Context-Free Grammars: Exploring Language Patterns
Can you find context-free grammars for the following languages?
(a) L={anbm:n≤m+3}
(b) L={anbm:n=m−1}
(c) L={anbm:n=2m}
(d) L={anbm:2n≤m≤3n}
(e) L={w∈{a,b}∗:na(w)=nb(w)}
(f) L={w∈{a,b}∗:na(v)≥nb(v), where v is any prefix of w}
(g) L={w∈{a,b}∗:na(w)=2nb(w)+1}
Answer:
In summary, the context-free grammars for the given languages are as follows:
(a) S → aSb | aSbb | ε
(b) S → aSb | ε
(c) S → aSb | aSbb | aaSbb | ε
(d) S → aaSbb | aSbbb | ε
(e) S → aSb | bSa | aS | bS | ε
(f) S → aSb | aSS | ε
(g) S → aSbb | aSb | ε
Explanation:
In order to find the context-free grammars for the given languages, we need to define the production rules for each language. Let's go through each language one by one:
(a) L={anbm:n≤m+3}
The production rules for this language can be defined as:
S → aSb | aSbb | ε
(b) L={anbm:n=m−1}
The production rules for this language can be defined as:
S → aSb | ε
(c) L={anbm:n=2m}
The production rules for this language can be defined as:
S → aSb | aSbb | aaSbb | ε
(d) L={anbm:2n≤m≤3n}
The production rules for this language can be defined as:
S → aaSbb | aSbbb | ε
(e) L={w∈{a,b}∗:na(w)=nb(w)}
The production rules for this language can be defined as:
S → aSb | bSa | aS | bS | ε
(f) L={w∈{a,b}∗:na(v)≥nb(v), where v is any prefix of w}
The production rules for this language can be defined as:
S → aSb | aSS | ε
(g) L={w∈{a,b}∗:na(w)=2nb(w)+1}
The production rules for this language can be defined as:
S → aSbb | aSb | ε