/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 31 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 109 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f5(A, B) -> Com_1(f5(-(1) + A, B)) :|: A >= 1 f5(A, B) -> Com_1(f1(A, C)) :|: 0 >= A f300(A, B) -> Com_1(f5(-(1) + A, B)) :|: A >= 1 f300(A, B) -> Com_1(f1(A, C)) :|: 0 >= A The start-symbols are:[f300_2] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, Ar_0 + 3) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f5(Ar_0, Ar_1) -> Com_1(f5(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Fresh_1)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f5(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Fresh_0)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 1 produces the following problem: 2: T: (Comp: ?, Cost: 1) f5(Ar_0, Ar_1) -> Com_1(f5(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Fresh_1)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f5(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Fresh_0)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f5) = 1 Pol(f1) = 0 Pol(f300) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f5(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Fresh_1)) [ 0 >= Ar_0 ] strictly and produces the following problem: 3: T: (Comp: ?, Cost: 1) f5(Ar_0, Ar_1) -> Com_1(f5(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f5(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Fresh_1)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f5(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Fresh_0)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f5) = V_1 Pol(f1) = V_1 Pol(f300) = V_1 Pol(koat_start) = V_1 orients all transitions weakly and the transition f5(Ar_0, Ar_1) -> Com_1(f5(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ] strictly and produces the following problem: 4: T: (Comp: Ar_0, Cost: 1) f5(Ar_0, Ar_1) -> Com_1(f5(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f5(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Fresh_1)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f5(Ar_0 - 1, Ar_1)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Fresh_0)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound Ar_0 + 3 Time: 0.043 sec (SMT: 0.039 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f300 0: f5 -> f5 : A'=-1+A, [ A>=1 ], cost: 1 1: f5 -> f1 : B'=free, [ 0>=A ], cost: 1 2: f300 -> f5 : A'=-1+A, [ A>=1 ], cost: 1 3: f300 -> f1 : B'=free_1, [ 0>=A ], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 2: f300 -> f5 : A'=-1+A, [ A>=1 ], cost: 1 Removed unreachable and leaf rules: Start location: f300 0: f5 -> f5 : A'=-1+A, [ A>=1 ], cost: 1 2: f300 -> f5 : A'=-1+A, [ A>=1 ], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 0. Accelerating the following rules: 0: f5 -> f5 : A'=-1+A, [ A>=1 ], cost: 1 Accelerated rule 0 with metering function A, yielding the new rule 4. Removing the simple loops: 0. Accelerated all simple loops using metering functions (where possible): Start location: f300 4: f5 -> f5 : A'=0, [ A>=1 ], cost: A 2: f300 -> f5 : A'=-1+A, [ A>=1 ], cost: 1 Chained accelerated rules (with incoming rules): Start location: f300 2: f300 -> f5 : A'=-1+A, [ A>=1 ], cost: 1 5: f300 -> f5 : A'=0, [ -1+A>=1 ], cost: A Removed unreachable locations (and leaf rules with constant cost): Start location: f300 5: f300 -> f5 : A'=0, [ -1+A>=1 ], cost: A ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: f300 5: f300 -> f5 : A'=0, [ -1+A>=1 ], cost: A Computing asymptotic complexity for rule 5 Solved the limit problem by the following transformations: Created initial limit problem: -1+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 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: n Rule cost: A Rule guard: [ -1+A>=1 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)