergents of the continued fraction.
$endgroup$
– Yuval Filmus
3 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
3 hours ago
add a comment |
$begingroup$
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
4 hours ago
$begingroup$
The usual construction is to take convergents of the continued fraction.
$endgroup$
– Yuval Filmus
3 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
3 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
4 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
4 hours ago
$begingroup$
The usual construction is to take convergents of the continued fraction.
$endgroup$
– Yuval Filmus
3 hours ago
$begingroup$
The usual construction is to take convergents of the continued fraction.
$endgroup$
– Yuval Filmus
3 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
3 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
3 hours ago
add a comment |
ChenyiShiwen is a new contributor. Be nice, and check out our Code of Conduct.
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Have you tried applying Parikh’s theorem?
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– Yuval Filmus
8 hours ago
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Why not show that it’s not semilinear directly? Use the definition.
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– Yuval Filmus
7 hours ago
3
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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.
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– Hendrik Jan
4 hours ago
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@ChenyiShiwen, so the pumping lemma does work.
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– Apass.Jack
4 hours ago
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@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?
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– ChenyiShiwen
1 hour ago