关于两类形式语言问题的技术咨询：非t#t串语言与CFG产生式集合

Hey there, let's break down your two formal language questions one by one, nice and clear:

1. Is the language of all {0,1,#}-strings not of the form t#t (t ∈ {0,1}*) context-free? #

Short answer: Yes, this language is context-free. Here's why:

First, let's formalize the language we're talking about:
$$ L = { w \in {0,1,#}^* \mid w \neq t#t \text{ for any } t \in {0,1}^* } $$

We can split $L$ into three disjoint subsets that cover all possible strings in $L$:

• Subset 1: All strings with no # character (these are just arbitrary {0,1}-strings, which is a regular language, hence context-free).
• Subset 2: All strings with a # where the length of the prefix before # is not equal to the length of the suffix after #.
• Subset 3: All strings with a # where the prefix and suffix have the same length but differ in at least one character.

Since context-free languages (CFLs) are closed under union, we just need to show each subset is context-free, then their union is too:

• For Subset 1: A trivial CFG works: $S \to 0S \mid 1S \mid \varepsilon$.
• For Subset 2: We can generate strings where the prefix is shorter than the suffix (or vice versa) with:
  $$
  S \to A#B \mid B#A \
  A \to 0A \mid 1A \mid \varepsilon \
  B \to 0B \mid 1B \mid 0 \mid 1
  $$
  Here, $A$ generates any {0,1}-string (including empty), and $B$ generates non-empty {0,1}-strings, ensuring length inequality.
• For Subset 3: We generate strings where prefix and suffix have the same length but differ somewhere, using recursive rules that preserve length and introduce a mismatch:
  $$
  C \to 0C0 \mid 1C1 \mid 0C1 \mid 1C0 \mid 0#1 \mid 1#0
  $$
  The base cases (0#1, 1#0) create explicit mismatches, and the recursive rules extend the string while keeping lengths equal (and retaining at least one mismatch).

Combining all these into a single CFG with start symbol $S$ (where $S \to \text{Subset1 rules} \mid \text{Subset2 rules} \mid \text{Subset3 rules}$) proves $L$ is context-free.

Note: It's okay that the complement of $L$ (the language of all $t#t$ strings) is not context-free—CFLs are not closed under complementation, so this doesn't contradict our conclusion.

2. Is the set of production rules of a context-free grammar (CFG) itself a regular set? #

First, let's clarify what this question means:
We can represent each CFG production rule as a string (e.g., using uppercase letters for non-terminals, lowercase for terminals, -> for the production arrow, and ε for empty right-hand sides). The question asks: is the collection of all such valid production strings a regular language? Or, for a specific CFG, is its finite set of productions a regular set?

The answer is Yes in both cases:

Case 1: All valid CFG productions #

We can describe the syntax of a valid production rule with a regular expression, which means the set is a regular language. For example, using standard symbol conventions:

• Non-terminals: [A-Z] (any uppercase letter)
• Terminals: [a-z] (any lowercase letter)
• Production arrow: "->"
• Empty right-hand side: "ε"

A valid production is either:

• A non-terminal followed by -> and one or more terminals/non-terminals: [A-Z]->[a-zA-Z]+
• A non-terminal followed by -> and the empty string symbol: [A-Z]->ε

Combining these gives the regular expression:
[A-Z]->([a-zA-Z]+|ε)

Since regular expressions define regular languages, the set of all valid CFG productions is regular.

Case 2: Productions of a specific CFG #

Any finite set of strings is a regular language. You can build a finite automaton that simply checks if the input string is exactly one of the production rules in the set—no recursion or counting needed. So even a single production rule is a regular set (a singleton language), and any finite collection of them is too.

内容的提问来源于stack exchange，提问作者Kitchen