/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^1)) proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 23 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 117 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f4(A, B, C) -> Com_1(f5(A, B, 1)) :|: 0 >= A && 0 >= B f0(A, B, C) -> Com_1(f2(A, B, 1)) :|: A >= 1 f4(A, B, C) -> Com_1(f4(A, B - 1, C)) :|: B >= 1 f0(A, B, C) -> Com_1(f4(A, B, 0)) :|: 0 >= A The start-symbols are:[f0_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, Ar_1 + 3) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f4(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1, 1)) [ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 1 produces the following problem: 2: T: (Comp: ?, Cost: 1) f4(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1, 1)) [ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f4) = 1 Pol(f5) = 0 Pol(f0) = 1 Pol(f2) = 0 Pol(koat_start) = 1 orients all transitions weakly and the transition f4(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1, 1)) [ 0 >= Ar_0 /\ 0 >= Ar_1 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) f4(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1, 1)) [ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f4) = V_2 Pol(f5) = V_2 Pol(f0) = V_2 Pol(f2) = V_2 Pol(koat_start) = V_2 orients all transitions weakly and the transition f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_1 >= 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) f4(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1, 1)) [ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, 1)) [ Ar_0 >= 1 ] (Comp: Ar_1, Cost: 1) f4(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1 - 1, Ar_2)) [ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f4(Ar_0, Ar_1, 0)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound Ar_1 + 3 Time: 0.041 sec (SMT: 0.032 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f0 0: f4 -> f5 : C'=1, [ 0>=A && 0>=B ], cost: 1 2: f4 -> f4 : B'=-1+B, [ B>=1 ], cost: 1 1: f0 -> f2 : C'=1, [ A>=1 ], cost: 1 3: f0 -> f4 : C'=0, [ 0>=A ], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 1: f0 -> f2 : C'=1, [ A>=1 ], cost: 1 Removed unreachable and leaf rules: Start location: f0 2: f4 -> f4 : B'=-1+B, [ B>=1 ], cost: 1 3: f0 -> f4 : C'=0, [ 0>=A ], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 0. Accelerating the following rules: 2: f4 -> f4 : B'=-1+B, [ B>=1 ], cost: 1 Accelerated rule 2 with metering function B, yielding the new rule 4. Removing the simple loops: 2. Accelerated all simple loops using metering functions (where possible): Start location: f0 4: f4 -> f4 : B'=0, [ B>=1 ], cost: B 3: f0 -> f4 : C'=0, [ 0>=A ], cost: 1 Chained accelerated rules (with incoming rules): Start location: f0 3: f0 -> f4 : C'=0, [ 0>=A ], cost: 1 5: f0 -> f4 : B'=0, C'=0, [ 0>=A && B>=1 ], cost: 1+B Removed unreachable locations (and leaf rules with constant cost): Start location: f0 5: f0 -> f4 : B'=0, C'=0, [ 0>=A && B>=1 ], cost: 1+B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: f0 5: f0 -> f4 : B'=0, C'=0, [ 0>=A && B>=1 ], cost: 1+B Computing asymptotic complexity for rule 5 Solved the limit problem by the following transformations: Created initial limit problem: 1-A (+/+!), 1+B (+), B (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {A==0,B==n} resulting limit problem: [solved] Solution: A / 0 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: [ 0>=A && B>=1 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)