/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1). (0) CpxIntTrs (1) Koat Proof [FINISHED, 8 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A, B, C) -> Com_1(f5(A, 0, C)) :|: A >= 128 f0(A, B, C) -> Com_1(f7(A, 0, D)) :|: 127 >= A f7(A, B, C) -> Com_1(f7(A, B + 1, C + 1)) :|: 35 >= B f7(A, B, C) -> Com_1(f5(A, B, C)) :|: B >= 36 The start-symbols are:[f0_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 39) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, 0, ar_2)) [ ar_0 >= 128 ] (Comp: ?, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f7(ar_0, 0, d)) [ 127 >= ar_0 ] (Comp: ?, Cost: 1) f7(ar_0, ar_1, ar_2) -> Com_1(f7(ar_0, ar_1 + 1, ar_2 + 1)) [ 35 >= ar_1 ] (Comp: ?, Cost: 1) f7(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1, ar_2)) [ ar_1 >= 36 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(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) f0(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, 0, ar_2)) [ ar_0 >= 128 ] (Comp: 1, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f7(ar_0, 0, d)) [ 127 >= ar_0 ] (Comp: ?, Cost: 1) f7(ar_0, ar_1, ar_2) -> Com_1(f7(ar_0, ar_1 + 1, ar_2 + 1)) [ 35 >= ar_1 ] (Comp: ?, Cost: 1) f7(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1, ar_2)) [ ar_1 >= 36 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f0) = 1 Pol(f5) = 0 Pol(f7) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f7(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1, ar_2)) [ ar_1 >= 36 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, 0, ar_2)) [ ar_0 >= 128 ] (Comp: 1, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f7(ar_0, 0, d)) [ 127 >= ar_0 ] (Comp: ?, Cost: 1) f7(ar_0, ar_1, ar_2) -> Com_1(f7(ar_0, ar_1 + 1, ar_2 + 1)) [ 35 >= ar_1 ] (Comp: 1, Cost: 1) f7(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1, ar_2)) [ ar_1 >= 36 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f0) = 36 Pol(f5) = -V_2 Pol(f7) = -V_2 + 36 Pol(koat_start) = 36 orients all transitions weakly and the transition f7(ar_0, ar_1, ar_2) -> Com_1(f7(ar_0, ar_1 + 1, ar_2 + 1)) [ 35 >= ar_1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, 0, ar_2)) [ ar_0 >= 128 ] (Comp: 1, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f7(ar_0, 0, d)) [ 127 >= ar_0 ] (Comp: 36, Cost: 1) f7(ar_0, ar_1, ar_2) -> Com_1(f7(ar_0, ar_1 + 1, ar_2 + 1)) [ 35 >= ar_1 ] (Comp: 1, Cost: 1) f7(ar_0, ar_1, ar_2) -> Com_1(f5(ar_0, ar_1, ar_2)) [ ar_1 >= 36 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2) -> Com_1(f0(ar_0, ar_1, ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 39 Time: 0.067 sec (SMT: 0.064 sec) ---------------------------------------- (2) BOUNDS(1, 1)