/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 10 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 219 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f300(A, B, C) -> Com_1(f300(-(1) + A, -(2) + A, C)) :|: A >= 1 && B + A >= 1 && B >= 1 f300(A, B, C) -> Com_1(f1(A, B, D)) :|: A >= 1 && 0 >= B + A && B >= 1 f300(A, B, C) -> Com_1(f1(A, B, D)) :|: B >= 1 && 0 >= A f300(A, B, C) -> Com_1(f1(A, B, D)) :|: 0 >= B f2(A, B, C) -> Com_1(f300(A, B, C)) :|: TRUE The start-symbols are:[f2_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, ar_0 + 3) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f300(ar_0 - 1, ar_0 - 2, ar_2)) [ ar_0 >= 1 /\ ar_1 + ar_0 >= 1 /\ ar_1 >= 1 ] (Comp: ?, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ ar_0 >= 1 /\ 0 >= ar_1 + ar_0 /\ ar_1 >= 1 ] (Comp: ?, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ ar_1 >= 1 /\ 0 >= ar_0 ] (Comp: ?, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ 0 >= ar_1 ] (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f300(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 Testing for reachability in the complexity graph removes the following transition from problem 1: f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ ar_0 >= 1 /\ 0 >= ar_1 + ar_0 /\ ar_1 >= 1 ] We thus obtain the following problem: 2: T: (Comp: ?, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ 0 >= ar_1 ] (Comp: ?, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ ar_1 >= 1 /\ 0 >= ar_0 ] (Comp: ?, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f300(ar_0 - 1, ar_0 - 2, ar_2)) [ ar_0 >= 1 /\ ar_1 + ar_0 >= 1 /\ ar_1 >= 1 ] (Comp: ?, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f300(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) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ 0 >= ar_1 ] (Comp: 1, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ ar_1 >= 1 /\ 0 >= ar_0 ] (Comp: ?, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f300(ar_0 - 1, ar_0 - 2, ar_2)) [ ar_0 >= 1 /\ ar_1 + ar_0 >= 1 /\ ar_1 >= 1 ] (Comp: 1, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f300(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(f300) = 1 Pol(f1) = 0 Pol(f2) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ 0 >= ar_1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ 0 >= ar_1 ] (Comp: 1, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ ar_1 >= 1 /\ 0 >= ar_0 ] (Comp: ?, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f300(ar_0 - 1, ar_0 - 2, ar_2)) [ ar_0 >= 1 /\ ar_1 + ar_0 >= 1 /\ ar_1 >= 1 ] (Comp: 1, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f300(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(f300) = V_1 Pol(f1) = V_1 Pol(f2) = V_1 Pol(koat_start) = V_1 orients all transitions weakly and the transition f300(ar_0, ar_1, ar_2) -> Com_1(f300(ar_0 - 1, ar_0 - 2, ar_2)) [ ar_0 >= 1 /\ ar_1 + ar_0 >= 1 /\ ar_1 >= 1 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ 0 >= ar_1 ] (Comp: 1, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f1(ar_0, ar_1, d)) [ ar_1 >= 1 /\ 0 >= ar_0 ] (Comp: ar_0, Cost: 1) f300(ar_0, ar_1, ar_2) -> Com_1(f300(ar_0 - 1, ar_0 - 2, ar_2)) [ ar_0 >= 1 /\ ar_1 + ar_0 >= 1 /\ ar_1 >= 1 ] (Comp: 1, Cost: 1) f2(ar_0, ar_1, ar_2) -> Com_1(f300(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 ar_0 + 3 Time: 0.084 sec (SMT: 0.079 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: f2 0: f300 -> f300 : A'=-1+A, B'=-2+A, [ A>=1 && A+B>=1 && B>=1 ], cost: 1 1: f300 -> f1 : C'=free, [ A>=1 && 0>=A+B && B>=1 ], cost: 1 2: f300 -> f1 : C'=free_1, [ B>=1 && 0>=A ], cost: 1 3: f300 -> f1 : C'=free_2, [ 0>=B ], cost: 1 4: f2 -> f300 : [], cost: 1 Removed unreachable and leaf rules: Start location: f2 0: f300 -> f300 : A'=-1+A, B'=-2+A, [ A>=1 && A+B>=1 && B>=1 ], cost: 1 4: f2 -> f300 : [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 0. Accelerating the following rules: 0: f300 -> f300 : A'=-1+A, B'=-2+A, [ A>=1 && A+B>=1 && B>=1 ], cost: 1 Accelerated rule 0 with backward acceleration, yielding the new rule 5. Accelerated rule 0 with backward acceleration, yielding the new rule 6. Accelerated rule 0 with backward acceleration, yielding the new rule 7. Removing the simple loops: 0. Also removing duplicate rules:. Accelerated all simple loops using metering functions (where possible): Start location: f2 6: f300 -> f300 : A'=0, B'=-1, [ 0>=1 ], cost: A 7: f300 -> f300 : A'=1, B'=0, [ A+B>=1 && B>=1 && -1+A>0 ], cost: -1+A 4: f2 -> f300 : [], cost: 1 Chained accelerated rules (with incoming rules): Start location: f2 4: f2 -> f300 : [], cost: 1 8: f2 -> f300 : A'=1, B'=0, [ A+B>=1 && B>=1 && -1+A>0 ], cost: A Removed unreachable locations (and leaf rules with constant cost): Start location: f2 8: f2 -> f300 : A'=1, B'=0, [ A+B>=1 && B>=1 && -1+A>0 ], cost: A ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: f2 8: f2 -> f300 : A'=1, B'=0, [ A+B>=1 && B>=1 && -1+A>0 ], cost: A Computing asymptotic complexity for rule 8 Simplified the guard: 8: f2 -> f300 : A'=1, B'=0, [ B>=1 && -1+A>0 ], cost: A Solved the limit problem by the following transformations: Created initial limit problem: -1+A (+/+!), A (+), B (+/+!) [not solved] applying transformation rule (C) using substitution {B==1} resulting limit problem: 1 (+/+!), -1+A (+/+!), A (+) [not solved] applying transformation rule (B), deleting 1 (+/+!) resulting 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 B / 1 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: [ B>=1 && -1+A>0 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)