Answers to the exercise
(i) If the string abbaab is derived from aba in one step, we must find a common beginning and end of the string, as both are derived from a. The only choice is a -> ab, and consequently b -> ba. That is,
aba -> ab ba ab = abbaab
For the second step we get
abbaab -> ab ba ba ab ab ba = abbabaababba
(ii) If the string ababaabab is derived from aba in one step, we have two choices:
a -> ab and b -> abaab
and
a -> abab and b -> a
Return to Simple L-Systems.