Pushdown Automata (PDA) - Transition Function, Acceptance, and CFG Conversion

1. Transition Function of PDA

A Pushdown Automaton (PDA) is defined as a 7-tuple:

$$PDA = (Q, \Sigma, \Gamma, \delta, q_0, Z_0, F)$$

Where:

• $Q$ = finite set of states
• $\Sigma$ = input alphabet
• $\Gamma$ = stack alphabet
• $\delta$ = transition function
• $q_0$ = initial state
• $Z_0$ = initial stack symbol
• $F$ = set of final/accepting states

Transition Function $\delta$:

$$\delta : Q \times (\Sigma \cup {\varepsilon}) \times \Gamma \rightarrow \mathcal{P}(Q \times \Gamma^*)$$

Interpretation: Given a current state $q \in Q$, a current input symbol $a \in \Sigma$ (or $\varepsilon$), and the top of stack symbol $X \in \Gamma$, the transition function returns a set of pairs $(p, \gamma)$ where:

• $p$ is the new state
• $\gamma \in \Gamma^*$ is the string that replaces $X$ on the stack

Notation: $\delta(q, a, X) = {(p, \gamma), \ldots}$

2. Two Ways PDA Accepts a String

A PDA can accept a string in two ways:

Way 1: Acceptance by Final State

• A string $w$ is accepted if, after reading the entire input $w$, the PDA is in a final (accepting) state $q \in F$.
• The stack content does not matter at the time of acceptance.
• Formally: $(q_0, w, Z_0) \vdash^* (q_f, \varepsilon, \gamma)$ for some $q_f \in F$ and $\gamma \in \Gamma^*$

Way 2: Acceptance by Empty Stack

• A string $w$ is accepted if, after reading the entire input $w$, the stack becomes empty.
• The state does not need to be a final state.
• Formally: $(q_0, w, Z_0) \vdash^* (q, \varepsilon, \varepsilon)$ for any state $q \in Q$

Note: Both methods are equivalent in power; any language accepted by one method can be accepted by the other.

3. Converting the Given CFG to PDA

Given Grammar:

$$S \rightarrow AS \mid \varepsilon$$ $$A \rightarrow Ab \mid Bb \mid ab$$ $$B \rightarrow \text{(not explicitly given, assumed terminal or derived)}$$

Note: Variable $B$ appears in $A \rightarrow Bb$ but no production for $B$ is given. We include it as-is and handle it in the PDA transitions. It is likely a terminal or a typo for a variable; we treat $B$ as a variable with no further productions (or it may be a terminal symbol in some interpretations). We proceed with the given grammar.

Method: CFG to PDA (Acceptance by Empty Stack)

Standard Construction:

For any CFG, we construct a PDA with:

• States: ${q_0, q_1, q_2}$ (typically three states: $q_0$ to push start symbol, $q_1$ as the main loop state, $q_2$ as accepting state)
• Or more commonly, a single-state PDA ${q}$ is used for simplicity.

Rules for conversion:

Step-by-Step PDA Construction

PDA: $M = ({q_0, q_1}, {a, b}, {S, A, B, a, b, Z_0}, \delta, q_0, Z_0, {q_1})$

Initialization transitions:

$$\delta(q_0, \varepsilon, Z_0) = {(q_1, SZ_0)}$$

This pushes the start symbol $S$ onto the stack and moves to the working state $q_1$.

Production Rules $\rightarrow$ PDA Transitions

For variable $S$:

So: $\delta(q_1, \varepsilon, S) = {(q_1, AS),\ (q_1, \varepsilon)}$

For variable $A$:

So: $\delta(q_1, \varepsilon, A) = {(q_1, Ab),\ (q_1, Bb),\ (q_1, ab)}$

For variable $B$: the grammar gives no production for $B$, so there is no transition $\delta(q_1, \varepsilon, B)$. Any derivation that uses $A \to Bb$ can therefore never remove $B$ from the stack and never finishes, so that alternative contributes no strings to the language; in effect only $A \to Ab \mid ab$ matter for what the PDA actually accepts.

Terminal matching transitions (match input symbol against stack top, then pop both):

$$\delta(q_1, a, a) = {(q_1, \varepsilon)}$$ $$\delta(q_1, b, b) = {(q_1, \varepsilon)}$$

Acceptance by empty stack:

$$\delta(q_1, \varepsilon, Z_0) = {(q_1, \varepsilon)}$$

Once $Z_0$ itself is popped with no more input to read, the stack is empty and the string is accepted.

Complete PDA

$$M = ({q_0, q_1}, {a, b}, {S, A, B, a, b, Z_0}, \delta, q_0, Z_0, {q_1})$$

with transition function

$$\delta(q_0, \varepsilon, Z_0) = {(q_1, SZ_0)}$$ $$\delta(q_1, \varepsilon, S) = {(q_1, AS),\ (q_1, \varepsilon)}$$ $$\delta(q_1, \varepsilon, A) = {(q_1, Ab),\ (q_1, ab)}$$ $$\delta(q_1, a, a) = {(q_1, \varepsilon)}$$ $$\delta(q_1, b, b) = {(q_1, \varepsilon)}$$ $$\delta(q_1, \varepsilon, Z_0) = {(q_1, \varepsilon)}$$

($B$ is kept in the stack alphabet only because it appears in the given grammar; since it has no production, it never actually gets pushed by any transition that leads to acceptance.)

Sample Acceptance Trace for $w = ab$

$$(q_0,\ ab,\ Z_0) \vdash (q_1,\ ab,\ SZ_0) \vdash (q_1,\ ab,\ ASZ_0) \vdash (q_1,\ ab,\ abSZ_0)$$ $$\vdash (q_1,\ b,\ bSZ_0) \vdash (q_1,\ \varepsilon,\ SZ_0) \vdash (q_1,\ \varepsilon,\ Z_0) \vdash (q_1,\ \varepsilon,\ \varepsilon)$$

The input is fully consumed and the stack is empty, so $w = ab$ is accepted, both by the empty-stack criterion and by the final-state criterion (since $q_1 \in F$). This completes the construction of a PDA equivalent to the given CFG.

Regular Operators and DFA Minimization Using Table Filling Algorithm

Part 1: Two Regular Operators

Regular operators are the operations used to construct regular expressions from smaller ones. Any two of the following are acceptable:

1. Union (| or +): If $R$ and $S$ are regular expressions, then $R + S$ denotes the set of strings belonging to either $R$ or $S$.

   • Example: $(a \mid b)$ denotes ${a, b}$

2. Concatenation (.): If $R$ and $S$ are regular expressions, then $RS$ denotes strings formed by a string of $R$ followed by a string of $S$.

   • Example: $ab$ denotes ${ab}$

3. Kleene Star (*): If $R$ is a regular expression, then $R^{*}$ denotes zero or more concatenations of $R$.

   • Example: $a^{*}$ denotes ${\varepsilon, a, aa, aaa, \ldots}$

Part 2: DFA Minimization Using Table Filling

Given Data (extracted transition table)

• Start state: $A$
• Final state(s): ${F}$

Note: The problem statement references a machine whose diagram/table is not fully reproduced in the prompt. The solution uses the transition table extracted above. If the intended machine differs, the extracted table above is what this solution assumes.

Step 1: Initialize the triangular table

Mark every pair where exactly one state is final. Final = ${F}$.

Marked pairs (one is $F$): $(A,F), (B,F), (C,F), (D,F), (E,F)$.

B | _ |
C | _ | _ |
D | _ | _ | _ |
E | _ | _ | _ | _ |
F | X | X | X | X | X |
  | A | B | C | D | E |

Step 2: Iterative marking

For each unmarked pair, check transitions on $0$ and $1$; mark if any successor pair is marked.

Pass 1:

• $(A,B)$: $0\to(B,A)$ unmarked, $1\to(C,D)$ unmarked → unmarked
• $(A,C)$: $1\to(C,F)$ marked → MARK
• $(A,D)$: $1\to(C,F)$ marked → MARK
• $(A,E)$: $1\to(C,F)$ marked → MARK
• $(B,C)$: $0\to(A,E)$ marked → MARK
• $(B,D)$: $0\to(A,E)$ marked → MARK
• $(B,E)$: $0\to(A,E)$ marked → MARK
• $(C,D)$: $0\to(E,E)$ same, $1\to(F,F)$ same → unmarked
• $(C,E)$: $0\to(E,E)$ same, $1\to(F,F)$ same → unmarked
• $(D,E)$: $0\to(E,E)$ same, $1\to(F,F)$ same → unmarked

Pass 2 (re-check remaining unmarked pairs):

• $(A,B)$: $0\to(A,B)$ still unmarked, $1\to(C,D)$ still unmarked → remains unmarked
• $(C,D)$: both transitions to same states → remains unmarked
• $(C,E)$: both transitions to same states → remains unmarked
• $(D,E)$: both transitions to same states → remains unmarked

No further changes → algorithm terminates.

Final table:

B | _ |
C | X | X |
D | X | X | _ |
E | X | X | _ | _ |
F | X | X | X | X | X |
  | A | B | C | D | E |

Step 3: Group equivalent (unmarked) states

Unmarked pairs: $(A,B)$, $(C,D)$, $(C,E)$, $(D,E)$.

• $A \equiv B$
• $C \equiv D \equiv E$
• $F$ (alone)

Equivalence classes: $${A, B}, \quad {C, D, E}, \quad {F}$$

Step 4: Minimized DFA

Let:

• $P = {A, B}$ (start state)
• $Q = {C, D, E}$
• $R = {F}$ (final state)

Compute transitions (using any representative):

• $P$: on $0$, $\delta(A,0)=B\in P$; on $1$, $\delta(A,1)=C\in Q$
• $Q$: on $0$, $\delta(C,0)=E\in Q$; on $1$, $\delta(C,1)=F\in R$
• $R$: on $0$, $\delta(F,0)=F\in R$; on $1$, $\delta(F,1)=F\in R$

Minimized transition table:

The minimized DFA has 3 states: ${A,B}$, ${C,D,E}$, ${F}$.

Turing Machine as Enumerator and Encoding

Part 1: Turing Machine as an Enumerator

A Turing machine used as an enumerator is a variant of the standard TM equipped with a work tape and a separate output tape (printer). Instead of taking an input and accepting/rejecting it, an enumerator $E$ starts on a blank tape and generates (lists) the strings of a language.

Working principle

• $E$ starts with all tapes blank (no input is provided).
• It computes, occasionally printing a string on the output tape followed by a separator symbol #.
• The set of all strings printed constitutes the language enumerated by $E$:

$$L(E) = {, w \mid E \text{ eventually prints } w \text{ on the output tape} ,}$$

• Strings may be printed in any order and may repeat.

Key theorem

A language $L$ is recursively enumerable (Turing recognizable) if and only if some enumerator $E$ enumerates it. Hence enumerators and recognizers define the same class of languages.

Part 2: Encoding the Given Turing Machine

Given data (extracted)

• States: $Q = {q_0, q_1, q_2}$
• Input alphabet: $\Sigma = {a, b}$
• Tape alphabet: $\Gamma = {a, b, B}$
• Start state: $q_2$
• Blank: $B$
• Final states: $F$ (not explicitly listed in question)
• Input: $w = ba$
• Transitions:
  • $\delta(q_0, b) \to (q_1, b, R)$
  • $\delta(q_1, a) \to (q_2, a, R)$
  • $\delta(q_2, a) \to (q_1, a, R)$
  • $\delta(q_2, b) \to (q_2, b, L)$

Note: the set $F$ of final states is not given, so it cannot be encoded. The encoding below covers states, symbols and transitions, which is what the data supports.

Step 1: Encode states (unary in 1's)

Step 2: Encode tape symbols

Step 3: Encode directions

Step 4: Encoding scheme

Each transition $\delta(q, s) \to (q', s', D)$ is encoded (fields separated by single $0$) as:

$$\text{code}(q),0,\text{code}(s),0,\text{code}(q'),0,\text{code}(s'),0,\text{code}(D)$$

T1: $\delta(q_0, b) \to (q_1, b, R)$ $$1,0,11,0,11,0,11,0,11$$

T2: $\delta(q_1, a) \to (q_2, a, R)$ $$11,0,1,0,111,0,1,0,11$$

T3: $\delta(q_2, a) \to (q_1, a, R)$ $$111,0,1,0,11,0,1,0,11$$

T4: $\delta(q_2, b) \to (q_2, b, L)$ $$111,0,11,0,111,0,11,0,1$$

Step 5: Full encoding (transitions separated by $00$)

$$\boxed{1,0,11,0,11,0,11,0,11 ; 00 ; 11,0,1,0,111,0,1,0,11 ; 00 ; 111,0,1,0,11,0,1,0,11 ; 00 ; 111,0,11,0,111,0,11,0,1}$$

Step 6: Trace on input $w = ba$ (start state $q_2$)

Tape: $B,b,a,B$, head on first symbol $b$, state $q_2$.

• Config 1: State $q_2$, read $b$. Apply $\delta(q_2,b)\to(q_2,b,L)$: write $b$, move Left. Head now on the blank $B$ to the left of $b$.
• Config 2: State $q_2$, read $B$. There is no transition for $(q_2, B)$, so the machine halts here.

Since $F$ is not specified, whether this halting configuration is accepting cannot be determined. With the given transitions, the computation halts after one move on the left blank, having never moved right into the $a$. Thus for start state $q_2$ the input $ba$ is not processed to completion in an accepting sense.

(Remark: if the intended start state were $q_0$, the trace would be $q_0 b a \Rightarrow q_1 a \Rightarrow q_2$ (blank) and halt for lack of a $(q_2,B)$ rule. The question explicitly gives start state $q_2$, so the trace above is the correct one for the stated data.)

Final result

• Enumerator: TM with output tape listing all strings of $L$, equivalent to recursive enumerability.
• Encoded string of the machine is the boxed binary string in Step 5.

Language of a Grammar and Leftmost Derivation of 00100

Step 1 - Given Data

• Grammar $G$ with productions: $$S \rightarrow 0S0 \mid 1 \mid \varepsilon$$
• Start symbol: $S$
• Terminals: ${0, 1}$
• Non-terminal: ${S}$
• Target string: $00100$

All data is present and readable.

Step 2 - Solution

Definition: Language of a Grammar

For a grammar $G = (V, T, P, S)$ where $V$ is the set of variables (non-terminals), $T$ the set of terminals, $P$ the set of productions, and $S$ the start symbol, the language of the grammar, written $L(G)$, is the set of all terminal strings derivable from $S$:

$$L(G) = {, w \in T^* \mid S \Rightarrow^* w ,}$$

where $\Rightarrow^*$ denotes derivation in zero or more steps. Only strings consisting entirely of terminals belong to $L(G)$.

Leftmost Derivation of 00100

In a leftmost derivation, the leftmost non-terminal is expanded at each step. Since there is only one non-terminal $S$, this is straightforward.

$$S \Rightarrow 0S0 \quad (S \rightarrow 0S0)$$ $$\Rightarrow 00S00 \quad (S \rightarrow 0S0)$$ $$\Rightarrow 00100 \quad (S \rightarrow 1)$$

Complete derivation:

$$S \Rightarrow 0S0 \Rightarrow 00S00 \Rightarrow 00100$$

Therefore $00100 \in L(G)$.

Verification of leaf string: outer pair $0\ldots0$, inner pair $0\ldots0$, centre $1$ gives $0,0,1,0,0 = 00100$ ✓

Parse Tree for 00100

          S
        / | \
       0  S  0
        / | \
       0  S  0
          |
          1

Reading the leaves left to right: $0,\ 0,\ 1,\ 0,\ 0 = 00100$ ✓

Summary Table

The string $00100$ is derived in 3 steps, confirming $00100 \in L(G)$.

ε-closure of a State and Difference Between Moore and Mealy Machine

Part 1: ε-closure of a State

Definition:

ε-closure of a state is the set of all states that can be reached from a particular state without reading any input symbol, i.e., by reading only ε (epsilon) transitions.

Formally, for a state q, the ε-closure(q) includes:

• The state q itself
• All states reachable from q by following one or more ε-transitions

Example:

If state A has ε-transitions to B, and B has an ε-transition to C, then:

ε-closure(A) = {A, B, C}

Key Points:

• Every state is always in its own ε-closure.
• ε-closure is used when converting ε-NFA to DFA, where the start state of the DFA is taken as ε-closure of the start state of the ε-NFA.

Part 2: Difference Between Moore Machine and Mealy Machine

Both are finite state machines (FSMs) that produce output, but they differ in how output is generated.

Summary

• ε-closure captures all states reachable via epsilon moves alone, and is fundamental to converting ε-NFA to DFA.
• Mealy machine produces output based on both state and input (faster, fewer states), while Moore machine produces output based only on the current state (slower, more states).

DFA Accepting Binary Strings Ending with '00'

STEP 1 - EXTRACT (Given Data)

• Alphabet: $\Sigma = {0, 1}$
• Language: $L = {w \in \Sigma^* \mid w \text{ ends with } 00}$
• Test string for extended transition function: $111000$

No numeric data missing; the problem is fully solvable.

STEP 2 - SOLVE

Part 1: DFA Design

We track the relevant suffix (last two symbols). Three states suffice:

Formal 5-tuple: $$M = ({q_0, q_1, q_2},\ {0,1},\ \delta,\ q_0,\ {q_2})$$

Transition Table:

Transition Function:

• $\delta(q_0, 0) = q_1,\quad \delta(q_0, 1) = q_0$
• $\delta(q_1, 0) = q_2,\quad \delta(q_1, 1) = q_0$
• $\delta(q_2, 0) = q_2,\quad \delta(q_2, 1) = q_0$

State Diagram (description):

      1                    1
    (loop)              (self on q2 via 1 back to q0)
   ┌────┐
   ↓    │
 →( q0 )──0──▶( q1 )──0──▶(( q2 ))──0──▶ (self loop 0)
   ▲            │             │
   └─────1──────┘             │
   ▲──────────────1───────────┘

• $q_0$: on $1$ self-loop; on $0 \to q_1$
• $q_1$: on $0 \to q_2$; on $1 \to q_0$
• $q_2$ (accepting): on $0$ self-loop; on $1 \to q_0$

Part 2: Extended Transition Function $\delta^*(q_0, 111000)$

Definition: $$\delta^(q, \varepsilon) = q, \qquad \delta^(q, wa) = \delta(\delta^*(q, w), a)$$

Result

$$\delta^*(q_0,\ 111000) = q_2$$

Since $q_2$ is the accepting state, the string $111000$ (which ends in $00$) is accepted. The DFA design is verified correct.

Turing Machine: Testing String "())))"

STEP 1 - EXTRACT (Given Data)

Input string: ()))) = ( ) ) ) ) (1 open bracket, 4 close brackets)

Transition Table:

• Start state: $q_0$
• Accept state: $q_4$
• $B$ = blank

Note on ambiguity: The table is malformed in the LaTeX. The $q_0$ column headings mix up which symbol triggers each action. The intended machine is the classic balanced-parenthesis TM: on $q_0$ reading ( write X and go right to $q_1$; skip over already-marked X's; halt-accept on blank. The result (accept/reject) does not depend on these cosmetic issues, so I trace using the standard interpretation.

STEP 2 - SOLVE

Initial tape (head at position 1, state $q_0$):

( ) ) ) )
^

Trace

At step 5, in state $q_0$ the head reads ). The machine needs to find another ( to pair with the remaining ) symbols, but there are none. State $q_0$ has no rule for reading ) (there is no unmarked ( left).

The machine halts without reaching the accept state $q_4$.

Result: STRING IS REJECTED ✗

Reason: The string ()))) has $1$ opening bracket and $4$ closing brackets: it is unbalanced. After the TM matches the single ( with one ), three unmatched ) remain with no ( to pair them. The machine crashes in $q_0$, so ()))) is not accepted.

Transition Diagram

        (→X,R                    )→X,L / Y→Y,L / B→Y,L
   ──►( q0 )───────────────►( q1 )───────────────►( q2 )
        │  ▲                                          │
        │  │      X→X,R (match found, restart)        │
        │  └──────────────────────────────────────────┘
        │
        │ Y→Y,R                                        │ Y→Y,R
        │                                              ▼
        │ B→(no move)                               ( q3 )
        ▼                                              │ B→R
      (( q4 ))  ◄───────── B→R ────────────────────────┘
       ACCEPT

• $q_4$ (double circle) is the accept state.
• Acceptance requires reaching $q_4$ on a blank after all brackets are matched.
• For ()))), the machine never reaches $q_4$.

Conclusion

The string ()))) is REJECTED because it is unbalanced (1 open vs 4 close); the machine halts in $q_0$ with no applicable transition on ).