/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 22 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 611 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: eval1(A, B) -> Com_1(eval2(A, B)) :|: A >= 1 && B >= 1 && A >= B + 1 eval1(A, B) -> Com_1(eval3(A, B)) :|: A >= 1 && B >= 1 && B >= A eval2(A, B) -> Com_1(eval2(A - 1, B)) :|: A >= 1 eval2(A, B) -> Com_1(eval1(A, B)) :|: 0 >= A eval3(A, B) -> Com_1(eval3(A, B - 1)) :|: B >= 1 eval3(A, B) -> Com_1(eval1(A, B)) :|: 0 >= B start(A, B) -> Com_1(eval1(A, B)) :|: TRUE The start-symbols are:[start_2] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, ar_0 + ar_1 + 7) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_0 >= ar_1 + 1 ] (Comp: ?, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_1 >= ar_0 ] (Comp: ?, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ] (Comp: ?, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ] (Comp: ?, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ] (Comp: ?, Cost: 1) start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(start(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: 1, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_0 >= ar_1 + 1 ] (Comp: 1, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_1 >= ar_0 ] (Comp: ?, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ] (Comp: ?, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ] (Comp: ?, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ] (Comp: 1, Cost: 1) start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(eval3) = 1 Pol(eval1) = 0 Pol(eval2) = 1 and size complexities S("koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ]", 0-0) = ar_0 S("koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ]", 0-1) = ar_1 S("start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1))", 0-0) = ar_0 S("start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1))", 0-1) = ar_1 S("eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ]", 0-0) = ar_0 S("eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ]", 0-1) = ar_1 S("eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ]", 0-0) = ar_0 S("eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ]", 0-1) = ar_1 S("eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ]", 0-0) = ar_0 S("eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ]", 0-1) = ar_1 S("eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ]", 0-0) = ar_0 S("eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ]", 0-1) = ar_1 S("eval1(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1)) [ ar_0 >= 1 /\\ ar_1 >= 1 /\\ ar_1 >= ar_0 ]", 0-0) = ar_0 S("eval1(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1)) [ ar_0 >= 1 /\\ ar_1 >= 1 /\\ ar_1 >= ar_0 ]", 0-1) = ar_1 S("eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1)) [ ar_0 >= 1 /\\ ar_1 >= 1 /\\ ar_0 >= ar_1 + 1 ]", 0-0) = ar_0 S("eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1)) [ ar_0 >= 1 /\\ ar_1 >= 1 /\\ ar_0 >= ar_1 + 1 ]", 0-1) = ar_1 orients the transitions eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ] eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ] eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ] eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ] weakly and the transitions eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ] eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_0 >= ar_1 + 1 ] (Comp: 1, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_1 >= ar_0 ] (Comp: ?, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ] (Comp: 2, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ] (Comp: 2, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ] (Comp: 1, Cost: 1) start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(eval1) = V_1 Pol(eval2) = V_1 Pol(eval3) = V_1 Pol(start) = V_1 Pol(koat_start) = V_1 orients all transitions weakly and the transition eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_0 >= ar_1 + 1 ] (Comp: 1, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_1 >= ar_0 ] (Comp: ar_0, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ] (Comp: 2, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ] (Comp: ?, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ] (Comp: 2, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ] (Comp: 1, Cost: 1) start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(eval1) = V_2 Pol(eval2) = V_2 Pol(eval3) = V_2 Pol(start) = V_2 Pol(koat_start) = V_2 orients all transitions weakly and the transition eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval2(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_0 >= ar_1 + 1 ] (Comp: 1, Cost: 1) eval1(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1)) [ ar_0 >= 1 /\ ar_1 >= 1 /\ ar_1 >= ar_0 ] (Comp: ar_0, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval2(ar_0 - 1, ar_1)) [ ar_0 >= 1 ] (Comp: 2, Cost: 1) eval2(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_0 ] (Comp: ar_1, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval3(ar_0, ar_1 - 1)) [ ar_1 >= 1 ] (Comp: 2, Cost: 1) eval3(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) [ 0 >= ar_1 ] (Comp: 1, Cost: 1) start(ar_0, ar_1) -> Com_1(eval1(ar_0, ar_1)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1) -> Com_1(start(ar_0, ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound ar_0 + ar_1 + 7 Time: 0.081 sec (SMT: 0.073 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: start 0: eval1 -> eval2 : [ A>=1 && B>=1 && A>=1+B ], cost: 1 1: eval1 -> eval3 : [ A>=1 && B>=1 && B>=A ], cost: 1 2: eval2 -> eval2 : A'=-1+A, [ A>=1 ], cost: 1 3: eval2 -> eval1 : [ 0>=A ], cost: 1 4: eval3 -> eval3 : B'=-1+B, [ B>=1 ], cost: 1 5: eval3 -> eval1 : [ 0>=B ], cost: 1 6: start -> eval1 : [], cost: 1 Simplified all rules, resulting in: Start location: start 0: eval1 -> eval2 : [ B>=1 && A>=1+B ], cost: 1 1: eval1 -> eval3 : [ A>=1 && B>=1 && B>=A ], cost: 1 2: eval2 -> eval2 : A'=-1+A, [ A>=1 ], cost: 1 3: eval2 -> eval1 : [ 0>=A ], cost: 1 4: eval3 -> eval3 : B'=-1+B, [ B>=1 ], cost: 1 5: eval3 -> eval1 : [ 0>=B ], cost: 1 6: start -> eval1 : [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 2: eval2 -> eval2 : A'=-1+A, [ A>=1 ], cost: 1 Accelerated rule 2 with metering function A, yielding the new rule 7. Removing the simple loops: 2. Accelerating simple loops of location 2. Accelerating the following rules: 4: eval3 -> eval3 : B'=-1+B, [ B>=1 ], cost: 1 Accelerated rule 4 with metering function B, yielding the new rule 8. Removing the simple loops: 4. Accelerated all simple loops using metering functions (where possible): Start location: start 0: eval1 -> eval2 : [ B>=1 && A>=1+B ], cost: 1 1: eval1 -> eval3 : [ A>=1 && B>=1 && B>=A ], cost: 1 3: eval2 -> eval1 : [ 0>=A ], cost: 1 7: eval2 -> eval2 : A'=0, [ A>=1 ], cost: A 5: eval3 -> eval1 : [ 0>=B ], cost: 1 8: eval3 -> eval3 : B'=0, [ B>=1 ], cost: B 6: start -> eval1 : [], cost: 1 Chained accelerated rules (with incoming rules): Start location: start 0: eval1 -> eval2 : [ B>=1 && A>=1+B ], cost: 1 1: eval1 -> eval3 : [ A>=1 && B>=1 && B>=A ], cost: 1 9: eval1 -> eval2 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 1+A 10: eval1 -> eval3 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 1+B 3: eval2 -> eval1 : [ 0>=A ], cost: 1 5: eval3 -> eval1 : [ 0>=B ], cost: 1 6: start -> eval1 : [], cost: 1 Eliminated locations (on tree-shaped paths): Start location: start 11: eval1 -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 2+A 12: eval1 -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 2+B 6: start -> eval1 : [], cost: 1 Accelerating simple loops of location 0. Accelerating the following rules: 11: eval1 -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 2+A 12: eval1 -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 2+B Accelerated rule 11 with NONTERM (after strengthening guard), yielding the new rule 13. Accelerated rule 12 with NONTERM (after strengthening guard), yielding the new rule 14. Removing the simple loops:. Accelerated all simple loops using metering functions (where possible): Start location: start 11: eval1 -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 2+A 12: eval1 -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 2+B 13: eval1 -> [6] : [ B>=1 && A>=1+B && A>=1 && 0>=1+B ], cost: INF 14: eval1 -> [6] : [ A>=1 && B>=1 && B>=A && 0>=A ], cost: INF 6: start -> eval1 : [], cost: 1 Chained accelerated rules (with incoming rules): Start location: start 6: start -> eval1 : [], cost: 1 15: start -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 3+A 16: start -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 3+B Removed unreachable locations (and leaf rules with constant cost): Start location: start 15: start -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 3+A 16: start -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 3+B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: start 15: start -> eval1 : A'=0, [ B>=1 && A>=1+B && A>=1 ], cost: 3+A 16: start -> eval1 : B'=0, [ A>=1 && B>=1 && B>=A ], cost: 3+B Computing asymptotic complexity for rule 15 Solved the limit problem by the following transformations: Created initial limit problem: 3+A (+), A-B (+/+!), B (+/+!) [not solved] applying transformation rule (C) using substitution {B==1} resulting limit problem: 1 (+/+!), -1+A (+/+!), 3+A (+) [not solved] applying transformation rule (C) using substitution {A==1+B} resulting limit problem: 1 (+/+!), B (+/+!), 4+B (+) [not solved] applying transformation rule (B), deleting 1 (+/+!) resulting limit problem: B (+/+!), 4+B (+) [not solved] applying transformation rule (D), replacing 4+B (+) by B (+) resulting limit problem: B (+) [solved] Solution: A / 1+n B / 1 Resulting cost 4+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: 4+n Rule cost: 3+A Rule guard: [ B>=1 && A>=1+B ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)