Download presentation

Presentation is loading. Please wait.

Published byAlison Cowdrey Modified over 2 years ago

1
Symbolic Model Checking Revision Slides Dr. Eng. Amr T. Abdel-Hamid NETW 703 Winter 2012 Network Protocols Slides based on slides of: Jim Kurose, Keith Ross, “Computer Networking: A Top Down Approach Featuring the Internet”, 2nd edition, Addison-Wesley, July 2002. Jiangchuan (JC) Liu, Assistant Professor, SFU & others

2
Dr. Amr Talaat Netw 703 Network Protocols Functionals Now, we can think of all temporal operators also as functions fr om sets of states to sets of states For example: or if we use the set notation AX p = (S - EX(S - p)) LogicSet p q p q p q p q p S – p False TrueS

3
Dr. Amr Talaat Netw 703 Network Protocols Fixpoint Characterizations Fixpoint CharacterizationEquivalences AG p = y. p AX y AG p = p AX AG p EG p = y. p EX y EG p = p EX EG p AF p = y. p AX y AF p = p AX AF p EF p = y. p EX y EF p = p EX EF p A(pUq) = y. q A (p X (y)) A(pUq)=q (p AX (p AU q)) E(pUq) = y. q E (p X (y)) E(pUq) = q (p EX (p EU q))

4
Dr. Amr Talaat Netw 703 Network Protocols EF Fixpoint Computation EF p = y. p EX y is the limit of the sequence: , p EX , p EX(p EX ), p EX(p EX(p EX )),... which is equivalent to , p, p EX p, p EX (p EX (p) ),...

5
Dr. Amr Talaat Netw 703 Network Protocols EF Fixpoint Computation s2s1s4s3 p p Start 1 st iteration p EX = {s1,s4} EX( )= {s1,s4} ={s1,s4} 2 nd iteration p EX(p EX ) = {s1,s4} EX({s1,s4})= {s1,s4} {s3}={s1,s3,s4} 3 rd iteration p EX(p EX(p EX )) = {s1,s4} EX({s1,s3,s4})= {s1,s4} {s2,s3,s4}={s1,s2,s3,s4} 4 th iteration p EX(p EX(p EX(p EX ))) = {s1,s4} EX({s1,s2,s3,s4})= {s1,s4} {s1,s2,s3,s4} = {s1,s2,s3,s4}

6
Dr. Amr Talaat Netw 703 Network Protocols EF Fixpoint Computation p EF(p)states that can reach p p EX(p) EX(EX(p))... EF(p) states that can reach p p EX(p) EX(EX(p)) ... EF(p)

7
Dr. Amr Talaat Netw 703 Network Protocols Greatest Fixpoint Given a monotonic function F, its greatest fixpoint is the least upp er bound (lub) of all the extensive elements: y. F y = { y | F y y } The greatest fixpoint y. F y is the limit of the following sequenc e (assuming F is -continuous): S, F S, F 2 S, F 3 S,... If S is finite, then we can compute the greatest fixpoint using the above sequence

8
Dr. Amr Talaat Netw 703 Network Protocols EG Fixpoint Computation Similarly, EG p = y. p EX y is the limit of the sequence: S, p EX S, p EX(p EX S), p EX(p EX (p EX S)),... which is equivalent to S, p, p EX p, p EX (p EX (p) ),...

9
Dr. Amr Talaat Netw 703 Network Protocols EG Fixpoint Computation s2s1s4s3 pp p Start S = {s1,s2,s3,s4} 1 st iteration p EX S = {s1,s3,s4} EX({s1,s2,s3,s4})= {s1,s3,s4} {s1,s2,s3,s4}={s1,s3,s4} 2 nd iteration p EX(p EX S) = {s1,s3,s4} EX({s1,s3,s4})= {s1,s3,s4} {s2,s3,s4}={s3,s4} 3 rd iteration p EX(p EX(p EX S)) = {s1,s3,s4} EX({s3,s4})= {s1,s3,s4} {s2,s3,s4}={s3,s4}

10
Dr. Amr Talaat Netw 703 Network Protocols EG Fixpoint Computation EG(p) EG(p) states that can avoid reaching pp EX(p) EX(EX(p))... EG(p) states that can avoid reaching p p EX(p) EX(EX(p)) ...

11
Dr. Amr Talaat Netw 703 Network Protocols Example 11/80 1 2 3 4 5 6 a,b c b,c a dc For the FSM below, formally check the following properties, using Fixpoint Theorm: AG(a ∨c ∨b) AF(a b) If failed show the subset of the design the property holds for as well as the counter example S = {1,2,3,4,5,6}, AP = {a,b,c,d}, R = {(1,2), (1,3),(2,3), (3,4), (4,4), (4,5), (5,2), (2,6), (6,1)} L(1) = {a,b}, L(2) = {c}, L(3) = {b,c}, L(4) = {a}, L(5) = {c}, L(6) = {d}

12
Dr. Amr Talaat Netw 703 Network Protocols Example (cont.) Remember that: H(a ∪ b) = H(a) ∪ H(b) ∪ H(c) ={1,4} ∪ {2,3,5} ∪ {1,3} = {1,2,3,4,5} AG(a ∨ c ∨ b) = AG p = y. p AX y = y. p AX y AX p = EX( p) I0 S = {1,2,3,4,5,6} I1 {1,2,3,4,5} ∩ S = {1,2,3,4,5} ∩ {1,3,4,5,6} = {1,2,3,4,5} I2 {1,2,3,4,5} ∩ AX(1,2,3,4,5) = {1,2,3,4,5} ∩ {1,3,4,5,6} = {1,3,4,5} This is because that : AX(1,2,3,4,5) = EX( (1,2,3,4,5)) = EX(6) = (2) = S- {2 } = {1,3,4,5,6} I3 {1,2,3,4,5} ∩ AX(1,3,4,5) = {1,3,4,5} This is because that : AX(1,3,4,5) = EX( (1,3,4,5)) = **** I3 = I2 H(AG(a ∨ b ∨ c)) = {1,3,4,5} The property does not hold, except for the above states, and it is clear that s tates {2,6} can be considered as counter examples. state 6 does not contain neither a,c,b and state 2 does not have a proceedin g one on one of its pathes path (2,6) 12/80

13
Dr. Amr Talaat Netw 703 Network Protocols Example (AF(ab)) 13/80

Similar presentations

OK

Web Science & Technologies University of Koblenz ▪ Landau, Germany Models of Definite Programs.

Web Science & Technologies University of Koblenz ▪ Landau, Germany Models of Definite Programs.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

By appt only movie Ppt on table tennis game Ppt on ministry of corporate affairs government Ppt on edge detection python Ppt on power grid corporation of india limited Ppt on bluetooth hacking windows Ppt on united nations organization Converter word para ppt online Ppt on improvement in food resources science Download ppt on oxidation and reduction for dummies