/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, nat(1 + Arg_1) + max(1, 2 + Arg_0)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 147 ms]
(2) BOUNDS(1, nat(1 + Arg_1) + max(1, 2 + Arg_0))
(3) Loat Proof [FINISHED, 433 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f1(A, B) -> Com_1(f2(A, B)) :|: A >= 1 && B >= 1
f2(A, B) -> Com_1(f2(A - 1, B)) :|: A >= 2 && B >= 1
f2(A, B) -> Com_1(f2(A, B - 1)) :|: A >= 1 && B >= 2

The start-symbols are:[f1_2]


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(1) Koat2 Proof (FINISHED)
YES( ?, max([0, 1+Arg_1])+max([1, 2+Arg_0]) {O(n)})



Initial Complexity Problem:

  Start:  f1  

  Program_Vars:  Arg_0, Arg_1  

  Temp_Vars:    

  Locations:  f1, f2  

  Transitions:

  0: f1->f2

  1: f2->f2

  2: f2->f2  



Timebounds: 

  Overall timebound: max([0, 1+Arg_1])+max([1, 2+Arg_0]) {O(n)}

  0: f1->f2: 1 {O(1)}

  1: f2->f2: max([0, 1+Arg_0]) {O(n)}

  2: f2->f2: max([0, 1+Arg_1]) {O(n)}



Costbounds:

  Overall costbound: max([0, 1+Arg_1])+max([1, 2+Arg_0]) {O(n)}

  0: f1->f2: 1 {O(1)}

  1: f2->f2: max([0, 1+Arg_0]) {O(n)}

  2: f2->f2: max([0, 1+Arg_1]) {O(n)}



Sizebounds:

`Lower:

  0: f1->f2, Arg_0: 1 {O(1)}

  0: f1->f2, Arg_1: 1 {O(1)}

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

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

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

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

`Upper:

  0: f1->f2, Arg_0: Arg_0 {O(n)}

  0: f1->f2, Arg_1: Arg_1 {O(n)}

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

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

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

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


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

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f1

      0: f1 -> f2 : [ A>=1 && B>=1 ], cost: 1

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

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



Checking for constant complexity:

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

      0: f1 -> f2 : [ A>=1 && B>=1 ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

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

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



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

   Accelerated rule 2 with metering function -1+B, yielding the new rule 4.

   Removing the simple loops: 1 2.



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

   Start location: f1

      0: f1 -> f2 : [ A>=1 && B>=1 ], cost: 1

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

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



Chained accelerated rules (with incoming rules):

   Start location: f1

      0: f1 -> f2 : [ A>=1 && B>=1 ], cost: 1

      5: f1 -> f2 : A'=1, [ B>=1 && A>=2 ], cost: A

      6: f1 -> f2 : B'=1, [ A>=1 && B>=2 ], cost: B



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

   Start location: f1

      5: f1 -> f2 : A'=1, [ B>=1 && A>=2 ], cost: A

      6: f1 -> f2 : B'=1, [ A>=1 && B>=2 ], cost: B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f1

      5: f1 -> f2 : A'=1, [ B>=1 && A>=2 ], cost: A

      6: f1 -> f2 : B'=1, [ A>=1 && B>=2 ], cost: B



Computing asymptotic complexity for rule 5

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      A / n

      B / n

   Resulting cost 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: n

   Rule cost:   A

   Rule guard:  [ B>=1 && A>=2 ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)