The following Context-Free Grammar (CFG) : S → aB | bA A → a | as | bAA B → b | bs | aBB will generate
equal numbers of a’s and b’s
Read the lowercase s in the productions as the start symbol S. This is the classic grammar in which A derives strings with one more a than b, and B derives strings with one more b than a.
Check A: A → a has one extra a; A → aS adds an a to a balanced S; A → bAA adds one b and two extra a’s, net one extra a. Similarly B → b, B → bS and B → aBB each give one more b than a.
Then S → aB gives a plus (one extra b) and S → bA gives b plus (one extra a), so every string from S has the same number of a’s and b’s. For example S ⇒ aB ⇒ ab and S ⇒ bA ⇒ baS ⇒ baab.
The counts need not be both odd or both even (ab has one of each, abab has two of each), and they are never different, so only the equal-count description fits.
Hence, the answer is equal numbers of a’s and b’s.
The context free grammar given by
\(S \to XYX\)
\(X \to aX \mid bX \mid \lambda\)
\(Y \to bbb\)
generates the language which is defined by regular expression :
A context free grammar for \(L = \{ w \mid n_{0}(w) \gt n_{1}(w) \}\) is given by :
If all the production rules have single non - terminal symbol on the left side, the grammar defined is :
Match the following :
| List - I | List - II |
| a. Chomsky Normal form | i. \(S \to b\,S\,S \mid a\,S \mid c\) |
| b. Greibach Normal form | ii. \(S \to a\,S\,b \mid ab\) |
| c. S-grammar | iii. \(S \to AS \mid a\) \(A \to SA \mid b\) |
| d. LL grammar | iv. \(S \to a\,B\,S\,B\) \(B \to b\) |
The context free grammar for the language \(L = \{a^{n}b^{m}c^{k} \mid k = |n - m|, n \geq 0, m \geq 0, k \geq 0\}\) is
For the above specified passage, which of the following represent the grammar for the language accepted the machine?
Which of the folowing CFG(s) is/are in chomsky Normal form (All capital letters are variables & lower case are terminals)
A. S → ABC / AB
A → a
B → b
C → d
B. X → RT / TR
T → t
R → XT / r
C. P → qP / sQ
Q → r / s
D. M → MN / MP
N → nm / n
P → p
Choose the correct answer from the options given below:
Let G = (V, T, S, P) be a context-free grammar such that every one of its productions is of the form A → v, with |v| = K > 1. The derivation tree for any W ∈ L(G) has a height h such that
| A | S $\rightarrow$ Ab aS$\rightarrow$ aA A$\rightarrow$ a | B | S $\rightarrow$ Ab A $\rightarrow$ Sa A $\rightarrow$ a |
| C | S $\rightarrow$ AS S $\rightarrow$ aA A$\rightarrow$ a | D | S $\rightarrow$ Ab S $\rightarrow$ aA A$\rightarrow$ a |
| E | S $\rightarrow$ Sb A $\rightarrow$ Aa A $\rightarrow$ $\varepsilon$ |
Which of the following grammars is (are)ambiguous ?
(1) S → SS | aSb | bSa |e
(2) S → aSbS | bSaS |e
(3) S → aAB
A → bBb
B → A Ι e
Consider the grammar,
G = S → abScB | e,
B→ bB | b.
What is the language generated by the given grammar ?
Consider the given grammar
S → Abc | ad
A → eS | Cr| ξ
C → f/p/ξ
Compute FIRST (A)
Select the sentence which is meaningful and grammatically correct.