/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(NON_POLY, ?)
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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(INF, INF).

(0) CpxIntTrs
(1) Loat Proof [FINISHED, 415 ms]
(2) BOUNDS(INF, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
eval(A, B) -> Com_1(eval(A - 1, C)) :|: A >= 0
eval(A, B) -> Com_1(eval(A, B - 1)) :|: B >= 0
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'=-1+A, B'=free, [ A>=0 ], cost: 1

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



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

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

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



   Accelerated rule 0 with metering function 1+A, yielding the new rule 3.

   Accelerated rule 1 with metering function 1+B, yielding the new rule 4.

   Nested simple loops 0 (outer loop) and 4 (inner loop) with metering function 1+A, resulting in the new rules: 5, 6.

   Removing the simple loops: 0 1.



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

   Start location: start

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

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

      5: eval -> eval : A'=-1, B'=-1, [ A>=0 && free>=0 ], cost: 2+free*(1+A)+2*A

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

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



Chained accelerated rules (with incoming rules):

   Start location: start

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

      7: start -> eval : A'=-1, B'=free, [ A>=0 ], cost: 2+A

      8: start -> eval : B'=-1, [ B>=0 ], cost: 2+B

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

     10: start -> eval : A'=-1, B'=-1, [ B>=0 && A>=0 && free>=0 ], cost: 4+free*(1+A)+2*A+B



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

   Start location: start

      7: start -> eval : A'=-1, B'=free, [ A>=0 ], cost: 2+A

      8: start -> eval : B'=-1, [ B>=0 ], cost: 2+B

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

     10: start -> eval : A'=-1, B'=-1, [ B>=0 && A>=0 && free>=0 ], cost: 4+free*(1+A)+2*A+B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: start

      7: start -> eval : A'=-1, B'=free, [ A>=0 ], cost: 2+A

      8: start -> eval : B'=-1, [ B>=0 ], cost: 2+B

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

     10: start -> eval : A'=-1, B'=-1, [ B>=0 && A>=0 && free>=0 ], cost: 4+free*(1+A)+2*A+B



Computing asymptotic complexity for rule 7

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      2+A (+), 1+A (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      A / n

   Resulting cost 2+n has complexity: Poly(n^1)



Found new complexity Poly(n^1).



Computing asymptotic complexity for rule 9

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      3+free*A+free+2*A (+), 1+free (+/+!), 1+A (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      free / n

      A / 0

   Resulting cost 3+n has complexity: Unbounded



Found new complexity Unbounded.



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

   Complexity:  Unbounded

   Cpx degree:  Unbounded

   Solved cost: 3+n

   Rule cost:   3+free*(1+A)+2*A

   Rule guard:  [ A>=0 && free>=0 ]



WORST_CASE(INF,?)


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(2)
BOUNDS(INF, INF)