Designing a Context-Free Grammar and PDA for a Language
Can you provide a context-free grammar for the language L={a^nb^m, n=m+3}, and design a Pushdown Automaton (PDA) that accepts it?
Context-Free Grammar for L={a^nb^m, n=m+3}:
Production rules:
- S -> aaaSB
- SB -> aSBb | ε
PDA for L={a^nb^m, n=m+3}:
States: q0, q1, q2, q3, q4Start state: q0
Accept state: q4
Stack symbol: Z0 (initial stack symbol)
Transition rules:
- q0, ε, Z0 -> q1, Z0 (Push 3 Z0 onto the stack)
- q1, a, Z0 -> q1, aZ0 (Push a onto the stack for each input a)
- q1, b, a -> q2, ε (Pop a from the stack for each input b)
- q2, b, a -> q2, ε (Pop a from the stack for each input b)
- q2, ε, Z0 -> q3, Z0 (Check if the stack is empty)
- q3, b, Z0 -> q3, ε (Pop Z0 from the stack for each remaining input b)
- q3, ε, Z0 -> q4, Z0 (Accept if the stack is empty) Answer (100-110 words):
The language L={a^nb^m, n=m+3} consists of strings that start with n 'a's followed by m 'b's, where n is equal to m plus 3. A context-free grammar for this language can be defined with a start symbol S and two production rules: S -> aaaSB and SB -> aSBb | ε. This grammar generates strings by adding three 'a's in the beginning and then appending 'b's after an arbitrary number of 'a's. A Pushdown Automaton (PDA) can be designed to accept this language by using five states and appropriate transition rules. The PDA starts with an empty stack, pushes three symbols onto the stack for every 'a', and pops 'a's for each 'b'. It checks if the stack is empty at the end to ensure the number of 'a's and 'b's are balanced, accepting the language if the stack is empty.