Consider the production rules of grammer G: S → AbB A → aAb ∣ λ B → bB ∣ λ Which of the following language L is generated by grammer G?
Understanding the Formal Grammar G
A formal grammar is a set of production rules that describe all possible strings in a formal language. The grammar G is defined by its production rules. We have the starting symbol S and non-terminal symbols A and B, along with terminal symbols 'a' and 'b', and the empty string 'λ'.
The production rules for grammar G are:
- \( S \rightarrow AbB \)
- \( A \rightarrow aAb \mid \lambda \)
- \( B \rightarrow bB \mid \lambda \)
To find the language L generated by grammar G, we need to see what strings can be derived starting from S.
Analyzing the Production Rules
Let's analyze what strings can be generated by each non-terminal symbol:
- Rule for A: \( A \rightarrow aAb \mid \lambda \). This rule allows us to generate pairs of 'a' followed by 'b', potentially nested, or the empty string.
- If we apply \( A \rightarrow \lambda \), we get the empty string.
- If we apply \( A \rightarrow aAb \) once and then \( A \rightarrow \lambda \), we get \( a\lambda b = ab \).
- If we apply \( A \rightarrow aAb \) twice and then \( A \rightarrow \lambda \), we get \( a(aAb)b = aaAbb = aa\lambda bb = aabb \).
- Applying \( A \rightarrow aAb \) k times and then \( A \rightarrow \lambda \) results in \( a^k b^k \).
- Rule for B: \( B \rightarrow bB \mid \lambda \). This rule allows us to generate sequences of 'b's or the empty string.
- If we apply \( B \rightarrow \lambda \), we get the empty string.
- If we apply \( B \rightarrow bB \) once and then \( B \rightarrow \lambda \), we get \( b\lambda = b \).
- If we apply \( B \rightarrow bB \) twice and then \( B \rightarrow \lambda \), we get \( b(bB) = bbB = bb\lambda = bb \).
- Applying \( B \rightarrow bB \) j times and then \( B \rightarrow \lambda \) results in \( b^j \).
Deriving the Language Generated by S
The starting rule is \( S \rightarrow AbB \). This means any string generated by S will consist of a string derived from A, followed by a literal 'b', followed by a string derived from B.
A string derived from A is of the form \( a^k b^k \) for some \( k \ge 0 \).
A string derived from B is of the form \( b^j \) for some \( j \ge 0 \).
Substituting these into the S production, a string generated by S is of the form \( (a^k b^k) b (b^j) \), where \( k \ge 0 \) and \( j \ge 0 \).
Combining the 'b' terms, we get \( a^k b^{k+1+j} \).
Let's represent the number of 'a's as \( n \) and the total number of 'b's as \( m \).
- Number of 'a's: \( n = k \)
- Number of 'b's: \( m = k + 1 + j \)
From \( n = k \), since \( k \ge 0 \), we have \( n \ge 0 \).
From \( m = k + 1 + j \), since \( k \ge 0 \) and \( j \ge 0 \), the minimum value for \( k+1+j \) occurs when \( k=0 \) and \( j=0 \), giving \( m = 0 + 1 + 0 = 1 \). In general, \( m \ge k+1 \). Substituting \( k=n \), we get \( m \ge n+1 \). This inequality is equivalent to \( m > n \).
Thus, the language L generated by grammar G consists of strings of the form \( a^n b^m \) where the number of 'a's is \( n \ge 0 \) and the number of 'b's is \( m > n \).
So, the language is \( L = \{a^n b^m \mid n \ge 0, m > n\} \).
Comparing with Options
Let's compare our derived language with the given options:
- \( L = \{a^n b^m \mid n \ge 0, m > n\} \) - This matches our result.
- \( L = \{a^n b^m \mid n \ge 0, m \ge 0\} \) - This includes cases where \( m \le n \), which our grammar cannot produce (e.g., 'a', 'aab', 'aaabbb').
- \( L = \{a^n b^m \mid n \ge m\} \) - This requires the number of 'a's to be greater than or equal to the number of 'b's, which is the opposite of our condition \( m > n \).
- \( L = \{a^n b^m \mid n \ge m, m > 0\} \) - Similar to option 3, this requires \( n \ge m \), which is incorrect.
Therefore, the language generated by grammar G is \( L = \{a^n b^m \mid n \ge 0, m > n\} \).
| Derivation Step | Resulting String Form | Conditions |
|---|---|---|
| \( A \Rightarrow^* a^k b^k \) | \( a^k b^k \) | \( k \ge 0 \) |
| \( B \Rightarrow^* b^j \) | \( b^j \) | \( j \ge 0 \) |
| \( S \rightarrow AbB \) | \( (a^k b^k) b (b^j) \) | \( k \ge 0, j \ge 0 \) |
| Simplify S derivation | \( a^k b^{k+1+j} \) | \( k \ge 0, j \ge 0 \) |
| Let \( n=k, m=k+1+j \) | \( a^n b^m \) | \( n \ge 0, m = n+1+j, j \ge 0 \) |
| Condition on m | \( m = n+1+j \implies m \ge n+1 \implies m > n \) | \( n \ge 0, m > n \) |
Revision Table: Formal Grammars and Languages
| Concept | Description |
|---|---|
| Formal Grammar | A set of rules for forming strings in a formal language. Typically defined by a tuple \( (V, \Sigma, P, S) \) where V is the set of non-terminal symbols, \( \Sigma \) is the set of terminal symbols, P is the set of production rules, and S is the start symbol. |
| Production Rule | A rule specifying how symbols can be replaced by other symbols (e.g., \( A \rightarrow aAb \)). Used to derive strings from the start symbol. |
| Terminal Symbols | Symbols that form the final strings of the language (e.g., 'a', 'b'). |
| Non-terminal Symbols | Symbols that appear on the left side of production rules and are eventually replaced by terminal symbols (e.g., S, A, B). |
| Empty String (λ) | A string with zero symbols. Can be part of the language or used in derivations. |
| Derivation | A sequence of applications of production rules starting from the start symbol to obtain a string of terminal symbols. |
| Language Generated | The set of all terminal strings that can be derived from the start symbol using the grammar's production rules. |
Additional Information: Types of Grammars and Languages
Formal grammars are classified into types based on the form of their production rules, forming the Chomsky hierarchy:
- Type 0 (Recursively Enumerable Grammars): No restrictions on production rules. Generate recursively enumerable languages.
- Type 1 (Context-Sensitive Grammars): Rules are of the form \( \alpha A \beta \rightarrow \alpha \gamma \beta \) where A is a non-terminal and \( \gamma \) is non-empty. Generate context-sensitive languages.
- Type 2 (Context-Free Grammars - CFG): Rules are of the form \( A \rightarrow \gamma \) where A is a non-terminal. The grammar G in this question is a Context-Free Grammar. Generate context-free languages.
- Type 3 (Regular Grammars): Rules are restricted to forms like \( A \rightarrow aB \) or \( A \rightarrow a \) (right-linear) or \( A \rightarrow Ba \) or \( A \rightarrow a \) (left-linear). Generate regular languages.
Context-free grammars are widely used in describing the syntax of programming languages and in parsing.
The language \( L = \{a^n b^m \mid n \ge 0, m > n\} \) generated by grammar G is a context-free language.
Important Questions from Regular Languages - Teaching
Consider the following regular expressions:
(a) r = a(b + a)*
(b) s = a(a + b) +
(c) t = aa*b
Choose the correct answer from the options given below based on the relation between the languages generated by the regular expressions above:
Let L 1 and L 2 be languages over ∑ = {a, b} represented by the regular expressions (a* + b)* and (a + b)* respectively.
Which of the following is true with respect to the two languages?
Consider the following grammar:
S → 0A|0BB
A → 00A|λ
B → 1B|11C
C → B
Which language does this grammar generate?
Consider ∑ = {w, x} and T = {x, y, z}. Define homomorphism h by:
h(x) = xzy
h(w) = zxyy
If L is the regular language denoted by r = (w + x*)(ww)*, then the regular language h(L) is given byMatch List I with List II:
L R: Regular language, LCF: Context free language
L REC : Recursive language, L RE : Recursively enumerable language.
List I
List II
(A) Recursively Enumerable language
(I) L̅ REC ∪ L RE
(B) Recursive language
(II) L̅ CF ∪ L REC
(C) Context Free language
(III) L R∩ L CF
Choose the correct answer from the options given below:
