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


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WORST_CASE(Omega(n^1), O(n^1))
proof of /export/starexec/sandbox/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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(2, 2 + Arg_0)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 38 ms]
(2) BOUNDS(1, max(2, 2 + Arg_0))
(3) Loat Proof [FINISHED, 145 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f2(A, B) -> Com_1(f3(A, 1)) :|: 0 >= A
f0(A, B) -> Com_1(f2(A, 0)) :|: TRUE
f2(A, B) -> Com_1(f2(A - 1, B)) :|: A >= 1

The start-symbols are:[f0_2]


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(1) Koat2 Proof (FINISHED)
YES( ?, max([2, 2+Arg_0]) {O(n)})



Initial Complexity Problem:

  Start:  f0  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  f0, f2, f3  

  Transitions:

  f0(Arg_0,Arg_1) -> f2(Arg_0,0):|:

  f2(Arg_0,Arg_1) -> f2(Arg_0-1,Arg_1):|:Arg_1 <= 0 && 0 <= Arg_1 && 1 <= Arg_0

  f2(Arg_0,Arg_1) -> f3(Arg_0,1):|:Arg_1 <= 0 && 0 <= Arg_1 && Arg_0 <= 0  



Timebounds: 

  Overall timebound: max([2, 2+Arg_0]) {O(n)}

  1: f0->f2: 1 {O(1)}

  0: f2->f3: 1 {O(1)}

  2: f2->f2: max([0, Arg_0]) {O(n)}



Costbounds:

  Overall costbound: max([2, 2+Arg_0]) {O(n)}

  1: f0->f2: 1 {O(1)}

  0: f2->f3: 1 {O(1)}

  2: f2->f2: max([0, Arg_0]) {O(n)}



Sizebounds:

`Lower:

  1: f0->f2, Arg_0: Arg_0 {O(n)}

  1: f0->f2, Arg_1: 0 {O(1)}

  0: f2->f3, Arg_0: min([0, Arg_0]) {O(n)}

  0: f2->f3, Arg_1: 1 {O(1)}

  2: f2->f2, Arg_0: 0 {O(1)}

  2: f2->f2, Arg_1: 0 {O(1)}

`Upper:

  1: f0->f2, Arg_0: Arg_0 {O(n)}

  1: f0->f2, Arg_1: 0 {O(1)}

  0: f2->f3, Arg_0: 0 {O(1)}

  0: f2->f3, Arg_1: 1 {O(1)}

  2: f2->f2, Arg_0: Arg_0 {O(n)}

  2: f2->f2, Arg_1: 0 {O(1)}


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

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f2 -> f3 : B'=1, [ 0>=A ], cost: 1

      2: f2 -> f2 : A'=-1+A, [ A>=1 ], cost: 1

      1: f0 -> f2 : B'=0, [], cost: 1



Removed unreachable and leaf rules:

   Start location: f0

      2: f2 -> f2 : A'=-1+A, [ A>=1 ], cost: 1

      1: f0 -> f2 : B'=0, [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      2: f2 -> f2 : A'=-1+A, [ A>=1 ], cost: 1



   Accelerated rule 2 with metering function A, yielding the new rule 3.

   Removing the simple loops: 2.



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

   Start location: f0

      3: f2 -> f2 : A'=0, [ A>=1 ], cost: A

      1: f0 -> f2 : B'=0, [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f0

      1: f0 -> f2 : B'=0, [], cost: 1

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



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

   Start location: f0

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



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

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



Computing asymptotic complexity for rule 4

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      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 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:  [ A>=1 ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)