/export/starexec/sandbox/solver/bin/starexec_run_complexity /export/starexec/sandbox/benchmark/theBenchmark.koat /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- WORST_CASE(?, O(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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 119 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: start(A, B, C, D) -> Com_1(lbl51(E, B, 0, D)) :|: A >= B && A <= B && C >= D && C <= D lbl51(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: C >= A && C >= 0 && 9 >= C lbl51(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: A >= C + 3 && C >= 0 && 9 >= C lbl51(A, B, C, D) -> Com_1(cut(A, B, C, D)) :|: A >= C + 1 && C + 2 >= A && 9 >= A && C >= 0 && 9 >= C lbl51(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: A >= 10 && A >= C + 1 && C + 2 >= A && C >= 0 && 9 >= C cut(A, B, C, D) -> Com_1(lbl51(E, B, A, D)) :|: C + 2 >= A && 9 >= A && C >= 0 && A >= C + 1 start0(A, B, C, D) -> Com_1(start(B, B, D, D)) :|: TRUE The start-symbols are:[start0_4] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 43) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl51(Fresh_1, Ar_1, 0, Ar_3)) [ Ar_0 = Ar_1 /\ Ar_2 = Ar_3 ] (Comp: ?, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: ?, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 3 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: ?, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(cut(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 1 /\ Ar_2 + 2 >= Ar_0 /\ 9 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: ?, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 10 /\ Ar_0 >= Ar_2 + 1 /\ Ar_2 + 2 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: ?, Cost: 1) cut(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl51(Fresh_0, Ar_1, Ar_0, Ar_3)) [ Ar_2 + 2 >= Ar_0 /\ 9 >= Ar_0 /\ Ar_2 >= 0 /\ Ar_0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_1, Ar_1, Ar_3, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start0(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) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl51(Fresh_1, Ar_1, 0, Ar_3)) [ Ar_0 = Ar_1 /\ Ar_2 = Ar_3 ] (Comp: ?, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: ?, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 3 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: ?, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(cut(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 1 /\ Ar_2 + 2 >= Ar_0 /\ 9 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: ?, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 10 /\ Ar_0 >= Ar_2 + 1 /\ Ar_2 + 2 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: ?, Cost: 1) cut(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl51(Fresh_0, Ar_1, Ar_0, Ar_3)) [ Ar_2 + 2 >= Ar_0 /\ 9 >= Ar_0 /\ Ar_2 >= 0 /\ Ar_0 >= Ar_2 + 1 ] (Comp: 1, Cost: 1) start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_1, Ar_1, Ar_3, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(start) = 1 Pol(lbl51) = 1 Pol(stop) = 0 Pol(cut) = 1 Pol(start0) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transitions lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 10 /\ Ar_0 >= Ar_2 + 1 /\ Ar_2 + 2 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 3 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl51(Fresh_1, Ar_1, 0, Ar_3)) [ Ar_0 = Ar_1 /\ Ar_2 = Ar_3 ] (Comp: 1, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: 1, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 3 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: ?, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(cut(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 1 /\ Ar_2 + 2 >= Ar_0 /\ 9 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: 1, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 10 /\ Ar_0 >= Ar_2 + 1 /\ Ar_2 + 2 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: ?, Cost: 1) cut(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl51(Fresh_0, Ar_1, Ar_0, Ar_3)) [ Ar_2 + 2 >= Ar_0 /\ 9 >= Ar_0 /\ Ar_2 >= 0 /\ Ar_0 >= Ar_2 + 1 ] (Comp: 1, Cost: 1) start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_1, Ar_1, Ar_3, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(start) = 19 Pol(lbl51) = -2*V_3 + 19 Pol(stop) = 1 Pol(cut) = -2*V_3 + 18 Pol(start0) = 19 Pol(koat_start) = 19 orients all transitions weakly and the transitions lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(cut(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 1 /\ Ar_2 + 2 >= Ar_0 /\ 9 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] cut(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl51(Fresh_0, Ar_1, Ar_0, Ar_3)) [ Ar_2 + 2 >= Ar_0 /\ 9 >= Ar_0 /\ Ar_2 >= 0 /\ Ar_0 >= Ar_2 + 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl51(Fresh_1, Ar_1, 0, Ar_3)) [ Ar_0 = Ar_1 /\ Ar_2 = Ar_3 ] (Comp: 1, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: 1, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 3 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: 19, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(cut(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 1 /\ Ar_2 + 2 >= Ar_0 /\ 9 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: 1, Cost: 1) lbl51(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 10 /\ Ar_0 >= Ar_2 + 1 /\ Ar_2 + 2 >= Ar_0 /\ Ar_2 >= 0 /\ 9 >= Ar_2 ] (Comp: 19, Cost: 1) cut(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl51(Fresh_0, Ar_1, Ar_0, Ar_3)) [ Ar_2 + 2 >= Ar_0 /\ 9 >= Ar_0 /\ Ar_2 >= 0 /\ Ar_0 >= Ar_2 + 1 ] (Comp: 1, Cost: 1) start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_1, Ar_1, Ar_3, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 43 Time: 0.113 sec (SMT: 0.093 sec) ---------------------------------------- (2) BOUNDS(1, 1)