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Let's say we have the language L = {w $\in$ {a,b}$^*$ : ($\exists n \in \mathbb{N} $)[$w|_b = 5^n$]}. I want to know if this is a regular language or not. How do I go about doing this? I'm familiar with the Myhill-Nerode theorem but I don't know how to apply it.

WindBreeze
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You don't need the Myhill-Nerode theorem. Suppose that $L$ was regular and let $c>0$ be its pumping length.

The word $b^{5^c}$ is in $L$ and therefore there is some positive constant $k \le c < 5^c$ such that $b^{5^c + ik} \in L$ for all integer values of $i \ge -1$. For $i =1$ you must have $b^{5^c + k} \in L$ which implies that $5^{c} + k$ is a power of $5$.

This is a contradiction since $ 5^ c < 5^c + k < 5^c + 5^c = 2 \cdot 5^c < 5^{c+1}. $

If you still want to use the Myhill-Nerode theorem you can proceed as follows: Pick any two non-negative integers $i,j$ with $i<j$ and consider the two words $x= b^{5^i}$ and $x = {5^j}$. Let $z = b^{4 \cdot 5^i}$. Notice that $xz = b^{5^i + 4 \cdot 5^i} = b^{5^{i+1}} \in L$ while $yz = b^{ 5^j + 4\cdot 5^i} \not\in L$ since $5^j < 5^j + 4 \cdot 5^i < 5^j + 4 \cdot 5^j = 5^{j+1} $. Therefore $z$ is a distinguishing extension for $x$ and $y$ and they cannot belong to the same equivalence class for the relation $R$ defined by $\alpha R \beta$ iff $\alpha \in \Sigma^*$ and $\beta \in \Sigma^*$ have no distinguishing extension. This shows that the number of equivalence classes in the quotient set $L_{/R}$ is not finite and, by the Myhill-Nerode theorem, $L$ is not regular.

Steven
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  • what does pumping length mean? – WindBreeze Jun 10 '20 at 22:14
  • It is just the minimum constant $c$ for which the pumping lemma holds, that is, such that any word $w \in L$ with $|w| \ge c$ can be written as $xyz$, with $|xy| \le c$, $|y| \ge 1$, and $xy^iz \in L$ for all values of $i \ge 0$. See the formal definition on Wikipedia – Steven Jun 10 '20 at 22:20
  • is there a way to prove this with the myhill-nerode theorem? – WindBreeze Jun 10 '20 at 23:21
  • See the edited answer. – Steven Jun 10 '20 at 23:48
  • Why do we have 4 in 4*5^i? – WindBreeze Jun 11 '20 at 15:40
  • $4 \cdot 5^i$ is the minimum number of additional $b$s that need to be added to $x=b^{5^i}$ in order to obtain a word that is still in $L$, i.e., if $z=b^{4 \cdot 5^i}$ then $xz \in L$. We want $xz \in L$ and $yz \not\in L$ because we are trying to show that $z$ is a distinguishing extension. (Other choices of $z$ are possible, but it suffices to find one). – Steven Jun 11 '20 at 15:55
  • I have trouble seeing why 4*5^i is the minimum number of bs to add for the word to still be in L. I thought that we needed 5^n bs ? I can't see what 4 has to do in all of this. – WindBreeze Jun 11 '20 at 16:52
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    Pick any natural number $i$. What is the first power of $5$ that is larger than $5^i$? It is $5^{i+1}$. The difference between $5^{i+1}$ and $5^i$ is $5^{i+1} - 5^i = 5^i (5 - 1) = 4 \cdot 5^i$. If you want to append a word $z$ containing the minimum number of $b$s to a string $x \in L$ consisting of $5^i$ $b$s in order to have $xz \in L$, then you will need to select $z = b^{4 \cdot 5^i}$. – Steven Jun 11 '20 at 17:16