348
I have personally enjoyed several Aha! moments from studying basic automata theory. NFAs and DFAs form a microcosm for theoretical computer science as a whole.
Does Non-determinism Lead to Efficiency? There are standard examples where the minimal deterministic automaton for a language is exponentially larger than a minimal non-deterministic automaton. ...
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 ...
34
There are many good theoretical reasons to study N/DFAs. Two that immediately come to mind are:
Turing machines (we think) capture everything that's computable. However, we can ask: What parts of a Turing machine are "essential"? What happens when you limit a Turing machine in various ways? DFAs are a very severe and natural limitation (taking away ...
34
Yes, there is. Define a context-free expression to be a term generated
by the following grammar:
$$
\begin{array}{lcll}
g & ::= & \epsilon & \mbox{Empty string}\\
& | & c & \mbox{Character $c$ in alphabet $\Sigma$} \\
& | & g \cdot g & \mbox{Concatenation} \\
& | & \bot & \mbox{...
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 ...
31
To add one more perspective to the rest of the answers: because you can actually do stuff with finite automata, in contrast with Turing machines.
Just about any interesting property of Turing machines are undecidable. On the contrary, with finite automata, just about everything is decidable. Language equality, inclusion, emptiness and universality are all ...
27
State. you need to learn that one can model the world (for certain problems) as a finite state space, and one can think about computation in this settings. This is a simple insight but extremely useful if you do any programming - you would encounter state again and again and again, and FA give you a way to think about them. I consider this to be a sufficient ...
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 ...
21
Although it is not really the reason they were originally studied, finite automata and the regular languages they recognize are tractable enough that they have been used as building blocks for more complicated mathematical theories. In this context see particularly automatic groups (groups in which the elements can be represented by strings in a regular ...
19
About Q1:
Both the ambiguity problem (given a CFG, whether it is ambiguous) and the inherent ambiguity problem (given a CFG, whether its language is inherently ambiguous, i.e. whether any equivalent CFG is ambiguous) are undecidable. Here are the original references:
The undecidability of ambiguity was proved by Cantor (1962), Floyd (1962), and Chomsky and ...
18
You are asking (at least) two different questions: (a) What parts of theory build on finite automata nowadays? (b) Why were finite automata developed in the first place? I think the best way to address the latter is to look at the old papers, such as:
Rabin, Scott, Finite Automata and Their Decision Problems, 1959
Here are the first two paragraphs:
...
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
Lower bounds for algebraic circuits
In the setting of algebraic circuits, where a lower bound on circuit size is analogous to a lower bound on time, many results are known, but there are only a few core techniques in the more modern results. I know you asked for time lower bounds, but I think in many cases the hope is that the algebraic lower bounds will one ...
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
Another reason is that they're relatively practical theoretical models. A Turing machine, apart from the impossibility of the infinite tape, is kind of an awkward fit for what it's like to program a computer (note that this is not a good analogy to begin with!). PDAs and DFAs however are quite amenable to being models of actual programs in the sense that a ...
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
Finite automata in which the initial state is also the unique accepting state have the form $r^∗$, where $r$ is some regular expression. However, as J.-E. Pin points out below, the converse is not true: there are languages of the form $r^*$ which are not accepted by a DFA with a unique accepting state.
Intuitively, given a sequence of states $q_0, \ldots, ...
14
An $\omega$-regular language is actually quite low in the Borel hierarchy (inside $\Delta_3$), a result due to
R. McNaughton, Testing and generating infinite sequences by a finite automaton, Information and Control 9 (1966), 521-530.
For a proof and more details, you can look at Chapter 3 of the following book
D. Perrin et J.-É. Pin, Infinite words, ...
14
Let $A = \{1, ..., k\}$ be an ordered alphabet. Then each word on $A^*$ can be viewed as a number in base $k + 1$ (note that $0$ is never used on purpose). Now define
$$
rank(u) = \begin{cases}
u &\text{if $u \in L$} \\
0 &\text{otherwise}
\end{cases}
$$
Then $rank$ preserves the shortlex (or radix) order, which is the order $\leqslant$ on $A^*$ ...
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
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
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
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 ...
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 ...
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