91影视

A homomorphism is function $\mathrm{h}:{\mathrm{危}}^{*}鈫�{\mathrm{螖}}^{*}$defined as follows:

鈥� $\mathrm{h}(鈭�)=鈭�$and for $\mathrm{a}鈭�\mathrm{危},\mathrm{h}\left(\mathrm{a}\right)$is any string in ${\mathrm{螖}}^{*}$.

鈥� For $\mathrm{a}={\mathrm{a}}_1{\mathrm{a}}_2...{\mathrm{a}}_{\mathrm{n}}鈭�\mathrm{危}*(\mathrm{n}鈮�2),\mathrm{h}\left(\mathrm{a}\right)=\mathrm{h}\left({\mathrm{a}}_1\right)\mathrm{h}\left({\mathrm{a}}_2\right)...\mathrm{h}\left({\mathrm{a}}_{\mathrm{n}}\right).$

鈥� A homomorphism h maps a string$\mathrm{a}鈭�\mathrm{危}$ * to a string in${\mathrm{螖}}^{*}$ by mapping each character of a to a string $\mathrm{h}\left(\mathrm{a}\right)$鈭�${\mathrm{螖}}^{*}$.

鈥� A homomorphism is a function from strings to strings that 鈥渞espects鈥� concatenation: for any$\mathrm{x},\mathrm{y}鈭�\mathrm{危}*,\mathrm{h}\left(\mathrm{xy}\right)=\mathrm{h}\left(\mathrm{x}\right)\mathrm{h}\left(\mathrm{y}\right).$ (Any such function is a homomorphism.)

a).

Regular languages are closed under homomorphism, i.e., if $\mathrm{L}$is a regular language and $\mathrm{h}$is a homomorphism,

then$\mathrm{h}\left(\mathrm{L}\right)$is also regular.

Proof:

Use the representation of regular languages in terms of regular expressions to argue this.

鈥� Define homomorphism as an operation on regular expressions

鈥� Show that $\mathrm{L}\left(\mathrm{h}\left(\mathrm{R}\right)\right)=\mathrm{h}\left(\mathrm{L}\left(\mathrm{R}\right)\right)$

鈥� Let $\left(\mathrm{R}\right)$be such that $\mathrm{L}=\mathrm{L}\left(\mathrm{R}\right).\mathrm{Let}\mathrm{R}0=\mathrm{h}\left(\mathrm{R}\right).\mathrm{Then}\mathrm{h}\left(\mathrm{L}\right)=\mathrm{L}\left(\mathrm{R}\right).$

鈥�$\mathrm{h}(鈭�)=鈭�$ and$\mathrm{for}$$\mathrm{a}鈭�\mathrm{危},\mathrm{h}\left(\mathrm{a}\right)$ is any string in${\mathrm{螖}}^{*}$ .

鈥� For $\mathrm{a}={\mathrm{a}}_1{\mathrm{a}}_2...{\mathrm{a}}_{\mathrm{n}}鈭�\mathrm{危}*(\mathrm{n}鈮�2),\mathrm{h}\left(\mathrm{a}\right)=\mathrm{h}\left({\mathrm{a}}_1\right)\mathrm{h}\left({\mathrm{a}}_2\right)...\mathrm{h}\left({\mathrm{a}}_{\mathrm{n}}\right).$

鈥� A homomorphism h maps a string $\mathrm{a}鈭�\mathrm{危}$* to a string in${\mathrm{螖}}^{*}$ by mapping each character of a to a string $\mathrm{h}\left(\mathrm{a}\right)$鈭�${\mathrm{螖}}^{*}$.

鈥� A homomorphism is a function from strings to strings that 鈥渞espects鈥� concatenation: for any $\mathrm{x},\mathrm{y}鈭�\mathrm{危}*,\mathrm{h}\left(\mathrm{xy}\right)=\mathrm{h}\left(\mathrm{x}\right)\mathrm{h}\left(\mathrm{y}\right).$(Any such function is a homomorphism.)

Hence, the class of regular languages is closed under homomorphism is proved.

b).

鈥� Show that $\mathrm{L}\left(\mathrm{h}\left(\mathrm{R}\right)\right)=\mathrm{h}\left(\mathrm{L}\left(\mathrm{R}\right)\right)$

鈥� Let $\left(\mathrm{R}\right)$be such that $\mathrm{L}=\mathrm{L}\left(\mathrm{R}\right).\mathrm{Let}\mathrm{R}0=\mathrm{h}\left(\mathrm{R}\right).\mathrm{Then}\mathrm{h}\left(\mathrm{L}\right)=\mathrm{L}\left(\mathrm{R}\right).$

鈥� $\mathrm{h}(鈭�)=鈭�$and for $\mathrm{a}鈭�\mathrm{危},\mathrm{h}\left(\mathrm{a}\right)$ is any string in ${\mathrm{螖}}^{*}$.

鈥� For $\mathrm{a}={\mathrm{a}}_1{\mathrm{a}}_2...{\mathrm{a}}_{\mathrm{n}}鈭�\mathrm{危}*(\mathrm{n}鈮�2),\mathrm{h}\left(\mathrm{a}\right)=\mathrm{h}\left({\mathrm{a}}_1\right)\mathrm{h}\left({\mathrm{a}}_2\right)...\mathrm{h}\left({\mathrm{a}}_{\mathrm{n}}\right).$

鈥� A homomorphism h maps a string $\mathrm{a}鈭�\mathrm{危}$* to a string in${\mathrm{螖}}^{*}$ by mapping each character of a to a string $\mathrm{h}\left(\mathrm{a}\right)$鈭� ${\mathrm{螖}}^{*}$.

鈥� A homomorphism is a function from strings to strings that 鈥渞espects鈥� concatenation: for any$\mathrm{x},\mathrm{y}鈭�\mathrm{危}*,\mathrm{h}\left(\mathrm{xy}\right)=\mathrm{h}\left(\mathrm{x}\right)\mathrm{h}\left(\mathrm{y}\right).$ (Any such function is a homomorphism.)

These properties does not follows by non-regular language so, it is not closed under homomorphism.

Hence, the class of non-regular languages is not closed under homomorphism.