/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, max(1, 2 + Arg_0)). (0) CpxIntTrs (1) Koat2 Proof [FINISHED, 47 ms] (2) BOUNDS(1, max(1, 2 + Arg_0)) (3) Loat Proof [FINISHED, 130 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f3(A) -> Com_1(f1(A)) :|: A >= 0 f1(A) -> Com_1(f1(A - 1)) :|: A >= 1 The start-symbols are:[f3_1] ---------------------------------------- (1) Koat2 Proof (FINISHED) YES( ?, max([1, 2+Arg_0]) {O(n)}) Initial Complexity Problem: Start: f3 Program_Vars: Arg_0 Temp_Vars: Locations: f1, f3 Transitions: 1: f1->f1 0: f3->f1 Timebounds: Overall timebound: max([1, 2+Arg_0]) {O(n)} 1: f1->f1: max([0, 1+Arg_0]) {O(n)} 0: f3->f1: 1 {O(1)} Costbounds: Overall costbound: max([1, 2+Arg_0]) {O(n)} 1: f1->f1: max([0, 1+Arg_0]) {O(n)} 0: f3->f1: 1 {O(1)} Sizebounds: `Lower: 1: f1->f1, Arg_0: 0 {O(1)} 0: f3->f1, Arg_0: 0 {O(1)} `Upper: 1: f1->f1, Arg_0: Arg_0 {O(n)} 0: f3->f1, Arg_0: Arg_0 {O(n)} ---------------------------------------- (2) BOUNDS(1, max(1, 2 + Arg_0)) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f3 0: f3 -> f1 : [ A>=0 ], cost: 1 1: f1 -> f1 : A'=-1+A, [ A>=1 ], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: f3 -> f1 : [ A>=0 ], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 1: f1 -> f1 : A'=-1+A, [ A>=1 ], cost: 1 Accelerated rule 1 with metering function A, yielding the new rule 2. Removing the simple loops: 1. Accelerated all simple loops using metering functions (where possible): Start location: f3 0: f3 -> f1 : [ A>=0 ], cost: 1 2: f1 -> f1 : A'=0, [ A>=1 ], cost: A Chained accelerated rules (with incoming rules): Start location: f3 0: f3 -> f1 : [ A>=0 ], cost: 1 3: f3 -> f1 : A'=0, [ A>=1 ], cost: 1+A Removed unreachable locations (and leaf rules with constant cost): Start location: f3 3: f3 -> f1 : A'=0, [ A>=1 ], cost: 1+A ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: f3 3: f3 -> f1 : A'=0, [ A>=1 ], cost: 1+A Computing asymptotic complexity for rule 3 Solved the limit problem by the following transformations: Created initial limit problem: A (+/+!), 1+A (+) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {A==n} resulting limit problem: [solved] Solution: A / n Resulting cost 1+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: 1+n Rule cost: 1+A Rule guard: [ A>=1 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)