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


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


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 320 ms]
(2) BOUNDS(INF, INF)


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

(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C, D, E, F, G) -> Com_1(f12(H, I, J, 0, E, F, G)) :|: TRUE
f12(A, B, C, D, E, F, G) -> Com_1(f12(A, B, C, D + 1, E, F, G)) :|: C >= D + 1
f25(A, B, C, D, E, F, G) -> Com_1(f25(A, B, C, D, E, F + 1, G)) :|: E >= F + 1
f25(A, B, C, D, E, F, G) -> Com_1(f34(A, B, C, D, E, F, G)) :|: F >= E
f12(A, B, C, D, E, F, G) -> Com_1(f25(A, B, C, D, A, 0, H)) :|: D >= C

The start-symbols are:[f0_7]


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

(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f0 -> f12 : A'=free_1, B'=free_2, C'=free, D'=0, [], cost: 1

      1: f12 -> f12 : D'=1+D, [ C>=1+D ], cost: 1

      4: f12 -> f25 : E'=A, F'=0, G'=free_3, [ D>=C ], cost: 1

      2: f25 -> f25 : F'=1+F, [ E>=1+F ], cost: 1

      3: f25 -> f34 : [ F>=E ], cost: 1



Checking for constant complexity:

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

      0: f0 -> f12 : A'=free_1, B'=free_2, C'=free, D'=0, [], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

      0: f0 -> f12 : A'=free_1, B'=free_2, C'=free, D'=0, [], cost: 1

      1: f12 -> f12 : D'=1+D, [ C>=1+D ], cost: 1

      4: f12 -> f25 : E'=A, F'=0, G'=free_3, [ D>=C ], cost: 1

      2: f25 -> f25 : F'=1+F, [ E>=1+F ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: f12 -> f12 : D'=1+D, [ C>=1+D ], cost: 1



   Accelerated rule 1 with metering function C-D, yielding the new rule 5.

   Removing the simple loops: 1.



Accelerating simple loops of location 2.

   Accelerating the following rules:

      2: f25 -> f25 : F'=1+F, [ E>=1+F ], cost: 1



   Accelerated rule 2 with metering function -F+E, yielding the new rule 6.

   Removing the simple loops: 2.



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

   Start location: f0

      0: f0 -> f12 : A'=free_1, B'=free_2, C'=free, D'=0, [], cost: 1

      4: f12 -> f25 : E'=A, F'=0, G'=free_3, [ D>=C ], cost: 1

      5: f12 -> f12 : D'=C, [ C>=1+D ], cost: C-D

      6: f25 -> f25 : F'=E, [ E>=1+F ], cost: -F+E



Chained accelerated rules (with incoming rules):

   Start location: f0

      0: f0 -> f12 : A'=free_1, B'=free_2, C'=free, D'=0, [], cost: 1

      7: f0 -> f12 : A'=free_1, B'=free_2, C'=free, D'=free, [ free>=1 ], cost: 1+free

      4: f12 -> f25 : E'=A, F'=0, G'=free_3, [ D>=C ], cost: 1

      8: f12 -> f25 : E'=A, F'=A, G'=free_3, [ D>=C && A>=1 ], cost: 1+A



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

   Start location: f0

      0: f0 -> f12 : A'=free_1, B'=free_2, C'=free, D'=0, [], cost: 1

      7: f0 -> f12 : A'=free_1, B'=free_2, C'=free, D'=free, [ free>=1 ], cost: 1+free

      8: f12 -> f25 : E'=A, F'=A, G'=free_3, [ D>=C && A>=1 ], cost: 1+A



Eliminated locations (on tree-shaped paths):

   Start location: f0

      9: f0 -> f25 : A'=free_1, B'=free_2, C'=free, D'=0, E'=free_1, F'=free_1, G'=free_3, [ 0>=free && free_1>=1 ], cost: 2+free_1

     10: f0 -> f25 : A'=free_1, B'=free_2, C'=free, D'=free, E'=free_1, F'=free_1, G'=free_3, [ free>=1 && free_1>=1 ], cost: 2+free+free_1



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

      9: f0 -> f25 : A'=free_1, B'=free_2, C'=free, D'=0, E'=free_1, F'=free_1, G'=free_3, [ 0>=free && free_1>=1 ], cost: 2+free_1

     10: f0 -> f25 : A'=free_1, B'=free_2, C'=free, D'=free, E'=free_1, F'=free_1, G'=free_3, [ free>=1 && free_1>=1 ], cost: 2+free+free_1



Computing asymptotic complexity for rule 9

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      2+free_1 (+), 1-free (+/+!), free_1 (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {free==-n,free_1==n}

      resulting limit problem:

      [solved]



   Solution:

      free / -n

      free_1 / n

   Resulting cost 2+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: 2+n

   Rule cost:   2+free_1

   Rule guard:  [ 0>=free && free_1>=1 ]



WORST_CASE(INF,?)


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

(2)
BOUNDS(INF, INF)