/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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_2 + max(Arg_1 + nat(1 + Arg_0 + -1 * Arg_1), Arg_1)) + max(2, 3 + Arg_0 + -1 * Arg_1)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 2244 ms]
(2) BOUNDS(1, nat(-1 * Arg_2 + max(Arg_1 + nat(1 + Arg_0 + -1 * Arg_1), Arg_1)) + max(2, 3 + Arg_0 + -1 * Arg_1))
(3) Loat Proof [FINISHED, 360 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f1(A, B, C) -> Com_1(f1(A, B + 1, C)) :|: A >= B + 1
f3(A, B, C) -> Com_1(f1(A, B, C)) :|: B >= C + 1
f1(A, B, C) -> Com_1(f1(A, B, C + 1)) :|: B >= C + 2 && B >= A

The start-symbols are:[f3_3]


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(1) Koat2 Proof (FINISHED)
YES( ?, max([0, -(Arg_2)+max([Arg_1, Arg_1+max([0, 1+Arg_0-Arg_1])])])+max([2, 3+Arg_0-Arg_1]) {O(n)})



Initial Complexity Problem:

  Start:  f3  

  Program_Vars:  Arg_0, Arg_1, Arg_2  

  Temp_Vars:    

  Locations:  f1, f3, n_f1___1  

  Transitions:

  135: f1->f1

  134: f1->n_f1___1

  1: f3->f1

  133: n_f1___1->n_f1___1  



Timebounds: 

  Overall timebound: max([0, -(Arg_2)+max([Arg_1, Arg_1+max([0, 1+Arg_0-Arg_1])])])+max([2, 3+Arg_0-Arg_1]) {O(n)}

  134: f1->n_f1___1: 1 {O(1)}

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

  1: f3->f1: 1 {O(1)}

  133: n_f1___1->n_f1___1: max([0, -(Arg_2)+max([Arg_1, Arg_1+max([0, 1+Arg_0-Arg_1])])]) {O(n)}



Costbounds:

  Overall costbound: max([0, -(Arg_2)+max([Arg_1, Arg_1+max([0, 1+Arg_0-Arg_1])])])+max([2, 3+Arg_0-Arg_1]) {O(n)}

  134: f1->n_f1___1: 1 {O(1)}

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

  1: f3->f1: 1 {O(1)}

  133: n_f1___1->n_f1___1: max([0, -(Arg_2)+max([Arg_1, Arg_1+max([0, 1+Arg_0-Arg_1])])]) {O(n)}



Sizebounds:

`Lower:

  134: f1->n_f1___1, Arg_0: Arg_0 {O(n)}

  134: f1->n_f1___1, Arg_1: Arg_1 {O(n)}

  134: f1->n_f1___1, Arg_2: 1+Arg_2 {O(n)}

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

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

  135: f1->f1, Arg_2: Arg_2 {O(n)}

  1: f3->f1, Arg_0: Arg_0 {O(n)}

  1: f3->f1, Arg_1: Arg_1 {O(n)}

  1: f3->f1, Arg_2: Arg_2 {O(n)}

  133: n_f1___1->n_f1___1, Arg_0: Arg_0 {O(n)}

  133: n_f1___1->n_f1___1, Arg_1: Arg_1 {O(n)}

  133: n_f1___1->n_f1___1, Arg_2: 1+Arg_2 {O(n)}

`Upper:

  134: f1->n_f1___1, Arg_0: Arg_0 {O(n)}

  134: f1->n_f1___1, Arg_1: max([Arg_1, Arg_1+max([0, 1+Arg_0-Arg_1])]) {O(n)}

  134: f1->n_f1___1, Arg_2: 1+Arg_2 {O(n)}

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

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

  135: f1->f1, Arg_2: Arg_2 {O(n)}

  1: f3->f1, Arg_0: Arg_0 {O(n)}

  1: f3->f1, Arg_1: Arg_1 {O(n)}

  1: f3->f1, Arg_2: Arg_2 {O(n)}

  133: n_f1___1->n_f1___1, Arg_0: Arg_0 {O(n)}

  133: n_f1___1->n_f1___1, Arg_1: max([Arg_1, Arg_1+max([0, 1+Arg_0-Arg_1])]) {O(n)}

  133: n_f1___1->n_f1___1, Arg_2: 1+Arg_2+max([0, -(Arg_2)+max([Arg_1, Arg_1+max([0, 1+Arg_0-Arg_1])])]) {O(n)}


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(2)
BOUNDS(1, nat(-1 * Arg_2 + max(Arg_1 + nat(1 + Arg_0 + -1 * Arg_1), Arg_1)) + max(2, 3 + Arg_0 + -1 * Arg_1))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f3

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

      2: f1 -> f1 : C'=1+C, [ B>=2+C && B>=A ], cost: 1

      1: f3 -> f1 : [ B>=1+C ], cost: 1



Checking for constant complexity:

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

      1: f3 -> f1 : [ B>=1+C ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

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

      2: f1 -> f1 : C'=1+C, [ B>=2+C && B>=A ], cost: 1



   Accelerated rule 0 with metering function A-B, yielding the new rule 3.

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

   Removing the simple loops: 0 2.



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

   Start location: f3

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

      4: f1 -> f1 : C'=-1+B, [ B>=2+C && B>=A ], cost: -1-C+B

      1: f3 -> f1 : [ B>=1+C ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f3

      1: f3 -> f1 : [ B>=1+C ], cost: 1

      5: f3 -> f1 : B'=A, [ B>=1+C && A>=1+B ], cost: 1+A-B

      6: f3 -> f1 : C'=-1+B, [ B>=2+C && B>=A ], cost: -C+B



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

   Start location: f3

      5: f3 -> f1 : B'=A, [ B>=1+C && A>=1+B ], cost: 1+A-B

      6: f3 -> f1 : C'=-1+B, [ B>=2+C && B>=A ], cost: -C+B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f3

      5: f3 -> f1 : B'=A, [ B>=1+C && A>=1+B ], cost: 1+A-B

      6: f3 -> f1 : C'=-1+B, [ B>=2+C && B>=A ], cost: -C+B



Computing asymptotic complexity for rule 5

   Solved the limit problem by the following transformations:

      Created initial limit problem:

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



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



      applying transformation rule (C) using substitution {C==0,A==n,B==1}

      resulting limit problem:

      [solved]



   Solution:

      C / 0

      A / n

      B / 1

   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:   1+A-B

   Rule guard:  [ B>=1+C && A>=1+B ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)