L(G) = { w E {a, b}* : every symbol in w is b, and length(w) = 2 }

G = ? (select all correct answers)

Question

L(G) = { w E {a, b}* : every symbol in w is b, and length(w) >= 2 }

G = ?  (select all correct answers)

BlackTom AI · Accepted Answer

The language is {b^n | n ≥ 2} — all strings consisting solely of b's with length at least 2.

1) Relevant knowledge points and concepts
- A context-free grammar G with start symbol S generates a language by recursive productions.
- Right-recursive rules (S → b S) and left-recursive rules (S → S b) can both generate arbitrarily many b's.
- A base case production (S → bb) ensures the minimum length 2.

2) Step-by-step reasoning for S → b S | bb
- Base: S ⇒ bb produces b^2.
- Induction/derivation for n > 2: apply S → b S repeatedly (k = n−2 times) then apply S → bb.
  Example n = 3: S ⇒ b S ⇒ b bb = b^3.
  Example n = 5: S ⇒ b S ⇒ b b S ⇒ b b b S ⇒ b b b bb = b^5.
- Therefore this grammar generates exactly b^n for all n ≥ 2 and nothing else (no a's appear).

3) Step-by-step reasoning for S → S b | bb
- Base: S ⇒ bb produces b^2.
- Induction/derivation for n > 2: apply S → S b repeatedly (k = n−2 times) on the right, then replace the leftmost S by bb.
  Example n = 3: S ⇒ S b ⇒ bb b = b^3.
  Example n = 6: S ⇒ S b ⇒ S b b ⇒ S b b b ⇒ S b b b b ⇒ bb b b b b = b^6.
- Thus this left-recursive grammar also generates exactly b^n for all n ≥ 2.

4) Why the correct answers are correct
- Both S → b S | bb and S → S b | bb have a base that enforces minimum length 2 and recursion that adds one b per recursion, so they produce every b^n with n ≥ 2 and no other strings.

5) Why the other options are incorrect
- S → b S | b: Base b allows deriving b (length 1), so the language includes b and thus fails the “length ≥ 2” requirement.
- S → b S b | bb: Base bb (length 2) is fine, but recursion adds two b's per application (wraps b ... b), so the grammar generates only even-length strings b^2, b^4, b^6, ... and misses odd lengths (e.g., b^3), so it does not match all n ≥ 2.

6) Conclusion
Both "S --> b S | bb" and "S --> S b | bb" generate exactly { w ∈ {a,b}* : every symbol is b and length(w) ≥ 2 }. This matches the provided correct answers.

L(G) = { w E {a, b}* : every symbol in w is b, and length(w) >= 2 } G = ? (select all correct answers)多项选择题

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