/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 110 ms]
(2) BOUNDS(INF, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f2(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f2(A, L, M, N, F, F, G, H, I, J, K)) :|: A >= 2
f2(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f1(A, L, M, D, E, F, N, H, I, J, K)) :|: 1 >= A
f300(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f2(A, L, M, N, O, O, G, H, P, Q, R)) :|: H >= 1
f300(A, B, C, D, E, F, G, H, I, J, K) -> Com_1(f1(A, L, M, D, N, P, O, H, Q, N, R)) :|: 0 >= H

The start-symbols are:[f300_11]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f300

      0: f2 -> f2 : B'=free_2, C'=free, D'=free_1, E'=F, [ A>=2 ], cost: 1

      1: f2 -> f1 : B'=free_5, C'=free_3, G'=free_4, [ 1>=A ], cost: 1

      2: f300 -> f2 : B'=free_12, C'=free_7, D'=free_9, E'=free_11, F'=free_11, Q'=free_6, J'=free_8, K'=free_10, [ H>=1 ], cost: 1

      3: f300 -> f1 : B'=free_19, C'=free_14, E'=free_16, F'=free_18, G'=free_13, Q'=free_15, J'=free_16, K'=free_17, [ 0>=H ], cost: 1



Checking for constant complexity:

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

      2: f300 -> f2 : B'=free_12, C'=free_7, D'=free_9, E'=free_11, F'=free_11, Q'=free_6, J'=free_8, K'=free_10, [ H>=1 ], cost: 1



Removed unreachable and leaf rules:

   Start location: f300

      0: f2 -> f2 : B'=free_2, C'=free, D'=free_1, E'=F, [ A>=2 ], cost: 1

      2: f300 -> f2 : B'=free_12, C'=free_7, D'=free_9, E'=free_11, F'=free_11, Q'=free_6, J'=free_8, K'=free_10, [ H>=1 ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: f2 -> f2 : B'=free_2, C'=free, D'=free_1, E'=F, [ A>=2 ], cost: 1



   Accelerated rule 0 with NONTERM, yielding the new rule 4.

   Removing the simple loops: 0.



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

   Start location: f300

      4: f2 -> [3] : [ A>=2 ], cost: NONTERM

      2: f300 -> f2 : B'=free_12, C'=free_7, D'=free_9, E'=free_11, F'=free_11, Q'=free_6, J'=free_8, K'=free_10, [ H>=1 ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f300

      2: f300 -> f2 : B'=free_12, C'=free_7, D'=free_9, E'=free_11, F'=free_11, Q'=free_6, J'=free_8, K'=free_10, [ H>=1 ], cost: 1

      5: f300 -> [3] : B'=free_12, C'=free_7, D'=free_9, E'=free_11, F'=free_11, Q'=free_6, J'=free_8, K'=free_10, [ H>=1 && A>=2 ], cost: NONTERM



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

   Start location: f300

      5: f300 -> [3] : B'=free_12, C'=free_7, D'=free_9, E'=free_11, F'=free_11, Q'=free_6, J'=free_8, K'=free_10, [ H>=1 && A>=2 ], cost: NONTERM



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f300

      5: f300 -> [3] : B'=free_12, C'=free_7, D'=free_9, E'=free_11, F'=free_11, Q'=free_6, J'=free_8, K'=free_10, [ H>=1 && A>=2 ], cost: NONTERM



Computing asymptotic complexity for rule 5

   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:  [ H>=1 && A>=2 ]



NO


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(2)
BOUNDS(INF, INF)