Construct a PDA that recognizes the following language: {x\in {a,b}* : #a(x) = #b(x)} Hint: To get the maximum number of points, use as few states and nonterminals as possible! Alphabet: (a, b} o Stack alphabet (the first symbol is the initial one): Z Y X o Acceptance condition: empty stack Deterministic (DPDA): false Hints apply +

Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
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Construct a PDA that recognizes the following language:
{x\in {a,b}* : #a(x) = #b(x)}
Hint: To get the maximum number of points, use as few states and nonterminals as possible!
Alphabet: (a, b)
o Stack alphabet (the first symbol is the initial one): Z Y X
o Acceptance condition: empty stack
Deterministic (DPDA): false
Hints
apply
+
Transcribed Image Text:Construct a PDA that recognizes the following language: {x\in {a,b}* : #a(x) = #b(x)} Hint: To get the maximum number of points, use as few states and nonterminals as possible! Alphabet: (a, b) o Stack alphabet (the first symbol is the initial one): Z Y X o Acceptance condition: empty stack Deterministic (DPDA): false Hints apply +
Expert Solution
Step 1

PDA for Number of a = Number of b

a) While scanning the input, if 'a' comes first we push it onto the stack.
Also if after 'a', again the letter 'a' comes then push it again.

b) If 'b' comes first, then we push it in the STACK (Assumption -> 'a' did not come in the stack yet)
Similar to previos step, if again 'b' comes then push it in the STACK.

c) Now if 'a' is present at the top of STACK and 'b' comes next then, pop 'a'
Similarily if 'b' is present in top of STACK and 'a' comes next then, pop 'b'

So after the complete string is scanned to the end, if nothing is left in the STACK (empty stack) then we can say that it is accepted by the PDA.

→ PUSH
la, z/az)
(b, Z/bz)
1
FINAL STATE
(a,ble) (,alaa)
(b, ale) (b, b/bb)
POP

Stack Transitions ->

δ(0, a, Z) = (0, aZ)     
δ(0, a, a) = (0, aa)     
δ(0, b, Z) = (0, bZ)     
δ(0, b, b) = (0, bb)     
δ(0, b, a) = (0, ε)     
δ(0, a, b) = (0, ε)
δ(0, ε, Z) = (1, Z)  
steps

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