Lecture Notes in Computer Science 3472

19 Model Checking 589
the new column label b (line 21). The result of this operation is shown as A 1 in
Table 19.3, now row(ε) �= row(a).
Next we check whether table A 1 is closed. The Learner discovers that there is
one row in the lower part of the table which is not in the upper part (lines 9–10).
In order to rectify this we add the row of a to the upper part and in the lower
part we add its successor states, shown in Table 19.3, Table A 2 . The extension
to the table gives rise to a non-closed table again. So for this reason we move
row of aa in the same manner to the upper part, see the result in Table A 3 .Now
the table is closed and automaton A 3 can be formed, shown in Figure 19.11. The
Learner queries the Oracle with this hypothesis, but the Oracle answers with a
counterexample.
A 0
A 3
ε a, b ε
a, b
a
aa
a, b
a, b
ε
b
A 4
Fig. 19.11. The machine’s approximations
The counterexample is t = ba, whereba ∈L(Mex ) but ba �∈ L(A 3 ). Finding
the breakpoint is the process of following A 3 along t and discovering where the
mistake of not accepting t is made. The breakpoint in this case is found for
i =0sinces0ba ∈L(Mex ) but s1a �∈ L(Mex ) where the states s0 and s1 are
represented by ε and a, respectively. We add the new column label a to ROT
and extend the lower part of the table.
In the loop of checking if ROT is closed, the Learner will discover that the
row of b does not exist in the upper part of the table. This row is moved to
the upper part of the table and the lower part is extended. The result shown in
A 4 , Table 19.3. This table is closed and with a final equivalence check with the
corresponding automaton A4 in Figure 19.11 as hypothesis the answer from the
Oracle is a ’yes’ and the algorithm terminates.
19.4.4 Discrimination Trees
In this section we will discuss a fourth approach to implementing the Learner
in the setting described earlier. Instead of using sets for storing the Learner’s
observations, as in the observation pack algorithm, we will show how it is possible
to use a tree.
The Learner’s data structure is in this case a binary tree with labeled nodes.
Definition 19.20. Given an alphabet Σ,adiscrimination tree is tuple DT =
(SD, ED, t) where
b
a
a
b
a
aa
a, b
a, b