please explain your answer in detail.

give reasons why other options are incorrect.

give reasons why other options are incorrect.

The Gateway to Computer Science Excellence

+1 vote

The following Context-Free Grammar (CFG) :

$S \rightarrow aB | bA$

$A \rightarrow a | as | bAA$

$B \rightarrow b | bs | aBB$

will generate

- Odd numbers of $a's$ and odd numbers of $b's$
- Even numbers of $a's$ and even numbers of $b's$
- Equal numbers of $a's$ and $b's$
- Different numbers of $a's$ and $b's$

closed as a duplicate of:
UGCNET-Dec2014-II-35

0 votes

Answer C: Equal number of a's and b's

Option A is false because we can get even number of a's and even number of b's.

Example1: S -> bA -> bbAA -> bbaA -> bbaa

option B is false because we can get odd number of a's and odd number of b's.

Example2: S -> aB -> ab

The above two examples are creating equal number of a's and b's. This makes option D false

Option A is false because we can get even number of a's and even number of b's.

Example1: S -> bA -> bbAA -> bbaA -> bbaa

option B is false because we can get odd number of a's and odd number of b's.

Example2: S -> aB -> ab

The above two examples are creating equal number of a's and b's. This makes option D false

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