/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 323 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) -> Com_1(f2(A, B, C, D, E, F, G)) :|: 0 >= A
f0(A, B, C, D, E, F, G) -> Com_1(f0(-(1) + A, C, -(1) + C, A, E, F, G)) :|: A >= 1
f1(A, B, C, D, E, F, G) -> Com_1(f0(-(1) + A, B, -(1) + C, D, C, A, G)) :|: A >= 1 && C >= 1
f2(A, B, C, D, E, F, G) -> Com_1(f4(A, B, C, D, E, F, H)) :|: 0 >= C
f2(A, B, C, D, E, F, G) -> Com_1(f0(H, B, C, D, E, F, G)) :|: H >= 1 && C >= 1
f3(A, B, C, D, E, F, G) -> Com_1(f2(H, B, I, D, E, F, G)) :|: TRUE

The start-symbols are:[f3_7]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f3

      0: f0 -> f2 : [ 0>=A ], cost: 1

      1: f0 -> f0 : A'=-1+A, B'=C, C'=-1+C, D'=A, [ A>=1 ], cost: 1

      2: f1 -> f0 : A'=-1+A, C'=-1+C, E'=C, F'=A, [ A>=1 && C>=1 ], cost: 1

      3: f2 -> f4 : G'=free, [ 0>=C ], cost: 1

      4: f2 -> f0 : A'=free_1, [ free_1>=1 && C>=1 ], cost: 1

      5: f3 -> f2 : A'=free_2, C'=free_3, [], cost: 1



Checking for constant complexity:

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

      5: f3 -> f2 : A'=free_2, C'=free_3, [], cost: 1



Removed unreachable and leaf rules:

   Start location: f3

      0: f0 -> f2 : [ 0>=A ], cost: 1

      1: f0 -> f0 : A'=-1+A, B'=C, C'=-1+C, D'=A, [ A>=1 ], cost: 1

      4: f2 -> f0 : A'=free_1, [ free_1>=1 && C>=1 ], cost: 1

      5: f3 -> f2 : A'=free_2, C'=free_3, [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      1: f0 -> f0 : A'=-1+A, B'=C, C'=-1+C, D'=A, [ A>=1 ], cost: 1



   Accelerated rule 1 with metering function A, yielding the new rule 6.

   Removing the simple loops: 1.



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

   Start location: f3

      0: f0 -> f2 : [ 0>=A ], cost: 1

      6: f0 -> f0 : A'=0, B'=1+C-A, C'=C-A, D'=1, [ A>=1 ], cost: A

      4: f2 -> f0 : A'=free_1, [ free_1>=1 && C>=1 ], cost: 1

      5: f3 -> f2 : A'=free_2, C'=free_3, [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f3

      0: f0 -> f2 : [ 0>=A ], cost: 1

      4: f2 -> f0 : A'=free_1, [ free_1>=1 && C>=1 ], cost: 1

      7: f2 -> f0 : A'=0, B'=1+C-free_1, C'=C-free_1, D'=1, [ free_1>=1 && C>=1 ], cost: 1+free_1

      5: f3 -> f2 : A'=free_2, C'=free_3, [], cost: 1



Eliminated locations (on tree-shaped paths):

   Start location: f3

      8: f2 -> f2 : A'=0, B'=1+C-free_1, C'=C-free_1, D'=1, [ free_1>=1 && C>=1 ], cost: 2+free_1

      5: f3 -> f2 : A'=free_2, C'=free_3, [], cost: 1



Accelerating simple loops of location 2.

   Accelerating the following rules:

      8: f2 -> f2 : A'=0, B'=1+C-free_1, C'=C-free_1, D'=1, [ free_1>=1 && C>=1 ], cost: 2+free_1



      During metering: Instantiating temporary variables by {free_1==1}

   Accelerated rule 8 with metering function C, yielding the new rule 9.

   Removing the simple loops: 8.



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

   Start location: f3

      9: f2 -> f2 : A'=0, B'=1, C'=0, D'=1, [ C>=1 ], cost: 3*C

      5: f3 -> f2 : A'=free_2, C'=free_3, [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f3

      5: f3 -> f2 : A'=free_2, C'=free_3, [], cost: 1

     10: f3 -> f2 : A'=0, B'=1, C'=0, D'=1, [ free_3>=1 ], cost: 1+3*free_3



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

   Start location: f3

     10: f3 -> f2 : A'=0, B'=1, C'=0, D'=1, [ free_3>=1 ], cost: 1+3*free_3



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f3

     10: f3 -> f2 : A'=0, B'=1, C'=0, D'=1, [ free_3>=1 ], cost: 1+3*free_3



Computing asymptotic complexity for rule 10

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      free_3 (+/+!), 1+3*free_3 (+) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {free_3==n}

      resulting limit problem:

      [solved]



   Solution:

      free_3 / n

   Resulting cost 1+3*n has complexity: Unbounded



Found new complexity Unbounded.



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

   Complexity:  Unbounded

   Cpx degree:  Unbounded

   Solved cost: 1+3*n

   Rule cost:   1+3*free_3

   Rule guard:  [ free_3>=1 ]



WORST_CASE(INF,?)


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(2)
BOUNDS(INF, INF)