/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 35 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 328 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: eval_ndecr_start(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_bb0_in(v_0, v_i_0, v_n)) :|: TRUE eval_ndecr_bb0_in(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_0(v_0, v_i_0, v_n)) :|: TRUE eval_ndecr_0(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_1(v_0, v_i_0, v_n)) :|: TRUE eval_ndecr_1(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_2(v_n - 1, v_i_0, v_n)) :|: TRUE eval_ndecr_2(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_3(v_0, v_i_0, v_n)) :|: TRUE eval_ndecr_3(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_4(v_0, v_i_0, v_n)) :|: TRUE eval_ndecr_4(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_bb1_in(v_0, v_0, v_n)) :|: TRUE eval_ndecr_bb1_in(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_bb2_in(v_0, v_i_0, v_n)) :|: v_i_0 > 1 eval_ndecr_bb1_in(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_bb3_in(v_0, v_i_0, v_n)) :|: v_i_0 <= 1 eval_ndecr_bb2_in(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_bb1_in(v_0, v_i_0 - 1, v_n)) :|: TRUE eval_ndecr_bb3_in(v_0, v_i_0, v_n) -> Com_1(eval_ndecr_stop(v_0, v_i_0, v_n)) :|: TRUE The start-symbols are:[eval_ndecr_start_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 2*Ar_1 + 11) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalndecrstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb0in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalndecrbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr0(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalndecr0(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr1(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalndecr1(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr2(Ar_1 - 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalndecr2(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr3(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalndecr3(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr4(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalndecr4(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb1in(Ar_0, Ar_1, Ar_0)) (Comp: ?, Cost: 1) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ] (Comp: ?, Cost: 1) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb3in(Ar_0, Ar_1, Ar_2)) [ 1 >= Ar_2 ] (Comp: ?, Cost: 1) evalndecrbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb1in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: ?, Cost: 1) evalndecrbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstart(Ar_0, Ar_1, Ar_2)) [ 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) evalndecrstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecrbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr0(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr1(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr2(Ar_1 - 1, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr2(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr3(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr4(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb1in(Ar_0, Ar_1, Ar_0)) (Comp: ?, Cost: 1) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ] (Comp: ?, Cost: 1) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb3in(Ar_0, Ar_1, Ar_2)) [ 1 >= Ar_2 ] (Comp: ?, Cost: 1) evalndecrbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb1in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: ?, Cost: 1) evalndecrbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalndecrstart) = 2 Pol(evalndecrbb0in) = 2 Pol(evalndecr0) = 2 Pol(evalndecr1) = 2 Pol(evalndecr2) = 2 Pol(evalndecr3) = 2 Pol(evalndecr4) = 2 Pol(evalndecrbb1in) = 2 Pol(evalndecrbb2in) = 2 Pol(evalndecrbb3in) = 1 Pol(evalndecrstop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalndecrbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstop(Ar_0, Ar_1, Ar_2)) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb3in(Ar_0, Ar_1, Ar_2)) [ 1 >= Ar_2 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalndecrstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecrbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr0(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr1(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr2(Ar_1 - 1, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr2(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr3(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr4(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb1in(Ar_0, Ar_1, Ar_0)) (Comp: ?, Cost: 1) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ] (Comp: 2, Cost: 1) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb3in(Ar_0, Ar_1, Ar_2)) [ 1 >= Ar_2 ] (Comp: ?, Cost: 1) evalndecrbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb1in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: 2, Cost: 1) evalndecrbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalndecrstart) = V_2 Pol(evalndecrbb0in) = V_2 Pol(evalndecr0) = V_2 Pol(evalndecr1) = V_2 Pol(evalndecr2) = V_1 Pol(evalndecr3) = V_1 Pol(evalndecr4) = V_1 Pol(evalndecrbb1in) = V_3 Pol(evalndecrbb2in) = V_3 - 1 Pol(evalndecrbb3in) = V_3 Pol(evalndecrstop) = V_3 Pol(koat_start) = V_2 orients all transitions weakly and the transition evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) evalndecrstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecrbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr0(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr1(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr2(Ar_1 - 1, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr2(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr3(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr4(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb1in(Ar_0, Ar_1, Ar_0)) (Comp: Ar_1, Cost: 1) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ] (Comp: 2, Cost: 1) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb3in(Ar_0, Ar_1, Ar_2)) [ 1 >= Ar_2 ] (Comp: ?, Cost: 1) evalndecrbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb1in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: 2, Cost: 1) evalndecrbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstart(Ar_0, Ar_1, Ar_2)) [ 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) evalndecrstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecrbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr0(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr1(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr2(Ar_1 - 1, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr2(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr3(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecr4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalndecr4(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb1in(Ar_0, Ar_1, Ar_0)) (Comp: Ar_1, Cost: 1) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 2 ] (Comp: 2, Cost: 1) evalndecrbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb3in(Ar_0, Ar_1, Ar_2)) [ 1 >= Ar_2 ] (Comp: Ar_1, Cost: 1) evalndecrbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrbb1in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: 2, Cost: 1) evalndecrbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalndecrstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 2*Ar_1 + 11 Time: 0.061 sec (SMT: 0.047 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalndecrstart 0: evalndecrstart -> evalndecrbb0in : [], cost: 1 1: evalndecrbb0in -> evalndecr0 : [], cost: 1 2: evalndecr0 -> evalndecr1 : [], cost: 1 3: evalndecr1 -> evalndecr2 : A'=-1+B, [], cost: 1 4: evalndecr2 -> evalndecr3 : [], cost: 1 5: evalndecr3 -> evalndecr4 : [], cost: 1 6: evalndecr4 -> evalndecrbb1in : C'=A, [], cost: 1 7: evalndecrbb1in -> evalndecrbb2in : [ C>=2 ], cost: 1 8: evalndecrbb1in -> evalndecrbb3in : [ 1>=C ], cost: 1 9: evalndecrbb2in -> evalndecrbb1in : C'=-1+C, [], cost: 1 10: evalndecrbb3in -> evalndecrstop : [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: evalndecrstart -> evalndecrbb0in : [], cost: 1 Removed unreachable and leaf rules: Start location: evalndecrstart 0: evalndecrstart -> evalndecrbb0in : [], cost: 1 1: evalndecrbb0in -> evalndecr0 : [], cost: 1 2: evalndecr0 -> evalndecr1 : [], cost: 1 3: evalndecr1 -> evalndecr2 : A'=-1+B, [], cost: 1 4: evalndecr2 -> evalndecr3 : [], cost: 1 5: evalndecr3 -> evalndecr4 : [], cost: 1 6: evalndecr4 -> evalndecrbb1in : C'=A, [], cost: 1 7: evalndecrbb1in -> evalndecrbb2in : [ C>=2 ], cost: 1 9: evalndecrbb2in -> evalndecrbb1in : C'=-1+C, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalndecrstart 16: evalndecrstart -> evalndecrbb1in : A'=-1+B, C'=-1+B, [], cost: 7 17: evalndecrbb1in -> evalndecrbb1in : C'=-1+C, [ C>=2 ], cost: 2 Accelerating simple loops of location 7. Accelerating the following rules: 17: evalndecrbb1in -> evalndecrbb1in : C'=-1+C, [ C>=2 ], cost: 2 Accelerated rule 17 with metering function -1+C, yielding the new rule 18. Removing the simple loops: 17. Accelerated all simple loops using metering functions (where possible): Start location: evalndecrstart 16: evalndecrstart -> evalndecrbb1in : A'=-1+B, C'=-1+B, [], cost: 7 18: evalndecrbb1in -> evalndecrbb1in : C'=1, [ C>=2 ], cost: -2+2*C Chained accelerated rules (with incoming rules): Start location: evalndecrstart 16: evalndecrstart -> evalndecrbb1in : A'=-1+B, C'=-1+B, [], cost: 7 19: evalndecrstart -> evalndecrbb1in : A'=-1+B, C'=1, [ -1+B>=2 ], cost: 3+2*B Removed unreachable locations (and leaf rules with constant cost): Start location: evalndecrstart 19: evalndecrstart -> evalndecrbb1in : A'=-1+B, C'=1, [ -1+B>=2 ], cost: 3+2*B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalndecrstart 19: evalndecrstart -> evalndecrbb1in : A'=-1+B, C'=1, [ -1+B>=2 ], cost: 3+2*B Computing asymptotic complexity for rule 19 Solved the limit problem by the following transformations: Created initial limit problem: -2+B (+/+!), 3+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 3+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: 3+2*n Rule cost: 3+2*B Rule guard: [ -1+B>=2 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)