/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 9 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 339 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: eval_abc_start(v_a, v_b, v_i_0) -> Com_1(eval_abc_bb0_in(v_a, v_b, v_i_0)) :|: TRUE eval_abc_bb0_in(v_a, v_b, v_i_0) -> Com_1(eval_abc_0(v_a, v_b, v_i_0)) :|: TRUE eval_abc_0(v_a, v_b, v_i_0) -> Com_1(eval_abc_1(v_a, v_b, v_i_0)) :|: TRUE eval_abc_1(v_a, v_b, v_i_0) -> Com_1(eval_abc_2(v_a, v_b, v_i_0)) :|: TRUE eval_abc_2(v_a, v_b, v_i_0) -> Com_1(eval_abc_3(v_a, v_b, v_i_0)) :|: TRUE eval_abc_3(v_a, v_b, v_i_0) -> Com_1(eval_abc_4(v_a, v_b, v_i_0)) :|: TRUE eval_abc_4(v_a, v_b, v_i_0) -> Com_1(eval_abc_bb1_in(v_a, v_b, v_a)) :|: TRUE eval_abc_bb1_in(v_a, v_b, v_i_0) -> Com_1(eval_abc_bb2_in(v_a, v_b, v_i_0)) :|: v_i_0 <= v_b eval_abc_bb1_in(v_a, v_b, v_i_0) -> Com_1(eval_abc_bb3_in(v_a, v_b, v_i_0)) :|: v_i_0 > v_b eval_abc_bb2_in(v_a, v_b, v_i_0) -> Com_1(eval_abc_bb1_in(v_a, v_b, v_i_0 + 1)) :|: TRUE eval_abc_bb3_in(v_a, v_b, v_i_0) -> Com_1(eval_abc_stop(v_a, v_b, v_i_0)) :|: TRUE The start-symbols are:[eval_abc_start_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 2*Ar_1 + 2*Ar_2 + 13) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalabcstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb0in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabcbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc0(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabc0(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc1(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabc1(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc2(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabc2(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc3(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabc3(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc4(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabc4(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb1in(Ar_1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 ] (Comp: ?, Cost: 1) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalabcbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb1in(Ar_0 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabcbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstart(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) evalabcstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabcbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc0(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc1(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc2(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc2(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc3(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc4(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb1in(Ar_1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 ] (Comp: ?, Cost: 1) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalabcbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb1in(Ar_0 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabcbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalabcstart) = 2 Pol(evalabcbb0in) = 2 Pol(evalabc0) = 2 Pol(evalabc1) = 2 Pol(evalabc2) = 2 Pol(evalabc3) = 2 Pol(evalabc4) = 2 Pol(evalabcbb1in) = 2 Pol(evalabcbb2in) = 2 Pol(evalabcbb3in) = 1 Pol(evalabcstop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalabcbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstop(Ar_0, Ar_1, Ar_2)) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 + 1 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalabcstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabcbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc0(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc1(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc2(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc2(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc3(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc4(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb1in(Ar_1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 ] (Comp: 2, Cost: 1) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalabcbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb1in(Ar_0 + 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalabcbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalabcstart) = -V_2 + V_3 + 1 Pol(evalabcbb0in) = -V_2 + V_3 + 1 Pol(evalabc0) = -V_2 + V_3 + 1 Pol(evalabc1) = -V_2 + V_3 + 1 Pol(evalabc2) = -V_2 + V_3 + 1 Pol(evalabc3) = -V_2 + V_3 + 1 Pol(evalabc4) = -V_2 + V_3 + 1 Pol(evalabcbb1in) = -V_1 + V_3 + 1 Pol(evalabcbb2in) = -V_1 + V_3 Pol(evalabcbb3in) = -V_1 + V_3 Pol(evalabcstop) = -V_1 + V_3 Pol(koat_start) = -V_2 + V_3 + 1 orients all transitions weakly and the transition evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) evalabcstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabcbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc0(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc1(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc2(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc2(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc3(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc4(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb1in(Ar_1, Ar_1, Ar_2)) (Comp: Ar_1 + Ar_2 + 1, Cost: 1) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 ] (Comp: 2, Cost: 1) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalabcbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb1in(Ar_0 + 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalabcbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstart(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) evalabcstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb0in(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabcbb0in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc0(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc0(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc1(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc2(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc2(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc3(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc3(Ar_0, Ar_1, Ar_2) -> Com_1(evalabc4(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalabc4(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb1in(Ar_1, Ar_1, Ar_2)) (Comp: Ar_1 + Ar_2 + 1, Cost: 1) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 ] (Comp: 2, Cost: 1) evalabcbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 + 1 ] (Comp: Ar_1 + Ar_2 + 1, Cost: 1) evalabcbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcbb1in(Ar_0 + 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalabcbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalabcstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 2*Ar_1 + 2*Ar_2 + 13 Time: 0.058 sec (SMT: 0.044 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalabcstart 0: evalabcstart -> evalabcbb0in : [], cost: 1 1: evalabcbb0in -> evalabc0 : [], cost: 1 2: evalabc0 -> evalabc1 : [], cost: 1 3: evalabc1 -> evalabc2 : [], cost: 1 4: evalabc2 -> evalabc3 : [], cost: 1 5: evalabc3 -> evalabc4 : [], cost: 1 6: evalabc4 -> evalabcbb1in : A'=B, [], cost: 1 7: evalabcbb1in -> evalabcbb2in : [ C>=A ], cost: 1 8: evalabcbb1in -> evalabcbb3in : [ A>=1+C ], cost: 1 9: evalabcbb2in -> evalabcbb1in : A'=1+A, [], cost: 1 10: evalabcbb3in -> evalabcstop : [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: evalabcstart -> evalabcbb0in : [], cost: 1 Removed unreachable and leaf rules: Start location: evalabcstart 0: evalabcstart -> evalabcbb0in : [], cost: 1 1: evalabcbb0in -> evalabc0 : [], cost: 1 2: evalabc0 -> evalabc1 : [], cost: 1 3: evalabc1 -> evalabc2 : [], cost: 1 4: evalabc2 -> evalabc3 : [], cost: 1 5: evalabc3 -> evalabc4 : [], cost: 1 6: evalabc4 -> evalabcbb1in : A'=B, [], cost: 1 7: evalabcbb1in -> evalabcbb2in : [ C>=A ], cost: 1 9: evalabcbb2in -> evalabcbb1in : A'=1+A, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalabcstart 16: evalabcstart -> evalabcbb1in : A'=B, [], cost: 7 17: evalabcbb1in -> evalabcbb1in : A'=1+A, [ C>=A ], cost: 2 Accelerating simple loops of location 7. Accelerating the following rules: 17: evalabcbb1in -> evalabcbb1in : A'=1+A, [ C>=A ], cost: 2 Accelerated rule 17 with metering function 1+C-A, yielding the new rule 18. Removing the simple loops: 17. Accelerated all simple loops using metering functions (where possible): Start location: evalabcstart 16: evalabcstart -> evalabcbb1in : A'=B, [], cost: 7 18: evalabcbb1in -> evalabcbb1in : A'=1+C, [ C>=A ], cost: 2+2*C-2*A Chained accelerated rules (with incoming rules): Start location: evalabcstart 16: evalabcstart -> evalabcbb1in : A'=B, [], cost: 7 19: evalabcstart -> evalabcbb1in : A'=1+C, [ C>=B ], cost: 9+2*C-2*B Removed unreachable locations (and leaf rules with constant cost): Start location: evalabcstart 19: evalabcstart -> evalabcbb1in : A'=1+C, [ C>=B ], cost: 9+2*C-2*B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalabcstart 19: evalabcstart -> evalabcbb1in : A'=1+C, [ C>=B ], cost: 9+2*C-2*B Computing asymptotic complexity for rule 19 Solved the limit problem by the following transformations: Created initial limit problem: 1+C-B (+/+!), 9+2*C-2*B (+) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {C==0,B==-n} resulting limit problem: [solved] Solution: C / 0 B / -n Resulting cost 9+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: 9+2*n Rule cost: 9+2*C-2*B Rule guard: [ C>=B ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)