/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(?, O(1)) proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).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, 20 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A, B, C, D, E, F, G) -> Com_1(f5(0, B, C, D, E, F, G)) :|: TRUE f5(A, B, C, D, E, F, G) -> Com_1(f5(A + 1, B, C, D, E, F, G)) :|: 99 >= A f17(A, B, C, D, E, F, G) -> Com_1(f17(A, B, C, D, E, F, G)) :|: TRUE f17(A, B, C, D, E, F, G) -> Com_1(f17(A, B + 1, C, D, E, F, G)) :|: 0 >= H + 1 f17(A, B, C, D, E, F, G) -> Com_1(f17(A, B + 1, C, D, E, F, G)) :|: TRUE f32(A, B, C, D, E, F, G) -> Com_1(f32(A, B, C, D, E, F, G)) :|: TRUE f32(A, B, C, D, E, F, G) -> Com_1(f32(A, B, C + 1, D, E, F, G)) :|: 0 >= H + 1 f32(A, B, C, D, E, F, G) -> Com_1(f32(A, B, C + 1, D, E, F, G)) :|: TRUE f32(A, B, C, D, E, F, G) -> Com_1(f13(A, B, C, C, C, F, G)) :|: TRUE f17(A, B, C, D, E, F, G) -> Com_1(f32(A, B, B, B, E, B, H)) :|: 0 >= I + 1 f17(A, B, C, D, E, F, G) -> Com_1(f32(A, B, B, B, E, B, H)) :|: TRUE f17(A, B, C, D, E, F, G) -> Com_1(f13(A, B, C, B, E, B, H)) :|: TRUE f5(A, B, C, D, E, F, G) -> Com_1(f13(A, B, C, A - 2, E, F, G)) :|: A >= 100 f5(A, B, C, D, E, F, G) -> Com_1(f17(A, A - 2, C, A - 2, E, F, G)) :|: 0 >= A + 1 && A >= 100 The start-symbols are:[f0_7] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 102) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f5(Ar_0 + 1, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 99 >= Ar_0 ] (Comp: ?, Cost: 1) f17(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f17(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) (Comp: ?, Cost: 1) f17(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f17(Ar_0, Ar_1 + 1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= H + 1 ] (Comp: ?, Cost: 1) f17(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f17(Ar_0, Ar_1 + 1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) (Comp: ?, Cost: 1) f32(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f32(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) (Comp: ?, Cost: 1) f32(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f32(Ar_0, Ar_1, Ar_2 + 1, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 >= H + 1 ] (Comp: ?, Cost: 1) f32(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f32(Ar_0, Ar_1, Ar_2 + 1, Ar_3, Ar_4, Ar_5, Ar_6)) (Comp: ?, Cost: 1) f32(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f13(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5, Ar_6)) (Comp: ?, Cost: 1) f17(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f32(Ar_0, Ar_1, Ar_1, Ar_1, Ar_4, Ar_1, Fresh_2)) [ 0 >= I + 1 ] (Comp: ?, Cost: 1) f17(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f32(Ar_0, Ar_1, Ar_1, Ar_1, Ar_4, Ar_1, Fresh_1)) (Comp: ?, Cost: 1) f17(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f13(Ar_0, Ar_1, Ar_2, Ar_1, Ar_4, Ar_1, Fresh_0)) (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f13(Ar_0, Ar_1, Ar_2, Ar_0 - 2, Ar_4, Ar_5, Ar_6)) [ Ar_0 >= 100 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f17(Ar_0, Ar_0 - 2, Ar_2, Ar_0 - 2, Ar_4, Ar_5, Ar_6)) [ 0 >= Ar_0 + 1 /\ Ar_0 >= 100 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_0]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f17(Ar_0)) [ 0 >= Ar_0 + 1 /\ Ar_0 >= 100 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f13(Ar_0)) [ Ar_0 >= 100 ] (Comp: ?, Cost: 1) f17(Ar_0) -> Com_1(f13(Ar_0)) (Comp: ?, Cost: 1) f17(Ar_0) -> Com_1(f32(Ar_0)) (Comp: ?, Cost: 1) f17(Ar_0) -> Com_1(f32(Ar_0)) [ 0 >= I + 1 ] (Comp: ?, Cost: 1) f32(Ar_0) -> Com_1(f13(Ar_0)) (Comp: ?, Cost: 1) f32(Ar_0) -> Com_1(f32(Ar_0)) (Comp: ?, Cost: 1) f32(Ar_0) -> Com_1(f32(Ar_0)) [ 0 >= H + 1 ] (Comp: ?, Cost: 1) f32(Ar_0) -> Com_1(f32(Ar_0)) (Comp: ?, Cost: 1) f17(Ar_0) -> Com_1(f17(Ar_0)) (Comp: ?, Cost: 1) f17(Ar_0) -> Com_1(f17(Ar_0)) [ 0 >= H + 1 ] (Comp: ?, Cost: 1) f17(Ar_0) -> Com_1(f17(Ar_0)) (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 99 >= Ar_0 ] (Comp: ?, Cost: 1) f0(Ar_0) -> Com_1(f5(0)) start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transitions from problem 2: f5(Ar_0) -> Com_1(f17(Ar_0)) [ 0 >= Ar_0 + 1 /\ Ar_0 >= 100 ] f17(Ar_0) -> Com_1(f13(Ar_0)) f17(Ar_0) -> Com_1(f32(Ar_0)) f17(Ar_0) -> Com_1(f32(Ar_0)) [ 0 >= I + 1 ] f32(Ar_0) -> Com_1(f13(Ar_0)) f32(Ar_0) -> Com_1(f32(Ar_0)) f32(Ar_0) -> Com_1(f32(Ar_0)) [ 0 >= H + 1 ] f32(Ar_0) -> Com_1(f32(Ar_0)) f17(Ar_0) -> Com_1(f17(Ar_0)) f17(Ar_0) -> Com_1(f17(Ar_0)) [ 0 >= H + 1 ] f17(Ar_0) -> Com_1(f17(Ar_0)) We thus obtain the following problem: 3: T: (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f13(Ar_0)) [ Ar_0 >= 100 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 99 >= Ar_0 ] (Comp: ?, Cost: 1) f0(Ar_0) -> Com_1(f5(0)) (Comp: 1, Cost: 0) koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 3 produces the following problem: 4: T: (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f13(Ar_0)) [ Ar_0 >= 100 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 99 >= Ar_0 ] (Comp: 1, Cost: 1) f0(Ar_0) -> Com_1(f5(0)) (Comp: 1, Cost: 0) koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f5) = 1 Pol(f13) = 0 Pol(f0) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f5(Ar_0) -> Com_1(f13(Ar_0)) [ Ar_0 >= 100 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 1) f5(Ar_0) -> Com_1(f13(Ar_0)) [ Ar_0 >= 100 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 99 >= Ar_0 ] (Comp: 1, Cost: 1) f0(Ar_0) -> Com_1(f5(0)) (Comp: 1, Cost: 0) koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f5) = -V_1 + 100 Pol(f13) = -V_1 Pol(f0) = 100 Pol(koat_start) = 100 orients all transitions weakly and the transition f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 99 >= Ar_0 ] strictly and produces the following problem: 6: T: (Comp: 1, Cost: 1) f5(Ar_0) -> Com_1(f13(Ar_0)) [ Ar_0 >= 100 ] (Comp: 100, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 99 >= Ar_0 ] (Comp: 1, Cost: 1) f0(Ar_0) -> Com_1(f5(0)) (Comp: 1, Cost: 0) koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 102 Time: 0.077 sec (SMT: 0.054 sec) ---------------------------------------- (2) BOUNDS(1, 1)