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Implementability of Timed Controllers Based on joint works with Karine Altisen, Patricia Bouyer, Martin De Wulf, Laurent Doyen, Jean-Fran cois Raskin, Pierre-Alain Reynier, and Stavros Tripakis Nicolas Markey Lab. Sp ecification et V

  1. Modeling the execution platform [Altisen & Tripakis, 2005] 1. Transforming A into Prog( A ). 2. Modeling the digital CPU. 3. Modeling the global clock. 4. Modeling the input/output variables. 5. Classical verification techniques on the product of those automata.

  2. Pros and cons of this approach Pros: Very expressive: the platform can be described with many details; Relies on classical techniques: the verification step is applied on standard timed automata. Existing tools can be used.

  3. Pros and cons of this approach Pros: Very expressive: the platform can be described with many details; Relies on classical techniques: the verification step is applied on standard timed automata. Existing tools can be used. Cons: Formal meaning?: if the model satisfies some property, what does it really mean? Faster is better?: we expect that a program proved to be implementable on a given platform remains implementable on a faster platform. This property fails to hold with this modeling.

  4. Outline of the talk Introduction 1 Modeling the execution platform [Altisen & Tripakis, 2005] 2 A semantical approach [De Wulf et al., 2004] 3 Conclusions 4

  5. A semantical approach [De Wulf et al. , 2004] 1. “Implementation” Semantics We consider a simple model of a platform, that repeatedly executes the following actions: store the value of the global clock; compute guards; fire one of the enabled transitions. We assume that one such loop takes at most ∆ P t.u. to execute; the global clock is updated every ∆ L t.u. � We write � A � Impl ∆ P , ∆ L for the set of executions of a timed automaton A under this semantics.

  6. A semantical approach [De Wulf et al. , 2004] 1. “Implementation” Semantics 2. Enlarged Semantics We define the enlarged semantics for timed automata, by enlarging guards on transitions by a small tolerance ∆: If � g � = [ a ; b ], then � g � AASAP = [ a − ∆ , b + ∆] . ∆ � We write � A � AASAP for the set of executions of a timed ∆ automaton A under this semantics.

  7. A semantical approach [De Wulf et al. , 2004] 1. “Implementation” Semantics 2. Enlarged Semantics We define the enlarged semantics for timed automata, by enlarging guards on transitions by a small tolerance ∆: If � g � = [ a ; b ], then � g � AASAP = [ a − ∆ , b + ∆] . ∆ � We write � A � AASAP for the set of executions of a timed ∆ automaton A under this semantics. Theorem ([DDR04]) If ∆ > 3∆ L + 4∆ P , then � A � Impl ∆ P , ∆ L ⊆ � A � AASAP . ∆

  8. A semantical approach [De Wulf et al. , 2004] We focus on safety properties for the implementation semantics: we want to ensure that an implementation will avoid bad states. � Reach ∆ ( A ) is the set of reachable states under the AASAP semantics. ∆ 1 ≤ ∆ 2 ⇒ Reach ∆ 1 ( A ) ⊆ Reach ∆ 2 ( A ) � R ( A ) = � ∆ > 0 Reach ∆ ( A ) is the set of reachable states under the AASAP semantics for any ∆ > 0.

  9. A semantical approach [De Wulf et al. , 2004] We focus on safety properties for the implementation semantics: we want to ensure that an implementation will avoid bad states. � Reach ∆ ( A ) is the set of reachable states under the AASAP semantics. ∆ 1 ≤ ∆ 2 ⇒ Reach ∆ 1 ( A ) ⊆ Reach ∆ 2 ( A ) � R ( A ) = � ∆ > 0 Reach ∆ ( A ) is the set of reachable states under the AASAP semantics for any ∆ > 0. Lemma For any timed automata A and for any set of zones B, R ( A ) ∩ B = ∅ iff ∃ ∆ > 0 . Reach ∆ ( A ) ∩ B = ∅ .

  10. An example: Standard semantics y 2 x ≤ 2 x :=0 x =0 x =1 a b c Bad y =2 y :=0 y ≥ 2 y :=0 1 x 0 1 2

  11. An example: Standard semantics y 2 x ≤ 2 x :=0 x =0 x =1 a b c Bad y =2 y :=0 y ≥ 2 y :=0 1 x 0 1 2

  12. An example: Standard semantics y 2 x ≤ 2 x :=0 x =0 x =1 a b c Bad y =2 y :=0 y ≥ 2 y :=0 1 x 0 1 2

  13. An example: Standard semantics y 2 x ≤ 2 x :=0 x =0 x =1 a b c Bad y =2 y :=0 y ≥ 2 y :=0 1 x 0 1 2

  14. An example: Standard semantics y 2 x ≤ 2 x :=0 x =0 x =1 a b c Bad y =2 y :=0 y ≥ 2 y :=0 1 x 0 1 2

  15. An example: Standard semantics y 2 x ≤ 2 x :=0 x =0 x =1 a b c Bad y =2 y :=0 y ≥ 2 y :=0 1 x 0 1 2

  16. An example: Standard semantics y 2 x ≤ 2 x :=0 x =0 x =1 a b c Bad y =2 y :=0 y ≥ 2 y :=0 1 x 0 1 2

  17. An example: Standard semantics y 2 x ≤ 2 x :=0 x =0 x =1 a b c Bad y =2 y :=0 y ≥ 2 y :=0 1 x 0 1 2

  18. An example: Standard semantics y 2 x ≤ 2 x :=0 x =0 x =1 a b c Bad y =2 y :=0 y ≥ 2 y :=0 1 x 0 1 2

  19. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  20. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  21. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − ∆ 1+∆

  22. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − ∆ 1+∆

  23. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad 2+2∆ y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  24. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad 2+2∆ y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  25. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − 3∆

  26. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − 3∆

  27. An example with ∆ > 0 y 2 2+4∆ x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  28. An example with ∆ > 0 y 2 2+4∆ x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  29. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − 5∆

  30. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − 5∆

  31. An example with ∆ > 0 y 2+6∆ 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  32. An example with ∆ > 0 y 2+6∆ 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  33. An example with ∆ > 0 y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  34. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  35. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  36. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − ∆ 1+∆

  37. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − ∆ 1+∆

  38. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ 2+2∆ y :=0 1 x 0 1 2

  39. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ 2+2∆ y :=0 1 x 0 1 2

  40. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − 3∆

  41. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − 3∆

  42. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ 2+4∆ y :=0 1 x 0 1 2

  43. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ 2+4∆ y :=0 1 x 0 1 2

  44. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − 5∆

  45. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2 1 − 5∆

  46. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 2+6∆ y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  47. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 2+6∆ y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  48. An example with ∆ very small y 2 x ≤ 2+∆ x :=0 x ∈ [1 − ∆;1+∆] x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 − ∆ y :=0 1 x 0 1 2

  49. Difference between � A � and R ( A ) Reach( � A � ) R ( A ) y y 2 2 1 1 x x 0 1 2 0 1 2 x ≤ 2 x :=0 x =1 x ≤ ∆ a b c Bad y :=0 y ∈ [2 − ∆ , 2+∆] y ≥ 2 y :=0

  50. An algorithm for computing R ( A ) Input: A Timed Automaton A Output: The set R ( A ) 1. build the region graph G of A ; 2. compute SCC ( G ) = the set of strongly connected components of G ; 3. J := [( q 0 )] ; 4. J := Reach ( G , J ) ; 5. while ∃ S ∈ SCC ( G ) . S �⊆ J and S ∩ J � = ∅ , J := J ∪ S ; J := Reach ( G , J ) ; 6. return( J );

  51. An algorithm for computing R ( A ) Input: A Timed Automaton A Output: The set R ( A ) 1. build the region graph G of A ; 2. compute SCC ( G ) = the set of strongly connected components of G ; 3. J := [( q 0 )] ; 4. J := Reach ( G , J ) ; 5. while ∃ S ∈ SCC ( G ) . S �⊆ J and S ∩ J � = ∅ , J := J ∪ S ; J := Reach ( G , J ) ; 6. return( J );

  52. An algorithm for computing R ( A ) Input: A Timed Automaton A Output: The set R ( A ) 1. build the region graph G of A ; 2. compute SCC ( G ) = the set of strongly connected components of G ; 3. J := [( q 0 )] ; 4. J := Reach ( G , J ) ; 5. while ∃ S ∈ SCC ( G ) . S �⊆ J and S ∩ J � = ∅ , J := J ∪ S ; J := Reach ( G , J ) ; 6. return( J );

  53. An algorithm for computing R ( A ) Input: A Timed Automaton A Output: The set R ( A ) 1. build the region graph G of A ; 2. compute SCC ( G ) = the set of strongly connected components of G ; 3. J := [( q 0 )] ; 4. J := Reach ( G , J ) ; 5. while ∃ S ∈ SCC ( G ) . S �⊆ J and S ∩ J � = ∅ , J := J ∪ S ; J := Reach ( G , J ) ; 6. return( J );

  54. An algorithm for computing R ( A ) Input: A Timed Automaton A Output: The set R ( A ) 1. build the region graph G of A ; 2. compute SCC ( G ) = the set of strongly connected components of G ; 3. J := [( q 0 )] ; 4. J := Reach ( G , J ) ; 5. while ∃ S ∈ SCC ( G ) . S �⊆ J and S ∩ J � = ∅ , J := J ∪ S ; J := Reach ( G , J ) ; 6. return( J );

  55. An algorithm for computing R ( A ) Input: A Timed Automaton A Output: The set R ( A ) 1. build the region graph G of A ; 2. compute SCC ( G ) = the set of strongly connected components of G ; 3. J := [( q 0 )] ; 4. J := Reach ( G , J ) ; 5. while ∃ S ∈ SCC ( G ) . S �⊆ J and S ∩ J � = ∅ , J := J ∪ S ; J := Reach ( G , J ) ; 6. return( J );

  56. J ⊆ R ∆ ( A ) Lemma Let A be a TA with n clocks, ∆ ∈ Q > 0 , and δ = ∆ / n. Let u be a valuation s.t. there exists a trajectory π [0 , T ] in � A � with π (0) = π ( T ) = u. Let v ∈ [ u ] ∩ B ( u , δ ) . Then there exists a trajectory from u to v in � A � ∆ . Proof: We build the new trajectory by slightly modifying the delay transitions in π . This crucially depends on the fact that all clocks are reset along the cycle. Corollary Let A be a TA and p = p 0 p 1 . . . p k be a cycle in the region graph (i.e. p k = p 0 ). For any ∆ > 0 and any x , y ∈ p 0 , there exists a trajectory from x to y.

  57. J ⊆ R ∆ ( A ) Lemma Let A be a TA with n clocks, ∆ ∈ Q > 0 , and δ = ∆ / n. Let u be a valuation s.t. there exists a trajectory π [0 , T ] in � A � with π (0) = π ( T ) = u. Let v ∈ [ u ] ∩ B ( u , δ ) . Then there exists a trajectory from u to v in � A � ∆ . Proof: We build the new trajectory by slightly modifying the delay transitions in π . This crucially depends on the fact that all clocks are reset along the cycle. Corollary Let A be a TA and p = p 0 p 1 . . . p k be a cycle in the region graph (i.e. p k = p 0 ). For any ∆ > 0 and any x , y ∈ p 0 , there exists a trajectory from x to y.

  58. J ⊇ R ∆ ( A ) Lemma Let A be a TA, δ ∈ R > 0 and k ∈ N . There exists D ∈ Q > 0 s.t. for all ∆ ≤ D, any k-step trajectory π ′ = ( q ′ 0 , t ′ 0 )( q ′ 1 , t ′ 1 ) . . . ( q ′ k , t ′ k ) in � A � ∆ can be approximated be a k-step trajectory π = ( q 0 , t 0 )( q 1 , t 1 ) . . . ( q k , t k ) in � A � with � q i − q ′ i � ≤ δ for all i. The proof involves parametric DBMs. Corollary Let A be a TA with n clocks and W regions, α < 1 / (2 n ) , and α ∆ < 2 2 W · (4 n +2) . Let x ∈ J and y s.t. there exists a trajectory from x to y in � A � ∆ . Then d ( J , y ) < α .

  59. J ⊇ R ∆ ( A ) Lemma Let A be a TA, δ ∈ R > 0 and k ∈ N . There exists D ∈ Q > 0 s.t. for all ∆ ≤ D, any k-step trajectory π ′ = ( q ′ 0 , t ′ 0 )( q ′ 1 , t ′ 1 ) . . . ( q ′ k , t ′ k ) in � A � ∆ can be approximated be a k-step trajectory π = ( q 0 , t 0 )( q 1 , t 1 ) . . . ( q k , t k ) in � A � with � q i − q ′ i � ≤ δ for all i. The proof involves parametric DBMs. Corollary Let A be a TA with n clocks and W regions, α < 1 / (2 n ) , and α ∆ < 2 2 W · (4 n +2) . Let x ∈ J and y s.t. there exists a trajectory from x to y in � A � ∆ . Then d ( J , y ) < α .

  60. Can we relax the assumption on cycles? Our algorithm does not work if we relax the “progress-cycle” constraint. For instance: y y x x 0 0 z z x =0 , x :=0 x ≤ ∆ , x :=0 y , z ≥ 1 − ∆ y = z =1 a b a b x ≤ ∆ x =0 c c x , y ≤ ∆ , z ≥ 1 − ∆ x = y =0 , z =1

  61. Can we relax the assumption on cycles? Our algorithm does not work if we relax the “progress-cycle” constraint. For instance: y y x x 0 0 z z x =0 , x :=0 x ≤ ∆ , x :=0 y , z ≥ 1 − ∆ y = z =1 a b a b x ≤ ∆ x =0 c c x , y ≤ ∆ , z ≥ 1 − ∆ x = y =0 , z =1

  62. Can we relax the assumption on cycles? Our algorithm does not work if we relax the “progress-cycle” constraint. For instance: y y x x 0 0 z z x =0 , x :=0 x ≤ ∆ , x :=0 y , z ≥ 1 − ∆ y = z =1 a b a b x ≤ ∆ x =0 c c x , y ≤ ∆ , z ≥ 1 − ∆ x = y =0 , z =1

  63. Can we relax the assumption on cycles? Our algorithm does not work if we relax the “progress-cycle” constraint. For instance: y y x x 0 0 z z x =0 , x :=0 x ≤ ∆ , x :=0 y , z ≥ 1 − ∆ y = z =1 a b a b x ≤ ∆ x =0 c c x , y ≤ ∆ , z ≥ 1 − ∆ x = y =0 , z =1

  64. Can we relax the assumption on cycles? Our algorithm does not work if we relax the “progress-cycle” constraint. For instance: y y x x 0 0 z z x =0 , x :=0 x ≤ ∆ , x :=0 y , z ≥ 1 − ∆ y = z =1 a b a b x ≤ ∆ x =0 c c x , y ≤ ∆ , z ≥ 1 − ∆ x = y =0 , z =1

  65. Can we relax the assumption on cycles? Our algorithm does not work if we relax the “progress-cycle” constraint. For instance: y y x x 0 0 z z x =0 , x :=0 x ≤ ∆ , x :=0 y , z ≥ 1 − ∆ y = z =1 a b a b x ≤ ∆ x =0 c c x , y ≤ ∆ , z ≥ 1 − ∆ x = y =0 , z =1

  66. Can we relax the assumption on cycles? Our algorithm does not work if we relax the “progress-cycle” constraint. For instance: y y x x 0 0 z z x =0 , x :=0 x ≤ ∆ , x :=0 y , z ≥ 1 − ∆ y = z =1 a b a b x ≤ ∆ x =0 c c x , y ≤ ∆ , z ≥ 1 − ∆ x = y =0 , z =1

  67. Can we relax the assumption on cycles? Our algorithm does not work if we relax the “progress-cycle” constraint. For instance: y y x x 0 0 z z x =0 , x :=0 x ≤ ∆ , x :=0 y , z ≥ 1 − ∆ y = z =1 a b a b x ≤ ∆ x =0 c c x , y ≤ ∆ , z ≥ 1 − ∆ x = y =0 , z =1

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