Consider the following languages over = {0, 1},
A = {2 : w contains 011 as a substring} B = the language matched by the regular expression 0(0 + 1)*1
Let M(B) be the DFA obtained by converting N(B) using the subset construction with all unnecessary states removed. Let ¬M(B) be the DFA obtained by swapping the final and non-final states of MB. Let M be the product of M(A) and ¬M(B) with all unnecessary state removed.
How many states would M have? How many final states would M have? How many states does M(B) have? This question has been boggling me, I have spent a few hours on jFlap putting together the intersection of these two DFA, for A and ¬B. No success. Thank you.
The language M has to satisfy 2 conditions,
1) M(A) - has to have a substring 011
2) ¬M(B) - should not start with 0
and end with 1
You can solve this by breaking the language M into 2 parts,
1) that starts with 0
but does not end with 1
and having substring 001
2) and another that starts with 1
and having substring 001
Therefore the DFA for M would be something like this
The states from 2-5 takes care of part 1, and states from 6-9 takes care of part 2