/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(Omega(n^1), ?)
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(n^1, INF).

(0) CpxIntTrs
(1) Loat Proof [FINISHED, 420 ms]
(2) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B, C) -> Com_1(f2(A, B, C)) :|: A >= 0 && B >= C
f2(A, B, C) -> Com_1(f2(A - 1, B, C - 1)) :|: A >= 1 && B + 1 >= C
f2(A, B, C) -> Com_1(f2(A, B + C - 1, C - 1)) :|: A >= 0 && B >= 0

The start-symbols are:[f0_3]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f0 -> f2 : [ A>=0 && B>=C ], cost: 1

      1: f2 -> f2 : A'=-1+A, C'=-1+C, [ A>=1 && 1+B>=C ], cost: 1

      2: f2 -> f2 : B'=-1+C+B, C'=-1+C, [ A>=0 && B>=0 ], cost: 1



Checking for constant complexity:

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

      0: f0 -> f2 : [ A>=0 && B>=C ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: f2 -> f2 : A'=-1+A, C'=-1+C, [ A>=1 && 1+B>=C ], cost: 1

      2: f2 -> f2 : B'=-1+C+B, C'=-1+C, [ A>=0 && B>=0 ], cost: 1



   Accelerated rule 1 with metering function A, yielding the new rule 3.

   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 -> f2 : [ A>=0 && B>=C ], cost: 1

      2: f2 -> f2 : B'=-1+C+B, C'=-1+C, [ A>=0 && B>=0 ], cost: 1

      3: f2 -> f2 : A'=0, C'=C-A, [ A>=1 && 1+B>=C ], cost: A



Chained accelerated rules (with incoming rules):

   Start location: f0

      0: f0 -> f2 : [ A>=0 && B>=C ], cost: 1

      4: f0 -> f2 : B'=-1+C+B, C'=-1+C, [ A>=0 && B>=C && B>=0 ], cost: 2

      5: f0 -> f2 : A'=0, C'=C-A, [ B>=C && A>=1 ], cost: 1+A



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

   Start location: f0

      5: f0 -> f2 : A'=0, C'=C-A, [ B>=C && A>=1 ], cost: 1+A



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

      5: f0 -> f2 : A'=0, C'=C-A, [ B>=C && A>=1 ], cost: 1+A



Computing asymptotic complexity for rule 5

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      1-C+B (+/+!), A (+/+!), 1+A (+) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {C==0,A==n,B==0}

      resulting limit problem:

      [solved]



   Solution:

      C / 0

      A / n

      B / 0

   Resulting cost 1+n has complexity: Poly(n^1)



Found new complexity Poly(n^1).



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

   Complexity:  Poly(n^1)

   Cpx degree:  1

   Solved cost: 1+n

   Rule cost:   1+A

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



WORST_CASE(Omega(n^1),?)


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(2)
BOUNDS(n^1, INF)