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


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WORST_CASE(NON_POLY, ?)
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 630 ms]
(2) BOUNDS(INF, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f11(A, B, C, D, E, F) -> Com_1(f11(A - 2, B - 1, C + 1, G, E, F)) :|: A >= 1 && G >= 1
f11(A, B, C, D, E, F) -> Com_1(f11(A - 2, B, C, G, E, F)) :|: 0 >= G && A >= 1 && A >= B + 1
f21(A, B, C, D, E, F) -> Com_1(f21(A, B, C, D, E, F)) :|: TRUE
f23(A, B, C, D, E, F) -> Com_1(f26(A, B, C, D, E, F)) :|: TRUE
f11(A, B, C, D, E, F) -> Com_1(f21(A, B, C, D, E, F)) :|: 0 >= A
f0(A, B, C, D, E, F) -> Com_1(f11(4, G, 0, D, G, 4)) :|: G >= 1

The start-symbols are:[f0_6]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f11 -> f11 : A'=-2+A, B'=-1+B, C'=1+C, D'=free, [ A>=1 && free>=1 ], cost: 1

      1: f11 -> f11 : A'=-2+A, D'=free_1, [ 0>=free_1 && A>=1 && A>=1+B ], cost: 1

      4: f11 -> f21 : [ 0>=A ], cost: 1

      2: f21 -> f21 : [], cost: 1

      3: f23 -> f26 : [], cost: 1

      5: f0 -> f11 : A'=4, B'=free_2, C'=0, E'=free_2, F'=4, [ free_2>=1 ], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

      0: f11 -> f11 : A'=-2+A, B'=-1+B, C'=1+C, D'=free, [ A>=1 && free>=1 ], cost: 1

      1: f11 -> f11 : A'=-2+A, D'=free_1, [ 0>=free_1 && A>=1 && A>=1+B ], cost: 1

      4: f11 -> f21 : [ 0>=A ], cost: 1

      2: f21 -> f21 : [], cost: 1

      5: f0 -> f11 : A'=4, B'=free_2, C'=0, E'=free_2, F'=4, [ free_2>=1 ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: f11 -> f11 : A'=-2+A, B'=-1+B, C'=1+C, D'=free, [ A>=1 && free>=1 ], cost: 1

      1: f11 -> f11 : A'=-2+A, D'=free_1, [ 0>=free_1 && A>=1 && A>=1+B ], cost: 1



   Accelerated rule 0 with metering function meter (where 2*meter==A), yielding the new rule 6.

   Accelerated rule 1 with backward acceleration, yielding the new rule 7.

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

   Removing the simple loops: 0 1.



Accelerating simple loops of location 1.

   Accelerating the following rules:

      2: f21 -> f21 : [], cost: 1



   Accelerated rule 2 with NONTERM, yielding the new rule 8.

   Removing the simple loops: 2.



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

   Start location: f0

      4: f11 -> f21 : [ 0>=A ], cost: 1

      6: f11 -> f11 : A'=-2*meter+A, B'=-meter+B, C'=meter+C, D'=free, [ A>=1 && free>=1 && 2*meter==A && meter>=1 ], cost: meter

      7: f11 -> f11 : A'=-2*k+A, D'=free_1, [ 0>=free_1 && A>=1 && A>=1+B && k>0 && 2-2*k+A>=1 && 2-2*k+A>=1+B ], cost: k

      8: f21 -> [6] : [], cost: INF

      5: f0 -> f11 : A'=4, B'=free_2, C'=0, E'=free_2, F'=4, [ free_2>=1 ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f0

      4: f11 -> f21 : [ 0>=A ], cost: 1

     11: f11 -> [6] : [ 0>=A ], cost: INF

      5: f0 -> f11 : A'=4, B'=free_2, C'=0, E'=free_2, F'=4, [ free_2>=1 ], cost: 1

      9: f0 -> f11 : A'=0, B'=-2+free_2, C'=2, D'=free, E'=free_2, F'=4, [ free_2>=1 && free>=1 ], cost: 3

     10: f0 -> f11 : A'=4-2*k, B'=free_2, C'=0, D'=free_1, E'=free_2, F'=4, [ free_2>=1 && 0>=free_1 && 4>=1+free_2 && k>0 && 6-2*k>=1 && 6-2*k>=1+free_2 ], cost: 1+k



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

   Start location: f0

     11: f11 -> [6] : [ 0>=A ], cost: INF

      5: f0 -> f11 : A'=4, B'=free_2, C'=0, E'=free_2, F'=4, [ free_2>=1 ], cost: 1

      9: f0 -> f11 : A'=0, B'=-2+free_2, C'=2, D'=free, E'=free_2, F'=4, [ free_2>=1 && free>=1 ], cost: 3

     10: f0 -> f11 : A'=4-2*k, B'=free_2, C'=0, D'=free_1, E'=free_2, F'=4, [ free_2>=1 && 0>=free_1 && 4>=1+free_2 && k>0 && 6-2*k>=1 && 6-2*k>=1+free_2 ], cost: 1+k



Eliminated locations (on tree-shaped paths):

   Start location: f0

     12: f0 -> [6] : A'=0, B'=-2+free_2, C'=2, D'=free, E'=free_2, F'=4, [ free_2>=1 && free>=1 ], cost: INF

     13: f0 -> [6] : A'=4-2*k, B'=free_2, C'=0, D'=free_1, E'=free_2, F'=4, [ free_2>=1 && 0>=free_1 && 4>=1+free_2 && k>0 && 6-2*k>=1 && 6-2*k>=1+free_2 && 0>=4-2*k ], cost: INF



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

     12: f0 -> [6] : A'=0, B'=-2+free_2, C'=2, D'=free, E'=free_2, F'=4, [ free_2>=1 && free>=1 ], cost: INF

     13: f0 -> [6] : A'=4-2*k, B'=free_2, C'=0, D'=free_1, E'=free_2, F'=4, [ free_2>=1 && 0>=free_1 && 4>=1+free_2 && k>0 && 6-2*k>=1 && 6-2*k>=1+free_2 && 0>=4-2*k ], cost: INF



Computing asymptotic complexity for rule 12

   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:  [ free_2>=1 && free>=1 ]



NO


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(2)
BOUNDS(INF, INF)