Memorylessness and optimality Fact In our PTGAs, optimal strategies do not always exist. Fact When optimal strategies exist, they might require some memory. Example x =1 � x ≤ 1 p =2 ˙ x < 1 , x :=0 x > 0 p =1 ˙ An optimal strategy depends on the date at which the blue state is entered. But there is a memoryless ε -optimal strategy.
Decidability of 1PTGAs Definition Given ε > 0 and N ∈ Z + , a strategy σ is ( ε, N ) acceptable if σ is ε -optimal and memoryless, there is a partition ( I n ) n ≤ N of [0 , M ] (where M is the maximal constant of the guards and invariants of the game) s.t., for any q ∈ Q c , x �→ σ ( q , x ) is constant on each I n .
Decidability of 1PTGAs Definition Given ε > 0 and N ∈ Z + , a strategy σ is ( ε, N ) acceptable if σ is ε -optimal and memoryless, there is a partition ( I n ) n ≤ N of [0 , M ] (where M is the maximal constant of the guards and invariants of the game) s.t., for any q ∈ Q c , x �→ σ ( q , x ) is constant on each I n . Main Theorem For every location, the optimal cost is computable and is piecewise affine. There exists N ∈ Z + s.t., for any ε > 0, we can effectively compute an ( ε, N )-acceptable (thus, almost-optimal and memoryless) strategy.
Simplifying the problem We restrict to TGAs with maximal constant 1 (in clock constraints)
Simplifying the problem We restrict to TGAs with maximal constant 1 (in clock constraints) Example x =1 x =1 x =1 x :=0 x :=0 x :=0 x < 3 x < 1 x =1 x =1 x =1 p =2 ˙ p =2 ˙ p =2 ˙ p =2 ˙ p =2 ˙ x ≤ 4 x ≤ 1 x ≤ 1 x ≤ 1 x ≤ 1 x :=0 x :=0 x :=0 x ≥ 2 x =1 x =1 x =1 x :=0 x :=0 x :=0
Simplifying the problem We restrict to strongly-connected TGAs without resets.
Simplifying the problem We restrict to strongly-connected TGAs without resets. Example x ≤ 1 p =4 ˙ p =1 ˙ p =3 ˙ p =1 ˙ x :=0 x :=0
Simplifying the problem We restrict to strongly-connected TGAs without resets. Example x ≤ 1 p =4 ˙ p =1 ˙ p =3 ˙ p =1 ˙ x :=0 x :=0
Simplifying the problem We restrict to strongly-connected TGAs without resets. Example x ≤ 1 p =4 ˙ p =1 ˙ x ≤ 1 p =4 ˙ p =1 ˙ p =3 ˙ p =1 ˙ x :=0 x :=0 p =3 ˙ p =1 ˙ x :=0 x :=0
Simplifying the problem We restrict to strongly-connected TGAs without resets. Example x ≤ 1 p =4 ˙ p =1 ˙ x ≤ 1 p =4 ˙ p =1 ˙ p =3 ˙ p =1 ˙ x :=0 p =3 ˙ p =1 ˙ x :=0 + ∞ x :=0
Simplifying the problem We restrict to strongly-connected TGAs without resets. m x :=0 n G m 1 G ′ + ∞ x :=0 n 2
Simplifying the problem We restrict to strongly-connected TGAs without resets. Theorem m x :=0 OptCost G ( q , x ) = OptCost G ′ ( q 1 , x ) . n G m 1 G ′ + ∞ x :=0 n 2
Simplifying the problem We restrict to strongly-connected TGAs without resets. Theorem m x :=0 OptCost G ( q , x ) = OptCost G ′ ( q 1 , x ) . n G Theorem If σ ′ is ( ε ′ , N ′ )-acceptable in G ′ , then m 1 σ ′ ( q 2 , x ) G ′ σ ( q , x ) = if Cost( q 2 , x ) ≤ Cost( q 1 , x ) σ ′ ( q 1 , x ) otherwise + ∞ x :=0 n 2 is (2 ε ′ , N ′ )-acceptable in G .
Simplifying the problem Reduced to x :=0 strongly-connected PTGAs clock is bounded by 1 x :=0 no resetting transitions.
Simplifying the problem Reduced to x :=0 strongly-connected PTGAs clock is bounded by 1 x :=0 no resetting transitions.
Simplifying the problem Reduced to x :=0 strongly-connected PTGAs clock is bounded by 1 x :=0 no resetting transitions.
Simplifying the problem Reduced to x :=0 strongly-connected PTGAs clock is bounded by 1 x :=0 no resetting transitions.
Simplifying the problem Reduced to x :=0 strongly-connected PTGAs clock is bounded by 1 x :=0 no resetting transitions.
Simplifying the problem Reduced to x :=0 strongly-connected PTGAs clock is bounded by 1 x :=0 no resetting transitions.
Simplifying the problem Reduced to x :=0 strongly-connected PTGAs clock is bounded by 1 x :=0 no resetting transitions.
Simplifying the problem Reduced to x :=0 strongly-connected PTGAs clock is bounded by 1 x :=0 no resetting transitions.
Simplifying the problem Reduced to x :=0 strongly-connected PTGAs clock is bounded by 1 no resetting transitions.
Simplifying the problem Reduced to x :=0 strongly-connected PTGAs clock is bounded by 1 no resetting transitions.
Main theorem with outside cost-functions Theorem Let G be a strongly-connected non- resetting 1PTGA with outside cost- functions. p =3 ˙ OptCost G is computable; p =1 ˙ in each location, function x ≤ 1 x �→ OptCost G ( q , x ) p =5 ˙ p =1 ˙ is decreasing, piecewise affine and continuous. Its finitely many segments either have slope − c where c is the price of some locations, or are fragments of the outside cost-functions; There exists N ∈ Z + s.t., for any ε > 0 , we can compute an ( ε, N ) -acceptable strategy σ .
Operations on cost functions: controllable locations p = 3 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: controllable locations p = 3 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: controllable locations p = 3 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: controllable locations p = 3 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: controllable locations p = 3 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: controllable locations p = 3 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: uncontrollable locations p = 2 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: uncontrollable locations p = 2 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: uncontrollable locations p = 2 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: uncontrollable locations p = 2 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: uncontrollable locations p = 2 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Operations on cost functions: uncontrollable locations p = 2 ˙ p = 5 ˙ p = 3 ˙ p = 2 ˙ p = 1 ˙
Inductive proof Ideas of the proof Induction on the number of non-urgent locations in the SCC base cases: all locations are urgent (thus uncontrollable); there is only one location, which is controllable (thus non-urgent). induction step: we consider one of the non-urgent locations having minimal cost rate: if it is controllable, we create two SCCs having one less non-urgent location; if it is uncontrollable, we make it urgent and add an extra outside cost function to which it can go. Skip proof
Inductive proof – base cases p =1 ˙ p =3 ˙ x ≤ 1 p =5 ˙ p =1 ˙
Inductive proof – base cases p =1 ˙ p =3 ˙ x ≤ 1 p =5 ˙ p =1 ˙
Inductive proof – base cases p =1 ˙ p =3 ˙ x ≤ 1 p =5 ˙ p =1 ˙
Inductive proof – base cases p =3 ˙
Inductive proof – base cases p =3 ˙
Inductive proof – base cases p =3 ˙
Inductive proof – base cases p =3 ˙
Inductive proof – base cases p =3 ˙
Inductive proof – inductive cases When q min is controllable: p =3 ˙ p =5 ˙ x ≤ 1 p =1 ˙ p =2 ˙
Inductive proof – inductive cases When q min is controllable: Let σ be a winning strategy. p =3 ˙ p =5 ˙ x ≤ 1 p =1 ˙ p =2 ˙
Inductive proof – inductive cases When q min is controllable: Let σ be a winning strategy. Assume there exists an outcome of σ s.t.: p =3 ˙ p =5 ˙ ( q min , u ) → ∗ ( q min , v ) → ∗ win x ≤ 1 with 0 ≤ u < v ≤ 1. p =1 ˙ p =2 ˙
Inductive proof – inductive cases When q min is controllable: Let σ be a winning strategy. Assume there exists an outcome of σ s.t.: p =3 ˙ p =5 ˙ ( q min , u ) → ∗ ( q min , v ) → ∗ win x ≤ 1 with 0 ≤ u < v ≤ 1. p =1 ˙ p =2 ˙ Then σ is not optimal: waiting in q min would have been cheaper.
Inductive proof – inductive cases When q min is controllable: p =3 ˙ p =5 ˙ x ≤ 1 p =1 ˙ p =2 ˙
Inductive proof – inductive cases When q min is controllable: p =3 ˙ p =5 ˙ p =3 ˙ p =5 ˙ x ≤ 1 x ≤ 1 p =1 ˙ p =1 ˙ p =2 ˙ p =2 ˙
Inductive proof – inductive cases When q min is controllable: p =3 ˙ p =5 ˙ p =3 ˙ p =5 ˙ x ≤ 1 x ≤ 1 + ∞ p =2 ˙ p =2 ˙ ˙ p =1
Inductive proof – inductive cases When q min is controllable: m n q min G m 1 n 1 G ′ q min m 2 + ∞ n 2
Inductive proof – inductive cases When q min is controllable: m Theorem OptCost G ′ ( q 1 , x ) = OptCost G ( q , x ) . n q min G m 1 n 1 G ′ q min m 2 + ∞ n 2
Inductive proof – inductive cases When q min is controllable: m Theorem OptCost G ′ ( q 1 , x ) = OptCost G ( q , x ) . n q min G Theorem Let σ ′ be an ( ε ′ , N ′ )-acceptable strategy m 1 for G ′ . Let σ ′ ( q 2 , x ) n 1 G ′ σ ( q , x ) = if Cost G ′ ( q 2 , x ) ≤ OptCost G ′ ( q min , x ) σ ′ ( q 1 , x ) otherwise q min m 2 Then σ is (3 ε ′ , N )-acceptable in G , for + ∞ n 2 some N independant of ε ′ .
Inductive proof – inductive cases When q min is uncontrollable: p =3 ˙ p =1 ˙ x ≤ 1 p =5 ˙ p =1 ˙
Inductive proof – inductive cases When q min is uncontrollable: Make q min urgent and apply I.H.: p =3 ˙ p =1 ˙ x ≤ 1 p =5 ˙ p =1 ˙
Inductive proof – inductive cases When q min is uncontrollable: Make q min urgent and apply I.H.: p =3 ˙ p =1 ˙ x ≤ 1 p =5 ˙ p =1 ˙
Inductive proof – inductive cases When q min is uncontrollable: Make q min urgent and apply I.H.: p =3 ˙ p =1 ˙ First instance where slope x ≤ 1 less than c min p =5 ˙ p =1 ˙
Inductive proof – inductive cases When q min is uncontrollable: Make q min urgent and apply I.H.: p =3 ˙ p =1 ˙ It’s better to x ≤ 1 wait in q min ... p =5 ˙ p =1 ˙
Inductive proof – inductive cases When q min is uncontrollable: Make q min urgent and apply I.H.: p =3 ˙ p =1 ˙ x ≤ 1 p =5 ˙ p =1 ˙
Inductive proof – inductive cases When q min is uncontrollable: Apply I.H. again: p =3 ˙ p =1 ˙ x ≤ 1 p =5 ˙ p =1 ˙
Inductive proof – inductive cases When q min is uncontrollable: Apply I.H. again: p =3 ˙ p =1 ˙ x ≤ 1 p =5 ˙ p =1 ˙
Inductive proof – inductive cases When q min is uncontrollable: Apply I.H. again: p =3 ˙ p =1 ˙ First instance where slope x ≤ 1 less than c min p =5 ˙ p =1 ˙
Inductive proof – inductive cases When q min is uncontrollable: Apply I.H. again: p =3 ˙ p =1 ˙ It’s better to wait in q min ... x ≤ 1 p =5 ˙ p =1 ˙
Inductive proof – inductive cases When q min is uncontrollable: p =3 ˙ p =1 ˙ This procedure terminates because x ≤ 1 fragments having slope strictly less than c min are fragments of outside p =5 ˙ p =1 ˙ functions.
Outline of the talk Introduction 1 Definitions and examples 2 Existence of optimal strategies in 1PTGAs is decidable 3 (Pseudo-)algorithm for computing the optimal cost 4 Conclusion 5
Recommend
More recommend
