/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 222 ms]
(2) BOUNDS(INF, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C, D) -> Com_1(f6(B, B, D, D)) :|: TRUE
f6(A, B, C, D) -> Com_1(f6(A - 1, B, C - 1, D)) :|: 0 >= A + 1
f6(A, B, C, D) -> Com_1(f6(A - 1, B, C - 1, D)) :|: A >= 1
f6(A, B, C, D) -> Com_1(f14(0, B, C, D)) :|: D >= B + 1 && A >= 0 && A <= 0
f6(A, B, C, D) -> Com_1(f14(0, B, C, D)) :|: B >= 1 + D && A >= 0 && A <= 0
f6(A, B, C, D) -> Com_1(f14(0, B, C, B)) :|: A >= 0 && A <= 0 && B >= D && B <= D

The start-symbols are:[f0_4]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f0 -> f6 : A'=B, C'=D, [], cost: 1

      1: f6 -> f6 : A'=-1+A, C'=-1+C, [ 0>=1+A ], cost: 1

      2: f6 -> f6 : A'=-1+A, C'=-1+C, [ A>=1 ], cost: 1

      3: f6 -> f14 : A'=0, [ D>=1+B && A==0 ], cost: 1

      4: f6 -> f14 : A'=0, [ B>=1+D && A==0 ], cost: 1

      5: f6 -> f14 : A'=0, D'=B, [ A==0 && B==D ], cost: 1



Checking for constant complexity:

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

      0: f0 -> f6 : A'=B, C'=D, [], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

      0: f0 -> f6 : A'=B, C'=D, [], cost: 1

      1: f6 -> f6 : A'=-1+A, C'=-1+C, [ 0>=1+A ], cost: 1

      2: f6 -> f6 : A'=-1+A, C'=-1+C, [ A>=1 ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: f6 -> f6 : A'=-1+A, C'=-1+C, [ 0>=1+A ], cost: 1

      2: f6 -> f6 : A'=-1+A, C'=-1+C, [ A>=1 ], cost: 1



   Accelerated rule 1 with NONTERM, yielding the new rule 6.

   Accelerated rule 2 with metering function A, yielding the new rule 7.

   Removing the simple loops: 1 2.



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

   Start location: f0

      0: f0 -> f6 : A'=B, C'=D, [], cost: 1

      6: f6 -> [3] : [ 0>=1+A ], cost: NONTERM

      7: f6 -> f6 : A'=0, C'=C-A, [ A>=1 ], cost: A



Chained accelerated rules (with incoming rules):

   Start location: f0

      0: f0 -> f6 : A'=B, C'=D, [], cost: 1

      8: f0 -> [3] : A'=B, C'=D, [ 0>=1+B ], cost: NONTERM

      9: f0 -> f6 : A'=0, C'=D-B, [ B>=1 ], cost: 1+B



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

   Start location: f0

      8: f0 -> [3] : A'=B, C'=D, [ 0>=1+B ], cost: NONTERM

      9: f0 -> f6 : A'=0, C'=D-B, [ B>=1 ], cost: 1+B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

      8: f0 -> [3] : A'=B, C'=D, [ 0>=1+B ], cost: NONTERM

      9: f0 -> f6 : A'=0, C'=D-B, [ B>=1 ], cost: 1+B



Computing asymptotic complexity for rule 8

   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+B ]



NO


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(2)
BOUNDS(INF, INF)