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模型检测方法

模型检测方法. 中国科学院软件研究所 张文辉 http://lcs.ios.ac.cn/~zwh/pv. BDD. a. (a  b)  (  b  c)  (a  d). b. b. c. c. c. c. d. d. d. d. d. d. d. d. 0. 0. 0. 0. 0. 1. 0. 0. 1. 1. 1. 1. 1. 1. 0. 0. BDD. a. (a  b)  (  b  c)  (a  d). b. b. c. c. c. c. d. d. d.

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模型检测方法

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  1. 模型检测方法 中国科学院软件研究所 张文辉 http://lcs.ios.ac.cn/~zwh/pv

  2. BDD a (ab)(bc)(ad) b b c c c c d d d d d d d d 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 2

  3. BDD a (ab)(bc)(ad) b b c c c c d d d d d d d d 0 1 3

  4. BDD a (ab)(bc)(ad) b b c c c c d d d d d d d d 0 1 4

  5. BDD a (ab)(bc)(ad) b b c c c c d d d d d 0 1 5

  6. BDD a (ab)(bc)(ad) b b c c c c d d d d d 0 1 6

  7. BDD a (ab)(bc)(ad) b b c c c c d d d 0 1 7

  8. BDD a (ab)(bc)(ad) b b c c c c d d d 0 1 8

  9. BDD a (ab)(bc)(ad) b b c c c c d 0 1 9

  10. BDD a (ab)(bc)(ad) b b c c c c d 0 1 10

  11. BDD a (ab)(bc)(ad) b b c c d 0 1 11

  12. BDD a (ab)(bc)(ad) b b c c c c d d d d d d d d 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 12

  13. BDD a (ab)(bc)(ad) b b c c c c d d d d d d d d 0 0 0 0 0 1 0 0 1 1 1 1 1 1 0 0 13

  14. BDD a (ab)(bc)(ad) b b 0 c 1 c d 0 1 0 0 1 14

  15. BDD a (ab)(bc)(ad) b b c c d 0 1 15

  16. 限界模型检测与验证 • 从模型的局部考察一个性质是否满足 • 对一些不满足的性质可能很快知道问题 • 对一些满足的性质也可能很快知道结论

  17. 限界模型检测与验证

  18. 限界模型检测与验证

  19. 限界模型检测与验证 • M,s |=  ，限界模型 M0, M1, …. 问题：是否存在k，Mk,s |=m  ? • 存在k, Mk,s |=m ,则 M,s |=  ==> 系统满足性质 可靠性 K 较小时， 较快验证系统性质

  20. 限界模型检测与验证 • M,s |=  ，限界模型 M0, M1, …. 问题：是否存在k，Mk,s |=m  ? • 存在k, Mk,s |=m ,则 M,s |=  则 M,s |=  ==> 系统存在问题 可靠性 K 较小时， 较快查出系统问题

  21. 自动售茶机 {p0,q0} s0 {p1,q0} {p2,q0} s1 s2 {p4,q1} s3 s4 {p2,q0} s5 {p3,q2}

  22. E(q0 U q2) vs A(q0 R q2) • P0: • s0 • P1: • s0 s1; • s0 s2; • P2: • s0 s1 s3; • s0 s1 s5; • s0 s2 s4; • s0 s2 s5; 我们有 M2, s0 s1 s5 |= (q0 U q2) 因此 M2满足 E(q0 U q2) M 满足 E(q0 U q2) M0|= E(q0 U q2), M1|= E(q0 U q2) M2=(S,P2,s0,L)是最小 可确定E(q0 U q2)是否满足的限界模型

  23. AG(q0q2) vs EF(q0q2) • P0: • s0 • P1: • s0 s1; • s0 s2; • P2: • s0 s1 s3; • s0 s1 s5; • s0 s2 s4; • s0 s2 s5; 我们有 M2, s0 s2 s4 |= F(q0q2) 因此 M2满足 EF(q0q2) M 不满足 AG(q0q2) M0|=EF(q0q2), M1|=EF(q0q2) M2=(S,P2,s0,L)是最小 可确定AG(q0q2)是否满足的限界模型

  24. 限界模型 • P0: • s0 • P1: • s0 s1; • s0 s2; • P2: • s0 s1 s3; • s0 s1 s5; • s0 s2 s4; • s0 s2 s5; • P4: • s0 s1 s3 s4 s5; • s0 s1 s3 s5 s0; • s0 s1 s5 s0 s1; • s0 s1 s5 s0 s2; • s0 s2 s4 s5 s0; • s0 s2 s5 s0 s1; • s0 s2 s5 s0 s2; • P3: • s0 s1 s3 s4; • s0 s1 s3 s5; • s0 s1 s5 s0; • s0 s2 s4 s5; • s0 s2 s5 s0;

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