/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, n^1).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, 27 ms] (2) BOUNDS(1, n^1) (3) Loat Proof [FINISHED, 286 ms] (4) BOUNDS(n^1, INF) ---------------------------------------- (0) Obligation: Complexity Int TRS consisting of the following rules: evalrandom1dstart(A, B) -> Com_1(evalrandom1dentryin(A, B)) :|: TRUE evalrandom1dentryin(A, B) -> Com_1(evalrandom1dbb5in(A, 1)) :|: A >= 1 evalrandom1dentryin(A, B) -> Com_1(evalrandom1dreturnin(A, B)) :|: 0 >= A evalrandom1dbb5in(A, B) -> Com_1(evalrandom1dbb1in(A, B)) :|: A >= B evalrandom1dbb5in(A, B) -> Com_1(evalrandom1dreturnin(A, B)) :|: B >= A + 1 evalrandom1dbb1in(A, B) -> Com_1(evalrandom1dbb5in(A, B + 1)) :|: 0 >= C + 1 evalrandom1dbb1in(A, B) -> Com_1(evalrandom1dbb5in(A, B + 1)) :|: C >= 1 evalrandom1dbb1in(A, B) -> Com_1(evalrandom1dbb5in(A, B + 1)) :|: TRUE evalrandom1dreturnin(A, B) -> Com_1(evalrandom1dstop(A, B)) :|: TRUE The start-symbols are:[evalrandom1dstart_2] ---------------------------------------- (1) Koat Proof (FINISHED) YES(?, 4*Ar_0 + 15) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) evalrandom1dstart(Ar_0, Ar_1) -> Com_1(evalrandom1dentryin(Ar_0, Ar_1)) (Comp: ?, Cost: 1) evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, 1)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ 0 >= C + 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ C >= 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) (Comp: ?, Cost: 1) evalrandom1dreturnin(Ar_0, Ar_1) -> Com_1(evalrandom1dstop(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(evalrandom1dstart(Ar_0, Ar_1)) [ 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) evalrandom1dstart(Ar_0, Ar_1) -> Com_1(evalrandom1dentryin(Ar_0, Ar_1)) (Comp: 1, Cost: 1) evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, 1)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ 0 >= C + 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ C >= 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) (Comp: ?, Cost: 1) evalrandom1dreturnin(Ar_0, Ar_1) -> Com_1(evalrandom1dstop(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(evalrandom1dstart(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalrandom1dstart) = 2 Pol(evalrandom1dentryin) = 2 Pol(evalrandom1dbb5in) = 2 Pol(evalrandom1dreturnin) = 1 Pol(evalrandom1dbb1in) = 2 Pol(evalrandom1dstop) = 0 Pol(koat_start) = 2 orients all transitions weakly and the transitions evalrandom1dreturnin(Ar_0, Ar_1) -> Com_1(evalrandom1dstop(Ar_0, Ar_1)) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) evalrandom1dstart(Ar_0, Ar_1) -> Com_1(evalrandom1dentryin(Ar_0, Ar_1)) (Comp: 1, Cost: 1) evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, 1)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ] (Comp: 2, Cost: 1) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ 0 >= C + 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ C >= 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) (Comp: 2, Cost: 1) evalrandom1dreturnin(Ar_0, Ar_1) -> Com_1(evalrandom1dstop(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(evalrandom1dstart(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(evalrandom1dbb5in) = V_1 - V_2 + 1 Pol(evalrandom1dbb1in) = V_1 - V_2 and size complexities S("koat_start(Ar_0, Ar_1) -> Com_1(evalrandom1dstart(Ar_0, Ar_1)) [ 0 <= 0 ]", 0-0) = Ar_0 S("koat_start(Ar_0, Ar_1) -> Com_1(evalrandom1dstart(Ar_0, Ar_1)) [ 0 <= 0 ]", 0-1) = Ar_1 S("evalrandom1dreturnin(Ar_0, Ar_1) -> Com_1(evalrandom1dstop(Ar_0, Ar_1))", 0-0) = Ar_0 S("evalrandom1dreturnin(Ar_0, Ar_1) -> Com_1(evalrandom1dstop(Ar_0, Ar_1))", 0-1) = ? S("evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1))", 0-0) = Ar_0 S("evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1))", 0-1) = ? S("evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ C >= 1 ]", 0-0) = Ar_0 S("evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ C >= 1 ]", 0-1) = ? S("evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ 0 >= C + 1 ]", 0-0) = Ar_0 S("evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ 0 >= C + 1 ]", 0-1) = ? S("evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]", 0-0) = Ar_0 S("evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ]", 0-1) = ? S("evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]", 0-0) = Ar_0 S("evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ]", 0-1) = ? S("evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 ]", 0-0) = Ar_0 S("evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 ]", 0-1) = Ar_1 S("evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, 1)) [ Ar_0 >= 1 ]", 0-0) = Ar_0 S("evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, 1)) [ Ar_0 >= 1 ]", 0-1) = 1 S("evalrandom1dstart(Ar_0, Ar_1) -> Com_1(evalrandom1dentryin(Ar_0, Ar_1))", 0-0) = Ar_0 S("evalrandom1dstart(Ar_0, Ar_1) -> Com_1(evalrandom1dentryin(Ar_0, Ar_1))", 0-1) = Ar_1 orients the transitions evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ] evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ C >= 1 ] evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ 0 >= C + 1 ] evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) weakly and the transition evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) evalrandom1dstart(Ar_0, Ar_1) -> Com_1(evalrandom1dentryin(Ar_0, Ar_1)) (Comp: 1, Cost: 1) evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, 1)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: Ar_0 + 2, Cost: 1) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ] (Comp: 2, Cost: 1) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ 0 >= C + 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ C >= 1 ] (Comp: ?, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) (Comp: 2, Cost: 1) evalrandom1dreturnin(Ar_0, Ar_1) -> Com_1(evalrandom1dstop(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(evalrandom1dstart(Ar_0, Ar_1)) [ 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) evalrandom1dstart(Ar_0, Ar_1) -> Com_1(evalrandom1dentryin(Ar_0, Ar_1)) (Comp: 1, Cost: 1) evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, 1)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) evalrandom1dentryin(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: Ar_0 + 2, Cost: 1) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb1in(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 ] (Comp: 2, Cost: 1) evalrandom1dbb5in(Ar_0, Ar_1) -> Com_1(evalrandom1dreturnin(Ar_0, Ar_1)) [ Ar_1 >= Ar_0 + 1 ] (Comp: Ar_0 + 2, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ 0 >= C + 1 ] (Comp: Ar_0 + 2, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) [ C >= 1 ] (Comp: Ar_0 + 2, Cost: 1) evalrandom1dbb1in(Ar_0, Ar_1) -> Com_1(evalrandom1dbb5in(Ar_0, Ar_1 + 1)) (Comp: 2, Cost: 1) evalrandom1dreturnin(Ar_0, Ar_1) -> Com_1(evalrandom1dstop(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(evalrandom1dstart(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 4*Ar_0 + 15 Time: 0.056 sec (SMT: 0.047 sec) ---------------------------------------- (2) BOUNDS(1, n^1) ---------------------------------------- (3) Loat Proof (FINISHED) ### Pre-processing the ITS problem ### Initial linear ITS problem Start location: evalrandom1dstart 0: evalrandom1dstart -> evalrandom1dentryin : [], cost: 1 1: evalrandom1dentryin -> evalrandom1dbb5in : B'=1, [ A>=1 ], cost: 1 2: evalrandom1dentryin -> evalrandom1dreturnin : [ 0>=A ], cost: 1 3: evalrandom1dbb5in -> evalrandom1dbb1in : [ A>=B ], cost: 1 4: evalrandom1dbb5in -> evalrandom1dreturnin : [ B>=1+A ], cost: 1 5: evalrandom1dbb1in -> evalrandom1dbb5in : B'=1+B, [ 0>=1+free ], cost: 1 6: evalrandom1dbb1in -> evalrandom1dbb5in : B'=1+B, [ free_1>=1 ], cost: 1 7: evalrandom1dbb1in -> evalrandom1dbb5in : B'=1+B, [], cost: 1 8: evalrandom1dreturnin -> evalrandom1dstop : [], cost: 1 Checking for constant complexity: The following rule is satisfiable with cost >= 1, yielding constant complexity: 0: evalrandom1dstart -> evalrandom1dentryin : [], cost: 1 Removed unreachable and leaf rules: Start location: evalrandom1dstart 0: evalrandom1dstart -> evalrandom1dentryin : [], cost: 1 1: evalrandom1dentryin -> evalrandom1dbb5in : B'=1, [ A>=1 ], cost: 1 3: evalrandom1dbb5in -> evalrandom1dbb1in : [ A>=B ], cost: 1 5: evalrandom1dbb1in -> evalrandom1dbb5in : B'=1+B, [ 0>=1+free ], cost: 1 6: evalrandom1dbb1in -> evalrandom1dbb5in : B'=1+B, [ free_1>=1 ], cost: 1 7: evalrandom1dbb1in -> evalrandom1dbb5in : B'=1+B, [], cost: 1 Simplified all rules, resulting in: Start location: evalrandom1dstart 0: evalrandom1dstart -> evalrandom1dentryin : [], cost: 1 1: evalrandom1dentryin -> evalrandom1dbb5in : B'=1, [ A>=1 ], cost: 1 3: evalrandom1dbb5in -> evalrandom1dbb1in : [ A>=B ], cost: 1 7: evalrandom1dbb1in -> evalrandom1dbb5in : B'=1+B, [], cost: 1 ### Simplification by acceleration and chaining ### Eliminated locations (on linear paths): Start location: evalrandom1dstart 9: evalrandom1dstart -> evalrandom1dbb5in : B'=1, [ A>=1 ], cost: 2 10: evalrandom1dbb5in -> evalrandom1dbb5in : B'=1+B, [ A>=B ], cost: 2 Accelerating simple loops of location 2. Accelerating the following rules: 10: evalrandom1dbb5in -> evalrandom1dbb5in : B'=1+B, [ A>=B ], cost: 2 Accelerated rule 10 with metering function 1+A-B, yielding the new rule 11. Removing the simple loops: 10. Accelerated all simple loops using metering functions (where possible): Start location: evalrandom1dstart 9: evalrandom1dstart -> evalrandom1dbb5in : B'=1, [ A>=1 ], cost: 2 11: evalrandom1dbb5in -> evalrandom1dbb5in : B'=1+A, [ A>=B ], cost: 2+2*A-2*B Chained accelerated rules (with incoming rules): Start location: evalrandom1dstart 9: evalrandom1dstart -> evalrandom1dbb5in : B'=1, [ A>=1 ], cost: 2 12: evalrandom1dstart -> evalrandom1dbb5in : B'=1+A, [ A>=1 ], cost: 2+2*A Removed unreachable locations (and leaf rules with constant cost): Start location: evalrandom1dstart 12: evalrandom1dstart -> evalrandom1dbb5in : B'=1+A, [ A>=1 ], cost: 2+2*A ### Computing asymptotic complexity ### Fully simplified ITS problem Start location: evalrandom1dstart 12: evalrandom1dstart -> evalrandom1dbb5in : B'=1+A, [ A>=1 ], cost: 2+2*A Computing asymptotic complexity for rule 12 Solved the limit problem by the following transformations: Created initial limit problem: A (+/+!), 2+2*A (+) [not solved] removing all constraints (solved by SMT) resulting limit problem: [solved] applying transformation rule (C) using substitution {A==n} resulting limit problem: [solved] Solution: A / n Resulting cost 2+2*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: 2+2*n Rule cost: 2+2*A Rule guard: [ A>=1 ] WORST_CASE(Omega(n^1),?) ---------------------------------------- (4) BOUNDS(n^1, INF)