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Results tagged with Search options user 303
11 results

Questions related to formal languages, grammars, and automata theory

You can choose the language of the halting problem $\qquad \displaystyle B_1 = \{\langle T \rangle \mid T \text{ halts on } \langle T \rangle\} \in B$ and its complement $\qquad \displaystyle C_1 = … answered Jun 10 '12 by David Lewis Yes,$A$is a CFL. Use the context free language (with the notation$|x|_0$meaning what you have as #0's in$x$):$B=\{x\in \{0,1\}^*:3\cdot|x|_0=|x|_1\}$and the morphism$f:\{a,b,c\}^*\rightarr …
answered Apr 28 '12 by David Lewis
Hint... There may be a direct application of the pumping lemma, but I suggest you take a look at closure properties. Consider the non-CFL in the first answer to the reference post, $P=\{a^p:p \text{ … answered May 14 '12 by David Lewis Below is a hint for working with the Turing Machine (TM) formalism for RE languages. But finishing that approach from the hint depends on how you've been working with TMs. You have a TM, say$T_L$to … answered May 8 '12 by David Lewis It is regular. Let's first work in binary, which will generalize to any base > 1. Let$M_{a,b}$be the language in question. For a = 1, b = 0 we get$M_{1,0} = \{1, 10, 11, 100, 101, ...\}$which … answered Mar 22 '12 by David Lewis Hints...$L_a$can be transformed into an well-known non-regular language by a very simple homomorphism that basically "loses" information. For$L_b$you might try transforming it into its "opposite … answered May 9 '12 by David Lewis Yes,$\mbox{Before}(\beta)$and$\mbox{After}(\beta)$are context-free languages. Here's how I would prove it. First, a lemma (which is the crux). If$L$is CF then:$\mbox{Before}(L,\beta) = \{ …
answered Mar 26 '12 by David Lewis
The first trick here is to think of the multiplication table as the transition table of an automaton $A$ with each state representing a letter in your multiplication table, but not worrying about acce …
answered Apr 24 '12 by David Lewis
As Kaveh says in a comment, Kleene bestowed the name way back when he kicked off automata theory and formal languages. I believe the term was arbitrary, though it has been many years since I read his …
answered May 10 '12 by David Lewis
One Turing machine is more powerful than one pushdown automaton -- that is a fundamental theorem of automata theory and can be proved in a number of ways. For example, the halting problem for TMs is …
answered Mar 22 '12 by David Lewis
Hint -- You need to show that your TM has the same power as a finite-state automaton (as the commenter Dave Clarke said), that is, given such a TM, construct a FSA that accepts the same language. B …
answered May 10 '12 by David Lewis