Necessary properties of formal langagues in certain classes that rely on closure against repetition of certain subwords.

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34 views

Use the pumping lemma to show that the language is not regular [on hold]

I am working on this problem : Use the pumping lemma to show that the language $\{0^n 1^{n} \mid n ≥ 1\}$ is not regular. May someone give me some suggestion about how to solve this problem?
0
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1answer
48 views

Proving a language isn't regular using the pumping lemma [on hold]

Let the language $$ L = \{ a^nb^m : m,n \text{ has the same integer-quotient, (ignoring the remainder) } \} $$ Show that $L$ isn't regular using the pumping-lemma. Let's assume by contradiction ...
1
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1answer
44 views

Proof that a language is not regular using pumping lemma

I have a language $L$ that I think is not regular: $L = \{w\in \{0,1,...,9\}^* \; | \enspace w \enspace \text{is a decimal representation of a number divisible by 3}\}$ I'm using pumping lemma in my ...
0
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0answers
22 views

Ogden’s lemma on CFG

I'm trying to understand Ogden's lemma, and I know there are 4 cases, but in the next example I can only find 3: Assume A = {$0^n1^m0^k$ | k = $max${n, m}} is CF: Choose z = $0^k1^k0^k ∈ A$, z = ...
1
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1answer
32 views

When to pump up and down?

When I'm solving a question I usually spent too much time testing whether I should pump or down? Is there any formula to know when to use which? Also, on proofing non context free grammar we use ...
-1
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1answer
40 views

pumping lemma for $L=\{a^n b^m c^k \mid n = m \vee m\neq k\}$ [duplicate]

Using pumping lemma, how can I prove that $L=\{a^n b^m c^k \mid n = m \vee m\neq k\}$ is not regular?. If I choose $w= a^m b^m c^m$ and pump up with $i=2$, if have $a^m=1 b^m c^m$ but the string is ...
3
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1answer
39 views

Is the language of all $a^n$ for which $n$ has an even number of digits in 10-base system regular?

Is the language $ L = \{a^n ~| ~n \text{ has even number of digits in 10-base system}\} $ regular? My approach: let the $ p $ be from the Pumping Lemma. Chose the smallest $ n $ which has even number ...
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2answers
72 views

Verification wanted: Show the language $L=\{0^m1^n \enspace | \enspace m \neq n\}$ is not regular [closed]

$$L=\{0^m1^n \enspace | \enspace m \neq n\}$$ I saw that this exact question exists elsewhere, but I couldn't understand what was being said there. My question does not mandate the use of the Pumping ...
0
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1answer
52 views

Proving that $L=\lbrace{ab^{n}ba^{n}|n\geq1}\rbrace$ is not regular with pumping lemma

I'm trying to understand the pumping lemma for regular languages and would like to prove that $L=\lbrace{ab^{n}ba^{n}|n\geq1}\rbrace$ is not regular. My suggestion is as follows: Assuming ...
3
votes
1answer
53 views

A non-regular language satisfying the pumping lemma

I got a problem to solve, which is to demostrate that the language $L$, given by: $L = \{ab^nc^n\mid n \geq 0\} \cup \{a^kw \mid k\geq 2 \wedge w \in \Sigma^*\}$ Satisfies the pumping lemma. Is not ...
0
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0answers
43 views

The pumping lemma for the context free languages [duplicate]

I am trying to use the pumping lemma to show this is not a context free language $$ L = \{a^n b^{2n} a^n\mid n\ge 0\} $$ My idea is fist assume it is a CFG language and let $n$ be the pumping lemma ...
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0answers
27 views

$L = \{a^{n^3} | \ge 0\}$ Use the Pumping Lemma to show that L is not regular [duplicate]

Use the Pumping Lemma to show that $L$ is not regular: $$ L = \{{a^{n^3} | \ge 0}\}$$ I feel like I have a good intuition of what the Pumping Lemma states; strings that belong to a regular language ...
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2answers
53 views

I think I have a regular expression for a non-regular language

Let $W = \{a^n b^m \mid n\ge m+5,m\le 5\}$, where $\Sigma=\{a, b\}$. I have proved that this language is irregular through pumping Lemma. But through regular expression it is proving that the ...
0
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1answer
41 views

Showing that $\{ c^n a^m b^{n+m} : n+m \geq 6\}$ is not regular [duplicate]

I'm trying to show that $L_6=\{c^n a^m b^p : n+m=p,p \geq 6\}$ is not regular. I need a little help, I was practicing the pumping lemma, and I encountered this language, I saw these conditions and got ...
0
votes
2answers
67 views

How to prove that these two languages are regular, or not regular? [duplicate]

I have these two languages $L_1={\{a^n b^m,n≥m+5,m>0}\}$ Where $∑=(a,b)$ $L_2={\{a^n b^m,n≥m+5,m≤5}\}$ Where $∑=(a,b)$ As you can see that there is only one difference, the condition of ...
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1answer
25 views

Pumping lemma of regular language [duplicate]

I was wondering in how to solve this question, I feel a bit confused: for $\Sigma = \{1,\#\}$, consider $$D=\{w \mid w=x_1 \# x_2 \# \cdots \# x_k \text{ for } k \geq 0, \text{ each } x_i \in 1^*, ...
3
votes
1answer
156 views

How does a regular language satisfies the second condition of the pumping lemma

I'm a little bit confused about the second condition of the pumping lemma which are: $|y|\geq1$ $|xy|\leq p$ $\forall i \geq 0:xyiz\in L$ I don't understand why the length of ...
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1answer
64 views

Pumping Lemma confusion

I have the following language... $$A=\{a^ib^i | i>0\}\cup\{a^jb^k|j>2, k>3\}$$ Now, pumping lemma states that a regular language can be written in the form $x=pq^ir$. What confuses me is ...
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1answer
49 views

Pumping Lemma for Regular Language seems to Fail

Let $L = \{ab^ncd \mid n \geq 0\}$. If we take $p = 5$ and $w = abbcd$ and write $w_i = xy^iz$, where $x = abb$, $y=c$, $z=d$, then $w_2 = abbccd$ which is not in $L$. We conclude that ...
0
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1answer
97 views

How do I show that $a^n w b^n$ is not regular?

Given $ \Sigma= \{a,b\} $, show that $ L:= \{a^nwb^n: m,n \in \mathbb N, m\geqslant n, w\in\Sigma^m\} $ is not regular. I'm trying to proof this with the Pumping Lemma, but I'm kind of confused ...
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1answer
107 views

Pumping Lemma Proof by contradiction [duplicate]

So, hi guys! I have a language which I am trying to prove that it is not regular using the Pumping Lemma. I am pretty new into the lemma so I would appreciate any help.The language is $$L = \{w | w ...
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2answers
75 views

L ={ $a^{m^n}$ | $m$>$n$ } is Regular or not by pumping Lemma [duplicate]

L ={ $a^{m^n}$ | $m$>$n$ } I am bit confuse whether to consider this language as L = $(a^{m})^{n}$ OR L = $a^{\left(m^n\right)}$. If it is considered as L = $(a^m)^{n}$ then want to check it ...
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0answers
146 views

Is the complement of this language Context-Free $\{(a^nb^n)^m \mid n>0,m>0\}$?

I've been asked to decide whether a given language is a Context-Free Language (CFL). If yes, I should find the grammar that creates her, and if not, I need to prove it (with the pumping lemma). The ...
2
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1answer
78 views

Help on using the pumping lemma?

I'm trying to prove that a language is not regular. That language is: {w ∈ {a, b}* | amount of a's in w is equivalent to the amount of b's in w, mod 2}. I have an inkling that this language is not ...
1
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1answer
65 views

Applying the context-free pumping lemma to a language with crossed nestings

For proving language $\{a^nb^mc^nd^m \mid n,m > 0\}$ is not context free. Do I have to use $z = a^pb^pc^pd^p$ as pumping lemma string where $p$ is pumping length? Or do I have to use a string that ...
1
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1answer
76 views

When using the Pumping lemma, how do I deal with different cases of y?

I want to prove L is not regular:$$L={\{www|w \in \Sigma^*\}}$$ $$\Sigma=\{a,b\}$$ I am sure I can do so using pumping lemma. I used $$ab^pab^pab^p$$as my chosen string but I am stuck. I do not know ...
2
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1answer
45 views

Pumping lemma for 0^n and n>0

When applying the pumping lemma to $L = \{ 0^n \mid n>0\}$ I do the following: $S = 0^p$ $x = \varepsilon$ $y = 0^p$ $z = \varepsilon$ so $S = xyz = (\varepsilon)0^p(\varepsilon)$ For $x y^i z$ ...
2
votes
2answers
121 views

Show that $0^i$ where $i$ is a power of 2 is not context free

I'm having difficulty trying to use the pumping lemma in order to show that $L= \{0^i \mid \ i \text{ is a power of 2 }\} $ is not context free. I"m starting by stating that $ s = 0^p$ and then $ s = ...
0
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1answer
79 views

Pumping Lemma for $L=\{a^{2k} b^n b^k \mid k\ge0, n\ge0\}$

$L=\{a^{2k}b^nb^k\mid k\geq0, n\geq0\}$ over alphabet $\{a,b\}$ How do I prove that $L$ is not regular using Pumping Lemma? All the examples I've come across had same exponents all around, and I'm a ...
0
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2answers
111 views

High Level Explanation of the Pumping Lemma

I have a problem that I cannot figure out regarding using the pumping lemma to prove that a language is not regular. I don't understand how I go about proving through contradiction that the language ...
0
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1answer
60 views

How to prove that $\{0^n 1^{5n} \mid n \ge 10000 \}$ is not a regular language?

I proved that $$ \{ 0^n 1^{5n} \mid n \geq 0 \}$$ is not a regular language using Pumping Lemma by following way. Solve by contradiction that $ L = \{0^n 1^{5n} \mid n \geq 0 \}$ is regular ...
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2answers
105 views

Show that for any natural number n, there is a regular language that is not recognized by any DFA with at most n final states

Just as the question asks, I am trying to understand the relationship between the number of accept states a DFA has (not necessarily the total number of states) and the languages it can accept. I ...
3
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1answer
161 views

Why does this pumping lemma application “prove” that 0*1* is not regular?

Here is a proof that $0^*1^*$ is not regular, even though it is regular. I'm having a hard time figuring out what is wrong with the proof. Assume $0^*1^*$ is regular. Let $p$ be the pumping length as ...
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1answer
156 views

Can $\{a^mb^nc^n\mid m,n \ge 1\}$ be proved non-regular using the pumping lemma?

$\{a^mb^nc^n\mid m,n \ge 1\}$ intuitively seems like a non-regular language. It looks like the machine needs to remember the number of $b$s (which isn't limited). The pumping lemma can be used to ...
3
votes
1answer
150 views

Prove that the language is not regular without using Pumping Lemma

I am practising problems on Regular Languages and I came across this question: Prove that the language $$\{a^m b^n : m ≥ 0, n ≥ 0, m \ne n\}$$ is not regular. (Using the pumping lemma for this ...
3
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3answers
367 views

Clearing a Confusion regarding the Proof of equal no of a's and b's not being a regular language

I was wondering about its proof. The direct use of pumping lemma here is not a viability. So a certain teacher of mine proved this with the notion that $a^{n}b^{n}$ being a subset of this language ...
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3answers
122 views

Unable to understand an inequality in an application of the pumping lemma for context-free languages

The problem Prove that the language $\qquad L = \{a^n b^j \mid n = j^2\}$ is not context free using pumping lemma. Approach taken by the book To prove such statements, the book takes the ...
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1answer
68 views

CFL, pumping lemma

I have difficulty with proving that the language $ L = \{ a^p b^q | p \ge 1 , q \ge 1 , p \ge q^2 \vee q \ge p^2\}$ $ w = uvxyz $ I've chosen word $ w = a^{N^2} b^N $ where $ N $ is a constant ...
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0answers
54 views

Context-free Language, Pumping lemma

I want to prove that $ L = {a^n b^m c^{ \lfloor \frac{n}{m} \rfloor } } $ isn't context free language, so I choose N - constant from lemma so the word is $ w = a^N b^N c $ and $ w = uvxyz $ 1 ...
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1answer
130 views

requirement for pumping lemma in regular language

I am a bit confused on the theory of the pumping lemma. As I know is used to decide if a language is regular or not. This is what I have understood so far though For a regular language $L$, there ...
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2answers
298 views

Using the Pumping Lemma to show that the language $a^n b a^n$ is not regular

I've seen a lot couple of questions regarding the pumping lemma that are pretty similar to each other and this one is unfortunately not the exception. Most likely will be this question marked as a ...
2
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1answer
85 views

Is there a Context-free grammar for this language?

Is there a Context-free grammar for the following language: $L=\{ x\#1^m|x \in \{0,1\}^* \space and \space the \space m^{th} \space char \space in \space x \space ...
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1answer
74 views

Prove not context free

How can we prove that: $$ L = \{ w_1\#w_2 \mid w_1 \in w_2;\; |w_2| > |w_1|;\; w_1 , w_2 \in \{0, 1\}^*\} $$ is not context-free? The language defines $w_1$ as a sub-string of $w_2$, and they ...
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1answer
193 views

Is the language $L = \{a^nb^m : n = 2^m\}$ context-free?

Is the language $L = \{a^nb^m : n = 2^m\}$ context-free? Assume L is a context-free language. Then $\ \exists p\in \mathbb{Z}^{+}:\forall s\in L\left | s \right |\geq p. s = uvxyz,\left | vy \right ...
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1answer
60 views

Show L is not context free using the CFL pumping lemma

I am trying to use the pumping lemma to show this language is not context free: $L = a^nb^{n+1}c^{2n} : n \ge 0$ So I took $z = a^mb^{m+1}c^{2m}$ where $|z| = 4m+1 > m$. We can decompose $z = ...
3
votes
1answer
128 views

Show that the pumping lemmas for context-free and regular languages are equivalent for unary languages

I want to show that for any language $L \subseteq \{ a \}^* $, $L$ satisfies the pumping lemma for context free languages if and only if it satisfies the pumping lemma for regular languages. I know ...
0
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1answer
67 views

Pumping lemma not regular [duplicate]

How would about proving this is not regular with the pumping lemma. Please include all steps and explain all steps. I am really new with this. $1^{2x}0^y$ and $y>= x$ Does it matter which side ...
0
votes
1answer
127 views

Using the pumping lemma for a proof by contradiction [duplicate]

I'm trying to prove that the set of even-length strings with the two middle symbols being equal cannot be accepted by finite automata. I can explain why it cannot be accepted intuitively, but I'm ...
0
votes
1answer
105 views

Pumping lemma on {a^n | n=3^k} — help finishing the proof [duplicate]

I am working on a pumping lemma question and trying to prove that the following is not regular, but I can't finish the proof, if someone can help me it will be great. So I am given this language: $L ...
1
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0answers
48 views

Proving a language is regular or non-regular [duplicate]

I'm struggling a bit to understand two of the problems we were given in class. Could someone look over my work and maybe give me a few hints? State whether the following languages are regular or not ...