Lecture Notes in Computer Science 3472

3 State Verification 75
0, and x cannot be a UIO sequence for s. This means that from every state
except s, input sequence a1a2 ...ak leads to an accepting state. This includes
the initial states of the automata, and we conclude that a1a2 ...ak is accepted
by all automata. If x contains an r before the first occurrence of f , a similar
argument shows that the part of x between the last occurrence of r before the
first f and the first f is a string accepted by all automata.
(2) To show that it is PSPACE-hard to find out whether all states of a given
machine M have UIO sequences, we again use the Finite Automata Intersection
problem, with a very similar reduction. The same transitions as above are used,
plus some more, as described below. Add a new symbol at to the input alphabet
for every state t of every automaton. State t gets a transition on input at to s
with output 0. Every state u �= t, belonging to automaton Ai gets a transition
to the initial state of Ai on input at , with output 0. State s gets a transition
with input at to the initial state of the automaton which t belongs to, also with
output 0.
If the automata do not accept a common string, state s does not have a UIO
sequence, by the same argument as for problem (1) above. On the other hand, if
all automata accept the string x = a1 ...ak ,thenra1 ...ak f is a UIO sequence
for s, and for every other state t, the sequence ata1 ...ak f is a UIO sequence.
(3) Finally, we show that it is PSPACE-hard to determine whether any state
of a given machine has a UIO sequence. The sameconstructionasintheproof
of claim (1) is used, except that two identical copies of each automaton Ai are
used. This means that only state s can have a UIO sequence, and as above, if
all automata accept string x = a1 ...ak then ra1 ...ak f is a UIO sequence for s,
and otherwise there is none. This finishes the proof of Theorem 3.2.
In fact, Theorem 3.2 holds even if we consider only Mealy machines with
binary input and output alphabets [LY94].
The proof of Theorem 3.2 also yields an upper bound on the length of the
shortest UIO sequence for a given state. Since paths in the exponentially large
graph constructed in the PSPACE membership proof correspond directly to
input-output sequences with attached output sequences, it follows that if there
is a path of length m in the constructed graph from vertex (s, S \{s}) tosome
vertex (s ′ , ∅), for some states s, s ′ ,thens has a UIO sequence of length m. Since
no simple path in the graph can be longer than the number of vertices in the
graph, this shows that if s has a UIO sequence, then it has a UIO sequence of
length at most |S|·2 |S| .
Lee and Yannakakis proved a lower bound on the length of UIO-sequences as
well, showing that even if a state has a UIO sequence, it does not always have
one of polynomial length.
Theorem 3.3 (Lee and Yannakakis [LY94]). There are machines with states
that have UIO sequences, but only of exponential length.