/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 154 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 526 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: evalfstart(A, B, C) -> Com_1(evalfentryin(A, B, C)) :|: TRUE evalfentryin(A, B, C) -> Com_1(evalfbb3in(0, 0, C)) :|: TRUE evalfbb3in(A, B, C) -> Com_1(evalfbbin(A, B, C)) :|: 99 >= B evalfbb3in(A, B, C) -> Com_1(evalfreturnin(A, B, C)) :|: B >= 100 evalfbbin(A, B, C) -> Com_1(evalfbb1in(A, B, C)) :|: C >= A + 1 evalfbbin(A, B, C) -> Com_1(evalfbb2in(A, B, C)) :|: A >= C evalfbb1in(A, B, C) -> Com_1(evalfbb3in(A + 1, B, C)) :|: TRUE evalfbb2in(A, B, C) -> Com_1(evalfbb3in(A, B + 1, C)) :|: TRUE evalfreturnin(A, B, C) -> Com_1(evalfstop(A, B, C)) :|: TRUE The start-symbols are:[evalfstart_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 3*Ar_2 + 607) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(0, 0, Ar_2)) (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbbin(Ar_0, Ar_1, Ar_2)) [ 99 >= Ar_1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 100 ] (Comp: ?, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 ] (Comp: ?, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1 + 1, Ar_2)) (Comp: ?, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(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) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(0, 0, Ar_2)) (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbbin(Ar_0, Ar_1, Ar_2)) [ 99 >= Ar_1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 100 ] (Comp: ?, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 ] (Comp: ?, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1 + 1, Ar_2)) (Comp: ?, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalfstart) = 2 Pol(evalfentryin) = 2 Pol(evalfbb3in) = 2 Pol(evalfbbin) = 2 Pol(evalfreturnin) = 1 Pol(evalfbb1in) = 2 Pol(evalfbb2in) = 2 Pol(evalfstop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 100 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(0, 0, Ar_2)) (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbbin(Ar_0, Ar_1, Ar_2)) [ 99 >= Ar_1 ] (Comp: 2, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 100 ] (Comp: ?, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 ] (Comp: ?, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1 + 1, Ar_2)) (Comp: 2, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalfstart) = V_3 Pol(evalfentryin) = V_3 Pol(evalfbb3in) = -V_1 + V_3 Pol(evalfbbin) = -V_1 + V_3 Pol(evalfreturnin) = -V_1 + V_3 Pol(evalfbb1in) = -V_1 + V_3 - 1 Pol(evalfbb2in) = -V_1 + V_3 Pol(evalfstop) = -V_1 + V_3 Pol(koat_start) = V_3 orients all transitions weakly and the transition evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 + 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(0, 0, Ar_2)) (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbbin(Ar_0, Ar_1, Ar_2)) [ 99 >= Ar_1 ] (Comp: 2, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 100 ] (Comp: Ar_2, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 ] (Comp: ?, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1 + 1, Ar_2)) (Comp: 2, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 4 produces the following problem: 5: T: (Comp: 1, Cost: 1) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(0, 0, Ar_2)) (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbbin(Ar_0, Ar_1, Ar_2)) [ 99 >= Ar_1 ] (Comp: 2, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 100 ] (Comp: Ar_2, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_2 ] (Comp: Ar_2, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1 + 1, Ar_2)) (Comp: 2, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 5 to obtain the following invariants: For symbol evalfbb1in: X_3 - 1 >= 0 /\ X_2 + X_3 - 1 >= 0 /\ -X_2 + X_3 + 98 >= 0 /\ X_1 + X_3 - 1 >= 0 /\ -X_1 + X_3 - 1 >= 0 /\ -X_2 + 99 >= 0 /\ X_1 - X_2 + 99 >= 0 /\ X_2 >= 0 /\ X_1 + X_2 >= 0 /\ X_1 >= 0 For symbol evalfbb2in: X_1 - X_3 >= 0 /\ -X_2 + 99 >= 0 /\ X_1 - X_2 + 99 >= 0 /\ X_2 >= 0 /\ X_1 + X_2 >= 0 /\ X_1 >= 0 For symbol evalfbb3in: X_2 >= 0 /\ X_1 + X_2 >= 0 /\ X_1 >= 0 For symbol evalfbbin: -X_2 + 99 >= 0 /\ X_1 - X_2 + 99 >= 0 /\ X_2 >= 0 /\ X_1 + X_2 >= 0 /\ X_1 >= 0 For symbol evalfreturnin: X_2 - 100 >= 0 /\ X_1 + X_2 - 100 >= 0 /\ X_1 >= 0 This yielded the following problem: 6: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) [ Ar_1 - 100 >= 0 /\ Ar_0 + Ar_1 - 100 >= 0 /\ Ar_0 >= 0 ] (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 - Ar_2 >= 0 /\ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 ] (Comp: Ar_2, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0 + 1, Ar_1, Ar_2)) [ Ar_2 - 1 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ -Ar_1 + Ar_2 + 98 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ -Ar_0 + Ar_2 - 1 >= 0 /\ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 ] (Comp: ?, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2)) [ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_0 >= Ar_2 ] (Comp: Ar_2, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= 100 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbbin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ 99 >= Ar_1 ] (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(0, 0, Ar_2)) (Comp: 1, Cost: 1) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 200 Pol(evalfstart) = 200 Pol(evalfreturnin) = -2*V_2 Pol(evalfstop) = -2*V_2 Pol(evalfbb2in) = -2*V_2 + 199 Pol(evalfbb3in) = -2*V_2 + 200 Pol(evalfbb1in) = -2*V_2 + 200 Pol(evalfbbin) = -2*V_2 + 200 Pol(evalfentryin) = 200 orients all transitions weakly and the transitions evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2)) [ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_0 >= Ar_2 ] evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 - Ar_2 >= 0 /\ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 ] strictly and produces the following problem: 7: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) [ Ar_1 - 100 >= 0 /\ Ar_0 + Ar_1 - 100 >= 0 /\ Ar_0 >= 0 ] (Comp: 200, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 - Ar_2 >= 0 /\ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 ] (Comp: Ar_2, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0 + 1, Ar_1, Ar_2)) [ Ar_2 - 1 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ -Ar_1 + Ar_2 + 98 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ -Ar_0 + Ar_2 - 1 >= 0 /\ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 ] (Comp: 200, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2)) [ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_0 >= Ar_2 ] (Comp: Ar_2, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= 100 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbbin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ 99 >= Ar_1 ] (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(0, 0, Ar_2)) (Comp: 1, Cost: 1) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 7 produces the following problem: 8: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) [ Ar_1 - 100 >= 0 /\ Ar_0 + Ar_1 - 100 >= 0 /\ Ar_0 >= 0 ] (Comp: 200, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1 + 1, Ar_2)) [ Ar_0 - Ar_2 >= 0 /\ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 ] (Comp: Ar_2, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0 + 1, Ar_1, Ar_2)) [ Ar_2 - 1 >= 0 /\ Ar_1 + Ar_2 - 1 >= 0 /\ -Ar_1 + Ar_2 + 98 >= 0 /\ Ar_0 + Ar_2 - 1 >= 0 /\ -Ar_0 + Ar_2 - 1 >= 0 /\ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 ] (Comp: 200, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2)) [ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_0 >= Ar_2 ] (Comp: Ar_2, Cost: 1) evalfbbin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ -Ar_1 + 99 >= 0 /\ Ar_0 - Ar_1 + 99 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= 100 ] (Comp: Ar_2 + 201, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbbin(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ 99 >= Ar_1 ] (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(0, 0, Ar_2)) (Comp: 1, Cost: 1) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) start location: koat_start leaf cost: 0 Complexity upper bound 3*Ar_2 + 607 Time: 0.196 sec (SMT: 0.161 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalfstart 0: evalfstart -> evalfentryin : [], cost: 1 1: evalfentryin -> evalfbb3in : A'=0, B'=0, [], cost: 1 2: evalfbb3in -> evalfbbin : [ 99>=B ], cost: 1 3: evalfbb3in -> evalfreturnin : [ B>=100 ], cost: 1 4: evalfbbin -> evalfbb1in : [ C>=1+A ], cost: 1 5: evalfbbin -> evalfbb2in : [ A>=C ], cost: 1 6: evalfbb1in -> evalfbb3in : A'=1+A, [], cost: 1 7: evalfbb2in -> evalfbb3in : B'=1+B, [], cost: 1 8: evalfreturnin -> evalfstop : [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: evalfstart -> evalfentryin : [], cost: 1 Removed unreachable and leaf rules: Start location: evalfstart 0: evalfstart -> evalfentryin : [], cost: 1 1: evalfentryin -> evalfbb3in : A'=0, B'=0, [], cost: 1 2: evalfbb3in -> evalfbbin : [ 99>=B ], cost: 1 4: evalfbbin -> evalfbb1in : [ C>=1+A ], cost: 1 5: evalfbbin -> evalfbb2in : [ A>=C ], cost: 1 6: evalfbb1in -> evalfbb3in : A'=1+A, [], cost: 1 7: evalfbb2in -> evalfbb3in : B'=1+B, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalfstart 9: evalfstart -> evalfbb3in : A'=0, B'=0, [], cost: 2 2: evalfbb3in -> evalfbbin : [ 99>=B ], cost: 1 10: evalfbbin -> evalfbb3in : A'=1+A, [ C>=1+A ], cost: 2 11: evalfbbin -> evalfbb3in : B'=1+B, [ A>=C ], cost: 2 Eliminated locations (on tree-shaped paths): Start location: evalfstart 9: evalfstart -> evalfbb3in : A'=0, B'=0, [], cost: 2 12: evalfbb3in -> evalfbb3in : A'=1+A, [ 99>=B && C>=1+A ], cost: 3 13: evalfbb3in -> evalfbb3in : B'=1+B, [ 99>=B && A>=C ], cost: 3 Accelerating simple loops of location 2. Accelerating the following rules: 12: evalfbb3in -> evalfbb3in : A'=1+A, [ 99>=B && C>=1+A ], cost: 3 13: evalfbb3in -> evalfbb3in : B'=1+B, [ 99>=B && A>=C ], cost: 3 Accelerated rule 12 with metering function C-A, yielding the new rule 14. Accelerated rule 13 with metering function 100-B, yielding the new rule 15. Removing the simple loops: 12 13. Accelerated all simple loops using metering functions (where possible): Start location: evalfstart 9: evalfstart -> evalfbb3in : A'=0, B'=0, [], cost: 2 14: evalfbb3in -> evalfbb3in : A'=C, [ 99>=B && C>=1+A ], cost: 3*C-3*A 15: evalfbb3in -> evalfbb3in : B'=100, [ 99>=B && A>=C ], cost: 300-3*B Chained accelerated rules (with incoming rules): Start location: evalfstart 9: evalfstart -> evalfbb3in : A'=0, B'=0, [], cost: 2 16: evalfstart -> evalfbb3in : A'=C, B'=0, [ C>=1 ], cost: 2+3*C 17: evalfstart -> evalfbb3in : A'=0, B'=100, [ 0>=C ], cost: 302 Removed unreachable locations (and leaf rules with constant cost): Start location: evalfstart 16: evalfstart -> evalfbb3in : A'=C, B'=0, [ C>=1 ], cost: 2+3*C ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalfstart 16: evalfstart -> evalfbb3in : A'=C, B'=0, [ C>=1 ], cost: 2+3*C Computing asymptotic complexity for rule 16 Solved the limit problem by the following transformations: Created initial limit problem: C (+/+!), 2+3*C (+) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {C==n} resulting limit problem: [solved] Solution: C / 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*C Rule guard: [ C>=1 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)