Match List I with List II: Choose the correct answer from the options given below:List I List II (A) Type 0 (I) Finite automata (B) Type 1 (II) Tuning machine (C) Type 2 (III) Linear bound automata (D) Type 3 (IV) Pushdown automata
(A) - (II), (B) - (III), (C) - (IV), (D) - (I)
The Chomsky Hierarchy is a classification of formal grammars. It groups grammars into four types (Type 0, Type 1, Type 2, and Type 3) based on their generative power. Each type of grammar corresponds to a specific class of formal languages, and these language classes, in turn, are recognized by specific types of automata.
Let's break down the correspondence between the grammar types listed in List I and the automata models listed in List II:
Based on the standard Chomsky Hierarchy correspondence, we can create the following matches:
| List I (Grammar Type) | List II (Automata) | Match |
|---|---|---|
| (A) Type 0 | (II) Turing Machine | (A) - (II) |
| (B) Type 1 | (III) Linear Bound Automata | (B) - (III) |
| (C) Type 2 | (IV) Pushdown Automata | (C) - (IV) |
| (D) Type 3 | (I) Finite Automata | (D) - (I) |
Putting these matches together, we get the combination: (A) - (II), (B) - (III), (C) - (IV), (D) - (I).
This arrangement correctly pairs each type of grammar from the Chomsky hierarchy with the corresponding abstract machine (automaton) that recognizes the languages generated by that type of grammar.
| Type | Grammar Type | Language Type | Automaton | Production Rule Form (\(\alpha \rightarrow \beta\)) |
|---|---|---|---|---|
| 0 | Unrestricted | Recursively Enumerable | Turing Machine | No restrictions on \(\alpha\) and \(\beta\) (except \(\alpha \neq \epsilon\)) |
| 1 | Context-Sensitive | Context-Sensitive | Linear Bounded Automaton (LBA) | \(|\alpha| \le |\beta|\) (except possible \(\epsilon\) from start symbol S under certain conditions) |
| 2 | Context-Free | Context-Free | Pushdown Automaton (PDA) | \(A \rightarrow \beta\), where \(A\) is a single non-terminal |
| 3 | Regular | Regular | Finite Automaton (FA) | \(A \rightarrow aB\) or \(A \rightarrow a\) (right-linear) or \(A \rightarrow Ba\) or \(A \rightarrow a\) (left-linear) |
The Chomsky Hierarchy not only classifies grammars and languages but also establishes a hierarchy of computational power among the automata. As we move from Type 3 to Type 0, the expressive power of the grammars and languages increases, and consequently, the complexity and capabilities of the required automata also increase.
This hierarchy demonstrates that adding memory or computational capability to an automaton allows it to recognize a wider class of languages.
The reduced grammar equivalent to the grammar, whose production rules are given below, is
S → AB | CA
B → BC | AB
A → a
C → a B | b
Consider the following grammars:
G 1: S → aSb|bSa|aa
G 2: S → aSb|bSa|SS|λ
G 3: S → aSb|bSa|SS|a
G 4: S → aSb|bSa|SS|SSS|λ
Which of the following is correct w.r.t. the above grammars?Consider L = L 1 ∩ L 2
Where L 1 = {0 m1 m20 n1 n |m, n >= 0}
L 2 = {0 m1 n2 k | m, n, k ≥ 0}
Then, the language L is
Consider the language \(L = \left\{ {{a^n}{b^{n - 3}}\ | n > 2} \right\}\) on ∑ = {a, b}. Which one of the following grammars generates the language L?
Consider the following grammar:
S → XY
X → YaY | a and y → bbX
Which of the following statements is/are true about the above grammar?
(a) Strings produced by the grammar can have consecutive three a's
(b) Every string produced by the grammar have alternate a and b
(c) Every string produced by the grammar have at least two a's
(d) Every string produced by the grammar have b's in multiple of 2.