/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^2))
proof of /export/starexec/sandbox2/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 + -100 * Arg_1) + max(2, 3 + -1 * Arg_1) + nat(100 + -1 * Arg_0 + Arg_1 * Arg_0 + -100 * Arg_1) + nat(1 + -1 * Arg_0)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 445 ms]
(2) BOUNDS(1, nat(1 + -100 * Arg_1) + max(2, 3 + -1 * Arg_1) + nat(100 + -1 * Arg_0 + Arg_1 * Arg_0 + -100 * Arg_1) + nat(1 + -1 * Arg_0))
(3) Loat Proof [FINISHED, 836 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
f300(A, B, C) -> Com_1(f300(-(99) + A, 0, C)) :|: 0 >= A + 1 && B + 1 >= 0 && B + 1 <= 0
f300(A, B, C) -> Com_1(f300(1 + A, 1 + B, C)) :|: 0 >= A + 1 && B >= 0
f300(A, B, C) -> Com_1(f300(1 + A, 1 + B, C)) :|: 0 >= A + 1 && 0 >= 2 + B
f300(A, B, C) -> Com_1(f1(A, B, D)) :|: A >= 0
f2(A, B, C) -> Com_1(f300(A, B, C)) :|: TRUE

The start-symbols are:[f2_3]


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(1) Koat2 Proof (FINISHED)
YES( ?, max([0, 1+-100*Arg_1])+max([2, 3-Arg_1])+max([0, (1-Arg_1)*(100-Arg_0)])+max([0, 1-Arg_0]) {O(n^2)})



Initial Complexity Problem:

  Start:  f2  

  Program_Vars:  Arg_0, Arg_1, Arg_2  

  Temp_Vars:  D  

  Locations:  f1, f2, f300  

  Transitions:

  4: f2->f300

  3: f300->f1

  0: f300->f300

  1: f300->f300

  2: f300->f300  



Timebounds: 

  Overall timebound: max([0, 1+-100*Arg_1])+max([2, 3-Arg_1])+max([0, (1-Arg_1)*(100-Arg_0)])+max([0, 1-Arg_0]) {O(n^2)}

  4: f2->f300: 1 {O(1)}

  0: f300->f300: max([0, 1-Arg_1]) {O(n)}

  1: f300->f300: max([0, (1-Arg_1)*(100-Arg_0)])+max([0, 1-Arg_0]) {O(n^2)}

  2: f300->f300: max([0, 1+-100*Arg_1]) {O(n)}

  3: f300->f1: 1 {O(1)}



Costbounds:

  Overall costbound: max([0, 1+-100*Arg_1])+max([2, 3-Arg_1])+max([0, (1-Arg_1)*(100-Arg_0)])+max([0, 1-Arg_0]) {O(n^2)}

  4: f2->f300: 1 {O(1)}

  0: f300->f300: max([0, 1-Arg_1]) {O(n)}

  1: f300->f300: max([0, (1-Arg_1)*(100-Arg_0)])+max([0, 1-Arg_0]) {O(n^2)}

  2: f300->f300: max([0, 1+-100*Arg_1]) {O(n)}

  3: f300->f1: 1 {O(1)}



Sizebounds:

`Lower:

  4: f2->f300, Arg_0: Arg_0 {O(n)}

  4: f2->f300, Arg_1: Arg_1 {O(n)}

  4: f2->f300, Arg_2: Arg_2 {O(n)}

  0: f300->f300, Arg_0: -99+Arg_0 {O(n)}

  0: f300->f300, Arg_1: 0 {O(1)}

  0: f300->f300, Arg_2: Arg_2 {O(n)}

  1: f300->f300, Arg_0: min([Arg_0, -(99-Arg_0)]) {O(n)}

  1: f300->f300, Arg_1: 1 {O(1)}

  1: f300->f300, Arg_2: Arg_2 {O(n)}

  2: f300->f300, Arg_0: Arg_0 {O(n)}

  2: f300->f300, Arg_1: Arg_1 {O(n)}

  2: f300->f300, Arg_2: Arg_2 {O(n)}

  3: f300->f1, Arg_0: 0 {O(1)}

  3: f300->f1, Arg_1: min([1, Arg_1]) {O(n)}

`Upper:

  4: f2->f300, Arg_0: Arg_0 {O(n)}

  4: f2->f300, Arg_1: Arg_1 {O(n)}

  4: f2->f300, Arg_2: Arg_2 {O(n)}

  0: f300->f300, Arg_0: -100 {O(1)}

  0: f300->f300, Arg_1: 0 {O(1)}

  0: f300->f300, Arg_2: Arg_2 {O(n)}

  1: f300->f300, Arg_0: 0 {O(1)}

  1: f300->f300, Arg_1: max([0, (1-Arg_1)*(100-Arg_0)])+max([0, 1-Arg_0])+max([0, Arg_1]) {O(n^2)}

  1: f300->f300, Arg_2: Arg_2 {O(n)}

  2: f300->f300, Arg_0: 0 {O(1)}

  2: f300->f300, Arg_1: -1 {O(1)}

  2: f300->f300, Arg_2: Arg_2 {O(n)}

  3: f300->f1, Arg_0: max([0, Arg_0]) {O(n)}

  3: f300->f1, Arg_1: max([-1, max([Arg_1, max([0, (1-Arg_1)*(100-Arg_0)])+max([0, 1-Arg_0])+max([0, Arg_1])])]) {O(n^2)}


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(2)
BOUNDS(1, nat(1 + -100 * Arg_1) + max(2, 3 + -1 * Arg_1) + nat(100 + -1 * Arg_0 + Arg_1 * Arg_0 + -100 * Arg_1) + nat(1 + -1 * Arg_0))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: f2

      0: f300 -> f300 : A'=-99+A, B'=0, [ 0>=1+A && 1+B==0 ], cost: 1

      1: f300 -> f300 : A'=1+A, B'=1+B, [ 0>=1+A && B>=0 ], cost: 1

      2: f300 -> f300 : A'=1+A, B'=1+B, [ 0>=1+A && 0>=2+B ], cost: 1

      3: f300 -> f1 : C'=free, [ A>=0 ], cost: 1

      4: f2 -> f300 : [], cost: 1



Checking for constant complexity:

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

      4: f2 -> f300 : [], cost: 1



Removed unreachable and leaf rules:

   Start location: f2

      0: f300 -> f300 : A'=-99+A, B'=0, [ 0>=1+A && 1+B==0 ], cost: 1

      1: f300 -> f300 : A'=1+A, B'=1+B, [ 0>=1+A && B>=0 ], cost: 1

      2: f300 -> f300 : A'=1+A, B'=1+B, [ 0>=1+A && 0>=2+B ], cost: 1

      4: f2 -> f300 : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 0.

   Accelerating the following rules:

      0: f300 -> f300 : A'=-99+A, B'=0, [ 0>=1+A && 1+B==0 ], cost: 1

      1: f300 -> f300 : A'=1+A, B'=1+B, [ 0>=1+A && B>=0 ], cost: 1

      2: f300 -> f300 : A'=1+A, B'=1+B, [ 0>=1+A && 0>=2+B ], cost: 1



   Accelerated rule 0 with metering function -B, yielding the new rule 5.

   Accelerated rule 1 with metering function -A, yielding the new rule 6.

   Accelerated rule 2 with metering function -1-A (after adding A>=B), yielding the new rule 7.

   Accelerated rule 2 with metering function -1-B (after adding A<=B), yielding the new rule 8.

   Removing the simple loops: 0 1 2.



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

   Start location: f2

      5: f300 -> f300 : A'=A+99*B, B'=0, [ 0>=1+A && 1+B==0 ], cost: -B

      6: f300 -> f300 : A'=0, B'=-A+B, [ 0>=1+A && B>=0 ], cost: -A

      7: f300 -> f300 : A'=-1, B'=-1-A+B, [ 0>=2+B && A>=B && -1-A>=1 ], cost: -1-A

      8: f300 -> f300 : A'=-1+A-B, B'=-1, [ 0>=1+A && 0>=2+B && A<=B ], cost: -1-B

      4: f2 -> f300 : [], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: f2

      4: f2 -> f300 : [], cost: 1

      9: f2 -> f300 : A'=A+99*B, B'=0, [ 0>=1+A && 1+B==0 ], cost: 1-B

     10: f2 -> f300 : A'=0, B'=-A+B, [ 0>=1+A && B>=0 ], cost: 1-A

     11: f2 -> f300 : A'=-1, B'=-1-A+B, [ 0>=2+B && A>=B && -1-A>=1 ], cost: -A

     12: f2 -> f300 : A'=-1+A-B, B'=-1, [ 0>=1+A && 0>=2+B && A<=B ], cost: -B



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

   Start location: f2

      9: f2 -> f300 : A'=A+99*B, B'=0, [ 0>=1+A && 1+B==0 ], cost: 1-B

     10: f2 -> f300 : A'=0, B'=-A+B, [ 0>=1+A && B>=0 ], cost: 1-A

     11: f2 -> f300 : A'=-1, B'=-1-A+B, [ 0>=2+B && A>=B && -1-A>=1 ], cost: -A

     12: f2 -> f300 : A'=-1+A-B, B'=-1, [ 0>=1+A && 0>=2+B && A<=B ], cost: -B



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: f2

      9: f2 -> f300 : A'=A+99*B, B'=0, [ 0>=1+A && 1+B==0 ], cost: 1-B

     10: f2 -> f300 : A'=0, B'=-A+B, [ 0>=1+A && B>=0 ], cost: 1-A

     11: f2 -> f300 : A'=-1, B'=-1-A+B, [ 0>=2+B && A>=B && -1-A>=1 ], cost: -A

     12: f2 -> f300 : A'=-1+A-B, B'=-1, [ 0>=1+A && 0>=2+B && A<=B ], cost: -B



Computing asymptotic complexity for rule 9

   Could not solve the limit problem.

   Resulting cost 0 has complexity: Unknown



Computing asymptotic complexity for rule 10

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      -A (+/+!), 1-A (+), 1+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 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:  [ 0>=1+A && B>=0 ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)