/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 523 ms]
(2) BOUNDS(1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval(A, B) -> Com_1(eval(D, B)) :|: A >= 0 && B >= 0 && B >= A + 1 && A + B >= 2 * C && 3 * C >= A + B + 1 && D >= C + 1 && A + B >= 2 * E && 3 * E >= A + B + 1 && E + 1 >= D
eval(A, B) -> Com_1(eval(A, D)) :|: A >= 0 && B >= 0 && B >= A + 1 && A + B >= 2 * C && 3 * C >= A + B + 1 && D >= C && A + B >= 2 * E && 3 * E >= A + B + 1 && E >= D
start(A, B) -> Com_1(eval(A, B)) :|: TRUE

The start-symbols are:[start_2]


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(1) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: start

      0: eval -> eval : A'=free_1, [ A>=0 && B>=0 && B>=1+A && A+B>=2*free && 3*free>=1+A+B && free_1>=1+free && A+B>=2*free_2 && 3*free_2>=1+A+B && 1+free_2>=free_1 ], cost: 1

      1: eval -> eval : B'=free_4, [ A>=0 && B>=0 && B>=1+A && A+B>=2*free_3 && 3*free_3>=1+A+B && free_4>=free_3 && A+B>=2*free_5 && 3*free_5>=1+A+B && free_5>=free_4 ], cost: 1

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



Checking for constant complexity:

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

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



Simplified all rules, resulting in:

   Start location: start

      0: eval -> eval : A'=free_1, [ A>=0 && B>=1+A && 3*free>=1+A+B && free_1>=1+free && A+B>=2*free_2 && 3*free_2>=1+A+B && 1+free_2>=free_1 ], cost: 1

      1: eval -> eval : B'=free_4, [ A>=0 && B>=1+A && 3*free_3>=1+A+B && free_4>=free_3 && A+B>=2*free_5 && 3*free_5>=1+A+B && free_5>=free_4 ], 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'=free_1, [ A>=0 && B>=1+A && 3*free>=1+A+B && free_1>=1+free && A+B>=2*free_2 && 3*free_2>=1+A+B && 1+free_2>=free_1 ], cost: 1

      1: eval -> eval : B'=free_4, [ A>=0 && B>=1+A && 3*free_3>=1+A+B && free_4>=free_3 && A+B>=2*free_5 && 3*free_5>=1+A+B && free_5>=free_4 ], cost: 1



      During metering: Instantiating temporary variables by {free==-1+free_1,free_1==1+free,free_2==-1+free_1}

   Accelerated rule 0 with metering function 1+free-free_1, yielding the new rule 3.

      During metering: Instantiating temporary variables by {free_4==free_5,free_5==free_4,free_3==free_4}

   Accelerated rule 1 with metering function free_4-free_5, yielding the new rule 4.

   Removing the simple loops: 0 1.



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

   Start location: start

      3: eval -> eval : A'=1+free, [ A>=0 && B>=1+A && -3+3*free_1>=1+A+B && A+B>=-2+2*free_1 && free_1>=1+free && 1+free-free_1>=1 ], cost: 1+free-free_1

      4: eval -> eval : B'=free_5, [ A>=0 && B>=1+A && 3*free_4>=1+A+B && free_5>=free_4 && A+B>=2*free_4 && free_4-free_5>=1 ], cost: free_4-free_5

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



Chained accelerated rules (with incoming rules):

   Start location: start

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



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

   Start location: start

     <empty>



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

     <empty>



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)