/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 123 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: evalDis2start(A, B, C) -> Com_1(evalDis2entryin(A, B, C)) :|: TRUE evalDis2entryin(A, B, C) -> Com_1(evalDis2bb3in(B, C, A)) :|: TRUE evalDis2bb3in(A, B, C) -> Com_1(evalDis2bbin(A, B, C)) :|: A >= C + 1 evalDis2bb3in(A, B, C) -> Com_1(evalDis2returnin(A, B, C)) :|: C >= A evalDis2bbin(A, B, C) -> Com_1(evalDis2bb1in(A, B, C)) :|: B >= C + 1 evalDis2bbin(A, B, C) -> Com_1(evalDis2bb2in(A, B, C)) :|: C >= B evalDis2bb1in(A, B, C) -> Com_1(evalDis2bb3in(A, B, C + 1)) :|: TRUE evalDis2bb2in(A, B, C) -> Com_1(evalDis2bb3in(A, B + 1, C)) :|: TRUE evalDis2returnin(A, B, C) -> Com_1(evalDis2stop(A, B, C)) :|: TRUE The start-symbols are:[evalDis2start_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 12*ar_1 + 6*ar_2 + 6*ar_0 + 7) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalDis2start(ar_0, ar_1, ar_2) -> Com_1(evalDis2entryin(ar_0, ar_1, ar_2)) (Comp: ?, Cost: 1) evalDis2entryin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_1, ar_2, ar_0)) (Comp: ?, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bbin(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 ] (Comp: ?, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2returnin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ] (Comp: ?, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb1in(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 + 1 ] (Comp: ?, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb2in(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 ] (Comp: ?, Cost: 1) evalDis2bb1in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1, ar_2 + 1)) (Comp: ?, Cost: 1) evalDis2bb2in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1 + 1, ar_2)) (Comp: ?, Cost: 1) evalDis2returnin(ar_0, ar_1, ar_2) -> Com_1(evalDis2stop(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalDis2start(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) evalDis2start(ar_0, ar_1, ar_2) -> Com_1(evalDis2entryin(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 1) evalDis2entryin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_1, ar_2, ar_0)) (Comp: ?, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bbin(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 ] (Comp: ?, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2returnin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ] (Comp: ?, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb1in(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 + 1 ] (Comp: ?, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb2in(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 ] (Comp: ?, Cost: 1) evalDis2bb1in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1, ar_2 + 1)) (Comp: ?, Cost: 1) evalDis2bb2in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1 + 1, ar_2)) (Comp: ?, Cost: 1) evalDis2returnin(ar_0, ar_1, ar_2) -> Com_1(evalDis2stop(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalDis2start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalDis2start) = 2 Pol(evalDis2entryin) = 2 Pol(evalDis2bb3in) = 2 Pol(evalDis2bbin) = 2 Pol(evalDis2returnin) = 1 Pol(evalDis2bb1in) = 2 Pol(evalDis2bb2in) = 2 Pol(evalDis2stop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalDis2returnin(ar_0, ar_1, ar_2) -> Com_1(evalDis2stop(ar_0, ar_1, ar_2)) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2returnin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalDis2start(ar_0, ar_1, ar_2) -> Com_1(evalDis2entryin(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 1) evalDis2entryin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_1, ar_2, ar_0)) (Comp: ?, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bbin(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 ] (Comp: 2, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2returnin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ] (Comp: ?, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb1in(ar_0, ar_1, ar_2)) [ ar_1 >= ar_2 + 1 ] (Comp: ?, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb2in(ar_0, ar_1, ar_2)) [ ar_2 >= ar_1 ] (Comp: ?, Cost: 1) evalDis2bb1in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1, ar_2 + 1)) (Comp: ?, Cost: 1) evalDis2bb2in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1 + 1, ar_2)) (Comp: 2, Cost: 1) evalDis2returnin(ar_0, ar_1, ar_2) -> Com_1(evalDis2stop(ar_0, ar_1, ar_2)) (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalDis2start(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 evalDis2bb1in: X_2 - X_3 - 1 >= 0 /\ X_1 - X_3 - 1 >= 0 For symbol evalDis2bb2in: X_1 - X_3 - 1 >= 0 /\ -X_2 + X_3 >= 0 /\ X_1 - X_2 - 1 >= 0 For symbol evalDis2bbin: X_1 - X_3 - 1 >= 0 For symbol evalDis2returnin: -X_1 + X_3 >= 0 This yielded the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalDis2start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalDis2returnin(ar_0, ar_1, ar_2) -> Com_1(evalDis2stop(ar_0, ar_1, ar_2)) [ -ar_0 + ar_2 >= 0 ] (Comp: ?, Cost: 1) evalDis2bb2in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1 + 1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ -ar_1 + ar_2 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ] (Comp: ?, Cost: 1) evalDis2bb1in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1, ar_2 + 1)) [ ar_1 - ar_2 - 1 >= 0 /\ ar_0 - ar_2 - 1 >= 0 ] (Comp: ?, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb2in(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ ar_2 >= ar_1 ] (Comp: ?, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb1in(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ ar_1 >= ar_2 + 1 ] (Comp: 2, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2returnin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ] (Comp: ?, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bbin(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 ] (Comp: 1, Cost: 1) evalDis2entryin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_1, ar_2, ar_0)) (Comp: 1, Cost: 1) evalDis2start(ar_0, ar_1, ar_2) -> Com_1(evalDis2entryin(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(evalDis2start) = 2*V_2 - 2*V_3 Pol(evalDis2returnin) = 2*V_1 - 2*V_2 Pol(evalDis2stop) = 2*V_1 - 2*V_2 Pol(evalDis2bb2in) = 2*V_1 - 2*V_2 - 1 Pol(evalDis2bb3in) = 2*V_1 - 2*V_2 Pol(evalDis2bb1in) = 2*V_1 - 2*V_2 Pol(evalDis2bbin) = 2*V_1 - 2*V_2 Pol(evalDis2entryin) = 2*V_2 - 2*V_3 orients all transitions weakly and the transitions evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb2in(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ ar_2 >= ar_1 ] evalDis2bb2in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1 + 1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ -ar_1 + ar_2 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalDis2start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalDis2returnin(ar_0, ar_1, ar_2) -> Com_1(evalDis2stop(ar_0, ar_1, ar_2)) [ -ar_0 + ar_2 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalDis2bb2in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1 + 1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ -ar_1 + ar_2 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ] (Comp: ?, Cost: 1) evalDis2bb1in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1, ar_2 + 1)) [ ar_1 - ar_2 - 1 >= 0 /\ ar_0 - ar_2 - 1 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb2in(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ ar_2 >= ar_1 ] (Comp: ?, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb1in(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ ar_1 >= ar_2 + 1 ] (Comp: 2, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2returnin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ] (Comp: ?, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bbin(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 ] (Comp: 1, Cost: 1) evalDis2entryin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_1, ar_2, ar_0)) (Comp: 1, Cost: 1) evalDis2start(ar_0, ar_1, ar_2) -> Com_1(evalDis2entryin(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_2 Pol(evalDis2start) = -2*V_1 + 2*V_2 Pol(evalDis2returnin) = 2*V_1 - 2*V_3 Pol(evalDis2stop) = 2*V_1 - 2*V_3 Pol(evalDis2bb2in) = 2*V_1 - 2*V_3 Pol(evalDis2bb3in) = 2*V_1 - 2*V_3 Pol(evalDis2bb1in) = 2*V_1 - 2*V_3 - 1 Pol(evalDis2bbin) = 2*V_1 - 2*V_3 Pol(evalDis2entryin) = -2*V_1 + 2*V_2 orients all transitions weakly and the transitions evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb1in(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ ar_1 >= ar_2 + 1 ] evalDis2bb1in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1, ar_2 + 1)) [ ar_1 - ar_2 - 1 >= 0 /\ ar_0 - ar_2 - 1 >= 0 ] strictly and produces the following problem: 6: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(evalDis2start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalDis2returnin(ar_0, ar_1, ar_2) -> Com_1(evalDis2stop(ar_0, ar_1, ar_2)) [ -ar_0 + ar_2 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalDis2bb2in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1 + 1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ -ar_1 + ar_2 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ] (Comp: 2*ar_0 + 2*ar_1, Cost: 1) evalDis2bb1in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1, ar_2 + 1)) [ ar_1 - ar_2 - 1 >= 0 /\ ar_0 - ar_2 - 1 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb2in(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ ar_2 >= ar_1 ] (Comp: 2*ar_0 + 2*ar_1, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb1in(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ ar_1 >= ar_2 + 1 ] (Comp: 2, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2returnin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ] (Comp: ?, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bbin(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 ] (Comp: 1, Cost: 1) evalDis2entryin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_1, ar_2, ar_0)) (Comp: 1, Cost: 1) evalDis2start(ar_0, ar_1, ar_2) -> Com_1(evalDis2entryin(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(evalDis2start(ar_0, ar_1, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalDis2returnin(ar_0, ar_1, ar_2) -> Com_1(evalDis2stop(ar_0, ar_1, ar_2)) [ -ar_0 + ar_2 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalDis2bb2in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1 + 1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ -ar_1 + ar_2 >= 0 /\ ar_0 - ar_1 - 1 >= 0 ] (Comp: 2*ar_0 + 2*ar_1, Cost: 1) evalDis2bb1in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_0, ar_1, ar_2 + 1)) [ ar_1 - ar_2 - 1 >= 0 /\ ar_0 - ar_2 - 1 >= 0 ] (Comp: 2*ar_1 + 2*ar_2, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb2in(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ ar_2 >= ar_1 ] (Comp: 2*ar_0 + 2*ar_1, Cost: 1) evalDis2bbin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb1in(ar_0, ar_1, ar_2)) [ ar_0 - ar_2 - 1 >= 0 /\ ar_1 >= ar_2 + 1 ] (Comp: 2, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2returnin(ar_0, ar_1, ar_2)) [ ar_2 >= ar_0 ] (Comp: 2*ar_0 + 4*ar_1 + 2*ar_2 + 1, Cost: 1) evalDis2bb3in(ar_0, ar_1, ar_2) -> Com_1(evalDis2bbin(ar_0, ar_1, ar_2)) [ ar_0 >= ar_2 + 1 ] (Comp: 1, Cost: 1) evalDis2entryin(ar_0, ar_1, ar_2) -> Com_1(evalDis2bb3in(ar_1, ar_2, ar_0)) (Comp: 1, Cost: 1) evalDis2start(ar_0, ar_1, ar_2) -> Com_1(evalDis2entryin(ar_0, ar_1, ar_2)) start location: koat_start leaf cost: 0 Complexity upper bound 12*ar_1 + 6*ar_2 + 6*ar_0 + 7 Time: 0.173 sec (SMT: 0.154 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalDis2start 0: evalDis2start -> evalDis2entryin : [], cost: 1 1: evalDis2entryin -> evalDis2bb3in : A'=B, B'=C, C'=A, [], cost: 1 2: evalDis2bb3in -> evalDis2bbin : [ A>=1+C ], cost: 1 3: evalDis2bb3in -> evalDis2returnin : [ C>=A ], cost: 1 4: evalDis2bbin -> evalDis2bb1in : [ B>=1+C ], cost: 1 5: evalDis2bbin -> evalDis2bb2in : [ C>=B ], cost: 1 6: evalDis2bb1in -> evalDis2bb3in : C'=1+C, [], cost: 1 7: evalDis2bb2in -> evalDis2bb3in : B'=1+B, [], cost: 1 8: evalDis2returnin -> evalDis2stop : [], cost: 1 Removed unreachable and leaf rules: Start location: evalDis2start 0: evalDis2start -> evalDis2entryin : [], cost: 1 1: evalDis2entryin -> evalDis2bb3in : A'=B, B'=C, C'=A, [], cost: 1 2: evalDis2bb3in -> evalDis2bbin : [ A>=1+C ], cost: 1 4: evalDis2bbin -> evalDis2bb1in : [ B>=1+C ], cost: 1 5: evalDis2bbin -> evalDis2bb2in : [ C>=B ], cost: 1 6: evalDis2bb1in -> evalDis2bb3in : C'=1+C, [], cost: 1 7: evalDis2bb2in -> evalDis2bb3in : B'=1+B, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalDis2start 9: evalDis2start -> evalDis2bb3in : A'=B, B'=C, C'=A, [], cost: 2 2: evalDis2bb3in -> evalDis2bbin : [ A>=1+C ], cost: 1 10: evalDis2bbin -> evalDis2bb3in : C'=1+C, [ B>=1+C ], cost: 2 11: evalDis2bbin -> evalDis2bb3in : B'=1+B, [ C>=B ], cost: 2 Eliminated locations (on tree-shaped paths): Start location: evalDis2start 9: evalDis2start -> evalDis2bb3in : A'=B, B'=C, C'=A, [], cost: 2 12: evalDis2bb3in -> evalDis2bb3in : C'=1+C, [ A>=1+C && B>=1+C ], cost: 3 13: evalDis2bb3in -> evalDis2bb3in : B'=1+B, [ A>=1+C && C>=B ], cost: 3 Accelerating simple loops of location 2. Accelerating the following rules: 12: evalDis2bb3in -> evalDis2bb3in : C'=1+C, [ A>=1+C && B>=1+C ], cost: 3 13: evalDis2bb3in -> evalDis2bb3in : B'=1+B, [ A>=1+C && C>=B ], 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+C-B, yielding the new rule 16. Removing the simple loops: 12 13. Accelerated all simple loops using metering functions (where possible): Start location: evalDis2start 9: evalDis2start -> evalDis2bb3in : A'=B, B'=C, C'=A, [], cost: 2 14: evalDis2bb3in -> evalDis2bb3in : C'=A, [ A>=1+C && B>=1+C && B>=A ], cost: -3*C+3*A 15: evalDis2bb3in -> evalDis2bb3in : C'=B, [ A>=1+C && B>=1+C && A>=B ], cost: -3*C+3*B 16: evalDis2bb3in -> evalDis2bb3in : B'=1+C, [ A>=1+C && C>=B ], cost: 3+3*C-3*B Chained accelerated rules (with incoming rules): Start location: evalDis2start 9: evalDis2start -> evalDis2bb3in : A'=B, B'=C, C'=A, [], cost: 2 17: evalDis2start -> evalDis2bb3in : A'=B, B'=C, C'=B, [ B>=1+A && C>=1+A && C>=B ], cost: 2-3*A+3*B 18: evalDis2start -> evalDis2bb3in : A'=B, B'=C, [ B>=1+A && C>=1+A && B>=C ], cost: 2+3*C-3*A 19: evalDis2start -> evalDis2bb3in : A'=B, B'=1+A, C'=A, [ B>=1+A && A>=C ], cost: 5-3*C+3*A Removed unreachable locations (and leaf rules with constant cost): Start location: evalDis2start 17: evalDis2start -> evalDis2bb3in : A'=B, B'=C, C'=B, [ B>=1+A && C>=1+A && C>=B ], cost: 2-3*A+3*B 18: evalDis2start -> evalDis2bb3in : A'=B, B'=C, [ B>=1+A && C>=1+A && B>=C ], cost: 2+3*C-3*A 19: evalDis2start -> evalDis2bb3in : A'=B, B'=1+A, C'=A, [ B>=1+A && A>=C ], cost: 5-3*C+3*A ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalDis2start 17: evalDis2start -> evalDis2bb3in : A'=B, B'=C, C'=B, [ B>=1+A && C>=1+A && C>=B ], cost: 2-3*A+3*B 18: evalDis2start -> evalDis2bb3in : A'=B, B'=C, [ B>=1+A && C>=1+A && B>=C ], cost: 2+3*C-3*A 19: evalDis2start -> evalDis2bb3in : A'=B, B'=1+A, C'=A, [ B>=1+A && A>=C ], cost: 5-3*C+3*A Computing asymptotic complexity for rule 17 Solved the limit problem by the following transformations: Created initial limit problem: 2-3*A+3*B (+), 1+C-B (+/+!), -A+B (+/+!), C-A (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {C==2*n,A==0,B==n} resulting limit problem: [solved] Solution: C / 2*n A / 0 B / n 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*A+3*B Rule guard: [ B>=1+A && C>=1+A && C>=B ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)