/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^2)) 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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, max(1, 1 + Arg_1 + -1 * Arg_2) + nat(Arg_0 + -1 * Arg_1) + nat(Arg_0 * Arg_1 + -1 * Arg_0 * Arg_2 + Arg_0 * nat(Arg_0 + -1 * Arg_1) + Arg_1 * Arg_2 + -1 * Arg_1 * nat(Arg_0 + -1 * Arg_1) + -1 * Arg_1^2)). (0) CpxIntTrs (1) Koat2 Proof [FINISHED, 248 ms] (2) BOUNDS(1, max(1, 1 + Arg_1 + -1 * Arg_2) + nat(Arg_0 + -1 * Arg_1) + nat(Arg_0 * Arg_1 + -1 * Arg_0 * Arg_2 + Arg_0 * nat(Arg_0 + -1 * Arg_1) + Arg_1 * Arg_2 + -1 * Arg_1 * nat(Arg_0 + -1 * Arg_1) + -1 * Arg_1^2)) (3) Loat Proof [FINISHED, 337 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f1(A, B, C) -> Com_1(f1(A, B + 1, C)) :|: A >= B + 1 f3(A, B, C) -> Com_1(f1(A, B, C)) :|: B >= C + 1 f1(A, B, C) -> Com_1(f1(A, B, C + 1)) :|: B >= C + 2 && B >= A The start-symbols are:[f3_3] ---------------------------------------- (1) Koat2 Proof (FINISHED) YES( ?, 1+max([0, Arg_1-Arg_2])+max([0, (Arg_0-Arg_1)*(Arg_1-Arg_2+max([0, Arg_0-Arg_1]))])+max([0, Arg_0-Arg_1]) {O(n^2)}) Initial Complexity Problem: Start: f3 Program_Vars: Arg_0, Arg_1, Arg_2 Temp_Vars: Locations: f1, f3 Transitions: f1(Arg_0,Arg_1,Arg_2) -> f1(Arg_0,Arg_1+1,Arg_2):|:1+Arg_2 <= Arg_1 && Arg_1+1 <= Arg_0 f1(Arg_0,Arg_1,Arg_2) -> f1(Arg_0,Arg_1,Arg_2+1):|:1+Arg_2 <= Arg_1 && Arg_2+2 <= Arg_1 && Arg_0 <= Arg_1 f3(Arg_0,Arg_1,Arg_2) -> f1(Arg_0,Arg_1,Arg_2):|:Arg_2+1 <= Arg_1 Timebounds: Overall timebound: 1+max([0, Arg_1-Arg_2])+max([0, (Arg_0-Arg_1)*(Arg_1-Arg_2+max([0, Arg_0-Arg_1]))])+max([0, Arg_0-Arg_1]) {O(n^2)} 0: f1->f1: max([0, Arg_0-Arg_1]) {O(n)} 2: f1->f1: max([0, Arg_1-Arg_2])+max([0, (Arg_0-Arg_1)*(Arg_1-Arg_2+max([0, Arg_0-Arg_1]))]) {O(n^2)} 1: f3->f1: 1 {O(1)} Costbounds: Overall costbound: 1+max([0, Arg_1-Arg_2])+max([0, (Arg_0-Arg_1)*(Arg_1-Arg_2+max([0, Arg_0-Arg_1]))])+max([0, Arg_0-Arg_1]) {O(n^2)} 0: f1->f1: max([0, Arg_0-Arg_1]) {O(n)} 2: f1->f1: max([0, Arg_1-Arg_2])+max([0, (Arg_0-Arg_1)*(Arg_1-Arg_2+max([0, Arg_0-Arg_1]))]) {O(n^2)} 1: f3->f1: 1 {O(1)} Sizebounds: `Lower: 0: f1->f1, Arg_0: Arg_0 {O(n)} 0: f1->f1, Arg_1: Arg_1 {O(n)} 0: f1->f1, Arg_2: Arg_2 {O(n)} 2: f1->f1, Arg_0: Arg_0 {O(n)} 2: f1->f1, Arg_1: Arg_1 {O(n)} 2: f1->f1, Arg_2: Arg_2 {O(n)} 1: f3->f1, Arg_0: Arg_0 {O(n)} 1: f3->f1, Arg_1: Arg_1 {O(n)} 1: f3->f1, Arg_2: Arg_2 {O(n)} `Upper: 0: f1->f1, Arg_0: Arg_0 {O(n)} 0: f1->f1, Arg_1: Arg_1+max([0, Arg_0-Arg_1]) {O(n)} 0: f1->f1, Arg_2: Arg_2 {O(n)} 2: f1->f1, Arg_0: Arg_0 {O(n)} 2: f1->f1, Arg_1: max([Arg_1, Arg_1+max([0, Arg_0-Arg_1])]) {O(n)} 2: f1->f1, Arg_2: Arg_2+max([0, Arg_1-Arg_2])+max([0, (Arg_0-Arg_1)*(Arg_1-Arg_2+max([0, Arg_0-Arg_1]))]) {O(n^2)} 1: f3->f1, Arg_0: Arg_0 {O(n)} 1: f3->f1, Arg_1: Arg_1 {O(n)} 1: f3->f1, Arg_2: Arg_2 {O(n)} ---------------------------------------- (2) BOUNDS(1, max(1, 1 + Arg_1 + -1 * Arg_2) + nat(Arg_0 + -1 * Arg_1) + nat(Arg_0 * Arg_1 + -1 * Arg_0 * Arg_2 + Arg_0 * nat(Arg_0 + -1 * Arg_1) + Arg_1 * Arg_2 + -1 * Arg_1 * nat(Arg_0 + -1 * Arg_1) + -1 * Arg_1^2)) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f3 0: f1 -> f1 : B'=1+B, [ A>=1+B ], cost: 1 2: f1 -> f1 : C'=1+C, [ B>=2+C && B>=A ], cost: 1 1: f3 -> f1 : [ B>=1+C ], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 0. Accelerating the following rules: 0: f1 -> f1 : B'=1+B, [ A>=1+B ], cost: 1 2: f1 -> f1 : C'=1+C, [ B>=2+C && B>=A ], cost: 1 Accelerated rule 0 with metering function A-B, yielding the new rule 3. Accelerated rule 2 with metering function -1-C+B, yielding the new rule 4. Removing the simple loops: 0 2. Accelerated all simple loops using metering functions (where possible): Start location: f3 3: f1 -> f1 : B'=A, [ A>=1+B ], cost: A-B 4: f1 -> f1 : C'=-1+B, [ B>=2+C && B>=A ], cost: -1-C+B 1: f3 -> f1 : [ B>=1+C ], cost: 1 Chained accelerated rules (with incoming rules): Start location: f3 1: f3 -> f1 : [ B>=1+C ], cost: 1 5: f3 -> f1 : B'=A, [ B>=1+C && A>=1+B ], cost: 1+A-B 6: f3 -> f1 : C'=-1+B, [ B>=2+C && B>=A ], cost: -C+B Removed unreachable locations (and leaf rules with constant cost): Start location: f3 5: f3 -> f1 : B'=A, [ B>=1+C && A>=1+B ], cost: 1+A-B 6: f3 -> f1 : C'=-1+B, [ B>=2+C && B>=A ], cost: -C+B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: f3 5: f3 -> f1 : B'=A, [ B>=1+C && A>=1+B ], cost: 1+A-B 6: f3 -> f1 : C'=-1+B, [ B>=2+C && B>=A ], cost: -C+B Computing asymptotic complexity for rule 5 Solved the limit problem by the following transformations: Created initial limit problem: -C+B (+/+!), 1+A-B (+), A-B (+/+!) [not solved] applying transformation rule (C) using substitution {B==1+C} resulting limit problem: 1 (+/+!), -1-C+A (+/+!), -C+A (+) [not solved] applying transformation rule (C) using substitution {A==1+B} resulting limit problem: 1 (+/+!), 1-C+B (+), -C+B (+/+!) [not solved] applying transformation rule (B), deleting 1 (+/+!) resulting limit problem: 1-C+B (+), -C+B (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {C==-2*n,B==-n} resulting limit problem: [solved] Solution: C / -2*n A / 1-n B / 1-2*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-B Rule guard: [ B>=1+C && A>=1+B ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)