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Let $L$ be a language. $w \in\ L$ , and $w$ could be broken in $xyz$.

Then if $L$ is regular, there exists a pumping length $p$ such that:

  1. $\vert y\vert>0$
  2. $\vert xy\vert \leq p$
  3. $\forall i \geq 0,\ xy^i z\ \in L$

I understand that this lemma is

  • based on pigeon hole principle
  • we use it to prove that a language is not regular by contradiction.

For condition (1), I understand that the string we are pumping should be $\geq 1$.

But I don't understand the significance of condition (2).

Why should the length of $xy$ be equal to or smaller than $p$ . Won't it work otherwise?

rsonx
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1 Answers1

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$p$ is the number of states in the automaton. As you say, the pumping lemma is about the pigeonhole principle. Suppose that $q$ is the first state that's repeated when you read input $w$. Then $x$ is the string that you read before you reach $q$ for the first time, and $y$ is the string you read between the first and second visits to $q$. Since there are only $p$ states including the start state, you must repeat a state after doing at most $p$ transitions, i.e., reading a string of length at most $p$. So the condition $|xy|\leq p$ comes from the proof.

In reality, when an automaton reads a particular string $w$, there might be many times when it returns to a state that has already been visited. If you take $w\in L$ any split $w=xyz$ with $|y|\geq 1$, such that the automaton is in the same state at the beginning and end of $y$ then $xy^iz$ will be in $L$ for all $i$, regardless of how long $xy$ is.

David Richerby
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