/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^1))
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, max(2, 3 + Arg_0)).

(0) CpxIntTrs
(1) Koat2 Proof [FINISHED, 2839 ms]
(2) BOUNDS(1, max(2, 3 + Arg_0))
(3) Loat Proof [FINISHED, 425 ms]
(4) BOUNDS(n^1, INF)


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(0)
Obligation:
Complexity Int TRS consisting of the following rules:
l0(A, B, F, X) -> Com_1(l1(A, B, F, X)) :|: TRUE
l1(A, B, F, X) -> Com_1(l1(A, B, F + X, X - 1)) :|: X >= 1
l1(A, B, F, X) -> Com_1(l2(A, B, F, X)) :|: X < 1
l2(A, B, F, X) -> Com_1(l3(F, F, F, X)) :|: F >= 1
l3(A, B, F, X) -> Com_1(l3(A + B, B - 1, F, X)) :|: A >= 1
l3(A, B, F, X) -> Com_1(l2(A, B, F - 1, X)) :|: 0 >= A

The start-symbols are:[l0_4]


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



Initial Complexity Problem:

  Start:  l0  

  Program_Vars:  Arg_0, Arg_1, Arg_2, Arg_3  

  Temp_Vars:    

  Locations:  l0, l1, l2, l3  

  Transitions:

  0: l0->l1

  1: l1->l1

  2: l1->l2

  4: l3->l3  



Timebounds: 

  Overall timebound: max([2, 3+Arg_0]) {O(n)}

  0: l0->l1: 1 {O(1)}

  1: l1->l1: max([0, 1+Arg_0]) {O(n)}

  2: l1->l2: 1 {O(1)}

  4: l3->l3: 0 {O(1)}



Costbounds:

  Overall costbound: max([2, 3+Arg_0]) {O(n)}

  0: l0->l1: 1 {O(1)}

  1: l1->l1: max([0, 1+Arg_0]) {O(n)}

  2: l1->l2: 1 {O(1)}

  4: l3->l3: 0 {O(1)}



Sizebounds:

`Lower:

  0: l0->l1, Arg_0: Arg_0 {O(n)}

  0: l0->l1, Arg_1: Arg_1 {O(n)}

  0: l0->l1, Arg_2: Arg_2 {O(n)}

  0: l0->l1, Arg_3: Arg_3 {O(n)}

  1: l1->l1, Arg_0: 0 {O(1)}

  1: l1->l1, Arg_1: Arg_1 {O(n)}

  1: l1->l1, Arg_2: Arg_2 {O(n)}

  1: l1->l1, Arg_3: Arg_3 {O(n)}

  2: l1->l2, Arg_0: min([0, Arg_0]) {O(n)}

  2: l1->l2, Arg_1: Arg_1 {O(n)}

  2: l1->l2, Arg_2: Arg_2 {O(n)}

  2: l1->l2, Arg_3: Arg_3 {O(n)}

  4: l3->l3, Arg_0: inf {Infinity}

  4: l3->l3, Arg_1: inf {Infinity}

  4: l3->l3, Arg_2: inf {Infinity}

  4: l3->l3, Arg_3: inf {Infinity}

`Upper:

  0: l0->l1, Arg_0: Arg_0 {O(n)}

  0: l0->l1, Arg_1: Arg_1 {O(n)}

  0: l0->l1, Arg_2: Arg_2 {O(n)}

  0: l0->l1, Arg_3: Arg_3 {O(n)}

  1: l1->l1, Arg_0: Arg_0 {O(n)}

  1: l1->l1, Arg_1: max([Arg_0, max([Arg_0, Arg_1])])+max([0, Arg_0*(1+Arg_0)]) {O(n^2)}

  1: l1->l1, Arg_2: Arg_2 {O(n)}

  1: l1->l1, Arg_3: Arg_3 {O(n)}

  2: l1->l2, Arg_0: 0 {O(1)}

  2: l1->l2, Arg_1: max([Arg_1, max([Arg_0, max([Arg_0, Arg_1])])+max([0, Arg_0*(1+Arg_0)])]) {O(n^2)}

  2: l1->l2, Arg_2: Arg_2 {O(n)}

  2: l1->l2, Arg_3: Arg_3 {O(n)}

  4: l3->l3, Arg_0: -(inf) {Infinity}

  4: l3->l3, Arg_1: -(inf) {Infinity}

  4: l3->l3, Arg_2: -(inf) {Infinity}

  4: l3->l3, Arg_3: -(inf) {Infinity}


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(2)
BOUNDS(1, max(2, 3 + Arg_0))

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(3) Loat Proof (FINISHED)


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



Initial linear ITS problem

   Start location: l0

      0: l0 -> l1 : [], cost: 1

      1: l1 -> l1 : A'=-1+A, B'=A+B, [ A>=1 ], cost: 1

      2: l1 -> l2 : [ 0>=A ], cost: 1

      3: l2 -> l3 : C'=B, D'=B, [ B>=1 ], cost: 1

      4: l3 -> l3 : C'=C+D, D'=-1+D, [ C>=1 ], cost: 1

      5: l3 -> l2 : B'=-1+B, [ 0>=C ], cost: 1



Checking for constant complexity:

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

      0: l0 -> l1 : [], cost: 1



### Simplification by acceleration and chaining ###



Accelerating simple loops of location 1.

   Accelerating the following rules:

      1: l1 -> l1 : A'=-1+A, B'=A+B, [ A>=1 ], cost: 1



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

   Removing the simple loops: 1.



Accelerating simple loops of location 3.

   Accelerating the following rules:

      4: l3 -> l3 : C'=C+D, D'=-1+D, [ C>=1 ], cost: 1



   Found no metering function for rule 4.

   Removing the simple loops:.



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

   Start location: l0

      0: l0 -> l1 : [], cost: 1

      2: l1 -> l2 : [ 0>=A ], cost: 1

      6: l1 -> l1 : A'=0, B'=1/2*A+1/2*A^2+B, [ A>=1 ], cost: A

      3: l2 -> l3 : C'=B, D'=B, [ B>=1 ], cost: 1

      4: l3 -> l3 : C'=C+D, D'=-1+D, [ C>=1 ], cost: 1

      5: l3 -> l2 : B'=-1+B, [ 0>=C ], cost: 1



Chained accelerated rules (with incoming rules):

   Start location: l0

      0: l0 -> l1 : [], cost: 1

      7: l0 -> l1 : A'=0, B'=1/2*A+1/2*A^2+B, [ A>=1 ], cost: 1+A

      2: l1 -> l2 : [ 0>=A ], cost: 1

      3: l2 -> l3 : C'=B, D'=B, [ B>=1 ], cost: 1

      8: l2 -> l3 : C'=2*B, D'=-1+B, [ B>=1 ], cost: 2

      5: l3 -> l2 : B'=-1+B, [ 0>=C ], cost: 1



Eliminated locations (on tree-shaped paths):

   Start location: l0

      9: l0 -> l2 : [ 0>=A ], cost: 2

     10: l0 -> l2 : A'=0, B'=1/2*A+1/2*A^2+B, [ A>=1 ], cost: 2+A



Applied pruning (of leafs and parallel rules):

   Start location: l0

     10: l0 -> l2 : A'=0, B'=1/2*A+1/2*A^2+B, [ A>=1 ], cost: 2+A



### Computing asymptotic complexity ###



Fully simplified ITS problem

   Start location: l0

     10: l0 -> l2 : A'=0, B'=1/2*A+1/2*A^2+B, [ A>=1 ], cost: 2+A



Computing asymptotic complexity for rule 10

   Solved the limit problem by the following transformations:

      Created initial limit problem:

      A (+/+!), 2+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 2+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: 2+n

   Rule cost:   2+A

   Rule guard:  [ A>=1 ]



WORST_CASE(Omega(n^1),?)


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(4)
BOUNDS(n^1, INF)