DFA complement/intersection problem

"Paul J. Lucas" <pjl@mac.com>
31 Mar 2002 23:30:06 -0500

          From comp.compilers

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DFA complement/intersection problem pjl@mac.com (Paul J. Lucas) (2002-03-31)
Re: DFA complement/intersection problem rcbilson@plg2.math.uwaterloo.ca (2002-04-06)
Re: DFA complement/intersection problem mickunas@cs.uiuc.edu (Dennis Mickunas) (2002-04-06)
Re: DFA complement/intersection problem heinrich@idirect.com (Kenn Heinrich) (2002-04-06)
DFA complement/intersection problem cfc@world.std.com (Chris F Clark) (2002-04-06)
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From: "Paul J. Lucas" <pjl@mac.com>
Newsgroups: comp.compilers
Date: 31 Mar 2002 23:30:06 -0500
Organization: Prodigy Internet http://www.prodigy.com
Keywords: lex, DFA, theory
Posted-Date: 31 Mar 2002 23:30:06 EST

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.

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.

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.

Then, to compute the intersection, you do a cross product of
all the states in L and M and relabel each pair of states (p,q)
in the intersected langauge N. (You also do stuff with the
transitions, but that's not relevant here.) Furthermore, a
state in N is accepting iff (p,q) is accepting.

Now, since ~M has no accepting states, N can't either; therefore
N is empty and this shows that L <= M.

However, if you do this for M <= L, you get the same result
because ~L has no accepting states so N' = intersect(M,~L)
doesn't either; therefore N' is empty and M <= L. But this
isn't right!

What am I missing? Where is the mistake.

- Paul

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