Minimum pumping length of regular language

Question

I know the method to the find the minimum pumping length of regular language by constructing minimal DFA and finding the number of states but I am not able to quite understand why is it working.

For example, $L = (10)^*$. Here minimum pumping length is 1 by constructing DFA. But by pumping lemma, how can there be a string of length 1 belonging to $L$? As pumping lemma says the string $S$ should be such that $|S| \le p$ and $S$ should belong to $L$, where $p$ is the pumping length, and here there is no string of length 1 belonging to $L$.

Answer 1

Your example in fact demonstrates that the pumping length can be smaller than the number of states in the minimal DFA. In your case the minimal DFA contains 3 states, but the pumping length is 2.

The pumping length of a language $L$ is the minimal $p$ such that every word $w \in L$ of length at least $p$ can be written as $w = xyz$, where $|xy| \leq p$, $y \neq \epsilon$, and $xy^iz \in L$ for all $i \geq 0$.

If $L$ contains no words of length $p$ that's not problematic at all. For example, the pumping length of $\{ w \}$ is $|w|+1$. In that case the pumping property holds vacuously.