/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 35 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 735 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: evalfstart(A, B, C) -> Com_1(evalfentryin(A, B, C)) :|: TRUE evalfentryin(A, B, C) -> Com_1(evalfbb5in(0, B, C)) :|: TRUE evalfbb5in(A, B, C) -> Com_1(evalfreturnin(A, B, C)) :|: A >= B evalfbb5in(A, B, C) -> Com_1(evalfbb6in(A, B, C)) :|: B >= A + 1 evalfbb6in(A, B, C) -> Com_1(evalfbb2in(A, B, A)) :|: 0 >= D + 1 evalfbb6in(A, B, C) -> Com_1(evalfbb2in(A, B, A)) :|: D >= 1 evalfbb6in(A, B, C) -> Com_1(evalfreturnin(A, B, C)) :|: TRUE evalfbb2in(A, B, C) -> Com_1(evalfbb4in(A, B, C)) :|: C >= B evalfbb2in(A, B, C) -> Com_1(evalfbb3in(A, B, C)) :|: B >= C + 1 evalfbb3in(A, B, C) -> Com_1(evalfbb1in(A, B, C)) :|: 0 >= D + 1 evalfbb3in(A, B, C) -> Com_1(evalfbb1in(A, B, C)) :|: D >= 1 evalfbb3in(A, B, C) -> Com_1(evalfbb4in(A, B, C)) :|: TRUE evalfbb1in(A, B, C) -> Com_1(evalfbb2in(A, B, C + 1)) :|: TRUE evalfbb4in(A, B, C) -> Com_1(evalfbb5in(C + 1, B, C)) :|: TRUE evalfreturnin(A, B, C) -> Com_1(evalfstop(A, B, C)) :|: TRUE The start-symbols are:[evalfstart_3] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 36*Ar_1 + 44) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb5in(0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb6in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_0)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_0)) [ D >= 1 ] (Comp: ?, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb4in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_1 ] (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ D >= 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb4in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2 + 1)) (Comp: ?, Cost: 1) evalfbb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb5in(Ar_2 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(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) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb5in(0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb6in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_0)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_0)) [ D >= 1 ] (Comp: ?, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb4in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_1 ] (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ D >= 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb4in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2 + 1)) (Comp: ?, Cost: 1) evalfbb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb5in(Ar_2 + 1, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalfstart) = 2 Pol(evalfentryin) = 2 Pol(evalfbb5in) = 2 Pol(evalfreturnin) = 1 Pol(evalfbb6in) = 2 Pol(evalfbb2in) = 2 Pol(evalfbb4in) = 2 Pol(evalfbb3in) = 2 Pol(evalfbb1in) = 2 Pol(evalfstop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb5in(0, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb6in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_0)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_0)) [ D >= 1 ] (Comp: 2, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb4in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_1 ] (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ D >= 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb4in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2 + 1)) (Comp: ?, Cost: 1) evalfbb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb5in(Ar_2 + 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalfstart) = 2*V_2 + 2 Pol(evalfentryin) = 2*V_2 + 2 Pol(evalfbb5in) = -2*V_1 + 2*V_2 + 2 Pol(evalfreturnin) = -2*V_1 + 2*V_2 Pol(evalfbb6in) = -2*V_1 + 2*V_2 + 1 Pol(evalfbb2in) = 2*V_2 - 2*V_3 + 1 Pol(evalfbb4in) = 2*V_2 - 2*V_3 Pol(evalfbb3in) = 2*V_2 - 2*V_3 Pol(evalfbb1in) = 2*V_2 - 2*V_3 - 1 Pol(evalfstop) = -2*V_1 + 2*V_2 Pol(koat_start) = 2*V_2 + 2 orients all transitions weakly and the transitions evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb6in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_0 + 1 ] evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb5in(0, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb6in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_0)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_0)) [ D >= 1 ] (Comp: 2, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb4in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ 0 >= D + 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ D >= 1 ] (Comp: ?, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb4in(Ar_0, Ar_1, Ar_2)) (Comp: ?, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2 + 1)) (Comp: ?, Cost: 1) evalfbb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb5in(Ar_2 + 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(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) evalfstart(Ar_0, Ar_1, Ar_2) -> Com_1(evalfentryin(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 1) evalfentryin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb5in(0, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= Ar_1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalfbb5in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb6in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_0 + 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_0)) [ 0 >= D + 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_0)) [ D >= 1 ] (Comp: 2, Cost: 1) evalfbb6in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfreturnin(Ar_0, Ar_1, Ar_2)) (Comp: 8*Ar_1 + 8, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb4in(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= Ar_1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalfbb2in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb3in(Ar_0, Ar_1, Ar_2)) [ Ar_1 >= Ar_2 + 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ 0 >= D + 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb1in(Ar_0, Ar_1, Ar_2)) [ D >= 1 ] (Comp: 2*Ar_1 + 2, Cost: 1) evalfbb3in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb4in(Ar_0, Ar_1, Ar_2)) (Comp: 4*Ar_1 + 4, Cost: 1) evalfbb1in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb2in(Ar_0, Ar_1, Ar_2 + 1)) (Comp: 10*Ar_1 + 10, Cost: 1) evalfbb4in(Ar_0, Ar_1, Ar_2) -> Com_1(evalfbb5in(Ar_2 + 1, Ar_1, Ar_2)) (Comp: 2, Cost: 1) evalfreturnin(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstop(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(evalfstart(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 36*Ar_1 + 44 Time: 0.097 sec (SMT: 0.074 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalfstart 0: evalfstart -> evalfentryin : [], cost: 1 1: evalfentryin -> evalfbb5in : A'=0, [], cost: 1 2: evalfbb5in -> evalfreturnin : [ A>=B ], cost: 1 3: evalfbb5in -> evalfbb6in : [ B>=1+A ], cost: 1 4: evalfbb6in -> evalfbb2in : C'=A, [ 0>=1+free ], cost: 1 5: evalfbb6in -> evalfbb2in : C'=A, [ free_1>=1 ], cost: 1 6: evalfbb6in -> evalfreturnin : [], cost: 1 7: evalfbb2in -> evalfbb4in : [ C>=B ], cost: 1 8: evalfbb2in -> evalfbb3in : [ B>=1+C ], cost: 1 9: evalfbb3in -> evalfbb1in : [ 0>=1+free_2 ], cost: 1 10: evalfbb3in -> evalfbb1in : [ free_3>=1 ], cost: 1 11: evalfbb3in -> evalfbb4in : [], cost: 1 12: evalfbb1in -> evalfbb2in : C'=1+C, [], cost: 1 13: evalfbb4in -> evalfbb5in : A'=1+C, [], cost: 1 14: evalfreturnin -> evalfstop : [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: evalfstart -> evalfentryin : [], cost: 1 Removed unreachable and leaf rules: Start location: evalfstart 0: evalfstart -> evalfentryin : [], cost: 1 1: evalfentryin -> evalfbb5in : A'=0, [], cost: 1 3: evalfbb5in -> evalfbb6in : [ B>=1+A ], cost: 1 4: evalfbb6in -> evalfbb2in : C'=A, [ 0>=1+free ], cost: 1 5: evalfbb6in -> evalfbb2in : C'=A, [ free_1>=1 ], cost: 1 7: evalfbb2in -> evalfbb4in : [ C>=B ], cost: 1 8: evalfbb2in -> evalfbb3in : [ B>=1+C ], cost: 1 9: evalfbb3in -> evalfbb1in : [ 0>=1+free_2 ], cost: 1 10: evalfbb3in -> evalfbb1in : [ free_3>=1 ], cost: 1 11: evalfbb3in -> evalfbb4in : [], cost: 1 12: evalfbb1in -> evalfbb2in : C'=1+C, [], cost: 1 13: evalfbb4in -> evalfbb5in : A'=1+C, [], cost: 1 Simplified all rules, resulting in: Start location: evalfstart 0: evalfstart -> evalfentryin : [], cost: 1 1: evalfentryin -> evalfbb5in : A'=0, [], cost: 1 3: evalfbb5in -> evalfbb6in : [ B>=1+A ], cost: 1 5: evalfbb6in -> evalfbb2in : C'=A, [], cost: 1 7: evalfbb2in -> evalfbb4in : [ C>=B ], cost: 1 8: evalfbb2in -> evalfbb3in : [ B>=1+C ], cost: 1 10: evalfbb3in -> evalfbb1in : [], cost: 1 11: evalfbb3in -> evalfbb4in : [], cost: 1 12: evalfbb1in -> evalfbb2in : C'=1+C, [], cost: 1 13: evalfbb4in -> evalfbb5in : A'=1+C, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalfstart 15: evalfstart -> evalfbb5in : A'=0, [], cost: 2 16: evalfbb5in -> evalfbb2in : C'=A, [ B>=1+A ], cost: 2 7: evalfbb2in -> evalfbb4in : [ C>=B ], cost: 1 8: evalfbb2in -> evalfbb3in : [ B>=1+C ], cost: 1 11: evalfbb3in -> evalfbb4in : [], cost: 1 17: evalfbb3in -> evalfbb2in : C'=1+C, [], cost: 2 13: evalfbb4in -> evalfbb5in : A'=1+C, [], cost: 1 Eliminated locations (on tree-shaped paths): Start location: evalfstart 15: evalfstart -> evalfbb5in : A'=0, [], cost: 2 16: evalfbb5in -> evalfbb2in : C'=A, [ B>=1+A ], cost: 2 19: evalfbb2in -> evalfbb2in : C'=1+C, [ B>=1+C ], cost: 3 20: evalfbb2in -> evalfbb5in : A'=1+C, [ C>=B ], cost: 2 21: evalfbb2in -> evalfbb5in : A'=1+C, [ B>=1+C ], cost: 3 Accelerating simple loops of location 4. Accelerating the following rules: 19: evalfbb2in -> evalfbb2in : C'=1+C, [ B>=1+C ], cost: 3 Accelerated rule 19 with metering function -C+B, yielding the new rule 22. Removing the simple loops: 19. Accelerated all simple loops using metering functions (where possible): Start location: evalfstart 15: evalfstart -> evalfbb5in : A'=0, [], cost: 2 16: evalfbb5in -> evalfbb2in : C'=A, [ B>=1+A ], cost: 2 20: evalfbb2in -> evalfbb5in : A'=1+C, [ C>=B ], cost: 2 21: evalfbb2in -> evalfbb5in : A'=1+C, [ B>=1+C ], cost: 3 22: evalfbb2in -> evalfbb2in : C'=B, [ B>=1+C ], cost: -3*C+3*B Chained accelerated rules (with incoming rules): Start location: evalfstart 15: evalfstart -> evalfbb5in : A'=0, [], cost: 2 16: evalfbb5in -> evalfbb2in : C'=A, [ B>=1+A ], cost: 2 23: evalfbb5in -> evalfbb2in : C'=B, [ B>=1+A ], cost: 2-3*A+3*B 20: evalfbb2in -> evalfbb5in : A'=1+C, [ C>=B ], cost: 2 21: evalfbb2in -> evalfbb5in : A'=1+C, [ B>=1+C ], cost: 3 Eliminated locations (on tree-shaped paths): Start location: evalfstart 15: evalfstart -> evalfbb5in : A'=0, [], cost: 2 24: evalfbb5in -> evalfbb5in : A'=1+A, C'=A, [ B>=1+A ], cost: 5 25: evalfbb5in -> evalfbb5in : A'=1+B, C'=B, [ B>=1+A ], cost: 4-3*A+3*B 26: evalfbb5in -> [11] : [ B>=1+A ], cost: 2-3*A+3*B Accelerating simple loops of location 2. Accelerating the following rules: 24: evalfbb5in -> evalfbb5in : A'=1+A, C'=A, [ B>=1+A ], cost: 5 25: evalfbb5in -> evalfbb5in : A'=1+B, C'=B, [ B>=1+A ], cost: 4-3*A+3*B Accelerated rule 24 with metering function -A+B, yielding the new rule 27. Found no metering function for rule 25. Removing the simple loops: 24. Accelerated all simple loops using metering functions (where possible): Start location: evalfstart 15: evalfstart -> evalfbb5in : A'=0, [], cost: 2 25: evalfbb5in -> evalfbb5in : A'=1+B, C'=B, [ B>=1+A ], cost: 4-3*A+3*B 26: evalfbb5in -> [11] : [ B>=1+A ], cost: 2-3*A+3*B 27: evalfbb5in -> evalfbb5in : A'=B, C'=-1+B, [ B>=1+A ], cost: -5*A+5*B Chained accelerated rules (with incoming rules): Start location: evalfstart 15: evalfstart -> evalfbb5in : A'=0, [], cost: 2 28: evalfstart -> evalfbb5in : A'=1+B, C'=B, [ B>=1 ], cost: 6+3*B 29: evalfstart -> evalfbb5in : A'=B, C'=-1+B, [ B>=1 ], cost: 2+5*B 26: evalfbb5in -> [11] : [ B>=1+A ], cost: 2-3*A+3*B Eliminated locations (on tree-shaped paths): Start location: evalfstart 30: evalfstart -> [11] : A'=0, [ B>=1 ], cost: 4+3*B 31: evalfstart -> [13] : [ B>=1 ], cost: 6+3*B 32: evalfstart -> [13] : [ B>=1 ], cost: 2+5*B ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalfstart 31: evalfstart -> [13] : [ B>=1 ], cost: 6+3*B 32: evalfstart -> [13] : [ B>=1 ], cost: 2+5*B Computing asymptotic complexity for rule 31 Solved the limit problem by the following transformations: Created initial limit problem: 6+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 6+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: 6+3*n Rule cost: 6+3*B Rule guard: [ B>=1 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)