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I have the following task and its solution.

Question

Given the language

$$ A \triangleq\left\{1^{n} 0^{n} \mid n \in \mathbb{N}\right\} \text { with } \Sigma_{A} \triangleq\{1,0\}, $$

give all equivalence classes of the Myhill-Nerode relation.

Solution

  • $[1^k]_{\equiv A} = \{1^k\}$ for $k \in \mathbb N$.
  • $[1^\ell0]_{\equiv A} = \{1^{\ell+i-1} 0^i \mid i \geq 1\}$ for $\ell \in \mathbb N^+$.
  • $[0]_{\equiv A} = \{ 0x, 1^n 0^m, x01 y \mid x,y \in \Sigma_A^* \land n,m \in \mathbb N^+ \land m > n \}$.

What are they doing in the second bullet of the solution?

The first and third bullets are clear. In the first they construct the equivalence classes of all $1$'s including $\lambda$.

In the third bullet they construct just one equivalence class with all the words which are not in the language $A$.

But what are they doing in the second bullet? What does the exponent $\ell+i-1$ mean? Why don't they write $[1^n0^m]_{\equiv A} = \{1^n 0^m \mid n,m \in \mathbb N^+\}$?

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I think it makes sense to answer your subquestions out of order.


Why they don't write $[1^n0^m] = \{1^n 0^m | n,m \in N^+\}$

They're supposed to be equivalence classes. $[1^n0^m]$ would break up lots of classes, and duplicate lots of the states from line 3.


What does the Exponent $n+i-1$ mean?

Firstly, $n+i-1 \ge i$, so there are at least as many $1$s as $0$s.

Secondly, $n$ is the parameter which defines the class.

So $\left[1^{\mathrm{l}}0\right]_{\equiv \mathrm{A}}$ is the set of states which need $0^{l-1}$ to reach the accepting state.

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There are infinitely many equivalence classes:

  • For each $k \geq 0$, the equivalence class of $1^k$, which consists only of $1^k$.
  • For each $\ell \geq 1$, the equivalence class of $1^\ell 0$, which consists of $1^{\ell-1+i} 0^i$ for all $i \geq 1$.
  • All other words (the equivalence class of $0$): words beginning with $0$, words containing a substring $01$, and words of the form $1^n0^m$ with $m > n$.

The last equivalence class is easiest to understand: it consists of those words which cannot be extended to words in $A$.

The equivalence class of $1^\ell 0$ consists of all words whose only extension in $A$ is by $0^{\ell-1}$.

Finally, $1^k$ can be extended to a word in $A$ by adding $1^j 0^{k+j}$ for arbitrary $j \geq 0$.

Using this, it is easy to check that all of these equivalence classes are actually distinct. In particular, it is definitely not true that all words of the form $1^n 0^m$ belong to the same equivalence class, if only since some of them belong to $A$ and some don't.

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  • $\begingroup$ thanks for your answer. If I understand this right than each prefix of a word only allows one suffix to be completed in that language. So $[1^0] = \{ \lambda \}$ or $[1^1] = \{ 1 \}$ and prefix $0$ or $[1^2] = \{ 1^2 \}$ and prefix $0^2$ and so on. But if we look at the second bullet , we have does equivalence class: $ [a b]=\quad L \quad \quad $ Suffix: $ \quad\{\epsilon\} $ $ \left[a^{2} b\right]=\left\{a^{2} b, a^{3} b^{2}, a^{4} b^{3}, \ldots\right\} \quad $ Suffix : $ \quad\{b\} $ . . . $[1^\ell0]_{\equiv A} = \{1^{\ell+i-1} 0^i \mid i \geq 1\}$ $\endgroup$ – Lisa.Neust Jan 30 at 14:32
  • $\begingroup$ But than we have two classes with the same suffix is this possible? I'm so sry for my bad English, I'm learning it since 4 month, but I hope you can understand. $\endgroup$ – Lisa.Neust Jan 30 at 14:35
  • $\begingroup$ We don't have two classes with the exact same set of suffixes. It's OK to have a suffix common to two classes. $\endgroup$ – Yuval Filmus Jan 30 at 14:40

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