16 December 2000
Consider the set of DFA’s which can be constructed over some alphabet
that equivalence of DFA’s is an equivalence relation on this set.
I show that the relation is reflexive, symmetric, and transitive.
First, a DFA is clearly equivalent to itself, because if M is the DFA, L(M)=L(M).
Second, if M is equivalent to M’ then L(M)=L(M’)==>L(M’)=L(M)==> M’ is
equivalent to M.
Third, if M is equivalent to M’ and M’ equivalent to M” then
L(M)=L(M’)=L(M”) ==> M is equivalent to M”.
All semester you have been asked to write out definitions I have given you.
ask you to create the definition of what is called a “lazy NFA.”
All of the FA’s we
have defined this semester “chew up”
or single symbols of an alphabet at each
A lazy NFA “chews up” strings, that is, moving from one state to another
consumes a string.
Provide a formal definition for a lazy NFA, a formal definition of
acceptance of a string by a lazy NFA, and a formal definition of
the language of
A lazy NFA, M, is a 5-tuple, M=(Q,
, s, F,
where Q is a finite set of states,
is an alphabet, s
Q is the start state, F
Q is a set of final states, and
*xQ is a transition relation.
A string w is accepted by M if we may write
and find states a
such that a
F, (s, a
for i=1, 2, …, n-1.
The language of M is the set of strings accepted by M.