Question

For $\Sigma = \{a,b\}$, let us consider the regular language $L = \{x \mid x = a^{2+3k} \text{ or } x = b^{10+12k},\; k \ge 0\}$. Which one of the following can be a pumping length (the constant guaranteed by the pumping lemma) for $L$?

• A. 3
• B. 5
• C. 9
• D. 24

Hint:

For union of regular languages, the pumping length must be large enough to work for {all} components. Choosing a value larger than the maximum cycle length of the DFA is always safe.

Answer 1

The Correct Option is D

Step 1: Understand the structure of the language.
The language $L$ is the union of two regular languages:
\[ L_1 = \{a^{2+3k} \mid k \ge 0\}, \quad L_2 = \{b^{10+12k} \mid k \ge 0\} \] Both $L_1$ and $L_2$ are regular since they are arithmetic progressions of string lengths over a single symbol.
Step 2: Recall the pumping lemma for regular languages.
For a regular language, there exists a pumping length $p$ such that every string $w \in L$ with $|w| \ge p$ can be written as $w = xyz$, satisfying:
$|xy| \le p$, $|y| \ge 1$, and $xy^iz \in L$ for all $i \ge 0$.
The value of $p$ depends on the structure (number of states) of a DFA recognizing $L$.
Step 3: Determine a safe pumping length.
For $L_1$, strings follow a cycle of length $3$ (due to $a^{2+3k}$).
For $L_2$, strings follow a cycle of length $12$ (due to $b^{10+12k}$).
A pumping length must be large enough to handle both components of the union language.
A safe choice is a number that is at least the larger cycle length and allows valid pumping for all sufficiently long strings.
Step 4: Evaluate the given options.
(A) 3: Too small to accommodate the $b^{10+12k}$ part.
(B) 5: Still smaller than the minimum length of strings in $L_2$.
(C) 9: Less than 10; strings like $b^{10}$ would violate pumping conditions.
(D) 24: Sufficiently large to handle both cyclic patterns (3 and 12). Correct.
Step 5: Conclusion.
A valid pumping length guaranteed by the pumping lemma for the language $L$ is \[ \boxed{24} \]