computation-theory

I have this grammar S->S+S|SS|(S)|S*|a I want to know how to eliminate the left recursion from this grammar because the S+S is really confusing... Let's see if we can simplify the given grammar. S -> S*|S+S|SS|(S)|a We can write it as; S -> S*|SQ|SS|B|a Q -> +S B -> (S) Now, you can eliminate left recursion in familiar territory. S -> BS'|aS' S' -> *S'|QS'|SS'|e Q -> +S B -> (S) Note that e is epsilon/lambda. We have removed the left recursion, so we no longer have need of Q and B. S -> (S)S'|aS' S' -> *S'|+SS'|SS'|e You'll find this useful when dealing with left recursion elimination . My

I know this might be a newbie question, but I'm trying to make sense of this sentence(from a paper on a meta language that uses EBNF): Logical and (&) binds stronger than logical or (|). Before that it says: Conditions are: condition ::= condition (`&´ | `|´ ) condition | `!´ condition | relation relation ::= expression ( `=´ | `#´ | `<´ | `<=´ | `>´ | `>=´ ) expression thanks This refers to precedence . In other words, if you have A & B | C , you really have (A & B) | C . Operations with higher precedence/that bind stronger are evaluated first. 来源： https://stackoverflow.com/questions/16876455

I understand that complete GPUs are behemoths of computing - including every step of calculation, and memory. So obviously a GPU can compute whatever we want - it's Turing complete. My question is in regard to a single shader on various GPUs ("Stream Processor"/"CUDA Core"): Is it Turing complete? Can I (in theory) compute an arbitrary function over arbitrary inputs by using a single shader? I'm trying to understand at what "scale" of computation shaders live. Kamil Czerski Did You mean shader as a program used to compute shading? On wiki talk I found: (...)Shader models 1.x and 2.0 are indeed

There is a well-known fact that C++ templates are turing-complete , CSS is turing-complete (!) and that the C# overload resolution is NP-hard (even without generics). But is C# 4.0 (with co/contravariance, generics etc) compile-time turing complete ? Terrance Unlike templates in C++, generics in C# (and other .net lang) are a runtime generated feature. The compiler does do some checking as to verify the types use but, actual substitution happens at runtime. Same goes for Co and contravariance if I'm not mistaken as well as even the preprocessor directives . Lots of CLR magic. (At the

I just took my midterm but couldn't answer this question. Can someone please give a couple of examples of the language and construct a grammar for the language or at least show me how i will go about it? Also how to write grammar for L : L = {a n b m | n,m = 0,1,2,..., n <= 2m } ? Thanks in advance. Grijesh Chauhan How to write grammar for formal language? Before read my this answer you should read first: Tips for creating Context free grammars . Grammar for {a n b m | n,m = 0,1,2,..., n <= 2m } What is you language L = {a n b m | n,m = 0,1,2,..., n <= 2m } description? Language description :

We know that C++ template metaprogramming is Turing complete , but preprocessor metaprogramming is not . C++11 gives us a new form of metaprogramming: computation of constexpr functions. Is this form of computation Turing-complete? I am thinking that since recursion and the conditional operator (?:) are allowed in constexpr functions, it would be, but I would like someone with more expertise to confirm. Richard Smith tl;dr: constexpr in C++11 was not Turing-complete, due to a bug in the specification of the language, but that bug has been addressed in later drafts of the standard, and clang