Minimum pumping length of a context-free language

Question

I was studying about the minimum pumping length of the language $L$ containing all palindromes over $\{a,b\}$ from this material about the pumping Lemma for CFLs.

The productions are as follows:

$$S\to aSa \mid bSb \mid a \mid b \mid \lambda$$

We have two cases to deal with,

• when the string $w\in L$ is of sufficient length and |$w$| is even,

the string is either of the form $w=uaau^R$ or $ubbu^R$.

I understood the explanation and why the minimum pumping length is 3.

• when the string $w\in L$ is of sufficient length and |$w$| is odd, there may be four possibilities,

$$uaaau^R$$ $$uabau^R$$ $$ubbbu^R$$ $$ubabu^R$$

I didn't understand the explanation given which goes like:

I didn't get why they left out $v$ or if $v$ is considered to be $\lambda$.

However, I tried to proceed like this:

If we decompose $uaaau^R$ as $w=uvxyz$ as required by the pumping lemma for CFLs, we get $u=u,\ v=a,\ x=a,\ y=a, z=u^R$, the first 3 conditions of the lemma are satisfied.

For the last condition, we see that $uv^ixy^iz=ua^iaa^iu^R\in L$ for any $i\geq0$

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Similarly, we can decompose $uabau^R$ as $u=u,\ v=a,\ x=b,\ y=a, z=u^R$.

So, $uv^ixy^iz=ua^iba^iu^R\in L$ for any $i\geq0$

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Could you please tell me whether my procedure is wrong and how to improve it?

I'm also very confused as to why they didn't mention $v$ in their explanation since the pumping lemma requires the string to be divided into 5 sub-strings $u,v,x,y,z$.

Answer 1

Please ignore the second example in that material you are studying since it contains some errors. Please see the following corrected version of that example.

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Let $L$ be the language consisting of all palindromes over $\{a, b\}$. The following is an unambiguous grammar for $L$. $$S \to aSa\mid bSb\mid a\mid b\mid\lambda$$ What is the minimum pumping length of $L$?

The answer is $1$.

If a palindrome $w$ has even length, the substring $aa$ or $bb$ is in the middle of the string. That is, $w = uaau^R$ or $w = ubbu^R$. Suppose $w = uaau^R$. We let $u = u$, $v = a$, $x = \lambda$, $y = \lambda$, and $z = au^R$. The first three conditions are obviously satisfied. For any $i\ge0$, $uv^ixy^i = ua^{i+1}u^R\in L$. The case that $w = ubbu^R$ is similar.

If $w$ has odd length, then there are two possibilities:

• $w = uau^R$.
  We let $u = u$, $v = a$, $x=\lambda$, $y = \lambda$, and $z = u^R$. The four conditions are satisfied.
• $w = ubu^R$.
  This case is similar to the case above.

The minimum pumping length cannot be $0$ since $1\le|v|+|y|\le|vxy|\le p$.