/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 146 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 545 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: evalDis1start(A, B, C, D) -> Com_1(evalDis1entryin(A, B, C, D)) :|: TRUE evalDis1entryin(A, B, C, D) -> Com_1(evalDis1bb3in(B, A, D, C)) :|: TRUE evalDis1bb3in(A, B, C, D) -> Com_1(evalDis1bbin(A, B, C, D)) :|: A >= B + 1 evalDis1bb3in(A, B, C, D) -> Com_1(evalDis1returnin(A, B, C, D)) :|: B >= A evalDis1bbin(A, B, C, D) -> Com_1(evalDis1bb1in(A, B, C, D)) :|: C >= D + 1 evalDis1bbin(A, B, C, D) -> Com_1(evalDis1bb2in(A, B, C, D)) :|: D >= C evalDis1bb1in(A, B, C, D) -> Com_1(evalDis1bb3in(A, B, C, D + 1)) :|: TRUE evalDis1bb2in(A, B, C, D) -> Com_1(evalDis1bb3in(A, B + 1, C, D)) :|: TRUE evalDis1returnin(A, B, C, D) -> Com_1(evalDis1stop(A, B, C, D)) :|: TRUE The start-symbols are:[evalDis1start_4] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 6*Ar_0 + 6*Ar_1 + 3*Ar_2 + 3*Ar_3 + 7) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3)) (Comp: ?, Cost: 1) evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_1, Ar_0, Ar_3, Ar_2)) (Comp: ?, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 1 ] (Comp: ?, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_0 ] (Comp: ?, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) (Comp: ?, Cost: 1) evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) (Comp: ?, Cost: 1) evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1stop(Ar_0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3)) [ 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) evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 1) evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_1, Ar_0, Ar_3, Ar_2)) (Comp: ?, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 1 ] (Comp: ?, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_0 ] (Comp: ?, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) (Comp: ?, Cost: 1) evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) (Comp: ?, Cost: 1) evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1stop(Ar_0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalDis1start) = 2 Pol(evalDis1entryin) = 2 Pol(evalDis1bb3in) = 2 Pol(evalDis1bbin) = 2 Pol(evalDis1returnin) = 1 Pol(evalDis1bb1in) = 2 Pol(evalDis1bb2in) = 2 Pol(evalDis1stop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1stop(Ar_0, Ar_1, Ar_2, Ar_3)) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_0 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 1) evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_1, Ar_0, Ar_3, Ar_2)) (Comp: ?, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 2, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_0 ] (Comp: ?, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) (Comp: ?, Cost: 1) evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) (Comp: 2, Cost: 1) evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1stop(Ar_0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalDis1start) = -V_3 + V_4 Pol(evalDis1entryin) = -V_3 + V_4 Pol(evalDis1bb3in) = V_3 - V_4 Pol(evalDis1bbin) = V_3 - V_4 Pol(evalDis1returnin) = V_3 - V_4 Pol(evalDis1bb1in) = V_3 - V_4 - 1 Pol(evalDis1bb2in) = V_3 - V_4 Pol(evalDis1stop) = V_3 - V_4 Pol(koat_start) = -V_3 + V_4 orients all transitions weakly and the transition evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_3 + 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 1) evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_1, Ar_0, Ar_3, Ar_2)) (Comp: ?, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 2, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_0 ] (Comp: Ar_2 + Ar_3, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) (Comp: ?, Cost: 1) evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) (Comp: 2, Cost: 1) evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1stop(Ar_0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3)) [ 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) evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 1) evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_1, Ar_0, Ar_3, Ar_2)) (Comp: ?, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 2, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_0 ] (Comp: Ar_2 + Ar_3, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: Ar_2 + Ar_3, Cost: 1) evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) (Comp: ?, Cost: 1) evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) (Comp: 2, Cost: 1) evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1stop(Ar_0, Ar_1, Ar_2, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 5 to obtain the following invariants: For symbol evalDis1bb1in: X_3 - X_4 - 1 >= 0 /\ X_1 - X_2 - 1 >= 0 For symbol evalDis1bb2in: -X_3 + X_4 >= 0 /\ X_1 - X_2 - 1 >= 0 For symbol evalDis1bbin: X_1 - X_2 - 1 >= 0 For symbol evalDis1returnin: -X_1 + X_2 >= 0 This yielded the following problem: 6: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ -Ar_0 + Ar_1 >= 0 ] (Comp: ?, Cost: 1) evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ Ar_0 - Ar_1 - 1 >= 0 ] (Comp: Ar_2 + Ar_3, Cost: 1) evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_2 - Ar_3 - 1 >= 0 /\ Ar_0 - Ar_1 - 1 >= 0 ] (Comp: ?, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 - Ar_1 - 1 >= 0 /\ Ar_3 >= Ar_2 ] (Comp: Ar_2 + Ar_3, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 - Ar_1 - 1 >= 0 /\ Ar_2 >= Ar_3 + 1 ] (Comp: 2, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_0 ] (Comp: ?, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_1, Ar_0, Ar_3, Ar_2)) (Comp: 1, Cost: 1) evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = -2*V_1 + 2*V_2 Pol(evalDis1start) = -2*V_1 + 2*V_2 Pol(evalDis1returnin) = 2*V_1 - 2*V_2 Pol(evalDis1stop) = 2*V_1 - 2*V_2 Pol(evalDis1bb2in) = 2*V_1 - 2*V_2 - 1 Pol(evalDis1bb3in) = 2*V_1 - 2*V_2 Pol(evalDis1bb1in) = 2*V_1 - 2*V_2 Pol(evalDis1bbin) = 2*V_1 - 2*V_2 Pol(evalDis1entryin) = -2*V_1 + 2*V_2 orients all transitions weakly and the transitions evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 - Ar_1 - 1 >= 0 /\ Ar_3 >= Ar_2 ] evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ Ar_0 - Ar_1 - 1 >= 0 ] strictly and produces the following problem: 7: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ -Ar_0 + Ar_1 >= 0 ] (Comp: 2*Ar_0 + 2*Ar_1, Cost: 1) evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ Ar_0 - Ar_1 - 1 >= 0 ] (Comp: Ar_2 + Ar_3, Cost: 1) evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_2 - Ar_3 - 1 >= 0 /\ Ar_0 - Ar_1 - 1 >= 0 ] (Comp: 2*Ar_0 + 2*Ar_1, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 - Ar_1 - 1 >= 0 /\ Ar_3 >= Ar_2 ] (Comp: Ar_2 + Ar_3, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 - Ar_1 - 1 >= 0 /\ Ar_2 >= Ar_3 + 1 ] (Comp: 2, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_0 ] (Comp: ?, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_1, Ar_0, Ar_3, Ar_2)) (Comp: 1, Cost: 1) evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3)) 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, Ar_3) -> Com_1(evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: 2, Cost: 1) evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ -Ar_0 + Ar_1 >= 0 ] (Comp: 2*Ar_0 + 2*Ar_1, Cost: 1) evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ -Ar_2 + Ar_3 >= 0 /\ Ar_0 - Ar_1 - 1 >= 0 ] (Comp: Ar_2 + Ar_3, Cost: 1) evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_2 - Ar_3 - 1 >= 0 /\ Ar_0 - Ar_1 - 1 >= 0 ] (Comp: 2*Ar_0 + 2*Ar_1, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb2in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 - Ar_1 - 1 >= 0 /\ Ar_3 >= Ar_2 ] (Comp: Ar_2 + Ar_3, Cost: 1) evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb1in(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 - Ar_1 - 1 >= 0 /\ Ar_2 >= Ar_3 + 1 ] (Comp: 2, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1returnin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_1 >= Ar_0 ] (Comp: Ar_2 + Ar_3 + 2*Ar_0 + 2*Ar_1 + 1, Cost: 1) evalDis1bb3in(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bbin(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1bb3in(Ar_1, Ar_0, Ar_3, Ar_2)) (Comp: 1, Cost: 1) evalDis1start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(evalDis1entryin(Ar_0, Ar_1, Ar_2, Ar_3)) start location: koat_start leaf cost: 0 Complexity upper bound 6*Ar_0 + 6*Ar_1 + 3*Ar_2 + 3*Ar_3 + 7 Time: 0.171 sec (SMT: 0.133 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalDis1start 0: evalDis1start -> evalDis1entryin : [], cost: 1 1: evalDis1entryin -> evalDis1bb3in : A'=B, B'=A, C'=D, D'=C, [], cost: 1 2: evalDis1bb3in -> evalDis1bbin : [ A>=1+B ], cost: 1 3: evalDis1bb3in -> evalDis1returnin : [ B>=A ], cost: 1 4: evalDis1bbin -> evalDis1bb1in : [ C>=1+D ], cost: 1 5: evalDis1bbin -> evalDis1bb2in : [ D>=C ], cost: 1 6: evalDis1bb1in -> evalDis1bb3in : D'=1+D, [], cost: 1 7: evalDis1bb2in -> evalDis1bb3in : B'=1+B, [], cost: 1 8: evalDis1returnin -> evalDis1stop : [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: evalDis1start -> evalDis1entryin : [], cost: 1 Removed unreachable and leaf rules: Start location: evalDis1start 0: evalDis1start -> evalDis1entryin : [], cost: 1 1: evalDis1entryin -> evalDis1bb3in : A'=B, B'=A, C'=D, D'=C, [], cost: 1 2: evalDis1bb3in -> evalDis1bbin : [ A>=1+B ], cost: 1 4: evalDis1bbin -> evalDis1bb1in : [ C>=1+D ], cost: 1 5: evalDis1bbin -> evalDis1bb2in : [ D>=C ], cost: 1 6: evalDis1bb1in -> evalDis1bb3in : D'=1+D, [], cost: 1 7: evalDis1bb2in -> evalDis1bb3in : B'=1+B, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalDis1start 9: evalDis1start -> evalDis1bb3in : A'=B, B'=A, C'=D, D'=C, [], cost: 2 2: evalDis1bb3in -> evalDis1bbin : [ A>=1+B ], cost: 1 10: evalDis1bbin -> evalDis1bb3in : D'=1+D, [ C>=1+D ], cost: 2 11: evalDis1bbin -> evalDis1bb3in : B'=1+B, [ D>=C ], cost: 2 Eliminated locations (on tree-shaped paths): Start location: evalDis1start 9: evalDis1start -> evalDis1bb3in : A'=B, B'=A, C'=D, D'=C, [], cost: 2 12: evalDis1bb3in -> evalDis1bb3in : D'=1+D, [ A>=1+B && C>=1+D ], cost: 3 13: evalDis1bb3in -> evalDis1bb3in : B'=1+B, [ A>=1+B && D>=C ], cost: 3 Accelerating simple loops of location 2. Accelerating the following rules: 12: evalDis1bb3in -> evalDis1bb3in : D'=1+D, [ A>=1+B && C>=1+D ], cost: 3 13: evalDis1bb3in -> evalDis1bb3in : B'=1+B, [ A>=1+B && D>=C ], cost: 3 Accelerated rule 12 with metering function C-D, yielding the new rule 14. Accelerated rule 13 with metering function A-B, yielding the new rule 15. Removing the simple loops: 12 13. Accelerated all simple loops using metering functions (where possible): Start location: evalDis1start 9: evalDis1start -> evalDis1bb3in : A'=B, B'=A, C'=D, D'=C, [], cost: 2 14: evalDis1bb3in -> evalDis1bb3in : D'=C, [ A>=1+B && C>=1+D ], cost: 3*C-3*D 15: evalDis1bb3in -> evalDis1bb3in : B'=A, [ A>=1+B && D>=C ], cost: 3*A-3*B Chained accelerated rules (with incoming rules): Start location: evalDis1start 9: evalDis1start -> evalDis1bb3in : A'=B, B'=A, C'=D, D'=C, [], cost: 2 16: evalDis1start -> evalDis1bb3in : A'=B, B'=A, C'=D, [ B>=1+A && D>=1+C ], cost: 2-3*C+3*D 17: evalDis1start -> evalDis1bb3in : A'=B, C'=D, D'=C, [ B>=1+A && C>=D ], cost: 2-3*A+3*B Removed unreachable locations (and leaf rules with constant cost): Start location: evalDis1start 16: evalDis1start -> evalDis1bb3in : A'=B, B'=A, C'=D, [ B>=1+A && D>=1+C ], cost: 2-3*C+3*D 17: evalDis1start -> evalDis1bb3in : A'=B, C'=D, D'=C, [ B>=1+A && C>=D ], cost: 2-3*A+3*B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalDis1start 16: evalDis1start -> evalDis1bb3in : A'=B, B'=A, C'=D, [ B>=1+A && D>=1+C ], cost: 2-3*C+3*D 17: evalDis1start -> evalDis1bb3in : A'=B, C'=D, D'=C, [ B>=1+A && C>=D ], cost: 2-3*A+3*B Computing asymptotic complexity for rule 16 Solved the limit problem by the following transformations: Created initial limit problem: -C+D (+/+!), 2-3*C+3*D (+), -A+B (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {C==0,D==n,A==-n,B==0} resulting limit problem: [solved] Solution: C / 0 D / n 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*D Rule guard: [ B>=1+A && D>=1+C ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)