Let $L$ be a regular language. Prove that the number of states of the minimum accepting automata $L$ can be equal to a constant from the pumping lemma, but any smaller number can't.
From what I understand if we have a word $w=xyz$ we can create an automata with $3$ states $S_1 \to S_2 \to S_3$ where $S_1 \to S_2$ when $x$, $S_2 \to S_2$ when $y$ and $S_2 \to S_3$ when $z$. So the constant would be equal to $3$ and we can't go down to $2$. But I don't know how to prove that for every regular language.
Let me first remind you the pumping lemma for regular languages.
Let $L$ be a regular language. Then there exists an integer $p > 0$ (called the pumping length) such that every word $w \in L$ of length at least $p$ can be written as $w = xyz$, in such a way that $|xy| \leqslant p$, $|y| \geqslant 1$ and $xy^*z \subseteq L$.
If $n$ is the number of states of the minimal automaton of $L$, then one can choose $n$ for the pumping length. However, it may happen that the mimimal possible pumping length is strictly smaller than $n$. For instance, for $L = a^* \cup b^*$, one can take $p = 1$ as the pumping length, but the minimal automaton of $L$ has three states.