/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 185 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 420 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: start(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: 0 >= A + 1 && B >= C && B <= C && D >= A && D <= A start(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: 0 >= C + 1 && B >= C && B <= C && D >= A && D <= A start(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: A >= 0 && A + 2 >= C && C >= 0 && C + 2 >= A && B >= C && B <= C && D >= A && D <= A start(A, B, C, D) -> Com_1(lbl81(A, B, C, 1 + D)) :|: A >= 0 && C >= A + 3 && B >= C && B <= C && D >= A && D <= A start(A, B, C, D) -> Com_1(lbl91(A, 1 + B, C, D)) :|: A >= C + 3 && C >= 0 && B >= C && B <= C && D >= A && D <= A lbl81(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: C >= 3 + A && A >= 0 && D + 2 >= C && D + 2 <= C && B >= C && B <= C lbl81(A, B, C, D) -> Com_1(lbl81(A, B, C, 1 + D)) :|: C >= D + 3 && D >= A + 1 && A >= 0 && C >= D + 2 && B >= C && B <= C lbl81(A, B, C, D) -> Com_1(lbl91(A, 1 + B, C, D)) :|: D >= C + 3 && D >= A + 1 && A >= 0 && C >= D + 2 && B >= C && B <= C lbl91(A, B, C, D) -> Com_1(stop(A, B, C, D)) :|: A >= 3 + C && C >= 0 && B + 2 >= A && B + 2 <= A && D >= A && D <= A lbl91(A, B, C, D) -> Com_1(lbl81(A, B, C, 1 + D)) :|: B >= A + 3 && A >= B + 2 && B >= C + 1 && C >= 0 && D >= A && D <= A lbl91(A, B, C, D) -> Com_1(lbl91(A, 1 + B, C, D)) :|: A >= B + 3 && A >= B + 2 && B >= C + 1 && C >= 0 && D >= A && D <= A start0(A, B, C, D) -> Com_1(start(A, C, C, A)) :|: TRUE The start-symbols are:[start0_4] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 2*Ar_0 + 2*Ar_2 + 8) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_0 + 1 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_2 + 1 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 0 /\ Ar_0 + 2 >= Ar_2 /\ Ar_2 >= 0 /\ Ar_2 + 2 >= Ar_0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_0 >= 0 /\ Ar_2 >= Ar_0 + 3 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 3 /\ Ar_2 >= 0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 + 3 /\ Ar_0 >= 0 /\ Ar_3 + 2 = Ar_2 /\ Ar_1 = Ar_2 ] (Comp: ?, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 3 /\ Ar_3 >= Ar_0 + 1 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_3 + 2 /\ Ar_1 = Ar_2 ] (Comp: ?, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_3 >= Ar_2 + 3 /\ Ar_3 >= Ar_0 + 1 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_3 + 2 /\ Ar_1 = Ar_2 ] (Comp: ?, Cost: 1) lbl91(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 /\ Ar_1 + 2 = Ar_0 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) lbl91(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_1 >= Ar_0 + 3 /\ Ar_0 >= Ar_1 + 2 /\ Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) lbl91(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 3 /\ Ar_0 >= Ar_1 + 2 /\ Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_0, Ar_2, Ar_2, Ar_0)) (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 Testing for reachability in the complexity graph removes the following transitions from problem 1: lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_3 >= Ar_2 + 3 /\ Ar_3 >= Ar_0 + 1 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_3 + 2 /\ Ar_1 = Ar_2 ] lbl91(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_1 >= Ar_0 + 3 /\ Ar_0 >= Ar_1 + 2 /\ Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_3 = Ar_0 ] We thus obtain the following problem: 2: T: (Comp: ?, Cost: 1) lbl91(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 3 /\ Ar_0 >= Ar_1 + 2 /\ Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) lbl91(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 /\ Ar_1 + 2 = Ar_0 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 3 /\ Ar_3 >= Ar_0 + 1 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_3 + 2 /\ Ar_1 = Ar_2 ] (Comp: ?, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 + 3 /\ Ar_0 >= 0 /\ Ar_3 + 2 = Ar_2 /\ Ar_1 = Ar_2 ] (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 3 /\ Ar_2 >= 0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_0 >= 0 /\ Ar_2 >= Ar_0 + 3 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 0 /\ Ar_0 + 2 >= Ar_2 /\ Ar_2 >= 0 /\ Ar_2 + 2 >= Ar_0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_2 + 1 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_0 + 1 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_0, Ar_2, Ar_2, Ar_0)) (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 2 produces the following problem: 3: T: (Comp: ?, Cost: 1) lbl91(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 3 /\ Ar_0 >= Ar_1 + 2 /\ Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) lbl91(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 /\ Ar_1 + 2 = Ar_0 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 3 /\ Ar_3 >= Ar_0 + 1 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_3 + 2 /\ Ar_1 = Ar_2 ] (Comp: ?, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 + 3 /\ Ar_0 >= 0 /\ Ar_3 + 2 = Ar_2 /\ Ar_1 = Ar_2 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 3 /\ Ar_2 >= 0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_0 >= 0 /\ Ar_2 >= Ar_0 + 3 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 0 /\ Ar_0 + 2 >= Ar_2 /\ Ar_2 >= 0 /\ Ar_2 + 2 >= Ar_0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_2 + 1 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_0 + 1 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_0, Ar_2, Ar_2, Ar_0)) (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(lbl91) = 1 Pol(stop) = 0 Pol(lbl81) = 1 Pol(start) = 1 Pol(start0) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transitions lbl91(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 /\ Ar_1 + 2 = Ar_0 /\ Ar_3 = Ar_0 ] lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 + 3 /\ Ar_0 >= 0 /\ Ar_3 + 2 = Ar_2 /\ Ar_1 = Ar_2 ] strictly and produces the following problem: 4: T: (Comp: ?, Cost: 1) lbl91(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 3 /\ Ar_0 >= Ar_1 + 2 /\ Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) lbl91(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 /\ Ar_1 + 2 = Ar_0 /\ Ar_3 = Ar_0 ] (Comp: ?, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 3 /\ Ar_3 >= Ar_0 + 1 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_3 + 2 /\ Ar_1 = Ar_2 ] (Comp: 1, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 + 3 /\ Ar_0 >= 0 /\ Ar_3 + 2 = Ar_2 /\ Ar_1 = Ar_2 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 3 /\ Ar_2 >= 0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_0 >= 0 /\ Ar_2 >= Ar_0 + 3 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 0 /\ Ar_0 + 2 >= Ar_2 /\ Ar_2 >= 0 /\ Ar_2 + 2 >= Ar_0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_2 + 1 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_0 + 1 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_0, Ar_2, Ar_2, Ar_0)) (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(lbl91) = 2*V_1 - V_2 + 2*V_3 - V_4 Pol(stop) = 2*V_1 - V_2 + 2*V_3 - V_4 Pol(lbl81) = 2*V_1 + V_2 - V_4 Pol(start) = V_1 - V_2 + 2*V_3 Pol(start0) = V_1 + V_3 Pol(koat_start) = V_1 + V_3 orients all transitions weakly and the transitions lbl91(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 3 /\ Ar_0 >= Ar_1 + 2 /\ Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_3 = Ar_0 ] lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 3 /\ Ar_3 >= Ar_0 + 1 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_3 + 2 /\ Ar_1 = Ar_2 ] strictly and produces the following problem: 5: T: (Comp: Ar_0 + Ar_2, Cost: 1) lbl91(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_1 + 3 /\ Ar_0 >= Ar_1 + 2 /\ Ar_1 >= Ar_2 + 1 /\ Ar_2 >= 0 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) lbl91(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 /\ Ar_1 + 2 = Ar_0 /\ Ar_3 = Ar_0 ] (Comp: Ar_0 + Ar_2, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 3 /\ Ar_3 >= Ar_0 + 1 /\ Ar_0 >= 0 /\ Ar_2 >= Ar_3 + 2 /\ Ar_1 = Ar_2 ] (Comp: 1, Cost: 1) lbl81(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= Ar_0 + 3 /\ Ar_0 >= 0 /\ Ar_3 + 2 = Ar_2 /\ Ar_1 = Ar_2 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl91(Ar_0, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= Ar_2 + 3 /\ Ar_2 >= 0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(lbl81(Ar_0, Ar_1, Ar_2, Ar_3 + 1)) [ Ar_0 >= 0 /\ Ar_2 >= Ar_0 + 3 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_0 >= 0 /\ Ar_0 + 2 >= Ar_2 /\ Ar_2 >= 0 /\ Ar_2 + 2 >= Ar_0 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_2 + 1 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(stop(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_0 + 1 /\ Ar_1 = Ar_2 /\ Ar_3 = Ar_0 ] (Comp: 1, Cost: 1) start0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(start(Ar_0, Ar_2, Ar_2, Ar_0)) (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 2*Ar_0 + 2*Ar_2 + 8 Time: 0.170 sec (SMT: 0.143 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: start0 0: start -> stop : [ 0>=1+A && B==C && D==A ], cost: 1 1: start -> stop : [ 0>=1+C && B==C && D==A ], cost: 1 2: start -> stop : [ A>=0 && 2+A>=C && C>=0 && 2+C>=A && B==C && D==A ], cost: 1 3: start -> lbl81 : D'=1+D, [ A>=0 && C>=3+A && B==C && D==A ], cost: 1 4: start -> lbl91 : B'=1+B, [ A>=3+C && C>=0 && B==C && D==A ], cost: 1 5: lbl81 -> stop : [ C>=3+A && A>=0 && 2+D==C && B==C ], cost: 1 6: lbl81 -> lbl81 : D'=1+D, [ C>=3+D && D>=1+A && A>=0 && C>=2+D && B==C ], cost: 1 7: lbl81 -> lbl91 : B'=1+B, [ D>=3+C && D>=1+A && A>=0 && C>=2+D && B==C ], cost: 1 8: lbl91 -> stop : [ A>=3+C && C>=0 && 2+B==A && D==A ], cost: 1 9: lbl91 -> lbl81 : D'=1+D, [ B>=3+A && A>=2+B && B>=1+C && C>=0 && D==A ], cost: 1 10: lbl91 -> lbl91 : B'=1+B, [ A>=3+B && A>=2+B && B>=1+C && C>=0 && D==A ], cost: 1 11: start0 -> start : B'=C, D'=A, [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 11: start0 -> start : B'=C, D'=A, [], cost: 1 Removed unreachable and leaf rules: Start location: start0 3: start -> lbl81 : D'=1+D, [ A>=0 && C>=3+A && B==C && D==A ], cost: 1 4: start -> lbl91 : B'=1+B, [ A>=3+C && C>=0 && B==C && D==A ], cost: 1 6: lbl81 -> lbl81 : D'=1+D, [ C>=3+D && D>=1+A && A>=0 && C>=2+D && B==C ], cost: 1 7: lbl81 -> lbl91 : B'=1+B, [ D>=3+C && D>=1+A && A>=0 && C>=2+D && B==C ], cost: 1 9: lbl91 -> lbl81 : D'=1+D, [ B>=3+A && A>=2+B && B>=1+C && C>=0 && D==A ], cost: 1 10: lbl91 -> lbl91 : B'=1+B, [ A>=3+B && A>=2+B && B>=1+C && C>=0 && D==A ], cost: 1 11: start0 -> start : B'=C, D'=A, [], cost: 1 Removed rules with unsatisfiable guard: Start location: start0 3: start -> lbl81 : D'=1+D, [ A>=0 && C>=3+A && B==C && D==A ], cost: 1 4: start -> lbl91 : B'=1+B, [ A>=3+C && C>=0 && B==C && D==A ], cost: 1 6: lbl81 -> lbl81 : D'=1+D, [ C>=3+D && D>=1+A && A>=0 && C>=2+D && B==C ], cost: 1 10: lbl91 -> lbl91 : B'=1+B, [ A>=3+B && A>=2+B && B>=1+C && C>=0 && D==A ], cost: 1 11: start0 -> start : B'=C, D'=A, [], cost: 1 Simplified all rules, resulting in: Start location: start0 3: start -> lbl81 : D'=1+D, [ A>=0 && C>=3+A && B==C && D==A ], cost: 1 4: start -> lbl91 : B'=1+B, [ A>=3+C && C>=0 && B==C && D==A ], cost: 1 6: lbl81 -> lbl81 : D'=1+D, [ C>=3+D && D>=1+A && A>=0 && B==C ], cost: 1 10: lbl91 -> lbl91 : B'=1+B, [ A>=3+B && B>=1+C && C>=0 && D==A ], cost: 1 11: start0 -> start : B'=C, D'=A, [], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 6: lbl81 -> lbl81 : D'=1+D, [ C>=3+D && D>=1+A && A>=0 && B==C ], cost: 1 Accelerated rule 6 with metering function -2+C-D, yielding the new rule 12. Removing the simple loops: 6. Accelerating simple loops of location 2. Accelerating the following rules: 10: lbl91 -> lbl91 : B'=1+B, [ A>=3+B && B>=1+C && C>=0 && D==A ], cost: 1 Accelerated rule 10 with metering function -2+A-B, yielding the new rule 13. Removing the simple loops: 10. Accelerated all simple loops using metering functions (where possible): Start location: start0 3: start -> lbl81 : D'=1+D, [ A>=0 && C>=3+A && B==C && D==A ], cost: 1 4: start -> lbl91 : B'=1+B, [ A>=3+C && C>=0 && B==C && D==A ], cost: 1 12: lbl81 -> lbl81 : D'=-2+C, [ C>=3+D && D>=1+A && A>=0 && B==C ], cost: -2+C-D 13: lbl91 -> lbl91 : B'=-2+A, [ A>=3+B && B>=1+C && C>=0 && D==A ], cost: -2+A-B 11: start0 -> start : B'=C, D'=A, [], cost: 1 Chained accelerated rules (with incoming rules): Start location: start0 3: start -> lbl81 : D'=1+D, [ A>=0 && C>=3+A && B==C && D==A ], cost: 1 4: start -> lbl91 : B'=1+B, [ A>=3+C && C>=0 && B==C && D==A ], cost: 1 14: start -> lbl81 : D'=-2+C, [ A>=0 && C>=3+A && B==C && D==A && C>=4+D ], cost: -2+C-D 15: start -> lbl91 : B'=-2+A, [ A>=3+C && C>=0 && B==C && D==A && A>=4+B ], cost: -2+A-B 11: start0 -> start : B'=C, D'=A, [], cost: 1 Removed unreachable locations (and leaf rules with constant cost): Start location: start0 14: start -> lbl81 : D'=-2+C, [ A>=0 && C>=3+A && B==C && D==A && C>=4+D ], cost: -2+C-D 15: start -> lbl91 : B'=-2+A, [ A>=3+C && C>=0 && B==C && D==A && A>=4+B ], cost: -2+A-B 11: start0 -> start : B'=C, D'=A, [], cost: 1 Eliminated locations (on tree-shaped paths): Start location: start0 16: start0 -> lbl81 : B'=C, D'=-2+C, [ A>=0 && C>=4+A ], cost: -1+C-A 17: start0 -> lbl91 : B'=-2+A, D'=A, [ C>=0 && A>=4+C ], cost: -1-C+A ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: start0 16: start0 -> lbl81 : B'=C, D'=-2+C, [ A>=0 && C>=4+A ], cost: -1+C-A 17: start0 -> lbl91 : B'=-2+A, D'=A, [ C>=0 && A>=4+C ], cost: -1-C+A Computing asymptotic complexity for rule 16 Solved the limit problem by the following transformations: Created initial limit problem: -3+C-A (+/+!), -1+C-A (+), 1+A (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {C==n,A==0} resulting limit problem: [solved] Solution: C / n A / 0 Resulting cost -1+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: -1+n Rule cost: -1+C-A Rule guard: [ A>=0 && C>=4+A ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)