User:Msiddalingaiah/NFA to CFG

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The following context-free_grammar describes the same language as the regular expression (a|b)*abb:

$A_0 \rightarrow a A_0 \quad | \quad b A_0 \quad | \quad a A_1$

$A_1 \rightarrow b A_2$

$A_2 \rightarrow b A_3$

$A_3 \rightarrow \epsilon$

From page 178 of Aho, Sethi, Ullman

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