/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 11 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 208 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f2(A, B, C, D, E, F) -> Com_1(f2(-(1) + A, -(1) + B, A, B, -(2) + A, F)) :|: A >= 1 && B >= 1 f3(A, B, C, D, E, F) -> Com_1(f2(A, B, C, D, E, F)) :|: TRUE f2(A, B, C, D, E, F) -> Com_1(f4(A, G, C, D, E, H)) :|: 0 >= B && 0 >= G f2(A, B, C, D, E, F) -> Com_1(f4(A, B, C, D, E, H)) :|: B >= 1 && 0 >= A The start-symbols are:[f3_6] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, Ar_0 + 3) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0 - 1, Ar_1 - 1, Ar_0, Ar_1, Ar_0 - 2, Ar_5)) [ Ar_0 >= 1 /\ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Fresh_1, Ar_2, Ar_3, Ar_4, Fresh_2)) [ 0 >= Ar_1 /\ 0 >= Fresh_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Fresh_0)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 1 produces the following problem: 2: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0 - 1, Ar_1 - 1, Ar_0, Ar_1, Ar_0 - 2, Ar_5)) [ Ar_0 >= 1 /\ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Fresh_1, Ar_2, Ar_3, Ar_4, Fresh_2)) [ 0 >= Ar_1 /\ 0 >= Fresh_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Fresh_0)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f2) = 1 Pol(f3) = 1 Pol(f4) = 0 Pol(koat_start) = 1 orients all transitions weakly and the transitions f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Fresh_1, Ar_2, Ar_3, Ar_4, Fresh_2)) [ 0 >= Ar_1 /\ 0 >= Fresh_1 ] f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Fresh_0)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ] strictly and produces the following problem: 3: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0 - 1, Ar_1 - 1, Ar_0, Ar_1, Ar_0 - 2, Ar_5)) [ Ar_0 >= 1 /\ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: 1, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Fresh_1, Ar_2, Ar_3, Ar_4, Fresh_2)) [ 0 >= Ar_1 /\ 0 >= Fresh_1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Fresh_0)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f2) = V_1 Pol(f3) = V_1 Pol(f4) = V_1 Pol(koat_start) = V_1 orients all transitions weakly and the transition f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0 - 1, Ar_1 - 1, Ar_0, Ar_1, Ar_0 - 2, Ar_5)) [ Ar_0 >= 1 /\ Ar_1 >= 1 ] strictly and produces the following problem: 4: T: (Comp: Ar_0, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0 - 1, Ar_1 - 1, Ar_0, Ar_1, Ar_0 - 2, Ar_5)) [ Ar_0 >= 1 /\ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: 1, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Fresh_1, Ar_2, Ar_3, Ar_4, Fresh_2)) [ 0 >= Ar_1 /\ 0 >= Fresh_1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f4(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Fresh_0)) [ Ar_1 >= 1 /\ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound Ar_0 + 3 Time: 0.089 sec (SMT: 0.073 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f3 0: f2 -> f2 : A'=-1+A, B'=-1+B, C'=A, D'=B, E'=-2+A, [ A>=1 && B>=1 ], cost: 1 2: f2 -> f4 : B'=free, F'=free_1, [ 0>=B && 0>=free ], cost: 1 3: f2 -> f4 : F'=free_2, [ B>=1 && 0>=A ], cost: 1 1: f3 -> f2 : [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 1: f3 -> f2 : [], cost: 1 Removed unreachable and leaf rules: Start location: f3 0: f2 -> f2 : A'=-1+A, B'=-1+B, C'=A, D'=B, E'=-2+A, [ A>=1 && B>=1 ], cost: 1 1: f3 -> f2 : [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 0. Accelerating the following rules: 0: f2 -> f2 : A'=-1+A, B'=-1+B, C'=A, D'=B, E'=-2+A, [ A>=1 && B>=1 ], cost: 1 Accelerated rule 0 with metering function B (after adding A>=B), yielding the new rule 4. Accelerated rule 0 with metering function A (after adding A<=B), yielding the new rule 5. Removing the simple loops: 0. Accelerated all simple loops using metering functions (where possible): Start location: f3 4: f2 -> f2 : A'=A-B, B'=0, C'=1+A-B, D'=1, E'=-1+A-B, [ A>=1 && B>=1 && A>=B ], cost: B 5: f2 -> f2 : A'=0, B'=-A+B, C'=1, D'=1-A+B, E'=-1, [ A>=1 && B>=1 && A<=B ], cost: A 1: f3 -> f2 : [], cost: 1 Chained accelerated rules (with incoming rules): Start location: f3 1: f3 -> f2 : [], cost: 1 6: f3 -> f2 : A'=A-B, B'=0, C'=1+A-B, D'=1, E'=-1+A-B, [ A>=1 && B>=1 && A>=B ], cost: 1+B 7: f3 -> f2 : A'=0, B'=-A+B, C'=1, D'=1-A+B, E'=-1, [ A>=1 && B>=1 && A<=B ], cost: 1+A Removed unreachable locations (and leaf rules with constant cost): Start location: f3 6: f3 -> f2 : A'=A-B, B'=0, C'=1+A-B, D'=1, E'=-1+A-B, [ A>=1 && B>=1 && A>=B ], cost: 1+B 7: f3 -> f2 : A'=0, B'=-A+B, C'=1, D'=1-A+B, E'=-1, [ A>=1 && B>=1 && A<=B ], cost: 1+A ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: f3 6: f3 -> f2 : A'=A-B, B'=0, C'=1+A-B, D'=1, E'=-1+A-B, [ A>=1 && B>=1 && A>=B ], cost: 1+B 7: f3 -> f2 : A'=0, B'=-A+B, C'=1, D'=1-A+B, E'=-1, [ A>=1 && B>=1 && A<=B ], cost: 1+A Computing asymptotic complexity for rule 6 Simplified the guard: 6: f3 -> f2 : A'=A-B, B'=0, C'=1+A-B, D'=1, E'=-1+A-B, [ B>=1 && A>=B ], cost: 1+B Solved the limit problem by the following transformations: Created initial limit problem: 1+B (+), 1+A-B (+/+!), B (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {A==n,B==n} resulting limit problem: [solved] Solution: A / n B / 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+B Rule guard: [ B>=1 && A>=B ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)