computation-theory

问题 In the Chomsky classification of formal languages, I need some examples of Non-Linear, Unambiguous and also Non-Deterministic Context-Free-Language(N-CFL)? Linear Language : For which Linear grammar is possible( ⊆ CFG) e.g. L 1 = {a n b n | n ≥ 0 } Deterministic Context Free Language(D-CFG) : For which Deterministic Push-Down-Automata(D-PDA) is possible e.g. L 2 = {a n b n c m | n ≥ 0, m ≥ 0 } L 2 is unambiguous. A CF grammar that is not linear is nonlinear. L nl = {w: n a (w) = n b (w)} is

问题 In the Chomsky classification of formal languages, I need some examples of Non-Linear, Unambiguous and also Non-Deterministic Context-Free-Language(N-CFL)? Linear Language : For which Linear grammar is possible( ⊆ CFG) e.g. L 1 = {a n b n | n ≥ 0 } Deterministic Context Free Language(D-CFG) : For which Deterministic Push-Down-Automata(D-PDA) is possible e.g. L 2 = {a n b n c m | n ≥ 0, m ≥ 0 } L 2 is unambiguous. A CF grammar that is not linear is nonlinear. L nl = {w: n a (w) = n b (w)} is

问题 I know a n b n for n > 0 is not regular by the pumping lemma but I would imagine a*b* to be regular since both a,b don't have to be the same length. Is there a proof for it being regular or not? 回答1: Answer to your question: imagine a*b* to be regular, Is there a proof for it being regular or not? No need to imagine, expression a*b* is called regular expression (re), and regular expressions are possible only for regular languages. If a language is not a regular then regular expression is also

问题 Is the union of a collection of context-free languages always context-free ? Justify your answer ..... I know that the answer is yes, but how can I prove it ? 回答1: To show that the finite union of context-free languages is context-free you just have to build a context-free grammar for the union language, exactly as you would do to prove that the union of two context-free languages is context-free. If G1,...,GN are the context-free grammars for the N context-free languages you have, rename all

问题 Its well known in theoretical computer science that the "Hello world tester" program is an undecidable problem.(Here is a link what i mean by hello world tester My question is since given a program as input we can't say what the program will do,can we solve the reverse problem: Given set of input and output,is there an algorithm for writing a program that writes a program to achieve a one to one mapping between the given input and output. I know about metaprogramming but my question is more

问题 The standard proof of of the impossibility of solving the halting problem is usually something like this //does_halt() takes a function as input and returns true if it will ever finish computing function paradox() {if does_halt(paradox()) { while(true){} } } This proof requires that you call the halting function recursively, but is it possible to make a halting function that always computes the correct result as long as it isn't called on itself? 回答1: That proof does not require recursion.

问题 I am wondering are all infinite languages undecidable? They must be right, as the TM trying to decide an infinite language would just loop forever, which makes it a recgonizer, not a decider. Thanks guys. 回答1: No, there are many infinite languages that are decidable. One trivial example is the language {n € N | a^n} , i.e. the language of words that only contain the letter "a". This language can be matched by the regular expression a* . so it is a regular language and thus decidable. 回答2: As

问题 For example, how does a PDA know how to read a palindrome in L = {a, b}*? PDA that accepts palindromes over {a,b}* : So, based on my drawing of the PDA: How does it know when the first half of the string is on the final terminal (letter of the alphabet), and therefore knows to go from state 0 to state 1 (and furthermore knowing to "pop" letters from the stack backwards, hence creating the palindrome)? 回答1: This is a nondeterministic pushdown automata. The answer to your question is that it

问题 A = {0^a 1^b 2^c | a < b < c} I need to show that A is not context-free. I'm guessing I have to use the Pumping Lemma for this, but how? 回答1: The goal is to prove that for any string with length >= a minimum pumping length, the string cannot be pumped. That is, if you split it into substrings uvxyz , the string that results from making copies (or removing copies) of v and y are still in language A . Note that you only have to show that one string in the language cannot be pumped (as long as

问题 Besides strings of any number of a's and b's like aa.. or bb.. ,Would regular expression (a*+b*) contain a string like ab or any string ending with b ? Is (a*+b*) same as (a* b*) ? I am a little bit confuse about the strings generated by regular expression (a*+b*) and would really appreciate if someone can help. 回答1: Unless you're working with a regex language which explicitly classifies the *+ as a special token which either has a special meaning, or is reserved for future extension (and