/export/starexec/sandbox2/solver/bin/starexec_run_complexity /export/starexec/sandbox2/benchmark/theBenchmark.koat /export/starexec/sandbox2/output/output_files


--------------------------------------------------------------------------------


WORST_CASE(NON_POLY, ?)
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(INF, INF).

(0) CpxIntTrs
(1) Loat Proof [FINISHED, 347 ms]
(2) BOUNDS(INF, INF)


----------------------------------------

(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B) -> Com_1(f4(0, 99)) :|: TRUE
f4(A, B) -> Com_1(f4(E, B)) :|: B >= A + 1 && C >= D + 1
f4(A, B) -> Com_1(f4(A, E)) :|: B >= A + 1
f4(A, B) -> Com_1(f11(A, B)) :|: A >= B

The start-symbols are:[f0_2]


----------------------------------------

(1) Loat Proof (FINISHED)


### Pre-processing the ITS problem ###



Initial linear ITS problem

   Start location: f0

      0: f0 -> f4 : A'=0, B'=99, [], cost: 1

      1: f4 -> f4 : A'=free_2, [ B>=1+A && free>=1+free_1 ], cost: 1

      2: f4 -> f4 : B'=free_3, [ B>=1+A ], cost: 1

      3: f4 -> f11 : [ A>=B ], cost: 1



Checking for constant complexity:

   The following rule is satisfiable with cost >= 1, yielding constant complexity:

      0: f0 -> f4 : A'=0, B'=99, [], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

      0: f0 -> f4 : A'=0, B'=99, [], cost: 1

      1: f4 -> f4 : A'=free_2, [ B>=1+A && free>=1+free_1 ], cost: 1

      2: f4 -> f4 : B'=free_3, [ B>=1+A ], cost: 1



Simplified all rules, resulting in:

   Start location: f0

      0: f0 -> f4 : A'=0, B'=99, [], cost: 1

      1: f4 -> f4 : A'=free_2, [ B>=1+A ], cost: 1

      2: f4 -> f4 : B'=free_3, [ B>=1+A ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: f4 -> f4 : A'=free_2, [ B>=1+A ], cost: 1

      2: f4 -> f4 : B'=free_3, [ B>=1+A ], cost: 1



   Accelerated rule 1 with NONTERM (after strengthening guard), yielding the new rule 4.

   Accelerated rule 2 with NONTERM (after strengthening guard), yielding the new rule 5.

   Removing the simple loops:.



Accelerated all simple loops using metering functions (where possible):

   Start location: f0

      0: f0 -> f4 : A'=0, B'=99, [], cost: 1

      1: f4 -> f4 : A'=free_2, [ B>=1+A ], cost: 1

      2: f4 -> f4 : B'=free_3, [ B>=1+A ], cost: 1

      4: f4 -> [3] : [ B>=1+A && B>=1+free_2 ], cost: NONTERM

      5: f4 -> [3] : [ B>=1+A && free_3>=1+A ], cost: NONTERM



Chained accelerated rules (with incoming rules):

   Start location: f0

      0: f0 -> f4 : A'=0, B'=99, [], cost: 1

      6: f0 -> f4 : A'=free_2, B'=99, [], cost: 2

      7: f0 -> f4 : A'=0, B'=free_3, [], cost: 2

      8: f0 -> [3] : A'=0, B'=99, [], cost: NONTERM

      9: f0 -> [3] : A'=0, B'=99, [], cost: NONTERM



Removed unreachable locations (and leaf rules with constant cost):

   Start location: f0

      8: f0 -> [3] : A'=0, B'=99, [], cost: NONTERM

      9: f0 -> [3] : A'=0, B'=99, [], cost: NONTERM



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

      9: f0 -> [3] : A'=0, B'=99, [], cost: NONTERM



Computing asymptotic complexity for rule 9

   Guard is satisfiable, yielding nontermination

   Resulting cost NONTERM has complexity: Nonterm



Found new complexity Nonterm.



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Nonterm

   Cpx degree:  Nonterm

   Solved cost: NONTERM

   Rule cost:   NONTERM

   Rule guard:  []



NO


----------------------------------------

(2)
BOUNDS(INF, INF)