如何判断一种语言是否是LL(1)?

发布于 2024-11-30 08:45:14 字数 84 浏览 8 评论 0原文

我有一个语法,我可以检查它是否是 LL(1)。然而,有没有办法检查文法生成的语言是否是LL(1)呢? LL(1) 语法和 LL(1) 语言到底有什么区别?

I have a grammar and I can check whether or not is is LL(1). However, is there any way to check if the language generated by the grammar is LL(1)? And what exactly is the difference between LL(1) grammars and LL(1) languages?

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装纯掩盖桑 2024-12-07 08:45:14

任何 LL(1) 语法都定义了 LL(1) 语言。根据定义,如果存在某种生成 LL(1) 语法的语言,则该语言就是 LL(1),因此该语言具有 LL(1) 语法这一事实自动意味着该语言是 LL(1) 。

详细地说,语言是一组字符串,该语言的语法是描述该语言的一种方法。有些语言具有 LL(1) 语法,而其他语言则没有。然而,语法不是 LL(1) 的事实并不意味着它描述的语言不是。例如,考虑以下语法:

A -> ab | ac

该语法不是 LL(1),因为当看到终端 a 时尝试预测 A 的产生式时,它包含 FIRST/FIRST 冲突。然而,它描述了一种 LL(1) 语言,因为该语言也是由语法描述的,

A -> aX
X -> b | c

所以这些语法生成的语言(只包含 ab 和 ac)确实是 LL(1)。

确定任意文法描述的语言是否为 LL(1) 要困难得多,据我所知,唯一的方法是显式地展示初始文法生成的语言的 LL(1) 文法(这很棘手)或从数学上证明不存在这样的语法。

希望这有帮助!

Any grammar that is LL(1) defines an LL(1) language. By definition, a language is LL(1) if there is some grammar that generates it that is LL(1), so the fact that you have an LL(1) grammar for the language automatically means that the language is LL(1).

To elaborate, a language is a set of strings and a grammar for that language is a means of describing that language. Some languages have LL(1) grammars while others do not. However, the fact that a grammar is not LL(1) does not mean that the language it describes is not. For example, consider this grammar:

A -> ab | ac

This grammar is not LL(1) because it contains a FIRST/FIRST conflict when trying to predict the production for A when seeing terminal a. However, it describes an LL(1) language, since the language is also described by the grammar

A -> aX
X -> b | c

So the language generated by these grammars (which just contains ab and ac) is indeed LL(1).

Determining whether the language described by an arbitrary grammar is LL(1) is much harder and to the best of my knowledge the only way to do it would be to either explicitly exhibit an LL(1) grammar for the language generated by the initial grammar (which is tricky) or to mathematically prove that no such grammar exists.

Hope this helps!

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