28
There's a significant difference between the question as you pose it and the question posed in the exercise. The question asks for an example of a set of regular languages $L_{1}, L_{2}, \ldots$ such that their union
$$
L = \bigcup_{i=1}^{\infty}L_{i}
$$
is not regular. Note the range of the union: $1$ to $\infty$. Regular languages are closed under finite ...
17
Note that the languages $\emptyset$, $\{\epsilon\}$ and $\Sigma^*$ are regular. Let $L_2$ be any non-regular language over $\Sigma$.
Union. $\emptyset \cup L_2 = L_2$, which is non-regular; $\Sigma^*\cup L_2 = \Sigma^*$, which is regular.
Intersection. $\Sigma^*\cap L_2 = L_2$; $\emptyset\cap L_2 = \emptyset$.
Subtraction. $\Sigma^*\setminus L_2 = \...
16
First of all, the question you are asking is open, since an affirmative answer shows that $\sf NP = coNP$. In fact it is one of the most prominent open problems in computer science.
If $\sf P= NP$, then the class $\sf NP$ is closed under complement since $\sf P$ is. If on the other hand $\sf P \not = NP$ then we cannot say whether $\sf NP = coNP$ or not. ...
15
"Effectively closed" means that the family is closed under the operation, and that the closure can be computed by giving an automaton/grammar for it (if the original languages are also given in such an effective representation). E.g., given a finite state automaton, we can actually find an automaton for the complement.
Then it is a natural question, whether ...
15
For all of the examples in this answer, I'm taking the alphabet to be $\{0,1\}$. Note that the languages $\emptyset$ and $\{0,1\}^*$ are definitely not NP-complete.
The class of NP-complete languages is not closed under intersection. For any NP-complete language $L$, let $L_0 = \{0w\mid w\in L\}$ and $L_1 = \{1w\mid w\in L\}$. $L_0$ and $L_1$ ...
13
This is a common misconception: complexity is not a measure of size. That is, it's not that "bigger" language are harder.
Intuitively, a language becomes harder when it's harder to describe it (TMs being a form of description). For example, as @Yuval Filmus points out in the comments, the language whose description is "everything" is very easy to decide.
...
13
I was posting only a hint, then I saw other full answers, so this is a full (hidden) succinct solution :-)
13
The statement is false. Consider the language $L = \{a^n b : n \geq 0\}$. Then $L' = \{ a^n b a^n b : n \geq 0 \}$ is not regular (exercise).
The invalid point in your reasoning is a confusion between the following two languages: $L' = \{ ww : w \in L \}$ and $L'' = \{ w_1w_2 : w_1,w_2 \in L \}$. It is $L''$ which is the concatenation of $L$ with itself. ...
12
There is another way to look at this problem.
Consider that the Language $L$ is a CFL. This means that there is a grammar $G=\{N,\sum,P,S\}$ that satisfies the CFL. We can assume that this is in Chomsky Normal Form.
If $\epsilon$ is part of the language, trivially $\epsilon^R$ is also part of the language.
Now for every production of the form $P_1 \...
12
This characterization of bounded context-free languages is due to Ginsburg ("The Mathematical Theory of Context-Free Languages"), and appears as Corollary 5.3.1 in his book. For general $k$ there are some restrictions on the semilinear sets, but for $k \leq 2$ these restrictions are always satisfied, and so it is straightforward to deduce that the complement ...
11
Hint:
Suppose it was. Let $L$ be a recognizable language and let $\overline{L}$ be its complement. If $\overline{L}$ was recognizable as well, let $M_1$ and $M_2$ be recognizers of $L$ and $\overline{L}$, respectively.
Can you now use $M_1$ and $M_2$ to construct a decider for $L$? What does this mean for undecidable and recognizable problems like the ...
11
The reason that $A$ is not defined as in NP is that we use a finer notion for NP-completeness, which is mapping (Karp) reductions.
Intuitively, in a Karp reduction, we not only want the problem $A$ to be solvable using an oracle for $B$, but we also require that this oracle is only used once, and only as the very last operation.
Why this notion and not ...
11
You cannot do it in general.
It would make the complement of your language semi-decidable, like the language itself. Hence they would be both decidable by running both machines in parallel.
Of course, this concerns only the case of semi -decidable languages for which the TM does not always halt for words not in the language.
If you have a decidable ...
11
Suppose that $L_2 = \Sigma^*$ and $L_1$ includes the empty string. Then $L_1 \cdot L_2 = \Sigma^*$, yet $L_1$ doesn't even have to be computable.
11
Yes you are on right track.
We can first define ($A$ avoids $B$) as ($A$ - ($A$ has $B$)), where ($A$ has $B$) are strings of $A$ which contain strings of $B$ as substrings. Then ($A$ avoids $B$) will be strings of $A$ that do not contain strings of $B$ as substrings.
We can define ($A$ has $B$) as $A \cap (\Sigma^* B \Sigma^*)$. Then ($A$ has $B$) are ...
10
As Shaull noted in the comments, $\{a^n b^n\}$ works. The language is trivially context-free but not regular, so I'll show the complement is context-free. A word which is not of the form $a^n b^n$ is either $a^n b^m$ where $n\neq m$, or not of the form $a^n b^m$ at all. So
$(a+b)^{\ast}-{a^n b^n}=\{a^i b^j: i \neq j\} \cup ((a+b)^{\ast}-a^{\ast} b^{\ast})$
...
10
I looked up Hopcroft and Ullman 1979 and it say on page 281 that it is not closed under reversal. But I found no proof in my very fast look at the relevant chapter.
Searching the web does also give a negative answer, with counter example, on stackoverflow by a member of CS (notation adapted):
$(a+b+c)^*WcW^R$, where $W \in (a+b)^+$; this is non-...
10
Note that the construction for the intersection and union ("and" and "or") of two automata is exactly the same, except for the definition of which states are accepting. The same principle applies to any Boolean combination of any finite set of languages: use the product construction and the appropriate definition of which states should be accepting.
10
The notion of a PDA can be generalized to an $S(n)$ auxiliary pushdown automaton ($S(n)$-AuxPDA). It consists of
a read-only input tape, surrounded by endmarkers,
a finite state control,
a read-write storage tape of length $S(n)$, where $n$ is the length of the input string, and
a stack
In "Hopcroft/Ullman (1979) Introduction to Automata Theory, Languages, ...
10
Going by the OP's comments, the real question here is not the one in the title, but "Why is the set of regular expressions a context-free (rather than regular) language?"
The reason is simply the occurrence of parentheses in more complex regular expressions like $(a+b)^*b(a+c)^*$. In order for such an expression to be well-formed, the parentheses must be ...
10
Because regular langauges are closed under complementation. That is, if $L$ is regular, so is $\overline{L}$. (Exercise: prove this.)
So, suppose that $L$ is non-regular. If its complement $\overline{L}$ were regular, then $\overline{\overline{L}}=L$ would also have to be regular.
10
Let $L_1, L_2$ be languages, then the concatenation $L_1\circ L_2=\{w\mid w=xy, x\in L_1, y\in L_2\}$. If $L_2=\varnothing$, then there is no string $y\in L_2$ and so there is no possible $w$ such that $w=xy$. Thus for any $L_1$, we'll have $L_1\circ\varnothing = \varnothing$.
10
The claim is that the intersection of a regular language and a context-free language is context-free. You've intersected a regular language ($\{ab\}$) and a context-free language ($\{a^nb^n\mid n\geq 0\}$) and the result was a context-free language ($\{a b\}$). Sure, that language is also regular but every regular language is also context-free. The statement ...
9
The first closure property, closure under intersection, is a DIY proof if you choose the right model for the context-sensitive languages. By defining them with the help of linear-bounded automata you can run two of these automata successively to test (nondeterministically) for acceptance of the intersection.
Second, closure under complement, is hard! It ...
9
That is possible, but perhaps not exactly in the way you ask. As a start take the Dyck language $D_2$ of all strings of matching brackets over two pairs of brackets, say $\{ [,], (,) \}$, or more abstractly $a_1,b_1,a_2,b_2$.
Then every context-free language can be obtained from $D_2$ using homomorphisms, inverse homomorphisms and intersection with regular ...
9
Hint: Use dynamic programming. If the input is $x_1 \ldots x_n$, compute inductively whether $x_1 \ldots x_i \in L^*$. Use the fact that you can check whether $x_{j+1} \ldots x_i \in L$ in polynomial time.
9
So when doing the transition table of the two automata, if there is no transition, should I just ignore it like in the 3rd automaton?
If there is no transition in one of the automata, then that one won't accept the input word. Therefore, the automaton for the intersection should not accept either; not having a transition is one way to make sure of that.
8
First of all, homomorphism is a special case of substitution. Now define the homomorphism as a substitution $h$ that replaces each symbol $a$ in an alphabet $\Sigma$ for a string $h(a) \in \Gamma^*$, where $\Gamma$ can be another alphabet (they can perfectly be the same). Formally the homomorphism $h(\epsilon) = \epsilon$ and for all strings in $ \Sigma^*$, $...
8
Let $R$ be a regular language. Then there is a finite automaton $M$ for which the language, $L(M)$, of $M$ is $R$. Without loss of generality, we may assume that $M$ has no unreachable states. Construct a new FA, $M'$, as follows:
For each state $q$ of $M$ which has a sequence of transitions to a final state of $M$, make $q$ a final state. Call the set of ...
8
Suppose there are two words in the language whose lengths are relatively prime. Let these lengths be $x$ and $y$. We know (see this) that by adding these numbers to each other repeatedly, we can get any number greater than $(x - 1)(y - 1) - 1$. So if $x$ and $y$ are $13$ and $7$, we can write any number greater than $72$ as a linear combination of $7$ and $...
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