/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 146 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 719 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: evalEx6start(A, B, C) -> Com_1(evalEx6entryin(A, B, C)) :|: TRUE evalEx6entryin(A, B, C) -> Com_1(evalEx6bb3in(B, A, C)) :|: TRUE evalEx6bb3in(A, B, C) -> Com_1(evalEx6bbin(A, B, C)) :|: C >= B + 1 evalEx6bb3in(A, B, C) -> Com_1(evalEx6returnin(A, B, C)) :|: B >= C evalEx6bbin(A, B, C) -> Com_1(evalEx6bb1in(A, B, C)) :|: A >= B + 1 evalEx6bbin(A, B, C) -> Com_1(evalEx6bb2in(A, B, C)) :|: B >= A evalEx6bb1in(A, B, C) -> Com_1(evalEx6bb3in(A, B + 1, C)) :|: TRUE evalEx6bb2in(A, B, C) -> Com_1(evalEx6bb3in(A + 1, B, C)) :|: TRUE evalEx6returnin(A, B, C) -> Com_1(evalEx6stop(A, B, C)) :|: TRUE The start-symbols are:[evalEx6start_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 6*ar_1 + 12*ar_2 + 6*ar_0 + 7) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2)) (Comp: ?, Cost: 1) evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2)) (Comp: ?, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] (Comp: ?, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ] (Comp: ?, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ ar_0 >= ar_1 + 1 ] (Comp: ?, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ ar_1 >= ar_0 ] (Comp: ?, Cost: 1) evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) (Comp: ?, Cost: 1) evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) (Comp: ?, Cost: 1) evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(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: 1, Cost: 1) evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 1) evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2)) (Comp: ?, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] (Comp: ?, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ] (Comp: ?, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ ar_0 >= ar_1 + 1 ] (Comp: ?, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ ar_1 >= ar_0 ] (Comp: ?, Cost: 1) evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) (Comp: ?, Cost: 1) evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) (Comp: ?, Cost: 1) evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalEx6start) = 2 Pol(evalEx6entryin) = 2 Pol(evalEx6bb3in) = 2 Pol(evalEx6bbin) = 2 Pol(evalEx6returnin) = 1 Pol(evalEx6bb1in) = 2 Pol(evalEx6bb2in) = 2 Pol(evalEx6stop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 1) evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2)) (Comp: ?, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] (Comp: 2, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ] (Comp: ?, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ ar_0 >= ar_1 + 1 ] (Comp: ?, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ ar_1 >= ar_0 ] (Comp: ?, Cost: 1) evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) (Comp: ?, Cost: 1) evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) (Comp: 2, Cost: 1) evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 3 to obtain the following invariants: For symbol evalEx6bb1in: -X_2 + X_3 - 1 >= 0 /\ X_1 - X_2 - 1 >= 0 For symbol evalEx6bb2in: -X_2 + X_3 - 1 >= 0 /\ -X_1 + X_3 - 1 >= 0 /\ -X_1 + X_2 >= 0 For symbol evalEx6bbin: -X_2 + X_3 - 1 >= 0 For symbol evalEx6returnin: X_2 - X_3 >= 0 This yielded the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) [ ar_1 - ar_2 >= 0 ] (Comp: ?, Cost: 1) evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ -ar_0 + ar_2 - 1 >= 0 /\ -ar_0 + ar_1 >= 0 ] (Comp: ?, Cost: 1) evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ] (Comp: ?, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_1 >= ar_0 ] (Comp: ?, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 >= ar_1 + 1 ] (Comp: 2, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ] (Comp: ?, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] (Comp: 1, Cost: 1) evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2)) (Comp: 1, Cost: 1) evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = -2*V_2 + 2*V_3 Pol(evalEx6start) = -2*V_2 + 2*V_3 Pol(evalEx6returnin) = -2*V_1 + 2*V_3 Pol(evalEx6stop) = -2*V_1 + 2*V_3 Pol(evalEx6bb2in) = -2*V_1 + 2*V_3 - 1 Pol(evalEx6bb3in) = -2*V_1 + 2*V_3 Pol(evalEx6bb1in) = -2*V_1 + 2*V_3 Pol(evalEx6bbin) = -2*V_1 + 2*V_3 Pol(evalEx6entryin) = -2*V_2 + 2*V_3 orients all transitions weakly and the transitions evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_1 >= ar_0 ] evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ -ar_0 + ar_2 - 1 >= 0 /\ -ar_0 + ar_1 >= 0 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) [ ar_1 - ar_2 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ -ar_0 + ar_2 - 1 >= 0 /\ -ar_0 + ar_1 >= 0 ] (Comp: ?, Cost: 1) evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_1 >= ar_0 ] (Comp: ?, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 >= ar_1 + 1 ] (Comp: 2, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ] (Comp: ?, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] (Comp: 1, Cost: 1) evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2)) (Comp: 1, Cost: 1) evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = -2*V_1 + 2*V_3 Pol(evalEx6start) = -2*V_1 + 2*V_3 Pol(evalEx6returnin) = -2*V_2 + 2*V_3 Pol(evalEx6stop) = -2*V_2 + 2*V_3 Pol(evalEx6bb2in) = -2*V_2 + 2*V_3 Pol(evalEx6bb3in) = -2*V_2 + 2*V_3 Pol(evalEx6bb1in) = -2*V_2 + 2*V_3 - 1 Pol(evalEx6bbin) = -2*V_2 + 2*V_3 Pol(evalEx6entryin) = -2*V_1 + 2*V_3 orients all transitions weakly and the transitions evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 >= ar_1 + 1 ] evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ] strictly and produces the following problem: 6: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) [ ar_1 - ar_2 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ -ar_0 + ar_2 - 1 >= 0 /\ -ar_0 + ar_1 >= 0 ] (Comp: 2*ar_0 + 2*ar_2, Cost: 1) evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_1 >= ar_0 ] (Comp: 2*ar_0 + 2*ar_2, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 >= ar_1 + 1 ] (Comp: 2, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ] (Comp: ?, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] (Comp: 1, Cost: 1) evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2)) (Comp: 1, Cost: 1) evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2)) start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 6 produces the following problem: 7: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalEx6start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalEx6returnin(ar_0, ar_1, ar_2) -> Com_1(evalEx6stop(ar_0, ar_1, ar_2)) [ ar_1 - ar_2 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalEx6bb2in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0 + 1, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ -ar_0 + ar_2 - 1 >= 0 /\ -ar_0 + ar_1 >= 0 ] (Comp: 2*ar_0 + 2*ar_2, Cost: 1) evalEx6bb1in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_0, ar_1 + 1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb2in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_1 >= ar_0 ] (Comp: 2*ar_0 + 2*ar_2, Cost: 1) evalEx6bbin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb1in(ar_0, ar_1, ar_2)) [ -ar_1 + ar_2 - 1 >= 0 /\ ar_0 >= ar_1 + 1 ] (Comp: 2, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6returnin(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 ] (Comp: 2*ar_0 + 4*ar_2 + 2*ar_1 + 1, Cost: 1) evalEx6bb3in(ar_0, ar_1, ar_2) -> Com_1(evalEx6bbin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 + 1 ] (Comp: 1, Cost: 1) evalEx6entryin(ar_0, ar_1, ar_2) -> Com_1(evalEx6bb3in(ar_1, ar_0, ar_2)) (Comp: 1, Cost: 1) evalEx6start(ar_0, ar_1, ar_2) -> Com_1(evalEx6entryin(ar_0, ar_1, ar_2)) start location: koat_start leaf cost: 0 Complexity upper bound 6*ar_1 + 12*ar_2 + 6*ar_0 + 7 Time: 0.180 sec (SMT: 0.159 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalEx6start 0: evalEx6start -> evalEx6entryin : [], cost: 1 1: evalEx6entryin -> evalEx6bb3in : A'=B, B'=A, [], cost: 1 2: evalEx6bb3in -> evalEx6bbin : [ C>=1+B ], cost: 1 3: evalEx6bb3in -> evalEx6returnin : [ B>=C ], cost: 1 4: evalEx6bbin -> evalEx6bb1in : [ A>=1+B ], cost: 1 5: evalEx6bbin -> evalEx6bb2in : [ B>=A ], cost: 1 6: evalEx6bb1in -> evalEx6bb3in : B'=1+B, [], cost: 1 7: evalEx6bb2in -> evalEx6bb3in : A'=1+A, [], cost: 1 8: evalEx6returnin -> evalEx6stop : [], cost: 1 Removed unreachable and leaf rules: Start location: evalEx6start 0: evalEx6start -> evalEx6entryin : [], cost: 1 1: evalEx6entryin -> evalEx6bb3in : A'=B, B'=A, [], cost: 1 2: evalEx6bb3in -> evalEx6bbin : [ C>=1+B ], cost: 1 4: evalEx6bbin -> evalEx6bb1in : [ A>=1+B ], cost: 1 5: evalEx6bbin -> evalEx6bb2in : [ B>=A ], cost: 1 6: evalEx6bb1in -> evalEx6bb3in : B'=1+B, [], cost: 1 7: evalEx6bb2in -> evalEx6bb3in : A'=1+A, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalEx6start 9: evalEx6start -> evalEx6bb3in : A'=B, B'=A, [], cost: 2 2: evalEx6bb3in -> evalEx6bbin : [ C>=1+B ], cost: 1 10: evalEx6bbin -> evalEx6bb3in : B'=1+B, [ A>=1+B ], cost: 2 11: evalEx6bbin -> evalEx6bb3in : A'=1+A, [ B>=A ], cost: 2 Eliminated locations (on tree-shaped paths): Start location: evalEx6start 9: evalEx6start -> evalEx6bb3in : A'=B, B'=A, [], cost: 2 12: evalEx6bb3in -> evalEx6bb3in : B'=1+B, [ C>=1+B && A>=1+B ], cost: 3 13: evalEx6bb3in -> evalEx6bb3in : A'=1+A, [ C>=1+B && B>=A ], cost: 3 Accelerating simple loops of location 2. Accelerating the following rules: 12: evalEx6bb3in -> evalEx6bb3in : B'=1+B, [ C>=1+B && A>=1+B ], cost: 3 13: evalEx6bb3in -> evalEx6bb3in : A'=1+A, [ C>=1+B && B>=A ], cost: 3 Accelerated rule 12 with backward acceleration, yielding the new rule 14. Accelerated rule 12 with backward acceleration, yielding the new rule 15. Accelerated rule 13 with metering function 1-A+B, yielding the new rule 16. Removing the simple loops: 12 13. Accelerated all simple loops using metering functions (where possible): Start location: evalEx6start 9: evalEx6start -> evalEx6bb3in : A'=B, B'=A, [], cost: 2 14: evalEx6bb3in -> evalEx6bb3in : B'=C, [ C>=1+B && A>=1+B && A>=C ], cost: 3*C-3*B 15: evalEx6bb3in -> evalEx6bb3in : B'=A, [ C>=1+B && A>=1+B && C>=A ], cost: 3*A-3*B 16: evalEx6bb3in -> evalEx6bb3in : A'=1+B, [ C>=1+B && B>=A ], cost: 3-3*A+3*B Chained accelerated rules (with incoming rules): Start location: evalEx6start 9: evalEx6start -> evalEx6bb3in : A'=B, B'=A, [], cost: 2 17: evalEx6start -> evalEx6bb3in : A'=B, B'=C, [ C>=1+A && B>=1+A && B>=C ], cost: 2+3*C-3*A 18: evalEx6start -> evalEx6bb3in : A'=B, [ C>=1+A && B>=1+A && C>=B ], cost: 2-3*A+3*B 19: evalEx6start -> evalEx6bb3in : A'=1+A, B'=A, [ C>=1+A && A>=B ], cost: 5+3*A-3*B Removed unreachable locations (and leaf rules with constant cost): Start location: evalEx6start 17: evalEx6start -> evalEx6bb3in : A'=B, B'=C, [ C>=1+A && B>=1+A && B>=C ], cost: 2+3*C-3*A 18: evalEx6start -> evalEx6bb3in : A'=B, [ C>=1+A && B>=1+A && C>=B ], cost: 2-3*A+3*B 19: evalEx6start -> evalEx6bb3in : A'=1+A, B'=A, [ C>=1+A && A>=B ], cost: 5+3*A-3*B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalEx6start 17: evalEx6start -> evalEx6bb3in : A'=B, B'=C, [ C>=1+A && B>=1+A && B>=C ], cost: 2+3*C-3*A 18: evalEx6start -> evalEx6bb3in : A'=B, [ C>=1+A && B>=1+A && C>=B ], cost: 2-3*A+3*B 19: evalEx6start -> evalEx6bb3in : A'=1+A, B'=A, [ C>=1+A && A>=B ], cost: 5+3*A-3*B Computing asymptotic complexity for rule 17 Solved the limit problem by the following transformations: Created initial limit problem: 1-C+B (+/+!), -A+B (+/+!), 2+3*C-3*A (+), C-A (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {C==0,A==-n,B==0} resulting limit problem: [solved] Solution: C / 0 A / -n B / 0 Resulting cost 2+3*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: 2+3*n Rule cost: 2+3*C-3*A Rule guard: [ C>=1+A && B>=1+A && B>=C ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)