Question

Show that $\{a^i b^j c^k \mid i>j>k>0\}$ is not a context free language by using pumping lemma

score 1.8k · Answer 1 · 2026-05-01 19:31:25

1.8k

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Show that $\{a^i b^j c^k \mid i>j>k>0\}$ is not a context free language by using pumping lemma

Published on 01 May 2026 - 19:31

score 1.5k · Answer 2 · 2026-05-01 19:32:27

1.5k

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Use Pumping Lemma to show that $L_7$ is not context-free

Published on 01 May 2026 - 19:32

score 276 · Answer 3 · 2026-04-13 23:38:09

276

Views

Context free grammar with or conditionals

Published on 13 Apr 2026 - 23:38

score 264 · Answer 4 · 2026-04-08 21:43:42

264

Views

Ambiguous CFG to Unambiguous CFG Transformation

Published on 08 Apr 2026 - 21:43

score 185 · Answer 5 · 2026-05-10 14:02:55

185

Views

Pumping Lemma for CFL

Published on 10 May 2026 - 14:02

score 150 · Answer 6 · 2026-04-15 18:16:10

150

Views

$L_4 = \{x: \#_{1}(x) = 2 \cdot \#_{10}(x) \}$ Find CFG given hints

Published on 15 Apr 2026 - 18:16

score 44 · Answer 7 · 2026-05-10 12:51:49

44

Views

Context Free grammar to Chomsky Normal Form

Published on 10 May 2026 - 12:51

score 86 · Answer 8 · 2026-04-15 06:47:14

86

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Let $L_{n} = \{x \in \Sigma^* | x = ywz, w^R = w, |w| \geq n, |y| = |z| \}$ Generate a cfg of $L_n$

Published on 15 Apr 2026 - 6:47

score 1.1k · Answer 9 · 2026-04-15 14:44:35

1.1k

Views

Show that $L=\{a^nb^m : m \neq n\}$ is context free language using closure under union.

Published on 15 Apr 2026 - 14:44

score 136 · Answer 10 · 2026-04-12 07:15:19

136

Views

Show that the following language is context-free/not context free by expressing the language as the union of three other languages.

Published on 12 Apr 2026 - 7:15

score 493 · Answer 11 · 2026-05-01 19:31:12

493

Views

Searching for a proof for a variant of the pumping lemma for context free languages

Published on 01 May 2026 - 19:31

score 2.7k · Answer 12 · 2026-05-10 12:34:21

2.7k

Views

Proof verification of the language of all palindromes as being context-free

Published on 10 May 2026 - 12:34

score 219 · Answer 13 · 2026-04-13 05:54:27

219

Views

How would I go about proving that the language accepted by a regular expression is subset of the language accepted by a context free grammar?

Published on 13 Apr 2026 - 5:54

score 130 · Answer 14 · 2026-04-13 20:45:37

130

Views

On left recursive context-free grammars

Published on 13 Apr 2026 - 20:45

score 56 · Answer 15 · 2026-04-12 19:27:50

56

Views

Is $L = \{a^{i+j}b^{j+k}c^{i+k} | i,j,k > 0\}$ context free?

Published on 12 Apr 2026 - 19:27

List Question

Show that $\{a^i b^j c^k \mid i>j>k>0\}$ is not a context free language by using pumping lemma

Use Pumping Lemma to show that $L_7$ is not context-free

Context free grammar with or conditionals

Ambiguous CFG to Unambiguous CFG Transformation

Pumping Lemma for CFL

$L_4 = \{x: \#_{1}(x) = 2 \cdot \#_{10}(x) \}$ Find CFG given hints

Context Free grammar to Chomsky Normal Form

Let $L_{n} = \{x \in \Sigma^* | x = ywz, w^R = w, |w| \geq n, |y| = |z| \}$ Generate a cfg of $L_n$

Show that $L=\{a^nb^m : m \neq n\}$ is context free language using closure under union.

Show that the following language is context-free/not context free by expressing the language as the union of three other languages.

Searching for a proof for a variant of the pumping lemma for context free languages

Proof verification of the language of all palindromes as being context-free

How would I go about proving that the language accepted by a regular expression is subset of the language accepted by a context free grammar?

On left recursive context-free grammars

Is $L = \{a^{i+j}b^{j+k}c^{i+k} | i,j,k > 0\}$ context free?

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