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00 Inbox/29622985 - 2026-04-28.md

@@ -16,3 +16,30 @@ tags:
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 ## 📝 Content
+
+A containment hierarchy of classes of formal grammars. Grammars are classified into four types with different limitations.
+
+### Type 0
+no restrictions
+
+### Type 1
+Each rule $sans(w)_1 -> sans(w_2)$ satisfies $sans(abs(w_1) <= abs(w_2))$ with the exception that $sans(S) -> epsilon$ is allowed if $sans(S)$ does not occur on any right hand side of rules.
+$sans(abs(w_1) <= abs(w_2))$ implies that no shortening rules are allowed.
+
+### Type 2 (context free)
+Same restriction as [[#Type 1|type 1]] and additionally for each rule $sans(w_1 -> w_2)$ the string $sans(w_1)$ contains only a single variable (i.e. $sans(w_1 in V)$)
+
+### Type 3 (regular)
+Same restriction as [[#Type 2|type 2]] and additionally $sans(w_2 in Sigma union Sigma V)$), i.e. the right hand side is either a single terminal symbol or a terminal symbol followed by a variable.
+
+> [!REMARK]
+> The order of the variable and the terminal symbol is chosen arbitrarily.
+> Important is just that within a regular grammar only a single order occurs. If for all rules $sans(w_2 in Sigma union V Sigma)$ it would also be a regular grammar.
+
+## Definition
+A language $sans(L subset Sigma^*)$ is said to be of type 0 (1,2,3) if there exist a type 0 (1,2,3) grammar $sans(G)$ for which $sans(L(G) = L)$.
+
+> [!REMARK]
+> In order to show that a language is op type 0 (1,2,3) one thus just has to find a corresponding grammar.
+> In order to show that a grammar is not of type 0 (1,2,3) one has to prove that no type 0 (1,2,3) grammar can exist generating the corresponding language.
+