/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(INF, INF).

(0) CpxIntTrs
(1) Loat Proof [FINISHED, 165 ms]
(2) BOUNDS(INF, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f1(A, B) -> Com_1(f0(A, B)) :|: TRUE
f0(A, B) -> Com_1(f0(A + B, B)) :|: A >= 1 && B >= 1
f0(A, B) -> Com_1(f0(A + B, B)) :|: A >= 1 && 0 >= B + 1

The start-symbols are:[f1_2]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f1

      0: f1 -> f0 : [], cost: 1

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

      2: f0 -> f0 : A'=A+B, [ A>=1 && 0>=1+B ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

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

      2: f0 -> f0 : A'=A+B, [ A>=1 && 0>=1+B ], cost: 1



   Accelerated rule 1 with NONTERM, yielding the new rule 3.

   Accelerated rule 2 with backward acceleration, yielding the new rule 4.

   Removing the simple loops: 1 2.



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

   Start location: f1

      0: f1 -> f0 : [], cost: 1

      3: f0 -> [2] : [ A>=1 && B>=1 ], cost: INF

      4: f0 -> f0 : A'=k*B+A, [ A>=1 && 0>=1+B && k>0 && (-1+k)*B+A>=1 ], cost: k



Chained accelerated rules (with incoming rules):

   Start location: f1

      0: f1 -> f0 : [], cost: 1

      5: f1 -> [2] : [ A>=1 && B>=1 ], cost: INF

      6: f1 -> f0 : A'=k*B+A, [ A>=1 && 0>=1+B && k>0 && (-1+k)*B+A>=1 ], cost: 1+k



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

   Start location: f1

      5: f1 -> [2] : [ A>=1 && B>=1 ], cost: INF

      6: f1 -> f0 : A'=k*B+A, [ A>=1 && 0>=1+B && k>0 && (-1+k)*B+A>=1 ], cost: 1+k



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f1

      5: f1 -> [2] : [ A>=1 && B>=1 ], cost: INF

      6: f1 -> f0 : A'=k*B+A, [ A>=1 && 0>=1+B && k>0 && (-1+k)*B+A>=1 ], cost: 1+k



Computing asymptotic complexity for rule 5

   Resulting cost INF 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: INF

   Rule cost:   INF

   Rule guard:  [ A>=1 && B>=1 ]



NO


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(2)
BOUNDS(INF, INF)