/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- WORST_CASE(Omega(n^3), O(n^3)) 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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^3, n^3). (0) CpxIntTrs (1) Koat Proof [FINISHED, 105 ms] (2) BOUNDS(1, n^3) (3) Loat Proof [FINISHED, 279 ms] (4) BOUNDS(n^3, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: l0(A, B, F, X, Y) -> Com_1(l1(A, B, F, X, Y)) :|: TRUE l1(A, B, F, X, Y) -> Com_1(l1(A, B, F + X, X + Y, Y - 1)) :|: Y >= 1 l1(A, B, F, X, Y) -> Com_1(l2(A, B, F, X, Y)) :|: Y < 1 l2(A, B, F, X, Y) -> Com_1(l2(A, B, F - 1, X, Y)) :|: F >= 1 l2(A, B, F, X, Y) -> Com_1(l3(A, B, F, X, Y)) :|: F < 1 l3(A, B, F, X, Y) -> Com_1(l3(A + B, B - 1, F, X)) :|: A >= 1 The start-symbols are:[l0_5] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 4*Ar_0 + 3*Ar_1 + 3*Ar_2 + 6*Ar_0^2 + 3*Ar_0*Ar_2 + 3*Ar_0^3 + 10) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ] (Comp: ?, Cost: 1) l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 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) l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ] (Comp: ?, Cost: 1) l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(l0) = 3 Pol(l1) = 3 Pol(l2) = 2 Pol(l3) = 1 Pol(l30) = 0 Pol(koat_start) = 3 orients all transitions weakly and the transitions l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ] l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ] l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: 3, Cost: 1) l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ] (Comp: 3, Cost: 1) l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ] (Comp: 3, Cost: 1) l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(l0) = V_1 Pol(l1) = V_1 Pol(l2) = V_1 Pol(l3) = V_1 Pol(l30) = V_1 Pol(koat_start) = V_1 orients all transitions weakly and the transition l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: Ar_0, Cost: 1) l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: 3, Cost: 1) l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ] (Comp: 3, Cost: 1) l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ] (Comp: 3, Cost: 1) l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(l2) = V_2 and size complexities S("koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]", 0-0) = Ar_0 S("koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]", 0-1) = Ar_1 S("koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]", 0-2) = Ar_2 S("koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]", 0-3) = Ar_3 S("koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ]", 0-4) = Ar_4 S("l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ]", 0-0) = Ar_0 S("l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ]", 0-1) = Ar_1 + Ar_2 + Ar_0 + 2*Ar_0^2 + Ar_0*Ar_2 + Ar_0^3 S("l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ]", 0-2) = Ar_2 + Ar_0 + Ar_0^2 S("l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ]", 0-3) = Ar_3 + Ar_4 S("l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ]", 0-4) = Ar_4 + 1 S("l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ]", 0-0) = Ar_0 S("l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ]", 0-1) = Ar_1 + Ar_2 + Ar_0 + 2*Ar_0^2 + Ar_0*Ar_2 + Ar_0^3 S("l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ]", 0-2) = Ar_2 + Ar_0 + Ar_0^2 S("l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ]", 0-3) = Ar_3 S("l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ]", 0-4) = Ar_4 S("l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ]", 0-0) = Ar_0 S("l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ]", 0-1) = Ar_1 + Ar_2 + Ar_0 + 2*Ar_0^2 + Ar_0*Ar_2 + Ar_0^3 S("l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ]", 0-2) = Ar_2 + Ar_0 + Ar_0^2 S("l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ]", 0-3) = Ar_3 S("l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ]", 0-4) = Ar_4 S("l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ]", 0-0) = Ar_0 S("l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ]", 0-1) = Ar_1 + Ar_2 + Ar_0 + 2*Ar_0^2 + Ar_0*Ar_2 + Ar_0^3 S("l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ]", 0-2) = Ar_2 + Ar_0 + Ar_0^2 S("l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ]", 0-3) = Ar_3 S("l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ]", 0-4) = Ar_4 S("l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ]", 0-0) = Ar_0 S("l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ]", 0-1) = Ar_1 + Ar_2 + Ar_0 + 2*Ar_0^2 + Ar_0*Ar_2 + Ar_0^3 S("l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ]", 0-2) = Ar_0 + Ar_2 + Ar_0^2 S("l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ]", 0-3) = Ar_3 S("l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ]", 0-4) = Ar_4 S("l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4))", 0-0) = Ar_0 S("l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4))", 0-1) = Ar_1 S("l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4))", 0-2) = Ar_2 S("l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4))", 0-3) = Ar_3 S("l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4))", 0-4) = Ar_4 orients the transitions l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ] weakly and the transition l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 1) l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: Ar_0, Cost: 1) l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l1(Ar_0 - 1, Ar_1 + Ar_2, Ar_2 + Ar_0, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: 3, Cost: 1) l1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_0 ] (Comp: 3*Ar_1 + 3*Ar_2 + 3*Ar_0 + 6*Ar_0^2 + 3*Ar_0*Ar_2 + 3*Ar_0^3, Cost: 1) l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l2(Ar_0, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 ] (Comp: 3, Cost: 1) l2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= Ar_1 ] (Comp: 3, Cost: 1) l3(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l30(Ar_0, Ar_1, Ar_2, Ar_3 + Ar_4, Ar_4 - 1)) [ Ar_3 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(l0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 4*Ar_0 + 3*Ar_1 + 3*Ar_2 + 6*Ar_0^2 + 3*Ar_0*Ar_2 + 3*Ar_0^3 + 10 Time: 0.186 sec (SMT: 0.162 sec) ---------------------------------------- (2) BOUNDS(1, n^3) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: l0 0: l0 -> l1 : [], cost: 1 1: l1 -> l1 : A'=-1+A, B'=C+B, C'=C+A, [ A>=1 ], cost: 1 2: l1 -> l2 : [ 0>=A ], cost: 1 3: l2 -> l2 : B'=-1+B, [ B>=1 ], cost: 1 4: l2 -> l3 : [ 0>=B ], cost: 1 5: l3 -> l30 : D'=D+E, E'=-1+E, [ D>=1 ], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: l0 -> l1 : [], cost: 1 Removed unreachable and leaf rules: Start location: l0 0: l0 -> l1 : [], cost: 1 1: l1 -> l1 : A'=-1+A, B'=C+B, C'=C+A, [ A>=1 ], cost: 1 2: l1 -> l2 : [ 0>=A ], cost: 1 3: l2 -> l2 : B'=-1+B, [ B>=1 ], cost: 1 ### Simplification by acceleration and chaining ### Accelerating simple loops of location 1. Accelerating the following rules: 1: l1 -> l1 : A'=-1+A, B'=C+B, C'=C+A, [ A>=1 ], cost: 1 Accelerated rule 1 with metering function A, yielding the new rule 6. Removing the simple loops: 1. Accelerating simple loops of location 2. Accelerating the following rules: 3: l2 -> l2 : B'=-1+B, [ B>=1 ], cost: 1 Accelerated rule 3 with metering function B, yielding the new rule 7. Removing the simple loops: 3. Accelerated all simple loops using metering functions (where possible): Start location: l0 0: l0 -> l1 : [], cost: 1 2: l1 -> l2 : [ 0>=A ], cost: 1 6: l1 -> l1 : A'=0, B'=-1/3*A+1/3*A^3+B+C*A, C'=C+1/2*A+1/2*A^2, [ A>=1 ], cost: A 7: l2 -> l2 : B'=0, [ B>=1 ], cost: B Chained accelerated rules (with incoming rules): Start location: l0 0: l0 -> l1 : [], cost: 1 8: l0 -> l1 : A'=0, B'=-1/3*A+1/3*A^3+B+C*A, C'=C+1/2*A+1/2*A^2, [ A>=1 ], cost: 1+A 2: l1 -> l2 : [ 0>=A ], cost: 1 9: l1 -> l2 : B'=0, [ 0>=A && B>=1 ], cost: 1+B Removed unreachable locations (and leaf rules with constant cost): Start location: l0 0: l0 -> l1 : [], cost: 1 8: l0 -> l1 : A'=0, B'=-1/3*A+1/3*A^3+B+C*A, C'=C+1/2*A+1/2*A^2, [ A>=1 ], cost: 1+A 9: l1 -> l2 : B'=0, [ 0>=A && B>=1 ], cost: 1+B Eliminated locations (on tree-shaped paths): Start location: l0 10: l0 -> l2 : B'=0, [ 0>=A && B>=1 ], cost: 2+B 11: l0 -> l2 : A'=0, B'=0, C'=C+1/2*A+1/2*A^2, [ A>=1 && -1/3*A+1/3*A^3+B+C*A>=1 ], cost: 2+2/3*A+1/3*A^3+B+C*A ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: l0 10: l0 -> l2 : B'=0, [ 0>=A && B>=1 ], cost: 2+B 11: l0 -> l2 : A'=0, B'=0, C'=C+1/2*A+1/2*A^2, [ A>=1 && -1/3*A+1/3*A^3+B+C*A>=1 ], cost: 2+2/3*A+1/3*A^3+B+C*A Computing asymptotic complexity for rule 10 Solved the limit problem by the following transformations: Created initial limit problem: 1-A (+/+!), B (+/+!), 2+B (+) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {A==0,B==n} resulting limit problem: [solved] Solution: A / 0 B / n Resulting cost 2+n has complexity: Poly(n^1) Found new complexity Poly(n^1). Computing asymptotic complexity for rule 11 Solved the limit problem by the following transformations: Created initial limit problem: 2+2/3*A+1/3*A^3+B+C*A (+), A (+/+!), -1/3*A+1/3*A^3+B+C*A (+/+!) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {C==-1+n,A==-1+n,B==-n} resulting limit problem: [solved] Solution: C / -1+n A / -1+n B / -n Resulting cost 2+1/3*n^3-4/3*n has complexity: Poly(n^3) Found new complexity Poly(n^3). Obtained the following overall complexity (w.r.t. the length of the input n): Complexity: Poly(n^3) Cpx degree: 3 Solved cost: 2+1/3*n^3-4/3*n Rule cost: 2+2/3*A+1/3*A^3+B+C*A Rule guard: [ A>=1 && -1/3*A+1/3*A^3+B+C*A>=1 ] WORST_CASE(Omega(n^3),?) ---------------------------------------- (4) BOUNDS(n^3, INF)