Question

Let $P=\left\{a,b\right\}.$Give a CFG generating the language of strings with twice as many$a\text{'}sasb\text{'}s$ . Prove that your grammar is correct.

Short answer

$P=\left\{a,b\right\}$,generating the language of strings with twice as many$a\text{'}sasb\text{'}s$.

Thecontext free grammar is,

$$\begin{gathered} S_O\to S_{1}aab\left|a{S}_{1}ab\right|aab{S}_1\left|S_{1}aba\right|a{S}_{1}ba\left|aba{S}_1\right| \\ {S}_{1}baa\left|b{S}_{1}aa\right|ba{S}_{1}a|baa{S}_1 \\ {S}_1\to S_0|\epsilon \end{gathered}$$

The generated language is proved correct using induction.

Step-by-step solution

Define Context free grammar

The context free language is generated by context free grammar.

These languages are accepted by Pushdown Automata. These are the superset of regular languages.

Consider context-free languages $L_1$described as$G_1=(V_1,S,R_1,S_1).$

Consider context-free language $L_2$described as$G_2=(V_2,S,R_2,S_2).$

Determine the grammar generating language with twice as many a’s as b’s

$P=\left\{a,b\right\}$,generating the language of strings with twice as many$a\text{'}sasb\text{'}s$ .

Thecontext free grammar is,

$$\begin{gathered} S_O\to S_{1}aab\left|a{S}_{1}ab\right|aab{S}_1\left|S_{1}aba\right|a{S}_{1}ba\left|aba{S}_1\right| \\ {S}_{1}baa\left|b{S}_{1}aa\right|ba{S}_{1}a|baa{S}_1 \\ {S}_1\to S_0|\epsilon \end{gathered}$$

Prove that the language generated is correct

Smallest strings possible are: $x_0\in \{aab,aba,baa\}$

all of which have$NA\left(x_o\right)=2NB\left(x_o\right),$$NA\left(x_o\right)=2NB\left(x_o\right),$

where ${\mathrm{N}}_A\left(x\right)$gives the number of a’s in string $x$and $N_B\left(x\right)$gives the number of b’s in string $x$. Assume $NA\left(x_n\right)=2NB\left(x_n\right),$holds. Show if$n$ is true $n+1$is also true, where $x_{n+1}$is string $x_n$with substring$s\in \left\{\epsilon ,aab,aba,baa\right\}$ inserted. Subsequent insertions of S₁ into strings produced, either add zero a’s and zero b’s or two a’s and one b .

$P=\left\{a,b\right\}$,generating the language of strings with twice as many$a\text{'}sasb\text{'}s$.

Thecontext free grammar is,

$$\begin{gathered} S_O\to S_{1}aab\left|a{S}_{1}ab\right|aab{S}_1\left|S_{1}aba\right|a{S}_{1}ba\left|aba{S}_1\right| \\ {S}_{1}baa\left|b{S}_{1}aa\right|ba{S}_{1}a|baa{S}_1 \\ {S}_1\to S_0|\epsilon \end{gathered}$$

The generated language is proved correct using induction.

Case when zero a’s and zero b’s are inserted.

$$\begin{gathered} N_A\left(x_{n+1}\right)=N_A\left(x_n\right)+0 \\ {N}_B\left(x_{n+1}\right)=N_B\left(x_n\right)+0 \\ {N}_A\left(x_{n+1}\right)=2N_B\left(x_{n+1}\right) \end{gathered}$$

Case when two a’s and one b are inserted.

$$\begin{gathered} 1NA\left(x_{n+1}\right)=NA\left(x_n\right)+2 \\ {N}_B\left(x_{n+1}\right)=N_B\left(x_n\right)+1 \\ {N}_A\left(x_n\right)+2=2\left(N_B\left(x_n\right)+1\right) \\ {N}_A\left(x_n\right)=2NB\left(x_n\right) \end{gathered}$$

Therefore, all strings generated using the grammar contain twice as many a’s as b’s.