91Ó°ÊÓ

Let $\mathit{D}\mathbf{=}\mathbf{\left\{}\mathit{x}\mathit{y}\mathbf{\right|}\mathit{x}\mathbf{,}\mathit{y}\mathbf{∈}\left\{0,1\right\}\mathbf{*}\mathit{a}\mathit{n}\mathit{d}\left|x\right|\mathbf{=}\left|y\right|\mathit{b}\mathit{u}\mathit{t}\mathit{x}\mathbf{≠}\mathit{y}\mathbf{\}}\mathbf{.}{\mathit{X}}_{\mathbf{1}}\mathbf{,}\mathbf{…}\mathbf{,}{\mathit{X}}_{\mathbf{n}}$Show that$\mathit{D}$ is a context-free language.

Let $\mathit{E}\mathbf{=}\left\{a^i{b}^j|i≠jand2i≠j\right\}$. Show that$\mathit{E}$ is context-free language.

For any language, let SUFFIX() =$\mathbf{\left\{}\mathit{v}\mathbf{\right|}\mathit{u}\mathit{v}\mathbf{∈}\mathit{A}\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{s}\mathit{o}\mathit{m}\mathit{e}\mathbf{}\mathit{s}\mathit{t}\mathit{r}\mathit{i}\mathit{n}\mathit{g}\mathbf{}\mathit{u}\mathbf{\}}\mathbf{.}$ Show that the class of context-free languages is closed under the SUFFIX operation

Let $\mathit{G}\mathbf{=}\mathbf{(}\mathit{V}\mathbf{,}\mathit{Î\pounds }\mathbf{,}\mathit{R}\mathbf{,}\mathbf{⟨}\mathbf{S}\mathbf{T}\mathbf{M}\mathbf{T}\mathbf{âŸ\copyright }\mathbf{)}$be the following grammar.

$\begin{array}{c}\mathbf{⟨}\mathbf{S}\mathbf{T}\mathbf{M}\mathbf{T}\mathbf{âŸ\copyright }\mathbf{→}\mathbf{⟨}\mathbf{A}\mathbf{S}\mathbf{S}\mathbf{I}\mathbf{G}\mathbf{N}\mathbf{âŸ\copyright }\mathbf{\left|}\mathbf{⟨}\mathbf{I}\mathbf{F}\mathbf{-}\mathbf{T}\mathbf{H}\mathbf{E}\mathbf{N}\mathbf{âŸ\copyright }\mathbf{\right|}\mathbf{⟨}\mathbf{I}\mathbf{F}\mathbf{-}\mathbf{T}\mathbf{H}\mathbf{E}\mathbf{N}\mathbf{-}\mathbf{E}\mathbf{L}\mathbf{S}\mathbf{E}\mathbf{âŸ\copyright }\\ \mathbf{⟨}\mathbf{I}\mathbf{F}\mathbf{-}\mathbf{T}\mathbf{H}\mathbf{E}\mathbf{N}\mathbf{âŸ\copyright }\mathbf{→}\mathbf{i}\mathbf{f}\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{d}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}\mathbf{⟨}\mathbf{S}\mathbf{T}\mathbf{M}\mathbf{T}\mathbf{âŸ\copyright }\\ \mathbf{⟨}\mathbf{I}\mathbf{F}\mathbf{-}\mathbf{T}\mathbf{H}\mathbf{E}\mathbf{N}\mathbf{-}\mathbf{E}\mathbf{L}\mathbf{S}\mathbf{E}\mathbf{âŸ\copyright }\mathbf{→}\mathbf{i}\mathbf{f}\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{d}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}\mathbf{⟨}\mathbf{S}\mathbf{T}\mathbf{M}\mathbf{T}\mathbf{âŸ\copyright }\mathbf{e}\mathbf{l}\mathbf{s}\mathbf{e}\mathbf{⟨}\mathbf{S}\mathbf{T}\mathbf{M}\mathbf{T}\mathbf{âŸ\copyright }\\ \mathbf{⟨}\mathbf{A}\mathbf{S}\mathbf{S}\mathbf{I}\mathbf{G}\mathbf{N}\mathbf{âŸ\copyright }\mathbf{→}\mathbf{a}\mathbf{:}\mathbf{=}\mathbf{1}\end{array}$

$\begin{array}{c}\mathbf{Î\pounds }\mathbf{=}\mathbf{\{}\mathbf{i}\mathbf{f}\mathbf{,}\mathbf{c}\mathbf{o}\mathbf{n}\mathbf{d}\mathbf{i}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{,}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{n}\mathbf{,}\mathbf{e}\mathbf{l}\mathbf{s}\mathbf{e}\mathbf{,}\mathbf{a}\mathbf{:}\mathbf{=}\mathbf{1}\mathbf{\}}\\ \mathbf{V}\mathbf{=}\mathbf{\{}\mathbf{⟨}\mathbf{S}\mathbf{T}\mathbf{M}\mathbf{T}\mathbf{âŸ\copyright }\mathbf{,}\mathbf{⟨}\mathbf{I}\mathbf{F}\mathbf{-}\mathbf{T}\mathbf{H}\mathbf{E}\mathbf{N}\mathbf{âŸ\copyright }\mathbf{,}\mathbf{⟨}\mathbf{I}\mathbf{F}\mathbf{-}\mathbf{T}\mathbf{H}\mathbf{E}\mathbf{N}\mathbf{-}\mathbf{E}\mathbf{L}\mathbf{S}\mathbf{E}\mathbf{âŸ\copyright }\mathbf{,}\mathbf{⟨}\mathbf{A}\mathbf{S}\mathbf{S}\mathbf{I}\mathbf{G}\mathbf{N}\mathbf{âŸ\copyright }\mathbf{\}}\end{array}$

G is a natural-looking grammar for a fragment of a programming language, but G is ambiguous.

a. Show that G is ambiguous.

b. Give a new unambiguous grammar for the same language

Give unambiguous CFGs for the following languages.

a. {$\mathbf{w}$ | in every prefix of w the number of a’s is at least the number of b’s}

b. {$\mathbf{w}$ | the number of a’s and the number of b’s in w are equal}

c. { $\mathbf{w}$| the number of a’s is at least the number of b’s in w}?

Show that the language A is inherently ambiguous.

$\mathbf{A}\mathbf{=}\mathbf{\left\{}{\mathbf{a}}^{\mathbf{i}}{\mathbf{b}}^{\mathbf{j}}{\mathbf{c}}^{\mathbf{k}}\mathbf{\right|}\mathbf{i}\mathbf{=}\mathbf{j}\mathbf{\text{ }}\mathbf{or}\mathbf{\text{ â¶ÄŠ}}\mathbf{j}\mathbf{=}\mathbf{k}\mathbf{\text{ â¶ÄŠ}}\mathbf{where}\mathbf{\text{ â¶ÄŠ}}\mathbf{i}\mathbf{,}\mathbf{j}\mathbf{,}\mathbf{k}\mathbf{â‰\yen }\mathbf{0}\mathbf{\}}$

1. Use the languages $\mathit{A}\mathbf{=}\mathbf{\{}{\mathbf{a}}^{\mathbf{m}}{\mathbf{b}}^{\mathbf{n}}{\mathbf{c}}^{\mathbf{n}}\mathbf{}\overline{)\mathbf{}\mathbf{m}\mathbf{,}\mathbf{n}\mathbf{â‰\yen }\mathbf{0}}\mathbf{\}}$and $\mathit{B}\mathbf{=}\mathbf{\{}{\mathbf{a}}^{\mathbf{n}}{\mathbf{b}}^{\mathbf{n}}{\mathbf{c}}^{\mathbf{m}}\mathbf{}\overline{)\mathbf{}\mathbf{m}\mathbf{,}\mathbf{n}\mathbf{â‰\yen }\mathbf{0}}\mathbf{\}}$together with Example 2.36 to show that the class of context-free languages is not closed under intersection.
2. Use part (a) and DeMorgan’s law (Theorem 0.20) to show that the class of context-free languages is not closed under complementation.

Use the pumping lemma to show that the following languages are not context free.

a. $\mathbf{\left\{}{\mathbf{0}}^{\mathbf{n}}{\mathbf{1}}^{\mathbf{n}}{\mathbf{0}}^{\mathbf{n}}{\mathbf{1}}^{\mathbf{n}}\mathbf{|}\mathbf{n}\mathbf{â‰\yen }\mathbf{0}\mathbf{\right\}}$

b.$\mathbf{\{}{\mathbf{0}}^{\mathbf{n}}\mathbf{\#}{\mathbf{0}}^{\mathbf{2}\mathbf{n}}\mathbf{\#}{\mathbf{0}}^{\mathbf{3}\mathbf{n}}\mathbf{|}\mathbf{n}\mathbf{â‰\yen }\mathbf{0}\mathbf{\}}$

c. role="math" localid="1659701399563" $\mathbf{\left\{}\mathbf{w}\mathbf{\#}\mathbf{t}\mathbf{|}\mathbf{w}\mathbf{}\mathbf{\text{ â¶ÄŠ}}\mathbf{is}\mathbf{}\mathbf{\text{ â¶ÄŠ}}\mathbf{a}\mathbf{\text{ â¶ÄŠ}}\mathbf{substring}\mathbf{}\mathbf{\text{ â¶ÄŠâ€Š}}\mathbf{of}\mathbf{}\mathbf{\text{ â¶ÄŠ}}\mathbf{t}\mathbf{,}\mathbf{where}\mathbf{\text{ â¶ÄŠ}}\mathbf{w}\mathbf{,}\mathbf{t}\mathbf{∈}{\mathbf{(}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{)}}^{\mathbf{*}}\mathbf{\right\}}$

d. $\mathbf{\left(}{\mathbf{t}}_{\mathbf{1}}{\mathbf{\#}}_{\mathbf{2}}\mathbf{\#}\mathbf{.}\mathbf{.....}\mathbf{\#}{\mathbf{t}}_{\mathbf{k}}\mathbf{|}\mathbf{k}\mathbf{â‰\yen }\mathbf{2}\mathbf{,}\mathbf{each}\mathbf{\text{ â¶Ä‰}}{\mathbf{t}}_{\mathbf{i}}\mathbf{∈}{\mathbf{\{}\mathbf{a}\mathbf{,}\mathbf{b}\mathbf{\}}}^{\mathbf{*}}\mathbf{and}\mathbf{\text{ â¶ÄŠ}}\mathbf{t}\mathbf{=}\mathbf{t}\mathbf{\text{ â¶ÄŠ}}\mathbf{for}\mathbf{\text{ â¶ÄŠ}}\mathbf{some}\mathbf{\text{ â¶ÄŠ}}\mathbf{i}\mathbf{≠}\mathbf{j}\mathbf{\right)}$

Let $\mathbf{∑}\mathbf{=}\mathbf{\{}\mathbf{1}\mathbf{,}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{4}\mathbf{\}}$and$\mathbf{C}\mathbf{=}\mathbf{\{}\mathbf{w}\mathbf{∈}\mathbf{Î\pounds }\mathbf{*}\mathbf{|}\mathbf{in}\mathbf{}\mathbf{w}\mathbf{,}$ the number of 1s equals the number of 2s, and the number of 3s equals the number of 4s} Show that$\mathbf{C}$is not context free.

Show that F = { ${\mathbf{a}}^{\mathbf{i}}{\mathbf{b}}^{\mathbf{j}}\mathbf{|}\mathbf{i}\mathbf{=}\mathbf{kj}$for some positive integer$\mathbf{k}$ } is not context free