Ericpony's Blog: Existence of sets that are regular but not computable

Thursday, April 6, 2017

Existence of sets that are regular but not computable

A regular set is called computable if one can construct a finite-state automaton to recognise it. Given a finite alphabet $\Sigma$, a rational transition system is a tuple $(I, T)$ comprised of a regular set $I\subseteq\Sigma^*$ as the initial states and a rational relation $T\subseteq\Sigma^*\times\Sigma^*$ as the transition relation. We will use $T^k$ to denote the rational relation formed by composing $T$ for $k$ times, and $T^*$ to denote the reflexive and transitive closure of $T$. In this way, $T^*(I)$ will represent the reachability set of system $(I,T)$, i.e., the set of words that can be reached from the words in $I$ through a finite number of transitions.
It is clear that $T^k(I)$ is regular and computable for any $0\le k<\infty.$ It however may be less obvious that the reachability set $T^*(I)$ is not necessarily regular. To see this, consider a system $\Sigma\equiv\{a,b\}$, $I\equiv\{(ab)^n:n\ge 1\}$ and $T\equiv\{(xbay, xaby): x,y\in \Sigma^*\}.$ Then $T^*(I) \cap a^*b^* =\{a^nb^n:n\ge1\}$ is a well-know non-regular set, which implies that $T^*(I)$ is non-regular as well. On the other hand, there are rational transition systems of which the reachability sets are guaranteed to be regular, such as pushdown systems. In fact, the reachability relation of a pushdown system is regular and poly-time computable [12]. Surprisingly, $T^*(I)$ it is not necessarily computable even when it is regular. Namely, given a regular language $A$ and a rational transition system $(I, T)$, it is undecidable to check where $T^*(I)=A$ even though the former is known to be a regular set. This post will try to present a self-contained argument about the existence of such systems.
Subword ordering. Our argument would centre around a reflexive and transitive relation called "a subword of": given $x,y\in\Sigma^*$, $x\preceq y$ holds iff $x$ can be obtained by removing some segments from $y.$
Lemma. (Higman) A set is finite if all elements in it are mutually incomparable w.r.t. $\preceq.$
Proof. For $x,y\in\Sigma^*$, let $x\sim y$ iff there are distinct alphabets $a_1,...,a_n$ in $\Sigma$ such that both $x$ and $y$ match $a_1^+a_2^+...a_n^+.$ It is clear that $\sim$ is an equivalence relation splitting $\Sigma^*$ into $\binom{N}{1}1! + \binom{N}{2}2! + ... + \binom{N}{N} N!$ classes where $N=|\Sigma|$, and two words are comparable iff they are in the same class. ☐
Given a set $X\subseteq \Sigma^*$, define the upward closure of $X$ as $uc(X)\equiv\{w:\exists w'\in X.\ w'\preceq w\}.$ We say $X$ is upward closed iff $X=uc(X).$ Higman's lemma tells us that every upward closed set have a finite representation.
Lemma 1. Every upward closed set is regular.
Proof. Note that for any $X\subseteq\Sigma^*$, elements in $min(X)\equiv\{w\in X: \forall w'\in X.\ w'\not\preceq w\}$ are mutually incomparable. It follows from Higman's lemma that $min(X)$ is finite. Also, $X$ is upward closed iff $X=uc(X)=\bigcup_{w\in min(X)}wc(w)$, which implies $X$ is regular since $min(X)$ is finite and $uc(w)$ is regular for every $w\in\Sigma^*.$ ☐
Another consequence of Higman's lemma is that every increasing (in the sense of set inclusion) sequence of upward closed sets is finite.
Lemma 2. If $X_0\subseteq X_1\subseteq X_2\subseteq...$ and $X_i=uc(X_i)$ for all $i\ge0$, then there exists a finite $n$ such that $X_k=X_n$ for all $k\ge n.$
Proof. Let $X\equiv \bigcup_{i\ge 0} X_i.$ Since $min(X)$ is finite by Higman's lemma, there exists $n$ such that $X_k\cap min(X)=\emptyset$ for all $k\ge n.$ But this means $X_k\subseteq X_n$ for all $k\ge n.$ ☐
Lemma 3. For every infinite sequence of words $w_0,w_1,...\in \Sigma^*$, there exists $i<j$ such that $w_i\preceq w_j.$
Lemma 4. For any sets $X_0,X_1,...\subseteq\Sigma^*$, $uc(\bigcup_{i\ge 0}X_i)=\bigcup_{i\ge 0}uc(X_i).$
Lossy channel systems. We briefly introduce the concept of lossy channel systems. A channel system with $n$ queues are like a pushdown system with $n$ stacks except that it uses FIFO queues instead of FILO stacks. Channel systems are more powerful than pushdown systems, because they only need one queue to simulate Turing machines while pushdown systems need two stacks. A lossy channel system is a channel system where contents in the queues can be spontaneously lost. Technically, the capability of losing data can be introduced to a channel system by adding a non-deterministic self loop to each control state that removes an arbitrary alphabet from the queues.
Note that channel systems (lossy or not) can be expressed as rational transition systems where a configuration with n queues is encoded as a word in form of [control state]#[contents of the 1st queue]#...#[contents of the nth queue]#. We can define a subword ordering $\preceq_L$ over the configurations of a channel system as follows:

$c$#$w_1$#$...$#$w_n$# $\preceq_L$ $c'$#$w_1'$#$...$#$w_n'$#$\quad$ iff $\quad c=c'$ and $w_i\preceq w_i'$ for $i=1,...,n.$

Since the number of control states is finite, Lemma 1~3 also holds for $\preceq_L.$
Lossy channel systems enjoy many good properties. Among them, two properties are of particular interest for our establishment:
Fact 1. If $u\rightarrow^* w$ then $u\rightarrow^* w'$ for all $w'\preceq w.$ It follows that the complement of the reachability set is upward closed w.r.t. $\preceq_L.$
Fact 2. For any set of configurations $C$, the set $pre^*(C)\equiv\{w:\exists w'\in C.\ w\rightarrow^* w'\}$ of its predecessors is upward closed w.r.t. $\preceq_L.$
Theorem 1. The reachability set of a lossy channel system is regular.
Proof. This theorem is an immediate consequence of Fact 1, Lemma 1, and the fact that regular sets are closed under complementation. ☐
Given a system $(I, T)$ and a regular set $B\subseteq\Sigma^*$, the reachability problem asks whether $T^*(I)\cap B \neq\emptyset$, i.e., words in $B$ are reachable from words in $I$ through a finite number of transitions.
Theorem 2. The reachability problem of a lossy channel system is decidable.
Proof. Historically, this result was first proved in [6] using backward analysis as follows. Define the backward transition function $pre(C)\equiv\{w:\exists w'\in C.\ w\rightarrow w'\}$, which would preserve regularity when the forward transition relation is regular. Given a regular set $B$ of the bad configurations, let $X_0\equiv B$ and $X_{n+1}\equiv X_n\cup pre(X_n)$ for $n\ge 0$. Define $X\equiv\bigcup_{i\ge 0}X_i$. Note that $X$ is the set of configurations backward reachable from $B$, and thus is upward closed by Fact 2. By Lemma 4, we can construct another sequence $X_0',X_1',...$ converging to $X$ by defining $X_n'\equiv uc(X_n).$ Each $X_{n+1}'$ can be computed from $X_n'$ for $n\ge0.$ Hence, by Lemma 2, the fixed point $X$ can be computed within a finite number of steps. Since $T^*(I)\cap B\neq\emptyset$ if and only if $X\cap I\neq\emptyset$, the reachability problem is decidable. ☐
Remark. The above proof actually gives an algorithm to construct the reachability set. However, the algorithm relies on Higman’s Lemma for termination and thus the time complexity is not primitive recursive [8,9].

Now we proceed to show that the reachability set of a lossy channel system, albeit regular, is not computable. Given a system $(I,T)$ and a word $w$, the recurrent state problem asks whether there exists an infinite trace from $I$ that visits $w$ infinitely often.
Theorem 3. [6] The recurrent state problem is undecidable for lossy channel systems.

To solve the recurrent state problem it suffices to check $w \in T^*(I)$ and $w \in T^*(\{w\})$. Hence, if these two questions are decidable then so is the recurrent state problem, leading to a contradiction with Theorem 3. Therefore, we can conclude that there exists no procedure that computes a finite-state automaton (or any symbolic representation allowing effective membership query) for $T^*(I)$.
Corollary. There exists a regular transition system $(I,T)$ such that $T^*(I)$ is regular but not computable.
Remark. Does the above corollary holds when $T$ is length-preserving? I didn't find any result about this at the time I was writing this post.

Discussion. The proof of Theorem 2 relies on properties of the subword ordering (or a well-quasi-ordering in general) to assert that backward search always terminates. In fact, reachability can be shown both r.e. and co-r.e. whenever a regular proof exists: just launch two procedures at the same time, one enumerating all possible counter-examples and one enumerating all possible proofs. It is clear that exactly one of the two procedures will terminate. (Here a counterexample is an execution path from $I$ to $B$, and a proof is a regular set $P$ satisfying the following three proof rules: i) $I\subseteq P$, ii) $T(P)\subseteq P$, and iii) $P\cap B=\emptyset.$) In particular, reachability is decidable for a transition system whenever the set of reachable configurations is regular.
More generally, the above argument works whenever a "regular" symbolic proof exists and the one-step image $T(X)$ or pre-image $T^{-1}(X)$ can be computed effectively from $X$. The "regularity" here refers to the existence of any finitary representations that are computable and closed under complement and intersection. This argument has been applied to prove decidability results for transition systems such as Petri nets and lossy counter systems [10]. See [11] for a general discussion on this kind of proof techniques.