/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).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) -> Com_1(f7(F, 0, 0, D, E)) :|: 0 >= F + 1 f0(A, B, C, D, E) -> Com_1(f7(F, 0, 0, D, E)) :|: F >= 1 f0(A, B, C, D, E) -> Com_1(f7(0, 1023, 0, D, E)) :|: TRUE f7(A, B, C, D, E) -> Com_1(f7(A, B, C + 1, D + 2, E)) :|: B >= C f7(A, B, C, D, E) -> Com_1(f21(A, B, C, D, E)) :|: E >= 0 && C >= 1 + B && 1022 >= E f7(A, B, C, D, E) -> Com_1(f21(A, B, C, D, E)) :|: C >= 1 + B && E >= 1023 f7(A, B, C, D, E) -> Com_1(f21(A, B, C, D, E)) :|: C >= 1 + B && 0 >= E + 1 The start-symbols are:[f0_5] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 1030) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Fresh_1, 0, 0, Ar_3, Ar_4)) [ 0 >= Fresh_1 + 1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Fresh_0, 0, 0, Ar_3, Ar_4)) [ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(0, 1023, 0, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f7(Ar_0, Ar_1, Ar_2 + 1, Ar_3 + 2, Ar_4)) [ Ar_1 >= Ar_2 ] (Comp: ?, Cost: 1) f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ] (Comp: ?, Cost: 1) f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ] (Comp: ?, Cost: 1) f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_1, Ar_2, Ar_4]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ] (Comp: ?, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ] (Comp: ?, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ] (Comp: ?, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f7(Ar_1, Ar_2 + 1, Ar_4)) [ Ar_1 >= Ar_2 ] (Comp: ?, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(1023, 0, Ar_4)) (Comp: ?, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ 0 >= Fresh_1 + 1 ] 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_1, Ar_2, Ar_4) -> Com_1(f0(Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ] (Comp: ?, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ] (Comp: ?, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ] (Comp: ?, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f7(Ar_1, Ar_2 + 1, Ar_4)) [ Ar_1 >= Ar_2 ] (Comp: 1, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(1023, 0, Ar_4)) (Comp: 1, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ Fresh_0 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ 0 >= Fresh_1 + 1 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 1 Pol(f0) = 1 Pol(f7) = 1 Pol(f21) = 0 orients all transitions weakly and the transitions f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ] f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ] f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ] (Comp: 1, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ] (Comp: 1, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ] (Comp: ?, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f7(Ar_1, Ar_2 + 1, Ar_4)) [ Ar_1 >= Ar_2 ] (Comp: 1, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(1023, 0, Ar_4)) (Comp: 1, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ Fresh_0 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ 0 >= Fresh_1 + 1 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 1024 Pol(f0) = 1024 Pol(f7) = V_1 - V_2 + 1 Pol(f21) = V_1 - V_2 orients all transitions weakly and the transition f7(Ar_1, Ar_2, Ar_4) -> Com_1(f7(Ar_1, Ar_2 + 1, Ar_4)) [ Ar_1 >= Ar_2 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 0) koat_start(Ar_1, Ar_2, Ar_4) -> Com_1(f0(Ar_1, Ar_2, Ar_4)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ 0 >= Ar_4 + 1 ] (Comp: 1, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_2 >= Ar_1 + 1 /\ Ar_4 >= 1023 ] (Comp: 1, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f21(Ar_1, Ar_2, Ar_4)) [ Ar_4 >= 0 /\ Ar_2 >= Ar_1 + 1 /\ 1022 >= Ar_4 ] (Comp: 1024, Cost: 1) f7(Ar_1, Ar_2, Ar_4) -> Com_1(f7(Ar_1, Ar_2 + 1, Ar_4)) [ Ar_1 >= Ar_2 ] (Comp: 1, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(1023, 0, Ar_4)) (Comp: 1, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ Fresh_0 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_1, Ar_2, Ar_4) -> Com_1(f7(0, 0, Ar_4)) [ 0 >= Fresh_1 + 1 ] start location: koat_start leaf cost: 0 Complexity upper bound 1030 Time: 0.052 sec (SMT: 0.044 sec) ---------------------------------------- (2) BOUNDS(1, 1)