/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 37 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 290 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f2(A, B, C) -> Com_1(f1(A, B, C)) :|: TRUE f1(A, B, C) -> Com_1(f1(1 + A, B, C)) :|: B >= 1 + A f1(A, B, C) -> Com_1(f1(1 + A, A, C)) :|: B >= D && A >= B && A <= B f1(A, B, C) -> Com_1(f300(A, B, D)) :|: A >= B && A >= B + 1 f1(A, B, C) -> Com_1(f300(A, B, D)) :|: A >= B && B >= A + 1 The start-symbols are:[f2_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 2*Ar_0 + 2*Ar_1 + 4) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_1, Ar_1 + 1, Ar_2)) [ Ar_0 >= D /\ Ar_1 = Ar_0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_1)) [ Ar_1 >= Ar_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= Ar_0 /\ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transition from problem 1: f1(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_0)) [ Ar_1 >= Ar_0 /\ Ar_0 >= Ar_1 + 1 ] We thus obtain the following problem: 2: T: (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_1)) [ Ar_1 >= Ar_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_1, Ar_1 + 1, Ar_2)) [ Ar_0 >= D /\ Ar_1 = Ar_0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 2 produces the following problem: 3: T: (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_1)) [ Ar_1 >= Ar_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_1, Ar_1 + 1, Ar_2)) [ Ar_0 >= D /\ Ar_1 = Ar_0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f1) = 1 Pol(f300) = 0 Pol(f2) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f1(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_1)) [ Ar_1 >= Ar_0 /\ Ar_1 >= Ar_0 + 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_1)) [ Ar_1 >= Ar_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_1, Ar_1 + 1, Ar_2)) [ Ar_0 >= D /\ Ar_1 = Ar_0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f1) = V_1 - V_2 + 1 Pol(f300) = V_1 - V_2 Pol(f2) = V_1 - V_2 + 1 Pol(koat_start) = V_1 - V_2 + 1 orients all transitions weakly and the transitions f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_1, Ar_1 + 1, Ar_2)) [ Ar_0 >= D /\ Ar_1 = Ar_0 ] f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_1)) [ Ar_1 >= Ar_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: Ar_0 + Ar_1 + 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_1, Ar_1 + 1, Ar_2)) [ Ar_0 >= D /\ Ar_1 = Ar_0 ] (Comp: Ar_0 + Ar_1 + 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 2*Ar_0 + 2*Ar_1 + 4 Time: 0.052 sec (SMT: 0.047 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f2 0: f2 -> f1 : [], cost: 1 1: f1 -> f1 : B'=1+B, [ A>=1+B ], cost: 1 2: f1 -> f1 : A'=B, B'=1+B, [ A>=free && B==A ], cost: 1 3: f1 -> f300 : C'=free_1, [ B>=A && B>=1+A ], cost: 1 4: f1 -> f300 : C'=free_2, [ B>=A && A>=1+B ], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: f2 -> f1 : [], cost: 1 Removed unreachable and leaf rules: Start location: f2 0: f2 -> f1 : [], cost: 1 1: f1 -> f1 : B'=1+B, [ A>=1+B ], cost: 1 2: f1 -> f1 : A'=B, B'=1+B, [ A>=free && B==A ], cost: 1 Simplified all rules, resulting in: Start location: f2 0: f2 -> f1 : [], cost: 1 1: f1 -> f1 : B'=1+B, [ A>=1+B ], cost: 1 2: f1 -> f1 : A'=B, B'=1+B, [ B==A ], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 1: f1 -> f1 : B'=1+B, [ A>=1+B ], cost: 1 2: f1 -> f1 : A'=B, B'=1+B, [ B==A ], cost: 1 Accelerated rule 1 with metering function A-B, yielding the new rule 5. Accelerated rule 2 with metering function 1+A-B, yielding the new rule 6. Removing the simple loops: 1 2. Accelerated all simple loops using metering functions (where possible): Start location: f2 0: f2 -> f1 : [], cost: 1 5: f1 -> f1 : B'=A, [ A>=1+B ], cost: A-B 6: f1 -> f1 : A'=A, B'=1+A, [ B==A ], cost: 1+A-B Chained accelerated rules (with incoming rules): Start location: f2 0: f2 -> f1 : [], cost: 1 7: f2 -> f1 : B'=A, [ A>=1+B ], cost: 1+A-B 8: f2 -> f1 : B'=1+A, [ B==A ], cost: 2+A-B Removed unreachable locations (and leaf rules with constant cost): Start location: f2 7: f2 -> f1 : B'=A, [ A>=1+B ], cost: 1+A-B 8: f2 -> f1 : B'=1+A, [ B==A ], cost: 2+A-B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: f2 7: f2 -> f1 : B'=A, [ A>=1+B ], cost: 1+A-B 8: f2 -> f1 : B'=1+A, [ B==A ], cost: 2+A-B Computing asymptotic complexity for rule 7 Solved the limit problem by the following transformations: Created initial limit problem: A-B (+/+!), 1+A-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+A-B Rule guard: [ A>=1+B ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)