Lecture Notes in Computer Science 3472

3 State Verification 73
Thus (s, Q) a −→(t, P) if and only if t is reachable from s with an a-transition in
M, andP is the subset of δ(Q, a) thatcanbereachedbyana-transition from
Q with the same output as the a-transition from s to t.
Consider a vertex (s, Q) ofG, and an input sequence x ∈ I ∗ . It is straightforward
to prove by induction on the length of x that if Q ′ is the subset of Q
that produces the same output as s on input x , then the unique path in G that
starts in (s, Q) and is labeled with x ends in (δ(s, x ),δ(Q ′ , x )).
If a vertex (s, S \{s}) inG can reach a vertex (t, ∅), for some state t, via
a directed path labeled x , then it follows from the above that it has a UIO
sequence. Indeed, the subset of S \{s} that gives the same output as s on input
x is empty. Thus s can be uniquely identified by its output on x . On the other
hand, if there is no such path, then there is no input sequence x such that s
gives a different output than all other states on x .Thuss does not have a UIO
sequence.
In problem (1) we are given a state s and want to know if it has a UIO
sequence. Thus what we need to do is to determine whether some vertex (t, ∅) is
reachable from the vertex (s, S \{s})inG. This can be done in nondeterministic
polynomial space. There is no need to explicitly compute the whole graph G,
which would require exponential space. Instead, the search procedure keeps track
only of the current vertex, and for each step computes all its successors and
nondeterministically chooses to proceed to one of them. Representing a vertex
only requires polynomial space, and each vertex has at most one successor for
each symbol in the input alphabet. As Chapter 1.4.2 explains, a consequence
of Savitch’s theorem is that PSPACE=NPSPACE, and the membership of
problem (1) in PSPACE follows. The result for problem (1) extends to problems
(2) and (3), since they can be solved by applying the nondeterministic polynomial
space procedure for problem (1) once for each state.
(1) We now proceed to showing that the problem of determining whether a
specific state s of a given machine M has a UIO sequence is PSPACE-hard.
This is achieved by reduction from the following problem. In Finite Automata
Intersection we are given k deterministic total finite automata, A1,...Ak, all
defined over the same alphabet Σ. The question is whether the intersection of
the languages accepted by the automata is nonempty, i.e., if there is some string
in Σ ∗ that is accepted by all k automata. As seen in Chapter 1.4.2, this problem
is PSPACE-complete.
The reduction composes the automata to a Mealy machine M in the following
way. The machine has the same input alphabet as the automata, plus two letters,
r (reset) and f (finish). Its output alphabet is {0, 1}. The states of the machine
are the states of the automata plus one additional state s. Every transition in one
of the automata is a transition of M, with the same input as in the automaton,
and output 0. Also, every state t of an automaton Ai gets a transition to the
initial state of Ai with input r and output 0, and a transition to s with input f .
This transition has output 1 if t is accepting in Ai and 0 if it is non-accepting.
Finally, s has a transition to itself for every input, including r and f ,with