/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(NON_POLY, ?) 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(INF, INF). (0) CpxIntTrs (1) Loat Proof [FINISHED, 337 ms] (2) BOUNDS(INF, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f9(A, B, C, D) -> Com_1(f15(A, 0, E, D)) :|: 0 >= A && E >= 1 f15(A, B, C, D) -> Com_1(f15(A, B, C, D)) :|: C >= 1 f23(A, B, C, D) -> Com_1(f23(A, B, C, D)) :|: TRUE f25(A, B, C, D) -> Com_1(f28(A, B, C, D)) :|: TRUE f15(A, B, C, D) -> Com_1(f9(E, B, C, 0)) :|: 0 >= C f9(A, B, C, D) -> Com_1(f23(A, B, C, D)) :|: A >= 1 f0(A, B, C, D) -> Com_1(f9(E, 0, C, 0)) :|: TRUE The start-symbols are:[f0_4] ---------------------------------------- (1) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f0 0: f9 -> f15 : B'=0, C'=free, [ 0>=A && free>=1 ], cost: 1 5: f9 -> f23 : [ A>=1 ], cost: 1 1: f15 -> f15 : [ C>=1 ], cost: 1 4: f15 -> f9 : A'=free_1, D'=0, [ 0>=C ], cost: 1 2: f23 -> f23 : [], cost: 1 3: f25 -> f28 : [], cost: 1 6: f0 -> f9 : A'=free_2, B'=0, D'=0, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 6: f0 -> f9 : A'=free_2, B'=0, D'=0, [], cost: 1 Removed unreachable and leaf rules: Start location: f0 0: f9 -> f15 : B'=0, C'=free, [ 0>=A && free>=1 ], cost: 1 5: f9 -> f23 : [ A>=1 ], cost: 1 1: f15 -> f15 : [ C>=1 ], cost: 1 4: f15 -> f9 : A'=free_1, D'=0, [ 0>=C ], cost: 1 2: f23 -> f23 : [], cost: 1 6: f0 -> f9 : A'=free_2, B'=0, D'=0, [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 1: f15 -> f15 : [ C>=1 ], cost: 1 Accelerated rule 1 with NONTERM, yielding the new rule 7. Removing the simple loops: 1. Accelerating simple loops of location 2. Accelerating the following rules: 2: f23 -> f23 : [], cost: 1 Accelerated rule 2 with NONTERM, yielding the new rule 8. Removing the simple loops: 2. Accelerated all simple loops using metering functions (where possible): Start location: f0 0: f9 -> f15 : B'=0, C'=free, [ 0>=A && free>=1 ], cost: 1 5: f9 -> f23 : [ A>=1 ], cost: 1 4: f15 -> f9 : A'=free_1, D'=0, [ 0>=C ], cost: 1 7: f15 -> [6] : [ C>=1 ], cost: NONTERM 8: f23 -> [7] : [], cost: NONTERM 6: f0 -> f9 : A'=free_2, B'=0, D'=0, [], cost: 1 Chained accelerated rules (with incoming rules): Start location: f0 0: f9 -> f15 : B'=0, C'=free, [ 0>=A && free>=1 ], cost: 1 5: f9 -> f23 : [ A>=1 ], cost: 1 9: f9 -> [6] : B'=0, C'=free, [ 0>=A && free>=1 ], cost: NONTERM 10: f9 -> [7] : [ A>=1 ], cost: NONTERM 4: f15 -> f9 : A'=free_1, D'=0, [ 0>=C ], cost: 1 6: f0 -> f9 : A'=free_2, B'=0, D'=0, [], cost: 1 Removed unreachable locations (and leaf rules with constant cost): Start location: f0 0: f9 -> f15 : B'=0, C'=free, [ 0>=A && free>=1 ], cost: 1 9: f9 -> [6] : B'=0, C'=free, [ 0>=A && free>=1 ], cost: NONTERM 10: f9 -> [7] : [ A>=1 ], cost: NONTERM 4: f15 -> f9 : A'=free_1, D'=0, [ 0>=C ], cost: 1 6: f0 -> f9 : A'=free_2, B'=0, D'=0, [], cost: 1 Eliminated locations (on linear paths): Start location: f0 9: f9 -> [6] : B'=0, C'=free, [ 0>=A && free>=1 ], cost: NONTERM 10: f9 -> [7] : [ A>=1 ], cost: NONTERM 6: f0 -> f9 : A'=free_2, B'=0, D'=0, [], cost: 1 Eliminated locations (on tree-shaped paths): Start location: f0 11: f0 -> [6] : A'=free_2, B'=0, C'=free, D'=0, [ 0>=free_2 && free>=1 ], cost: NONTERM 12: f0 -> [7] : A'=free_2, B'=0, D'=0, [ free_2>=1 ], cost: NONTERM ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: f0 11: f0 -> [6] : A'=free_2, B'=0, C'=free, D'=0, [ 0>=free_2 && free>=1 ], cost: NONTERM 12: f0 -> [7] : A'=free_2, B'=0, D'=0, [ free_2>=1 ], cost: NONTERM Computing asymptotic complexity for rule 11 Guard is satisfiable, yielding nontermination Resulting cost NONTERM has complexity: Nonterm Found new complexity Nonterm. Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Nonterm Cpx degree: Nonterm Solved cost: NONTERM Rule cost: NONTERM Rule guard: [ 0>=free_2 && free>=1 ] NO ---------------------------------------- (2) BOUNDS(INF, INF)