/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(?, O(1)) proof of /export/starexec/sandbox2/benchmark/theBenchmark.koat # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 11 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A, B, C, D, E, F, G, H, I, J) -> Com_1(f5(K, 0, 0, D, E, F, G, H, I, J)) :|: TRUE f5(A, B, C, D, E, F, G, H, I, J) -> Com_1(f5(A, B + 1, C + 1, 1, E, F, G, H, I, J)) :|: 15 >= C f5(A, B, C, D, E, F, G, H, I, J) -> Com_1(f5(A, B, C + 1, 0, E, F, G, H, I, J)) :|: 15 >= C f5(A, B, C, D, E, F, G, H, I, J) -> Com_1(f27(A, B, C, D, B, B, K, L, L, L)) :|: C >= 16 The start-symbols are:[f0_10] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 34) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f5(Fresh_2, 0, 0, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f5(Ar_0, Ar_1 + 1, Ar_2 + 1, 1, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) [ 15 >= Ar_2 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f5(Ar_0, Ar_1, Ar_2 + 1, 0, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) [ 15 >= Ar_2 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_1, Ar_1, Fresh_0, Fresh_1, Fresh_1, Fresh_1)) [ Ar_2 >= 16 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_2]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f5(Ar_2) -> Com_1(f27(Ar_2)) [ Ar_2 >= 16 ] (Comp: ?, Cost: 1) f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ] (Comp: ?, Cost: 1) f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ] (Comp: ?, Cost: 1) f0(Ar_2) -> Com_1(f5(0)) start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 2 produces the following problem: 3: T: (Comp: 1, Cost: 0) koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f5(Ar_2) -> Com_1(f27(Ar_2)) [ Ar_2 >= 16 ] (Comp: ?, Cost: 1) f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ] (Comp: ?, Cost: 1) f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ] (Comp: 1, Cost: 1) f0(Ar_2) -> Com_1(f5(0)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 1 Pol(f0) = 1 Pol(f5) = 1 Pol(f27) = 0 orients all transitions weakly and the transition f5(Ar_2) -> Com_1(f27(Ar_2)) [ Ar_2 >= 16 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f5(Ar_2) -> Com_1(f27(Ar_2)) [ Ar_2 >= 16 ] (Comp: ?, Cost: 1) f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ] (Comp: ?, Cost: 1) f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ] (Comp: 1, Cost: 1) f0(Ar_2) -> Com_1(f5(0)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 16 Pol(f0) = 16 Pol(f5) = -V_1 + 16 Pol(f27) = -V_1 orients all transitions weakly and the transition f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 0) koat_start(Ar_2) -> Com_1(f0(Ar_2)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f5(Ar_2) -> Com_1(f27(Ar_2)) [ Ar_2 >= 16 ] (Comp: 16, Cost: 1) f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ] (Comp: 16, Cost: 1) f5(Ar_2) -> Com_1(f5(Ar_2 + 1)) [ 15 >= Ar_2 ] (Comp: 1, Cost: 1) f0(Ar_2) -> Com_1(f5(0)) start location: koat_start leaf cost: 0 Complexity upper bound 34 Time: 0.051 sec (SMT: 0.048 sec) ---------------------------------------- (2) BOUNDS(1, 1)