/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^2), O(n^2))
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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^2, max(3, 1 + 2 * Arg_2) + max(2, 2 * Arg_2)^2).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 347 ms]
(2) BOUNDS(1, max(3, 1 + 2 * Arg_2) + max(2, 2 * Arg_2)^2)
(3) Loat Proof [FINISHED, 337 ms]
(4) BOUNDS(n^2, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f4(A, B, C) -> Com_1(f4(A, B + 1, C)) :|: A >= B + 1
f4(A, B, C) -> Com_1(f4(A + 1, 0, C)) :|: C >= A + 2 && B >= A
f0(A, B, C) -> Com_1(f4(0, 0, C)) :|: C >= 1

The start-symbols are:[f0_3]


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(1) Koat2 Proof (FINISHED)
YES( ?, 1+max([2, 2*Arg_2])*max([2, 2*Arg_2])+max([2, 2*Arg_2]) {O(n^2)})



Initial Complexity Problem:

  Start:  f0  

  Program_Vars:  Arg_0, Arg_1, Arg_2  

  Temp_Vars:    

  Locations:  f0, f4  

  Transitions:

  f0(Arg_0,Arg_1,Arg_2) -> f4(0,0,Arg_2):|:1 <= Arg_2

  f4(Arg_0,Arg_1,Arg_2) -> f4(Arg_0,Arg_1+1,Arg_2):|:1 <= Arg_2 && 1 <= Arg_1+Arg_2 && 1+Arg_1 <= Arg_2 && 1 <= Arg_0+Arg_2 && 1+Arg_0 <= Arg_2 && Arg_1 <= Arg_0 && 0 <= Arg_1 && 0 <= Arg_0+Arg_1 && 0 <= Arg_0 && Arg_1+1 <= Arg_0

  f4(Arg_0,Arg_1,Arg_2) -> f4(Arg_0+1,0,Arg_2):|:1 <= Arg_2 && 1 <= Arg_1+Arg_2 && 1+Arg_1 <= Arg_2 && 1 <= Arg_0+Arg_2 && 1+Arg_0 <= Arg_2 && Arg_1 <= Arg_0 && 0 <= Arg_1 && 0 <= Arg_0+Arg_1 && 0 <= Arg_0 && Arg_0+2 <= Arg_2 && Arg_0 <= Arg_1  



Timebounds: 

  Overall timebound: 1+max([2, 2*Arg_2])*max([2, 2*Arg_2])+max([2, 2*Arg_2]) {O(n^2)}

  2: f0->f4: 1 {O(1)}

  0: f4->f4: max([2, 2*Arg_2])*max([2, 2*Arg_2]) {O(n^2)}

  1: f4->f4: max([2, 2*Arg_2]) {O(n)}



Costbounds:

  Overall costbound: 1+max([2, 2*Arg_2])*max([2, 2*Arg_2])+max([2, 2*Arg_2]) {O(n^2)}

  2: f0->f4: 1 {O(1)}

  0: f4->f4: max([2, 2*Arg_2])*max([2, 2*Arg_2]) {O(n^2)}

  1: f4->f4: max([2, 2*Arg_2]) {O(n)}



Sizebounds:

`Lower:

  2: f0->f4, Arg_0: 0 {O(1)}

  2: f0->f4, Arg_1: 0 {O(1)}

  2: f0->f4, Arg_2: 1 {O(1)}

  0: f4->f4, Arg_0: 1 {O(1)}

  0: f4->f4, Arg_1: 1 {O(1)}

  0: f4->f4, Arg_2: 2 {O(1)}

  1: f4->f4, Arg_0: 1 {O(1)}

  1: f4->f4, Arg_1: 0 {O(1)}

  1: f4->f4, Arg_2: 2 {O(1)}

`Upper:

  2: f0->f4, Arg_0: 0 {O(1)}

  2: f0->f4, Arg_1: 0 {O(1)}

  2: f0->f4, Arg_2: Arg_2 {O(n)}

  0: f4->f4, Arg_0: max([2, 2*Arg_2]) {O(n)}

  0: f4->f4, Arg_1: max([2, 2*Arg_2])*max([2, 2*Arg_2]) {O(n^2)}

  0: f4->f4, Arg_2: Arg_2 {O(n)}

  1: f4->f4, Arg_0: max([2, 2*Arg_2]) {O(n)}

  1: f4->f4, Arg_1: 0 {O(1)}

  1: f4->f4, Arg_2: Arg_2 {O(n)}


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(2)
BOUNDS(1, max(3, 1 + 2 * Arg_2) + max(2, 2 * Arg_2)^2)

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f0

      0: f4 -> f4 : B'=1+B, [ A>=1+B ], cost: 1

      1: f4 -> f4 : A'=1+A, B'=0, [ C>=2+A && B>=A ], cost: 1

      2: f0 -> f4 : A'=0, B'=0, [ C>=1 ], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: f4 -> f4 : B'=1+B, [ A>=1+B ], cost: 1

      1: f4 -> f4 : A'=1+A, B'=0, [ C>=2+A && B>=A ], cost: 1



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

   Found no metering function for rule 1.

   Nested simple loops 1 (outer loop) and 3 (inner loop) with metering function -1+C-A, resulting in the new rules: 4, 5.

   Removing the simple loops: 0 1.



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

   Start location: f0

      3: f4 -> f4 : B'=A, [ A>=1+B ], cost: A-B

      4: f4 -> f4 : A'=-1+C, B'=-1+C, [ C>=2+A && B>=A && 1+A>=1 ], cost: -3/2+3/2*C+1/2*(-1+C-A)^2-3/2*A+(-1+C-A)*A

      5: f4 -> f4 : A'=-1+C, B'=-1+C, [ A>=1+B && C>=2+A && 1+A>=1 ], cost: -3/2+3/2*C+1/2*(-1+C-A)^2-1/2*A+(-1+C-A)*A-B

      2: f0 -> f4 : A'=0, B'=0, [ C>=1 ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f0

      2: f0 -> f4 : A'=0, B'=0, [ C>=1 ], cost: 1

      6: f0 -> f4 : A'=-1+C, B'=-1+C, [ C>=2 ], cost: -1/2+3/2*C+1/2*(-1+C)^2



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

   Start location: f0

      6: f0 -> f4 : A'=-1+C, B'=-1+C, [ C>=2 ], cost: -1/2+3/2*C+1/2*(-1+C)^2



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f0

      6: f0 -> f4 : A'=-1+C, B'=-1+C, [ C>=2 ], cost: -1/2+3/2*C+1/2*(-1+C)^2



Computing asymptotic complexity for rule 6

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      1/2*C^2+1/2*C (+), -1+C (+/+!) [not solved]



      removing all constraints (solved by SMT)

      resulting limit problem: [solved]



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

      resulting limit problem:

      [solved]



   Solution:

      C / n

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



Found new complexity Poly(n^2).



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

   Complexity:  Poly(n^2)

   Cpx degree:  2

   Solved cost: 1/2*n+1/2*n^2

   Rule cost:   -1/2+3/2*C+1/2*(-1+C)^2

   Rule guard:  [ C>=2 ]



WORST_CASE(Omega(n^2),?)


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(4)
BOUNDS(n^2, INF)