Lecture Notes in Computer Science 3472

58 Moez Krichen
I = {s1, s2, s3, s4, s5, s6}
C = {s1, s2, s3, s4, s5, s6}
a
0
u2 I = {s1, s3, s5}
C = {s2, s4, s6}
a
1
u4 I = {s1, s3, s5}
C = {s1, s3, s5}
b
0
u5 I = {s1, s3, s5}
C = {s1, s4, s6}
a
u6
u12
0
C = {s2}
I = {s5}
1
I = {s3}
C = {s5}
u7
u8
u9
u10
u11
1
I = {s1, s3}
C = {s1, s5}
a
0
I = {s1, s3}
C = {s2, s6}
a
1
I = {s1, s3}
C = {s1, s3}
b
0
I = {s1, s3}
C = {s1, s4}
a
0
I = {s1}
C = {s2}
It is constructed as follows:
u1
1
I = {s2, s4, s6} u3
C = {s1, s3, s5}
a
0
I = {s2, s4, s6} u13
C = {s2, s4, s6}
a
1
I = {s2, s4, s6} u14
C = {s1, s3, s5}
b
0
I = {s2, s4, s6} u15
C = {s1, s4, s6}
a
0 1
I = {s4}
C = {s2}
u16 I = {s2, s6}
C = {s1, s5}
a
1
I = {s2}
C = {s5}
u22
Fig. 2.13. Another ADS for machine M6.
0
I = {s2, s6} u18
C = {s2, s6}
a
1
I = {s2, s6} u19
C = {s1, s3}
b
0
I = {s2, s6} u20
C = {s1, s4}
a
0
I = {s6}
C = {s2}
• Step 0: T is initialized to the one node tree. The initial and current sets of
the root of T are assigned {s1, s2, s3, s4, s5, s6}. At this step, T contains only
node u1 of the tree shown in Fig. 2.13.
• Step 1: T due to step 0 is not closed. According to the execution of Algorithm
6, the current set {s1, s2, s3, s4, s5, s6} of node u1 canberefinedby
applying a. Thus, node u1 is labeled by a and nodes u2 and u3 are added to
T (with their corresponding current and initial sets).
• Step 2: T due to step 1 is closed since the current set of u2 equals the initial
set of u3 and vice versa. At this step, we have
π(T )={{s1, s3, s5}, {s2, s4, s6}},
u17
u21