/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), O(n^1))
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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(n^1, max(1, 1 + Arg_0 + Arg_1)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 35 ms]
(2) BOUNDS(1, max(1, 1 + Arg_0 + Arg_1))
(3) Loat Proof [FINISHED, 640 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f0(A, B) -> Com_1(f1(A, B)) :|: A >= 1
f1(A, B) -> Com_1(f1(A, B - A)) :|: A >= 1 && B >= 0

The start-symbols are:[f0_2]


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(1) Koat2 Proof (FINISHED)
YES( ?, max([1, 1+Arg_0+Arg_1]) {O(n)})



Initial Complexity Problem:

  Start:  f0  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  f0, f1  

  Transitions:

  f0(Arg_0,Arg_1) -> f1(Arg_0,Arg_1):|:1 <= Arg_0

  f1(Arg_0,Arg_1) -> f1(Arg_0,Arg_1-Arg_0):|:1 <= Arg_0 && 1 <= Arg_0 && 0 <= Arg_1  



Timebounds: 

  Overall timebound: max([1, 1+Arg_0+Arg_1]) {O(n)}

  0: f0->f1: 1 {O(1)}

  1: f1->f1: max([0, Arg_0+Arg_1]) {O(n)}



Costbounds:

  Overall costbound: max([1, 1+Arg_0+Arg_1]) {O(n)}

  0: f0->f1: 1 {O(1)}

  1: f1->f1: max([0, Arg_0+Arg_1]) {O(n)}



Sizebounds:

`Lower:

  0: f0->f1, Arg_0: 1 {O(1)}

  0: f0->f1, Arg_1: Arg_1 {O(n)}

  1: f1->f1, Arg_0: 1 {O(1)}

  1: f1->f1, Arg_1: -(Arg_0) {O(n)}

`Upper:

  0: f0->f1, Arg_0: Arg_0 {O(n)}

  0: f0->f1, Arg_1: Arg_1 {O(n)}

  1: f1->f1, Arg_0: Arg_0 {O(n)}

  1: f1->f1, Arg_1: Arg_1 {O(n)}


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(2)
BOUNDS(1, max(1, 1 + Arg_0 + Arg_1))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f0 -> f1 : [ A>=1 ], cost: 1

      1: f1 -> f1 : B'=-A+B, [ A>=1 && B>=0 ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: f1 -> f1 : B'=-A+B, [ A>=1 && B>=0 ], cost: 1



   Accelerated rule 1 with backward acceleration, yielding the new rule 2.

   Removing the simple loops: 1.



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

   Start location: f0

      0: f0 -> f1 : [ A>=1 ], cost: 1

      2: f1 -> f1 : B'=-k*A+B, [ A>=1 && B>=0 && k>0 && -(-1+k)*A+B>=0 ], cost: k



Chained accelerated rules (with incoming rules):

   Start location: f0

      0: f0 -> f1 : [ A>=1 ], cost: 1

      3: f0 -> f1 : B'=-k*A+B, [ A>=1 && B>=0 && k>0 && -(-1+k)*A+B>=0 ], cost: 1+k



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

   Start location: f0

      3: f0 -> f1 : B'=-k*A+B, [ A>=1 && B>=0 && k>0 && -(-1+k)*A+B>=0 ], cost: 1+k



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

      3: f0 -> f1 : B'=-k*A+B, [ A>=1 && B>=0 && k>0 && -(-1+k)*A+B>=0 ], cost: 1+k



Computing asymptotic complexity for rule 3

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {k==n,A==1,B==2*n}

      resulting limit problem:

      [solved]



   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      applying transformation rule (C) using substitution {A==1}

      resulting limit problem:

      1 (+/+!), 2-k+B (+/+!), 1+B (+/+!), k (+/+!), 1+k (+) [not solved]



      applying transformation rule (B), deleting 1 (+/+!)

      resulting limit problem:

      2-k+B (+/+!), 1+B (+/+!), k (+/+!), 1+k (+) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      k / n

      A / 1

      B / 2*n

   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+k

   Rule guard:  [ A>=1 && B>=0 && k>0 && -(-1+k)*A+B>=0 ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)