/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(?, O(1)) proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).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, 74 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A, B, C) -> Com_1(f8(0, B, C)) :|: TRUE f8(A, B, C) -> Com_1(f14(A, A, C)) :|: 9 >= A && 9 >= D f8(A, B, C) -> Com_1(f14(A, A, C)) :|: 9 >= A f23(A, B, C) -> Com_1(f28(A, B, D)) :|: 9 >= A && 0 >= E + 1 f23(A, B, C) -> Com_1(f28(A, B, D)) :|: 9 >= A f23(A, B, C) -> Com_1(f23(A + 1, B, C)) :|: 9 >= A f28(A, B, C) -> Com_1(f23(A + 1, B, C)) :|: TRUE f28(A, B, C) -> Com_1(f23(A + 1, B, C)) :|: 8 >= D f23(A, B, C) -> Com_1(f38(A, B, C)) :|: A >= 10 f8(A, B, C) -> Com_1(f8(A + 1, A, C)) :|: 9 >= A f14(A, B, C) -> Com_1(f8(A + 1, B, C)) :|: TRUE f14(A, B, C) -> Com_1(f8(A + 1, B, C)) :|: 8 >= D f8(A, B, C) -> Com_1(f23(0, B, C)) :|: A >= 10 The start-symbols are:[f0_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 145) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(ar_0, ar_1, ar_2) -> Com_1(f8(0, ar_1, ar_2)) (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f14(ar_0, ar_0, ar_2)) [ 9 >= ar_0 /\ 9 >= d ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f14(ar_0, ar_0, ar_2)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0, ar_1, ar_2) -> Com_1(f28(ar_0, ar_1, d)) [ 9 >= ar_0 /\ 0 >= e + 1 ] (Comp: ?, Cost: 1) f23(ar_0, ar_1, ar_2) -> Com_1(f28(ar_0, ar_1, d)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0, ar_1, ar_2) -> Com_1(f23(ar_0 + 1, ar_1, ar_2)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f28(ar_0, ar_1, ar_2) -> Com_1(f23(ar_0 + 1, ar_1, ar_2)) (Comp: ?, Cost: 1) f28(ar_0, ar_1, ar_2) -> Com_1(f23(ar_0 + 1, ar_1, ar_2)) [ 8 >= d ] (Comp: ?, Cost: 1) f23(ar_0, ar_1, ar_2) -> Com_1(f38(ar_0, ar_1, ar_2)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0 + 1, ar_0, ar_2)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f14(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0 + 1, ar_1, ar_2)) (Comp: ?, Cost: 1) f14(ar_0, ar_1, ar_2) -> Com_1(f8(ar_0 + 1, ar_1, ar_2)) [ 8 >= d ] (Comp: ?, Cost: 1) f8(ar_0, ar_1, ar_2) -> Com_1(f23(0, ar_1, ar_2)) [ ar_0 >= 10 ] (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 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [ar_0]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f23(0)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) [ 8 >= d ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f38(ar_0)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) [ 8 >= d ] (Comp: ?, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f23(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 /\ 0 >= e + 1 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 /\ 9 >= d ] (Comp: ?, Cost: 1) f0(ar_0) -> Com_1(f8(0)) start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 2 produces the following problem: 3: T: (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f23(0)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) [ 8 >= d ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f38(ar_0)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) [ 8 >= d ] (Comp: ?, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f23(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 /\ 0 >= e + 1 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 /\ 9 >= d ] (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f8(0)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 2 Pol(f0) = 2 Pol(f8) = 2 Pol(f23) = 1 Pol(f14) = 2 Pol(f38) = 0 Pol(f28) = 1 orients all transitions weakly and the transitions f8(ar_0) -> Com_1(f23(0)) [ ar_0 >= 10 ] f23(ar_0) -> Com_1(f38(ar_0)) [ ar_0 >= 10 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] (Comp: 2, Cost: 1) f8(ar_0) -> Com_1(f23(0)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) [ 8 >= d ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: 2, Cost: 1) f23(ar_0) -> Com_1(f38(ar_0)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) [ 8 >= d ] (Comp: ?, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f23(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 /\ 0 >= e + 1 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 /\ 9 >= d ] (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f8(0)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 10 Pol(f0) = 10 Pol(f8) = 10 Pol(f23) = -V_1 + 10 Pol(f14) = 10 Pol(f38) = -V_1 Pol(f28) = -V_1 + 9 orients all transitions weakly and the transitions f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 /\ 0 >= e + 1 ] f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 ] f23(ar_0) -> Com_1(f23(ar_0 + 1)) [ 9 >= ar_0 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] (Comp: 2, Cost: 1) f8(ar_0) -> Com_1(f23(0)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) [ 8 >= d ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: 2, Cost: 1) f23(ar_0) -> Com_1(f38(ar_0)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) [ 8 >= d ] (Comp: ?, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f23(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 /\ 0 >= e + 1 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 /\ 9 >= d ] (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f8(0)) start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 5 produces the following problem: 6: T: (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] (Comp: 2, Cost: 1) f8(ar_0) -> Com_1(f23(0)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) [ 8 >= d ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: 2, Cost: 1) f23(ar_0) -> Com_1(f38(ar_0)) [ ar_0 >= 10 ] (Comp: 20, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) [ 8 >= d ] (Comp: 20, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f23(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 /\ 0 >= e + 1 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 /\ 9 >= d ] (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f8(0)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f8) = -V_1 + 10 Pol(f14) = -V_1 + 9 and size complexities S("f0(ar_0) -> Com_1(f8(0))", 0-0) = 0 S("f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 /\\ 9 >= d ]", 0-0) = ? S("f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 ]", 0-0) = ? S("f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 /\\ 0 >= e + 1 ]", 0-0) = 50 S("f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 ]", 0-0) = 50 S("f23(ar_0) -> Com_1(f23(ar_0 + 1)) [ 9 >= ar_0 ]", 0-0) = 50 S("f28(ar_0) -> Com_1(f23(ar_0 + 1))", 0-0) = 50 S("f28(ar_0) -> Com_1(f23(ar_0 + 1)) [ 8 >= d ]", 0-0) = 50 S("f23(ar_0) -> Com_1(f38(ar_0)) [ ar_0 >= 10 ]", 0-0) = 50 S("f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 9 >= ar_0 ]", 0-0) = ? S("f14(ar_0) -> Com_1(f8(ar_0 + 1))", 0-0) = ? S("f14(ar_0) -> Com_1(f8(ar_0 + 1)) [ 8 >= d ]", 0-0) = ? S("f8(ar_0) -> Com_1(f23(0)) [ ar_0 >= 10 ]", 0-0) = 0 S("koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ]", 0-0) = ar_0 orients the transitions f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 9 >= ar_0 ] f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 /\ 9 >= d ] f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 ] f14(ar_0) -> Com_1(f8(ar_0 + 1)) [ 8 >= d ] f14(ar_0) -> Com_1(f8(ar_0 + 1)) weakly and the transitions f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 9 >= ar_0 ] f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 /\ 9 >= d ] f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 ] strictly and produces the following problem: 7: T: (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] (Comp: 2, Cost: 1) f8(ar_0) -> Com_1(f23(0)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) [ 8 >= d ] (Comp: ?, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) (Comp: 10, Cost: 1) f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: 2, Cost: 1) f23(ar_0) -> Com_1(f38(ar_0)) [ ar_0 >= 10 ] (Comp: 20, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) [ 8 >= d ] (Comp: 20, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f23(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 /\ 0 >= e + 1 ] (Comp: 10, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 /\ 9 >= d ] (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f8(0)) 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) -> Com_1(f0(ar_0)) [ 0 <= 0 ] (Comp: 2, Cost: 1) f8(ar_0) -> Com_1(f23(0)) [ ar_0 >= 10 ] (Comp: 20, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) [ 8 >= d ] (Comp: 20, Cost: 1) f14(ar_0) -> Com_1(f8(ar_0 + 1)) (Comp: 10, Cost: 1) f8(ar_0) -> Com_1(f8(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: 2, Cost: 1) f23(ar_0) -> Com_1(f38(ar_0)) [ ar_0 >= 10 ] (Comp: 20, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) [ 8 >= d ] (Comp: 20, Cost: 1) f28(ar_0) -> Com_1(f23(ar_0 + 1)) (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f23(ar_0 + 1)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f23(ar_0) -> Com_1(f28(ar_0)) [ 9 >= ar_0 /\ 0 >= e + 1 ] (Comp: 10, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f8(ar_0) -> Com_1(f14(ar_0)) [ 9 >= ar_0 /\ 9 >= d ] (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f8(0)) start location: koat_start leaf cost: 0 Complexity upper bound 145 Time: 0.136 sec (SMT: 0.127 sec) ---------------------------------------- (2) BOUNDS(1, 1)