Lecture Notes in Computer Science 3472

50 Moez Krichen
(5) Edges emanating from a common node are labeled with distinct output symbols,
(6) For each leaf u of T ,ifsu is the label of u, xu and yu are respectively the
input and output sequences formed by the concatenation of the node and
edge labels on the path from the root to the leaf u, thenλ(su, xu) =yu.
The length of an ADS is the depth of its corresponding tree T . ⊓⊔
The execution of an ADS given as mentioned in the definition above is done
as follows: first, we execute the input symbol label of the root of the tree. Then,
we observe the response of the machine. Among the outgoing edges of the current
node of the tree we choose the one labeled by the observed output. Then, we
execute the label of the node to which this edge leads. We keep on repeating this
until we meet a node (leaf) labeled by a state of the machine. In that case, we
terminate the experiment by declaring that the initial state of the machine was
the label of that node.
Let us consider an example of an ADS for better understanding this.
s1
0
b
a
1
a
0
0
b
1
s5
a
0
a
0
s3 s2 s4
1
s6 b
1 0 1 0
Fig. 2.9. An ADS for machine M6 ([LY94]).
Example. Consider machine M6 shown in Fig. 2.8. A possible ADS for this machine
is given in Fig. 2.9. Suppose that the initial state of M6 is s6. Then, the
execution of this ADS will be as follows. We first apply a which is the label of the
root of the ADS. The machine returns 1 and moves to s1. Since 1 is observed,
the tester follows the right branch and applies b. The machine returns 0 and
1
b
0
a
a
0