Why is the minimal pumping length for $(01)^*$ equal to 1?

Question

I see why the minimum pumping length is at most 2 (since 01 can be pumped). But why is this counterexample not valid?

Let $A=(01)^*$ and assume it has pumping lenght $p=1$. Then lets consider the string $01\in A$, since its lenght is greater than 1 it must hold the pumping lemma. But given $p=1$ necessarily $x=\varepsilon, y=0, z=1$ and $xy^2z=001\notin A$. Therefore minumum pumping length must be greater than one and equal or smaller than two.

Answer 1

The minimum pumping length depends on it's definition, there are two forms of pumping lemma. The weaker form says that forall $w$ in language exists $p$ s.t. if $p \leq |w|$ then there exists $x$, $y$, and $z$ s.t. followings hold

• $w = xyz$
• $|y| \geq 0$
• $\forall m, xy^mz \in A$.

With this weak form $p=1$ is enough for $A=(01)^*$, but the second (stronger) form tell us an additional fact which says

• $|xy| \leq p$.

for satisfying this fact we need to change the pumping length, as you could provide counterexample correctly, $p=1$ doesn't satisfies this new fact.