/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 13 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 136 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f3(A, B, C, D) -> Com_1(f1(0, B, C, D)) :|: TRUE f1(A, B, C, D) -> Com_1(f2(A, B, C, E)) :|: B >= C f1(A, B, C, D) -> Com_1(f2(1, 1 + B, C, E)) :|: B + 1 >= C && B + 1 <= C && A >= 0 && A <= 0 f1(A, B, C, D) -> Com_1(f1(0, 1 + B, -(1) + C, D)) :|: C >= 2 + B && C >= 1 + B && A >= 0 && A <= 0 The start-symbols are:[f3_4] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, Ar_1 + Ar_2 + 3) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_2 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(1, Ar_1 + 1, Ar_2, Fresh_0)) [ Ar_1 + 1 = Ar_2 /\ Ar_0 = 0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1 + 1, Ar_2 - 1, Ar_3)) [ Ar_2 >= Ar_1 + 2 /\ Ar_2 >= Ar_1 + 1 /\ Ar_0 = 0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 1 produces the following problem: 2: T: (Comp: 1, Cost: 1) f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_2 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(1, Ar_1 + 1, Ar_2, Fresh_0)) [ Ar_1 + 1 = Ar_2 /\ Ar_0 = 0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1 + 1, Ar_2 - 1, Ar_3)) [ Ar_2 >= Ar_1 + 2 /\ Ar_2 >= Ar_1 + 1 /\ Ar_0 = 0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f3) = 1 Pol(f1) = 1 Pol(f2) = 0 Pol(koat_start) = 1 orients all transitions weakly and the transitions f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(1, Ar_1 + 1, Ar_2, Fresh_0)) [ Ar_1 + 1 = Ar_2 /\ Ar_0 = 0 ] f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_2 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_2 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(1, Ar_1 + 1, Ar_2, Fresh_0)) [ Ar_1 + 1 = Ar_2 /\ Ar_0 = 0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1 + 1, Ar_2 - 1, Ar_3)) [ Ar_2 >= Ar_1 + 2 /\ Ar_2 >= Ar_1 + 1 /\ Ar_0 = 0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f3) = -V_2 + V_3 Pol(f1) = -V_2 + V_3 Pol(f2) = -V_2 + V_3 Pol(koat_start) = -V_2 + V_3 orients all transitions weakly and the transition f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1 + 1, Ar_2 - 1, Ar_3)) [ Ar_2 >= Ar_1 + 2 /\ Ar_2 >= Ar_1 + 1 /\ Ar_0 = 0 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) f3(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Fresh_1)) [ Ar_1 >= Ar_2 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f2(1, Ar_1 + 1, Ar_2, Fresh_0)) [ Ar_1 + 1 = Ar_2 /\ Ar_0 = 0 ] (Comp: Ar_1 + Ar_2, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f1(0, Ar_1 + 1, Ar_2 - 1, Ar_3)) [ Ar_2 >= Ar_1 + 2 /\ Ar_2 >= Ar_1 + 1 /\ Ar_0 = 0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f3(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound Ar_1 + Ar_2 + 3 Time: 0.098 sec (SMT: 0.093 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f3 0: f3 -> f1 : A'=0, [], cost: 1 1: f1 -> f2 : D'=free, [ B>=C ], cost: 1 2: f1 -> f2 : A'=1, B'=1+B, D'=free_1, [ 1+B==C && A==0 ], cost: 1 3: f1 -> f1 : A'=0, B'=1+B, C'=-1+C, [ C>=2+B && C>=1+B && A==0 ], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: f3 -> f1 : A'=0, [], cost: 1 Removed unreachable and leaf rules: Start location: f3 0: f3 -> f1 : A'=0, [], cost: 1 3: f1 -> f1 : A'=0, B'=1+B, C'=-1+C, [ C>=2+B && C>=1+B && A==0 ], cost: 1 Simplified all rules, resulting in: Start location: f3 0: f3 -> f1 : A'=0, [], cost: 1 3: f1 -> f1 : A'=0, B'=1+B, C'=-1+C, [ C>=2+B && A==0 ], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 3: f1 -> f1 : A'=0, B'=1+B, C'=-1+C, [ C>=2+B && A==0 ], cost: 1 Accelerated rule 3 with metering function meter (where 2*meter==-1+C-B), yielding the new rule 4. Removing the simple loops: 3. Accelerated all simple loops using metering functions (where possible): Start location: f3 0: f3 -> f1 : A'=0, [], cost: 1 4: f1 -> f1 : A'=0, B'=meter+B, C'=C-meter, [ C>=2+B && A==0 && 2*meter==-1+C-B && meter>=1 ], cost: meter Chained accelerated rules (with incoming rules): Start location: f3 0: f3 -> f1 : A'=0, [], cost: 1 5: f3 -> f1 : A'=0, B'=meter+B, C'=C-meter, [ C>=2+B && 2*meter==-1+C-B && meter>=1 ], cost: 1+meter Removed unreachable locations (and leaf rules with constant cost): Start location: f3 5: f3 -> f1 : A'=0, B'=meter+B, C'=C-meter, [ C>=2+B && 2*meter==-1+C-B && meter>=1 ], cost: 1+meter ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: f3 5: f3 -> f1 : A'=0, B'=meter+B, C'=C-meter, [ C>=2+B && 2*meter==-1+C-B && meter>=1 ], cost: 1+meter Computing asymptotic complexity for rule 5 Solved the limit problem by the following transformations: Created initial limit problem: 1+meter (+), C-2*meter-B (+/+!), 2-C+2*meter+B (+/+!), -1+C-B (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {C==-1,meter==-1+n,B==-2*n} resulting limit problem: [solved] Solution: C / -1 meter / -1+n B / -2*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: 1+meter Rule guard: [ C>=2+B && 2*meter==-1+C-B ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)