3.89/1.85 WORST_CASE(Omega(n^1), O(n^1)) 3.89/1.86 proof of /export/starexec/sandbox/benchmark/theBenchmark.koat 3.89/1.86 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.89/1.86 3.89/1.86 3.89/1.86 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, max(1, 1 + Arg_0 + -1 * Arg_1)). 3.89/1.86 3.89/1.86 (0) CpxIntTrs 3.89/1.86 (1) Koat2 Proof [FINISHED, 29 ms] 3.89/1.86 (2) BOUNDS(1, max(1, 1 + Arg_0 + -1 * Arg_1)) 3.89/1.86 (3) Loat Proof [FINISHED, 225 ms] 3.89/1.86 (4) BOUNDS(n^1, INF) 3.89/1.86 3.89/1.86 3.89/1.86 ---------------------------------------- 3.89/1.86 3.89/1.86 (0) 3.89/1.86 Obligation: 3.89/1.86 Complexity Int TRS consisting of the following rules: 3.89/1.86 eval(A, B) -> Com_1(eval(A - C, B + D)) :|: A >= B + 1 && C >= 0 && D >= 1 3.89/1.86 start(A, B) -> Com_1(eval(A, B)) :|: TRUE 3.89/1.86 3.89/1.86 The start-symbols are:[start_2] 3.89/1.86 3.89/1.86 3.89/1.86 ---------------------------------------- 3.89/1.86 3.89/1.86 (1) Koat2 Proof (FINISHED) 3.89/1.86 YES( ?, max([1, 1+Arg_0-Arg_1]) {O(n)}) 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Initial Complexity Problem: 3.89/1.86 3.89/1.86 Start: start 3.89/1.86 3.89/1.86 Program_Vars: Arg_0, Arg_1 3.89/1.86 3.89/1.86 Temp_Vars: C, D 3.89/1.86 3.89/1.86 Locations: eval, start 3.89/1.86 3.89/1.86 Transitions: 3.89/1.86 3.89/1.86 eval(Arg_0,Arg_1) -> eval(Arg_0-C,Arg_1+D):|:Arg_1+1 <= Arg_0 && 0 <= C && 1 <= D 3.89/1.86 3.89/1.86 start(Arg_0,Arg_1) -> eval(Arg_0,Arg_1):|: 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Timebounds: 3.89/1.86 3.89/1.86 Overall timebound: max([1, 1+Arg_0-Arg_1]) {O(n)} 3.89/1.86 3.89/1.86 0: eval->eval: max([0, Arg_0-Arg_1]) {O(n)} 3.89/1.86 3.89/1.86 1: start->eval: 1 {O(1)} 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Costbounds: 3.89/1.86 3.89/1.86 Overall costbound: max([1, 1+Arg_0-Arg_1]) {O(n)} 3.89/1.86 3.89/1.86 0: eval->eval: max([0, Arg_0-Arg_1]) {O(n)} 3.89/1.86 3.89/1.86 1: start->eval: 1 {O(1)} 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Sizebounds: 3.89/1.86 3.89/1.86 `Lower: 3.89/1.86 3.89/1.86 0: eval->eval, Arg_1: Arg_1 {O(n)} 3.89/1.86 3.89/1.86 1: start->eval, Arg_0: Arg_0 {O(n)} 3.89/1.86 3.89/1.86 1: start->eval, Arg_1: Arg_1 {O(n)} 3.89/1.86 3.89/1.86 `Upper: 3.89/1.86 3.89/1.86 0: eval->eval, Arg_0: Arg_0 {O(n)} 3.89/1.86 3.89/1.86 1: start->eval, Arg_0: Arg_0 {O(n)} 3.89/1.86 3.89/1.86 1: start->eval, Arg_1: Arg_1 {O(n)} 3.89/1.86 3.89/1.86 3.89/1.86 ---------------------------------------- 3.89/1.86 3.89/1.86 (2) 3.89/1.86 BOUNDS(1, max(1, 1 + Arg_0 + -1 * Arg_1)) 3.89/1.86 3.89/1.86 ---------------------------------------- 3.89/1.86 3.89/1.86 (3) Loat Proof (FINISHED) 3.89/1.86 3.89/1.86 3.89/1.86 ### Pre-processing the ITS problem ### 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Initial linear ITS problem 3.89/1.86 3.89/1.86 Start location: start 3.89/1.86 3.89/1.86 0: eval -> eval : A'=A-free, B'=free_1+B, [ A>=1+B && free>=0 && free_1>=1 ], cost: 1 3.89/1.86 3.89/1.86 1: start -> eval : [], cost: 1 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 ### Simplification by acceleration and chaining ### 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Accelerating simple loops of location 0. 3.89/1.86 3.89/1.86 Accelerating the following rules: 3.89/1.86 3.89/1.86 0: eval -> eval : A'=A-free, B'=free_1+B, [ A>=1+B && free>=0 && free_1>=1 ], cost: 1 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 During metering: Instantiating temporary variables by {free_1==1,free==0} 3.89/1.86 3.89/1.86 Accelerated rule 0 with metering function A-B, yielding the new rule 2. 3.89/1.86 3.89/1.86 Removing the simple loops: 0. 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Accelerated all simple loops using metering functions (where possible): 3.89/1.86 3.89/1.86 Start location: start 3.89/1.86 3.89/1.86 2: eval -> eval : A'=A, B'=A, [ A>=1+B ], cost: A-B 3.89/1.86 3.89/1.86 1: start -> eval : [], cost: 1 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Chained accelerated rules (with incoming rules): 3.89/1.86 3.89/1.86 Start location: start 3.89/1.86 3.89/1.86 1: start -> eval : [], cost: 1 3.89/1.86 3.89/1.86 3: start -> eval : B'=A, [ A>=1+B ], cost: 1+A-B 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Removed unreachable locations (and leaf rules with constant cost): 3.89/1.86 3.89/1.86 Start location: start 3.89/1.86 3.89/1.86 3: start -> eval : B'=A, [ A>=1+B ], cost: 1+A-B 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 ### Computing asymptotic complexity ### 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Fully simplified ITS problem 3.89/1.86 3.89/1.86 Start location: start 3.89/1.86 3.89/1.86 3: start -> eval : B'=A, [ A>=1+B ], cost: 1+A-B 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Computing asymptotic complexity for rule 3 3.89/1.86 3.89/1.86 Solved the limit problem by the following transformations: 3.89/1.86 3.89/1.86 Created initial limit problem: 3.89/1.86 3.89/1.86 1+A-B (+), A-B (+/+!) [not solved] 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 removing all constraints (solved by SMT) 3.89/1.86 3.89/1.86 resulting limit problem: [solved] 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 applying transformation rule (C) using substitution {A==0,B==-n} 3.89/1.86 3.89/1.86 resulting limit problem: 3.89/1.86 3.89/1.86 [solved] 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Solution: 3.89/1.86 3.89/1.86 A / 0 3.89/1.86 3.89/1.86 B / -n 3.89/1.86 3.89/1.86 Resulting cost 1+n has complexity: Poly(n^1) 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Found new complexity Poly(n^1). 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 Obtained the following overall complexity (w.r.t. the length of the input n): 3.89/1.86 3.89/1.86 Complexity: Poly(n^1) 3.89/1.86 3.89/1.86 Cpx degree: 1 3.89/1.86 3.89/1.86 Solved cost: 1+n 3.89/1.86 3.89/1.86 Rule cost: 1+A-B 3.89/1.86 3.89/1.86 Rule guard: [ A>=1+B ] 3.89/1.86 3.89/1.86 3.89/1.86 3.89/1.86 WORST_CASE(Omega(n^1),?) 3.89/1.86 3.89/1.86 3.89/1.86 ---------------------------------------- 3.89/1.86 3.89/1.86 (4) 3.89/1.86 BOUNDS(n^1, INF) 3.96/1.89 EOF