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


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MAYBE
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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(1, INF).

(0) CpxIntTrs
(1) Loat Proof [FINISHED, 14.7 s]
(2) BOUNDS(1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval(A, B, C) -> Com_1(eval(A + 1, B + A, C)) :|: A >= B
eval(A, B, C) -> Com_1(eval(A - C, B + C * C, C - 1)) :|: A >= B
start(A, B, C) -> Com_1(eval(A, B, C)) :|: TRUE

The start-symbols are:[start_3]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: start

      0: eval -> eval : A'=1+A, B'=A+B, [ A>=B ], cost: 1

      1: eval -> eval : A'=-C+A, B'=C^2+B, C'=-1+C, [ A>=B ], cost: 1

      2: start -> eval : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: eval -> eval : A'=1+A, B'=A+B, [ A>=B ], cost: 1

      1: eval -> eval : A'=-C+A, B'=C^2+B, C'=-1+C, [ A>=B ], cost: 1



   Found no metering function for rule 0.

   Accelerated rule 1 with backward acceleration, yielding the new rule 3.

   Removing the simple loops: 1.



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

   Start location: start

      0: eval -> eval : A'=1+A, B'=A+B, [ A>=B ], cost: 1

      3: eval -> eval : A'=-1+1/2*k+A-k*C+1/2*k^2, B'=1-k^2*C-11/6*k+2*C+k*C^2+1/3*k^3-k*C+1/2*k^2+B, C'=-k+C, [ A>=B && k>0 && -3/2-(-1+k)*C+1/2*k+1/2*(-1+k)^2+A>=17/6-(-1+k)*C-11/6*k+2*C+(-1+k)*C^2+1/2*(-1+k)^2+1/3*(-1+k)^3-(-1+k)^2*C+B ], cost: k

      2: start -> eval : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: start

      2: start -> eval : [], cost: 1

      4: start -> eval : A'=1+A, B'=A+B, [ A>=B ], cost: 2

      5: start -> eval : A'=-1+1/2*k+A-k*C+1/2*k^2, B'=1-k^2*C-11/6*k+2*C+k*C^2+1/3*k^3-k*C+1/2*k^2+B, C'=-k+C, [ A>=B && k>0 && -3/2-(-1+k)*C+1/2*k+1/2*(-1+k)^2+A>=17/6-(-1+k)*C-11/6*k+2*C+(-1+k)*C^2+1/2*(-1+k)^2+1/3*(-1+k)^3-(-1+k)^2*C+B ], cost: 1+k



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

   Start location: start

      5: start -> eval : A'=-1+1/2*k+A-k*C+1/2*k^2, B'=1-k^2*C-11/6*k+2*C+k*C^2+1/3*k^3-k*C+1/2*k^2+B, C'=-k+C, [ A>=B && k>0 && -3/2-(-1+k)*C+1/2*k+1/2*(-1+k)^2+A>=17/6-(-1+k)*C-11/6*k+2*C+(-1+k)*C^2+1/2*(-1+k)^2+1/3*(-1+k)^3-(-1+k)^2*C+B ], cost: 1+k



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

      5: start -> eval : A'=-1+1/2*k+A-k*C+1/2*k^2, B'=1-k^2*C-11/6*k+2*C+k*C^2+1/3*k^3-k*C+1/2*k^2+B, C'=-k+C, [ A>=B && k>0 && -3/2-(-1+k)*C+1/2*k+1/2*(-1+k)^2+A>=17/6-(-1+k)*C-11/6*k+2*C+(-1+k)*C^2+1/2*(-1+k)^2+1/3*(-1+k)^3-(-1+k)^2*C+B ], cost: 1+k



Computing asymptotic complexity for rule 5

   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:  Unknown

   Cpx degree:  ?

   Solved cost: 0

   Rule cost:   0

   Rule guard:  []



WORST_CASE(Omega(0),?)


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(2)
BOUNDS(1, INF)