Understanding the Pushdown Automaton and its Language

A Pushdown Automaton (PDA) is a type of automaton that uses a stack in addition to finite state control. It is used to recognize context-free languages. The behavior of a PDA is determined by its states, input alphabet, stack alphabet, transition function, initial state, initial stack symbol, and set of final states.

Components of the Pushdown Automaton M

The given Pushdown Automaton \(M\) is defined as \(M = (Q, \Sigma, \Gamma, \delta, q_0, Z_0, F)\), where:

• States \(Q = \{q_0, q_1, q_2\}\)
• Input Alphabet \(\Sigma = \{a, b\}\)
• Stack Alphabet \(\Gamma = \{a, b, z\}\)
• Transition Function \(\delta\) (defined below)
• Initial State \(q_0\)
• Initial Stack Symbol \(Z_0 = z\)
• Final States \(F = \{q_2\}\)

Analysis of Pushdown Automaton Transitions

Let's examine the transition function \(\delta\) to understand how the PDA processes input strings. The transitions are given as:

• \(\delta(q_0, a, a) = \{(q_0, aa)\}\)
• \(\delta(q_0, b, a) = \{(q_0, ba)\}\)
• \(\delta(q_0, a, b) = \{(q_0, ab)\}\)
• \(\delta(q_0, b, b) = \{(q_0, bb)\}\)
• \(\delta(q_0, a, z) = \{(q_0, az)\}\)
• \(\delta(q_0, b, z) = \{(q_0, bz)\}\)
• \(\delta(q_0, \lambda, b) = \{(q_1, b)\}\)
• \(\delta(q_0, \lambda, a) = \{(q_1, a)\}\)
• \(\delta(q_1, a, a) = \{(q_1, \lambda)\}\)
• \(\delta(q_1, b, b) = \{(q_1, \lambda)\}\)
• \(\delta(q_1, \lambda, z) = \{(q_2, z)\}\)

Let's interpret these transitions:

State \(q_0\): Pushing phase

The transitions in state \(q_0\) with input symbols (\(\delta(q_0, \text{symbol}, \text{stack top})\)):

• If the input is 'a' and stack top is 'a', replace 'a' with 'aa' (push 'a').
• If the input is 'b' and stack top is 'a', replace 'a' with 'ba' (push 'b').
• If the input is 'a' and stack top is 'b', replace 'b' with 'ab' (push 'a').
• If the input is 'b' and stack top is 'b', replace 'b' with 'bb' (push 'b').
• If the input is 'a' and stack top is 'z', replace 'z' with 'az' (push 'a' onto 'z').
• If the input is 'b' and stack top is 'z', replace 'z' with 'bz' (push 'b' onto 'z').

These transitions effectively push the current input symbol onto the stack, regardless of the stack top (as long as it's not the initial 'z', where it pushes on top of 'z'). The state remains \(q_0\). This state is used to process the first part of the input string and store it on the stack, effectively in reverse order (last pushed symbol is at the top).

Transition from \(q_0\) to \(q_1\): Midpoint transition

The transitions \(\delta(q_0, \lambda, b) = \{(q_1, b)\}\) and \(\delta(q_0, \lambda, a) = \{(q_1, a)\}\) are \(\lambda\)-transitions (consuming no input). They allow the PDA to move from state \(q_0\) to state \(q_1\) non-deterministically if the stack top is 'a' or 'b'. The stack top symbol ('a' or 'b') remains on the stack. This transition point marks the boundary between the first part of the string (processed in \(q_0\)) and the second part (to be processed in \(q_1\)). Since the stack top must be 'a' or 'b' for this transition, it implies that at least one symbol must have been pushed onto the stack before this transition, meaning the first part of the string must be non-empty.

State \(q_1\): Popping/Matching phase

The transitions in state \(q_1\) with input symbols:

• \(\delta(q_1, a, a) = \{(q_1, \lambda)\}\): If input is 'a' and stack top is 'a', pop the stack.
• \(\delta(q_1, b, b) = \{(q_1, \lambda)\}\): If input is 'b' and stack top is 'b', pop the stack.

In state \(q_1\), the PDA reads the input symbol and matches it against the top of the stack. If they match, the stack symbol is popped. This state is used to process the second part of the input string, comparing it against the symbols previously pushed onto the stack in \(q_0\).

Transition from \(q_1\) to \(q_2\): Acceptance transition

The transition \(\delta(q_1, \lambda, z) = \{(q_2, z)\}\) is a \(\lambda\)-transition. It allows the PDA to move from state \(q_1\) to the final state \(q_2\) when the stack top is 'z'. This indicates that all symbols pushed onto the stack in \(q_0\) have been successfully matched and popped in \(q_1\), and the initial stack symbol 'z' is now at the top.

Identifying the Accepted Language

Let the input string be \(w\). The PDA works by splitting \(w\) into two parts, say \(w = uv\). The first part, \(u\), is processed in state \(q_0\). For every symbol read in \(u\), that symbol is pushed onto the stack. If \(u = s_1 s_2 \dots s_k\), after processing \(u\), the stack (on top of the initial \(z\)) will contain \(s_k s_{k-1} \dots s_1\).

The PDA then transitions to state \(q_1\) via a \(\lambda\)-transition, which occurs at the boundary between \(u\) and \(v\). This transition requires the stack top to be 'a' or 'b', implying \(u\) must be non-empty.

The second part, \(v\), is processed in state \(q_1\). For each symbol in \(v\), the PDA expects the top of the stack to be the same symbol and pops it. If \(v = t_1 t_2 \dots t_m\), the PDA will successfully process \(v\) if and only if \(t_1\) matches the current stack top (which is \(s_k\)), \(t_2\) matches the next stack top (which is \(s_{k-1}\)), and so on, until the entire \(v\) is read and all symbols \(s_k, \dots, s_1\) are popped from the stack, leaving \(z\) at the top. This condition is met if and only if \(m=k\) and \(t_i = s_{k-i+1}\) for all \(1 \le i \le k\). This is the definition of \(v = u^R\).

After processing \(v\) successfully, the stack top is \(z\). The PDA transitions from \(q_1\) to the final state \(q_2\) using the \(\lambda\)-transition on \(z\).

Therefore, the PDA accepts strings \(w\) that can be split into \(w=uv\) such that \(u\) is processed in \(q_0\), \(v\) is processed in \(q_1\), \(v=u^R\), and \(u\) is non-empty (since the \(q_0 \to q_1\) transition requires stack top 'a' or 'b').

The language accepted is thus the set of strings of the form \(uu^R\) where \(u\) is a non-empty string from \(\{a, b\}^{+}\).

Comparing with Options

Let's compare our deduced language with the given options:

The language accepted is \(L = \{uu^R \mid u \in \{a, b\}^{+}\}\).

Option 1: \(L = \{w \mid n_a(w) = n_b(w), w \in \{a, b\}^{+}\}\). This language contains strings with an equal number of 'a's and 'b's, like 'aabb', 'abab'. Our PDA does not count symbol occurrences.

Option 2: \(L = \{w \mid n_a(w) \le n_b(w), w \in \{a, b\}^{+}\}\). This language contains strings where the count of 'a's is less than or equal to the count of 'b's. Our PDA does not count symbol occurrences.

Option 3: \(L = \{w \mid n_b(w) \le n_a(w), w \in \{a, b\}^{+}\}\). This language contains strings where the count of 'b's is less than or equal to the count of 'a's. Our PDA does not count symbol occurrences.

Option 4: \(L = \{ww^R \mid w \in \{a, b\}^{+}\}\). This language contains strings formed by a non-empty string \(w\) followed by its reverse \(w^R\). This matches our deduction exactly.

Thus, the language accepted by the given Pushdown Automaton is \(L = \{ww^R \mid w \in \{a, b\}^{+}\}\).

Revision Table: Pushdown Automata Concepts

Additional Information: Pushdown Automata and Context-Free Languages

Pushdown Automata (PDAs) are more powerful than Finite Automata because of their stack memory. This stack allows PDAs to recognize context-free languages (CFLs), which finite automata cannot. CFLs include languages like \(\{a^n b^n \mid n \ge 0\}\) or languages involving matching parentheses, or palindromes. The stack provides the memory needed to keep track of nested structures or to remember the first part of a string to compare it with the reverse of the second part, as seen in this problem.

Languages accepted by PDAs are exactly the set of context-free languages. There are two common ways for a PDA to accept a string:

1. Acceptance by final state: As in this problem, the PDA reaches a designated final state after reading the entire input string. The stack content at the end might or might not matter, depending on the specific definition used (sometimes required to be empty). In this case, reaching the final state \(q_2\) with input consumed is the acceptance condition.
2. Acceptance by empty stack: The PDA accepts a string if it reaches a configuration with an empty stack after reading the entire input, regardless of the current state.

For any language accepted by final state, there exists a PDA that accepts the same language by empty stack, and vice versa. The PDA in this problem uses acceptance by final state.

The language \(L = \{ww^R \mid w \in \{a, b\}^{+}\}\) is a standard example of a context-free language that is not regular. The stack is crucial for remembering the first part \(w\) so that the second part \(w^R\) can be matched against its reverse.

The structure \(ww^R\) represents non-empty even-length palindromes. Examples include "aa", "bb", "abba", "baab", "aabbaa", etc. The language \(\{w w^R \mid w \in \Sigma^*\}\) would include the empty string \(\lambda\) (\(\lambda\lambda^R = \lambda\)), corresponding to even-length palindromes including the empty string. The '+' in \(\{a, b\}^{+}\) means \(w\) must be non-empty, resulting in only non-empty even-length palindromes.