Question

Question 1 Consider CFG: A => OA1A01 epsilon, Which of these strings can be generated with this grammar? 001100 0011 00100 0110

Answer 1

All the options are correct we can derive each option using the grammar given.

Option 1-

Rule	Application	Result
Start → A	Start	A
A → 0A	A	0A
A → 0A	0A	00A
A → 1A	00A	001A
A → 1A	001A	0011A
A → 0A	0011A	00110A
A → 0	00110A	001100

OR

Rule	Application	Result
Start → A	Start	A
A → 0A	A	0A
A → 0A	0A	00A
A → 1A	00A	001A
A → 1A	001A	0011A
A → 0A	0011A	00110A
A → 0A	00110A	001100A
A → ɛ	001100A	001100

Option 2-

Rule	Application	Result
Start → A	Start	A
A → 0A	A	0A
A → 0A	0A	00A
A → 1A	00A	001A
A → 1	001A	0011

Or

Rule	Application	Result
Start → A	Start	A
A → 0A	A	0A
A → 0A	0A	00A
A → 1A	00A	001A
A → 1A	001A	0011A
A → ɛ	0011A	0011

Option3-

Rule	Application	Result
Start → A	Start	A
A → 0A	A	0A
A → 0A	0A	00A
A → 1A	00A	001A
A → 0A	001A	0010A
A → 0	0010A	00100

Or

Rule	Application	Result
Start → A	Start	A
A → 0A	A	0A
A → 0A	0A	00A
A → 1A	00A	001A
A → 0A	001A	0010A
A → 0A	0010A	00100A
A → ɛ	00100A	00100

Option 4-

Rule	Application	Result
Start → A	Start	A
A → 0A	A	0A
A → 1A	0A	01A
A → 1A	01A	011A
A → 0	011A	0110

Or

Rule	Application	Result
Start → A	Start	A
A → 0A	A	0A
A → 1A	0A	01A
A → 1A	01A	011A
A → 0A	011A	0110A
A → ɛ	0110A	0110

Note: In each case, ambiguity arises as we can derive each option in two ways.