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


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WORST_CASE(NON_POLY, ?)
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 326 ms]
(2) BOUNDS(INF, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f5(A, B, C, D, E, F, G, H, I, J, K)) :|: TRUE
f0(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f12(M, B, C, D, E, F, G, H, I, J, K)) :|: 15 >= L
f12(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f20(A, M, M, D, E, F, G, H, I, J, K)) :|: 0 >= A + 1
f12(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f20(A, M, M, D, E, F, G, H, I, J, K)) :|: A >= 1
f0(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f5(A, B, C, 1, 1, F, G, H, I, J, K)) :|: 15 >= M
f20(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f12(A - 1, B, C, D, E, 0, 0, H, I, J, K)) :|: TRUE
f12(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f5(0, B, C, 0, 0, F, G, M, M, 0, 0)) :|: A >= 0 && A <= 0

The start-symbols are:[f0_11]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f0 -> f5 : [], cost: 1

      1: f0 -> f12 : A'=free, [ 15>=free_1 ], cost: 1

      4: f0 -> f5 : D'=1, E'=1, [ 15>=free_4 ], cost: 1

      2: f12 -> f20 : B'=free_2, C'=free_2, [ 0>=1+A ], cost: 1

      3: f12 -> f20 : B'=free_3, C'=free_3, [ A>=1 ], cost: 1

      6: f12 -> f5 : A'=0, D'=0, E'=0, H'=free_5, Q'=free_5, J'=0, K'=0, [ A==0 ], cost: 1

      5: f20 -> f12 : A'=-1+A, F'=0, G'=0, [], cost: 1



Checking for constant complexity:

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

      0: f0 -> f5 : [], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

      1: f0 -> f12 : A'=free, [ 15>=free_1 ], cost: 1

      2: f12 -> f20 : B'=free_2, C'=free_2, [ 0>=1+A ], cost: 1

      3: f12 -> f20 : B'=free_3, C'=free_3, [ A>=1 ], cost: 1

      5: f20 -> f12 : A'=-1+A, F'=0, G'=0, [], cost: 1



Simplified all rules, resulting in:

   Start location: f0

      1: f0 -> f12 : A'=free, [], cost: 1

      2: f12 -> f20 : B'=free_2, C'=free_2, [ 0>=1+A ], cost: 1

      3: f12 -> f20 : B'=free_3, C'=free_3, [ A>=1 ], cost: 1

      5: f20 -> f12 : A'=-1+A, F'=0, G'=0, [], cost: 1



### Simplification by acceleration and chaining ###



Eliminated locations (on tree-shaped paths):

   Start location: f0

      1: f0 -> f12 : A'=free, [], cost: 1

      7: f12 -> f12 : A'=-1+A, B'=free_2, C'=free_2, F'=0, G'=0, [ 0>=1+A ], cost: 2

      8: f12 -> f12 : A'=-1+A, B'=free_3, C'=free_3, F'=0, G'=0, [ A>=1 ], cost: 2



Accelerating simple loops of location 1.

   Accelerating the following rules:

      7: f12 -> f12 : A'=-1+A, B'=free_2, C'=free_2, F'=0, G'=0, [ 0>=1+A ], cost: 2

      8: f12 -> f12 : A'=-1+A, B'=free_3, C'=free_3, F'=0, G'=0, [ A>=1 ], cost: 2



   Accelerated rule 7 with NONTERM, yielding the new rule 9.

   Accelerated rule 8 with metering function A, yielding the new rule 10.

   Removing the simple loops: 7 8.



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

   Start location: f0

      1: f0 -> f12 : A'=free, [], cost: 1

      9: f12 -> [4] : [ 0>=1+A ], cost: NONTERM

     10: f12 -> f12 : A'=0, B'=free_3, C'=free_3, F'=0, G'=0, [ A>=1 ], cost: 2*A



Chained accelerated rules (with incoming rules):

   Start location: f0

      1: f0 -> f12 : A'=free, [], cost: 1

     11: f0 -> [4] : A'=free, [ 0>=1+free ], cost: NONTERM

     12: f0 -> f12 : A'=0, B'=free_3, C'=free_3, F'=0, G'=0, [ free>=1 ], cost: 1+2*free



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

   Start location: f0

     11: f0 -> [4] : A'=free, [ 0>=1+free ], cost: NONTERM

     12: f0 -> f12 : A'=0, B'=free_3, C'=free_3, F'=0, G'=0, [ free>=1 ], cost: 1+2*free



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

     11: f0 -> [4] : A'=free, [ 0>=1+free ], cost: NONTERM

     12: f0 -> f12 : A'=0, B'=free_3, C'=free_3, F'=0, G'=0, [ free>=1 ], cost: 1+2*free



Computing asymptotic complexity for rule 11

   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:  [ 0>=1+free ]



NO


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(2)
BOUNDS(INF, INF)