Context-Free Grammar - Definition and Construction

STEP 1 - EXTRACT: Given Data

• Alphabet: $\Sigma = {a, b}$
• Language: all palindromes over ${a,b}$ that do NOT contain substring $aa$
• Target string for derivation and parse tree: babbbab
• Marks: 10

All required data is present.

STEP 2 - SOLVE

1. Definition of CFG

A Context-Free Grammar is a 4-tuple $G = (V, T, P, S)$ where:

• $V$ = finite set of variables (non-terminals)
• $T$ = finite set of terminals, with $V \cap T = \varnothing$
• $P$ = finite set of productions of the form $A \to \alpha$, where $A \in V$ and $\alpha \in (V \cup T)^*$
• $S \in V$ = the start symbol

Every production has exactly one non-terminal on the left, hence "context-free". CFGs generate the class of Context-Free Languages.

2. Language Analysis

$$L = {, w \in {a,b}^* \mid w = w^R \text{ and } aa \text{ is not a substring of } w ,}$$

Constraints:

• Palindrome: $w = w^R$
• No $aa$: two $a$'s must never be adjacent

Because $w$ is a palindrome, when we place an $a$ at both ends the inner string must not begin or end with $a$ (otherwise $aa$ would form at the junction). So an $a$-wrapped layer must contain an inner block that begins and ends with $b$ (or is a single $b$).

Sample valid strings: $\varepsilon,\ a,\ b,\ aba,\ bab,\ bbb,\ babab,\ babbbab$. Invalid: $aa,\ baab,\ aabaa$.

3. Constructing the CFG

Use two variables:

Productions $P$:

$$ \begin{aligned} S &\to \varepsilon \mid a \mid b \ S &\to bSb \ S &\to aBa \ B &\to b \mid bSb \end{aligned} $$

Grammar: $G = ({S,B},\ {a,b},\ P,\ S)$.

Why $aa$ never occurs:

• $S \to aBa$: the inner $B$ must start and end with $b$, giving $a,b\ldots b,a$, so no adjacent $a$'s at the junction.
• $S \to bSb$ and $B \to bSb$: wrap with $b$'s, always safe.
• No production ever places two $a$'s next to each other.

The palindrome property holds because each recursive rule adds the same terminal symmetrically on both ends.

4. Leftmost Derivation for babbbab

Structural decomposition:

$$ \underbrace{b}{\text{outer}}\ \underbrace{a}{}\ \underbrace{b\ b\ b}{\text{middle}}\ \underbrace{a}{}\ \underbrace{b}_{\text{outer}} $$

• Outer $b\ldots b$ → $S \to bSb$
• Inner $abbba$: $a\ldots a$ → $S \to aBa$
• Inner $bbb$: $b\ldots b$ → $B \to bSb$
• Center $b$ → $S \to b$

Leftmost derivation (always expand leftmost non-terminal):

$$S \Rightarrow bSb \Rightarrow baBab \Rightarrow babSbab \Rightarrow babbbab$$

Result: $babbbab$ ✓ (matches target)

5. Parse Tree for babbbab

                 S
              /  |  \
             b   S   b
               / | \
              a  B  a
               / | \
              b  S  b
                 |
                 b

Reading the leaves left to right: $b,a,b,b,b,a,b = babbbab$ ✓

The tree is symmetric about its center, confirming both the palindrome property and the absence of substring $aa$ (each $a$ is bounded by $b$'s).

Final Answer: The CFG $G = ({S,B},{a,b},P,S)$ with productions $S \to \varepsilon \mid a \mid b \mid bSb \mid aBa,\ B \to b \mid bSb$ generates the required language. The leftmost derivation for $babbbab$ uses the sequence $S \Rightarrow bSb \Rightarrow baBab \Rightarrow babSbab \Rightarrow babbbab$, with the parse tree shown above.

PDA for $L = {a^n b^n \mid n \geq 1}$

STEP 1 - Given Data

• Alphabet: $\Sigma = {a, b}$
• Language: strings with equal number of $a$'s followed by equal number of $b$'s
• Strings to test: $aabb$ and $aab$

$$L = {a^n b^n \mid n \geq 1} = {ab, aabb, aaabbb, \ldots}$$

STEP 2 - Construction and Solution

Strategy

1. Push each $a$ onto the stack.
2. On the first $b$, start popping $a$'s.
3. On subsequent $b$'s, keep popping $a$'s.
4. Accept when only $Z_0$ remains after all input consumed (empty count).

Formal Definition

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

• $Q = {q_0, q_1, q_2}$
• $\Sigma = {a, b}$
• $\Gamma = {a, Z_0}$
• Start state $q_0$, initial stack symbol $Z_0$
• $F = {q_2}$

Transition Table

Notation: input, top / pushed-string.

Acceptance of $aabb$

Input consumed, final state $q_2$ reached. $aabb$ is ACCEPTED.

Acceptance of $aab$

Input exhausted while stack still holds an unmatched $a$ (top $\neq Z_0$), so $q_2$ cannot be reached. $aab$ is REJECTED (2 a's, 1 b: unequal).

Summary

The PDA correctly accepts $a^n b^n$ and rejects strings with unequal counts.

Keep the transition table precise: write the $q_2$ transitions so that the empty-stack case is distinguished rather than folded into a,ε/a, and avoid an extra initialization state. The verdicts (accept $aabb$, reject $aab$) are unaffected.

Intractability and Time/Space Complexity of Turing Machine

Intractability

Intractability is a concept used to classify problems that cannot be solved in polynomial time but instead require exponential time algorithms.

• Problems that can be solved within reasonable time and space constraints are called tractable problems.
• Problems that cannot be solved in polynomial time but require exponential time algorithms are called intractable (or hard) problems.

An algorithm whose complexity measure increases with input size n no more rapidly than a polynomial in n is said to be polynomially bounded. An algorithm whose complexity grows exponentially is said to be exponentially bounded, and such problems fall under intractability.

Example: Problems like the Travelling Salesman Problem (decision version) are considered intractable as no known polynomial-time algorithm exists for them.

Time and Space Complexity of a Turing Machine

When a Turing Machine (TM) answers a specific instance of a decision problem:

• Time is measured as the number of moves made during computation.
• Space is measured as the number of tape squares used during computation.

The most natural measure of input size is the length of the input string. The worst case is considered, i.e., the maximum time or space required for any input string of that length.

Time Complexity

Let T be a Turing Machine. The time complexity of T is the function T_t defined on the natural numbers as:

For n ∈ N, T_t(n) is the maximum number of moves T can make on any input string of length n.

• If there exists an input string x such that |x| = n and T loops forever on input x, then T_t(n) is undefined.

Space Complexity

The space complexity of T is the function S_t defined as:

S_t(n) is the maximum number of tape squares used by T for any input string of length n.

• If T is a multi-tape TM, the number of tape squares means the maximum of the number of squares used across individual tapes.
• If for some input of length n, T loops forever, then S_t(n) is undefined.

Summary Table

Note: In complexity theory, we are always more interested in the growth rate of these functions rather than their absolute values.