Non-regular Maximal Prefix-Free Subsets of Regular Languages

Abstract

We investigate non-regular maximal prefix-free subsets (MPFS) of regular languages. We give a method to decide whether or not a regular language has any non-regular MPFS.

Next, we prove that if a regular language has any non-regular MPFS, then it also has a MPFS which is context-sensitive but not context-free, it has a MPFS which is recursive but not context-sensitive, and it has a MPFS which is not recursively enumerable.

We show that no regular language has a MPFS which is recursively enumerable but not recursive. Finally, for any regular language we can decide whether or not it has a context-free non-regular MPFS.

Appendix

Lemma 14. Let L be a regular language with the property $P_2$. Then L has a maximal prefix-free subset which is context-free, but not regular.

Proof

Here we show that the set C obtained as described in Lemma 14 is a MPFS of L. The proof of the lemma in the article shows that $C \in \mathrm {CF}\setminus \mathrm {Reg}$.

Let $D_1 = \{ xu^i \mid i \ge 0 \} \cdot C_1$. In Case 1, let $D_2 = \{ xu^iyv^i \mid i \ge 0 \}$, $D_3 = \{ xu^iyv^j \mid i \ge 1, 0 \le j < i \} \cdot C_2$. In Case 2, let $D_2 = \{ xu^iyv^i \mid i \ge 0 \} \cdot C_2$, $D_3 = \{ xu^iyv^j \mid i, j \ge 0; i \ne j \} \cdot C_3$.

(1) $C \subseteq L$: $C_0 \subseteq L$. We have $s \cdot xu^i = p$, and $C_1 \subseteq L_p$. Next, $s \cdot xu^iyv^j = q$ for $i, j \ge 0$. In Case 1, $q \in F$ and $C_2 \subseteq L_q$. In Case 2, $C_2, C_3 \subseteq L_q$. Therefore $C \subseteq L$.

(2) C is prefix-free: $C_0$ is prefix-free. $C_0$ does not contain any string in [x], therefore strings in $C_0$ are incomparable with any string in $D_1 \cup D_2 \cup D_3$.

Since $C_1$ is prefix-free and does not contain any strings in [u], $D_1$ is prefix-free as well. $C_1$ does not contain any string in [u] or [y], so strings in $D_1$ are incomparable with any string in $D_2 \cup D_3$.

Since $C_2$ and $C_3$ are prefix-free and do not contain any string in [v], for a given $i, j \ge 0$ the languages $xu^iyv^j \cdot C_2$ and $xu^iv^j \cdot C_3$ are also prefix-free. Let $i_1 < i_2$. Then $xu^{i_1}yv$ and $xu^{i_2}yv$ are incomparable, since u and y differ in the first symbol. Let $j_1 < j_2$. Then any string in $ux^iyv^{j_1} \cdot C_2$, resp. $C_3$ is incomparable with any string in $ux^iyv^{j_2} \cdot C_2$, resp. $C_3$, since no string in $C_2$, resp. $C_3$ is in [v].

(3) C is maximal. Let $w \in L$. Consider the following cases:

• $w \notin [x]$. Then w is comparable to a string in $C_0$.

• $w \le _p x$. Then in Case 1 $w \le _p xy \in D_2$, in Case 2 $w \le _p xyz \in D_2$.

• $w = xu^iw_1, i \ge 0, w_1 \notin [u] \cup [y]$. Then $w_1$ is comparable to a string $w_1'$ in $C_1$ and w is comparable to $xu^iw_1' \in D_1$.

• $w = xu^iw_1, i \ge 0, w_1 \le _p u$. Then in Case 1 $w \le _p xu^{i+1}yv^{i+1} \in D_2$, in Case 2 $w \le _p xu^{i+1}yv^{i+1}z \in D_2$.

• $w = xu^iw_1, i \ge 0, w_1 \le _p y$. Then in Case 1 $w \le _p xu^iyv^i \in D_2$, in Case 2 $w \le _p xu^iyv^iz \in D_2$.

• $w = xu^iyv^jw_2, i, j \ge 0, w_2 \notin [v]$.

  • Case 1: If $j \ge i$, then $w \ge _p xu^iyv^i \in D_2$. Otherwise $w_2$ is comparable to a $w_2' \in C_2$ and w is comparable to $xu^iyv^jw_2' \in D_3$.

  • Case 2: If $j = i$, then $w_2$ is comparable to a $w_2' \in C_2$ and w is comparable to $xu^iyv^iw_2' \in D_2$. Otherwise $w_2$ is comparable to a $w_2'' \in C_3$ and w is comparable to $xu^iyv^jw_2'' \in D_3$.

• $w = xu^iyv^jw_2, i, j \ge 0, w_2 \le _p v$.

  • Case 1: If $j \ge i$, then $w \ge _p xu^iyv^i \in D_2$. Otherwise $w \le _p xu^iyv^i \in D_2$.

  • Case 2: $w \le _p xu^iyv^{i+j+1}z' \in D_3$.

Therefore for any $w \in L$ there is a $w' \in C$ such that w is comparable to $w'$. Then C is a MPFS. $\square$

In the following, we use the notation introduced in Lemma 15.

Lemma 16. For every state $q_i, 0 \le i \le k$, there is at most one non-empty string $s_{q_i}$ such that $q_i \cdot s_{q_i} = q_i$, and the computation $q_i \xrightarrow []{s_{q_i}} q_i$ does not pass through $q_i$ except for the first and last state.

Proof

For a contradiction, let $q_i \cdot s_1 = q_i$ and $q_i \cdot s_2 = q_i$ for non-empty $s_1 \ne s_2$, where the computations on $s_1$ and $s_2$ do not pass through $q_i$. Then $s_1$ and $s_2$ are not comparable, and we have $s_1 = s'as_1'$ and $s_2 = s'bs_2'$ for $s, s_1', s_2' \in \Sigma ^*$ and $a, b \in \Sigma , a \ne b$. Let $q_0 \cdot x' = q_i$ and $q_i \cdot z' = q_k$.

Let $p = q = q_i \cdot s'$, $x = x's'$, , $u = v = as_1's'$, $y = bs_2's'$. Then L has the property $P_2$ since either $q = q_i$ is a final state if $i = k$ (Case 1), or $L_q \setminus [v]$ is not prefix-free, since $q \cdot bs_2's'z' = q_f$ and $q_f$ is a final non-$\varepsilon$ state (Case 2).

Lemma 17. Let $\mathrm {AC}(q_k) = \{ w' \in \Sigma ^* \mid s \cdot w' = q_k$, and the computation $s \xrightarrow []{w'} q_k$ does not contain any state more than once$\}$. Let $\ell$ be the first index such that $q_\ell$ occurs in the computation on w more than once, if such a state exists. Then $w = rs_{q_\ell }^i t$, where $r, t \in \Sigma ^*$, $i \ge 0$, and $rt \in \mathrm {AC}(q_k)$.

Proof

Let $q_\ell$ be the first and $q_{\ell '}$ be the last occurrence of the state $q_\ell$ in the computation on w. By Lemma 16, the only possible string that can be read between two consecutive passes through $q_\ell$ must be $s_{q_\ell }$. The computation therefore looks like this:

$$ q_0 \xrightarrow []{a_1} q_1 \xrightarrow []{a_2} \cdots \xrightarrow []{a_\ell } q_\ell \xrightarrow []{s_{q_\ell }} q_\ell \xrightarrow []{s_{q_\ell }} \cdots \xrightarrow []{s_{q_\ell }} q_{\ell '} \xrightarrow []{a_{\ell '+1}} q_{\ell '+1} \xrightarrow []{a_{\ell '+2}} \cdots \xrightarrow []{a_k} q_k $$

Let $r = a_1a_2 \cdots a_\ell$ and $t = a_{\ell '+1}a_{\ell '+2} \cdots a_k$. Let us show that $rt \in \mathrm {AC}(q_k)$; that is, the states $q_0, q_1, \dots , q_{\ell -1}, q_{\ell '+1}, \dots , q_k$ are all distinct.

We know that $q_\ell$ is the first state which occurs in the computation more than once, therefore if two states $q_i$ and $q_j$ for $i < j$ among the above are equivalent, it must be that $\ell '< i < j \le k$.

Thus the computation goes through a cycle $q_\ell \xrightarrow []{s_{q_\ell }} q_\ell$, potentially several times. After that, the computation goes through the cycle $q_i \xrightarrow []{s_{q_i}} q_i$. This cycle does not contain $q_\ell$, since $q_\ell '$ is the last occurrence of this state and $\ell ' < i$. The computation must therefore “leave” the $q_\ell$ cycle at some point. Let $q_p$ be the last state in the computation that follows this cycle. Without further technical details, let us observe that we have $p = q_p$, $q = q_i$, $x = a_1a_2 \cdots a_p$, $u = s_{q_p}$, $y = a_{p+1}a_{p+2} \cdots a_i$, $v = s_{q_i}$ and L has the property P since either $q = q_i = q_k$ is final, or the final non-$\varepsilon$ state $q_k$ is reachable from $q_i$ and thus $L_{q_i}$ is not prefix-free.

This is a contradiction with the initial assumption that the language L does not have the property P. $\square$

Lemma 18. Let R, S, and T be finite languages. Let $L \subseteq \{ rs^it \mid r \in R, s \in S, t \in T, i \ge 0 \}$ be a context-free language. Then L is regular.

Proof

It holds that:

$$ L = \bigcup _{\begin{array}{c} r \in R\\ s \in S\\ t \in T \end{array}} \{ rs^it \mid rs^it \in L \}, $$

that is, L is a union of finitely many languages of the form $\{ rs^it \mid rs^it \in L \}$ for some specific r, s, t. For each of these languages we have $\{ rs^it \mid rs^it \in L \} = r \cdot \{s^i \mid rs^it \in L \} \cdot t = r \cdot ( r \backslash L / t ) \cdot t$, where $\backslash$ and / are the left and right quotient operation, respectively. This set is context-free, since L is context-free and $\mathrm {CF}$ is closed under concatenation and left and right quotients by regular languages.

It follows that $r \backslash \{ rs^it \mid rs^it \in L \} / t = \{ s^i \mid rs^it \in L \}$ is also context-free, and since CF is closed under inverse homomorphism, the set $\{a^i \mid rs^it \in L \}$ is context-free as well. However, the latter language is unary, and every unary context-free language is also regular. Therefore every language $\{ rs^it \mid rs^it \in L \} = r \cdot \{s^i \mid rs^it \in L \} \cdot t$ is regular as well, and L is a union of finitely many regular languages. Hence L is regular. $\square$