Step ep-bounded bounded Bisimu imulation lation
Nishanthan Kamaleson
University of Birmingham HIERATIC meeting, Jena, September 2014
Joint work with: David Parker, Jonathan Rowe, Mate Puljiz and Chris Good
Step ep-bounded bounded Bisimu imulation lation Nishanthan - - PowerPoint PPT Presentation
Step ep-bounded bounded Bisimu imulation lation Nishanthan Kamaleson University of Birmingham Joint work with: David Parker, Jonathan Rowe, Mate Puljiz and Chris Good HIERATIC meeting, Jena, September 2014 Overview Markov chains &
Joint work with: David Parker, Jonathan Rowe, Mate Puljiz and Chris Good
1
1
{a}
1 0.2 0.1 0.3 1 1 0.4 0. 7 0.1 0.9
B1
6
7
Π0: B1={s0,s1,s2,s3,s4,s5}, B2={s6} B2 is a splitter for B1 (since e.g. P(s1,B2)=0≠1=P(s2,B2))
1
1
{a}
1 0.2 0.1 0.3 1 1 0.4 0.7 0.1 0.9
B2 B1
0.3
8
Π0: B1={s0,s1,s2,s3,s4,s5}, B2={s6} B2 is a splitter for B1 (since e.g. P(s1,B2)=0≠1=P(s2,B2)) Π1: B1={s0,s1,s4,s5}, B2={s6}, B3={s2,s3} B3 is a splitter for B1 (since e.g. P(s1,B3)=0≠0.3=P(s0,B3)) B2 B1
1
1
{a}
1 0.2 0.1 0.3 1 0.4 0.7 0.1 0.9
B3
1 0.3
9
Π0: B1={s0,s1,s2,s3,s4,s5}, B2={s6} B2 is a splitter for B1 (since e.g. P(s1,B2)=0≠1=P(s2,B2)) Π1: B1={s0,s1,s4,s5}, B2={s6}, B3={s2,s3} B3 is a splitter for B1 (since e.g. P(s1,B3)=0≠0.3=P(s0,B3)) Π2: B1={s1,s5,s4}, B2={s6}, B3={s2,s3}, B4={s0} B1 is a splitter for B4 (since e.g. P(s4,B1)=0.3≠1.0=P(s1,B1))
0.3
B2 B1 B3 B4
1
1
{a}
1 0.2 0.1 0.3 1 0.4 0.7 0.1 0.9 1
10
Π0: B1={s0,s1,s2,s3,s4,s5}, B2={s6} B2 is a splitter for B1 (since e.g. P(s1,B2)=0≠1=P(s2,B2)) Π1: B1={s0,s1,s4,s5}, B2={s6}, B3={s2,s3} B3 is a splitter for B1 (since e.g. P(s1,B3)=0≠0.3=P(s0,B3)) Π2: B1={s1,s5,s4}, B2={s6}, B3={s2,s3}, B4={s0} B1 is a splitter for B4 (since e.g. P(s4,B1)=0.3≠1.0=P(s1,B1)) No more splitters S/~ = { {s1,s5}, {s6}, {s2,s3}, {s0},{s4} } B2 B4
1
1
{a}
1 0.2 0.1 0.3 1 0.4 0.7 0.1 0.9
B3 B1 B5
0.3 1
11
S/~ = { {s1,s5}, {s6}, {s2,s3}, {s0},{s4} }
1
1
{a}
1 0.2 0.1 0.3 1 0.4 0.7 0.1 0.9
1 s6 {a}
1 0.3 0.3
s1,s5 s2,s3
1
s0 s4
1 0.7 0.7 0.3 1
12
13
Π0: B1={s0,s1,s2,s3,s4,s5}, B2={s6}
Similarly signatures will computed for other states.
1
1
{a}
1 0.2 0.1 0.3 1 1 0.4 0.7 0.1 0.9
B2 B1
0.3
14
Π0: B1={s0,s1,s2,s3,s4,s5}, B2={s6}
Similarly signatures will computed for all the states. Π1: B1={s0,s1,s4,s5}, B2={s6}, B3={s2,s3}
B2 B1
1
1
{a}
1 0.2 0.1 0.3 1 0.4 0.7 0.1 0.9
B3
1 0.3
15
Π0: B1={s0,s1,s2,s3,s4,s5}, B2={s6}
Similarly signatures will computed for all the states. Π1: B1={s0,s1,s4,s5}, B2={s6}, B3={s2,s3}
Π2: B4={s1,s5, s4}, B2={s6}, B3={s2,s3}, B1={s0}
B2 B1 B3 B4
1
1
{a}
1 0.2 0.1 0.3 1 0.4 0.7 0.1 0.9 1 0.3
16
S/~ = { {s1,s5}, {s6}, {s2,s3}, {s0},{s4} } B2 B1
1
1
{a}
1 0.2 0.1 0.3 1 0.4 0.7 0.1 0.9
B3 B4 B5
0.3
Π0: B1={s0,s1,s2,s3,s4,s5}, B2={s6}
Similarly signatures will computed for all the states. Π1: B1={s0,s1,s4,s5}, B2={s6}, B3={s2,s3}
Π2: B4={s1,s5, s4}, B2={s6}, B3={s2,s3}, B1={s0}
Further splitting is not possible
17
S/~ = { {s1,s5}, {s6}, {s2,s3}, {s0},{s4} }
1
1
{a}
1 0.2 0.1 0.3 1 0.4 0.7 0.1 0.9
1 s6 {a}
1 0.3 0.3
s1,s5 s2,s3
1
s0 s4
1 0.7 0.7 0.3 1
18
19
20
(“s6”)]
21
Π0: B1={s0,s1,s2,s3,s4,s5}, B2={s6}
Similarly signatures will computed for all the states. Π1: B1={s0,s1,s4,s5}, B2={s6}, B3={s2,s3}
B2 B1
1
1
{a}
1 0.2 0.1 0.3 1 0.4 0.7 0.1 0.9
B3
1 0.3
22
1
1
{a}
1 0.2 0.1 0.3 1 0.4 0.7 0.1 0.9 1 0.3
S/~ = { {s0,s1,s4,s5}, {s6}, {s2,s3}} 1 s6 {a}
1 1.0
s0,s1,s4,s5 s2,s3
1
23
24
25