/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), O(n^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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 146 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 923 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: eval_start_start(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_bb0_in(v__0, v__0_sink, v_1, v_3, v_n)) :|: TRUE eval_start_bb0_in(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_0(v__0, v__0_sink, v_1, v_3, v_n)) :|: TRUE eval_start_0(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_1(v__0, v__0_sink, v_1, v_3, v_n)) :|: TRUE eval_start_1(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_2(v__0, v__0_sink, v_1, v_3, v_n)) :|: TRUE eval_start_2(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_bb1_in(v_n, v__0_sink, v_1, v_3, v_n)) :|: TRUE eval_start_bb1_in(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_bb2_in(v__0, v__0, v_1, v_3, v_n)) :|: v__0 > 0 eval_start_bb1_in(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_bb4_in(v__0, v__0_sink, v_1, v_3, v_n)) :|: v__0 <= 0 eval_start_bb2_in(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_bb3_in(v__0, v__0_sink, v__0_sink - 1, v_3, v_n)) :|: v__0_sink - 1 > 0 eval_start_bb2_in(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_bb1_in(v__0_sink - 1, v__0_sink, v_1, v_3, v_n)) :|: v__0_sink - 1 <= 0 eval_start_bb3_in(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_5(v__0, v__0_sink, v_1, v_3, v_n)) :|: TRUE eval_start_5(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_6(v__0, v__0_sink, v_1, nondef_0, v_n)) :|: TRUE eval_start_6(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_bb1_in(v_1, v__0_sink, v_1, v_3, v_n)) :|: v_3 > 0 eval_start_6(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_bb2_in(v__0, v_1, v_1, v_3, v_n)) :|: v_3 <= 0 eval_start_bb4_in(v__0, v__0_sink, v_1, v_3, v_n) -> Com_1(eval_start_stop(v__0, v__0_sink, v_1, v_3, v_n)) :|: TRUE The start-symbols are:[eval_start_start_5] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 16*Ar_1 + 9) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_1, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_2 - 1, Ar_4)) [ Ar_2 >= 2 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_2 - 1, Ar_1, Ar_2, Ar_3, Ar_4)) [ 1 >= Ar_2 ] (Comp: ?, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Fresh_0)) (Comp: ?, Cost: 1) evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_3, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_4 >= 1 ] (Comp: ?, Cost: 1) evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_3, Ar_3, Ar_4)) [ 0 >= Ar_4 ] (Comp: ?, Cost: 1) evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 1 produces the following problem: 2: T: (Comp: 1, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_1, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_2 - 1, Ar_4)) [ Ar_2 >= 2 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_2 - 1, Ar_1, Ar_2, Ar_3, Ar_4)) [ 1 >= Ar_2 ] (Comp: ?, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Fresh_0)) (Comp: ?, Cost: 1) evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_3, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_4 >= 1 ] (Comp: ?, Cost: 1) evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_3, Ar_3, Ar_4)) [ 0 >= Ar_4 ] (Comp: ?, Cost: 1) evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalstartstart) = 2 Pol(evalstartbb0in) = 2 Pol(evalstart0) = 2 Pol(evalstart1) = 2 Pol(evalstart2) = 2 Pol(evalstartbb1in) = 2 Pol(evalstartbb2in) = 2 Pol(evalstartbb4in) = 1 Pol(evalstartbb3in) = 2 Pol(evalstart5) = 2 Pol(evalstart6) = 2 Pol(evalstartstop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_1, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: 2, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_2 - 1, Ar_4)) [ Ar_2 >= 2 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_2 - 1, Ar_1, Ar_2, Ar_3, Ar_4)) [ 1 >= Ar_2 ] (Comp: ?, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Fresh_0)) (Comp: ?, Cost: 1) evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_3, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_4 >= 1 ] (Comp: ?, Cost: 1) evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_3, Ar_3, Ar_4)) [ 0 >= Ar_4 ] (Comp: 2, Cost: 1) evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalstartstart) = 2*V_2 Pol(evalstartbb0in) = 2*V_2 Pol(evalstart0) = 2*V_2 Pol(evalstart1) = 2*V_2 Pol(evalstart2) = 2*V_2 Pol(evalstartbb1in) = 2*V_1 Pol(evalstartbb2in) = 2*V_3 - 1 Pol(evalstartbb4in) = 2*V_1 Pol(evalstartbb3in) = 2*V_4 Pol(evalstart5) = 2*V_4 Pol(evalstart6) = 2*V_4 Pol(evalstartstop) = 2*V_1 Pol(koat_start) = 2*V_2 orients all transitions weakly and the transitions evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_2 - 1, Ar_4)) [ Ar_2 >= 2 ] evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_1, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 2*Ar_1, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: 2, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] (Comp: 2*Ar_1, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_2 - 1, Ar_4)) [ Ar_2 >= 2 ] (Comp: ?, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_2 - 1, Ar_1, Ar_2, Ar_3, Ar_4)) [ 1 >= Ar_2 ] (Comp: ?, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Fresh_0)) (Comp: ?, Cost: 1) evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_3, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_4 >= 1 ] (Comp: ?, Cost: 1) evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_3, Ar_3, Ar_4)) [ 0 >= Ar_4 ] (Comp: 2, Cost: 1) evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 4 produces the following problem: 5: T: (Comp: 1, Cost: 1) evalstartstart(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb0in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstartbb0in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 1) evalstart2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_1, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 2*Ar_1, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: 2, Cost: 1) evalstartbb1in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] (Comp: 2*Ar_1, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_2 - 1, Ar_4)) [ Ar_2 >= 2 ] (Comp: 4*Ar_1, Cost: 1) evalstartbb2in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_2 - 1, Ar_1, Ar_2, Ar_3, Ar_4)) [ 1 >= Ar_2 ] (Comp: 2*Ar_1, Cost: 1) evalstartbb3in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 2*Ar_1, Cost: 1) evalstart5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Fresh_0)) (Comp: 2*Ar_1, Cost: 1) evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb1in(Ar_3, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_4 >= 1 ] (Comp: 2*Ar_1, Cost: 1) evalstart6(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartbb2in(Ar_0, Ar_1, Ar_3, Ar_3, Ar_4)) [ 0 >= Ar_4 ] (Comp: 2, Cost: 1) evalstartbb4in(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstop(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(evalstartstart(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 16*Ar_1 + 9 Time: 0.126 sec (SMT: 0.093 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 -> evalstartbb1in : A'=B, [], cost: 1 5: evalstartbb1in -> evalstartbb2in : C'=A, [ A>=1 ], cost: 1 6: evalstartbb1in -> evalstartbb4in : [ 0>=A ], cost: 1 7: evalstartbb2in -> evalstartbb3in : D'=-1+C, [ C>=2 ], cost: 1 8: evalstartbb2in -> evalstartbb1in : A'=-1+C, [ 1>=C ], cost: 1 9: evalstartbb3in -> evalstart5 : [], cost: 1 10: evalstart5 -> evalstart6 : E'=free, [], cost: 1 11: evalstart6 -> evalstartbb1in : A'=D, [ E>=1 ], cost: 1 12: evalstart6 -> evalstartbb2in : C'=D, [ 0>=E ], cost: 1 13: evalstartbb4in -> 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 -> evalstartbb1in : A'=B, [], cost: 1 5: evalstartbb1in -> evalstartbb2in : C'=A, [ A>=1 ], cost: 1 7: evalstartbb2in -> evalstartbb3in : D'=-1+C, [ C>=2 ], cost: 1 8: evalstartbb2in -> evalstartbb1in : A'=-1+C, [ 1>=C ], cost: 1 9: evalstartbb3in -> evalstart5 : [], cost: 1 10: evalstart5 -> evalstart6 : E'=free, [], cost: 1 11: evalstart6 -> evalstartbb1in : A'=D, [ E>=1 ], cost: 1 12: evalstart6 -> evalstartbb2in : C'=D, [ 0>=E ], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalstartstart 17: evalstartstart -> evalstartbb1in : A'=B, [], cost: 5 5: evalstartbb1in -> evalstartbb2in : C'=A, [ A>=1 ], cost: 1 8: evalstartbb2in -> evalstartbb1in : A'=-1+C, [ 1>=C ], cost: 1 19: evalstartbb2in -> evalstart6 : D'=-1+C, E'=free, [ C>=2 ], cost: 3 11: evalstart6 -> evalstartbb1in : A'=D, [ E>=1 ], cost: 1 12: evalstart6 -> evalstartbb2in : C'=D, [ 0>=E ], cost: 1 Eliminated locations (on tree-shaped paths): Start location: evalstartstart 17: evalstartstart -> evalstartbb1in : A'=B, [], cost: 5 5: evalstartbb1in -> evalstartbb2in : C'=A, [ A>=1 ], cost: 1 8: evalstartbb2in -> evalstartbb1in : A'=-1+C, [ 1>=C ], cost: 1 20: evalstartbb2in -> evalstartbb1in : A'=-1+C, D'=-1+C, E'=free, [ C>=2 && free>=1 ], cost: 4 21: evalstartbb2in -> evalstartbb2in : C'=-1+C, D'=-1+C, E'=free, [ C>=2 && 0>=free ], cost: 4 Accelerating simple loops of location 6. Accelerating the following rules: 21: evalstartbb2in -> evalstartbb2in : C'=-1+C, D'=-1+C, E'=free, [ C>=2 && 0>=free ], cost: 4 Accelerated rule 21 with metering function -1+C, yielding the new rule 22. Removing the simple loops: 21. Accelerated all simple loops using metering functions (where possible): Start location: evalstartstart 17: evalstartstart -> evalstartbb1in : A'=B, [], cost: 5 5: evalstartbb1in -> evalstartbb2in : C'=A, [ A>=1 ], cost: 1 8: evalstartbb2in -> evalstartbb1in : A'=-1+C, [ 1>=C ], cost: 1 20: evalstartbb2in -> evalstartbb1in : A'=-1+C, D'=-1+C, E'=free, [ C>=2 && free>=1 ], cost: 4 22: evalstartbb2in -> evalstartbb2in : C'=1, D'=1, E'=free, [ C>=2 && 0>=free ], cost: -4+4*C Chained accelerated rules (with incoming rules): Start location: evalstartstart 17: evalstartstart -> evalstartbb1in : A'=B, [], cost: 5 5: evalstartbb1in -> evalstartbb2in : C'=A, [ A>=1 ], cost: 1 23: evalstartbb1in -> evalstartbb2in : C'=1, D'=1, E'=free, [ A>=2 && 0>=free ], cost: -3+4*A 8: evalstartbb2in -> evalstartbb1in : A'=-1+C, [ 1>=C ], cost: 1 20: evalstartbb2in -> evalstartbb1in : A'=-1+C, D'=-1+C, E'=free, [ C>=2 && free>=1 ], cost: 4 Eliminated locations (on tree-shaped paths): Start location: evalstartstart 17: evalstartstart -> evalstartbb1in : A'=B, [], cost: 5 24: evalstartbb1in -> evalstartbb1in : A'=-1+A, C'=A, [ A>=1 && 1>=A ], cost: 2 25: evalstartbb1in -> evalstartbb1in : A'=-1+A, C'=A, D'=-1+A, E'=free, [ A>=2 && free>=1 ], cost: 5 26: evalstartbb1in -> evalstartbb1in : A'=0, C'=1, D'=1, E'=free, [ A>=2 && 0>=free ], cost: -2+4*A 27: evalstartbb1in -> [13] : [ A>=2 && 0>=free ], cost: -3+4*A Accelerating simple loops of location 5. Simplified some of the simple loops (and removed duplicate rules). Accelerating the following rules: 24: evalstartbb1in -> evalstartbb1in : A'=-1+A, C'=A, [ 1-A==0 ], cost: 2 25: evalstartbb1in -> evalstartbb1in : A'=-1+A, C'=A, D'=-1+A, E'=free, [ A>=2 && free>=1 ], cost: 5 26: evalstartbb1in -> evalstartbb1in : A'=0, C'=1, D'=1, E'=free, [ A>=2 && 0>=free ], cost: -2+4*A Accelerated rule 24 with metering function A, yielding the new rule 28. Accelerated rule 25 with metering function -1+A, yielding the new rule 29. Found no metering function for rule 26. Removing the simple loops: 24 25. Accelerated all simple loops using metering functions (where possible): Start location: evalstartstart 17: evalstartstart -> evalstartbb1in : A'=B, [], cost: 5 26: evalstartbb1in -> evalstartbb1in : A'=0, C'=1, D'=1, E'=free, [ A>=2 && 0>=free ], cost: -2+4*A 27: evalstartbb1in -> [13] : [ A>=2 && 0>=free ], cost: -3+4*A 28: evalstartbb1in -> evalstartbb1in : A'=0, C'=1, [ 1-A==0 ], cost: 2*A 29: evalstartbb1in -> evalstartbb1in : A'=1, C'=2, D'=1, E'=free, [ A>=2 && free>=1 ], cost: -5+5*A Chained accelerated rules (with incoming rules): Start location: evalstartstart 17: evalstartstart -> evalstartbb1in : A'=B, [], cost: 5 30: evalstartstart -> evalstartbb1in : A'=0, C'=1, D'=1, E'=free, [ B>=2 && 0>=free ], cost: 3+4*B 31: evalstartstart -> evalstartbb1in : A'=0, C'=1, [ 1-B==0 ], cost: 5+2*B 32: evalstartstart -> evalstartbb1in : A'=1, C'=2, D'=1, E'=free, [ B>=2 && free>=1 ], cost: 5*B 27: evalstartbb1in -> [13] : [ A>=2 && 0>=free ], cost: -3+4*A Eliminated locations (on tree-shaped paths): Start location: evalstartstart 33: evalstartstart -> [13] : A'=B, [ B>=2 && 0>=free ], cost: 2+4*B 34: evalstartstart -> [15] : [ B>=2 && 0>=free ], cost: 3+4*B 35: evalstartstart -> [15] : [ 1-B==0 ], cost: 5+2*B 36: evalstartstart -> [15] : [ B>=2 && free>=1 ], cost: 5*B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalstartstart 34: evalstartstart -> [15] : [ B>=2 && 0>=free ], cost: 3+4*B 35: evalstartstart -> [15] : [ 1-B==0 ], cost: 5+2*B 36: evalstartstart -> [15] : [ B>=2 && free>=1 ], cost: 5*B Computing asymptotic complexity for rule 34 Solved the limit problem by the following transformations: Created initial limit problem: -1+B (+/+!), 3+4*B (+), 1-free (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {free==-n,B==n} resulting limit problem: [solved] Solution: free / -n B / n Resulting cost 3+4*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: 3+4*n Rule cost: 3+4*B Rule guard: [ B>=2 && 0>=free ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)