/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^2), O(n^2)) 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^2, n^2). (0) CpxIntTrs (1) Koat Proof [FINISHED, 13 ms] (2) BOUNDS(1, n^2) (3) Loat Proof [FINISHED, 502 ms] (4) BOUNDS(n^2, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: evalwhile2start(A, B, C) -> Com_1(evalwhile2entryin(A, B, C)) :|: TRUE evalwhile2entryin(A, B, C) -> Com_1(evalwhile2bb4in(B, B, C)) :|: TRUE evalwhile2bb4in(A, B, C) -> Com_1(evalwhile2bb2in(A, B, B)) :|: A >= 1 evalwhile2bb4in(A, B, C) -> Com_1(evalwhile2returnin(A, B, C)) :|: 0 >= A evalwhile2bb2in(A, B, C) -> Com_1(evalwhile2bb1in(A, B, C)) :|: C >= 1 evalwhile2bb2in(A, B, C) -> Com_1(evalwhile2bb3in(A, B, C)) :|: 0 >= C evalwhile2bb1in(A, B, C) -> Com_1(evalwhile2bb2in(A, B, C - 1)) :|: TRUE evalwhile2bb3in(A, B, C) -> Com_1(evalwhile2bb4in(A - 1, B, C)) :|: TRUE evalwhile2returnin(A, B, C) -> Com_1(evalwhile2stop(A, B, C)) :|: TRUE The start-symbols are:[evalwhile2start_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 9*Ar_1 + 2*Ar_1^2 + 13) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: ?, Cost: 1) evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(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) evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: ?, Cost: 1) evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalwhile2start) = 2 Pol(evalwhile2entryin) = 2 Pol(evalwhile2bb4in) = 2 Pol(evalwhile2bb2in) = 2 Pol(evalwhile2returnin) = 1 Pol(evalwhile2bb1in) = 2 Pol(evalwhile2bb3in) = 2 Pol(evalwhile2stop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2)) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ] (Comp: 2, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: ?, Cost: 1) evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalwhile2start) = V_2 + 1 Pol(evalwhile2entryin) = V_2 + 1 Pol(evalwhile2bb4in) = V_1 + 1 Pol(evalwhile2bb2in) = V_1 Pol(evalwhile2returnin) = V_1 Pol(evalwhile2bb1in) = V_1 Pol(evalwhile2bb3in) = V_1 Pol(evalwhile2stop) = V_1 Pol(koat_start) = V_2 + 1 orients all transitions weakly and the transition evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2)) (Comp: Ar_1 + 1, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ] (Comp: 2, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: ?, Cost: 1) evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalwhile2bb3in) = 1 Pol(evalwhile2bb4in) = 0 Pol(evalwhile2bb2in) = 2 Pol(evalwhile2bb1in) = 2 and size complexities S("koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]", 0-0) = Ar_0 S("koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]", 0-1) = Ar_1 S("koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]", 0-2) = Ar_2 S("evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2))", 0-0) = ? S("evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2))", 0-1) = Ar_1 S("evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2))", 0-2) = ? S("evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2))", 0-0) = ? S("evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2))", 0-1) = Ar_1 S("evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2))", 0-2) = ? S("evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1))", 0-0) = ? S("evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1))", 0-1) = Ar_1 S("evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1))", 0-2) = ? S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]", 0-0) = ? S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]", 0-1) = Ar_1 S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]", 0-2) = ? S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]", 0-0) = ? S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]", 0-1) = Ar_1 S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]", 0-2) = ? S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ]", 0-0) = ? S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ]", 0-1) = Ar_1 S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ]", 0-2) = ? S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ]", 0-0) = ? S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ]", 0-1) = Ar_1 S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ]", 0-2) = Ar_1 S("evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2))", 0-0) = Ar_1 S("evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2))", 0-1) = Ar_1 S("evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2))", 0-2) = Ar_2 S("evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2))", 0-0) = Ar_0 S("evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2))", 0-1) = Ar_1 S("evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2))", 0-2) = Ar_2 orients the transitions evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2)) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1)) weakly and the transitions evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2)) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 1) evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2)) (Comp: Ar_1 + 1, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ] (Comp: 2, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: 2*Ar_1 + 2, Cost: 1) evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalwhile2bb2in) = V_3 + 1 Pol(evalwhile2bb1in) = V_3 and size complexities S("koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]", 0-0) = Ar_0 S("koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]", 0-1) = Ar_1 S("koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ]", 0-2) = Ar_2 S("evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2))", 0-0) = 3*Ar_1 + 162 S("evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2))", 0-1) = Ar_1 S("evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2))", 0-2) = ? S("evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2))", 0-0) = 3*Ar_1 + 18 S("evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2))", 0-1) = Ar_1 S("evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2))", 0-2) = ? S("evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1))", 0-0) = 3*Ar_1 + 18 S("evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1))", 0-1) = Ar_1 S("evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1))", 0-2) = ? S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]", 0-0) = 3*Ar_1 + 18 S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]", 0-1) = Ar_1 S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ]", 0-2) = ? S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]", 0-0) = 3*Ar_1 + 18 S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]", 0-1) = Ar_1 S("evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ]", 0-2) = ? S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ]", 0-0) = 3*Ar_1 + 54 S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ]", 0-1) = Ar_1 S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ]", 0-2) = ? S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ]", 0-0) = 3*Ar_1 + 18 S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ]", 0-1) = Ar_1 S("evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ]", 0-2) = Ar_1 S("evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2))", 0-0) = Ar_1 S("evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2))", 0-1) = Ar_1 S("evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2))", 0-2) = Ar_2 S("evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2))", 0-0) = Ar_0 S("evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2))", 0-1) = Ar_1 S("evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2))", 0-2) = Ar_2 orients the transitions evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1)) weakly and the transition evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] strictly and produces the following problem: 6: T: (Comp: 1, Cost: 1) evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2)) (Comp: Ar_1 + 1, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ] (Comp: 2, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] (Comp: Ar_1^2 + 2*Ar_1 + 1, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: 2*Ar_1 + 2, Cost: 1) evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 6 produces the following problem: 7: T: (Comp: 1, Cost: 1) evalwhile2start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2entryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalwhile2entryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_1, Ar_1, Ar_2)) (Comp: Ar_1 + 1, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_1)) [ Ar_0 >= 1 ] (Comp: 2, Cost: 1) evalwhile2bb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2returnin(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] (Comp: Ar_1^2 + 2*Ar_1 + 1, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb1in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalwhile2bb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb3in(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: Ar_1^2 + 2*Ar_1 + 1, Cost: 1) evalwhile2bb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb2in(Ar_0, Ar_1, Ar_2 - 1)) (Comp: 2*Ar_1 + 2, Cost: 1) evalwhile2bb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2bb4in(Ar_0 - 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalwhile2returnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2stop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalwhile2start(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 9*Ar_1 + 2*Ar_1^2 + 13 Time: 0.087 sec (SMT: 0.071 sec) ---------------------------------------- (2) BOUNDS(1, n^2) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalwhile2start 0: evalwhile2start -> evalwhile2entryin : [], cost: 1 1: evalwhile2entryin -> evalwhile2bb4in : A'=B, [], cost: 1 2: evalwhile2bb4in -> evalwhile2bb2in : C'=B, [ A>=1 ], cost: 1 3: evalwhile2bb4in -> evalwhile2returnin : [ 0>=A ], cost: 1 4: evalwhile2bb2in -> evalwhile2bb1in : [ C>=1 ], cost: 1 5: evalwhile2bb2in -> evalwhile2bb3in : [ 0>=C ], cost: 1 6: evalwhile2bb1in -> evalwhile2bb2in : C'=-1+C, [], cost: 1 7: evalwhile2bb3in -> evalwhile2bb4in : A'=-1+A, [], cost: 1 8: evalwhile2returnin -> evalwhile2stop : [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: evalwhile2start -> evalwhile2entryin : [], cost: 1 Removed unreachable and leaf rules: Start location: evalwhile2start 0: evalwhile2start -> evalwhile2entryin : [], cost: 1 1: evalwhile2entryin -> evalwhile2bb4in : A'=B, [], cost: 1 2: evalwhile2bb4in -> evalwhile2bb2in : C'=B, [ A>=1 ], cost: 1 4: evalwhile2bb2in -> evalwhile2bb1in : [ C>=1 ], cost: 1 5: evalwhile2bb2in -> evalwhile2bb3in : [ 0>=C ], cost: 1 6: evalwhile2bb1in -> evalwhile2bb2in : C'=-1+C, [], cost: 1 7: evalwhile2bb3in -> evalwhile2bb4in : A'=-1+A, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalwhile2start 9: evalwhile2start -> evalwhile2bb4in : A'=B, [], cost: 2 2: evalwhile2bb4in -> evalwhile2bb2in : C'=B, [ A>=1 ], cost: 1 10: evalwhile2bb2in -> evalwhile2bb2in : C'=-1+C, [ C>=1 ], cost: 2 11: evalwhile2bb2in -> evalwhile2bb4in : A'=-1+A, [ 0>=C ], cost: 2 Accelerating simple loops of location 3. Accelerating the following rules: 10: evalwhile2bb2in -> evalwhile2bb2in : C'=-1+C, [ C>=1 ], cost: 2 Accelerated rule 10 with metering function C, yielding the new rule 12. Removing the simple loops: 10. Accelerated all simple loops using metering functions (where possible): Start location: evalwhile2start 9: evalwhile2start -> evalwhile2bb4in : A'=B, [], cost: 2 2: evalwhile2bb4in -> evalwhile2bb2in : C'=B, [ A>=1 ], cost: 1 11: evalwhile2bb2in -> evalwhile2bb4in : A'=-1+A, [ 0>=C ], cost: 2 12: evalwhile2bb2in -> evalwhile2bb2in : C'=0, [ C>=1 ], cost: 2*C Chained accelerated rules (with incoming rules): Start location: evalwhile2start 9: evalwhile2start -> evalwhile2bb4in : A'=B, [], cost: 2 2: evalwhile2bb4in -> evalwhile2bb2in : C'=B, [ A>=1 ], cost: 1 13: evalwhile2bb4in -> evalwhile2bb2in : C'=0, [ A>=1 && B>=1 ], cost: 1+2*B 11: evalwhile2bb2in -> evalwhile2bb4in : A'=-1+A, [ 0>=C ], cost: 2 Eliminated locations (on tree-shaped paths): Start location: evalwhile2start 9: evalwhile2start -> evalwhile2bb4in : A'=B, [], cost: 2 14: evalwhile2bb4in -> evalwhile2bb4in : A'=-1+A, C'=B, [ A>=1 && 0>=B ], cost: 3 15: evalwhile2bb4in -> evalwhile2bb4in : A'=-1+A, C'=0, [ A>=1 && B>=1 ], cost: 3+2*B Accelerating simple loops of location 2. Accelerating the following rules: 14: evalwhile2bb4in -> evalwhile2bb4in : A'=-1+A, C'=B, [ A>=1 && 0>=B ], cost: 3 15: evalwhile2bb4in -> evalwhile2bb4in : A'=-1+A, C'=0, [ A>=1 && B>=1 ], cost: 3+2*B Accelerated rule 14 with metering function A, yielding the new rule 16. Accelerated rule 15 with metering function A, yielding the new rule 17. Removing the simple loops: 14 15. Accelerated all simple loops using metering functions (where possible): Start location: evalwhile2start 9: evalwhile2start -> evalwhile2bb4in : A'=B, [], cost: 2 16: evalwhile2bb4in -> evalwhile2bb4in : A'=0, C'=B, [ A>=1 && 0>=B ], cost: 3*A 17: evalwhile2bb4in -> evalwhile2bb4in : A'=0, C'=0, [ A>=1 && B>=1 ], cost: 2*A*B+3*A Chained accelerated rules (with incoming rules): Start location: evalwhile2start 9: evalwhile2start -> evalwhile2bb4in : A'=B, [], cost: 2 18: evalwhile2start -> evalwhile2bb4in : A'=0, C'=0, [ B>=1 ], cost: 2+3*B+2*B^2 Removed unreachable locations (and leaf rules with constant cost): Start location: evalwhile2start 18: evalwhile2start -> evalwhile2bb4in : A'=0, C'=0, [ B>=1 ], cost: 2+3*B+2*B^2 ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalwhile2start 18: evalwhile2start -> evalwhile2bb4in : A'=0, C'=0, [ B>=1 ], cost: 2+3*B+2*B^2 Computing asymptotic complexity for rule 18 Solved the limit problem by the following transformations: Created initial limit problem: 2+3*B+2*B^2 (+), B (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {B==n} resulting limit problem: [solved] Solution: B / n Resulting cost 2+2*n^2+3*n has complexity: Poly(n^2) Found new complexity Poly(n^2). Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Poly(n^2) Cpx degree: 2 Solved cost: 2+2*n^2+3*n Rule cost: 2+3*B+2*B^2 Rule guard: [ B>=1 ] WORST_CASE(Omega(n^2),?) ---------------------------------------- (4) BOUNDS(n^2, INF)