Lecture Notes in Computer Science 3472

580 Therese Berg and Harald Raffelt
that exactly accepts U, otherwise we receive a word that behaves different then
the constructed automaton but agrees with M; a so called counterexample. Let
us study how to process a counterexample.
Counterexample A counterexample is used to correct the hypothesis A we
have about the machine. From the counterexample we will get a word that,
when added to the pack, will escape.
Let t be the counterexample of length m, t = a0 ...am−1. For0≤ i ≤ m let
ui be the prefix of length i of t and vi the corresponding suffix, i.e. t = uivi.
Let si = δ(γ(ε), ui) be the state of the automaton based on the pack that is
reached by computing on ui, the initial state be s0 = γ(ε) andsi+1 = δ(si, ai).
The acceptance of t by this automaton is given by whether the final state sm is
accepting, which corresponds to whether sm ∈ U.
Since the set of strings that are accepted from a state distinguishes a state
from another we look upon suffix vi as an experiment on corresponding state
si and through membership queries we find out whether sivi ∈ U. Thefact
that t is a counterexample means that t ∈ U ⇐⇒ sm �∈ U, wheret = s0v0 and
sm = smvm. So consequently there must exist one or more breakpoint positions
i such that sivi ∈ U ⇐⇒ si+1vi+1 �∈ U. Sincesivi = siaivi+1, thesuffixvi+1
is an experiment that distinguishes siai from si+1 = δ(si, ai) =γ(siai). Now
add si+1vi+1 with the appropriate label to the component Csi+1. Consequently
γ(siai) is not anymore si+1, hence siai escapes and we can go on with the pack
expansion process.
Example of Observation Pack Let us study the observation pack algorithm
applied to the example of the DFA Mex with the alphabet {a, b}, shownin
Figure 19.8. Initially we conduct an experiment for the empty string. The result
of a membership query for ε is + and the first component is initialized with
this observation, C0 = {(ε, +)}. The component’s corresponding suffix-set is
E0 = {ε}. The evolution of O for this example can be viewed in Table 19.1. The
sign − in the table means that the component is unchanged.
In the next step we want to close the pack, therefore we ask membership
queries for εa and εb. The results are the observations (a, +) and (b, +), whose
strings are like ε, the access string for component C0. The pack is now closed
and we can construct a corresponding automaton A 0 , which can be seen in
Figure 19.9.
b
a
a
b
a, b
a, b
Fig. 19.8. The machine Mex