/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(INF, INF). (0) CpxIntTrs (1) Loat Proof [FINISHED, 338 ms] (2) BOUNDS(INF, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A, B, C, D, E, F, G) -> Com_1(f12(H, I, J, 0, E, F, G)) :|: TRUE f12(A, B, C, D, E, F, G) -> Com_1(f12(A, B, C, D + 1, E, F, G)) :|: C >= D + 1 f25(A, B, C, D, E, F, G) -> Com_1(f25(A, B, C, D, E, F + 1, G)) :|: E >= F + 1 f25(A, B, C, D, E, F, G) -> Com_1(f34(A, B, C, D, E, F, G)) :|: F >= E f12(A, B, C, D, E, F, G) -> Com_1(f25(A, B, C, D, A, 0, H)) :|: D >= C The start-symbols are:[f0_7] ---------------------------------------- (1) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f0 0: f0 -> f12 : A'=free_2, B'=free, C'=free_1, D'=0, [], cost: 1 1: f12 -> f12 : D'=1+D, [ C>=1+D ], cost: 1 4: f12 -> f25 : E'=A, F'=0, G'=free_3, [ D>=C ], cost: 1 2: f25 -> f25 : F'=1+F, [ E>=1+F ], cost: 1 3: f25 -> f34 : [ F>=E ], cost: 1 Removed unreachable and leaf rules: Start location: f0 0: f0 -> f12 : A'=free_2, B'=free, C'=free_1, D'=0, [], cost: 1 1: f12 -> f12 : D'=1+D, [ C>=1+D ], cost: 1 4: f12 -> f25 : E'=A, F'=0, G'=free_3, [ D>=C ], cost: 1 2: f25 -> f25 : F'=1+F, [ E>=1+F ], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 1: f12 -> f12 : D'=1+D, [ C>=1+D ], cost: 1 Accelerated rule 1 with metering function C-D, yielding the new rule 5. Removing the simple loops: 1. Accelerating simple loops of location 2. Accelerating the following rules: 2: f25 -> f25 : F'=1+F, [ E>=1+F ], cost: 1 Accelerated rule 2 with metering function -F+E, yielding the new rule 6. Removing the simple loops: 2. Accelerated all simple loops using metering functions (where possible): Start location: f0 0: f0 -> f12 : A'=free_2, B'=free, C'=free_1, D'=0, [], cost: 1 4: f12 -> f25 : E'=A, F'=0, G'=free_3, [ D>=C ], cost: 1 5: f12 -> f12 : D'=C, [ C>=1+D ], cost: C-D 6: f25 -> f25 : F'=E, [ E>=1+F ], cost: -F+E Chained accelerated rules (with incoming rules): Start location: f0 0: f0 -> f12 : A'=free_2, B'=free, C'=free_1, D'=0, [], cost: 1 7: f0 -> f12 : A'=free_2, B'=free, C'=free_1, D'=free_1, [ free_1>=1 ], cost: 1+free_1 4: f12 -> f25 : E'=A, F'=0, G'=free_3, [ D>=C ], cost: 1 8: f12 -> f25 : E'=A, F'=A, G'=free_3, [ D>=C && A>=1 ], cost: 1+A Removed unreachable locations (and leaf rules with constant cost): Start location: f0 0: f0 -> f12 : A'=free_2, B'=free, C'=free_1, D'=0, [], cost: 1 7: f0 -> f12 : A'=free_2, B'=free, C'=free_1, D'=free_1, [ free_1>=1 ], cost: 1+free_1 8: f12 -> f25 : E'=A, F'=A, G'=free_3, [ D>=C && A>=1 ], cost: 1+A Eliminated locations (on tree-shaped paths): Start location: f0 9: f0 -> f25 : A'=free_2, B'=free, C'=free_1, D'=0, E'=free_2, F'=free_2, G'=free_3, [ 0>=free_1 && free_2>=1 ], cost: 2+free_2 10: f0 -> f25 : A'=free_2, B'=free, C'=free_1, D'=free_1, E'=free_2, F'=free_2, G'=free_3, [ free_1>=1 && free_2>=1 ], cost: 2+free_1+free_2 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: f0 9: f0 -> f25 : A'=free_2, B'=free, C'=free_1, D'=0, E'=free_2, F'=free_2, G'=free_3, [ 0>=free_1 && free_2>=1 ], cost: 2+free_2 10: f0 -> f25 : A'=free_2, B'=free, C'=free_1, D'=free_1, E'=free_2, F'=free_2, G'=free_3, [ free_1>=1 && free_2>=1 ], cost: 2+free_1+free_2 Computing asymptotic complexity for rule 9 Solved the limit problem by the following transformations: Created initial limit problem: 2+free_2 (+), 1-free_1 (+/+!), free_2 (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {free_1==-n,free_2==n} resulting limit problem: [solved] Solution: free_1 / -n free_2 / n Resulting cost 2+n has complexity: Unbounded Found new complexity Unbounded. Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Unbounded Cpx degree: Unbounded Solved cost: 2+n Rule cost: 2+free_2 Rule guard: [ 0>=free_1 && free_2>=1 ] WORST_CASE(INF,?) ---------------------------------------- (2) BOUNDS(INF, INF)