/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^1), O(n^1)) proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 21 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 636 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: evalNestedSinglestart(A, B, C) -> Com_1(evalNestedSingleentryin(A, B, C)) :|: TRUE evalNestedSingleentryin(A, B, C) -> Com_1(evalNestedSinglebb5in(0, B, C)) :|: TRUE evalNestedSinglebb5in(A, B, C) -> Com_1(evalNestedSinglebb2in(A, B, A)) :|: B >= A + 1 evalNestedSinglebb5in(A, B, C) -> Com_1(evalNestedSinglereturnin(A, B, C)) :|: A >= B evalNestedSinglebb2in(A, B, C) -> Com_1(evalNestedSinglebb4in(A, B, C)) :|: C >= B evalNestedSinglebb2in(A, B, C) -> Com_1(evalNestedSinglebb3in(A, B, C)) :|: B >= C + 1 evalNestedSinglebb3in(A, B, C) -> Com_1(evalNestedSinglebb1in(A, B, C)) :|: 0 >= D + 1 evalNestedSinglebb3in(A, B, C) -> Com_1(evalNestedSinglebb1in(A, B, C)) :|: D >= 1 evalNestedSinglebb3in(A, B, C) -> Com_1(evalNestedSinglebb4in(A, B, C)) :|: TRUE evalNestedSinglebb1in(A, B, C) -> Com_1(evalNestedSinglebb2in(A, B, C + 1)) :|: TRUE evalNestedSinglebb4in(A, B, C) -> Com_1(evalNestedSinglebb5in(C + 1, B, C)) :|: TRUE evalNestedSinglereturnin(A, B, C) -> Com_1(evalNestedSinglestop(A, B, C)) :|: TRUE The start-symbols are:[evalNestedSinglestart_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 24*Ar_1 + 30) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalNestedSinglestart(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSingleentryin(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalNestedSingleentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb5in(0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_0)) [ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_1 ] (Comp: ?, Cost: 1) evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2)) [ D >= 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2 + 1)) (Comp: ?, Cost: 1) evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb5in(Ar_2 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestart(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) evalNestedSinglestart(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSingleentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalNestedSingleentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb5in(0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_0)) [ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_1 ] (Comp: ?, Cost: 1) evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2)) [ D >= 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2 + 1)) (Comp: ?, Cost: 1) evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb5in(Ar_2 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalNestedSinglestart) = 2 Pol(evalNestedSingleentryin) = 2 Pol(evalNestedSinglebb5in) = 2 Pol(evalNestedSinglebb2in) = 2 Pol(evalNestedSinglereturnin) = 1 Pol(evalNestedSinglebb4in) = 2 Pol(evalNestedSinglebb3in) = 2 Pol(evalNestedSinglebb1in) = 2 Pol(evalNestedSinglestop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestop(Ar_0, Ar_1, Ar_2)) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalNestedSinglestart(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSingleentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalNestedSingleentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb5in(0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_0)) [ Ar_1 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_1 ] (Comp: ?, Cost: 1) evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2)) [ D >= 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2 + 1)) (Comp: ?, Cost: 1) evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb5in(Ar_2 + 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalNestedSinglestart) = 2*V_2 + 2 Pol(evalNestedSingleentryin) = 2*V_2 + 2 Pol(evalNestedSinglebb5in) = -2*V_1 + 2*V_2 + 2 Pol(evalNestedSinglebb2in) = 2*V_2 - 2*V_3 + 1 Pol(evalNestedSinglereturnin) = -2*V_1 + 2*V_2 Pol(evalNestedSinglebb4in) = 2*V_2 - 2*V_3 Pol(evalNestedSinglebb3in) = 2*V_2 - 2*V_3 Pol(evalNestedSinglebb1in) = 2*V_2 - 2*V_3 - 1 Pol(evalNestedSinglestop) = -2*V_1 + 2*V_2 Pol(koat_start) = 2*V_2 + 2 orients all transitions weakly and the transitions evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_0)) [ Ar_1 >= Ar_0 + 1 ] evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) evalNestedSinglestart(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSingleentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalNestedSingleentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb5in(0, Ar_1, Ar_2)) (Comp: 2*Ar_1 + 2, Cost: 1) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_0)) [ Ar_1 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2)) [ D >= 1 ] (Comp: ?, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2 + 1)) (Comp: ?, Cost: 1) evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb5in(Ar_2 + 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 4 produces the following problem: 5: T: (Comp: 1, Cost: 1) evalNestedSinglestart(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSingleentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalNestedSingleentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb5in(0, Ar_1, Ar_2)) (Comp: 2*Ar_1 + 2, Cost: 1) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_0)) [ Ar_1 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) evalNestedSinglebb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] (Comp: 4*Ar_1 + 4, Cost: 1) evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2)) [ 0 >= D + 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2)) [ D >= 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalNestedSinglebb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2)) (Comp: 4*Ar_1 + 4, Cost: 1) evalNestedSinglebb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb2in(Ar_0, Ar_1, Ar_2 + 1)) (Comp: 6*Ar_1 + 6, Cost: 1) evalNestedSinglebb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglebb5in(Ar_2 + 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalNestedSinglereturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalNestedSinglestart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 24*Ar_1 + 30 Time: 0.078 sec (SMT: 0.062 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalNestedSinglestart 0: evalNestedSinglestart -> evalNestedSingleentryin : [], cost: 1 1: evalNestedSingleentryin -> evalNestedSinglebb5in : A'=0, [], cost: 1 2: evalNestedSinglebb5in -> evalNestedSinglebb2in : C'=A, [ B>=1+A ], cost: 1 3: evalNestedSinglebb5in -> evalNestedSinglereturnin : [ A>=B ], cost: 1 4: evalNestedSinglebb2in -> evalNestedSinglebb4in : [ C>=B ], cost: 1 5: evalNestedSinglebb2in -> evalNestedSinglebb3in : [ B>=1+C ], cost: 1 6: evalNestedSinglebb3in -> evalNestedSinglebb1in : [ 0>=1+free ], cost: 1 7: evalNestedSinglebb3in -> evalNestedSinglebb1in : [ free_1>=1 ], cost: 1 8: evalNestedSinglebb3in -> evalNestedSinglebb4in : [], cost: 1 9: evalNestedSinglebb1in -> evalNestedSinglebb2in : C'=1+C, [], cost: 1 10: evalNestedSinglebb4in -> evalNestedSinglebb5in : A'=1+C, [], cost: 1 11: evalNestedSinglereturnin -> evalNestedSinglestop : [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: evalNestedSinglestart -> evalNestedSingleentryin : [], cost: 1 Removed unreachable and leaf rules: Start location: evalNestedSinglestart 0: evalNestedSinglestart -> evalNestedSingleentryin : [], cost: 1 1: evalNestedSingleentryin -> evalNestedSinglebb5in : A'=0, [], cost: 1 2: evalNestedSinglebb5in -> evalNestedSinglebb2in : C'=A, [ B>=1+A ], cost: 1 4: evalNestedSinglebb2in -> evalNestedSinglebb4in : [ C>=B ], cost: 1 5: evalNestedSinglebb2in -> evalNestedSinglebb3in : [ B>=1+C ], cost: 1 6: evalNestedSinglebb3in -> evalNestedSinglebb1in : [ 0>=1+free ], cost: 1 7: evalNestedSinglebb3in -> evalNestedSinglebb1in : [ free_1>=1 ], cost: 1 8: evalNestedSinglebb3in -> evalNestedSinglebb4in : [], cost: 1 9: evalNestedSinglebb1in -> evalNestedSinglebb2in : C'=1+C, [], cost: 1 10: evalNestedSinglebb4in -> evalNestedSinglebb5in : A'=1+C, [], cost: 1 Simplified all rules, resulting in: Start location: evalNestedSinglestart 0: evalNestedSinglestart -> evalNestedSingleentryin : [], cost: 1 1: evalNestedSingleentryin -> evalNestedSinglebb5in : A'=0, [], cost: 1 2: evalNestedSinglebb5in -> evalNestedSinglebb2in : C'=A, [ B>=1+A ], cost: 1 4: evalNestedSinglebb2in -> evalNestedSinglebb4in : [ C>=B ], cost: 1 5: evalNestedSinglebb2in -> evalNestedSinglebb3in : [ B>=1+C ], cost: 1 7: evalNestedSinglebb3in -> evalNestedSinglebb1in : [], cost: 1 8: evalNestedSinglebb3in -> evalNestedSinglebb4in : [], cost: 1 9: evalNestedSinglebb1in -> evalNestedSinglebb2in : C'=1+C, [], cost: 1 10: evalNestedSinglebb4in -> evalNestedSinglebb5in : A'=1+C, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalNestedSinglestart 12: evalNestedSinglestart -> evalNestedSinglebb5in : A'=0, [], cost: 2 2: evalNestedSinglebb5in -> evalNestedSinglebb2in : C'=A, [ B>=1+A ], cost: 1 4: evalNestedSinglebb2in -> evalNestedSinglebb4in : [ C>=B ], cost: 1 5: evalNestedSinglebb2in -> evalNestedSinglebb3in : [ B>=1+C ], cost: 1 8: evalNestedSinglebb3in -> evalNestedSinglebb4in : [], cost: 1 13: evalNestedSinglebb3in -> evalNestedSinglebb2in : C'=1+C, [], cost: 2 10: evalNestedSinglebb4in -> evalNestedSinglebb5in : A'=1+C, [], cost: 1 Eliminated locations (on tree-shaped paths): Start location: evalNestedSinglestart 12: evalNestedSinglestart -> evalNestedSinglebb5in : A'=0, [], cost: 2 2: evalNestedSinglebb5in -> evalNestedSinglebb2in : C'=A, [ B>=1+A ], cost: 1 15: evalNestedSinglebb2in -> evalNestedSinglebb2in : C'=1+C, [ B>=1+C ], cost: 3 16: evalNestedSinglebb2in -> evalNestedSinglebb5in : A'=1+C, [ C>=B ], cost: 2 17: evalNestedSinglebb2in -> evalNestedSinglebb5in : A'=1+C, [ B>=1+C ], cost: 3 Accelerating simple loops of location 3. Accelerating the following rules: 15: evalNestedSinglebb2in -> evalNestedSinglebb2in : C'=1+C, [ B>=1+C ], cost: 3 Accelerated rule 15 with metering function -C+B, yielding the new rule 18. Removing the simple loops: 15. Accelerated all simple loops using metering functions (where possible): Start location: evalNestedSinglestart 12: evalNestedSinglestart -> evalNestedSinglebb5in : A'=0, [], cost: 2 2: evalNestedSinglebb5in -> evalNestedSinglebb2in : C'=A, [ B>=1+A ], cost: 1 16: evalNestedSinglebb2in -> evalNestedSinglebb5in : A'=1+C, [ C>=B ], cost: 2 17: evalNestedSinglebb2in -> evalNestedSinglebb5in : A'=1+C, [ B>=1+C ], cost: 3 18: evalNestedSinglebb2in -> evalNestedSinglebb2in : C'=B, [ B>=1+C ], cost: -3*C+3*B Chained accelerated rules (with incoming rules): Start location: evalNestedSinglestart 12: evalNestedSinglestart -> evalNestedSinglebb5in : A'=0, [], cost: 2 2: evalNestedSinglebb5in -> evalNestedSinglebb2in : C'=A, [ B>=1+A ], cost: 1 19: evalNestedSinglebb5in -> evalNestedSinglebb2in : C'=B, [ B>=1+A ], cost: 1-3*A+3*B 16: evalNestedSinglebb2in -> evalNestedSinglebb5in : A'=1+C, [ C>=B ], cost: 2 17: evalNestedSinglebb2in -> evalNestedSinglebb5in : A'=1+C, [ B>=1+C ], cost: 3 Eliminated locations (on tree-shaped paths): Start location: evalNestedSinglestart 12: evalNestedSinglestart -> evalNestedSinglebb5in : A'=0, [], cost: 2 20: evalNestedSinglebb5in -> evalNestedSinglebb5in : A'=1+A, C'=A, [ B>=1+A ], cost: 4 21: evalNestedSinglebb5in -> evalNestedSinglebb5in : A'=1+B, C'=B, [ B>=1+A ], cost: 3-3*A+3*B 22: evalNestedSinglebb5in -> [10] : [ B>=1+A ], cost: 1-3*A+3*B Accelerating simple loops of location 2. Accelerating the following rules: 20: evalNestedSinglebb5in -> evalNestedSinglebb5in : A'=1+A, C'=A, [ B>=1+A ], cost: 4 21: evalNestedSinglebb5in -> evalNestedSinglebb5in : A'=1+B, C'=B, [ B>=1+A ], cost: 3-3*A+3*B Accelerated rule 20 with metering function -A+B, yielding the new rule 23. Found no metering function for rule 21. Removing the simple loops: 20. Accelerated all simple loops using metering functions (where possible): Start location: evalNestedSinglestart 12: evalNestedSinglestart -> evalNestedSinglebb5in : A'=0, [], cost: 2 21: evalNestedSinglebb5in -> evalNestedSinglebb5in : A'=1+B, C'=B, [ B>=1+A ], cost: 3-3*A+3*B 22: evalNestedSinglebb5in -> [10] : [ B>=1+A ], cost: 1-3*A+3*B 23: evalNestedSinglebb5in -> evalNestedSinglebb5in : A'=B, C'=-1+B, [ B>=1+A ], cost: -4*A+4*B Chained accelerated rules (with incoming rules): Start location: evalNestedSinglestart 12: evalNestedSinglestart -> evalNestedSinglebb5in : A'=0, [], cost: 2 24: evalNestedSinglestart -> evalNestedSinglebb5in : A'=1+B, C'=B, [ B>=1 ], cost: 5+3*B 25: evalNestedSinglestart -> evalNestedSinglebb5in : A'=B, C'=-1+B, [ B>=1 ], cost: 2+4*B 22: evalNestedSinglebb5in -> [10] : [ B>=1+A ], cost: 1-3*A+3*B Eliminated locations (on tree-shaped paths): Start location: evalNestedSinglestart 26: evalNestedSinglestart -> [10] : A'=0, [ B>=1 ], cost: 3+3*B 27: evalNestedSinglestart -> [12] : [ B>=1 ], cost: 5+3*B 28: evalNestedSinglestart -> [12] : [ B>=1 ], cost: 2+4*B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalNestedSinglestart 27: evalNestedSinglestart -> [12] : [ B>=1 ], cost: 5+3*B 28: evalNestedSinglestart -> [12] : [ B>=1 ], cost: 2+4*B Computing asymptotic complexity for rule 27 Solved the limit problem by the following transformations: Created initial limit problem: 5+3*B (+), B (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {B==n} resulting limit problem: [solved] Solution: B / n Resulting cost 5+3*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: 5+3*n Rule cost: 5+3*B Rule guard: [ B>=1 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)