Lecture Notes in Computer Science 3472

44 Moez Krichen
In this proposition, the discrete partition of S means the partition in which
all blocks are singletons. The proof of the proposition follows immediately from
the definitions of a PDS and the initial uncertainty of x .
Now let us explain a bit about the current uncertainty. First, it provides us
with the possible final states of the machine at the end of the distinguishing
experiment. For example for machine M5 (Fig. 2.6), we have:
σ(ab) ={{s3}01, {s1}00, {s2}11}
But, that is not very important since, here, our goal is to identify the initial
state of the machine and not its final state.
However, our knowledge about the current uncertainty of the intermediary
steps of the experiment may help in computing a PDS. Let us consider machine
M5 again for better explaining this. After executing the input symbol a, wehave:
σ(a) ={{s1, s2}0, {s3}1}
That tells us that if on a M5 outputs 0 then we deduce that the machine is
currently occupying either s1 or s2. Similarly, if it outputs 1 we deduce that M5
is currently occupying s3. Thus, computing a PDS for M5 becomes a bit easier.
All what remains to do is to find a sequence (b for example) which distinguishes
s1 and s2. If such a sequence is found then a possible PDS of the machine will
be the input sequence obtained by appending this sequence to a.
From the preceding, we deduce that computing a PDS for a given machine
can be done recursively by consecutive refinements of the current uncertainty:
at each step, we choose an input symbol which refines at least one block of the
current uncertainty. We keep on doing that till reaching an uncertainty made
only by singletons. In that case, a PDS for the machine is the one obtained by
appending the consecutive used input symbols.
Now, the difficulty is with how to choose the suitable input symbol at each
step. The idea is that there are some “bad” inputs that must not be used. Let
us consider machine M3 (Fig. 2.3) for explaining this. M3 is a minimal machine,
however, it has no PDS. The problem is that both a and b cause irrevocable loss
of information about its initial state. For example, s1 and s2 are merged when
applying input symbol a. This means that independently on whether the initial
state is s1 or s2, on input a M3 outputs 0 and moves to the same final state
s1. Consequently, whatever we will apply next, we will be unable to determine
whether M3 was initially at s1 or s2. The input a is said to be invalid for the
block {s1, s2}. The following gives the mathematical definition of valid inputs.
Definition 2.5. For a given Mealy machine M =(I , O, S,δ,λ):
• An input a ∈ I is a valid input for a set of states C ⊆ S if:
∀ s, s ′ ∈ C :s �= s ′ ⇒ λ(s, a) �= λ(s ′ , a) orδ(s, a) �= δ(s ′ , a).
i.e., on input a, thestatess and s ′ either produce distinct output symbols
or move to distinct states.