/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).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, 123 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A, B, C, D, E, F, G) -> Com_1(f9(0, 0, H, D, E, F, G)) :|: TRUE f9(A, B, C, D, E, F, G) -> Com_1(f10(A, B, C, C, E, F, G)) :|: 0 >= C + 1 f9(A, B, C, D, E, F, G) -> Com_1(f10(A, B, C, C, E, F, G)) :|: C >= 1 f10(A, B, C, D, E, F, G) -> Com_1(f9(A + 1, A + 1, H, D, E, F, G)) :|: 9 >= A f16(A, B, C, D, E, F, G) -> Com_1(f28(A, B, C, D, E, F, G)) :|: A >= 10 f16(A, B, C, D, E, F, G) -> Com_1(f16(A + 1, B, C, D, A, H, H)) :|: 9 >= A && 0 >= H + 1 f16(A, B, C, D, E, F, G) -> Com_1(f16(A + 1, B, C, D, A, H, H)) :|: 9 >= A && H >= 1 f16(A, B, C, D, E, F, G) -> Com_1(f28(A, B, C, D, A, 0, 0)) :|: 9 >= A f10(A, B, C, D, E, F, G) -> Com_1(f16(0, B, C, D, E, F, G)) :|: A >= 10 f9(A, B, C, D, E, F, G) -> Com_1(f16(0, B, 0, 0, E, F, G)) :|: C >= 0 && C <= 0 The start-symbols are:[f0_7] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 61) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f9(0, 0, h, ar_3, ar_4, ar_5, ar_6)) (Comp: ?, Cost: 1) f9(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f10(ar_0, ar_1, ar_2, ar_2, ar_4, ar_5, ar_6)) [ 0 >= ar_2 + 1 ] (Comp: ?, Cost: 1) f9(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f10(ar_0, ar_1, ar_2, ar_2, ar_4, ar_5, ar_6)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) f10(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f9(ar_0 + 1, ar_0 + 1, h, ar_3, ar_4, ar_5, ar_6)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f28(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f16(ar_0 + 1, ar_1, ar_2, ar_3, ar_0, h, h)) [ 9 >= ar_0 /\ 0 >= h + 1 ] (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f16(ar_0 + 1, ar_1, ar_2, ar_3, ar_0, h, h)) [ 9 >= ar_0 /\ h >= 1 ] (Comp: ?, Cost: 1) f16(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f28(ar_0, ar_1, ar_2, ar_3, ar_0, 0, 0)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f10(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f16(0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f9(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f16(0, ar_1, 0, 0, ar_4, ar_5, ar_6)) [ ar_2 = 0 ] (Comp: 1, Cost: 0) koat_start(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6) -> Com_1(f0(ar_0, ar_1, ar_2, ar_3, ar_4, ar_5, ar_6)) [ 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, ar_2]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_2) -> Com_1(f0(ar_0, ar_2)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f16(0, 0)) [ ar_2 = 0 ] (Comp: ?, Cost: 1) f10(ar_0, ar_2) -> Com_1(f16(0, ar_2)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ h >= 1 ] (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ 0 >= h + 1 ] (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ 0 >= ar_2 + 1 ] (Comp: ?, Cost: 1) f0(ar_0, ar_2) -> Com_1(f9(0, h)) 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, ar_2) -> Com_1(f0(ar_0, ar_2)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f16(0, 0)) [ ar_2 = 0 ] (Comp: ?, Cost: 1) f10(ar_0, ar_2) -> Com_1(f16(0, ar_2)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ h >= 1 ] (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ 0 >= h + 1 ] (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ 0 >= ar_2 + 1 ] (Comp: 1, Cost: 1) f0(ar_0, ar_2) -> Com_1(f9(0, h)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 2 Pol(f0) = 2 Pol(f9) = 2 Pol(f16) = 1 Pol(f10) = 2 Pol(f28) = 0 orients all transitions weakly and the transitions f9(ar_0, ar_2) -> Com_1(f16(0, 0)) [ ar_2 = 0 ] f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ 9 >= ar_0 ] f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ ar_0 >= 10 ] f10(ar_0, ar_2) -> Com_1(f16(0, ar_2)) [ ar_0 >= 10 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_2) -> Com_1(f0(ar_0, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) f9(ar_0, ar_2) -> Com_1(f16(0, 0)) [ ar_2 = 0 ] (Comp: 2, Cost: 1) f10(ar_0, ar_2) -> Com_1(f16(0, ar_2)) [ ar_0 >= 10 ] (Comp: 2, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ h >= 1 ] (Comp: ?, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ 0 >= h + 1 ] (Comp: 2, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ 0 >= ar_2 + 1 ] (Comp: 1, Cost: 1) f0(ar_0, ar_2) -> Com_1(f9(0, h)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 10 Pol(f0) = 10 Pol(f9) = 10 Pol(f16) = -V_1 + 10 Pol(f10) = 10 Pol(f28) = -V_1 orients all transitions weakly and the transitions f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ h >= 1 ] f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ 0 >= h + 1 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_2) -> Com_1(f0(ar_0, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) f9(ar_0, ar_2) -> Com_1(f16(0, 0)) [ ar_2 = 0 ] (Comp: 2, Cost: 1) f10(ar_0, ar_2) -> Com_1(f16(0, ar_2)) [ ar_0 >= 10 ] (Comp: 2, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ h >= 1 ] (Comp: 10, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ 0 >= h + 1 ] (Comp: 2, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ ar_0 >= 10 ] (Comp: ?, Cost: 1) f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ 0 >= ar_2 + 1 ] (Comp: 1, Cost: 1) f0(ar_0, ar_2) -> Com_1(f9(0, h)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f9) = -V_1 + 10 Pol(f10) = -V_1 + 10 and size complexities S("f0(ar_0, ar_2) -> Com_1(f9(0, h))", 0-0) = 0 S("f0(ar_0, ar_2) -> Com_1(f9(0, h))", 0-1) = ? S("f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ 0 >= ar_2 + 1 ]", 0-0) = 10 S("f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ 0 >= ar_2 + 1 ]", 0-1) = ? S("f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ ar_2 >= 1 ]", 0-0) = 10 S("f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ ar_2 >= 1 ]", 0-1) = ? S("f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ]", 0-0) = 10 S("f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ]", 0-1) = ? S("f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ ar_0 >= 10 ]", 0-0) = 10 S("f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ ar_0 >= 10 ]", 0-1) = ? S("f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\\ 0 >= h + 1 ]", 0-0) = 10 S("f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\\ 0 >= h + 1 ]", 0-1) = ? S("f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\\ h >= 1 ]", 0-0) = 10 S("f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\\ h >= 1 ]", 0-1) = ? S("f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ 9 >= ar_0 ]", 0-0) = 10 S("f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ 9 >= ar_0 ]", 0-1) = ? S("f10(ar_0, ar_2) -> Com_1(f16(0, ar_2)) [ ar_0 >= 10 ]", 0-0) = 0 S("f10(ar_0, ar_2) -> Com_1(f16(0, ar_2)) [ ar_0 >= 10 ]", 0-1) = ? S("f9(ar_0, ar_2) -> Com_1(f16(0, 0)) [ ar_2 = 0 ]", 0-0) = 0 S("f9(ar_0, ar_2) -> Com_1(f16(0, 0)) [ ar_2 = 0 ]", 0-1) = 0 S("koat_start(ar_0, ar_2) -> Com_1(f0(ar_0, ar_2)) [ 0 <= 0 ]", 0-0) = ar_0 S("koat_start(ar_0, ar_2) -> Com_1(f0(ar_0, ar_2)) [ 0 <= 0 ]", 0-1) = ar_2 orients the transitions f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ ar_2 >= 1 ] f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ 0 >= ar_2 + 1 ] f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ] weakly and the transition f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ] strictly and produces the following problem: 6: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_2) -> Com_1(f0(ar_0, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) f9(ar_0, ar_2) -> Com_1(f16(0, 0)) [ ar_2 = 0 ] (Comp: 2, Cost: 1) f10(ar_0, ar_2) -> Com_1(f16(0, ar_2)) [ ar_0 >= 10 ] (Comp: 2, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ h >= 1 ] (Comp: 10, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ 0 >= h + 1 ] (Comp: 2, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ ar_0 >= 10 ] (Comp: 10, Cost: 1) f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ ar_2 >= 1 ] (Comp: ?, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ 0 >= ar_2 + 1 ] (Comp: 1, Cost: 1) f0(ar_0, ar_2) -> Com_1(f9(0, h)) start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 6 produces the following problem: 7: T: (Comp: 1, Cost: 0) koat_start(ar_0, ar_2) -> Com_1(f0(ar_0, ar_2)) [ 0 <= 0 ] (Comp: 2, Cost: 1) f9(ar_0, ar_2) -> Com_1(f16(0, 0)) [ ar_2 = 0 ] (Comp: 2, Cost: 1) f10(ar_0, ar_2) -> Com_1(f16(0, ar_2)) [ ar_0 >= 10 ] (Comp: 2, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ 9 >= ar_0 ] (Comp: 10, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ h >= 1 ] (Comp: 10, Cost: 1) f16(ar_0, ar_2) -> Com_1(f16(ar_0 + 1, ar_2)) [ 9 >= ar_0 /\ 0 >= h + 1 ] (Comp: 2, Cost: 1) f16(ar_0, ar_2) -> Com_1(f28(ar_0, ar_2)) [ ar_0 >= 10 ] (Comp: 10, Cost: 1) f10(ar_0, ar_2) -> Com_1(f9(ar_0 + 1, h)) [ 9 >= ar_0 ] (Comp: 11, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ ar_2 >= 1 ] (Comp: 11, Cost: 1) f9(ar_0, ar_2) -> Com_1(f10(ar_0, ar_2)) [ 0 >= ar_2 + 1 ] (Comp: 1, Cost: 1) f0(ar_0, ar_2) -> Com_1(f9(0, h)) start location: koat_start leaf cost: 0 Complexity upper bound 61 Time: 0.132 sec (SMT: 0.122 sec) ---------------------------------------- (2) BOUNDS(1, 1)