source: papers/FDL2012/abstraction_refinement.tex @ 62

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[52]1
2
[56]3\subsection{Properties of good refinement}
4When a counterexample is found to be spurious, it means that the current abstract model $\widehat{M}_i$ is too coarse and has to be refined.
5In this section, we will discuss about the refinement technique based on the integration of more verified properties of the concrete model's components in the abstract model to be generated. Moreover, the refinement step from $\widehat{M}_i$ to $\widehat{M}_{i+1}$ respects the properties below:
[52]6
7%\medskip
8
[58]9\begin{definition} An efficient \emph{refinement} verified the following properties:
[56]10\begin{enumerate}
11\item The new refinment is an over-approximation of the concrete model: $\widehat{M}_{i+1} \sqsubseteq \widehat{M}$.
12\item The new refinment is more concrete than the previous one:
13$\widehat{M}_{i} \sqsubseteq \widehat{M}_{i+1}$.
14\item The spurious counter-example in $\widehat{M}_i$ is  removed from
15$\widehat{M_{i+1}}$.
16\end{enumerate}
[58]17\end{definition}
[52]18
[56]19Moreover, the refinement steps should be easy to compute and ensure a fast
20convergence by the minimizing the number of iteration of the CEGAR loop.
[52]21
22
[56]23A possible refinement : concretization of selected abstract variables. How to choose variables and instants of concretization : introduce new CTL properties. The question is : how to select pertinent CTL properties ???
[52]24
[56]25\TODO{discussion sur comment garantir les points 1/2/3 et le reste du bon
26rafinement}
[58]27\subsection{Refinment by negation of the counterexample}
[52]28
[58]29\TODO{Mettre la def avant ?}
[56]30\TODO{Rafinement par négation du contre-exemple}
31The counterexample at a refinement step $i$, $\sigma$ is a path in the
32abstract model $\widehat{M}_i$ which dissatisfy $\Phi$.  In the counterexample given by the model-checker, the variables' value in each states are boolean.
33The spurious counter-example $\sigma$ is defined such that :
[52]34\begin{definition}
[56]35\textbf{\emph{The counterexample $\sigma$ :}} \\
[52]36\\
[56]37Let $\widehat{M}_i =\langle \widehat{AP}_i, \widehat{S}_i, \widehat{S}_{0i},
38\widehat{L}_i, \widehat{R}_i, \widehat{F}_i \rangle$ and let the length of the
39counterexample, $|\sigma| = n$: $ \sigma = s_{0} \rightarrow s_{1} \ldots
40s_{n}$ with $(s_{k}, s_{k+1}) \in \widehat{R}_i$ $\forall k \in [0..n-1]$.
41\begin{itemize}
42\item All its variables are concrete: $\forall s_i$ and $\forall p\in
43\widehat{AP}_i$, $p$ is either true or false
44(not {\it unknown}).
45\item  $\sigma$ is a counter-example in  $\widehat{M}_i$: $s_0\not\models \Phi$.
46\item  $\sigma$ is not a path of the concrete system $M$: $\exists k$ such
47that $(s_{k}, s_{k+1}) \not\in R$.
[57]48\end{itemize}
[52]49\end{definition}
50
[58]512. Negation of states in an AKS
52
53a) An (abstract) configuration in a state of the AKS represents a (convex ?) set of states of the concrete component.
54
55b) The negation of an configuration may be represented by a set of abstract configurations
56
57c) building the AKS of a spurious counter-example may lead to a blow-up of the number of states of the AKS
58
59
[52]60%\bigskip
61
[56]62%\begin{definition}
63%\textbf{\emph{Spurious counterexample :}} \\
64%\\
65%Let $\sigma_c = \langle s_{c,0}, s_{c,1}, s_{c,2}, ... , s_{c,k}, s_{c,k+1}, ... , s_{c,n}\rangle$ a path of length $n$ in the concrete model $M$ and in each state of $\sigma_c$ we have $s_{c,k} = \langle v_{c,k}^1, v_{c,k}^2, ... ,  v_{c,k}^{p'}, ... , v_{c,k}^{q'} \rangle$ with $\forall p' \in [1,q'], ~v_{i,k}^{p'} \in V_{c,k}$ and $V_{c,k} \in 2^{q'}$.\\
66%
67%\smallskip
68%
69%If $\forall k$ we have $\widehat{V}_{i,k} \subseteq V_{c,k}$ and $\forall v_{\bar{a}i,k} \in \widehat{V}_{i,k}, ~s_{i,k}|_{v_{\bar{a}i,k}} = s_{c,k}|_{v_{c,k}} $ then $M \nvDash \phi$ else $\sigma_i$ is \emph{spurious}.
70%
71%\end{definition}
[52]72
73
74
[56]75\subsection{Ordering of properties}
[52]76
77Before generating an abstract model to verify a global property $\phi$, the verified properties of all the components in the concrete model are ordered according to their pertinency in comparison to a global property $\phi$. In order to do so, the variable dependency of the variables present in global property $\phi$ has to be analysed. After this point, we refer to the variables present in the global property $\phi$ as \emph{primary variables}.
78
79%\bigskip
80
81The ordering of the properties will be based on the variable dependency graph.
82The variables in the model are weighted according to their dependency level
83\emph{vis-à-vis} primary variables and the properties will be weighted according to the sum of the weights of the variables present in it. We have decided to allocate a supplementary weight for variables which are present at the interface of a component whereas variables which do not interfere in the obtention of a primary variable will be weighted 0. Here is how we proceed:
84
85
86\begin{enumerate}
87
88\item {\emph{Establishment of primary variables' dependency and maximum graph depth}\\
89Each primary variable will be examined and their dependency graph is elobarated. The maximum graph depth among the primary variable dependency graphs will be identified and used to calibrate the weight of all the variables related to the global property.
90Given the primary variables of $\phi$, $V_{\phi} =  \langle v_{\phi_0}, v_{\phi_1}, ... , v_{\phi_k}, ... , v_{\phi_n} \rangle$ and $G{\_v_{\phi_k}}$ the dependency graph of primary variable $v_{\phi_k}$, we have the maximum graph depth $max_{d} = max(depth(Gv_{\phi_0}), depth(Gv_{\phi_1}), ... , depth(Gv_{\phi_k}), ... ,$\\$ depth(Gv_{\phi_n})) $.
91
92}
93
94\item {\emph{Weight allocation for each variables} \\
95Let's suppose $max_d$ is the maximum dependency graph depth calculated and $p$ is the unit weight. We allocate the variable weight as follows:
96\begin{itemize}
97\item{All the variables at degree $max_d$ of every dependency graph will be allocated the weight of $p$.}
98 \\ \hspace*{20mm} $Wv_{max_d} = p$
99\item{All the variables at degree $max_d - 1$ of every dependency graph will be allocated the weight of $2Wv_{max_d}$.}
100\\ \hspace*{20mm} $Wv_{max_d - 1} = 2Wv_{max_d}$
101\item{...}
102\item{All the variables at degree $1$ of every dependency graph will be allocated the weight of $2Wv_{2}$.}
103 \\ \hspace*{20mm} $Wv_{1} = 2Wv_{2}$
104\item{All the variables at degree $0$ (i.e. the primary variables) will be allocated the weight of $10Wv_{1}$.}
105 \\ \hspace*{20mm} $Wv_{0} = 10Wv_{1}$
106\end{itemize}
107
108We can see here that the primary variables are given a considerable
109ponderation due to their pertinency \emph{vis-à-vis} global  property. Furthermore, we will allocate a supplementary weight of $3Wv_{1}$ to variables at the interface of a component as they are the variables which assure the connection between the components if there is at least one variable in the dependency graph established in the previous step in the property. All other non-related variables have a weight equals to $0$.
110}
111
112
113\item {\emph{Ordering of the properties} \\
114Properties will be ordered according to the sum of the weight of the variables in it. Therefore, given a property $\varphi_i$ which contains $n+1$ variables, $V_{\varphi_i} =  \langle v_{\varphi_{i0}}, v_{\varphi_{i1}}, ... , v_{\varphi_{ik}}, ... , v_{\varphi_{in}} \rangle$, the weight  of $\varphi_i$ , $W_{\varphi_i} = \sum_{k=0}^{n} Wv_{\varphi_{ik}}$ .
115After this stage, we will check all the properties with weight $>0$ and allocate a supplementary weight of $3Wv_{1}$ for every variable at the interface present in the propery. After this process, the final weight of a property is obtained and the properties will be ordered in a list with the weight  decreasing (the heaviest first). We will refer to the ordered list of properties related to the global property $\phi$ as $L_\phi$.
116
117
118}
119
120\end{enumerate}
121
122%\bigskip
123
124\emph{\underline{Example:}}  \\
125
126For example, if a global property $\phi$ consists of 3 variables: $ p, q, r $ where:
127\begin{itemize}
128\item{$p$ is dependent of $a$ and $b$}
129\item{$b$ is dependent of $c$}
130\item{$q$ is dependent of $x$}
131\item{$r$ is independent}
132\end{itemize}
133
134Example with unit weight= 50.
135The primary variables: $p$, $q$ and $r$ are weighted $100x10=1000$ each. \\
136The secondary level variables : $a$, $b$ and $x$ are weighted $50x2=100$ each. \\
137The tertiary level variable $c$ is weighted $50$. \\
138The weight of a non-related variable is $0$.
139
140So each verified properties available pertinency will be evaluated by adding the weights of all the variables in it. It is definitely not an exact pertinency calculation of properties but provides a good indicator of their possible impact on the global property.
141
142\bigskip
143\begin{figure}[h!]
144   \centering
145%   \includegraphics[width=1.2\textwidth]{Dependency_graph_weight_PNG}
146%     \hspace*{-15mm}
147     \includegraphics{Dependency_graph_weight_PNG}
148   \caption{\label{DepGraphWeight} Example of weighting}
149\end{figure}
150
151%Dans la figure~\ref{étiquette} page~\pageref{étiquette},  
152
153
154
155After this pre-processing phase, we will have a list of properties $L_\phi  $ ordered according to their pertinency in comparison to the global property.
156
157
158
159
160
161
162
[58]163\subsection{Filtering properties}
[52]164
165The refinement process from $\widehat{M}_i$ to $\widehat{M}_{i+1}$ can be seperated into 2 steps:
166
167\begin{enumerate}
168
169\item {\emph{\underline{Step 1:}} \\
170
171As we would like to ensure the elimination of the counterexample previously found, we filter out properties that don't have an impact on the counterexample $\sigma_i$ thus won't eliminate it. In order to reach this obective, a Kripke Structure of the counterexample $\sigma_i$, $K(\sigma_i)$ is generated. $K(\sigma_i)$ is a succession of states corresponding to the counterexample path which dissatisfies the global property $\phi$.
172
173\bigskip
174
175\begin{definition}
176\textbf{\emph{The counterexample $\sigma_i$ Kripke Structure $K(\sigma_i)$ :}} \\
177Let a counterexample of length $n$, $ \sigma_i = \langle s_{\bar{a}i,0}, s_{\bar{a}i,1},\\ s_{\bar{a}i,2}, ... , s_{\bar{a}i,k}, s_{\bar{a}i,k+1}, ... , s_{\bar{a}i,n}\rangle $ with $ \forall k \in [0,n-1]$, we have \\
178$K(\sigma_i) = (AP_{\sigma_i}, S_{\sigma_i}, S_{0\sigma_i}, L_{\sigma_i}, R_{\sigma_i})$ a 5-tuple consisting of :
179
180\begin{itemize}
181\item { $AP_{\sigma_i}$ : a finite set of atomic propositions which corresponds to the variables in the abstract model $\widehat{V}_{i}$ }     
182\item { $S_{\sigma_i} = \{s_{\bar{a}i,0}, s_{\bar{a}i,1}, s_{\bar{a}i,2}, ... , s_{\bar{a}i,k}, s_{\bar{a}i,k+1}, ... , s_{\bar{a}i,n}\}$}
183\item { $S_{0\sigma_i} = \{s_{\bar{a}i,0}\}$}
184\item { $L_{\sigma_i}$ : $S_{\sigma_i} \rightarrow 2^{AP_{\sigma_i}}$ : a labeling function which labels each state with the set of atomic propositions true in that state. }
185\item { $R_{\sigma_i}$ = $ (s_{\bar{a}i,k}, s_{\bar{a}i,k+1})$ }
186\end{itemize}
187\end{definition}
188
189%\bigskip
190All the properties available are then model-checked on $K(\sigma_i)$.
191
192If:
193\begin{itemize}
194\item {\textbf{$K(\sigma_i) \vDash \varphi  \Rightarrow \varphi $ will not eliminate $\sigma_i$}}
195\item {\textbf{$K(\sigma_i) \nvDash \varphi  \Rightarrow \varphi $ will eliminate $\sigma_i$}}
196\end{itemize}
197
198%\bigskip
199
200
201\begin{figure}[h!]
202   \centering
203%   \includegraphics[width=1.2\textwidth]{K_sigma_i_S_PNG}
204%     \hspace*{-15mm}
205     \includegraphics{K_sigma_i_S_PNG}
206   \caption{\label{AKSNegCex} Kripke Structure of counterexample $\sigma_i$, $K(\sigma_i)$}
207\end{figure}
208
209%Dans la figure~\ref{étiquette} page~\pageref{étiquette},  
210
211%\bigskip
212
213
214\begin{figure}[h!]
215   \centering
216
217\begin{tikzpicture}[->,>=stealth',shorten >=1.5pt,auto,node distance=1.8cm,
218                    thick]
219  \tikzstyle{every state}=[fill=none,draw=blue,text=black]
220
221  \node[initial,state] (A)                            {$s_{\bar{a}i,0}$};
222  \node[state]           (B) [below of=A]     {$s_{\bar{a}i,1}$};
223
224  \node[state]           (C) [below of=B]        {$s_{\bar{a}i,k}$};
225
226  \node[state]           (D) [below of=C]       {$s_{\bar{a}i,n-1}$};
227  \node[state]           (E) [below of=D]       {$s_{\bar{a}i,n}$};
228
229  \path (A) edge              node {} (B)
230            (B) edge       node {} (C)
231            (C) edge             node {} (D)
232            (D) edge             node {} (E);
233
234\end{tikzpicture}
235
236   \caption{\label{AKSNegCex} Kripke Structure of counterexample $\sigma_i$, $K(\sigma_i)$}
237\end{figure}
238
239
240Therefore all properties that are satisfied won't be chosen to be integrated in the next step of refinement. At this stage, we already have a list of potential properties that will definitely eliminate the current counterexample $\sigma_i$ and might converge the abstract model towards a model sufficient to verify the global property $\phi$.
241
242}
243%\bigskip
244
245\item {\emph{\underline{Step 2:}} \\
246
247The property at the top of the list (not yet selected and excluding the properties which are satisfied by $K(\sigma_i)$) is selected to be integrated in the generation of $\widehat{M}_{i+1}$.
248%\bigskip
249
250}
251\end{enumerate}
252
253$\widehat{M}_{i+1}$ is model-checked and the refinement process is repeated until the model satisfies the global property or there is no property left to be integrated in next abstraction.
254
255
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