Re: DFA complement/intersection problem

Paul J. Lucas <pjl@mac.com> wrote:
>If I have two languages:
>
> L = a*
> M = (a|b)*
>
>it's obvious that L <= M (L is a subset of M) because all
>sequences of A* are also sequences of (a|b)*. However, when I
>write code for this, I get L <= M and M <= L which is wrong.
>From:
>
> L <= M === intersect(L,~M) = 0
>
>I first have two minimal DFA for both languages:
>
> L: (s0) --{a}-> (s0)
> M: (t0) --{a|b}-> (t0)
>
>That is: each DFA has a single state that is accepting. For L,
>there is a single transition on 'a' back to s0; for M, there
>are two transitions: one each for 'a' and 'b' back to t0.


What happens to machine L when it reads a 'b'? You haven't defined
the transition function in this case. I assume that this is because
you have an implicit "error" state. I think that making this "error"
state explicit will solve most of your problems.


>If I understand Hopcroft, et al, correctly, to compute ~M, the
>complement of M, you simply flip all the accepting and non-
>accepting states. For M, since there is only one state, it
>becomes non-accepting.


This makes sense, since M accepts any strings from the alphabet {a,b}.
Therefore, the complementary machine will accept nothing.


Applying the analogous construction for L as you state doesn't work the
same way, though -- L doesn't accept every string in {a,b}, just those
without 'b'. It makes sense that the complementary machine would accept
strings with a 'b' somewhere.


Richard

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