/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), O(n^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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 22 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 645 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: eval_start_start(v__0, v_flag_0, v_n) -> Com_1(eval_start_bb0_in(v__0, v_flag_0, v_n)) :|: TRUE eval_start_bb0_in(v__0, v_flag_0, v_n) -> Com_1(eval_start_0(v__0, v_flag_0, v_n)) :|: TRUE eval_start_0(v__0, v_flag_0, v_n) -> Com_1(eval_start_1(v__0, v_flag_0, v_n)) :|: TRUE eval_start_1(v__0, v_flag_0, v_n) -> Com_1(eval_start_2(v__0, v_flag_0, v_n)) :|: TRUE eval_start_2(v__0, v_flag_0, v_n) -> Com_1(eval_start_3(v__0, v_flag_0, v_n)) :|: TRUE eval_start_3(v__0, v_flag_0, v_n) -> Com_1(eval_start_4(v__0, v_flag_0, v_n)) :|: TRUE eval_start_4(v__0, v_flag_0, v_n) -> Com_1(eval_start_5(v__0, v_flag_0, v_n)) :|: TRUE eval_start_5(v__0, v_flag_0, v_n) -> Com_1(eval_start_bb1_in(v_n, 1, v_n)) :|: TRUE eval_start_bb1_in(v__0, v_flag_0, v_n) -> Com_1(eval_start_bb2_in(v__0, v_flag_0, v_n)) :|: v_flag_0 > 0 eval_start_bb1_in(v__0, v_flag_0, v_n) -> Com_1(eval_start_bb3_in(v__0, v_flag_0, v_n)) :|: v_flag_0 <= 0 eval_start_bb2_in(v__0, v_flag_0, v_n) -> Com_1(eval_start_bb1_in(v__0 - 1, 1, v_n)) :|: v__0 > 0 eval_start_bb2_in(v__0, v_flag_0, v_n) -> Com_1(eval_start_bb1_in(v__0, 1, v_n)) :|: v__0 > 0 && v__0 <= 0 eval_start_bb2_in(v__0, v_flag_0, v_n) -> Com_1(eval_start_bb1_in(v__0 - 1, 0, v_n)) :|: v__0 <= 0 && v__0 > 0 eval_start_bb2_in(v__0, v_flag_0, v_n) -> Com_1(eval_start_bb1_in(v__0, 0, v_n)) :|: v__0 <= 0 eval_start_bb3_in(v__0, v_flag_0, v_n) -> Com_1(eval_start_stop(v__0, v_flag_0, v_n)) :|: TRUE The start-symbols are:[eval_start_start_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 2*Ar_1 + 14) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart3(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstart3(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart4(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstart4(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_1, Ar_1, 1)) (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 /\ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0 - 1, Ar_1, 0)) [ 0 >= Ar_0 /\ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transitions from problem 1: evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 /\ 0 >= Ar_0 ] evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0 - 1, Ar_1, 0)) [ 0 >= Ar_0 /\ Ar_0 >= 1 ] We thus obtain the following problem: 2: T: (Comp: ?, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_1, Ar_1, 1)) (Comp: ?, Cost: 1) evalstart4(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstart3(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart4(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart3(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 2 produces the following problem: 3: T: (Comp: ?, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 1, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_1, Ar_1, 1)) (Comp: 1, Cost: 1) evalstart4(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart3(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalstartbb3in) = 1 Pol(evalstartstop) = 0 Pol(evalstartbb1in) = 2 Pol(evalstartbb2in) = 2 Pol(evalstart5) = 2 Pol(evalstart4) = 2 Pol(evalstart3) = 2 Pol(evalstart2) = 2 Pol(evalstart1) = 2 Pol(evalstart0) = 2 Pol(evalstartbb0in) = 2 Pol(evalstartstart) = 2 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalstartbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2)) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] strictly and produces the following problem: 4: T: (Comp: 2, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 1, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_1, Ar_1, 1)) (Comp: 1, Cost: 1) evalstart4(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart3(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalstartbb3in) = V_3 Pol(evalstartstop) = V_3 Pol(evalstartbb1in) = V_3 Pol(evalstartbb2in) = 1 Pol(evalstart5) = 1 Pol(evalstart4) = 1 Pol(evalstart3) = 1 Pol(evalstart2) = 1 Pol(evalstart1) = 1 Pol(evalstart0) = 1 Pol(evalstartbb0in) = 1 Pol(evalstartstart) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transition evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] strictly and produces the following problem: 5: T: (Comp: 2, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: 1, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 1, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_1, Ar_1, 1)) (Comp: 1, Cost: 1) evalstart4(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart3(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalstartbb3in) = V_1 Pol(evalstartstop) = V_1 Pol(evalstartbb1in) = V_1 Pol(evalstartbb2in) = V_1 Pol(evalstart5) = V_2 Pol(evalstart4) = V_2 Pol(evalstart3) = V_2 Pol(evalstart2) = V_2 Pol(evalstart1) = V_2 Pol(evalstart0) = V_2 Pol(evalstartbb0in) = V_2 Pol(evalstartstart) = V_2 Pol(koat_start) = V_2 orients all transitions weakly and the transition evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 ] strictly and produces the following problem: 6: T: (Comp: 2, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: 1, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: Ar_1, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 1, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_1, Ar_1, 1)) (Comp: 1, Cost: 1) evalstart4(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart3(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 6 produces the following problem: 7: T: (Comp: 2, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: 1, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: Ar_1, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_0 - 1, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: Ar_1 + 1, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 1, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb1in(Ar_1, Ar_1, 1)) (Comp: 1, Cost: 1) evalstart4(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart3(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 2*Ar_1 + 14 Time: 0.090 sec (SMT: 0.069 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalstartstart 0: evalstartstart -> evalstartbb0in : [], cost: 1 1: evalstartbb0in -> evalstart0 : [], cost: 1 2: evalstart0 -> evalstart1 : [], cost: 1 3: evalstart1 -> evalstart2 : [], cost: 1 4: evalstart2 -> evalstart3 : [], cost: 1 5: evalstart3 -> evalstart4 : [], cost: 1 6: evalstart4 -> evalstart5 : [], cost: 1 7: evalstart5 -> evalstartbb1in : A'=B, C'=1, [], cost: 1 8: evalstartbb1in -> evalstartbb2in : [ C>=1 ], cost: 1 9: evalstartbb1in -> evalstartbb3in : [ 0>=C ], cost: 1 10: evalstartbb2in -> evalstartbb1in : A'=-1+A, C'=1, [ A>=1 ], cost: 1 11: evalstartbb2in -> evalstartbb1in : C'=1, [ A>=1 && 0>=A ], cost: 1 12: evalstartbb2in -> evalstartbb1in : A'=-1+A, C'=0, [ 0>=A && A>=1 ], cost: 1 13: evalstartbb2in -> evalstartbb1in : C'=0, [ 0>=A ], cost: 1 14: evalstartbb3in -> evalstartstop : [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: evalstartstart -> evalstartbb0in : [], cost: 1 Removed unreachable and leaf rules: Start location: evalstartstart 0: evalstartstart -> evalstartbb0in : [], cost: 1 1: evalstartbb0in -> evalstart0 : [], cost: 1 2: evalstart0 -> evalstart1 : [], cost: 1 3: evalstart1 -> evalstart2 : [], cost: 1 4: evalstart2 -> evalstart3 : [], cost: 1 5: evalstart3 -> evalstart4 : [], cost: 1 6: evalstart4 -> evalstart5 : [], cost: 1 7: evalstart5 -> evalstartbb1in : A'=B, C'=1, [], cost: 1 8: evalstartbb1in -> evalstartbb2in : [ C>=1 ], cost: 1 10: evalstartbb2in -> evalstartbb1in : A'=-1+A, C'=1, [ A>=1 ], cost: 1 11: evalstartbb2in -> evalstartbb1in : C'=1, [ A>=1 && 0>=A ], cost: 1 12: evalstartbb2in -> evalstartbb1in : A'=-1+A, C'=0, [ 0>=A && A>=1 ], cost: 1 13: evalstartbb2in -> evalstartbb1in : C'=0, [ 0>=A ], cost: 1 Removed rules with unsatisfiable guard: Start location: evalstartstart 0: evalstartstart -> evalstartbb0in : [], cost: 1 1: evalstartbb0in -> evalstart0 : [], cost: 1 2: evalstart0 -> evalstart1 : [], cost: 1 3: evalstart1 -> evalstart2 : [], cost: 1 4: evalstart2 -> evalstart3 : [], cost: 1 5: evalstart3 -> evalstart4 : [], cost: 1 6: evalstart4 -> evalstart5 : [], cost: 1 7: evalstart5 -> evalstartbb1in : A'=B, C'=1, [], cost: 1 8: evalstartbb1in -> evalstartbb2in : [ C>=1 ], cost: 1 10: evalstartbb2in -> evalstartbb1in : A'=-1+A, C'=1, [ A>=1 ], cost: 1 13: evalstartbb2in -> evalstartbb1in : C'=0, [ 0>=A ], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalstartstart 21: evalstartstart -> evalstartbb1in : A'=B, C'=1, [], cost: 8 8: evalstartbb1in -> evalstartbb2in : [ C>=1 ], cost: 1 10: evalstartbb2in -> evalstartbb1in : A'=-1+A, C'=1, [ A>=1 ], cost: 1 13: evalstartbb2in -> evalstartbb1in : C'=0, [ 0>=A ], cost: 1 Eliminated locations (on tree-shaped paths): Start location: evalstartstart 21: evalstartstart -> evalstartbb1in : A'=B, C'=1, [], cost: 8 22: evalstartbb1in -> evalstartbb1in : A'=-1+A, C'=1, [ C>=1 && A>=1 ], cost: 2 23: evalstartbb1in -> evalstartbb1in : C'=0, [ C>=1 && 0>=A ], cost: 2 Accelerating simple loops of location 8. Accelerating the following rules: 22: evalstartbb1in -> evalstartbb1in : A'=-1+A, C'=1, [ C>=1 && A>=1 ], cost: 2 23: evalstartbb1in -> evalstartbb1in : C'=0, [ C>=1 && 0>=A ], cost: 2 Accelerated rule 22 with metering function A, yielding the new rule 24. Found no metering function for rule 23. Removing the simple loops: 22. Accelerated all simple loops using metering functions (where possible): Start location: evalstartstart 21: evalstartstart -> evalstartbb1in : A'=B, C'=1, [], cost: 8 23: evalstartbb1in -> evalstartbb1in : C'=0, [ C>=1 && 0>=A ], cost: 2 24: evalstartbb1in -> evalstartbb1in : A'=0, C'=1, [ C>=1 && A>=1 ], cost: 2*A Chained accelerated rules (with incoming rules): Start location: evalstartstart 21: evalstartstart -> evalstartbb1in : A'=B, C'=1, [], cost: 8 25: evalstartstart -> evalstartbb1in : A'=B, C'=0, [ 0>=B ], cost: 10 26: evalstartstart -> evalstartbb1in : A'=0, C'=1, [ B>=1 ], cost: 8+2*B Removed unreachable locations (and leaf rules with constant cost): Start location: evalstartstart 26: evalstartstart -> evalstartbb1in : A'=0, C'=1, [ B>=1 ], cost: 8+2*B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalstartstart 26: evalstartstart -> evalstartbb1in : A'=0, C'=1, [ B>=1 ], cost: 8+2*B Computing asymptotic complexity for rule 26 Solved the limit problem by the following transformations: Created initial limit problem: B (+/+!), 8+2*B (+) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {B==n} resulting limit problem: [solved] Solution: B / n Resulting cost 8+2*n has complexity: Poly(n^1) Found new complexity Poly(n^1). Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Poly(n^1) Cpx degree: 1 Solved cost: 8+2*n Rule cost: 8+2*B Rule guard: [ B>=1 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)