Lecture Notes in Computer Science 3472

1 Homing and Synchronizing Sequences 25
i.e., a trivial synchronizing sequence). In this application, the state space is
{0, 1} k and typically very big. Rho, Somenzi and Pixley [RSP93] suggested a
more practical algorithm for this special case based on binary decision diagrams
(BDDs).
1.4 Complexity
This section shows two hardness results for related problems. First, Section 1.4.1
shows that it is NP-hard to find a shortest homing or synchronizing sequence.
Second, Section 1.4.2 shows that it is PSPACE-complete to determine if there
is a homing or synchronizing sequence when the initial state is known to be in
a specified subset Q ⊆ S of the states.
1.4.1 Computing Shortest Homing and Synchronizing Sequences Is
NP-hard
If a homing or synchronizing sequence is going to be used in practice, it is natural
to ask for it to be as short as possible. We saw in Section 1.3.1 that we can always
find a homing sequence of length at most n(n−1)/2 if one exists and the machine
is minimized, and Sections 1.3.2 and 1.3.3 explain how to find synchronizing
or homing sequences of length O(n 3 ), for general machines. The algorithms
for computing these sequences run in polynomial time. But the algorithms of
Section 1.3.5 that compute minimal-length homing and synchronizing sequences
are exponential. In this section, we explain this exponential running time by
proving that the problems of finding homing and synchronizing sequences of
minimal length are significantly harder than those of finding just any sequence:
the problems are NP-hard, meaning they are unlikely to have polynomial-time
algorithms.
The Reduction Since only decision problems can be NP-complete, formally
it does not make sense to talk about NP-completeness of computing homing or
sequences. Instead, we look at the decision version of the problems: is there a
sequence of length at most k?
Theorem 1.21 ([Epp90]). The following problems, taking as input a Mealy machine
M and a positive integer k, are NP-complete:
(1) Does M have a homing sequence of length ≤ k?
(2) Does M have a synchronizing sequence of length ≤ k?
Proof. To show that the problems belong to NP, note that a nondeterministic
algorithm easily guesses a sequence of length ≤ k (where k is polynomial in the
size of the machine) and verifies that it is homing or synchronizing in polynomial
time.
To show that the problems are NP-hard, we reduce from the NP-complete
problem 3SAT [GJ79]. Recall that in a boolean formula, a literal is either a