# Tag Info

44

To complete the other answers: I think that Turing Machine are a better abstraction of what computers do than finite automata. Indeed, the main difference between the two models is that with finite automata, we expect to treat data that is bigger than the state space, and Turing Machine are a model for the other way around (state space >> data) by making the ...

32

There are two approaches when considering this question: historical that pertains to how concepts were discovered and technical which explains why certain concepts were adopted and others abandoned or even forgotten. Historically, the Turing Machine is perhaps the most intuitive model of several developed trying to answer the Entscheidungsproblem. This is ...

27

Every context-free language has either polynomial growth or exponential growth. In the notation of the question poser: Either there is a polynomial $p$ so that $w_n\le p(n)$ for all $n$ Or there exists a $c>1$, so that $w_n\ge c^n$ for infinitely many $n$. This has been shown for instance in: Roberto Incitti: "The growth function of context-free ...

18

There is the notion of primality of a language. It asks whether L can be written as $L_1 \cdot L_2$ where neither factor contains the empty word. A language is prime if it cannot be written in this form. For a given regular language, represented by a DFA, it is shown in [MNS] that it is PSPACE-complete to decide primality. [MNS] Wim Martens, Matthias ...

17

Take $S_5$ as alphabet and $$L= \{ \sigma_1\cdots \sigma_n \in S_5^*\mid \sigma_1\circ\cdots\circ\sigma_n = \text{Id}\}$$ Barrington proved in [2] that $L$ is $\textrm{NC}^1$-complete for $\textrm{AC}^0$ reduction (and even with a more restrictive reduction actually). In particular this shows that regular languages are not in $\textrm{TC}^0$ if $\textrm{... 16 There is even a stronger result than your request: There are exponentially-ambiguous NFAs for which the minimal polynomially-ambiguous NFAs are exponentially larger, and in particular the minimal UFAs. Check this paper by Hing Leung. 16 Visibly pushdown automata (or nested word automata, if you prefer working with nested words instead of finite words) extend the expressive power of deterministic finite automata: the class of regular languages is strictly contained within the class of visibly pushdown languages. For deterministic visibly pushdown automata, the language inclusion problem can ... 16 It's discussed in one of the very first papers about strings and complexity, namely, Dana Angluin, Finding patterns common to a set of strings, J. Comput. System Sci. 21 (1980), 46-62. Look at Theorem 3.6. The problem is NP-complete. It's also in A. Ehrenfeucht, G. Rozenberg, Finding a homomorphism between two words is NP-complete, Inform. Process. Lett. ... 16 They are typically called AND-functions. (I'm not joking.) Indeed, this concept has been considered before, and that's what people call them. See, for example, the book by Kobler, Schoning, and Toran on Graph Iso, where they talk about AND- and OR-functions for GI. And, by the way, there is an OR-function for GI (ibid.). The question of an AND-function for ... 15 Regular languages with unsolvable syntactic monoids are$\mathrm{NC}^1$-complete (due to Barrington; this is the underlying reason behind the more commonly quoted result that$\mathrm{NC}^1$equals uniform width-5 branching programs). Thus, any such language is not in$\mathrm{TC}^0$unless$\mathrm{TC}^0=\mathrm{NC}^1$. My favorite$\mathrm{NC}^1$-... 14 I think the IJFCS'05 paper by Leung: Descriptional complexity of nfa of different ambiguity provides an example with a family of NFA accepting finite languages that involve an exponential blowup for "disambiguation" (in the proof of Theorem 5). What is more, those automata have a special structure (DFA with multiple initial states). 14 If infinite words are in your scope, you can generalize DFA (with parity condition) to the so-called Good-for-Games automata (GFG), that still have polynomial containment. A NFA is GFG if there is a strategy$\sigma:A^*\times Q\times A\to \Delta$, that given the prefix read so far and the current state and letter, chooses a transition to go to the next ... 14 Essentially the same argument is made by Andries P.J. van der Walt (1976, Lemma 2.3 and Example 2.9) for the variant of the pumping lemma where$N$letters are marked and all three of$x$,$y$,$z$must contain marked letters. See also Autebert, Boasson, and Cousineau (1978) for more properties of abstract families of languages satisfying this variant of ... 14 Another paper to look at: Kai Salomaa, "Language Decompositions, Primality, and Trajectory-Based Operations", 2008. 13 This question generated a lot of literature in the 80's, partly due to a bad approach to the problem. This is a rather long story that I will try to summarize in this answer. 1. The case of finite words One can find two definitions of a minimal DFA in the literature. The first one is to define the minimal DFA of a regular language as the complete DFA ... 13 (I guess the important word in the original question is published''.) There is such an encoding of context-free parsing (more exactly of CYK-style parsing) in Roland Axelsson, Keijo Heljanko, and Martin Lange, Analyzing Context-Free Grammars Using an Incremental SAT Solver, ICALP 2008, Lecture Notes in Computer Science vol. 5126, pp. 410--422, doi:10.... 13 This question is related to the so called insertion systems. An insertion system is a special type of rewriting system whose rules are of the form$1 \rightarrow r$for all$r$in a given language$R$. Let us write$u \rightarrow_R v$if$u = u'u''$and$v = u'ru''$for some$r \in R$. Let us denote by$\buildrel{*}\over\rightarrow_R$the reflexive ... 13 The answer is yes. Suppose we have a factorization$Q = A\cdot B$. One easy observation is that$A$and$B$must be disjoint (since for$w\in A\cap B$we get$w^2\in Q$). In particular, only one of$A,B$can contain$\epsilon$. We can assume wlog (since the other case is completely symmetric) that$\epsilon\in B$. Then since$a$and$b$cannot be factored ... 13 Yes, every regular expression can be converted into an unambiguous one by converting to a DFA and then to a regular expression. And no, there aren't any inherently ambiguous regular languages in the sense described in the question. This is a classic result in automata theory: R. Book, S. Even, S. Greibach and G. Ott, Ambiguity in graphs and expressions, ... 12 The bounds... We have in fact$NFA(L) \ge Cov(M) + Cov(N)$, see Theorem 4 in (Gruber & Holzer 2006). For an upper bound, we have$2^{Cov(M)+Cov(N)} \ge DFA(L) \ge NFA(L)$, see Theorem 11 in the same paper. ...cannot be substantially improved There can be a subexponential gap between$Cov(M)+Cov(N)$and$NFA(L)$. The following example, and the proof ... 12 You can show$ |L|^i $is a tight upper bound by using the following language:$ L = \{ ab,aab,aaab,\ldots,a^kb \mid k \geq 1 \}. $Any concatenation gives a new string. For a lower bound, I can suggest the following unary language:$ U = \{a,aa,aaa,\ldots,a^k \mid k \geq 1 \} $. Then,$ U^i = \{ a^i,a^{i+1},\ldots,a^{ki} \} $and so$ |U^i| = i|U|-...

12

In general, $\omega$-regular languages may not have a unique minimal DBW. For example, the language "infinitely many a's and infinitely many b's" has two 3-state DBWs (in the picture replace $\neg a$ by $b$): As you can see, they are not topologically equivalent. Hence, the minimization problem is harder than the finite case, and in fact, it is NP-complete....

12

If $\mathrm{P\subseteq CSL}$, then $\mathrm{P\subseteq DSPACE}(n^2)$. By a padding argument, this implies $$\mathrm{DTIME}(t(n))\subseteq\mathrm{DSPACE}\bigl(t(n)^\epsilon\bigr)$$ for every superpolynomial well-behaved function $t(n)$ and every $\epsilon>0$. I believe such a strong advantage of space over time is not expected to be true. The best ...

12

No, the exponential lower bound for determinization holds already for unambiguous NFAs. This is obtained as follows: Consider the alphabet $\{a,b\}$, and the language: $$L_k=\{w\in \{a,b\}^*:\text{the k-th before last letter in }w\text{ is }b\}$$ It's easy to construct an unambiguous NFA for $L_k$: the NFA guesses when the $k$ before last letter is, and ...

11

The answer is no. I'll give an example of a language $L$ which is regular in binary but not in unary: Consider $L=\{10^k|k\in \mathbb{N}\}$. The corresponding language in unary is $L'=\{1^{2^k}|k\in \mathbb{N}\}$. It's easy to see that $L$ is regular while $L'$ is not even context free. L'' also isn't regular either, by the link @Sylvain posted in his ...

11

A Non deterministic XOR automaton (NXA) fits your question. A NXA $M$ is essentially an NFA, but a word $w\in \Sigma^*$ is said to be in $L(M)$ if it is accepted by an odd number of paths (Xor relation) instead of being accepted if there exists an accepting path for it (Or relation). NXAs are used for creating small representations of regular languages as ...

11

There exists a FPRAS (Fully Polynomial Randomized Approximation Scheme) for the problem of counting the words of length $n$ accepted by a NFA in the general case (without restricting to the acyclic NFA case). The result was published this year on STOC, by Arenas, Croquevielle, Jayaram and Riveros. Here is the talk https://youtu.be/tyK-uujHMLU and the paper ...

11

It seems there is a paper answering this exact question, and even in the more general case of $\omega$-regular languages, but I cannot find an open-access version. If somebody finds a link without paywall it would be great. I requested the full-text on ResearchGate. Title: Which Finite Monoids are Syntactic Monoids of Rational omega-Languages. Authors: ...

11

In a more elementary way than Denis's answer, the following is extracted from Pippenger's "Theories of Computability", p.87, and immediate to check. Definition: Let $M$ be a monoid, and $Y \subseteq M$. Define the congruence relation $\equiv_Y$ over $M$ by $x \equiv_Y y$ iff $\big[\forall w, z \in M$, $wxz \in Y \Leftrightarrow wyz \in Y\big]$. Definition:...

11

The terminology rigid seems to be relatively new compared to the term disjunctive used in the late 70's (and probably before, I didn't check for earlier references). A subset $P$ of a monoid $M$ is disjunctive if and only if the syntactic congruence of $P$ in $M$ is the equality relation. Thus a monoid is the syntactic monoid of a language if and only if it ...

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