/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), O(n^2)) proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, max(4, 4 + 2 * Arg_0) + nat(2 + 2 * Arg_0) + nat(Arg_0 + Arg_0^2)). (0) CpxIntTrs (1) Koat2 Proof [FINISHED, 856 ms] (2) BOUNDS(1, max(4, 4 + 2 * Arg_0) + nat(2 + 2 * Arg_0) + nat(Arg_0 + Arg_0^2)) (3) Loat Proof [FINISHED, 318 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: evalfstart(A, B, C) -> Com_1(evalfentryin(A, B, C)) :|: TRUE evalfentryin(A, B, C) -> Com_1(evalfbb4in(B, A, C)) :|: TRUE evalfbb4in(A, B, C) -> Com_1(evalfbb2in(A, B, A)) :|: B >= 1 evalfbb4in(A, B, C) -> Com_1(evalfreturnin(A, B, C)) :|: 0 >= B evalfbb2in(A, B, C) -> Com_1(evalfbb1in(A, B, C)) :|: C >= A evalfbb2in(A, B, C) -> Com_1(evalfbb3in(A, B, C)) :|: A >= C + 1 evalfbb1in(A, B, C) -> Com_1(evalfbb2in(A, B, C - 1)) :|: TRUE evalfbb3in(A, B, C) -> Com_1(evalfbb4in(A, B - 1, C)) :|: TRUE evalfreturnin(A, B, C) -> Com_1(evalfstop(A, B, C)) :|: TRUE The start-symbols are:[evalfstart_3] ---------------------------------------- (1) Koat2 Proof (FINISHED) YES( ?, 4+2*max([0, Arg_0])+max([0, 1+Arg_0])+max([0, Arg_0*(1+Arg_0)])+max([0, 1+Arg_0]) {O(n^2)}) Initial Complexity Problem: Start: evalfstart Program_Vars: Arg_0, Arg_1, Arg_2 Temp_Vars: Locations: evalfbb1in, evalfbb2in, evalfbb3in, evalfbb4in, evalfentryin, evalfreturnin, evalfstart, evalfstop Transitions: evalfbb1in(Arg_0,Arg_1,Arg_2) -> evalfbb2in(Arg_0,Arg_1,Arg_2-1):|:Arg_2 <= Arg_0 && Arg_0 <= Arg_2 && 1 <= Arg_1 evalfbb2in(Arg_0,Arg_1,Arg_2) -> evalfbb1in(Arg_0,Arg_1,Arg_2):|:Arg_2 <= Arg_0 && 1 <= Arg_1 && Arg_0 <= Arg_2 evalfbb2in(Arg_0,Arg_1,Arg_2) -> evalfbb3in(Arg_0,Arg_1,Arg_2):|:Arg_2 <= Arg_0 && 1 <= Arg_1 && Arg_2+1 <= Arg_0 evalfbb3in(Arg_0,Arg_1,Arg_2) -> evalfbb4in(Arg_0,Arg_1-1,Arg_2):|:1+Arg_2 <= Arg_0 && 1 <= Arg_1 evalfbb4in(Arg_0,Arg_1,Arg_2) -> evalfbb2in(Arg_0,Arg_1,Arg_0):|:1 <= Arg_1 evalfbb4in(Arg_0,Arg_1,Arg_2) -> evalfreturnin(Arg_0,Arg_1,Arg_2):|:Arg_1 <= 0 evalfentryin(Arg_0,Arg_1,Arg_2) -> evalfbb4in(Arg_1,Arg_0,Arg_2):|: evalfreturnin(Arg_0,Arg_1,Arg_2) -> evalfstop(Arg_0,Arg_1,Arg_2):|:Arg_1 <= 0 evalfstart(Arg_0,Arg_1,Arg_2) -> evalfentryin(Arg_0,Arg_1,Arg_2):|: Timebounds: Overall timebound: 4+2*max([0, Arg_0])+max([0, 1+Arg_0])+max([0, Arg_0*(1+Arg_0)])+max([0, 1+Arg_0]) {O(n^2)} 6: evalfbb1in->evalfbb2in: max([0, Arg_0*(1+Arg_0)]) {O(n^2)} 4: evalfbb2in->evalfbb1in: max([0, 1+Arg_0]) {O(n)} 5: evalfbb2in->evalfbb3in: max([0, Arg_0]) {O(n)} 7: evalfbb3in->evalfbb4in: max([0, Arg_0]) {O(n)} 2: evalfbb4in->evalfbb2in: max([0, 1+Arg_0]) {O(n)} 3: evalfbb4in->evalfreturnin: 1 {O(1)} 1: evalfentryin->evalfbb4in: 1 {O(1)} 8: evalfreturnin->evalfstop: 1 {O(1)} 0: evalfstart->evalfentryin: 1 {O(1)} Costbounds: Overall costbound: 4+2*max([0, Arg_0])+max([0, 1+Arg_0])+max([0, Arg_0*(1+Arg_0)])+max([0, 1+Arg_0]) {O(n^2)} 6: evalfbb1in->evalfbb2in: max([0, Arg_0*(1+Arg_0)]) {O(n^2)} 4: evalfbb2in->evalfbb1in: max([0, 1+Arg_0]) {O(n)} 5: evalfbb2in->evalfbb3in: max([0, Arg_0]) {O(n)} 7: evalfbb3in->evalfbb4in: max([0, Arg_0]) {O(n)} 2: evalfbb4in->evalfbb2in: max([0, 1+Arg_0]) {O(n)} 3: evalfbb4in->evalfreturnin: 1 {O(1)} 1: evalfentryin->evalfbb4in: 1 {O(1)} 8: evalfreturnin->evalfstop: 1 {O(1)} 0: evalfstart->evalfentryin: 1 {O(1)} Sizebounds: `Lower: 6: evalfbb1in->evalfbb2in, Arg_0: Arg_1 {O(n)} 6: evalfbb1in->evalfbb2in, Arg_1: 1 {O(1)} 6: evalfbb1in->evalfbb2in, Arg_2: -1+Arg_1 {O(n)} 4: evalfbb2in->evalfbb1in, Arg_0: Arg_1 {O(n)} 4: evalfbb2in->evalfbb1in, Arg_1: 1 {O(1)} 4: evalfbb2in->evalfbb1in, Arg_2: Arg_1 {O(n)} 5: evalfbb2in->evalfbb3in, Arg_0: Arg_1 {O(n)} 5: evalfbb2in->evalfbb3in, Arg_1: 1 {O(1)} 5: evalfbb2in->evalfbb3in, Arg_2: -1+Arg_1 {O(n)} 7: evalfbb3in->evalfbb4in, Arg_0: Arg_1 {O(n)} 7: evalfbb3in->evalfbb4in, Arg_1: 0 {O(1)} 7: evalfbb3in->evalfbb4in, Arg_2: -1+Arg_1 {O(n)} 2: evalfbb4in->evalfbb2in, Arg_0: Arg_1 {O(n)} 2: evalfbb4in->evalfbb2in, Arg_1: 1 {O(1)} 2: evalfbb4in->evalfbb2in, Arg_2: Arg_1 {O(n)} 3: evalfbb4in->evalfreturnin, Arg_0: Arg_1 {O(n)} 3: evalfbb4in->evalfreturnin, Arg_1: min([0, Arg_0]) {O(n)} 3: evalfbb4in->evalfreturnin, Arg_2: min([Arg_2, -(1-Arg_1)]) {O(n)} 1: evalfentryin->evalfbb4in, Arg_0: Arg_1 {O(n)} 1: evalfentryin->evalfbb4in, Arg_1: Arg_0 {O(n)} 1: evalfentryin->evalfbb4in, Arg_2: Arg_2 {O(n)} 8: evalfreturnin->evalfstop, Arg_0: Arg_1 {O(n)} 8: evalfreturnin->evalfstop, Arg_1: min([0, Arg_0]) {O(n)} 8: evalfreturnin->evalfstop, Arg_2: min([Arg_2, -(1-Arg_1)]) {O(n)} 0: evalfstart->evalfentryin, Arg_0: Arg_0 {O(n)} 0: evalfstart->evalfentryin, Arg_1: Arg_1 {O(n)} 0: evalfstart->evalfentryin, Arg_2: Arg_2 {O(n)} `Upper: 6: evalfbb1in->evalfbb2in, Arg_0: Arg_1 {O(n)} 6: evalfbb1in->evalfbb2in, Arg_1: Arg_0 {O(n)} 6: evalfbb1in->evalfbb2in, Arg_2: -1+Arg_1 {O(n)} 4: evalfbb2in->evalfbb1in, Arg_0: Arg_1 {O(n)} 4: evalfbb2in->evalfbb1in, Arg_1: Arg_0 {O(n)} 4: evalfbb2in->evalfbb1in, Arg_2: Arg_1 {O(n)} 5: evalfbb2in->evalfbb3in, Arg_0: Arg_1 {O(n)} 5: evalfbb2in->evalfbb3in, Arg_1: Arg_0 {O(n)} 5: evalfbb2in->evalfbb3in, Arg_2: -1+Arg_1 {O(n)} 7: evalfbb3in->evalfbb4in, Arg_0: Arg_1 {O(n)} 7: evalfbb3in->evalfbb4in, Arg_1: Arg_0 {O(n)} 7: evalfbb3in->evalfbb4in, Arg_2: -1+Arg_1 {O(n)} 2: evalfbb4in->evalfbb2in, Arg_0: Arg_1 {O(n)} 2: evalfbb4in->evalfbb2in, Arg_1: Arg_0 {O(n)} 2: evalfbb4in->evalfbb2in, Arg_2: Arg_1 {O(n)} 3: evalfbb4in->evalfreturnin, Arg_0: Arg_1 {O(n)} 3: evalfbb4in->evalfreturnin, Arg_1: 0 {O(1)} 3: evalfbb4in->evalfreturnin, Arg_2: max([Arg_2, -1+Arg_1]) {O(n)} 1: evalfentryin->evalfbb4in, Arg_0: Arg_1 {O(n)} 1: evalfentryin->evalfbb4in, Arg_1: Arg_0 {O(n)} 1: evalfentryin->evalfbb4in, Arg_2: Arg_2 {O(n)} 8: evalfreturnin->evalfstop, Arg_0: Arg_1 {O(n)} 8: evalfreturnin->evalfstop, Arg_1: 0 {O(1)} 8: evalfreturnin->evalfstop, Arg_2: max([Arg_2, -1+Arg_1]) {O(n)} 0: evalfstart->evalfentryin, Arg_0: Arg_0 {O(n)} 0: evalfstart->evalfentryin, Arg_1: Arg_1 {O(n)} 0: evalfstart->evalfentryin, Arg_2: Arg_2 {O(n)} ---------------------------------------- (2) BOUNDS(1, max(4, 4 + 2 * Arg_0) + nat(2 + 2 * Arg_0) + nat(Arg_0 + Arg_0^2)) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalfstart 0: evalfstart -> evalfentryin : [], cost: 1 1: evalfentryin -> evalfbb4in : A'=B, B'=A, [], cost: 1 2: evalfbb4in -> evalfbb2in : C'=A, [ B>=1 ], cost: 1 3: evalfbb4in -> evalfreturnin : [ 0>=B ], cost: 1 4: evalfbb2in -> evalfbb1in : [ C>=A ], cost: 1 5: evalfbb2in -> evalfbb3in : [ A>=1+C ], cost: 1 6: evalfbb1in -> evalfbb2in : C'=-1+C, [], cost: 1 7: evalfbb3in -> evalfbb4in : B'=-1+B, [], cost: 1 8: evalfreturnin -> evalfstop : [], cost: 1 Removed unreachable and leaf rules: Start location: evalfstart 0: evalfstart -> evalfentryin : [], cost: 1 1: evalfentryin -> evalfbb4in : A'=B, B'=A, [], cost: 1 2: evalfbb4in -> evalfbb2in : C'=A, [ B>=1 ], cost: 1 4: evalfbb2in -> evalfbb1in : [ C>=A ], cost: 1 5: evalfbb2in -> evalfbb3in : [ A>=1+C ], cost: 1 6: evalfbb1in -> evalfbb2in : C'=-1+C, [], cost: 1 7: evalfbb3in -> evalfbb4in : B'=-1+B, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalfstart 9: evalfstart -> evalfbb4in : A'=B, B'=A, [], cost: 2 2: evalfbb4in -> evalfbb2in : C'=A, [ B>=1 ], cost: 1 10: evalfbb2in -> evalfbb2in : C'=-1+C, [ C>=A ], cost: 2 11: evalfbb2in -> evalfbb4in : B'=-1+B, [ A>=1+C ], cost: 2 Accelerating simple loops of location 3. Accelerating the following rules: 10: evalfbb2in -> evalfbb2in : C'=-1+C, [ C>=A ], cost: 2 Accelerated rule 10 with metering function 1+C-A, yielding the new rule 12. Removing the simple loops: 10. Accelerated all simple loops using metering functions (where possible): Start location: evalfstart 9: evalfstart -> evalfbb4in : A'=B, B'=A, [], cost: 2 2: evalfbb4in -> evalfbb2in : C'=A, [ B>=1 ], cost: 1 11: evalfbb2in -> evalfbb4in : B'=-1+B, [ A>=1+C ], cost: 2 12: evalfbb2in -> evalfbb2in : C'=-1+A, [ C>=A ], cost: 2+2*C-2*A Chained accelerated rules (with incoming rules): Start location: evalfstart 9: evalfstart -> evalfbb4in : A'=B, B'=A, [], cost: 2 2: evalfbb4in -> evalfbb2in : C'=A, [ B>=1 ], cost: 1 13: evalfbb4in -> evalfbb2in : C'=-1+A, [ B>=1 ], cost: 3 11: evalfbb2in -> evalfbb4in : B'=-1+B, [ A>=1+C ], cost: 2 Eliminated locations (on tree-shaped paths): Start location: evalfstart 9: evalfstart -> evalfbb4in : A'=B, B'=A, [], cost: 2 14: evalfbb4in -> evalfbb4in : B'=-1+B, C'=-1+A, [ B>=1 ], cost: 5 Accelerating simple loops of location 2. Accelerating the following rules: 14: evalfbb4in -> evalfbb4in : B'=-1+B, C'=-1+A, [ B>=1 ], cost: 5 Accelerated rule 14 with metering function B, yielding the new rule 15. Removing the simple loops: 14. Accelerated all simple loops using metering functions (where possible): Start location: evalfstart 9: evalfstart -> evalfbb4in : A'=B, B'=A, [], cost: 2 15: evalfbb4in -> evalfbb4in : B'=0, C'=-1+A, [ B>=1 ], cost: 5*B Chained accelerated rules (with incoming rules): Start location: evalfstart 9: evalfstart -> evalfbb4in : A'=B, B'=A, [], cost: 2 16: evalfstart -> evalfbb4in : A'=B, B'=0, C'=-1+B, [ A>=1 ], cost: 2+5*A Removed unreachable locations (and leaf rules with constant cost): Start location: evalfstart 16: evalfstart -> evalfbb4in : A'=B, B'=0, C'=-1+B, [ A>=1 ], cost: 2+5*A ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalfstart 16: evalfstart -> evalfbb4in : A'=B, B'=0, C'=-1+B, [ A>=1 ], cost: 2+5*A Computing asymptotic complexity for rule 16 Solved the limit problem by the following transformations: Created initial limit problem: 2+5*A (+), 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 2+5*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: 2+5*n Rule cost: 2+5*A Rule guard: [ A>=1 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)