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


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MAYBE
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(1, INF).

(0) CpxIntTrs
(1) Loat Proof [FINISHED, 316 ms]
(2) BOUNDS(1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A) -> Com_1(f4(B)) :|: TRUE
f4(A) -> Com_1(f4(A + 1)) :|: 3 >= A && A >= 1
f4(A) -> Com_1(f4(1)) :|: 0 >= A && 3 >= A
f4(A) -> Com_1(f12(A)) :|: A >= 4

The start-symbols are:[f0_1]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f0 -> f4 : A'=free, [], cost: 1

      1: f4 -> f4 : A'=1+A, [ 3>=A && A>=1 ], cost: 1

      2: f4 -> f4 : A'=1, [ 0>=A && 3>=A ], cost: 1

      3: f4 -> f12 : [ A>=4 ], cost: 1



Checking for constant complexity:

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

      0: f0 -> f4 : A'=free, [], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

      0: f0 -> f4 : A'=free, [], cost: 1

      1: f4 -> f4 : A'=1+A, [ 3>=A && A>=1 ], cost: 1

      2: f4 -> f4 : A'=1, [ 0>=A && 3>=A ], cost: 1



Simplified all rules, resulting in:

   Start location: f0

      0: f0 -> f4 : A'=free, [], cost: 1

      1: f4 -> f4 : A'=1+A, [ 3>=A && A>=1 ], cost: 1

      2: f4 -> f4 : A'=1, [ 0>=A ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: f4 -> f4 : A'=1+A, [ 3>=A && A>=1 ], cost: 1

      2: f4 -> f4 : A'=1, [ 0>=A ], cost: 1



   Accelerated rule 1 with metering function 4-A, yielding the new rule 4.

   Found no metering function for rule 2.

   Removing the simple loops: 1.



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

   Start location: f0

      0: f0 -> f4 : A'=free, [], cost: 1

      2: f4 -> f4 : A'=1, [ 0>=A ], cost: 1

      4: f4 -> f4 : A'=4, [ 3>=A && A>=1 ], cost: 4-A



Chained accelerated rules (with incoming rules):

   Start location: f0

      0: f0 -> f4 : A'=free, [], cost: 1

      5: f0 -> f4 : A'=1, [], cost: 2

      6: f0 -> f4 : A'=4, [ 3>=free && free>=1 ], cost: 5-free



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

   Start location: f0

      6: f0 -> f4 : A'=4, [ 3>=free && free>=1 ], cost: 5-free



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

      6: f0 -> f4 : A'=4, [ 3>=free && free>=1 ], cost: 5-free



Computing asymptotic complexity for rule 6

   Could not solve the limit problem.

   Resulting cost 0 has complexity: Unknown



Obtained the following overall complexity (w.r.t. the length of the input n):

   Complexity:  Constant

   Cpx degree:  0

   Solved cost: 1

   Rule cost:   1

   Rule guard:  []



WORST_CASE(Omega(1),?)


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(2)
BOUNDS(1, INF)