/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1).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, 9 ms] (2) BOUNDS(1, 1) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: f0(A) -> Com_1(f5(B)) :|: B >= 1 f5(A) -> Com_1(f5(A + 1)) :|: 19 >= A f0(A) -> Com_1(f12(B)) :|: 0 >= B f5(A) -> Com_1(f12(A)) :|: A >= 20 The start-symbols are:[f0_1] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 46) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 19 >= Ar_0 ] (Comp: ?, Cost: 1) f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 20 ] (Comp: 1, Cost: 0) koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 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) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 19 >= Ar_0 ] (Comp: 1, Cost: 1) f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 20 ] (Comp: 1, Cost: 0) koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f0) = 1 Pol(f5) = 1 Pol(f12) = 0 Pol(koat_start) = 1 orients all transitions weakly and the transition f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 20 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) f0(Ar_0) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ 19 >= Ar_0 ] (Comp: 1, Cost: 1) f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ] (Comp: 1, Cost: 1) f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 >= 20 ] (Comp: 1, Cost: 0) koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 3 to obtain the following invariants: For symbol f5: X_1 - 1 >= 0 This yielded the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\ Ar_0 >= 20 ] (Comp: 1, Cost: 1) f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ] (Comp: 1, Cost: 1) f0(Ar_0) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ] start location: koat_start leaf cost: 0 By chaining the transition koat_start(Ar_0) -> Com_1(f0(Ar_0)) [ 0 <= 0 ] with all transitions in problem 4, the following new transitions are obtained: koat_start(Ar_0) -> Com_1(f5(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 ] koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\ 0 >= Fresh_0 ] We thus obtain the following problem: 5: T: (Comp: 1, Cost: 1) koat_start(Ar_0) -> Com_1(f5(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 ] (Comp: 1, Cost: 1) koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\ 0 >= Fresh_0 ] (Comp: 1, Cost: 1) f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\ Ar_0 >= 20 ] (Comp: 1, Cost: 1) f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ] (Comp: 1, Cost: 1) f0(Ar_0) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ] start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transitions from problem 5: f0(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 >= Fresh_0 ] f0(Ar_0) -> Com_1(f5(Fresh_1)) [ Fresh_1 >= 1 ] We thus obtain the following problem: 6: T: (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ] (Comp: 1, Cost: 1) f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\ Ar_0 >= 20 ] (Comp: 1, Cost: 1) koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\ 0 >= Fresh_0 ] (Comp: 1, Cost: 1) koat_start(Ar_0) -> Com_1(f5(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 ] start location: koat_start leaf cost: 0 By chaining the transition koat_start(Ar_0) -> Com_1(f5(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 ] with all transitions in problem 6, the following new transitions are obtained: koat_start(Ar_0) -> Com_1(f5(Fresh_1 + 1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ 19 >= Fresh_1 ] koat_start(Ar_0) -> Com_1(f12(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ Fresh_1 >= 20 ] We thus obtain the following problem: 7: T: (Comp: 1, Cost: 2) koat_start(Ar_0) -> Com_1(f5(Fresh_1 + 1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ 19 >= Fresh_1 ] (Comp: 1, Cost: 2) koat_start(Ar_0) -> Com_1(f12(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ Fresh_1 >= 20 ] (Comp: ?, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ] (Comp: 1, Cost: 1) f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\ Ar_0 >= 20 ] (Comp: 1, Cost: 1) koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\ 0 >= Fresh_0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f5) = -V_1 + 20 and size complexities S("koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\\ 0 >= Fresh_0 ]", 0-0) = ? S("f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\\ Ar_0 >= 20 ]", 0-0) = 20 S("f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\\ 19 >= Ar_0 ]", 0-0) = 20 S("koat_start(Ar_0) -> Com_1(f12(Fresh_1)) [ 0 <= 0 /\\ Fresh_1 >= 1 /\\ Fresh_1 - 1 >= 0 /\\ Fresh_1 >= 20 ]", 0-0) = ? S("koat_start(Ar_0) -> Com_1(f5(Fresh_1 + 1)) [ 0 <= 0 /\\ Fresh_1 >= 1 /\\ Fresh_1 - 1 >= 0 /\\ 19 >= Fresh_1 ]", 0-0) = 20 orients the transitions f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ] weakly and the transition f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ] strictly and produces the following problem: 8: T: (Comp: 1, Cost: 2) koat_start(Ar_0) -> Com_1(f5(Fresh_1 + 1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ 19 >= Fresh_1 ] (Comp: 1, Cost: 2) koat_start(Ar_0) -> Com_1(f12(Fresh_1)) [ 0 <= 0 /\ Fresh_1 >= 1 /\ Fresh_1 - 1 >= 0 /\ Fresh_1 >= 20 ] (Comp: 40, Cost: 1) f5(Ar_0) -> Com_1(f5(Ar_0 + 1)) [ Ar_0 - 1 >= 0 /\ 19 >= Ar_0 ] (Comp: 1, Cost: 1) f5(Ar_0) -> Com_1(f12(Ar_0)) [ Ar_0 - 1 >= 0 /\ Ar_0 >= 20 ] (Comp: 1, Cost: 1) koat_start(Ar_0) -> Com_1(f12(Fresh_0)) [ 0 <= 0 /\ 0 >= Fresh_0 ] start location: koat_start leaf cost: 0 Complexity upper bound 46 Time: 0.085 sec (SMT: 0.072 sec) ---------------------------------------- (2) BOUNDS(1, 1)